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  • Frank R. Spellman Handbook of Water and Wastewater Treatment Plant Operations LEWIS PUBLISHERS A CRC Press Company Boca Raton London New York Washington, D.C.© 2003 by CRC Press LLC
  • Library of Congress Cataloging-in-Publication Data Spellman, Frank R. Handbook of water & wastewater treatment plant operations / by Frank R. Spellman. p. cm. Includes bibliographical references and index. ISBN 1-56670-627-0 (alk. paper) 1. Water—treatment plants—Handbooks, manuals, etc. 2. Sewage disposal plants—Handbooks, manuals, etc. 3. Water—PuriÞcation—Handbooks, manuals, etc. 4. Sewage—PuriÞcation—Handbooks, manuals, etc. I. Title: Handbook of water and wastewater treatment plant operations. II. Title. TD434.S64 2003 628.1¢62—dc21 2003040119This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources areindicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and thepublisher cannot assume responsibility for the validity of all materials or for the consequences of their use.Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying,microÞlming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher.The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. SpeciÞcpermission must be obtained in writing from CRC Press LLC for such copying.Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431.Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identiÞcation and explanation,without intent to infringe. Visit the CRC Press Web site at www.crcpress.com © 2003 by CRC Press LLC Lewis Publishers is an imprint of CRC Press LLC No claim to original U.S. Government works International Standard Book Number 1-56670-627-0 Library of Congress Card Number 2003040119 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper © 2003 by CRC Press LLC
  • Preface Water does not divide; it connects. With simplicity it links privatization and the benchmarking process in this text. all aspects of our existence. On the other hand, how many of us thought security was a big deal prior to September 11? Some of us did, while David Rothenberg and Marta Ulvaenus some of us did not give it any thought at all. Today, things are different; we must adjust or fall behind. In the presentIn Handbook of Water and Wastewater Treatment Plant climate, falling behind on the security of our potable waterOperations, the intent of the author is twofold. The Þrst supplies is not an option. We must aggressively protectintent is to consolidate the information and experience in our precious water sources and those ancillaries that arewaterworks and wastewater treatment plant operations critical to maintaining and protecting water quality. Wethat have evolved as a result of technological advances in cover plant security concerns in this text.the Þeld, and as a result of the concepts and policies There are other current issues. For example, arsenicpromulgated by the environmental laws and the subse- in drinking water received a lot of coverage in the pressquent guidelines. The second intent is to discuss step-by- recently. We all know that arsenic is a deadly poison,step procedures for the correct and efÞcient operation of depending on dose, of course. Headlines stating thatwater and wastewater treatment systems. Tertiary to this arsenic has been found in certain municipal drinking watertwofold intent is the proper preparation of operators to supplies are a red ßag issue to many people. But is it reallyqualify for state licensure and certiÞcation examinations. an issue? We cover arsenic in drinking water in this text. With the impetus given to water quality improvement Another red ßag issue that has received some pressthrough the Municipal Construction Grants Program, the and the attention of regulators is the presence of patho-United States has undertaken an unprecedented building genic protozoans, such as Giardia and Cryptosporidium,program for new and improved water and wastewater treat- in drinking water supplies. We cover both of these proto-ment systems. To date, much emphasis has been placed on zoans in this text.training engineers to plan, design, and construct treatmentfacilities. At present, many programs in various engineer- In wastewater treatment (as well as water treatment),ing disciplines at many universities offer courses in water a lot of attention has been focused on disinfection by-and wastewater treatment plant design and operation. products in water efßuents outfalled into receiving water This text is not about the planning, designing, or con- bodies. We cover disinfection by-products in this text.struction of water and wastewater treatment facilities. Water and wastewater treatment is about mitigatingWhile these tasks are paramount to conception and con- the problems mentioned above. However, treatment oper-struction of needed facilities and needed infrastructure, ations are about much more. To handle today’s problems,many excellent texts are available that cover these impor- water and wastewater treatment system operators must betant areas. This text is not about engineering at all. Instead, generalists. Herein lies the problem. Many of the textsit is about operations and is designed for the operator. We presently available for water and wastewater operator useoften forget the old axiom: someone must build it, but are limited in scope and narrowly focused in content. Mostonce built, someone must operate it. It is the operation of of these texts take a bare bones approach to presentation.“it” that concerns us here. That is, the basics of each unit process are usually ade- Several excellent texts have been written on water and quately covered, but this is the extent of the coverage.wastewater treatment plant operations. Thus, the logical At present, available texts either ignore, avoid, or payquestion is, why a new text covering a well-trodden road? cursory attention to such important areas as the multiple- The compound answer is a text that is comprehensive barrier concept, maintaining infrastructure, benchmarking,in scope, current, and deals with real world problems plant security, operator roles, water hydraulics, microbi-involved with plant operations is needed. The simple ology, water ecology, basic electrical principles, pumping,answer is that after September 11, things have changed. conveyance, ßow measurement, basic water chemistry, Many of these changes were apparent before Septem- water quality issues, biomonitoring, sampling and testing,ber 11; at the same time, many of our present needs were water sources, and watershed protection. All of thesenot so apparent. Consider, for example, the need for plants important topics are thoroughly discussed in Handbook ofto become more efÞcient in operation and more economical Water and Wastewater Treatment Plant Operations.in practice. This is not new, but it now takes on added Though directed at water and wastewater operators,importance because of the threat of privatization. We cover this book will serve the needs of students; teachers; con- © 2003 by CRC Press LLC
  • sulting engineers; and technical personnel in city, state, the pertinent information for any problems you missed. Ifand federal organizations who must review operations and you miss many items, review the whole chapter.operating procedures. In order to maximize the usefulness The indented notes displayed in various locationsof the material contained in the test, it has been presented throughout this text indicate or emphasize importantin plain English in a simpliÞed and concise format. Many points to study carefully.tables have been developed, using a variety of sources. This text is accessible to those who have no experience with water and wastewater operations. If you work To assure correlation to modern practice and design, through the text systematically, you can acquire an under-illustrative problems are presented in terms of commonly standing of and skill in water and wastewater operations.used operational parameters. This will add a critical component to your professional Each chapter ends with a chapter review test to help knowledge.evaluate mastery of the concepts presented. Before goingon to the next chapter, take the review test, compare your Frank R. Spellmananswers to the key provided in Appendix A, and review Norfolk, VA © 2003 by CRC Press LLC
  • ContentsPART I Water and Wastewater Operations: An OverviewChapter 1 Problems Facing Water and Wastewater Treatment Operations1.1 Introduction1.2 The Paradigm Shift 1.2.1 A Change in the Way Things are Understood and Done1.3 Multiple-Barrier Concept 1.3.1 Multiple-Barrier Approach: Wastewater Operations1.4 Management Problems Facing Water and Wastewater Operations 1.4.1 Compliance with New, Changing, and Existing Regulations 1.4.2 Maintaining Infrastructure 1.4.3 Privatizing and/or Reengineering 1.4.4 Benchmarking 1.4.4.1 Benchmarking: The Process 1.4.5 The Bottom Line on Privatization1.5 Upgrading Security 1.5.1 The Bottom Line on Security1.6 Technical Management vs. Professional Management1.7 Chapter Review Questions and ProblemsReferencesChapter 2 Water and Wastewater Operators and Their Roles2.1 Water and Wastewater Operators2.2 Setting the Record Straight 2.2.1 The Computer-Literate Jack 2.2.2 Plant Operators as Emergency Responders 2.2.3 Operator Duties, Numbers, and Working Conditions2.3 Operator CertiÞcation/Licensure2.4 Chapter Review Questions and ProblemsReferencesChapter 3 Water and Wastewater References, Models, and Terminology3.1 Setting the Stage3.2 Treatment Process Models3.3 Key Terms Used in Waterworks and Wastewater Operations 3.3.1 Terminology and DeÞnitions3.4 Chapter Review Question and ProblemsReferences © 2003 by CRC Press LLC
  • PART II Water/Wastewater Operations: Math and Technical AspectsChapter 4 Water and Wastewater Math Operations4.1 Introduction4.2 Calculation Steps4.3 Table of Equivalents, Formulae, and Symbols4.4 Typical Water and Wastewater Math Operations 4.4.1 Arithmetic Average (or Arithmetic Mean) and Median 4.4.2 Ratio 4.4.3 Percent 4.4.3.1 Practical Percentage Calculations 4.4.4 Units and Conversions 4.4.4.1 Temperature Conversions 4.4.4.2 Milligrams per Liter (Parts per Million)4.5 Measurements: Areas and Volumes 4.5.1 Area of a Rectangle 4.5.2 Area of a Circle 4.5.3 Area of a Circular or Cylindrical Tank 4.5.4 Volume Calculations 4.5.4.1 Volume of Rectangular Tank 4.5.4.2 Volume of a Circular or Cylindrical Tank 4.5.4.3 Example Volume Problems4.6 Force, Pressure, and Head4.7 Flow 4.7.1 Flow Calculations 4.7.1.1 Instantaneous Flow Rates 4.7.1.2 Flow through a Full Pipeline 4.7.2 Velocity Calculations 4.7.3 Average Flow Rate Calculations 4.7.4 Flow Conversion Calculations4.8 Detention Time 4.8.1 Hydraulic Detention Time 4.8.1.1 Detention Time in Days 4.8.1.2 Detention Time in Hours 4.8.1.3 Detention Time in Minutes4.9 Chemical Dosage Calculations 4.9.1 Chlorine Dosage 4.9.2 Hypochlorite Dosage4.10 Percent Removal4.11 Population Equivalent or Unit Loading Factor4.12 SpeciÞc Gravity4.13 Percent Volatile Matter Reduction in Sludge4.14 Horsepower 4.14.1 Water Horsepower 4.14.2 Brake Horsepower 4.14.3 Motor Horsepower4.15 Electrical Power4.16 Chemical Coagulation and Sedimentation 4.16.1 Calculating Feed Rate 4.16.2 Calculating Solution Strength4.17 Filtration 4.17.1 Calculating the Rate of Filtration 4.17.2 Filter Backwash © 2003 by CRC Press LLC
  • 4.18 Practical Water Distribution System Calculations 4.18.1 Water Flow Velocity 4.18.2 Storage Tank Calculations 4.18.3 Distribution System Disinfection Calculations4.19 Complex Conversions 4.19.1 Concentration to Quantity 4.19.1.1 Concentration (Milligrams per Liter) to Pounds 4.19.1.2 Concentration (Milligrams per Liter) to Pounds/Day 4.19.1.3 Concentration (Milligrams per Liter) to Kilograms per Day 4.19.1.4 Concentration (milligrams/kilogram) to pounds/ton 4.19.2 Quantity to Concentration 4.19.2.1 Pounds to Concentration (Milligrams per Liter) 4.19.2.2 Pounds per Day to Concentration (Milligrams per Liter) 4.19.2.3 Kilograms per Day to Concentration (Milligrams per Liter) 4.19.3 Quantity to Volume or Flow Rate 4.19.3.1 Pounds to Tank Volume (Million Gallons) 4.19.3.2 Pounds per Day to Flow (Million Gallons per Day) 4.19.3.3 Kilograms per Day to Flow (Million Gallons per Day)4.20 Chapter Review Questions and ProblemsReferenceChapter 5 Water Hydraulics5.1 What is Water Hydraulics?5.2 Basic Concepts 5.2.1 Stevin’s Law5.3 Properties of Water 5.3.1 Density and SpeciÞc Gravity5.4 Force and Pressure 5.4.1 Hydrostatic Pressure 5.4.2 Effects of Water under Pressure5.5 Head 5.5.1 Static Head 5.5.2 Friction Head 5.5.3 Velocity Head 5.5.4 Total Dynamic Head (Total System Head) 5.5.5 Pressure/Head 5.5.6 Head/Pressure5.6 Flow/Discharge Rate: Water in Motion 5.6.1 Area/Velocity 5.6.2 Pressure/Velocity5.7 Piezometric Surface and Bernoulli’s Theorem 5.7.1 Law of Conservation of Energy 5.7.2 Energy Head 5.7.3 Piezometric Surface 5.7.3.1 Head Loss 5.7.3.2 Hydraulic Grade Line 5.7.4 Bernoulli’s Theorem 5.7.4.1 Bernoulli’s Equation5.8 Hydraulic Machines (Pumps) 5.8.1 Pumping Hydraulics5.9 Well and Wet Well Hydraulics 5.9.1 Well Hydraulics 5.9.2 Wet Well Hydraulics © 2003 by CRC Press LLC
  • 5.10 Friction Head Loss 5.10.1 Flow in Pipelines 5.10.2 Pipe and Open Flow Basics 5.10.3 Major Head Loss 5.10.3.1 Components of Major Head Loss 5.10.3.2 Calculating Major Head Loss 5.10.4 Minor Head Loss5.11 Basic Piping Hydraulics 5.11.1 Piping Networks 5.11.1.1 Energy Losses in Pipe Networks 5.11.1.2 Pipes in Series 5.11.1.3 Pipes in Parallel5.12 Open-Channel Flow 5.12.1 Characteristics of Open-Channel Flow 5.12.1.1 Laminar and Turbulent Flow 5.12.1.2 Uniform and Varied Flow 5.12.1.3 Critical Flow 5.12.1.4 Parameters Used in Open-Channel Flow 5.12.2 Open-Channel Flow Calculations 5.12.3 Open-Channel Flow: The Bottom Line5.13 Flow Measurement 5.13.1 Flow Measurement: The Old-Fashioned Way 5.13.2 Basis of Traditional Flow Measurement 5.13.3 Flow Measuring Devices 5.13.3.1 Differential Pressure Flowmeters 5.13.3.2 Magnetic Flowmeters 5.13.3.3 Ultrasonic Flowmeters 5.13.3.4 Velocity Flowmeters 5.13.3.5 Positive-Displacement Flowmeters 5.13.4 Open-Channel Flow Measurement 5.13.4.1 Weirs 5.13.4.2 Flumes5.14 Chapter Review Questions and ProblemsReferencesChapter 6 Fundamentals of Electricity6.1 Electricity: What Is It?6.2 Nature of Electricity6.3 The Structure of Matter6.4 Conductors, Semiconductors, and Insulators6.5 Static Electricity 6.5.1 Charged Bodies 6.5.2 Coulomb’s Law 6.5.3 Electrostatic Fields6.6 Magnetism 6.6.1 Magnetic Materials 6.6.2 Magnetic Earth6.7 Difference in Potential 6.7.1 The Water Analogy 6.7.2 Principal Methods of Producing Voltage6.8 Current6.9 Resistance6.10 Battery-Supplied Electricity © 2003 by CRC Press LLC
  • 6.10.1 The Voltaic Cell 6.10.2 Primary and Secondary Cells 6.10.3 Battery 6.10.3.1 Battery Operation 6.10.3.2 Combining Cells 6.10.4 Types of Batteries 6.10.4.1 Dry Cell 6.10.4.2 Lead-Acid Battery 6.10.4.3 Alkaline Cell 6.10.4.4 Nickel-Cadmium Cell 6.10.4.5 Mercury Cell 6.10.4.6 Battery Characteristics6.11 The Simple Electrical Circuit 6.11.1 Schematic Representation6.12 Ohm’s law6.13 Electrical Power 6.13.1 Electrical Power Calculations6.14 Electrical Energy6.15 Series DC Circuit Characteristics 6.15.1 Series Circuit Resistance 6.15.2 Series Circuit Current 6.15.3 Series Circuit Voltage 6.15.4 Series Circuit Power 6.15.5 Summary of the Rules for Series DC Circuits 6.15.6 General Series Circuit Analysis 6.15.6.1 Kirchhoff’s Voltage Law6.16 Ground6.17 Open and Short Circuits6.18 Parallel DC Circuits 6.18.1 Parallel Circuit Characteristics 6.18.2 Voltage in Parallel Circuits 6.18.3 Current in Parallel Circuits 6.18.4 Parallel Circuits and Kirchhoff’s Current Law 6.18.5 Parallel Circuit Resistance 6.18.5.1 Reciprocal Method 6.18.5.2 Product over the Sum Method 6.18.5.3 Reduction to an Equivalent Circuit 6.18.6 Power in Parallel Circuits 6.18.7 Rules for Solving Parallel DC Circuits6.19 Series-Parallel Circuits 6.19.1 Solving a Series-Parallel Circuit6.20 Conductors 6.20.1 Unit Size of Conductors 6.20.1.1 Square Mil 6.20.1.2 Circular Mil 6.20.1.3 Circular-Mil-Foot 6.20.1.4 Resistivity 6.20.1.5 Wire Measurement 6.20.2 Factors Governing the Selection of Wire Size 6.20.2.1 Copper vs. Other Metal Conductors 6.20.2.2 Temperature CoefÞcient 6.20.3 Conductor Insulation 6.20.4 Conductor Splices and Terminal Connections 6.20.5 Soldering Operations © 2003 by CRC Press LLC
  • 6.20.6 Solderless Connections 6.20.7 Insulation Tape6.21 Electromagnetism 6.21.1 Magnetic Field around a Single Conductor 6.21.2 Polarity of a Single Conductor 6.21.3 Field around Two Parallel Conductors 6.21.4 Magnetic Field of a Coil 6.21.4.1 Polarity of an Electromagnetic Coil 6.21.4.2 Strength of an Electromagnetic Field 6.21.5 Magnetic Units 6.21.6 Properties of Magnetic Materials 6.21.6.1 Permeability 6.21.6.2 Hysteresis 6.21.7 Electromagnets6.22 AC Theory 6.22.1 Basic AC Generator 6.22.1.1 Cycle 6.22.1.2 Frequency, Period, and Wavelength 6.22.2 Characteristic Values of AC Voltage and Current 6.22.2.1 Peak Amplitude 6.22.2.2 Peak-to-Peak Amplitude 6.22.2.3 Instantaneous Amplitude 6.22.2.4 Effective or Root-Mean-Square Value 6.22.2.5 Average Value 6.22.3 Resistance in AC Circuits 6.22.4 Phase Relationships6.23 Inductance 6.23.1 Self-Inductance 6.23.2 Mutual Inductance 6.23.3 Calculation of Total Inductance6.24 Practical Electrical Applications 6.24.1 Electrical Power Generation 6.24.2 DC Generators 6.24.3 AC Generators 6.24.4 Motors 6.24.4.1 DC Motors 6.24.4.2 AC Motors 6.24.5 Transformers 6.24.6 Power Distribution System Protection 6.24.6.1 Fuses 6.24.6.2 Circuit Breakers 6.24.6.3 Control Devices6.25 Chapter Review Questions and ProblemsChapter 7 Hydraulic Machines: Pumps7.1 Introduction7.2 Archimedes’ Screw7.3 Pumping Hydraulics 7.3.1 DeÞnitions7.4 Basic Principles of Water Hydraulics 7.4.1 Weight of Air 7.4.2 Weight of Water 7.4.3 Weight of Water Related to the Weight of Air 7.4.4 Water at Rest © 2003 by CRC Press LLC
  • 7.4.5 Gauge Pressure 7.4.6 Water in Motion 7.4.6.1 Discharge 7.4.6.2 The Law of Continuity 7.4.7 Pipe Friction7.5 Basic Pumping Calculations 7.5.1 Pumping Rates 7.5.2 Calculating Head Loss 7.5.3 Calculating Head 7.5.4 Calculating Horsepower and EfÞciency 7.5.4.1 Hydraulic Horsepower 7.5.4.2 Pump EfÞciency and Brake Horsepower 7.5.5 SpeciÞc Speed7.6 Pump Characteristic Curves 7.6.1 Head-Capacity Curve 7.6.2 The Power-Capacity Curve 7.6.3 The EfÞciency-Capacity (E-Q) Curve7.7 Pumps in Series and Parallel7.8 Considerations for Pumping Wastewater7.9 Types of Pumps Used in Water and Wastewater Treatment7.10 Introduction to Centrifugal Pumps 7.10.1 Description 7.10.2 Theory 7.10.3 Types of Centrifugal Pumps 7.10.3.1 Radial Flow Impeller Pumps 7.10.3.2 Mixed Flow Impeller Pumps 7.10.3.3 Axial Flow Impeller Pumps (Propeller Pump) 7.10.4 Characteristics and Performance Curves 7.10.4.1 Head-Capacity Curve 7.10.4.2 EfÞciency Curve 7.10.4.3 Brake Horsepower Curves 7.10.5 Advantages and Disadvantages of a Centrifugal Pump 7.10.5.1 Advantages 7.10.5.2 Disadvantages 7.10.6 Water and Wastewater Applications7.11 Centrifugal Pump Components 7.11.1 Casing 7.11.1.1 Solid Casing 7.11.1.2 Split Casings 7.11.2 Impeller 7.11.2.1 Semiopen Impeller 7.11.2.2 Open Impeller 7.11.2.3 Closed Impeller 7.11.3 Wear Rings 7.11.4 Shafts, Sleeves, and Couplings 7.11.4.1 Shafting 7.11.4.2 Sleeves 7.11.4.3 Couplings 7.11.5 StufÞng Box and Seals 7.11.5.1 StufÞng Box or Packing Assembly 7.11.5.2 Mechanical Seals 7.11.6 Bearings 7.11.6.1 Self-Aligning Double-Row Ball Bearing 7.11.6.2 Single- or Double-Row Antifriction Ball Bearing 7.11.6.3 Angular Contact Bearings © 2003 by CRC Press LLC
  • 7.11.6.4 Self-Aligning Spherical Roller Bearings 7.11.6.5 Single-Row Tapered Roller Bearings 7.11.6.6 Bearing Installation, Maintenance and Lubrication7.12 Centrifugal Pump: Operational Procedures 7.12.1 Installation 7.12.2 Start-Up 7.12.2.1 Start-Up Procedure 7.12.3 Normal Operation 7.12.4 Shutdown 7.12.5 Priming 7.12.5.1 Priming Procedure 7.12.6 Backßushing 7.12.6.1 Backßush Procedure 7.12.7 Manual Removal Procedure7.13 Centrifugal Pump: Maintenance Procedures 7.13.1 Pump and Motor Lubrication 7.13.2 Packing and Seal Replacement 7.13.2.1 Packing Procedure 7.13.2.2 Mechanical Seal Installation Procedure 7.13.3 Pump and Motor Bearing Inspection 7.13.4 Shaft and Coupling Alignment 7.13.4.1 Alignment Procedure 7.13.4.2 Removal of Obstructions7.14 Centrifugal Pumps Preventive Maintenance 7.14.1 Daily Maintenance 7.14.2 Weekly Maintenance 7.14.3 Monthly Maintenance 7.14.4 Quarterly Maintenance 7.14.5 Semiannual Maintenance7.15 Centrifugal Pump Lubrication 7.15.1 Purpose of Lubrication 7.15.1.1 Separates Surfaces 7.15.1.2 Prevents Wear 7.15.1.3 Cushions Shock 7.15.1.4 Transfers Heat 7.15.1.5 Corrosion Protection 7.15.1.6 Protective Seal 7.15.2 Lubrication Requirements 7.15.3 Lubrication Procedures 7.15.3.1 Motor Bearing Lubrication 7.15.3.2 Pump Bearing Lubrication7.16 Centrifugal Pump: Troubleshooting 7.16.1 The Troubleshooter 7.16.2 Troubleshooting: What Is It? 7.16.3 Goals of Troubleshooting 7.16.4 The Troubleshooting Process 7.16.5 Troubleshooting the Centrifugal Pump 7.16.5.1 Pump Fails to Prime or Loses its Prime 7.16.5.2 Pump Does Not Discharge 7.16.5.3 Pump Does Not Deliver Rated Capacity 7.16.5.4 Pump Does Not Deliver SufÞcient Pressure 7.16.5.5 Pump Starts and Stops Pumping 7.16.5.6 Pump Overloads Driver or Consumes Excessive Power 7.16.5.7 Pump Is Noisy or Has Extensive Vibration 7.16.5.8 Packing Has a Short Life © 2003 by CRC Press LLC
  • 7.16.5.9 Mechanical Seal Has a Short Life 7.16.5.10 Mechanical Seal Leaks Excessively 7.16.5.11 Bearings Have a Short Life 7.16.5.12 Pump Overheats or Seizes7.17 Centrifugal Pump ModiÞcations 7.17.1 Submersible Pumps 7.17.1.1 Applications 7.17.1.2 Advantages 7.17.1.3 Disadvantages 7.17.2 Recessed Impeller or Vortex Pumps 7.17.2.1 Applications 7.17.2.2 Advantages 7.17.2.3 Disadvantages 7.17.3 Turbine Pumps 7.17.3.1 Application 7.17.3.2 Advantages 7.17.3.3 Disadvantages7.18 Positive-Displacement Pumps 7.18.1 Reciprocating Pumps 7.18.1.1 Diaphragm Pumps 7.18.1.2 Metering Pumps 7.18.1.3 Rotary Pumps 7.18.1.4 Progressive-Cavity Pump 7.18.1.5 Special Purpose Pumps7.19 Chapter Review Questions and ProblemsReferencesChapter 8 Water and Wastewater Conveyance8.1 Delivering the Lifeblood of Civilization8.2 Conveyance Systems 8.2.1 DeÞnitions 8.2.2 Fluids vs. Liquids 8.2.3 Maintaining Fluid Flow in Piping Systems 8.2.3.1 Scaling 8.2.4 Piping System Maintenance 8.2.5 Valves 8.2.6 Piping System Accessories 8.2.7 Piping Systems: Temperature Effects 8.2.8 Piping Systems: Insulation8.3 Metallic Piping 8.3.1 Piping Materials 8.3.2 Piping: The Basics 8.3.2.1 Pipe Sizes 8.3.2.2 Pipe Wall Thickness 8.3.2.3 Piping ClassiÞcation 8.3.3 Types of Piping Systems 8.3.3.1 Code for IdentiÞcation of Pipelines 8.3.4 Metallic Piping Materials 8.3.4.1 Characteristics of Metallic Materials 8.3.5 Maintenance Characteristics of Metallic Piping 8.3.5.1 Expansion and Flexibility 8.3.5.2 Pipe Support Systems 8.3.5.3 Valve Selection 8.3.5.4 Isolation © 2003 by CRC Press LLC
  • 8.3.5.5 Preventing Backßow 8.3.5.6 Water Hammer 8.3.5.7 Air Binding 8.3.5.8 Corrosion Effects 8.3.6 Joining Metallic Pipe 8.3.6.1 Bell-and-Spigot Joints 8.3.6.2 Screwed or Threaded Joints 8.3.6.3 Flanged Joints 8.3.6.4 Welded Joints 8.3.6.5 Soldered and Brazed Joints8.4 Nonmetallic Piping 8.4.1 Nonmetallic Piping Materials 8.4.1.1 Clay Pipe 8.4.1.2 Concrete Pipe 8.4.1.3 Plastic Pipe8.5 Tubing 8.5.1 Tubing vs. Piping: The Difference 8.5.1.1 Tubing 8.5.2 Advantages of Tubing 8.5.2.1 Tubing: Mechanical Advantages 8.5.2.2 Chemical Advantages 8.5.3 Connecting Tubing 8.5.3.1 Cutting Tubing 8.5.3.2 Soldering Tubing 8.5.3.3 Connecting Flared/Nonßared Joints 8.5.4 Bending Tubing 8.5.5 Types of Tubing 8.5.5.1 Typical Tubing Applications8.6 Industrial Hoses 8.6.1 Hose Nomenclature 8.6.2 Factors Governing Hose Selection 8.6.3 Standards, Codes, and Sizes 8.6.3.1 Hose Size 8.6.4 Hose ClassiÞcations 8.6.4.1 Nonmetallic Hoses 8.6.4.2 Metallic Hoses 8.6.5 Hose Couplings 8.6.6 Hose Maintenance8.7 Pipe and Tube Fittings 8.7.1 Fittings 8.7.2 Functions of Fittings 8.7.2.1 Changing the Direction of Flow 8.7.2.2 Providing Branch Connections 8.7.2.3 Changing the Sizes of Lines 8.7.2.4 Sealing Lines 8.7.2.5 Connecting Lines 8.7.3 Types of Connections 8.7.3.1 Screwed Fittings 8.7.3.2 Flanged Connections 8.7.3.3 Connections 8.7.4 Tubing Fittings and Connections8.8 Valves 8.8.1 Valve Construction 8.8.2 Types of Valves © 2003 by CRC Press LLC
  • 8.8.2.1 Ball Valves 8.8.2.2 Gate Valves 8.8.2.3 Globe Valves 8.8.2.4 Needle Valves 8.8.2.5 Butterßy Valves 8.8.2.6 Plug Valves 8.8.2.7 Check Valves 8.8.2.8 Quick-Opening Valves 8.8.2.9 Diaphragm Valves 8.8.2.10 Regulating Valves 8.8.2.11 Relief Valves 8.8.2.12 Reducing Valves 8.8.3 Valve Operators 8.8.3.1 Pneumatic and Hydraulic Valve Operators 8.8.3.2 Magnetic Valve Operators 8.8.4 Valve Maintenance8.9 Piping System: Protective Devices 8.9.1 Applications 8.9.2 Strainers 8.9.3 Filters 8.9.4 Traps 8.9.4.1 Trap Maintenance and Testing8.10 Piping Ancillaries 8.10.1 Gauges 8.10.1.1 Pressure Gauges 8.10.2 Vacuum Breakers 8.10.3 Accumulators 8.10.4 Air Receivers 8.10.5 Heat Exchangers8.11 Chapter Review Questions and ProblemsReferencesChapter 9 Flow Measurement9.1 Introduction9.2 Methods of Measuring Flow 9.2.1 Weirs 9.2.2 The Oscillating Disk Water Meter 9.2.3 Flumes 9.2.4 Venturi Meter 9.2.5 Magnetic Flowmeter9.3 Flow Measurement Calculations 9.3.1 Calculation Method Used for Fill and Draw Technique 9.3.2 Calculation Method Used for Velocity/Area Technique 9.3.3 Calculation Method Used for V-Notch Weirs 9.3.4 Weir Overßow (Weir Loading Rate) 9.3.5 Calculation Method for Parshall Flume 9.3.6 Typical Flow Measurement Practice Calculations9.4 Flow Measurement Operational Problems9.5 Chapter Review Questions and ProblemsReferences © 2003 by CRC Press LLC
  • Part III Characteristics of WaterChapter 10 Basic Water Chemistry10.1 Introduction10.2 Chemistry Concepts and DeÞnitions 10.2.1 Concepts 10.2.2 DeÞnitions10.3 Water Chemistry Fundamentals 10.3.1 Matter 10.3.1.1 The Content of Matter: The Elements 10.3.2 Compound Substances10.4 The Water Molecule10.5 Water Solutions10.6 Water Constituents 10.6.1 Solids 10.6.2 Turbidity 10.6.3 Color 10.6.4 Dissolved Oxygen 10.6.5 Metals 10.6.6 Organic Matter 10.6.7 Inorganic Matter 10.6.7.1 Acids 10.6.7.2 Bases 10.6.7.3 Salts10.7 pH10.8 Alkalinity10.9 Hardness10.10 Water and Wastewater Chemicals and Chemical Processes 10.10.1 Odor Control (Wastewater Treatment) 10.10.2 Disinfection 10.10.3 Chemical Precipitation 10.10.4 Adsorption 10.10.5 Coagulation 10.10.6 Taste and Odor Removal 10.10.7 Water Softening 10.10.8 Recarbonation 10.10.9 Ion Exchange Softening 10.10.10 Scaling and Corrosion Control10.11 Chapter Review Questions and ProblemsReferencesChapter 11 Water Microbiology11.1 Introduction11.2 Microbiology: What Is It?11.3 Water and Wastewater Microorganisms 11.3.1 Key Terms 11.3.2 Microorganisms (in General) 11.3.3 ClassiÞcation 11.3.4 Differentiation 11.3.5 The Cell 11.3.5.1 Structure of the Bacterial Cell © 2003 by CRC Press LLC
  • 11.4 Bacteria 11.4.1 Bacterial Growth Factors 11.4.2 Destruction of Bacteria 11.4.3 Waterborne Bacteria11.5 Protozoa11.6 Microscopic Crustaceans11.7 Viruses11.8 Algae11.9 Fungi11.10 Nematodes and Flatworms (Worms)11.11 Pathogenic Protozoa and Helminths (Water) 11.11.1 Pathogenic Protozoa 11.11.1.1 Giardia 11.11.1.2 Cryptosporidium 11.11.1.3 Cyclospora 11.11.2 Helminths11.12 Biological Aspects and Processes (Wastewater) 11.12.1 Aerobic Process 11.12.2 Anaerobic Process 11.12.3 Anoxic Process 11.12.4 Photosynthesis 11.12.5 Growth Cycles 11.12.6 Biogeochemical Cycles 11.12.6.1 Carbon Cycle 11.12.6.2 Nitrogen Cycle 11.12.6.3 Sulfur Cycle 11.12.6.4 Phosphorus Cycle11.13 Chapter Review Questions and ProblemsReferencesChapter 12 Water Ecology12.1 Introduction12.2 Setting the Stage12.3 Ecology Terms 12.3.1 DeÞnition of Terms12.4 Levels of Organization12.5 Ecosystem12.6 Energy Flow in the Ecosystem12.7 Food Chain EfÞciency12.8 Ecological Pyramids12.9 Productivity12.10 Population Ecology12.11 Stream Genesis and Structure 12.11.1 Water Flow in a Stream 12.11.2 Stream Water Discharge 12.11.3 Transport of Material 12.11.4 Characteristics of Stream Channels 12.11.5 Stream ProÞles 12.11.6 Sinuosity 12.11.7 Bars, Rifßes, and Pools 12.11.8 The Floodplain 12.11.9 Adaptations to Stream Current © 2003 by CRC Press LLC
  • 12.11.10 Types of Adaptive Changes 12.11.11 SpeciÞc Adaptations12.12 Benthic Life: An Overview 12.12.1 Benthic Plants and Animals12.13 Benthic Macroinvertebrates 12.13.1 IdentiÞcation of Benthic Macroinvertebrates 12.13.2 Macroinvertebrates and the Food Web 12.13.3 Units of Organization 12.13.4 Typical Running Water Benthic Macroinvertebrates12.14 Insect Macroinvertebrates 12.14.1 Mayßies (Order: Ephemeroptera) 12.14.2 Stoneßies (Order: Plecoptera) 12.14.3 Caddisßies (Order: Trichoptera) 12.14.4 True Flies (Order: Diptera) 12.14.5 Beetles (Order: Coleoptera) 12.14.6 Water Strider (Jesus bugs; Order: Hemiptera) 12.14.7 Alderßies and Dobsonßies (Order: Megaloptera) 12.14.8 Dragonßies and Damselßies (Order: Odonata)12.15 Noninsect Macroinvertebrates 12.15.1 Oligochaeta (Family: TuiÞcidae; Genus: Tubifex) 12.15.2 Hirudinea (Leeches) 12.14.3 Gastropoda (Lung-Breathing Snail)12.16 Chapter Review Questions and ProblemsReferencesChapter 13 Water Quality13.1 Introduction13.2 The Water Cycle13.3 Water Quality Standards 13.3.1 Clean Water Act (1972) 13.3.2 Safe Drinking Water Act (1974)13.4 Water Quality Characteristics of Water and Wastewater 13.4.1 Physical Characteristics of Water and Wastewater 13.4.1.1 Solids 13.4.1.2 Turbidity 13.4.1.3 Color 13.4.1.4 Taste and Odor 13.4.1.5 Temperature 13.4.2 Chemical Characteristics of Water 13.4.2.1 Total Dissolved Solids (TDS) 13.4.2.2 Alkalinity 13.4.2.3 Hardness 13.4.2.4 Fluoride 13.4.2.5 Metals 13.4.2.6 Organics 13.4.2.7 Nutrients 13.4.3 Chemical Characteristics of Wastewater 13.4.3.1 Organic Substances 13.4.3.2 Inorganic Substances 13.4.4 Biological Characteristics of Water and Wastewater 13.4.4.1 Bacteria 13.4.4.2 Viruses 13.4.4.3 Protozoa 13.4.4.4 Worms (Helminths) © 2003 by CRC Press LLC
  • 13.5 Chapter Review Questions and ProblemsReferencesChapter 14 Biomonitoring, Monitoring, Sampling, and Testing14.1 What Is Biomonitoring? 14.1.1 Biotic Indices (Streams) 14.1.1.1 Benthic Macroinvertebrate Biotic Index14.2 Biological Sampling (Streams) 14.2.1 Biological Sampling: Planning 14.2.2 Sampling Stations 14.2.3 Sample Collection 14.2.3.1 Macroinvertebrate Sampling Equipment 14.2.3.2 Macroinvertebrate Sampling: Rocky-Bottom Streams 14.2.3.3 Macroinvertebrate Sampling: Muddy-Bottom Streams 14.2.4 Postsampling Routine 14.2.4.1 Sampling Devices 14.2.5 The Bottom Line on Biological Sampling14.3 Water Quality Monitoring (Drinking Water) 14.3.1 Is the Water Good or Bad? 14.3.2 State Water Quality Standards Programs 14.3.3 Designing a Water Quality Monitoring Program 14.3.4 General Preparation and Sampling Considerations 14.3.4.1 Method A: General Preparation of Sampling Containers 14.3.4.2 Method B: Acid Wash Procedures 14.3.5 Sample Types 14.3.6 Collecting Samples from a Stream 14.3.6.1 Whirl-pak® Bags 14.3.6.2 Screw-Cap Bottles 14.3.7 Sample Preservation and Storage 14.3.8 Standardization of Methods14.4 Test Methods (Drinking Water and Wastewater) 14.4.1 Titrimetric Methods 14.4.2 Colorimetric Methods 14.4.3 Visual Methods 14.4.4 Electronic Methods 14.4.5 Dissolved Oxygen Testing 14.4.5.1 Sampling and Equipment Considerations 14.4.5.2 Dissolved Oxygen Test Methods 14.4.6 Biochemical Oxygen Demand Testing 14.4.6.1 Sampling Considerations 14.4.6.2 BOD Sampling, Analysis, and Testing 14.4.7 Temperature Measurement 14.4.7.1 Sampling and Equipment Considerations 14.4.8 Hardness Measurement 14.4.8.1 Measuring Hardness 14.4.9 pH Measurement 14.4.9.1 Analytical and Equipment Considerations 14.4.9.2 pH Meters 14.4.9.3 pH Pocket Pals and Color Comparators © 2003 by CRC Press LLC
  • 14.4.10 Turbidity Measurement 14.4.10.1 Sampling and Equipment Considerations 14.4.10.2 Using a Secchi Disk 14.4.10.3 Transparency Tube 14.4.11 Orthophosphate Measurement 14.4.11.1 Forms of Phosphorus 14.4.11.2 The Phosphorus Cycle 14.4.11.3 Testing Phosphorus 14.4.11.4 Sampling and Equipment Considerations 14.4.11.5 Ascorbic Acid Method for Determining Orthophosphate 14.4.12 Nitrates Measurement 14.4.12.1 Sampling and Equipment Considerations 14.4.12.2 Cadmium Reduction Method 14.4.12.3 Nitrate Electrode Method 14.4.13 Solids Measurement 14.4.13.1 Solids Sampling and Equipment Considerations 14.4.13.2 Total Suspended Solids 14.4.13.3 Volatile Suspended Solids Testing 14.4.14 Conductivity Testing 14.4.14.1 Sampling, Testing, and Equipment Considerations 14.4.15 Total Alkalinity 14.4.15.1 Analytical and Equipment Considerations 14.4.15.2 Burets, Titrators, and Digital Titrators for Measuring Alkalinity 14.4.16 Fecal Coliform Bacteria Testing 14.4.16.1 Fecal Coliforms: General Information 14.4.16.2 Fecal Coliforms 14.4.16.3 Sampling Requirements 14.4.16.4 Sampling and Equipment Considerations 14.4.16.5 Fecal Coliform Testing 14.4.17 Apparent Color Testing/Analysis 14.4.18 Odor Analysis of Water 14.4.19 Chlorine Residual Testing/Analysis 14.4.19.1 DPD-Spectrophotometric 14.4.19.2 DPD-FAS Titration 14.4.19.3 Titrimetric–Amperometric Direct Titration 14.4.20 Fluorides14.5 Chapter Review Questions and ProblemsReferencesPart IV Water and Water TreatmentChapter 15 Potable Water Sources15.1 Introduction 15.1.1 Key Terms and DeÞnitions 15.1.2 Hydrologic Cycle15.2 Sources of Water15.3 Surface Water 15.3.1 Advantages and Disadvantages of Surface Water 15.3.2 Surface Water Hydrology 15.3.3 Raw Water Storage 15.3.4 Surface Water Intakes 15.3.5 Surface Water Screens 15.3.6 Surface Water Quality © 2003 by CRC Press LLC
  • 15.4 Groundwater 15.4.1 Groundwater Quality15.5 GUDISW15.6 Surface Water Quality and Treatment Requirements15.7 Public Water System Use Requirements15.8 Well Systems 15.8.1 Well Site Requirements 15.8.2 Types of Wells 15.8.2.1 Shallow Wells 15.8.2.2 Deep Wells 15.8.3 Components of a Well 15.8.3.1 Well Casing 15.8.3.2 Grout 15.8.3.3 Well Pad 15.8.3.4 Sanitary Seal 15.8.3.5 Well Screen 15.8.3.6 Casing Vent 15.8.3.7 Drop Pipe 15.8.3.8 Miscellaneous Well Components 15.8.4 Well Evaluation 15.8.5 Well Pumps 15.8.6 Routine Operation and Record Keeping Requirements 15.8.6.1 Well Log 15.8.7 Well Maintenance 15.8.7.1 Troubleshooting Well Problems 15.8.8 Well Abandonment15.9 Chapter Review Questions and ProblemsReferenceChapter 16 Watershed Protection16.1 Introduction16.2 Current Issues in Water Management16.3 What is a Watershed?16.4 Water Quality Impact16.5 Watershed Protection and Regulations16.6 A Watershed Protection Plan16.7 Reservoir Management Practices16.8 Watershed Management Practices16.9 Chapter Review Questions and ProblemsReferenceChapter 17 Water Treatment Operations and Unit Processes17.1 Introduction17.2 Waterworks Operators17.3 Purpose of Water Treatment17.4 Stages of Water Treatment17.5 Pretreatment 17.5.1 Aeration 17.5.2 Screening 17.5.3 Chemical Addition 17.5.3.1 Chemical Solutions 17.5.3.2 Chemical Feeders 17.5.3.3 Chemical Feeder Calibration © 2003 by CRC Press LLC
  • 17.5.3.4 Iron and Manganese Removal 17.5.3.5 Hardness Treatment 17.5.3.6 Corrosion Control17.6 Coagulation 17.6.1 Jar Testing Procedure17.7 Flocculation17.8 Sedimentation17.9 Filtration 17.9.1 Types of Filter Technologies 17.9.1.1 Slow Sand Filters 17.9.1.2 Rapid Sand Filters 17.9.1.3 Pressure Filter Systems 17.9.1.4 Diatomaceous Earth Filters 17.9.1.5 Direct Filtration 17.9.1.6 Alternate Filters 17.9.2 Common Filter Problems 17.9.3 Filtration and Compliance with Turbidity Requirements (IESWTR) 17.9.3.1 Regulatory Requirements 17.9.3.2 Individual Filter Monitoring 17.9.3.3 Reporting and Record Keeping 17.9.3.4 Additional Compliance Issues17.10 Disinfection 17.10.1 Need for Disinfection in Water Treatment. 17.10.2 Pathogens of Primary Concern 17.10.2.1 Bacteria 17.10.2.2 Viruses 17.10.2.3 Protozoa 17.10.3 Recent Waterborne Outbreaks 17.10.3.1 E. coli 17.10.3.2 Giardia lamblia 17.10.3.3 Cryptosporidium 17.10.3.4 Legionella pneumophila 17.10.4 Mechanism of Pathogen Inactivation 17.10.5 Other Uses of Disinfectants in Water Treatment 17.10.5.1 Minimization of DBP Formation 17.10.5.2 Control of Nuisance Asiatic Clams and Zebra Mussels 17.10.5.3 Oxidation of Iron and Manganese 17.10.5.4 Prevention of Regrowth in the Distribution System and Maintenance of Biological Stability 17.10.5.5 Removal of Taste and Odors through Chemical Oxidation 17.10.5.6 Improvement of Coagulation and Filtration EfÞciency 17.10.5.7 Prevention of Algal Growth in Sedimentation Basins and Filters 17.10.5.8 Removal of Color 17.10.6 Types of DBPs and Disinfection Residuals 17.10.6.1 Disinfection By-Product Formation 17.10.6.2 DBP Control Strategies 17.10.6.3 CT Factor 17.10.7 Pathogen Inactivation vs. DBP Formation 17.10.8 Disinfectant Residual Regulatory Requirements 17.10.9 Summary of Current National Disinfection Practices 17.10.10 Summary of Methods of Disinfection 17.10.11 Chlorination 17.10.11.1 Chlorine Terms 17.10.11.2 Chlorine Chemistry 17.10.11.3 Breakpoint Chlorination © 2003 by CRC Press LLC
  • 17.10.11.4 Gas Chlorination 17.10.11.5 Hypochlorination 17.10.11.6 Determining Chlorine Dosage 17.10.11.7 Chlorine Generation 17.10.11.8 Primary Uses and Points of Application of Chlorine 17.10.11.9 Factors Affecting Chlorination 17.10.11.10 Measuring Chlorine Residual 17.10.11.11 Pathogen Inactivation and Disinfection EfÞcacy 17.10.11.12 Disinfection By-Products 17.10.11.13 Operational Considerations 17.10.11.14 Advantages and Disadvantagesof Chlorine Use 17.10.11.15 Chlorine Summary Table17.11 Arsenic Removal from Drinking Water 17.11.1 Arsenic and Water 17.11.2 Arsenic Removal Technologies 17.11.2.1 Prescriptive Processes 17.11.2.2 Adsorptive Processes 17.11.2.3 Membrane Processes 17.11.2.4 Alternative Technologies17.12 Who is Ultimately Responsible for Drinking Water Quality?17.13 Chapter Review Questions and ProblemsReferencesChapter 18 Wastewater Treatment18.1 Wastewater Operators 18.1.1 The Wastewater Treatment Process: The Model18.2 Wastewater Terminology and DeÞnitions 18.2.1 Terminology and DeÞnitions18.3 Measuring Plant Performance 18.3.1 Plant Performance and EfÞciency 18.3.2 Unit Process Performance and EfÞciency 18.3.3 Percent Volatile Matter Reduction in Sludge18.4 Hydraulic Detention Time 18.4.1 Detention Time in Days 18.4.2 Detention Time in Hours 18.4.3 Detention Time in Minutes18.5 Wastewater Sources and Characteristics 18.5.1 Wastewater Sources 18.5.1.1 Generation of Wastewater 18.5.2 ClassiÞcation of Wastewater 18.5.3 Wastewater Characteristics 18.5.3.1 Physical Characteristics 18.5.3.2 Chemical Characteristics 18.5.3.3 Biological Characteristics and Processes18.6 Wastewater Collection Systems 18.6.1 Gravity Collection System 18.6.2 Force Main Collection System 18.6.3 Vacuum System 18.6.4 Pumping Stations 18.6.4.1 Wet Well–Dry Well Pumping Stations 18.6.4.2 Wet Well Pumping Stations 18.6.4.3 Pneumatic Pumping Stations 18.6.4.4 Pumping Station Wet Well Calculations © 2003 by CRC Press LLC
  • 18.7 Preliminary Treatment 18.7.1 Screening 18.7.1.1 Manually Cleaned Screens 18.7.1.2 Mechanically Cleaned Screens 18.7.1.3 Safety 18.7.1.4 Screenings Removal Computations 18.7.2 Shredding 18.7.2.1 Comminution 18.7.2.2 Barminution 18.7.3 Grit Removal 18.7.3.1 Gravity and Velocity Controlled Grit Removal 18.7.3.2 Grit Removal Calculations 18.7.4 Preaeration 18.7.4.1 Operational Observations, Problems, and Troubleshooting 18.7.5 Chemical Addition 18.7.5.1 Operational Observations, Problems, and Troubleshooting 18.7.6 Equalization 18.7.6.1 Operational Observations, Problems, andTroubleshooting 18.7.7 Aerated Systems 18.7.8 Cyclone Degritter 18.7.9 Preliminary Treatment Sampling and Testing 18.7.10 Other Preliminary Treatment Process Control Calculations18.8 Primary Treatment (Sedimentation) 18.8.1 Process Description 18.8.1.1 Overview of Primary Treatment 18.8.2 Types of Sedimentation Tanks 18.8.2.1 Septic Tanks 18.8.2.2 Two-Story (Imhoff) Tank 18.8.2.3 Plain Settling Tanks (ClariÞers) 18.8.3 Operator Observations, Process Problems, and Troubleshooting 18.8.3.1 Primary ClariÞcation: Normal Operation 18.8.3.2 Primary ClariÞcation: Operational Parameters (Normal Observations) 18.8.4 Process Control Calculations 18.8.4.1 Percent Removal 18.8.4.2 Detention Time 18.8.4.3 Surface Loading Rate (Surface Settling Rate and Surface Overßow Rate) 18.8.4.4 Weir Overßow Rate (Weir Loading Rate) 18.8.4.5 Sludge Pumping 18.8.4.6 BOD and Suspended Solids Removal 18.8.5 Problem Analysis 18.8.6 Efßuent from Settling Tanks18.9 Secondary Treatment 18.9.1 Treatment Ponds 18.9.1.1 Types of Ponds 18.9.1.2 Process Control Calculations (Stabilization Ponds) 18.9.2 Trickling Filters 18.9.2.1 Trickling Filter DeÞnitions 18.9.2.2 Trickling Filter Equipment 18.9.2.3 Filter ClassiÞcations 18.9.2.4 Standard Operating Procedures 18.9.2.5 General Process Description 18.9.2.6 Operator Observations, Process Problems, and Troubleshooting 18.9.2.7 Process Calculations © 2003 by CRC Press LLC
  • 18.9.3 Rotating Biological Contactors 18.9.3.1 RBC Equipment 18.9.3.2 RBC Operation 18.9.3.3 RBC: Expected Performance 18.9.3.4 Operator Observations, Process Problems, and Troubleshooting 18.9.3.5 RBC: Process Control Calculations18.10 Activated Sludge 18.10.1 Activated Sludge Terminology 18.10.2 Activated Sludge Process: Equipment 18.10.2.1 Aeration Tank 18.10.2.2 Aeration 18.10.2.3 Settling Tank 18.10.2.4 Return Sludge 18.10.2.5 Waste Sludge 18.10.3 Overview of Activated Sludge Process 18.10.4 Activated Sludge Process: Factors Affecting Operation 18.10.4.1 Growth Curve 18.10.5 Activated Sludge Formation 18.10.6 Activated Sludge: Performance-Controlling Factors 18.10.6.1 Aeration 18.10.6.2 Alkalinity 18.10.6.3 Nutrients 18.10.6.4 pH 18.10.6.5 Temperature 18.10.6.6 Toxicity 18.10.6.7 Hydraulic Loading 18.10.6.8 Organic Loading 18.10.7 Activated Sludge ModiÞcations 18.10.7.1 Conventional Activated Sludge 18.10.7.2 Step Aeration 18.10.7.3 Complete Mix 18.10.7.4 Pure Oxygen 18.10.7.5 Contact Stabilization 18.10.7.6 Extended Aeration 18.10.7.7 Oxidation Ditch 18.10.8 Activated Sludge: Process Control Parameters 18.10.8.1 Alkalinity 18.10.8.2 Dissolved Oxygen 18.10.8.3 pH 18.10.8.4 Mixed Liquor Suspended Solids, Mixed Liquor Volatile Suspended Solids, and Mixed Liquor Total Suspended Solids 18.10.8.5 Return Activated Sludge Rate and Concentration 18.10.8.6 Waste Activated Sludge Flow Rate 18.10.8.7 Temperature 18.10.8.8 Sludge Blanket Depth 18.10.9 Operational Control Levels 18.10.9.1 Inßuent Characteristics 18.10.9.2 Industrial Contributions 18.10.9.3 Process Sidestreams 18.10.9.4 Seasonal Variations 18.10.9.5 Control Levels at Start-Up 18.10.10 Operator Observations: Inßuent and Aeration Tank 18.10.10.1 Visual Indicators: Inßuent and Aeration Tank 18.10.10.2 Final Settling Tank (ClariÞer) Observations © 2003 by CRC Press LLC
  • 18.10.11 Process Control Testing and Sampling 18.10.11.1 Aeration Inßuent Sampling 18.10.11.2 Aeration Tank 18.10.11.3 Settling Tank Inßuent 18.10.11.4 Settling Tank 18.10.11.5 Settling Tank Efßuent 18.10.11.6 Return Activated Sludge and Waste Activated Sludge 18.10.12 Process Control Adjustments 18.10.13 Troubleshooting Operational Problems 18.10.14 Process Control Calculations 18.10.14.1 Settled Sludge Volume 18.10.14.2 Estimated Return Rate 18.10.14.3 Sludge Volume Index 18.10.14.4 Waste Activated Sludge 18.10.14.5 Food to Microorganism Ratio (F:M Ratio) 18.10.14.6 Mean Cell Residence Time (MCRT) 18.10.14.7 Mass Balance 18.10.15 Solids Concentration: Secondary ClariÞer 18.10.16 Activated Sludge Process Record Keeping Requirements18.11 Disinfection of Wastewater 18.11.1 Chlorine Disinfection 18.11.1.1 Chlorination Terminology 18.11.1.2 Wastewater Chlorination: Facts and Process Description 18.11.1.3 Chlorination Equipment 18.11.1.4 Chlorination: Operation 18.11.1.5 Troubleshooting Operational Problems 18.11.1.6 Dechlorination 18.11.1.7 Chlorination Environmental Hazards and Safety 18.11.1.8 Chlorine: Safe Work Practice 18.11.1.9 Chlorination Process Calculations 18.11.2 UV Irradiation 18.11.3 Ozonation 18.11.4 Bromine Chloride 18.11.5 No Disinfection18.12 Advanced Wastewater Treatment 18.12.1 Chemical Treatment 18.12.1.1 Operation, Observation, and Troubleshooting Procedures 18.12.2 Microscreening 18.12.2.1 Operation, Observation, and Troubleshooting Procedures 18.12.3 Filtration 18.12.3.1 Filtration Process Description 18.12.3.2 Operation, Observation, and Troubleshooting Procedures 18.12.4 Biological NitriÞcation 18.12.4.1 Operation, Observation, and Troubleshooting Procedures 18.12.5 Biological Denitrifcation 18.12.5.1 Observation, Operation, and Troubleshooting Procedures 18.12.6 Carbon Adsorption 18.12.6.1 Operation, Observation, and Troubleshooting Procedures 18.12.7 Land Application 18.12.7.1 Types or Modes of Land Application 18.12.8 Biological Nutrient Removal18.13 Solids (Sludge or Biosolids) Handling 18.13.1 Sludge: Background Information 18.13.1.1 Sources of Sludge 18.13.1.2 Sludge Characteristics © 2003 by CRC Press LLC
  • 18.13.1.3 Sludge Pumping Calculations 18.13.1.4 Sludge Treatment: An Overview 18.13.2 Sludge Thickening 18.13.2.1 Gravity Thickening 18.13.2.2 Flotation Thickening 18.13.2.3 Solids Concentrators 18.13.3 Sludge Stabilization 18.13.3.1 Aerobic Digestion 18.13.3.2 Anaerobic Digestion 18.13.3.3 Other Sludge Stabilization Processes 18.13.4 Sludge Dewatering 18.13.4.1 Sand Drying Beds 18.13.4.2 Rotary Vacuum Filtration 18.13.4.3 Pressure Filtration 18.13.4.4 Centrifugation 18.13.4.5 Sludge Incineration 18.13.4.6 Land Application of Biosolids18.14 Permits, Records, and Reports 18.14.1 DeÞnitions 18.14.2 NPDES Permits 18.14.2.1 Reporting 18.14.2.2 Sampling and Testing 18.14.2.3 Efßuent Limitations 18.14.2.4 Compliance Schedules 18.14.2.5 Special Conditions 18.14.2.6 Licensed Operator Requirements 18.14.2.7 Chlorination or Dechlorination Reporting 18.14.2.8 Reporting Calculations18.15 Chapter Review Questions and ProblemsReferencesAppendix A Answers to Chapter Review Questions and ProblemsAppendix B Formulae © 2003 by CRC Press LLC
  • PART IWater and Wastewater Operations:An Overview© 2003 by CRC Press LLC
  • Problems Facing Water and 1 Wastewater Treatment Operations What is of all things most yielding, 1. Protection against protozoan and virus contam- Can overcome that which is most hard, ination Being substanceless, it can enter in 2. Implementation of the multiple barrier approach even where there is no crevice. to microbial control 3. New requirements of the Ground Water Disin- That is how I know the value fection Rule, the Total Coliform Rule and of action which is actionless. Distribution System, and the Lead and Copper Rule Lao Tzu, 5th Century B.C. 4. Regulations for trihalomethanes and disinfec- tion by-products (DBPs)1.1 INTRODUCTION We discuss this important shift momentarily but firstAlthough not often thought of as a commodity (or, for that it is important to abide by Voltaire’s advice: that is, “Ifmatter, not thought about at all), water is a commodity — you wish to converse with me, please define your terms.”a very valuable commodity. In this text, it is our position For those not familiar with the term paradigm, it canthat with the passage of time, potable water will become be defined in the following ways. A paradigm is the con-even more valuable. Moreover, with the passage of even sensus of the scientific community — “concrete problemmore time, potable water will be even more valuable than solutions that the profession has come to accept.”1 Thomaswe might imagine. It may be possibly comparable in pric- Kuhn coined the term paradigm. He outlined it in termsing, gallon for gallon, to what we pay for gasoline, or even of the scientific process. He felt that “one sense of para-more. digm, is global, embracing all the shared commitments of Earth was originally allotted a finite amount of water — a scientific group; the other isolates a particularly impor-we have no more or no less than that original allotment tant sort of commitment and is thus a subset of the first.”1today. It logically follows that, in order to sustain life as The concept of paradigm has two general levels. The firstwe know it, we must do everything we can to preserve is the encompassing whole, the summation of parts. Itand protect our water supply. We also must purify and consists of the theories, laws, rules, models, concepts, andreuse the water we presently waste (i.e., wastewater). definitions that go into a generally accepted fundamental theory of science. Such a paradigm is global in character. The other level of paradigm is that it can also be just one1.2 THE PARADIGM SHIFT of these laws, theories, models, etc. that combine to for- mulate a global paradigm. These have the property ofHistorically, the purpose of water supply systems has been being local. For instance, Galileo’s theory that the earthto provide pleasant drinking water that is free of disease rotated around the sun became a paradigm in itself, namelyorganisms and toxic substances. In addition, the purpose a generally accepted law in astronomy. Yet, on the otherof wastewater treatment has been to protect the health and hand, his theory combined with other local paradigms inwell being of our communities. Water and wastewater areas such as religion and politics to transform culture. Atreatment operations have accomplished this goal by paradigm can also be defined as a pattern or point of view(1) prevention of disease and nuisance conditions; that determines what is seen as reality.(2) avoidance of contamination of water supplies and nav- We use the latter definition in this text.igable waters; (3) maintenance of clean water for survival A paradigm shift is defined as a major change in theof fish, bathing, and recreation; and (4) generally conser- way things are thought about, especially scientifically. Oncevation of water quality for future use. a problem can no longer be solved in the existing paradigm, The purpose of water supply systems and wastewater new laws and theories emerge and form a new paradigm,treatment processes has not changed. However, primarily overthrowing the old if it is accepted. Paradigm shifts arebecause of new regulations the paradigm has shifted. the “occasional, discontinuous, revolutionary changes inThese include: tacitly shared points of view and preconceptions.”2 Simply, © 2003 by CRC Press LLC
  • a paradigm shift represents “a profound change in the Source Protectionthoughts, perceptions, and values that form a particularvision of reality.”3 For our purposes, we use the term ↓paradigm shift to mean a change in the way things are Optimization of Treatment Processunderstood and done. Trained & Certified Plant Operators1.2.1 A CHANGE IN THE WAY THINGS ARE ↓ Sound Distribution System Management UNDERSTOOD AND DONE A Second Dose of DisinfectantIn water supply systems, the historical focus, or traditional ↓approach, has been to control turbidity, iron and manga- Cross-Connection Controlnese, taste and odor, color, and coliforms. New regulationsprovided new focus, and thus a paradigm shift. Today the ↓traditional approach is no longer sufficient. Providing Continuous Monitoring & Testingacceptable water has become more sophisticated and FIGURE 1.1 Multiple-barrier approach.costly. In order to meet the requirements of the new para- SDWA, passed in 1974, amended in 1986, and reau-digm, a systems approach must be employed. In the sys- thorized in 1996, gives the U.S. Environmental Protectiontems approach, all components are interrelated. What Agency (EPA) the authority to set drinking water stan-affects one impacts others. The focus has shifted to mul- dards. This document is important for many reasons, buttiple requirements (i.e., new regulations require the pro- is even more important because it describes how the EPAcess to be modified or the plant upgraded). establishes these standards. To illustrate the paradigm shift in the operation of Drinking water standards are regulations that EPA setswater supply systems, let us look back on the traditional to control the level of contaminants in the nation’s drinkingapproach of disinfection. Disinfection was used in water water. These standards are part of SDWA’s multiple-barrierto destroy harmful organisms. It is still used in water to approach to drinking water protection (see Figure 1.1).destroy harmful organisms, but is now only one part of As shown in Figure 1.1, the multiple barrier approachthe multiple-barrier approach. Moreover, disinfection has includes the following elements:traditionally been used to treat for coliforms only. Cur-rently, because of the paradigm shift, disinfection now 1. Assessing and protecting drinking water(and in the future) is used against coliforms, Legionella, sources — This means doing everything possi-Giardia, Cryptosporidium, and others. Another example ble to prevent microbes and other contaminantsof the traditional vs. current practices is seen in the tradi- from entering water supplies. Minimizingtional approach to particulate removal in water to lessen human and animal activity around our water-turbidity and improve aesthetics. Current practice is still sheds is one part of this barrier.to decrease turbidity to improve aesthetics, but now micro- 2. Optimizing treatment processes — This pro-bial removal plus disinfection is practical. vides a second barrier and usually means filter- Another significant factor that contributed to the par- ing and disinfecting the water. It also meansadigm shift in water supply systems was the introduction making sure that the people who are responsibleof the Surface Water Treatment Rule (SWTR) in 1989. for our water are properly trained and certifiedSWTR requires water treatment plants to achieve 99.9% and knowledgeable of the public health issues(3 log) removal activation/inactivation of Giardia and involved.99.99% (4 log) removal/inactivation of viruses. SWTR 3. Ensuring the integrity of distribution systems —applies to all surface waters and ground waters under This consists of maintaining the quality ofdirect influence. water as it moves through the system on its way to the customer’s tap.1.3 MULTIPLE-BARRIER CONCEPT 4. Effecting correct cross-connection control pro- cedures — This is a critical fourth element inOn August 6, 1996, during the Safe Drinking Water Act the barrier approach. It is critical because the(SDWA) Reauthorization signing ceremony, President Bill greatest potential hazard in water distributionClinton stated, “A fundamental promise we must make to systems is associated with cross-connections toour people is that the food they eat and the water they nonpotable waters. There are many connectionsdrink are safe.” No rational person could doubt the impor- between potable and nonpotable systems —tance of the promise made in this statement. every drain in a hospital constitutes such a © 2003 by CRC Press LLC
  • connection, but cross-connections are those hidden part out of services delivered by water and waste- through which backflow can occur.4 water professionals. 5. Continuous monitoring and testing of the water Water service professionals provide water for typical before it reaches the tap — Monitoring water urban domestic and commercial uses, eliminate wastes, quality is a critical element in the barrier protect the public health and safety, and help control many approach. It should include having specific pro- forms of pollution. Wastewater service professionals treat cedures to follow should potable water ever fail the urban wastestream to remove pollutants before dis- to meet quality standards. charging the effluent into the environment. Water and wastewater treatment services are the urban circulatory With the involvement of EPA, local governments, system.6 In addition, like the human circulatory system,drinking water utilities, and citizens, these multiple barri- the urban circulatory system is less than effective if flowers ensure that the tap water in the U.S. and territories is is not maintained.safe to drink. Simply, in the multiple-barrier concept, we Maintaining flow is what water and wastewater oper-employ a holistic approach to water management that ations is all about. This seems easy enough; water hasbegins at the source and continues with treatment, through been flowing literally for eons. However, this is not to saydisinfection and distribution. that water and wastewater operations are not without prob- lems and/or challenges. The dawn of the 21st century brought with it, for many of us, aspirations of good things1.3.1 MULTIPLE-BARRIER APPROACH: ahead in the constant struggle to provide quality food and WASTEWATER OPERATIONS water for humanity. However, the only way in which weNot shown in Figure 1.1 is the fate of the used water. What can hope to accomplish this is to stay on the cutting edge of technology and to face all challenges head on. Somehappens to the wastewater produced? Wastewater is of these other challenges are addressed in the followingtreated via the multiple-barrier treatment train, which is sections.the combination of unit processes used in the system. Theprimary mission of the wastewater treatment plant (andthe operator/practitioner) is to treat the wastestream to a 1.4 MANAGEMENT PROBLEMS FACINGlevel of purity acceptable to return it to the environment WATER AND WASTEWATER OPERATIONSor for immediate reuse (i.e., reuse in such applications asirrigation of golf courses, etc.). Problems come and go, shifting from century to century, Water and wastewater operators maintain a continuous decade to decade, year to year, and site to site. They rangeurban water cycle on a daily basis. B.D. Jones sums up from the problems caused by natural forces (storms, earth-this point as follows: quakes, fires, floods, and droughts) to those caused by social forces, currently including terrorism. Delivering services is the primary function of municipal In general, five areas are of concern to many water government. It occupies the vast bulk of the time and effort and wastewater management personnel. of most city employees, is the source of most contacts that citizens have with local governments, occasionally 1. Complying with regulations and coping with becomes the subject of heated controversy, and is often new and changing regulations surrounded by myth and misinformation. Yet, service deliv- ery remains the “hidden function” of local government.5 2. Maintaining infrastructure 3. Privatization and/or reengineering In Handbook of Water and Wastewater Treatment 4. BenchmarkingPlant Operations, we focus on sanitary (or environmental) 5. Upgrading securityservices (excluding solid-waste disposal) — water andwastewater treatment — because they have been and 1.4.1 COMPLIANCE WITH NEW, CHANGING,remain indispensable for the functioning and growth of AND EXISTING REGULATIONS7cities. Next to air, water is the most important life-sustain-ing product on earth. Yet it is its service delivery (and all Adapting the workforce to the challenges of meetingthat it entails) that remains a “hidden function” of local changing regulations and standards for both water andgovernment.5 This hidden function is what this text is all wastewater treatment is a major concern. As mentioned,about. We present our discussion in a completely new and drinking water standards are regulations that EPA sets tounique dual manner — in what we call the new paradigm control the level of contaminants in the nation’s drinkingshift in water management and in the concept of the mul- water. Again, these standards are part of SDWA’s multiple-tiple barrier approach. Essentially, the Handbook takes the barrier approach to drinking water protection. © 2003 by CRC Press LLC
  • There are two categories of drinking water standards: 1.4.2 MAINTAINING INFRASTRUCTURE 1. A National Primary Drinking Water Regulation During the 1950s and 1960s, the U.S. government encour- (primary standard) — This is a legally enforce- aged the prevention of pollution by providing funds for able standard that applies to public water systems. the construction of municipal wastewater treatment plants, Primary standards protect drinking water quality water-pollution research, and technical training and assis- by limiting the levels of specific contaminants tance. New processes were developed to treat sewage, that can adversely affect public health and are analyze wastewater, and evaluate the effects of pollution known or anticipated to occur in water. They on the environment. In spite of these efforts, expanding take the form of Maximum Contaminant Levels population and industrial and economic growth caused the or Treatment Techniques. pollution and health difficulties to increase. 2. A National Secondary Drinking Water Regula- In response to the need to make a coordinated effort tion (secondary standard) — This is a nonen- to protect the environment, the National Environmental forceable guideline regarding contaminants that Policy Act was signed into law on January 1, 1970. In may cause cosmetic effects (e.g., skin or tooth December of that year, a new independent body — EPA — discoloration) or aesthetic effects (e.g., taste, was created to bring under one roof all of the pollution- odor, or color) in drinking water. USEPA rec- control programs related to air, water, and solid wastes. ommends secondary standards to water systems, In 1972, the Water Pollution Control Act Amendments but does not require systems to comply. How- expanded the role of the federal government in water ever, states may choose to adopt them as pollution control and significantly increased federal fund- enforceable standards. This information ing for construction of wastewater treatment plants. focuses on national primary standards. Many of the wastewater treatment plants in operation today are the result of federal grants made over the years. Drinking water standards apply to public water sys- For example, because of the 1977 Clean Water Acttems that provide water for human consumption through Amendment to the Federal Water Pollution Control Actat least 15 service connections or regularly serve at least of 1972 and the 1987 Clean Water Act Reauthorization25 individuals. Public water systems include municipal Bill, funding for wastewater treatment plants was provided.water companies, homeowner associations, schools, busi- Many large sanitation districts, with their multiplenesses, campgrounds and shopping malls. plant operations, and an even larger number of single plant More recent requirements, including the Clean Water operations in smaller communities in operation today are aAct Amendments that went into effect in February 2001, result of these early environmental laws. Because of theserequire water treatment plants to meet tougher standards. laws, the federal government provided grants of severalThey have presented new problems for treatment facilities hundred million dollars to finance construction of waste-to deal with and have offered some possible solutions to water treatment facilities throughout the country.the problems of meeting the new standards. These regula- Many of these locally or federally funded treatmenttions provide for communities to upgrade existing treatment plants are aging; based on our experience, we rate some assystems, replacing aging and outdated infrastructure with dinosaurs. The point is many facilities are facing problemsnew process systems. Their purpose is to ensure that facil- caused by aging equipment, facilities, and infrastructure.ities are able to filter out higher levels of impurities from Complicating the problems associated with natural agingdrinking water, reducing the health risk from bacteria, is the increasing pressure on inadequate older systems toprotozoa, and viruses, and that they are able to decrease meet demands of increased population and urban growth.levels of turbidity and reduce concentrations of chlorine Facilities built in the 1960s and 1970s are now 30 toby-products in drinking water. 40 years old; not only are they showing signs of wear and In regards to wastewater collection and treatment, the tear, but they simply were not designed to handle the levelNational Pollution Discharge Elimination System program of growth that has occurred in many municipalities.established by the Clean Water Act, issues permits that Regulations often necessitate a need to upgrade. Bycontrol wastewater treatment plant discharges. Meeting per- matching funds or providing federal money to cover somemit is always a concern for wastewater treatment managers of the costs, municipalities can take advantage of a win-because the effluent discharged into water bodies affects dow of opportunity to improve their facility at a lowerthose downstream of the release point. Individual point direct cost to the community. Those federal dollars, ofsource dischargers must use the best available technology course, do come with strings attached; they are to be spentto control the levels of pollution in the effluent they dis- on specific projects in specific areas. On the other hand,charge into streams. As systems age, and best available many times new regulatory requirements are put in placetechnology changes, meeting permit with existing equip- without the financial assistance needed to implement.ment and unit processes becomes increasingly difficult. When this occurs, either the local community ignores the © 2003 by CRC Press LLC
  • new requirements (until caught and forced to comply) or Our experience has shown that few words conjure upthey face the situation and implement through local tax more fear among municipal plant managers than privati-hikes to pay the cost of compliance. zation or reengineering. Privatization means allowing An example of how a change in regulations can force private enterprise to compete with government in providingthe issue is demonstrated by the demands made by the public services, such as water and wastewater operations.Occupational Safety and Health Administration (OSHA) Existing management, on the other hand, can accomplishand EPA in their Process Safety Management (PSM)/Risk reengineering internally or it can be used (and usually is)Management Planning (RMP) regulations. These regula- during the privatization process. Reengineering is the sys-tions put the use of elemental chlorine (and other listed tematic transformation of an existing system into a newhazardous materials) under scrutiny. Moreover, because form to realize quality improvements in operation, systemof these regulations, plant managers throughout the coun- capability, functionality, performance, or evolvability at atry are forced to choose which side of a double-edged lower cost, schedule, or risk to the customer.sword cuts their way the most. One edge calls for full Many on-site managers consider privatization and/orcompliance with the regulations (analogous to stuffing the reengineering schemes threatening. In the worse case sce-regulation through the eye of a needle). The other edge nario, a private contractor could bid the entire staff out ofcalls for substitution. This means replacing elemental their jobs. In the best case, privatization and/or re-engi-chlorine with a nonlisted hazardous chemical (e.g., neering is often a very real threat that forces on-sitehypochlorite) or a physical (ultraviolet irradiation) disin- managers into workforce cuts, improving efficiency andfectant — a very costly undertaking either way. cutting costs. (At the same time, on-site managers work to ensure the community receives safe drinking water andNote: Many of us who have worked in water and the facility meets standards and permits. This is done with wastewater treatment for years characterize fewer workers and without injury or accident to workers, PSM and RMP as the elemental chlorine killer. the facility, or the environment.) You have probably heard the old saying: “If you There are a number of reasons causing local officials can’t do away with something in one way, then to take a hard look at privatization and/or re-engineering. regulate it to death.” 1. Decaying infrastructures — Many water andNote: Changes resulting because of regulatory pressure wastewater operations include water and waste- sometimes mean replacing or changing existing water infrastructures that date back to the early equipment, increased chemical costs (e.g., sub- 1900s. The most recent systems were built with stituting hypochlorite for chlorine typically federal funds during the 1970s, and even these increases costs threefold), and could easily now need upgrading or replacing. The EPA involve increased energy and personnel costs. recently estimated that the nation’s 75,000+ Equipment condition, new technology, and drinking water systems alone would require financial concerns are all considerations when more than $100 billion in investments over the upgrades or new processes are chosen. In addi- next 20 years. Wastewater systems will require tion, the safety of the process must be considered a similar level of investment. because of the demands made by EPA and 2. Mandates — The federal government has OSHA. The potential of harm to workers, the reduced its contributions to local water and community, and the environment are all under wastewater systems over the past 30 years, study, as are the possible long-term effects of while at the same time imposing stricter water chlorination on the human population. quality and effluent standards under the Clean Water Act and SDWA. Moreover, as previously1.4.3 PRIVATIZING AND/OR REENGINEERING8 mentioned, new unfunded mandated safety reg- ulations, such as OSHA’s PSM and EPA’s RMP,As mentioned, water and wastewater treatment operations are expensive to implement using local sourcesare undergoing a new paradigm shift. We explained that of revenues or state revolving loan funds.this paradigm shift focused on the holistic approach to 3. Hidden function — Earlier we stated thattreating water. The shift is, however, more inclusive. It much of the work of water and wastewater treat-also includes thinking outside the box. In order to remain ment is a hidden function. Because of this lackefficient and therefore competitive in the real world of of visibility, it is often difficult for local officialsoperations, water and wastewater facilities have either to commit to making the necessary investmentsbought into the new paradigm shift, or been forcibly in community water and wastewater systems.“shifted” to doing other things (often these other things Simply, the local politicians lack the politicalhave little to do with water and wastewater operations). will — water pipes and interceptors are not © 2003 by CRC Press LLC
  • Start → Plan →Research →Observe → Analyze → AdaptFIGURE 1.2 Benchmarking process. visible and not perceived as immediately criti- performance vs. best-in-class operations, and using the cal for adequate funding. It is easier for elected analysis to meet and exceed the best in class. officials to ignore them in favor of expenditures of more visible services, such as police and fire. What benchmarking is: Additionally, raising water and sewage rates to cover operations and maintenance is not always 1. Benchmarking vs. best practices gives water effected because it is an unpopular move for and wastewater operations a way to evaluate elected officials. This means that water and their operations overall. sewer rates do not adequately cover the actual a. How effective cost of providing services in many municipalities. b. How cost effective 2. Benchmarking shows plants both how well In many locations throughout the U.S., expenditures their operations stack up, and how well thoseon water and wastewater services are the largest facing operations are implemented.local governments today. (This is certainly the case for 3. Benchmarking is an objective-setting process.those municipalities struggling to implement the latest 4. Benchmarking is a new way of doing business.storm water requirements). Thus, this area presents a great 5. Benchmarking forces an external view toopportunity for cost savings. Through privatization, water ensure correctness of objective-setting.and wastewater companies can take advantage of advanced 6. Benchmarking forces internal alignment totechnology, more flexible management practices, and achieve plant goals.streamlined procurement and construction practices to 7. Benchmarking promotes teamwork by direct-lower costs and make the critical improvements more ing attention to those practices necessary toquickly. remain competitive.1.4.4 BENCHMARKING Potential results of benchmarking:Primarily out of self-preservation (to retain their lucrative 1. Benchmarking may indicate direction ofpositions), many utility directors work against the trend required change rather than specific metricsto privatize water, wastewater, and other public operations. a. Costs must be reducedUsually the real work to prevent privatization is delegated b. Customer satisfaction must be increasedto the individual managers in charge of each specific oper- c. Return on assets must be increasedation. Moreover, it can be easily seen that working against d. Improved maintenanceprivatization by these local managers is also in their own e. Improved operational practicesself-interest and in the interest of their workers; their jobs 2. Best practices are translated into operationalmay be at stake. units of measure. The question is, of course, how does one go aboutpreventing his water and wastewater operation from being Targets:privatized? The answer is rather straightforward and clear:Efficiency must be improved at reduced cost. In the real 1. Consideration of available resources convertsworld, this is easier said than done, but is not impossible. benchmark findings to targets.For example, for those facilities under Total Quality Man- 2. A target represents what can realistically beagement (TQM), the process can be much easier. accomplished in a given timeframe. The advantage TQM offers the plant manager is the 3. A target can show progress toward benchmarkvariety of tools to help plan, develop, and implement water practices and metrics.and wastewater efficiency measures. These tools include 4. Quantification of precise targets should beself-assessments, statistical process control, International based on achieving benchmark.Organization for Standards 9000 and 14000, process anal-ysis, quality circle, and benchmarking (see Figure 1.2). Note: Benchmarking can be performance based, pro- Our focus in this text is on use of the benchmarking cess based, or strategy based and can comparetool to improve water and wastewater operation’s efficiency. financial or operational performance measures,Benchmarking is a process for rigorously measuring your methods or practices, or strategic choices. © 2003 by CRC Press LLC
  • 1.4.4.1 Benchmarking: The Process 1.4.4.1.2 Benchmarking: An Example To gain better understanding of the benchmarking process,When forming a benchmarking team, the goal should be we have provided the following limited example. It is into provide a benchmark that evaluates and compares priva- outline and summary form only — discussion of a full-tized and reengineered water and wastewater treatment blown study is beyond the scope of this text. (Althoughoperations to your operation. This helps your operation to the details described below come from a real study, webe more efficient, remain competitive, and make continual have provided a fictitious name for the sanitation district.)improvements. It is important to point out that benchmark-ing is more than simply setting a performance referenceor comparison; it is a way to facilitate learning for continualimprovements. The key to the learning process is looking Rachel’s Creek Sanitation Districtoutside one’s own plant to other plants that have discovered Introductionbetter ways of achieving improved performance. In January 1997, Rachel’s Creek Sanitation District1.4.4.1.1 Benchmarking Steps formed a benchmarking team with the goal of pro- viding a benchmark that evaluates and comparesAs shown in Figure 1.2, the benchmarking process con- privatized and re-engineered wastewater treatmentsists of five steps. operations to Rachel’s Creek operations in order to be more efficient and remain competitive. After three 1. Planning — Managers must select a process (or months of evaluating wastewater facilities using the processes) to be benchmarked. A benchmarking benchmarking tool, our benchmarking is complete. team should be formed. The process of bench- This report summarizes our findings and should serve marking must be thoroughly understood and as a benchmark by which to compare and evaluate documented. The performance measure for the Rachel’s Creek Sanitation District operations. process should be established (i.e., cost, time, and quality). Facilities 2. Research — Information on the best-in-class 41 wastewater treatment plants throughout the U.S. performer must be determined through research. The benchmarking team focused on the following target areas for comparison: The information can be derived from the indus- 1. Reengineering try’s network, industry experts, industry and 2. Organization trade associations, publications, public informa- 3. Operations and maintenance tion, and other award-winning operations. a. Contractual services 3. Observation — The observation step is a study b. Materials and supplies of the benchmarking subject’s performance c. Sampling and data collection level, processes, and practices that have achieved d. Maintenance those levels, and other enabling factors. 4. Operational directives 4. Analysis — In this phase, comparisons in per- 5. Utilities formance levels among facilities are determined. 6. Chemicals The root causes for the performance gaps are 7. Technology studied. To make accurate and appropriate com- 8. Permits parisons, the comparison data must be sorted, a. Water quality controlled for quality, and normalized. b. Solids quality 5. Adaptation — This phase is putting what is c. Air quality learned throughout the benchmarking process d. Odor quality into action. The findings of the benchmarking 9. Safety study must be communicated to gain accep- 10. Training and development tance, functional goals must be established, and 11. Process a plan must be developed. Progress should be 12. Communication monitored and, as required, corrections in the 13. Public relations process made. 14. Reuse 15. Support servicesNote: Benchmarking should be interactive. It should a. Pretreatment also recalibrate performance measures and b. Collection systems improve the process. c. Procurement © 2003 by CRC Press LLC
  • d. Finance and administration equipment, and kept the plant clean. Due to their e. Laboratory efficiency and low staff, we felt that most of the f. Human resources privately operated plants were better than ours. We agreed this needs to be changed. Using what weSummary of Findings: learned during our benchmarking effort, we can beOur overall evaluation of Rachel’s Creek Sanitation just as efficient as a privately operated plant and stillDistrict as compared to our benchmarking targets is maintain our standards of quality.a good one; that is, we are in good standing as com-pared to the 41 target facilities we benchmarked with.In the area of safety, we compare quite favorably.Only plant 34, with its own full time safety manager, 1.4.5 The Bottom Line on Privatizationappeared to be better than we are. We were verycompetitive with the privatized plants in our usage of Privatization is becoming of greater and greater concern.chemicals and far ahead of many public plants. We Governance boards see privatization as a potential way towere also competitive in the use of power. Our survey shift liability and responsibility from the municipality’sof what other plants are doing to cut power costs shoulders, with the attractive bonus of cutting costs. Bothshowed that we clearly identified those areas of water and wastewater facilities face constant pressure toimprovement and our current effort to further reduce work more efficiently and more cost-effectively withpower costs is on track. We were far ahead in the fewer workers to produce a higher quality product; alloptimization of our unit processes and we were lead- functions must be value-added. Privatization is increasing,ers in the area of odor control. and many municipalities are seriously considering out- There were also areas that we need to improve. sourcing part or all of their operations to contractors.To the Rachel’s Creek employee, reengineeringapplies to only the treatment department and has been 1.5 UPGRADING SECURITYlimited to cutting staff while plant practices and orga-nizational practices are outdated and inefficient. You may say Homeland Security is a Y2K problem thatUnder the reengineering section of this report, we doesn’t end January 1 of any given year.have provided a summary of reengineering efforts atthe reengineered plants visited. The experiences ofthese plants can be used to improve our own re- Governor Tom Ridge9engineering effort. Next is our organization and staff-ing levels. A private company could reduce the entire One consequence of the events of September 11 was EPA’streatment department staff by about 18 to 24%. The directive to establish a Water Protection Task Force to18 to 24% are based on number of employees and ensure that activities to protect and secure water sup-not costs. In the organization section of this report, ply/wastewater treatment infrastructure are comprehen-organizational models and their staffing levels are sive and carried out expeditiously. Another consequenceprovided as guidelines to improving our organization is a heightened concern among citizens in the U.S. overand determining optimum staffing levels. The last big the security of their critical water and wastewater infra-area that we need to improve is in the way we accom- structure. The nation’s water and wastewater infrastructureplish the work we perform. Our people are not used consisting of several thousand publicly owned water andefficiently because of outdated and inefficient policies wastewater treatment plants, more than 100,000 pumpingand work practices. Methods to improve the way we stations, hundreds of thousands of miles of water distri-do work are found throughout this report. We noted bution and sanitary sewers, and another 200,000 miles ofthat efficient work practices used by private compa- storm sewers is one of America’s most valuable resources,nies allow plants to operate with small staffs. with treatment and distribution/collection systems valued Overall, Rachel’s Creek Sanitation District treat- at more than $2.5 trillion. Wastewater treatment operationsment plants are much better than other public plants. taken alone include the sanitary and storm sewers, formingAlthough some plants may have better equipment, an extensive network that runs near or beneath key build-better technology, and cleaner effluents, the costs in ings and roads, and is contiguous to many communicationlabor and materials is much higher than ours. Several and transportation networks. Significant damage to theof the public plants were in bad condition. Contrary nation’s wastewater facilities or collection systems wouldto popular belief, the privately operated plants had result in loss of life; catastrophic environmental damagegood to excellent operations. These plants met permit, to rivers, lakes, and wetlands; contamination of drinkingcomplied with safety regulations, maintained plant water supplies; long-term public health impacts; destruction© 2003 by CRC Press LLC
  • of fish and shellfish production; and disruption of com- ment, public health, environmental protection,merce, the economy, and our normal way of life. and emergency response organizations. Governor Tom Ridge points out the security role for d. Ensure that employees are fully aware of thethe public professional (we interpret this to include water importance of vigilance and the seriousnessand wastewater professionals): of breaches in security, and make note of unaccompanied strangers on the site and Americans should find comfort in knowing that millions immediately notify designated security of their fellow citizens are working every day to ensure officers or local law enforcement agencies. our security at every level — federal, state, county, munic- e. Consider varying the timing of operational ipal. These are dedicated professionals who are good at procedures if possible in case someone is what they do. I’ve seen it up close, as Governor of Penn- watching the pattern changes. sylvania … but there may be gaps in the system. The job of the Office of Homeland Security will be to identify f. Upon the dismissal of an employee, change those gaps and work to close them.10 passcodes and make sure keys and access cards are returned. It is to shore up the gaps in the system that has driven g. Provide customer service staff with trainingmany water and wastewater facilities to increase security. and checklists of how to handle a threat if itIn its “Water Protection Task Force Alert #IV: What Waste- is called in.water Utilities Can Do Now to Guard Against Terrorist 3. Coordinating actions for effective emergencyand Security Threats,”11 EPA made several recommenda- responsetions to increase security and reduce threats from terrorism. a. Review existing emergency response plans,The recommendations include: and ensure they are current and relevant. b. Make sure employees have necessary train- 1. Guarding against unplanned physical intrusion ing in emergency operating procedures. (water and wastewater) c. Develop clear protocols and chains-of-com- a. Lock all doors and set alarms at your office, mand for reporting and responding to threats pumping stations, treatment plants, and along with relevant emergency, law enforce- vaults, and make it a rule that doors are ment, environmental, public health officials, locked and alarms are set. consumers, and the media. Practice the b. Limit access to facilities and control access emergency protocols regularly. to pumping stations, chemical and fuel stor- d. Ensure key utility personnel (both on and off age areas, giving close scrutiny to visitors duty) have access to crucial telephone num- and contractors. bers and contact information at all times. c. Post guards at treatment plants, and post Keep the call list up to date. “employee only” signs in restricted areas. e. Develop close relationships with local law d. Control access to storm sewers. enforcement agencies, and make sure they e. Secure hatches, metering vaults, manholes, know where critical assets are located. and other access points to the sanitary col- Request they add your facilities to their rou- lection system. tine rounds. f. Increase lighting in parking lots, treatment f. Work with local industries to ensure that bays, and other areas with limited staffing. their pretreatment facilities are secure. g. Control access to computer networks and g. Report to county or state health officials any control systems, and change the passwords illness among the employees that might be frequently. associated with wastewater contamination. h. Do not leave keys in equipment or vehicles h. Report criminal threats, suspicious behavior, at any time. or attacks on wastewater utilities immedi- 2. Making security a priority for employees ately to law enforcement officials and the a. Conduct background security checks on relevant field office of the Federal Bureau of employees at hiring and periodically Investigation. thereafter. 4. Investing in security and infrastructure b. Develop a security program with written improvements plans and train employees frequently. a. Assess the vulnerability of collection/distri- c. Ensure all employees are aware of communi- bution system, major pumping stations, cations protocols with relevant law enforce- water and wastewater treatment plants, © 2003 by CRC Press LLC
  • chemical and fuel storage areas, outfall able, introduction of techniques to make more water avail- pipes, and other key infrastructure elements. able through watershed management, cloud seeding, b. Assess the vulnerability of the storm water desalination of saline or brackish water, or area-wide edu- collection system. Determine where large cational programs to teach conservation or reuse of pipes run near or beneath government build- water.12 ings, banks, commercial districts, industrial Many of the management techniques employed in facilities, or are contiguous with major com- water treatment operations are also employed in wastewa- munication and transportation networks. ter treatment. In addition, wastewater treatment operations c. Move as quickly as possible with the most employ management techniques that may include upgrad- obvious and cost-effective physical improve- ing present systems for nutrient removal, reuse of process ments, such as perimeter fences, security residuals in an earth-friendly manner, and area-wide lighting, tamper-proofing manhole covers educational programs to teach proper domestic and indus- and valve boxes, etc. trial waste disposal practices. d. Improve computer system and remote oper- Whether managing a waterworks or wastewater treat- ational security. ment plant, the manager, in regards to expertise, must be e. Use local citizen watches. a well-rounded, highly skilled individual. No one ques- f. Seek financing for more expensive and com- tions the need for incorporation of these highly-trained prehensive system improvements. practitioners — well-versed in the disciplines of sanitary engineering, biology, chemistry, hydrology, environmental1.5.1 THE BOTTOM LINE ON SECURITY science, safety principles, accountants, auditors, technical aspects, and operations — in both professions. Based onAgain, when it comes to the security of our nation and personal experience, however, engineers, biologists,even of water and wastewater treatment facilities, few have chemists, and others with no formal management trainingsummed it better than Governor Ridge: are often hindered (limited) in their ability to solve the complex management problems currently facing both Now, obviously, the further removed we get from Sep- industries. tember 11, I think the natural tendency is to let down our There are those who will view this opinion with some guard. Unfortunately, we cannot do that. The government disdain. However, in the current environment where priva- will continue to do everything we can to find and stop tization, the need for upgrading security, and other pressing those who seek to harm us. And I believe we owe it to the American people to remind them that they must be concerns are present, skilled management professionals vigilant, as well.10 are needed to manage and mitigate these problems.1.6 TECHNICAL MANAGEMENT VS. 1.7 CHAPTER REVIEW QUESTIONS PROFESSIONAL MANAGEMENT AND PROBLEMSWater treatment operations management is management Answers to chapter review questions are found in Appen-that is directed toward providing water of the right quality, dix A.in the right quantity, at the right place, at the right time,and at the right price to meet various demands. Wastewater 1.1. Define paradigm as used in this text.treatment management is directed toward providing treat- 1.2. Define paradigm shift as used in this text.ment of incoming raw influent (no matter what the quan- 1.3. List five elements of the multiple-barriertity), at the right time, to meet regulatory requirements, approach.and at the right price to meet various requirements. 1.4. Explain the following: Water service delivery The techniques of management are manifold both in remains one of the hidden functions of localwater resource management and wastewater treatment government.operations. In water treatment operations, for example,management techniques may include: 1.5. Fill in the blank: __________ drinking water standards are not enforceable. Storage to detain surplus water available at one timeof the year for use later, transportation facilities to move 1.6. Explain the difference between privatizationwater from one place to another, manipulation of the pric- and reengineering.ing structure for water to reduce demand, use of changes 1.7. Define benchmarking.in legal systems to make better use of the supplies avail- 1.8. List the five benchmarking steps. © 2003 by CRC Press LLC
  • REFERENCES 7. This section adapted from Drinan, J.E., Water & Waste- water Treatment: A Guide for the Nonengineering Pro- 1. Holyningen-Huene, P., Reconstructing Scientific Revo- fessional, Technomic Publ., Lancaster, PA, 2001, pp. 2–3. lutions, University Chicago Press, 1993, p. 134. 8. Adapted from Johnson, R. and Moore, A., Policy Brief 2. Daly, H.E., Economics: Introduction to the steady-state 17: Opening the floodgates: why water privatization will economy, in Ecology, Ethics: Essays Toward a Steady continue, Reason Public Inst., January 2002, pp. 1–3, State Economy, W.H. Freeman & Company, San Fran- [www.rppi.org/pbrief17.html]. Accessed May 14, 2002. cisco, 1980, p.1. 9. Henry, K., New face of security. Gov. Security, Apr. 3. Capra, F., Turning Point: Science, Society and the Rising 2002, pp. 30–37. Culture, Simon and Schuster, New York, 1982, p. 30. 4. Angele, F.J., Sr., Cross Connections and Backflow Pro- 10. Henry, K., New face of security. Gov. Security, Apr. tection, 2nd ed., American Water Association, Denver, 2002, p. 33. 1974. 11. U.S. Environmental Protection Agency, Water Protec- 5. Jones, B.D., Service Delivery in the City: Citizen tion Task Force Alert #IV: What Wastewater Utilites Can Demand and Bureaucratic Rules, New York: Longman, Do Now to Guard Against Terrorist and Security New York, 1980, p. 2. Threats, Washington, D.C., Oct. 24, 2001. 6. A national movement for cleaner cities, Am. J. Public 12. Mather, J.R., Water Resources: Distribution, Use, and Health, 20, 296–97, 1930; Cox, G.W., Sanitary services Management, John Wiley & Sons, New York: 1984, of municipalities, Texas Municipalities, 26, 1939, 218. p. 384.© 2003 by CRC Press LLC
  • Water and Wastewater Operators 2 and Their Roles Our planet is shrouded in water, and yet 8 million children distribution/collection system is one thing; taking sam- under the age of five will die this year from lack of safe ples, operating equipment, monitoring conditions, and water. determining settings for chemical feed systems and high- pressure pumps, along with performing laboratory tests United Nations Environmental Program and recording the results in the plant daily operating log is another.2.1 WATER AND WASTEWATER It is, however, the non-typical functions, the diverse functions, and the off-the-wall functions that cause us to OPERATORS describe operators as jacks-of-all-trades. For example, inTo begin our discussion of water and wastewater opera- addition to their normal, routine, daily operating duties,tors, it is important that we point out a few significant operators may be called upon to make emergency repairsfactors. to systems (e.g., making a welding repair to a vital piece of machinery to keep the plant or unit process on line), • Employment as a water and wastewater operator perform material handling operations, make chemical is concentrated in local government and private additions to process flow; respond to hazardous materials water supply and sanitary services companies. emergencies, make confined space entries, perform site • Postsecondary training is increasingly an asset landscaping duties, and carry out several other assorted as the number of regulated contaminants grows functions. Remember, the plant operator’s job is to keep and treatment unit processes become more the plant running and to make permit. Keeping the plant complex. running, the flow flowing, and making permit — no matter • Operators must pass examinations certifying what — requires not only talent but also the performance that they are capable of overseeing various of a wide range of functions, many of which are not called treatment processes. for in written job descriptions.1 • Operators have a relatively high incidence of on-the-job (OTJ) injuries. 2.2 SETTING THE RECORD STRAIGHT To properly operate a water treatment and distribution Based on our experience, we have found that most peopleand/or a wastewater treatment and collection system usu- either have a preconceived notion as to what water andally requires a team of highly skilled personnel filling a wastewater operations are all about, or they have nary avariety of job classifications. Typical positions include clue. On the one hand, we understand that clean water isplant manager/plant superintendent, chief operator, lead essential for everyday life. Moreover, we have at least aoperator, operator, maintenance operator, distribution vague concept that water treatment plants and water oper-and/or interceptor system technicians, assistant operators, ators treat water to make it safe for consumption. On thelaboratory professionals, and clerical personnel, to list just other hand, when it comes to wastewater treatment anda few. system operations, many of us have an ingrained image Beyond the distinct job classification titles, over the of a sewer system managed and run by a bunch of seweryears those operating water and wastewater plants have rats. Others give wastewater and its treatment and the folksbeen called by a variety of titles. These include water who treat it no thought at all (that is, unless they are iratejockey, practitioner of water, purveyor of water, sewer rat, ratepayers upset at the cost of wastewater service).or just plain water or wastewater operator. Based on our Typically, the average person has other misconcep-experience we have come up with a title that perhaps more tions about water and wastewater operations. For example,closely characterizes what the water and wastewater oper- very few people can identify the exact source supply ofator really is: a jack-of-all-trades. This characterization their drinking water. Is it pumped from wells, rivers, orseems only fitting when you take into account the knowl- streams to water treatment plants? Similarly, where is itedge and skills required of operators to properly perform treated and distributed to customers? The average persontheir assigned duties. Moreover, operating the plant or is clueless as to the ultimate fate of wastewater. Once the © 2003 by CRC Press LLC
  • toilet is flushed, it is out of sight out of mind and that is Typical examples of the computer-literate operator’sthat. work (for illustrative purposes only) are provided as Beyond the few functions we have pointed out up to follows:2this point, what exactly is it those water and wastewateroperators, the 90,000+ jacks-of-all-trades in the U.S. do? • Monitors, adjusts, starts, and stops automatedOperators in both water and wastewater treatment systems water treatment processes and emergencycontrol unit processes and equipment to remove or destroy response systems to maintain a safe and efficientharmful materials, chemical compounds, and microorgan- water treatment operation; monitors treatmentisms from the water. They also control pumps, valves, and plant processing equipment and systems to identify malfunctions and their probable causeother processing equipment (including a wide array of following prescribed procedures; places equip-computerized systems) to convey the water or wastewater ment in or out of service or redirects processesthrough the various treatment processes (unit processes), around failed equipment; following prescribedand dispose (or reuse) of the removed solids (waste mate- procedures monitors and starts process relatedrials: sludge or biosolids). Operators also read, interpret, equipment, such as boilers, to maintain processand adjust meters and gauges to make sure plant equip- and permit objectives; refers difficult equip-ment and processes are working properly. They operate ment maintenance problems and malfunctionschemical-feeding devices, take samples of the water or to supervisor; monitors the system through awastewater, perform chemical and biological laboratory process integrated control terminal or remoteanalyses, and adjust the amount of chemicals, such as station terminal to assure control devices arechlorine, in the water and wastestream. They use a variety making proper treatment adjustments; operatesof instruments to sample and measure water quality, and the central control terminal keyboard to performcommon hand and power tools to make repairs and adjust- backup adjustments to such treatment processesments. Operators also make minor repairs to valves, as influent and effluent pumping, chemical feed,pumps, basic electrical equipment and other equipment. sedimentation, and disinfection; monitors spe-(Electrical work should only be accomplished by qualified cific treatment processes and security systemspersonnel.) at assigned remote plant stations; observes and As mentioned, water and wastewater system operators reviews terminal screen display of graphs,increasingly rely on computers to help monitor equipment, grids, charts and digital readouts to determinestore sampling results, make process-control decisions, process efficiency; responds to visual and audibleschedule and record maintenance activities, and produce alarms and indicators that indicate deviationsreports. Computer-operated automatic sampling devices from normal treatment processes and chemical hazards; identifies false alarms and other indi-are beginning to gain widespread acceptance and use in cators that do not require immediate response;both industries, especially at the larger facilities. When a alerts remote control locations to respond tosystem malfunction occurs, operators may use system alarms indicating trouble in that area; performscomputers to determine the cause and the solution to the alarm investigations.problem. • Switches over to semiautomatic or manual con- trol when the computer control system is not2.2.1 THE COMPUTER-LITERATE JACK properly controlling the treatment process; off- scans a malfunctioning field sensor point andAt many modern water and wastewater treatment plants inserts data obtained from field in order tooperators are required to perform skilled treatment plant maintain computer control; controls automatedoperations work and to monitor, operate, adjust and regulate mechanical and electrical treatment processesa computer-based treatment process. In addition, the opera- through the computer keyboard when computertor is also required to operate and monitor electrical, programs have failed; performs field tours tomechanical, and electronic processing and security equip- take readings when problems cannot be cor-ment through central and remote terminal locations in a rected through the computer keyboard; makessolids processing, water purification or wastewater treatment regular field tours of the plant to observe physicalplant. In those treatment facilities that are not completely or conditions; manually controls processes whenpartially automated, computer-controlled computers are necessary.used in other applications, such as in clerical applications • Determines and changes the amount of chemicalsand in a computer maintenance management system to be added for the amount of water, waste-(CMMS). The operator must be qualified to operate and water, or biosolids to be treated; takes periodicnavigate such computer systems. samples of treated residuals, biosolids processing © 2003 by CRC Press LLC
  • products and by-products, clean water, or amounts of liquid waste flow into sewers, exceeding a wastewater for laboratory analysis; receives, plant’s treatment capacity. Emergencies can also be caused stores, handles and applies chemicals and other by conditions inside a plant, such as oxygen deficiency supplies needed for operation of assigned sta- within a confined space or exposure to toxic and/or explo- tion; maintains inventory records of suppliers sive off-gases such as hydrogen sulfide and methane. To on hand and quantities used; prepares and sub- handle these conditions, operators are trained to make an mits daily shift operational reports; records emergency management response and use special safety daily activities in plant operation log, computer equipment and procedures to protect co-workers, public database or from a computer terminal; changes health, the facility, and the environment. During emergen- chemical feed tanks, chlorine cylinders, and cies, operators may work under extreme pressure to correct feed systems; flushes clogged feed and sam- problems as quickly as possible. These periods may create pling lines. dangerous working conditions; operators must be • Notes any malfunctioning equipment; makes extremely careful and cautious. minor adjustments when required; reports Operators who must aggressively respond to hazard- major malfunctions to higher-level operator and ous chemical leaks or spills (e.g., enter a chlorine gas filled enters maintenance and related task information room and install chlorine repair kit B on a damaged 1-ton into a CMMS and processes work requests for cylinder to stop the leak) must possess a Hazardous Mate- skilled maintenance personnel. rials (HAZMAT) emergency response technician 24-hour • Performs routine mechanical maintenance such certification. Additionally, many facilities, where elemen- as packing valves, adjusting belts, and replacing tal chlorine is used for disinfection, odor control, or other shear pins and air filters; lubricates equipment process applications, require operators to possess an by applying grease and adding oil; changes and appropriate certified pesticide applicator training comple- cleans strainers; drains condensate from pres- tion certificate. Because of OSHA’s specific confined sure vessels, gearboxes, and drip traps; performs space requirement whereby a standby rescue team for minor electrical maintenance such as replacing entrants must be available, many plants require operators bulbs and resetting low voltage circuit switches; to hold and maintain cardiopulmonary resuscitation/first prepares equipment for maintenance crews by aid certification. unblocking pipelines, pumps, and isolating and draining tanks; checks equipment as part of a Note: It is important to point out that many waste- preventive and predictive maintenance program; water facilities have substituted elemental chlo- reports more complex mechanical-electrical rine with sodium or calcium hypochlorite, problems to supervisors. ozone, or ultraviolet irradiation because of the • Responds, in a safe manner, to chlorine leaks stringent requirements of OSHA’s Process and chemical spills in compliance with the Safety Management Standard (29 CFR Occupational Safety and Health Admimistra- 1910.119) and the U.S. Environmental Protection tion’s (OSHA) Hazardous Waste Operational Agency’s (EPA) Risk Management Program. and Emergency Response (29 CFR 1910.120) This is not the case in most water treatment requirements and with plant specific emergency operations. In water treatment systems, elemen- response procedures; participates in chlorine tal chlorine is still employed because it provides and other chemical emergency response drills. chlorine residual that is important in maintain- • Prepares operational and maintenance reports ing safe drinking water supplies, especially as required, including flow and treatment infor- throughout lengthy distribution systems. mation; changes charts and maintains recording equipment; utilizes system and other software 2.2.3 OPERATOR DUTIES, NUMBERS, packages to generate reports and charts and AND WORKING CONDITIONS graphs of flow and treatment status and trends; maintains workplace housekeeping. The specific duties of plant operators depend on the type and size of plant. In smaller plants, one operator may2.2.2 PLANT OPERATORS AS EMERGENCY RESPONDERS control all machinery, perform sampling and lab analyses, keep records, handle customer complaints, troubleshootAs mentioned, occasionally operators must work under and make repairs, or perform routine maintenance. Inemergency conditions. Sometimes these emergency con- some locations, operators may handle both water treat-ditions are operational and not necessarily life threatening. ment and wastewater treatment operations. In larger plantsA good example occurs during a rain event when there with many employees, operators may be more specializedmay be a temporary loss of electrical power and large and only monitor one unit process (e.g., a solids handling © 2003 by CRC Press LLC
  • operator who operates and monitors an incinerator). Along or rotating shifts. Some overtime is occasionally requiredwith treatment operators, plant staffing may include envi- in emergencies.ronmentalists, biologists, chemists, engineers, laboratory Over the years, statistical reports have related histor-technicians, maintenance operators, supervisors, clerical ical evidence showing that the water and wastewaterhelp, and various assistants. industry is an extremely unsafe occupational field. This In the U.S., notwithstanding a certain amount of less than stellar safety performance has continued to dete-downsizing brought on by privatization activites, employ- riorate even in the age of the Occupational Safety andment opportunities for water and wastewater operators Health Act of 1970.have increased in number. The number of operators has The question is why is the water and wastewater treat-increased because of the ongoing construction of new ment industry’s OTJ injury rate so high? Several reasonswater and wastewater and solids handling facilities. In help to explain this high injury rate. First, all of the majoraddition, operator jobs have increased because of water classifications or hazards exist at water and wastewaterpollution standards that have become increasingly more treatment plants (typical exception radioactivity):stringent since adoption of two major federal environmentalregulations: The Clean Water Act of 1972 (and subsequent • Oxygen deficiencyamendments), which implemented a national system of • Physical injuriesregulation on the discharge of pollutants, and the Safe • Toxic gases and vaporsDrinking Water Act (SDWA) of 1974, which established • Infectionsstandards for drinking water. • Fire Operators are often hired in industrial facilities to • Explosionmonitor or pretreat wastes before discharge to municipal • Electrocutiontreatment plants. These wastes must meet certain mini-mum standards to ensure that they have been adequately Along with all the major classifications of hazards,pretreated and will not damage municipal treatment facil- other factors cause the high incidence of injury in theities. Municipal water treatment plants also must meet water and wastewater industry. Some of these can bestringent drinking water standards. This often means that attributed to:additional qualified staff members must be hired to monitorand treat/remove specific contaminants. Complicating the • Complex treatment systemsproblem is the fact that the list of contaminants regulated • Shift workby these regulations has grown over time. For example,the 1996 SDWA Amendments include standards for mon- • New employeesitoring Giardia and Cryptosporidium, two biological • Liberal workers’ compensation lawsorganisms (protozoans) that cause health problems. Oper- • Absence of safety lawsators must be familiar with the guidelines established by • Absence of safe work practices and safetyfederal regulations and how they affect their plant. In programsaddition to federal regulations, operators must be awareof any guidelines imposed by the state or locality in which Experience has shown that a lack of well-managedthe treatment process operates. safety programs and safe work practices are major factors Another unique factor related to water and wastewater causing the water and wastewater industry’s high inci-operators is their working conditions. Water and waste- dence of OTJ injuries.water treatment plant operators work indoors and outdoorsin all kinds of weather. Operators’ work is physically 2.3 OPERATOR CERTIFICATION/LICENSUREdemanding and often is performed in unclean locations(hence, the emanation of the descriptive but inappropriate A high school diploma or its equivalency usually istitle, sewer rat). They are exposed to slippery walkways; required as the entry-level credential to become a water orvapors; odors; heat; dust; and noise from motors, pumps, wastewater treatment plant operator-in-training. Operatorsengines, and generators. They work with hazardous chem- need mechanical aptitude and should be competent in basicicals. In water and wastewater plants, operators may be mathematics, chemistry, and biology. They must have theexposed to many bacterial and viral conditions. As men- ability to apply data to formulas of treatment requirements,tioned, dangerous gases, such as methane and hydrogen flow levels, and concentration levels. Some basic familiar-sulfide, could be present so they need to use proper safety ity with computers also is necessary because of the presentgear. trend toward computer-controlled equipment and more Operators generally work a 5-day, 40-hour week. sophisticated instrumentation. Certain operator positions,However, many treatment plants are in operation 24/7, and particularly in larger cities, are covered by civil serviceoperators may have to work nights, weekends, holidays, regulations. Applicants for these positions may be required © 2003 by CRC Press LLC
  • to pass a written examination testing mathematics skills, state licensure examinations are provided by various statemechanical aptitude, and general intelligence. and local agencies. Many employers provide tuition assis- Because treatment operations are becoming more tance for formal college training.complex, completion of an associate’s degree or 1-year Whether received from formal or informal sources,certificate program in water quality and wastewater treat- training provided for or obtained by water and wastewaterment technology is highly recommended. These creden- operators must include coverage of very specific sub-tials increase an applicant’s chances for both employmentand promotion. Advanced training programs are offered ject/topic areas. Though much of their training is similarthroughout the country. They provide a good general or the same, Tables 2.1 and 2.2 list many of the specificthrough advanced training on water and wastewater treat- specialized topics waterworks and wastewater operatorsment processes, as well as basic preparation for becoming are expected to have a fundamental knowledge.a licensed operator. They also offer a wide range of com-puter training courses. Note: It is important to note that both water and wastewater operators must have fundamental New water and wastewater operators-in-training typ-ically start out as attendants or assistants and learn the knowledge of basic science and math operations.practical aspects of their job under the direction of an Note: For many water and wastewater operators,experienced operator. They learn by observing, show-and- crossover training or overlapping training istell, and doing routine tasks. These tasks can include common practice.recording meter readings; taking samples of liquid wasteand sludge; and performing simple maintenance and repairwork on pumps, electrical motors, valves, and other plantor system equipment. Larger treatment plants generally TABLE 2.1combine this OTJ training with formal classroom or self- Specialized Topics for Wastewater Operatorspaced study programs. Some large sanitation districts Wastewater math Fecal coliform testingoperate their own 3- to 4-year apprenticeship schools. In Troubleshooting techniques Recordkeepingsome of these programs, each year of apprenticeship Preliminary treatment Flow measurementschool completed not only prepares the operator for the Sedimentation Sludge dewateringnext level of certification or licensure, but also satisfies a Ponds Drying bedsrequirement for advancement to the next higher pay grade. Trickling filters Centrifuges Rotating biological contactors Vacuum filtration The SDWA Amendments of 1996, enforced by the Activated sludge Pressure filtrationEPA, specify national minimum standards for certification Chemical treatment Sludge incineration(licensure) and recertification of operators of community Disinfection Land application of biosolidsand nontransient, noncommunity water systems. As a Solids thickening Laboratory proceduresresult, operators must pass an examination to certify that Solids stabilization General safetythey are capable of overseeing water and wastewater treat-ment operations. There are different levels of certificationdepending on the operator’s experience and training.Higher certification levels qualify the operator for a wider TABLE 2.2variety of treatment processes. Certification requirements Specialized Topics for Waterworks Operatorsvary by state and by size of treatment plants. Although Chemical addition Hydraulics — mathrelocation may mean having to become certified in a new Chemical feeders Laboratory practiceslocation, many states accept other states’ certifications. Chemical feeders — math Measuring and control In an attempt to ensure the currentness of training and Clarification Piping and valves Coagulation — flocculation Public healthqualifications and to improve operators’ skills and knowl- Corrosion control Pumpsedge, most state drinking water and water pollution con- Disinfection Recordkeepingtrol agencies offer on-going training courses. These Disinfection — math General sciencecourses cover principles of treatment processes and pro- Basic electricity and controls Electric motorscess control methods, laboratory practices, maintenance Filtration Financesprocedures, management skills, collection system opera- Filtration — math Storagetion, general safe work practices, chlorination procedures, Fluoridation Leak detectionsedimentation, biological treatment, sludge/biosolids Fluoridation — math Hydrantstreatment, biosolids land application and disposal, and General safe work practices Cross connection control and backflowflow measurements. Correspondence courses covering Bacteriology Stream ecologyboth water and wastewater operations and preparation for © 2003 by CRC Press LLC
  • 2.4 CHAPTER REVIEW QUESTIONS REFERENCES AND PROBLEMS 1. Spellman, F.R., Safe Work Practices for Wastewater Treatment Plants, 2nd ed., Technomic Publ., Lancaster, 2.1. Briefly explain the causal factors behind the PA, 2001, p. 2. high incidence of OTJ injuries for water and 2. Job description and requirements taken from compilation wastewater operators. of many requirements; many are from those described in 2.2. Why is computer literacy so important in [http://www.phila.gov/personnel/specs/7e45.htm]. Access- operating a modern water and wastewater ed May 15, 2002. treatment system? 2.3. Define CMMS. 2.4. List the necessary training requirement for HAZMAT responders. 2.5. Specify the national minimum standard for certification (licensure) and recertification for water and wastewater operators.© 2003 by CRC Press LLC
  • Water and Wastewater References, 3 Models, and Terminology Living things depend on water, but water does not depend vious processes. In essence, what we are doing is starting on living things. It has a life of its own. with a blank diagram and filling in unit processes as we progress. We have found that beginning certain chapters E.C. Pielou1 in this manner is important because the water and waste- water treatment process is a series of individual steps (unit3.1 SETTING THE STAGE processes) that treat the raw water or wastewater as it makes its way through the entire process. This formatThis handbook is a compilation or summary of informa- simply provides a pictorial presentation along with perti-tion available in many expert sources. While we have nent written information, enhancing the learning process.attempted to cover all aspects of water and wastewatertreatment system operation, let us point out that no onesingle handbook has all the information or all the answers. 3.3 KEY TERMS USED IN WATERWORKSMoreover, because of the physical limits of any written AND WASTEWATER OPERATIONStext, some topics are only given cursory exposure andlimited coverage. For those individuals seeking a more in In order to learn water and wastewater treatment opera-depth treatment of specific topics germane to water and tions (or any other technology for that matter), you mustwastewater treatment system operations, we recommend master the language associated with the technology.consulting one or more of the references listed in Each technology has its own terms with its own accom-Table 3.1, and any of the many other outstanding refer- panying definitions. Many of the terms used in water andences referred to throughout this text. wastewater treatment are unique; others combine words from many different technologies and professions. One thing is certain: water and wastewater operators without a clear3.2 TREATMENT PROCESS MODELS understanding of the terms related to their profession are ill-Figure 3.1 shows a basic schematic of the water treatment equipped to perform their duties in the manner required.process. Other unit processes used in treatment of water Usually a handbook or text like this one includes a(e.g., fluoridation) are not represented in Figure 3.1; we glossary of terms at the end of the work. In this handbookdiscuss many of the other unit processes in detail within we list and define many of the terms used right up front.the handbook. Figure 3.2 shows a basic schematic or Our experience shows that an early introduction to key-model of a wastewater treatment process that provides words is a benefit to readers. An up-front introduction toprimary and secondary treatment using the activated key terms facilitates a more orderly, logical, systematicsludge process. In secondary treatment (which provides learning activity. Those terms not defined in this sectionbiochemical oxygen demand removal beyond what is are defined as they appear in the text.achievable by simple sedimentation), three approaches arecommonly used: trickling filter, activated sludge, and oxi- Note: A short quiz on many of the following termsdation ponds. We discuss these systems in detail later in follows the end of this chapter.the text. We also discuss biological nutrient removal andstandard tertiary or advanced wastewater treatment. 3.3.1 TERMINOLOGY AND DEFINITIONS The purpose of the models shown in the figures andtheir subsequent renditions in later chapters is to allow Absorb to take in. Many things absorb water.readers to visually follow the water and wastewater treat- Acid rain the acidic rainfall that results when rainment process step-by-step as we present it later in the combines with sulfur oxides emissions fromwritten material. Figure 3.1 and Figure 3.2 help you under- combustion of fossil fuels (e.g., coal).stand how all the various unit processes sequentially fol- Acre-feet (acre-foot) an expression of water quantity.low and tie into each other. Thus, in chapters dealing with One acre-foot will cover one acre of ground 1 ftunit processes, we show Figures 3.1 and Figure 3.2 with deep. An acre-foot contains 43,560 ft3, 1,233 m3,the applicable process under discussion added to any pre- or 325,829 gal (U.S.). Also abbreviated as ac-ft. © 2003 by CRC Press LLC
  • TABLE 3.1 Recommended Reference Material 1. Kerri, K.D. et al., Small Water System Operation and Maintenance, 4th ed., California State University, Sacramento, 1997. 2. Kerri, K.D. et al., Water Distribution System Operation and Maintenance, California State University, Sacramento, 1994. 3. Kerri, K.D. et al., Water Treatment Plant Operation, Vols. 1 and 2, California State University, Sacramento, 1990. 4. Centers for Disease Control, Basic Mathematics, self-study course #3011-G, Atlanta, GA, 1998. 5. Centers for Disease Control, Waterborne Disease Control, self-study course #3014-G, Atlanta, GA, 1998. 6. Centers for Disease Control, Water Fluoridation, self-study course #3017-G, Atlanta, GA, 1998. 7. Introduction to Water Sources and Transmission, Vol. 1, American Water Works Association, Denver, CO, 1983. 8. Introduction to Water Treatment, Vol. 2, American Water Works Association, Denver, CO, 1983. 9. Introduction to Water Distribution, Vol. 3, American Water Works Association, Denver, CO, 1983. 10. Introduction to Water Quality Analysis, Vol. 4, American Water Works Association, Denver, CO, 1983. 11. Reference Handbook: Basic Science Concepts and Applications, American Water Works Association, Denver, CO. 12. Handbook of Water Analysis, 2nd ed., HACH Chemical Company, Loveland, CO., 1992. 13. Methods for Chemical Analysis of Water and Wastes, U.S. Environmental Protection Agency, Environmental Monitoring System Laboratory-Cincinnati (ESSL-CL), EPA-6000/4–79–020, revised March 1983 and 1979 (where applicable). 14. Standard Methods for the Examination of Water and Wastewater, 20th ed., American Public Health Association, Washington, D.C., 1998. 15. Price, J.K., Basic Math Concepts: For Water and Wastewater Plant Operators, Technomic Publ., Lancaster, PA, 1991. 16. Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1, 2, and 3, Technomic Publ., Lancaster, PA, 1999–2000. 17. Spellman, F.R. and Drinan, J., The Handbook for Waterworks Operator Certification, Vols. 1, 2, and 3, Technomic Publ., Lancaster, PA, 2001. 18. Spellman, F.R. and Drinan, J., Fundamentals for the Water and Wastewater Maintenance Operator Series: Electricity, Electronics, Pumping, Water Hydraulics, Piping and Valves, and Blueprint Reading, Technomic Publ., Lancaster, PA (distributed by CRC Press, Boca Raton, FL), 2000–2002. 19. Wastewater Treatment Plants: Planning, Design, and Operation, 2nd ed. Qasim, S.R. Lancaster, PA: Technomic Publ., Inc., 1999. 20. Haller, E., Simplified Wastewater Treatment Plant Operations, Technomic Publ., Lancaster, PA, 1999. 21. Kerri, K.D. et al., Operation of Wastewater Treatment Plants: A Field Study Program, Vol. 1, 4th ed., California State University, Sacramento, 1993. 22. Kerri, K.D. et al., Operation of Wastewater Treatment Plants, A Field Study Program, Vol. 2, 4th ed., California State University, Sacramento, 1993. Addition of Coagulant Water Mixing Flocculation Settling Sand To Storage and Supply Tank Basin Tank Filter Distribution Screening Sludge Disinfection ProcessingFIGURE 3.1 The water treatment model. (From Spellman, F.R., The Handbook for Waterworks Operator Certification, Vol. 1,Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Primary Treatment Secondary Treatment Influent Air Chlorine Effluent Collection Screening and Grit Primary Aeration Secondary Chlorine System Comminution Chamber Settling Settling Contact Tank Activated Sludge Screenings Grit Anaerobic Sludge Thickener Digester Dewatering Sludge DisposalFIGURE 3.2 Schematic of conventional wastewater treatment process. (From Spellman, F.R., Spellman’s Standard Handbook forWastewater Operators, Vol. 1, Technomic Publ., Lancaster, PA, 1999.)Activated carbon derived from vegetable or animal Aerobic conditions in which free, elemental oxygen materials by roasting in a vacuum furnace. Its is present. Also used to describe organisms, porous nature gives it a very high surface area biological activity, or treatment processes that per unit mass, as much as 1,000 m2/g, which is require free oxygen. 10 million times the surface area of 1 g of water Agglomeration floc particles colliding and gathering in an open container. Used in adsorption (see into a larger settleable mass. definition), activated carbon adsorbs substances Air gap the air space between the free-flowing dis- that are not or are only slightly adsorbed by charge end of a supply pipe and an unpressur- other methods. ized receiving vessel.Activated sludge the solids formed when micro- Algae bloom a phenomenon whereby excessive nutri- organisms are used to treat wastewater using ents within a river, stream, or lake causes an the activated sludge treatment process. It explosion of plant life that results in the deple- includes organisms, accumulated food materi- tion of the oxygen in the water needed by fish als, and waste products from the aerobic and other aquatic life. Algae bloom is usually decomposition process. the result of urban runoff (of lawn fertilizers, etc.). The potential tragedy is that of a “fish kill,”Advanced wastewater treatment treatment tech- where the stream life dies in one mass execution. nology to produce an extremely high-quality Alum aluminum sulfate, a standard coagulant used in discharge. water treatment.Adsorption the adhesion of a substance to the surface Ambient the expected natural conditions that occur in of a solid or liquid. Adsorption is often used to water unaffected or uninfluenced by human extract pollutants by causing them to attach to activities. such adsorbents as activated carbon or silica Anaerobic conditions in which no oxygen (free or gel. Hydrophobic (water-repulsing) adsorbents combined) is available. Also used to describe are used to extract oil from waterways in oil organisms, biological activity, or treatment pro- spills. cesses that function in the absence of oxygen.Aeration the process of bubbling air through a solu- Anoxic conditions in which no free, elemental oxygen tion, sometimes cleaning water of impurities by is present. The only source of oxygen is com- exposure to air. bined oxygen, such as that found in nitrate com- © 2003 by CRC Press LLC
  • pounds. Also used to describe biological activity view the term sludge as an ugly, inappropriate of treatment processes that function only in the four-letter word to describe biosolids. Biosolids presence of combined oxygen. is a product that can be reused; it has someAquifer a water-bearing stratum of permeable rock, value. Because biosolids has value, it certainly sand, or gravel. should not be classified as a waste product, andAquifer system a heterogeneous body of introduced when biosolids for beneficial reuse is permeable and less permeable material that addressed, it is made clear that it is not. acts as a water-yielding hydraulic unit of Biota all the species of plants and animals indigenous regional extent. to a certain area.Artesian water a well tapping a confined or artesian Boiling point the temperature at which a liquid boils. aquifer in which the static water level stands The temperature at which the vapor pressure above the top of the aquifer. The term is some- of a liquid equals the pressure on its surface. If times used to include all wells tapping confined the pressure of the liquid varies, the actual boil- water. Wells with water level above the water ing point varies. The boiling point of water is table are said to have positive artesian head 212°F or 100°C. (pressure), and those with water level below the Breakpoint point at which chlorine dosage satisfies water table have negative artesian head. chlorine demand.Average monthly discharge limitation the highest Breakthrough in filtering, when unwanted materials allowable discharge over a calendar month. start to pass through the filter.Average weekly discharge limitation t h e h i g h e s t Buffer a substance or solution that resists changes in allowable discharge over a calendar week. pH.Backflow reversal of flow when pressure in a service Calcium carbonate compound principally responsi- connection exceeds the pressure in the distribu- ble for hardness. tion main. Calcium hardness portion of total hardness causedBackwash fluidizing filter media with water, air, or a by calcium compounds. combination of the two so that individual grains Carbonaceous biochemical oxygen demand (CBOD5) can be cleaned of the material that has accu- the amount of biochemical oxygen demand that mulated during the filter run. can be attributed to carbonaceous material.Bacteria any of a number of one-celled organisms, Carbonate hardness caused primarily by compounds some of which cause disease. containing carbonate.Bar screen a series of bars formed into a grid used to Chemical oxygen demand (COD) the amount of screen out large debris from influent flow. chemically oxidizable materials present in theBase a substance that has a pH value between 7 and 14. wastewater.Basin a groundwater reservoir defined by the overly- Chlorination disinfection of water using chlorine as ing land surface and underlying aquifers that the oxidizing agent. contain water stored in the reservoir. Clarifier a device designed to permit solids to settleBeneficial use of water the use of water for any ben- or rise and be separated from the flow. Also eficial purpose. Such uses include domestic use, known as a settling tank or sedimentation basin. irrigation, recreation, fish and wildlife, fire pro- Coagulation the neutralization of the charges of col- tection, navigation, power, industrial use, etc. loidal matter. The benefit varies from one location to another Coliform a type of bacteria used to indicate possible and by custom. What constitutes beneficial use human or animal contamination of water. is often defined by statute or court decisions. Combined sewer a collection system that carries bothBiochemical oxygen demand (BOD) t h e o x y g e n wastewater and stormwater flows. used in meeting the metabolic needs of aerobic Comminution a process to shred solids into smaller, microorganisms in water rich in organic matter. less harmful particles.Biosolids from (1977) solid organic matter recovered Composite sample a combination of individual sam- from a sewage treatment process and used espe- ples taken in proportion to flow. cially as fertilizer (or soil amendment); usually Connate water pressurized water trapped in the pore used in plural (from Merriam-Webster’s Colle- spaces of sedimentary rock at the time it was giate Dictionary, 10th ed., 1998). deposited. It is usually highly mineralized. Note: In this text, biosolids is used in many Consumptive use (1) the quantity of water absorbed places (activated sludge being the exception) to by crops and transpired or used directly in the replace the standard term sludge. The authors building of plant tissue, together with the water © 2003 by CRC Press LLC
  • evaporated from the cropped area; (2) the quan- offensive odors. Excessive organic matter tity of water transpired and evaporated from a present in water because of inadequate waste cropped area or the normal loss of water from treatment and runoff from agricultural or urban the soil by evaporation and plant transpiration; land generally causes low DO. (3) the quantity of water discharged to the atmo- Dissolved solids the total amount of dissolved inor- sphere or incorporated in the products of the ganic material contained in water or wastes. process in connection with vegetative growth, Excessive dissolved solids make water unsuit- food processing, or an industrial process. able for drinking or industrial uses.Contamination (water) damage to the quality of Domestic consumption (use) water used for house- water sources by sewage, industrial waste, or hold purposes such as washing, food prepara- other material. tion, and showers. The quantity (or quantity perCross-connection a connection between a storm drain capita) of water consumed in a municipality or system and a sanitary collection system, district for domestic uses or purposes during a between two sections of a collection system to given period, it sometimes encompasses all uses, handle anticipated overloads of one system, or including the quantity wasted, lost, or otherwise between drinking (potable) water and an unsafe unaccounted for. water supply or sanitary collection system. Drawdown lowering the water level by pumping. It isDaily discharge the discharge of a pollutant measured measured in feet for a given quantity of water during a calendar day or any 24-hour period pumped during a specified period, or after the that reasonably represents a calendar day for pumping level has become constant. the purposes of sampling. Limitations Drinking water standards established by state agen- expressed as weight are total mass (weight) dis- cies, U.S. Public Health Service, and Environ- charged over the day. Limitations expressed in mental Protection Agency (EPA) for drinking other units are average measurement of the water in the U.S.. day. Effluent something that flows out, usually a pollutingDaily maximum discharge the highest allowable val- gas or liquid discharge. ues for a daily discharge. Effluent limitation any restriction imposed by theDarcy’s Law an equation for the computation of the regulatory agency on quantities, discharge quantity of water flowing through porous rates, or concentrations of pollutants discharged media. Darcy’s Law assumes that the flow is from point sources into state waters. laminar and that inertia can be neglected. The Energy in scientific terms, the ability or capacity of law states that the rate of viscous flow of homog- doing work. Various forms of energy include enous fluids through isotropic porous media is kinetic, potential, thermal, nuclear, rotational, proportional to, and in the direction of, the and electromagnetic. One form of energy may hydraulic gradient. be changed to another, as when coal is burnedDetention time the theoretical time water remains in to produce steam to drive a turbine that pro- a tank at a give flow rate. duces electric energy.Dewatering the removal or separation of a portion of Erosion the wearing away of the land surface by wind, water present in a sludge or slurry. water, ice or other geologic agents. ErosionDiffusion the process by which both ionic and molec- occurs naturally from weather or runoff, but is ular species dissolved in water move from areas often intensified by human land use practices. of higher concentration to areas of lower con- Eutrophication the process of enrichment of water centration. bodies by nutrients. Eutrophication of a lake nor-Discharge monitoring report (DMR) the monthly mally contributes to its slow evolution into a bog report required by the treatment plant’s or marsh and ultimately to dry land. Eutrophi- National Pollutant Discharge Elimination Sys- cation may be accelerated by human activities, tem (NPDES) discharge permit. thereby speeding up the aging process.Disinfection water treatment process that kills patho- Evaporation the process by which water becomes a genic organisms. vapor at a temperature below the boiling point.Disinfection by-products (DBPs) c h e m i c a l c o m - Facultative organisms that can survive and function pounds formed by the reaction of disinfectant in the presence or absence of free, elemental with organic compounds in water. oxygen.Dissolved oxygen (DO) the amount of oxygen dis- Fecal coliform the portion of the coliform bacteria solved in water or sewage. Concentrations of group that is present in the intestinal tracts and less than 5 ppm can limit aquatic life or cause feces of warm-blooded animals. © 2003 by CRC Press LLC
  • Field capacity the capacity of soil to hold water. It is Headloss amount of energy used by water in moving measured as the ratio of the weight of water from one point to another. retained by the soil to the weight of the dry Heavy metals metallic elements with high atomic soil. weights (e.g., mercury, chromium, cadmium,Filtration the mechanical process that removes par- arsenic, and lead). They can damage living ticulate matter by separating water from solid things at low concentrations and tend to accu- material, usually by passing it through sand. mulate in the food chain.Floc solids that join to form larger particles that will Holding pond a small basin or pond designed to hold settle better. sediment laden or contaminated water until itFlocculation slow mixing process in which particles can be treated to meet water quality standards are brought into contact with the intent of pro- or used in some other way. moting their agglomeration. Hydraulic cleaning cleaning pipe with water underFlume a flow rate measurement device. enough pressure to produce high water velocities.Fluoridation chemical addition to water to reduce Hydraulic gradient a measure of the change in incidence of dental caries in children. groundwater head over a given distance.Food-to-microorganisms ratio (F/M) an activated Hydraulic head the height above a specific datum sludge process control calculation based upon (generally sea level) that water will rise in a the amount of food (BOD5 or COD) available well. per pound of mixed liquor volatile suspended Hydrologic cycle (water cycle) the cycle of water solids. movement from the atmosphere to the earth and back to the atmosphere through various pro-Force main a pipe that carries wastewater under pres- cesses. These processes include: precipitation, sure from the discharge side of a pump to a infiltration, percolation, storage, evaporation, point of gravity flow downstream. transpiration and condensation.Grab sample an individual sample collected at a ran- Hydrology the science dealing with the properties, domly selected time. distribution, and circulation of water.Graywater water that has been used for showering, Impoundment a body of water, such as a pond, con- clothes washing, and faucet uses. Kitchen sink fined by a dam, dike, floodgate, or other barrier, and toilet water is excluded. This water has that is used to collect and store water for future excellent potential for reuse as irrigation for use. yards. Industrial wastewater wastes associated with indus-Grit heavy inorganic solids, such as sand, gravel, egg- trial manufacturing processes. shells, or metal filings. Infiltration the gradual downward flow of water fromGroundwater the supply of fresh water found beneath the surface into soil material. the earth’s surface (usually in aquifers) often Infiltration/inflow extraneous flows in sewers; sim- used for supplying wells and springs. Because ply, inflow is water discharged into sewer pipes groundwater is a major source of drinking or service connections from such sources as water, concern is growing over areas where foundation drains, roof leaders, cellar and yard leaching agricultural or industrial pollutants or area drains, cooling water from air conditioners, substances from leaking underground storage and other clean-water discharges from commer- tanks (USTs) are contaminating groundwater. cial and industrial establishments. Defined byGroundwater hydrology the branch of hydrology that Metcalf & Eddy as follows:2 deals with groundwater. It consists of its occur- • Infiltration water entering the collection rence and movements, its replenishment and system through cracks, joints, or breaks. depletion, the properties of rocks that control • Steady inflow water discharged from cellar groundwater movement and storage, and the and foundation drains, cooling water dis- methods of investigation and use of groundwater. charges, and drains from springs andGroundwater recharge the inflow to a groundwater swampy areas. This type of inflow is steady reservoir. and is identified and measured along withGroundwater runoff a portion of runoff that has infiltration. passed into the ground, has become groundwa- • Direct flow those types of inflow that have ter, and has been discharged into a stream chan- a direct stormwater runoff connection to the nel as spring or seepage water. sanitary sewer and cause an almost immedi-Hardness the concentration of calcium and magne- ate increase in wastewater flows. Possible sium salts in water. sources are roof leaders, yard and areaway © 2003 by CRC Press LLC
  • drains, manhole covers, cross connections Membrane process a process that draws a measured from storm drains and catch basins, and volume of water through a filter membrane with combined sewers. small enough openings to take out contaminants. • Total inflow the sum of the direct inflow at Metering pump a chemical solution feed pump that any point in the system plus any flow dis- adds a measured amount of solution with each charged from the system upstream through stroke or rotation of the pump. overflows, pumping station bypasses, and Mixed liquor the suspended solids concentration of the like. the mixed liquor. • Delayed inflow stormwater that may Mixed liquor volatile suspended solids (MLVSS) the require several days or more to drain through concentration of organic matter in the mixed the sewer system. This category can include liquor suspended solids. the discharge of sump pumps from cellar Milligrams/liter (mg/L) a measure of concentration drainage as well as the slowed entry of sur- equivalent to parts per million. face water through manholes in ponded National Pollutant Discharge Elimination System areas. permit a permit that authorizes the dischargeInfluent wastewater entering a tank, channel, or treat- of treated wastes and specifies the conditions ment process. that must be met for discharge.Inorganic chemical/compounds c h e m i c a l s u b - Nephelometric turbidity unit (NTU) a u n i t t h a t stances of mineral origin, not of carbon struc- indicates the amount of turbidity in a water ture. These include metals such as lead, iron sample. (ferric chloride), and cadmium. Nitrogenous oxygen demand (NOD) a measure ofIon exchange process a process used to remove hard- the amount of oxygen required to biologically ness from water. oxidize nitrogen compounds under specifiedJar Test laboratory procedure used to estimate proper conditions of time and temperature. coagulant dosage. Nonpoint Source (NPS) Pollution forms of pollu-Langelier saturation index (L.I.) a numerical index tion caused by sediment, nutrients, organic and that indicates whether calcium carbonate will be toxic substances originating from land use deposited or dissolved in a distribution system. activities that are carried to lakes and streamsLeaching the process by which soluble materials in by surface runoff. Nonpoint source pollution the soil, such as nutrients, pesticide chemicals occurs when the rate of materials entering these or contaminants, are washed into a lower layer water bodies exceeds natural levels. of soil or are dissolved and carried away by Nutrients substances required to support living organ- water. isms. Usually refers to nitrogen, phosphorus,License a certificate issued by the State Board of iron, and other trace metals. Waterworks/Wastewater Works Operators Organic chemicals/compounds animal or plant-pro- authorizing the holder to perform the duties of duced substances containing mainly carbon, a wastewater treatment plant operator. hydrogen, and oxygen, such as benzene andLift station a wastewater pumping station designed to toluene. “lift” the wastewater to a higher elevation. A Parts per million (ppm) the number of parts by lift station normally employs pumps or other weight of a substance per million parts of water. mechanical devices to pump the wastewater This unit is commonly used to represent pollut- and discharges into a pressure pipe called a ant concentrations. Large concentrations are force main. expressed in percentages.Maximum contaminant level (MCL) an enforce- Pathogenic disease causing. A pathogenic organism is able standard for protection of human health. capable of causing illness.Mean cell residence time (MCRT) t h e a v e r a g e Percolation the movement of water through the sub- length of time mixed liquor suspended solids surface soil layers, usually continuing downward particle remains in the activated sludge process. to the groundwater or water table reservoirs. May also be known as sludge retention time. pH a way of expressing both acidity and alkalinity onMechanical cleaning clearing pipe by using equip- a scale of 0 to 14, with 7 representing neutrality; ment (bucket machines, power rodders, or hand numbers less than 7 indicate increasing acidity rods) that scrapes, cuts, pulls, or pushes the and numbers greater than 7 indicate increasing material out of the pipe. alkalinity. © 2003 by CRC Press LLC
  • Photosynthesis a process in green plants in which Return activated sludge solids (RASS) the concen- water, carbon dioxide, and sunlight combine to tration of suspended solids in the sludge flow form sugar. being returned from the settling tank to the headPiezometric surface an imaginary surface that coin- of the aeration tank. cides with the hydrostatic pressure level of River basin a term used to designate the area drained water in an aquifer. by a river and its tributaries.Point source pollution a type of water pollution Sanitary wastewater wastes discharged from resi- resulting from discharges into receiving waters dences and from commercial, institutional, and from easily identifiable points. Common point similar facilities that include both sewage and sources of pollution are discharges from facto- industrial wastes. ries and municipal sewage treatment plants. Schmutzdecke layer of solids and biological growthPollution the alteration of the physical, thermal, that forms on top of a slow sand filter, allowing chemical, or biological quality of, or the con- the filter to remove turbidity effectively without tamination of, any water in the state that renders chemical coagulation. the water harmful, detrimental, or injurious to Scum the mixture of floatable solids and water humans, animal life, vegetation, property, pub- removed from the surface of the settling tank. lic health, safety, or welfare, or impairs the use- Sediment transported and deposited particles derived fulness or the public enjoyment of the water for from rocks, soil, or biological material. any lawful or reasonable purpose. Sedimentation a process that reduces the velocity ofPorosity that part of a rock that contains pore spaces water in basins so that suspended material can without regard to size, shape, interconnection, settle out by gravity. or arrangement of openings. It is expressed as Seepage the appearance and disappearance of water at percentage of total volume occupied by spaces. the ground surface. Seepage designates move-Potable water water satisfactorily safe for drinking ment of water in saturated material. It differs purposes from the standpoint of its chemical, from percolation, which is predominantly the physical, and biological characteristics. movement of water in unsaturated material.Precipitate a deposit on the earth of hail, rain, mist, Septic tanks used to hold domestic wastes when a sleet, or snow. The common process by which sewer line is not available to carry them to a atmospheric water becomes surface or subsur- treatment plant. The wastes are piped to under- face water, the term precipitation is also com- ground tanks directly from a home or homes. monly used to designate the quantity of water Bacteria in the wastes decompose some of the precipitated. organic matter, the sludge settles on the bottomPreventive maintenance (PM) regularly scheduled of the tank, and the effluent flows out of the servicing of machinery or other equipment tank into the ground through drains. using appropriate tools, tests, and lubricants. Settleability a process control test used to evaluate the This type of maintenance can prolong the useful settling characteristics of the activated sludge. life of equipment and machinery and increase Readings taken at 30 to 60 min are used to its efficiency by detecting and correcting prob- calculate the settled sludge volume (SSV) and lems before they cause a breakdown of the the sludge volume index (SVI). equipment. Settled sludge volume (SSV) the volume (in percent)Purveyor an agency or person that supplies potable occupied by an activated sludge sample after water. 30 to 60 minutes of settling. Normally writtenRadon a radioactive, colorless, odorless gas that as SSV with a subscript to indicate the time of occurs naturally in the earth. When trapped in the reading used for calculation (SSV60 or buildings, concentrations build up and can SSV30). cause health hazards such as lung cancer. Sludge the mixture of settleable solids and waterRecharge the addition of water into a groundwater removed from the bottom of the settling tank. system. Sludge retention time (SRT) see mean cell residenceReservoir a pond, lake, tank, or basin (natural or time. human made) where water is collected and used Sludge volume index (SVI) a process control calcu- for storage. Large bodies of groundwater are lation used to evaluate the settling quality of the called groundwater reservoirs; water behind a activated sludge. Requires the SSV30 and mixed dam is also called a reservoir of water. liquor suspended solids test results to calculate.Reverse osmosis a process in which almost pure water Soil moisture (soil water) water diffused in the soil. is passed through a semipermeable membrane. It is found in the upper part of the zone of © 2003 by CRC Press LLC
  • aeration from which water is discharged by by molecules underlying the layer of surface transpiration from plants or by soil evaporation. molecules.Specific heat the heat capacity of a material per unit Thermal pollution the degradation of water quality by mass. The amount of heat (in calories) required the introduction of a heated effluent. Primarily to raise the temperature of 1 g of a substance the result of the discharge of cooling waters from 1°C; the specific heat of water is 1 cal. industrial processes (particularly from electricalStorm sewer a collection system designed to carry power generation), waste heat eventually results only stormwater runoff. from virtually every energy conversion.Stormwater runoff resulting from rainfall and Titrant a solution of known strength of concentration; snowmelt. used in titration.Stream a general term for a body of flowing water. In Titration a process whereby a solution of known hydrology, the term is generally applied to the strength (titrant) is added to a certain volume of water flowing in a natural channel as distinct treated sample containing an indicator. A color from a canal. More generally, it is applied to change shows when the reaction is complete. the water flowing in any channel, natural or Titrator an instrument (usually a calibrated cylinder artificial. Some types of streams include: (tube-form)) used in titration to measure the (1) Ephemeral: a stream that flows only in amount of titrant being added to the sample. direct response to precipitation, and whose Total dissolved solids the amount of material (inor- channel is at all times above the water table; ganic salts and small amounts of organic mate- (2) Intermittent or Seasonal: a stream that flows rial) dissolved in water and commonly only at certain times of the year when it receives expressed as a concentration in terms of milli- water from springs, rainfall, or from surface grams per liter. sources such as melting snow; (3) Perennial: a Total suspended solids (TSS) total suspended solids stream that flows continuously; (4) Gaining: a in water, commonly expressed as a concentra- stream or reach of a stream that receives water tion in terms of milligrams per liter. from the zone of saturation. It is an effluent Toxicity the occurrence of lethal or sublethal adverse stream; (5) Insulated: a stream or reach of a affects on representative sensitive organisms due stream that neither contributes water to the zone to exposure to toxic materials. Adverse effects of saturation nor receives water from it. It is caused by conditions of temperature, dissolved separated from the zones of saturation by an oxygen, or nontoxic dissolved substances are impermeable bed; (6) Losing: a stream or reach excluded from the definition of toxicity. of a stream that contributes water to the zone Transpiration the process by which water vapor of saturation. An influent stream; (7) Perched: escapes from the living plant, principally the a perched stream is either a losing stream or an leaves, and enters the atmosphere. insulated stream that is separated from the Vaporization the change of a substance from a liquid underlying groundwater by a zone of aeration. or solid state to a gaseous state.Supernatant the liquid standing above a sediment or Volatile Organic Compound (VOC) a n y o r g a n i c precipitate. compound that participates in atmospheric pho-Surface water lakes, bays, ponds, impounding reser- tochemical reactions except for those desig- voirs, springs, rivers, streams, creeks, estuaries, nated by the EPA administrator as having neg- wetlands, marshes, inlets, canals, gulfs inside ligible photochemical reactivity. the territorial limits of the state, and all other Wastewater the water supply of a community after it bodies of surface water, natural or artificial, has been soiled by use. inland or coastal, fresh or salt, navigable or Waste activated sludge solids (WASS) the concen- nonnavigable, and including the beds and tration of suspended solids in the sludge being banks of all watercourses and bodies of surface removed from the activated sludge process. water, that are wholly or partially inside or bor- Water cycle the process by which water travels in a dering the state or subject to the jurisdiction of sequence from the air (condensation) to the the state; except that waters in treatment sys- earth (precipitation) and returns to the atmo- tems that are authorized by state or federal law, sphere (evaporation). It is also referred to as regulation, or permit, and that are created for the hydrologic cycle. the purpose of water treatment are not consid- Water quality standard a plan for water quality man- ered to be waters in the state. agement containing four major elements: waterSurface tension the free energy produced in a liquid use, criteria to protect uses, implementation surface by the unbalanced inward pull exerted plans, and enforcement plans. An antidegradation © 2003 by CRC Press LLC
  • statement is sometimes prepared to protect exist- atmospheric pressure and would not flow into ing high quality waters. a well.Water quality a term used to describe the chemical, Zoogleal slime the biological slime that forms on physical, and biological characteristics of water fixed film treatment devices. It contains a wide variety of organisms essential to the treatment with respect to its suitability for a particular use. process.Water supply any quantity of available water.Waterborne disease a disease caused by a microor- ganism that is carried from one person or ani- 3.4 CHAPTER REVIEW QUESTIONS mal to another by water. AND PROBLEMSWatershed the area of land that contributes surface 3.1. Matching exercise: Match the definitions runoff to a given point in a drainage system. listed in part A with the terms listed in Part BWeir a device used to measure wastewater flow. by placing the correct letter in the blank. AfterZone of aeration a region in the earth above the water completing this exercise, check your answers table. Water in the zone of aeration is under with those provided in Appendix A. Part A Part B 1. A nonchemical turbidity removal layer in a slow sand filter. _______ a. pH 2. Region in earth (soil) above the water table. _______ b. algae bloom 3. Compound associated with photochemical reaction. _______ c. zone of aeration 4. Oxygen used in water rich inorganic matter. _______ d. hydrological cycle 5. A stream that receives water from the zone of saturation. _______ e. point source pollution 6. The addition of water into a groundwater system. _______ f. perennial 7. The natural water cycle. _______ g. organic 8. Present in intestinal tracts and feces of animals and humans. _______ h. connate water 9. Discharge from a factory or municipal sewage treatment plant. _______ i. fecal coliform 10. Common to fixed film treatment devices. _______ j. BOD 11. Identified water that is safe to drink. _______ k. field capacity 12. The capacity of soil to hold water. _______ l. transpiration 13. Used to measure acidity and alkalinity. _______ m. biota 14. Rain mixed with sulfur oxides. _______ n. specific heat 15. Enrichment of water bodies by nutrients. _______ o. Schmutzdecke 16. A solution of known strength of concentration. _______ p. recharge 17. Water lost by foliage. _______ q. Zoogleal slime 18. Another name for a wastewater pumping station. _______ r. eutrophication 19. Plants and animals indigenous to an area. _______ s. gaining 20. The amount of oxygen dissolved in water. _______ t. VOC 21. A stream that flows continuously. _______ u. potable 22. A result of excessive nutrients within a water body. _______ v. acid rain 23. Change in groundwater head over a given distance. _______ w. titrant 24. Water trapped in sedimentary rocks. _______ x. lift station 25. Heat capacity of a material per unit mass. _______ y. DO 26. A compound derived from material that once lived. ______ z. hydraulic gradientREFERENCES 2. Metcalf & Eddy, Inc. Wastewater Engineering: Treat- ment, Disposal, Reuse, 3rd ed., Tchobanoglous, G. 1. Pielou, E.C., Fresh Water, University of Chicago, Chi- (Ed.), McGraw-Hill, New York, 1991, pp. 29–31. cago, 1998, preface. © 2003 by CRC Press LLC
  • PART IIWater/Wastewater Operations:Math and Technical Aspects© 2003 by CRC Press LLC
  • Water and Wastewater 4 Math Operations To operate a waterworks and/or a wastewater treatment solve the problem, along with, “What are they plant, and to pass the examination for an operator’s license, really looking for?” Writing down exactly what you must know how to perform certain mathematical oper- is being looked for is always smart. Sometimes ations. However, do not panic, as Price points out, “Those the answer has more than one unknown. For who have difficulty in math often do not lack the ability instance, you may need to find X and then find Y. for mathematical calculation, they merely have not learned, or have not been taught, the ‘language of math.”1 4. If the calculation calls for an equation, write it down. 5. Fill in the data in the equation and look to see4.1 INTRODUCTION what is missing.Without the ability to perform mathematical calculations, 6. Rearrange or transpose the equation, if necessary.operators would have difficulty in properly operating water 7. If available, use a calculator.and wastewater systems. In reality, most of the calculations 8. Always write down the answer.operators need to perform are not difficult. Generally, math 9. Check any solution obtained.ability through basic algebra is all that is needed. Experi-ence has shown that skill with math operations used inwater and wastewater system operations is an acquired skill 4.3 TABLE OF EQUIVALENTS, FORMULAE,that is enhanced and strengthened with practice. AND SYMBOLSNote: Keep in mind that mathematics is a language — In order to work mathematical operations to solve a universal language. Mathematical symbols problems (for practical application or for taking licensure have the same meaning to people speaking examinations), it is essential to understand the language, many different languages throughout the globe. equivalents, symbols, and terminology used. The key to learning mathematics is to learn the Because this handbook is designed for use in practical language — the symbols, definitions and terms, on-the-job applications, equivalents, formulae, and symbols of mathematics that allow you to understand the are included, as a ready reference, in Table 4.1. concepts necessary to perform the operations. In this chapter, we assume the reader is well grounded 4.4 TYPICAL WATER AND WASTEWATERin basic math principles. We do not cover basic operationssuch as addition, subtraction, multiplication, and division. MATH OPERATIONSHowever, we do include, for review purposes, a few basic 4.4.1 ARITHMETIC AVERAGE (OR ARITHMETIC MEAN)math calculations in the Chapter Review Questions/Prob- AND MEDIANlems at the end of the chapter. During the day-to-day operation of a wastewater treatment4.2 CALCULATION STEPS plant, considerable mathematical data are collected. The data, if properly evaluated, can provide useful informationAs with all math operations, many methods can be suc- for trend analysis and indicate how well the plant or unitcessfully used to solve water and wastewater system prob- process is operating. However, because there may be muchlems. We provide one of the standard methods of problem variation in the data, it is often difficult to determine trendssolving in the following: in performance. Arithmetic average refers to a statistical calculation 1. If appropriate, make a drawing of the informa- used to describe a series of numbers such as test results. tion in the problem. By calculating an average, a group of data is represented 2. Place the given data on the drawing. by a single number. This number may be considered typical 3. Determine what the question is. This is the first of the group. The arithmetic mean is the most commonly thing you should determine as you begin to used measurement of average value. © 2003 by CRC Press LLC
  • TABLE 4.1 Equivalents, Formulae, and Symbols Equivalents 12 in. = 1 ft 36 in. = 1 yd 144 in.2 = 1 ft2 9 ft2 = 1 yd2 43,560 ft2 = 1 ac 1 ft3 = 1728 in.3 1 ft3 H20 = 7.48 gal 1 ft3 H20 = 62.4 lb 1 gal of H20 = 8.34 lb 1L = 1.000 mL 1g = 1.000 mg 1 MGD (million gal[MG]/d) = 694 gal/min, 1.545 ft3/sec average BOD/capita/day = .17 lb average SS/capita/day = .20 average daily flow = assume 100 gal/capita/day Symbols A = Area V = Velocity t = Time SVI = Sludge Volume Index v = Volume # = Pounds eff = Effluent W = Width D = Depth L = Length H = Height Q = Flow C = Circumference r = Radius p = pi (3.14) WAS = Waste activated sludge RAS = Return activated sludge MLSS = Mixed liquor suspended solids MLVSS = Mixed liquor volatile suspended solids Formulae v SVI = ¥ 100 Concentration Q=A¥V Detention time = v/Q v=L¥W¥D area = W ¥ L Circular area = p ¥ Diameter2 C = pd Hydraulic loading rate = Q/A # MLSS in aeration tank Sludge age = # SS in primary eff d # SS in secondary system (aeration tank + sec. clarifier ) Mean cell residence time = # WAS d + SS in eff d # BOD d Organic loading rate = v Source: From Spellman, F.R., Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.© 2003 by CRC Press LLC
  • Note: When evaluating information based on aver- EXAMPLE 4.3 ages, remember that the average reflects the Problem: general nature of the group and does not nec- essarily reflect any one element of that group. A waterworks operator takes a chlorine residual measure- ment every day. We show part of the operating log in Arithmetic average is calculated by dividing the sum Table 4.2.of all of the available data points (test results) by thenumber of test results: TABLE 4.2 Test 1 + Test 2 + Test 3 + L + Test N Daily Chlorine Residual Results (4.1) Number of Tests Performed ( N ) Day Chlorine Residual (mg/L) EXAMPLE 4.1 Monday 0.9 Tuesday 1.0 Problem: Wednesday 1.2 Thursday 1.3 Effluent biochemical oxygen demand (BOD) test results Friday 1.4 for the treatment plant during the month of September are Saturday 1.1 shown below: Sunday 0.9 Test 1 20 mg/L Test 2 31 mg/L Source: From Spellman, F.R., Standard Test 3 22 mg/L Handbook for Wastewater Operators, Vol. Test 4 15 mg/L 1–3, Technomic Publ., Lancaster, PA, 1999. What is the average effluent BOD for the month of Find the mean. September? Solution: Solution: Add up the seven chlorine residual readings: 0.9 + 1.0 + 20 mg L + 31 mg L + 22 mg L + 15 mg L 1.2 + 1.3 + 1.4 + 1.1 + 0.9 = 7.8. Next, divide by the Average = number of measurements (in this case 7): 7.8 ∏ 7 = 1.11. 4 The mean chlorine residual for the week was 1.11 mg/L. = 22 mg L Definition: The median is defined as the value of EXAMPLE 4.2 the central item when the data are arrayed by size. First, arrange all of the readings in either Problem: ascending or descending order. Then find the For the primary influent flow, the following composite- middle value. sampled solids concentrations were recorded for the week: EXAMPLE 4.4 Monday 300 mg/L SS Tuesday 312 mg/L SS Problem: Wednesday 315 mg/L SS In our chlorine residual example, what is the median? Thursday 320 mg/L SS Friday 311 mg/L SS Saturday 320 mg/L SS Solution: Sunday 310 mg/L SS Arrange the values in ascending order: Total 2188 mg/L SS 0.9, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4 Solution: The middle value is the fourth one — 1.1. Therefore, the Sum of All Measurements median chlorine residual is 1.1 mg/L. (Usually, the median Average SS = will be a different value than the mean.) Number of Measurements Used 2188 mg L SS If the data contain an even number of values, you must = 7 add one more step, since no middle value is present. You must find the two values in the middle, and then find the = 312.6 mg L SS mean of those two values. © 2003 by CRC Press LLC
  • EXAMPLE 4.5 4.4.2 RATIO Problem: Recall that a ratio is the comparison of two numbers by A water system has four wells with the following capac- division or an indicated division. A ratio is usually stated ities: 115, 100, 125, and 90 gal/min. What are the mean in the form A is to B as C is to D, and is written as two and the median pumping capacities? fractions that are equal to each other: Solution: A C = (4.2) The mean is: B D We solve ratio problems by cross-multiplying; we 115 gal min + 100 gal min + 125 gal min + 90 gal min = multiply the left numerator (A) by the right denominator 4 (D) and say that A is equal to the left denominator (B) 107.5 gal min times the right numerator (C): To find the median, arrange the values in order: A¥D=B¥C 90 gal/min, 100 gal/min, 115 gal/min, 125 gal/min AD = BC (4.3) With four values, there is no single middle value, so we If one of the four items is unknown, we solve the ratio must take the mean of the two middle values: by dividing the two known items that are multiplied together by the known item that is multiplied by the 100 gpm + 115 gpm unknown. This is best shown by a couple of examples: = 107.5 gpm 2 EXAMPLE 4.7Note: At times, determining what the original num- Problem: bers were like is difficult (if not impossible) when dealing only with averages. If we need 4 lb of alum to treat 1000 gal of H2O, how many pounds of alum will we need to treat 12,000 gal- EXAMPLE 4.6 lons? Problem: Solution: A water system has four storage tanks. Three of them have We state this as a ratio: 4 lb of alum is to 1000 gallons of a capacity of 100,000 gal each, while the fourth has a H2O as pounds of alum (or x) is to 12,000 gal. We set this capacity of 1 million gallons (MG). What is the mean up this way: capacity of the storage tanks? Solution: 4 lb alum x lb alum = 1000 gal H 2 O 12, 000 gal H 2 O The mean capacity of the storage tanks is: Cross-multiplying: 100, 000 + 100, 000 + 100, 000 + 1, 000, 000 = 325, 000 gal 4 1000 ¥ x = 4 ¥ 12, 000 Note: Notice that no tank in Example 4.6 has a 4 ¥ 12, 000 x= capacity anywhere close to the mean. The 1000 median capacity requires us to take the mean of the two middle values; since they are both x = 48 lb alum 100,000 gal, the median is 100,000 gal. Although three of the tanks have the same EXAMPLE 4.8 capacity as the median, these data offer no Problem: indication that one of these tanks holds 1 million gal — information that could be If 10 gal of fuel oil costs $5.25, how much does 18 gal important for the operator to know. cost? © 2003 by CRC Press LLC
  • 10 gal 18 gal EXAMPLE 4.10 = $5.25 $y Problem: 10 ¥ y = 18 ¥ $5.25 Sludge contains 5.3% solids. What is the concentration of solids in decimal percent? 18 ¥ $5.25 y= 10 Solution: 94.5 y= Decimal Percent = 5.3% = 0.053 10 100 y = $9.45 Note: Unless otherwise noted all calculations in the handbook using percent values require the4.4.3 PERCENT percent be converted to a decimal before use.Percent (like fractions) is another way of expressing a part Note: To determine what quantity a percent equalsof a whole. The term percent means per hundred, so a first convert the percent to a decimal thenpercentage is the number out of 100. For example, 22% multiply by the total quantity.means 22 out of 100, or if we divide 22 by 100, we get Quantity = Total ¥ Decimal Percent (4.5)the decimal 0.22: EXAMPLE 4.11 22 22% = = 0.22 Problem: 100 Sludge drawn from the settling tank is 8% solids. If 24004.4.3.1 Practical Percentage Calculations gal of sludge are withdrawn, how many gallons of solids are removed?Percentage is often designated by the symbol %. Thus,10% means 10 percent, 10/100, or 0.10. These equivalents Solution:may be written in the reverse order: 0.10 = 10/100 = 10%. 8% gallons = ¥ 2400 gal = 192 galIn water and wastewater treatment, percent is frequently 100used to express plant performance and for control ofsludge treatment processes. EXAMPLE 4.12 Problem:Note: To determine percent divide the quantity you wish to express as a percent by the total quantity Calcium hypochlorite (HTH) contains 65% available chlorine. What is the decimal equivalent of 65%? then multiply by 100. Solution: Quantity ¥ 100 Percent = (4.4) Since 65% means 65 per hundred, divide 65 by 100 Total (65/100), which is 0.65. EXAMPLE 4.9 EXAMPLE 4.13 Problem: Problem: The plant operator removes 6000 gal of sludge from the If a 50-ft high water tank has 32 ft of water in it, how full is the tank in terms of the percentage of its capacity? settling tank. The sludge contains 320 gal of solids. What is the percent of solids in the sludge? Solution: Solution: 32 ft = 0.64 (decimal equivalent ) 50 ft 320 gal Percent = ¥ 100 0.64 ¥ 100 = 64% 6000 gal = 5.3% Thus, the tank is 64% full. © 2003 by CRC Press LLC
  • 4.4.4 UNITS AND CONVERSIONS TABLE 4.3Most of the calculations made in the water and wastewater Common Conversionsoperations involve using units. While the number tells ushow many, the units tell us what we have. Examples of Linear Measurementsunits include: inches, feet, square feet, cubic feet, gallons, 1 in. = 2.54 cmpounds, milliliters, milligrams per liter, pounds per square 1 ft = 30.5 cm 1 m = 100 cm = 3.281 ft = 39.4 in.inch, miles per hour, and so on. 1 ac = 43,560 ft2 Conversions are a process of changing the units of a 1 yd = 3 ftnumber to make the number usable in a specific instance. VolumeMultiplying or dividing into another number to change the 1 gal = 3.78 Lunits of the number accomplishes conversions. Common 1 ft3 = 7.48 galconversions in water and wastewater operations are: 1 L = 1000 mL 1 ac-ft = 43,560 ft3 1. Gallons per minute to cubic feet per second 1 gal = 32 cups 2. Million gallons to acre-feet 1 lb = 16 oz dry wt. 3. Cubic foot per second to acre-feet Weight 4. Cubic foot per second of water to weight 1 ft3 of water = 62.4 lb 5. Cubic foot of H2O to gallons 1 gal = 8.34 lb 1 lb = 453.6 g 6. Gallons of water to weight 1 kg = 1000 g = 2.2 lb 7. Gallons per minute to million gallons per day 1% = 10,000 mg/L 8. Pounds to feet of head (the measure of the pres- Pressure sure of water expressed as height of water in 1 ft of head = 0.433 psi feet — 1 psi = 2.31 feet of head) 1 psi = 2.31 ft of head Flow In many instances, the conversion factor cannot be 1 ft3/sec = 448 gal/minderived — it must be known. Therefore, we use tables 1 gal/min = 1440 gal/dsuch as the one below (Table 4.3) to determine the com-mon conversions. Source: From Spellman, F.R., Standard Handbook for Wastewater Operators, Vol.Note: Conversion factors are used to change measure- 1–3, Technomic Publ., Lancaster, PA, 1999. ments or calculated values from one unit of measure to another. In making the conversion from one unit to another, you must know two ∞C = 5 9 (∞F - 32∞) (4.6) things: (1) the exact number that relates the two units, and (2) whether to multiply or divide by ∞F = 5 9 (∞C + 32∞) (4.7) that number Most operators memorize some standard conversions. These conversions are not difficult to perform. TheThis happens because of using the conversions, not difficulty arises when we must recall these formulas frombecause of attempting to memorize them. memory. Probably the easiest way to recall these important4.4.4.1 Temperature Conversions formulas is to remember three basic steps for both Fahr- enheit and Celsius conversions:An example of a type of conversion typical in water andwastewater operations is provided in this section on tem- 1. Add 40°.perature conversions. 2. Multiply by the appropriate fraction (5/9 or 9/5). 3. Subtract 40°.Note: Operators should keep in mind that temperature conversions are only a small part of the many conversions that must be made in real world Obviously, the only variable in this method is the systems operations. choice of 5/9 or 9/5 in the multiplication step. To make the proper choice, you must be familiar with two scales. Most water and wastewater operators are familiar with On the Fahrenheit scale, the freezing point of water is 32°,the formulas used for Fahrenheit and Celsius temperature and 0° on the Celsius scale. The boiling point of water isconversions: 212° on the Fahrenheit scale and 100° on the Celsius scale. © 2003 by CRC Press LLC
  • What does this mean? Why is it important? Obviously, knowing how to make these temperature Note, for example, that at the same temperature, conversion calculations is useful. However, in practicalhigher numbers are associated with the Fahrenheit scale (real world) operations, you may wish to use a temperatureand lower numbers with the Celsius scale. This important conversion table.relationship helps you decide whether to multiply by 5/9or 9/5. Let us look at a few conversion problems to see 4.4.4.2 Milligrams per Liter (Parts per Million)how the three-step process works. One of the most common terms for concentration is milli- EXAMPLE 4.14 grams per liter (mg/L). For example, if a mass of 15 mg of oxygen is dissolved in a volume of 1 L of water, the con- Problem: centration of that solution is expressed simply as 15 mg/L. Convert 220°F to Celsius. Very dilute solutions are more conveniently expressed in terms of micrograms per liter (µg/L). For example, a Solution: concentration of 0.005 mg/L is preferably written as its equivalent, 5 µg/L. Since 1000 µg = 1 mg, simply move Using the three-step process, we proceed as follows: the decimal point three places to the right when converting from mg/L to µL. Move the decimal three places to the Step 1: Add 40°F: left when converting from µg/L, to mg/L. For example, a 220°F + 40°F = 260°F concentration of 1250 µ/L is equivalent to 1.25 mg/L. One liter of water has a mass of 1 kg. But 1 kg is Step 2: 260°F must be multiplied by either 5/9 or 9/5. Since equivalent to 1000 g or 1,000,000 mg. Therefore, if we the conversion is to the Celsius scale, you will be moving dissolve 1 mg of a substance in 1 L of H2O, we can say to number smaller than 260. Through reason and observa- that there is 1 mg of solute per 1 million mg of water, or tion, obviously we see that multiplying 260 by 9/5 would almost be the same as multiplying by 2, which would in other words, 1 part per million (ppm). double 260, rather than make it smaller. On the other hand, Note: For comparative purposes, we like to say that multiplying by 5/9 is about the same as multiplying by 1/2, 1 ppm is analogous to a full shot glass of water which would cut 260 in half. Because we wish to move to a smaller number, we should multiply by 5/9: sitting in the bottom of a full standard swim- ming pool. 5/9 ¥ 260°F = 144.4°C Neglecting the small change in the density of water as substances are dissolved in it, we can say that, in gen- Step 3: Now subtract 40°: eral, a concentration of 1 mg/L is equivalent to 1 ppm. Conversions are very simple; for example, a concentration 144.4°C ¥ 40°C = 104.4°C of 18.5 mg/L is identical to 18.5 ppm. Therefore, 220°F = 104.4°C The expression mg/L is preferred over ppm, just as the expression µg/L is preferred over its equivalent of parts per billion (ppb). However, both types of units are still EXAMPLE 4.15 used, and the waterworks/wastewater operator should be Problem: familiar with both. Convert 22°C to Fahrenheit. 4.5 MEASUREMENTS: AREAS Step 1: Add 40°F: AND VOLUMES 22°F + 40°F = 62°F Water and wastewater operators are often required to cal- culate surface areas and volumes. Step 2: Because we are converting from Celsius to Fahr- enheit, we are moving from a smaller to larger number, Area is a calculation of the surface of an object. For and should use 9/5 in the multiplication: example, the length and the width of a water tank can be measured, but the surface area of the water in the tank 9/5 ¥ 62°F = 112°F must be calculated. An area is found by multiplying two length measurements, so the result is a square measure- Step 3: Subtract 40°: ment. For example, when multiplying feet by feet, we get square feet, which is abbreviated ft2. Volume is the calcu- 112°F – 40°F = 72°F lation of the space inside a three-dimensional object, and Thus, 22°C = 72°F is calculated by multiplying three length measurements, © 2003 by CRC Press LLC
  • L W 8 ft D = 25 ft r = 12.5 ft C=? 12 ft FIGURE 4.1 Rectangular shape showing calculation of sur- face area. (From Spellman, F.R., Spellman’s Standard Hand- book for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) FIGURE 4.2 Circular shape showing diameter and radius.or an area by a length measurement. The result is a cubic (From Spellman, F.R., Spellman’s Standard Handbook formeasurement, such as cubic feet (abbreviated ft3). Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)4.5.1 AREA OF A RECTANGLE EXAMPLE 4.16The area of square or rectangular figures (such as the one Problem:shown in Figure 4.1) is calculated by multiplying themeasurements of the sides. Find the area of the circle shown in Figure 4.2. A=L¥W (4.8) Solution: To determine the area of the rectangle shown in A = p ¥ r2Figure 4.1, we proceed as follows: A = p ¥ 12.5 ft ¥ 12.5 ft A=L¥W A = 3.1416 ¥ 12.5 ft ¥ 12.5 ft A = 12 ft ¥ 8 ft A = 490.9 ft2 A = 96 ft2 At times, finding the diameter of a circular object is necessary, under circumstances that allow you to measure4.5.2 AREA OF A CIRCLE only the circumference (e.g., a pump shaft). The diameter and the circumference are related by the constant p:The diameter of a circle is the distance across the circlethrough its center, and is shown in calculations and on C C = p ¥ Diameter or (4.10)drawings by the letter D (see Figure 4.2). Half of the pdiameter — the distance from the center to the outsideedge — is called the radius (r). The distance around the 4.5.3 AREA OF A CIRCULAR OR CYLINDRICAL TANKoutside of the circle is called the circumference (C). In calculating the area of a circle, the radius must be If you were supervising a work team assigned to paint amultiplied by itself (or the diameter by itself); this process water or chemical storage tank, you would need to knowis called squaring, and is indicated by the superscript the surface area of the walls of the tank. To determine thenumber 2 following the item to be squared. For example, tank’s surface area, visualize the cylindrical walls as athe radius squared is written as r 2, which indicates to rectangle wrapped around a circular base. The area of amultiply the radius by the radius. rectangle is found by multiplying the length by the width; in this case, the width of the rectangle is the height of the When making calculations involving circular objects, wall and the length of the rectangle is the distance arounda special number is required — referred to by the Greek the circle, the circumference.letter pi (pronounced pie); the symbol for pi is p. Pi alwayshas the value 3.1416. The area of the sidewalls of a circular tank is found by multiplying the circumference of the base (C = p ¥ The area of a circle is equal to the radius squared times Diameter) times the height of the wall (H):the number pi. A = p ¥ r2 (4.9) A = p ¥ Diameter ¥ H (4.11) © 2003 by CRC Press LLC
  • For a tank with Diameter = 20 ft and H = 25 ft: r = 10 ft A=p¥D¥H A = p ¥ 20 ft ¥ 25 ft A = 3.1416 ¥ 20 ft ¥ 25 ft A = 1570.8 ft 2 H = 25 ft To determine the amount of paint needed, rememberto add the surface area of the top of the tank, which inthis case we will say is 314 ft2. Thus, the amount of paintneeded must cover 1570.8 ft2 + 314 ft2 = 1884.8 or 1885ft2. If the tank floor should be painted, add another 314 ft2.4.5.4 VOLUME CALCULATIONS4.5.4.1 Volume of Rectangular Tank FIGURE 4.4 Circular or cylindrical water tank. (From Spell- man, F.R., Spellman’s Standard Handbook for WastewaterThe volume of a rectangular object (such as a settling tank Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)like the one shown in Figure 4.3) is calculated by multi-plying together the length, the width, and the depth. To Note: For many calculations involving water, we needcalculate the volume, you must remember that the length to know the volume of the tank in gallons rathertimes the width is the surface area, which is then multi- than 1 ft3 contains 7.48 gal.plied by the depth. v=L¥W¥D (4.12) 4.5.4.2 Volume of a Circular or Cylindrical Tank EXAMPLE 4.17 A circular tank consists of a circular floor surface with a cylinder rising above it (see Figure 4.4). The volume of a Problem: circular tank is calculated by multiplying the surface area Using the dimensions given in Figure 4.3, determine the times the height of the tank walls. volume. EXAMPLE 4.18 Solution: Problem: v=L ¥ W ¥ D If a tank is 20 ft in diameter and 25 ft deep, how many v=A ¥ D gallons of water will it hold? v = 12 ft ¥ 6 ft ¥ 6 ft Solution: v = 432 ft 3 In this type of problem, calculate the surface area first, multiply by the height and then convert to gallons. 12 ft W 6 ft r = Diameter ∏ 2 = 20 ft ∏ 2 = 10 ft 6 ft A = p ¥ r2 D A = p ¥ 10 ft ¥ 10 ft A = 314 ft 2 L v=A ¥ H FIGURE 4.3 Rectangular settling tank illustrating calcula- tion of volume. (From Spellman, F.R., Spellman’s Standard v = 314 ft 2 ¥ 25 ft Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) v = 7850 ft 3 ¥ 7.5 gal ft 3 = 58.875 gal © 2003 by CRC Press LLC
  • 50 ft 70 ft 8 ft 30 ft 12 ft FIGURE 4.5 Rectangular tank. (From Spellman, F.R., Spell- FIGURE 4.6 Circular tank. (From Spellman, F.R., Spell- man’s Standard Handbook for Wastewater Operators, Vol. man’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) 1–3, Technomic Publ., Lancaster, PA, 1999.)4.5.4.3 Example Volume Problems EXAMPLE 4.21: CYLINDRICAL TANK VOLUME EXAMPLE 4.19: TANK VOLUME (RECTANGULAR) Problem: Problem: A cylindrical tank is 12 ft in diameter and 24 ft in height. What is the approximate capacity in liters? Calculate the volume of the rectangular tank shown in Figure 4.5. Note: Draw a rough diagram of the tank and labelNote: If a drawing is not supplied with the problem, dimensions. draw a rough picture or diagram. Make sure you label or identify the parts of the diagram Solution: from the information given in the question (see Figure 4.5). p ¥ d2 v( tan k ) = ¥H 4 Solution: p ¥ (12 ft ) 2 = ¥ 24 ft ( ) v ft 3 = L ¥ W ¥ D 4 = 2713 ft 3 = 50 ft ¥ 12 ft ¥ 8 ft = 4800 ft 3 Note: Remember, the problem asked for the tank capacity in liters. EXAMPLE 4.20: TANK VOLUME (CIRCULAR) Convert 2713 ft3 to liters: Problem: The diameter of a tank is 70 ft. When the water depth is = 2713 ¥ 7.48 gal ft 3 ¥ 3.785 L gal 30 ft, what is the volume of wastewater in the tank in = 76, 810 l gallons? Note: Draw a diagram similar to Figure 4.6. EXAMPLE 4.22: CHANNEL VOLUME CALCULATIONS Note: Channels are commonly used in water and Solution: wastewater treatment operations. Channels are typically rectangular or trapezoidal in ( v (gal ) = 0.785 ¥ Diameter ¥ D ¥ 7.48 gal ft 2 3 ) shape. For rectangular channels, use: 3 = 0.785 ¥ 70 ft ¥ 70 ft ¥ 30 ft ¥ 7.48 gal ft v=L¥W¥D = 863, 155 gal Problem: Note: Remember, the solution requires the result Determine the volume of wastewater (in ft3) in the section in gallons; thus, we must include 7.48 ft3 in of rectangular channel shown in Figure 4.7 when the the operation to ensure the result in gallons. wastewater is 5 ft deep. © 2003 by CRC Press LLC
  • 1000 ft 1800 ft 5 ft 10 in. 4 ftFIGURE 4.7 Open channel. (From Spellman, F.R., Spell- FIGURE 4.8 Circular pipe. (From Spellman, F.R., Spell-man’s Standard Handbook for Wastewater Operators, Vol. man’s Standard Handbook for Wastewater Operators, Vol.1–3, Technomic Publ., Lancaster, PA, 1999.) 1–3, Technomic Publ., Lancaster, PA, 1999.) Solution: Problem: ( ) = L ¥ W ¥ D = 600 ft ¥ 6 ft ¥ 5 ft v ft 3 What is the capacity in gallons of wastewater of a 10-in. diameter, 1800-ft section of pipeline (see Figure 4.8)? = 18, 000 ft 3 Note: Convert 10 in. to feet (10 in./12 in./ft = .833 ft). EXAMPLE 4.23: CHANNEL VOLUME (TRAPEZOIDAL) Solution: ( ) ( 2 )D¥L v (gal) = 0785 ¥ Diameter 2 ¥ L ¥ 7.48 gal ft 3 B +B v ft 3 = 1 2 = 0.785 ¥ .833 ¥ .833 ¥ 1800 ft ¥ 7.48 gal ft 3 where = 7334 gal B1 = distance across the bottom B2 = distance across water surface L = channel length EXAMPLE 4.25: PIPE VOLUME D = depth of water and wastewater Problem: Approximately how many gallons of wastewater would Problem: 800 ft of 8-in pipe hold? Determine the volume of wastewater (in gallons) in a section of trapezoidal channel when the wastewater depth Note: Convert 8 in. to feet (8 in./12 in./ft = .67 ft). is 5 ft. Solution: Given: B1 = 4 ft across the bottom p ¥ Diameter 2 v= ¥L B2 = 10 ft across water surface 4 L = 1000 ft p ¥ 0.672 = ¥ 800 ft 4 Solution: = 282 ft 3 v (gal) = (B1 + B2 ) D ¥ L ¥ 7.48 gal ft 3 Convert: 282 ft3 converted to gallons: 2 = (4 ft + 10 ft ) 5 ft ¥ 1000 ft ¥ 7.48 gal ft 3 = 282 ft 3 ¥ 7.48 gal ft 3 2 = 2110 gal = 7 ¥ 5 ¥ 1000 ¥ 7.48 EXAMPLE 4.26: PIT OR TRENCH VOLUME = 261, 800 gal Note: Pits and trenches are commonly used in water EXAMPLE 4.24: VOLUME OF CIRCULAR PIPELINE and wastewater plant operations. Thus, it is important to be able to determine their vol- umes. The calculation used in determining pit ( ) v ft 3 = 0785 ¥ Diameter 2 ¥ L or trench volume is similar to tank and channel© 2003 by CRC Press LLC
  • volume calculations with one difference — EXAMPLE 4.27: TRENCH VOLUME the volume is often expressed as cubic yards Problem: rather than cubic feet or gallons. What is the cubic yard volume of a trench 600 ft long, In calculating cubic yards, typically two approaches are 2.5 ft wide, and 4 ft deep? used: 1. Calculate the cubic feet volume, then convert to cubic Solution: yard volume. Convert dimensions in ft to yds before beginning the volume calculation: ( ) v ft 3 = L ¥ W ¥ D 600 ft ft 3 L= = 200 yd yd 3 = 3 3 ft yd 27 ft yd 2.5 ft 2. Express all dimensions in yards so that the resulting W= = 0.83 yd 3 ft yd volume calculated will be cubic yards. 5 ft D= = 1.67 yd (yds ¥ yds ¥ yds) = yd 3 3 ft yd ( ) v yd 3 = L ¥ W ¥ D Problem: = 200 yd ¥ 0.83 yd ¥ 1.67 yd A trench is to be excavated 3 ft wide, 5 ft deep, and 800 ft long. What is the cubic yards volume of the trench? = 277 yd 3 Note: Remember, draw a diagram similar to the EXAMPLE 4.28: POND VOLUMES one shown in Figure 4.9. Ponds and/or oxidation ditches are commonly used in Solution: wastewater treatment operations. To determine the volume of a pond (or ditch) it is necessary to determine if all four ( ) v ft 3 = L ¥ W ¥ D sides slope or if just two sides slope. This is important because the means used to determine volume would vary = 800 ft ¥ 3 ft ¥ 5 ft depending on the number of sloping sides. = 12, 000 ft 3 If only two of the sides slope and the ends are vertical, we calculate the volume using the equation: Now convert ft3 volume to yd3: ( ) ( 2 )D¥L B +B 12, 000 ft 3 v ft 3 = 1 2 = 27 ft 3 yd 3 However, when all sides slope as shown in Figure 4.10, = 444 yd 3 the equation we use must include average length and average width dimensions. The equation: 800 ft 5 ft v (gal) = (B1 + B2 ) D ¥ L ¥ 7.48 gal ft 3 23 ft Problem:FIGURE 4.9 Trench. (From Spellman, F.R., Spellman’s A pond is 6 ft deep with side slopes of 2:1 (2 ft horizontal:Standard Handbook for Wastewater Operators, Vol. 1–3, 1 ft vertical). Using the data supplied in Figure 4.10,Technomic Publ., Lancaster, PA, 1999.) calculate the volume of the pond in cubic feet.© 2003 by CRC Press LLC
  • Pond Bottom 320 ft 350 ft 600 ft 650 ft FIGURE 4.10 Pond. (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) Solution: v= (L 1 + L2 ) ¥ (W + W ) ¥ D 1 2 2 2 62.4 lbs (600 ft + 650 ft ) (320 ft + 350 ft ) 1 ft of = ¥ ¥ 6 ft 2 2 water = 625 ft ¥ 335 ft ¥ 6 ft 1 ft = 1, 256, 250 ft 3 1 ft FIGURE 4.11 One cubic foot of water weighs 62.4 lbs. (From4.6 FORCE, PRESSURE, AND HEAD Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)Force, pressure, and head are important parameters inwater and wastewater operations. Before we study calcu- 1 ft2 = 12 in. ¥ 12 in. = 144 in.2 (4.13)lations involving the relationship between force, pressure,and head, we must first define these terms. Therefore the pressure in pounds per square inch (psi) is: 1. Force — the push exerted by water on any 62.4 lb ft 2 62.4 lb ft 2 = = 0.433 psi (4.14) confining surface. Force can be expressed in 1 ft 2 144 in.2 ft 2 pounds, tons, grams, or kilograms. 2. Pressure — the force per unit area. The most If we use the bottom of the container as our reference common way of expressing pressure is in point, the head would be one foot. From this we can see pounds per square inch (psi). that one foot of head is equal to 0.433 psi. 3. Head — the vertical distance or height of water Note: In water and wastewater operations, 0.433 psi above a reference point. Head is usually is an important parameter. expressed in feet. In the case of water, head and pressure are related. Figure 4.12 illustrates some other important relation- ships between pressure and head. Figure 4.11 illustrates these terms. A cubical container Note: Force acts in a particular direction. Water in ameasuring one foot on each side can hold one cubic foot tank exerts force down on the bottom and outof water. A basic fact of science states that 1 ft3 H2O of the sides. Pressure acts in all directions. Aweights 62.4 lb. The force acting on the bottom of the marble at a water depth of one foot would havecontainer would be 62.4 lb. The pressure acting on the 0.433 psi of pressure acting inward on all sides.bottom of the container would be 62.4 lb/ft2. The area ofthe bottom in square inches is: Key Point: 1 ft of head = 0.433 psi. © 2003 by CRC Press LLC
  • 1 lb of water 2.31 ft 1 ft 0.433 lb of water 1 sq. in. AREA 1 ft water = 0.433 psi 1 sq. in. AREA 1 psi = 2.31 ft water FIGURE 4.12 Shows the relationship between pressure and head. (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) The parameter, 1 ft of head = 0.433 psi, is valuable Problem:and should be committed to memory. You should also Convert 50 ft to psi.know the relationship between pressure and feet of head —in other words, how many feet of head 1-psi represents.This is determined by dividing 1 by 0.433. Solution: 1 ft 50 psi 0.433 psi feet of head = = 2.31 ft psi ¥ = 21.7 psi 0.433 psi 1 1 ft What we are saying here is that if a pressure gauge were As the above examples demonstrate, when attemptingreading 12 psi, the height of the water necessary to represent to convert psi to ft, we divide by 0.433; when attemptingthis pressure would be 12 psi ¥ 2.31 ft/psi = 27.7 ft. to convert feet to psi, we multiply by 0.433. The aboveNote: Again, the key points: 1 ft = 0.433 psi, and process can be most helpful in clearing up the confusion 1 psi = 2.31 ft on whether to multiply or divide. Another way, however, Having two conversion methods for the same thing is may be more beneficial and easier for many operators tooften confusing. Thus, memorizing one and staying with use. Notice that the relationship between psi and feet isit is best. The most accurate conversion is: 1 ft = almost two to one. It takes slightly more than 2 ft to make0.433 psi — the standard conversion used throughout this 1 psi. Therefore, when looking at a problem where thehandbook. data are in pressure and the result should be in feet, the answer will be at least twice as large as the starting num- EXAMPLE 4.29 ber. For example, if the pressure were 25 psi, we intu- Problem: itively know that the head is over 50 ft. Therefore, we Convert 50 psi to ft of head. must divide by 0.433 to obtain the correct answer. Solution: EXAMPLE 4.30 50 psi 1 ft Problem: ¥ = 115.5 ft 1 0.433 psi Convert a pressure of 55 psi to ft of head. © 2003 by CRC Press LLC
  • Solution: EXAMPLE 4.35 55 psi 1 ft Problem: ¥ = 127 ft 1 0.433 psi What is the pressure (static pressure) 4 mi beneath the ocean surface? EXAMPLE 4.31 Problem: Solution: Convert 14 psi to ft of head. Change mi to ft, then to psi. Solution: 5, 380 ft mi ¥ 4 = 21, 120 ft 14 psi 1 ft ¥ = 32.3 ft 1 0.433 psi 21,120 ft = 9, 143 psi ( rounded ) 2.31 ft psi EXAMPLE 4.32 Problem: EXAMPLE 4.36 Between the top of a reservoir and the watering point, the Problem: elevation is 115 ft. What will the static pressure be at the watering point? A 50-ft diameter cylindrical tank contains 2.0 MG H2O. What is the water depth? At what pressure would a gauge Solution: at the bottom read in psi? 115 ft 0.433 psi ¥ = 49.8 psi Step 1: Change MG to ft3: 1 1 ft Using the preceding information, we can develop the 2, 000, 000 galfollowing equations for calculating pressure and head. = 267, 380 ft 3 7.48 Pressure ( psi) = 0.433 ¥ Head (ft ) Step 2: Using volume, solve for depth. Head (ft ) = 2.31 ¥ Pressure ( psi) v = 0.785 ¥ Diameter 2 ¥ D EXAMPLE 4.33 267, 380 ft 3 = 0.785 ¥ 150 2 ¥ D Problem: D = 15.1 ft Find the pressure (psi) in a 12-ft deep tank at a point 15 ft below the water surface. EXAMPLE 4.37 Solution: Problem: Pressure ( psi ) = 0.433 ¥ 5 ft The pressure in a pipe is 70 psi. What is the pressure in feet of water? What is the pressure in lb/ft2? = 2.17 psi ( rounded ) EXAMPLE 4.34 Solution: Problem: Step 1: Convert pressure to feet of water: A pressure gauge at the bottom of a tank reads 12.2 psi. How deep is the water in the tank? 70 psi ¥ 2.31 ft psi = 161.7 ft of water Solution: Step 2: Convert psi to psf: Head (ft ) = 2.31 ¥ 12.2 psi = 28.2 ft ( rounded ) 70 psi ¥ 144 in.2 ft 2 = 10, 080 lb ft 2 © 2003 by CRC Press LLC
  • EXAMPLE 4.38 Step 2: Now convert to gal/min by dividing by the number of minutes in a day (24 hrs per day X 60 min per hour) = Problem: 1440 min/day: The pressure in a pipeline is 6,476 lb/ft2. What is the head 135, 000 gal d on the pipe? = 93.8 or 94 gal min 1, 440 min d Head on pipe = ft of pressure In determining flow through a pipeline, channel or Pressure = Weight ¥ H stream, we use the following equation: 6476 lb ft 2 = 62.4 lb ft 3 ¥ H Q=A¥V H = 104 ft ( rounded ) where Q = cubic foot per second (ft3/sec)4.7 FLOW A = area in square feet (ft2) V = velocity in feet per second (ft/sec)Flow is expressed in many different terms (English Systemof measurements). The most common flow terms are: EXAMPLE 4.40 1. Gallons per minute (gal/min) Problem: 2. Cubic foot per second (ft3/sec) Find the flow in ft3/sec in an 8-in. line, if the velocity is 3. Gallons per day (gal/d) 3 ft/sec. 4. Million gallons per day (MGD) Solution: In converting flow rates, the most common flow conver- Step 1: Determine the cross-sectional area of the line insions are: 1 ft3/sec = 448 gal/min and 1 gal/min = 1440 gal/d. square feet. Start by converting the diameter of the pipe to inches. To convert gal/d to MGD, divide the gal/d by1,000,000. For instance, convert 150,000 gallons to MGD: Step 2: The diameter is 8 in.; therefore, the radius is 4 in. 4 in. is 4/12 of a foot or 0.33 ft. 150, 000 gal d = 0.150 MGD Step 3: Find the area in square feet: 1, 000, 000 A = p ¥ r2 In some instances, flow is given in MGD, but neededin gal/min. To make the conversion (MGD to gal/min) A = p ¥ 0.33 ft2requires two steps. A = p ¥ 0.109 ft2 1. Convert the gal/d by multiplying by 1,000,000 A = 0.342 ft2 2. Convert to gal/min by dividing by the number of minutes in a day (1440 min/d) Step 4: Use the equation, Q = A ¥ V: EXAMPLE 4.39 Q=A ¥ V Problem: Q = 3 ft sec ¥ 0.342 ft 2 Convert 0.135 MGD to gal/min Q = 1.03 ft 3 sec Solution: EXAMPLE 4.41 Step 1: First, convert the flow in MGD to gal/d: Problem: Find the flow in gal/min when the total flow for the day 0.135 MGD ¥ 1, 000, 000 = 135, 000 gal d is 75,000 gal/d. © 2003 by CRC Press LLC
  • Solution: 4.7.1.1.1 Instantaneous Flow into and out of a Rectangular Tank 75, 000 gal d = 52 gal min One of the primary flow measurements water and waste- 1, 440 min d water operators are commonly required to calculate is flow through a tank. This measurement can be determined EXAMPLE 4.42 using the Q = A ¥ V equation. For example, if the dis- Problem: charge valve to a tank were closed, the water level would begin to rise. If you time how fast the water rises, this Find the flow in gal/min when the flow is 0.45 ft3/sec. would give you an indication of the velocity of flow into the tank. This information can be plugged into Q = V ¥ A Solution: to determine the flow rate through the tank. Let us look at an example. 0.45 ft 3 sec 448 gal d ¥ = 202 gal min 1 1 ft 3 sec EXAMPLE 4.444.7.1 FLOW CALCULATIONS Problem:In water and wastewater treatment, one of the major con- A tank is 8 ft wide and 12 ft long. With the dischargecerns of the operator is not only to maintain flow, but also valve closed, the influent to the tank causes the water level to rise 1.5 ft in 1 min. What is the gal/min flow into theto measure it. Normally, flow measurements are deter- tank?mined by metering devices. These devices measure waterflow at a particular moment (instantaneous flow) or overa specified time (total flow). Instantaneous flow can also Solution:be determined mathematically. In this section, we discuss Step 1: First, calculate the ft3/min flow rate:how to mathematically determine instantaneous and aver-age flow rates and how to make flow conversions. ( ) Q ft 3 min = A ¥ V (ft min )4.7.1.1 Instantaneous Flow Rates = (8 ft ¥ 12 ft ¥ 1.5 ft min )In determining instantaneous flows rates through channels,tanks and pipelines, we can use Q = A ¥ V. = 144 ft 3 minNote: It is important to remember that when using an Step 2: Convert ft3/min flow rate to gal/min flow rate: equation such as Q = A ¥ V the units on the left side of the equation must match those units on the right side of the equation (A and V) with 144 ft 3 min ¥ 7.48 gal ft 3 = 1077 gal min respect to volume (cubic feet or gallons) and time (seconds, minutes, hours, or days). How do we compute flow rate from a tank when the influent valve is closed and the discharge pump remains EXAMPLE 4.43 on, lowering the wastewater level in the tank? Problem: First, we time the rate of this drop in wastewater level so that the velocity of flow from the tank can be calculated. A 4-ft wide channel has water flowing to a depth of 2 ft. Then we use the Q = A ¥ V equation to determine the If the velocity through the channel is 2 ft/sec, what are flow rate out of the tank, as illustrated in Example 4.45. the cubic feet per second (ft3/sec) flow rate through the channel? EXAMPLE 4.45 Solution: Problem: ( ) Q ft 3 sec = A ¥ V (ft sec) A tank is 9 ft wide and 11 ft long. The influent valve to the tank is closed and the water level drops 2.5 ft in 2 = 4 ft ¥ 2 ft ¥ 2 ft sec min. What is the gal/min flow from the tank? = 16 ft 3 sec Drop rate = 2.5 ft/2 min = 1.25 ft/min © 2003 by CRC Press LLC
  • Step 1: Calculate the ft3/min flow rate: EXAMPLE 4.47 Problem: ( ) Q ft 3 min = A ¥ V (ft min ) The flow through an 8-in. diameter pipeline is moving at = 9 ft ¥ 12 ft ¥ 1.25 ft min a velocity of 4 ft/sec. What is the ft3/sec flow rate through the full pipeline? = 124 ft 3 min Convert 8 in. to feet: 8 in./12 in. = 0.67 ft 3 Step 2: Convert ft /min flow rate to gal/min flow rate: ( ) Q ft 3 sec = A ¥ V (ft sec) 124 ft min ¥ 7.48 gal ft = 928 gal min 3 3 = 0.785 ¥ 0.67 ft ¥ 0.67 ft ¥ 4 ft sec4.7.1.1.2 Flow Rate into a Cylindrical Tank = 1.4 ft 3 secWe can use the same basic method to determine the flowrate when the tank is cylindrical in shape, as shown in 4.7.2 VELOCITY CALCULATIONSExample 4.46. To determine the velocity of flow in a channel or pipeline we use the Q = A ¥ V equation. However, to use the EXAMPLE 4.46 equation correctly we must transpose it. We simply write Problem: into the equation the information given and then transpose for the unknown (V in this case), as illustrated in Example The discharge valve to a 25-ft diameter cylindrical 4.48 for channels and 4.49 for pipelines.tank is closed. If the water rises at a rate of 12 inches in4 minutes, what is the gal/min flow into the tank? EXAMPLE 4.48 Solution: Problem: A channel has a rectangular cross section. The channel is Rise = 12 in. = 1 ft 5 ft wide with wastewater flowing to a depth of 2 ft. If = 1 ft 4 min the flow rate through the channel is 8500 gal/min, what is the velocity of the wastewater in the channel (ft/sec)? = 0.25 ft min Solution Step 1: Calculate the ft3/min flow into the tank: Convert gal/min to ft3/sec: ( ) Q ft 3 min = A ¥ V (ft min ) 8500 gal min = 18.9 ft 3 sec 7.48 gal ¥ 60 sec = 0.785 ¥ 25 ft ¥ 25 ft ¥ 0.25 ft min = 123 ft 3 min ( ) Q ft 3 s = A ¥ V (ft sec) Step 2: Then convert ft3/min flow rate to gal/min flow rate: 18.9 cfs = 5 ft ¥ 2 ft ¥ x ft sec 125 ft 3 min ¥ 7.48 gal ft 3 = 920 gal min V ( x ft sec) = 18.9 = 1.89 ft sec 5 ¥ 24.7.1.2 Flow through a Full Pipeline EXAMPLE 4.49Flow through pipelines is of considerable interest to waterdistribution operators and wastewater collection workers. Problem:The flow rate can be calculated using the Q = A ¥ V. The A full 8-in. diameter pipe delivers 250 gal/min. What iscross-sectional area of a round pipe is a circle, so the area, the velocity of flow in the pipeline (ft/sec)?A, is represented by 0.785 ¥ diameter2. Solution:Note: To avoid errors in terms, it is prudent to express pipe diameters as feet. Convert: 8 in/12 in to feet = 0.67 ft © 2003 by CRC Press LLC
  • Convert gal/min to ft3/sec flow: 60 1440 cfs cfm cfd 250 gal min = 0.56 ft 3 sec 7.48 gal ft 3 ¥ 60 sec min 7.48 7.48 7.48 60 1440 0.56 ft 3 sec = 0.785 ¥ 0.67 ft ¥ 0.67 ft ¥ x ft sec gps gpm gpd 0.56 ft 3 sec 8.34 8.34 8.34 V = = 1.6 ft sec 0.785 ¥ 0.67 ¥ 0.67 60 14404.7.3 AVERAGE FLOW RATE CALCULATIONS lbs/sec lbs/min lbs/dayFlow rates in water and wastewater systems vary consid-erably during the course of a day, week, month, or year.Therefore, when computing flow rates for trend analysis cfs = cubic feet per second gps = gallons per second cfm = cubic feet per minute gpm = gallons per minuteor for other purposes, an average flow rate is used to cfd = cubic feet per day gpd = gallons per daydetermine the typical flow rate. Example 4.50 illustratesone way to calculate an average flow rate. FIGURE 4.13 Flow conversions using the box method. The factors shown in the diagram have the following units associ- EXAMPLE 4.50 ated with them: 60 sec/min, 1440 min/day, 7.48 gal./ft3, and 8.34 lbs/gal. (Adapted from Price, J.K., Applied Math for Problem: Wastewater Plant Operators, Technomic Publ., Lancaster, PA, The following flows were recorded for the week: 1991, p. 32. With permission.) Monday 8.2 MGD multiplication by the factor indicated. Moving from a larger Tuesday 8.0 MGD box to a smaller box requires division by the factor indicated. Wednesday 7.3 MGD From Figure 4.13 it should be obvious that memoriz- Thursday 7.6 MGD ing the 9 boxes and the units in each box is not that Friday 8.2 MGD difficult. The values of 60, 1440, 7.48, and 8.34 are not Saturday 8.9 MGD Sunday 7.7 MGD that difficult to remember either — it is a matter of remem- bering the exact placement of the units and the values. What was the average daily flow rate for the week? Once this is accomplished, you have obtained a powerful tool that will enable you to make flow conversions in a Solution: relatively easy manner. Total of All Sample Flows Average Daily Flow = 4.8 DETENTION TIME Number of Days 55.9 MGD Detention time (DT) is the length of time water is retained = in a vessel or basin, or the period from the time the water 7d enters a settling basin until it flows out the other end. To = 8.0 MGD calculate the detention period of a basin, the volume of the basin must be first obtained. Using a basin 70 ft long,4.7.4 FLOW CONVERSION CALCULATIONS 25 ft wide and 12 ft deep, the volume would be:One of the tasks involving calculations that the wastewateroperator is typically called on to perform involves con- v=L ¥ W ¥ Dverting one expression of flow to another. The ability todo this is also a necessity for those preparing for licensure v = 70 ft ¥ 25 ft ¥ 12 ftexaminations. Probably the easiest way in which to accomplish flow v = 21, 000 ft 3conversions is to employ the box method illustrated inFigure 4.13. Gallons = V ¥ 7.48 gal ft 3 When using the box method it is important to rememberthat moving from a smaller box to a larger box requires Gallons = 21, 000 ¥ 7.48 = 157, 080 gal © 2003 by CRC Press LLC
  • If we assume that, the plant filters 300 gal/min, Solution:157,080 ÷ 300 = 524 min (rounded), or roughly 9 h ofdetention time. Stated another way, the detention time is 55, 000 gal DT = = 734 minthe length of time theoretically required for the coagulated 75 gal minwater to flow through the basin. 634 min If chlorine were added to the water as it entered the or = 12 h 60 min hbasin, the chlorine contact time would be 9 h. That is, todetermine the contact time (CT) used to determine the EXAMPLE 4.53effectiveness of chlorine, we must calculate detentiontime. Problem:Key point: True detention time is the “T” portion of If the fuel consumption to the boiler is 30 gal/d, how many the CT value. days will the 1000-gal tank last?Note: Detention time is also important when evaluat- Solution: ing the sedimentation and flocculation basins of a water treatment plant. 1000 gal Days = = 33.3 d 30 gal d Detention time is expressed in units of time. The mostcommon are: seconds, minutes, hours, and days. 4.8.1 HYDRAULIC DETENTION TIME The simplest way to calculate detention time is todivide the volume of the container by the flow rate into The term detention time or hydraulic detention time (HDT)the container. The theoretical detention time of a container refers to the average length of time (theoretical time) ais the same as the amount of time it would take to fill the drop of water, wastewater, or suspended particles remainscontainer if it were empty. in a tank or channel. It is calculated by dividing the water For volume, the most common units used are gallons. and wastewater in the tank by the flow rate through theHowever, on occasion, cubic feet may also be used. tank. The units of flow rate used in the calculation are Time units will be in whatever units are used to dependent on whether the detention time is to be calcu- lated in seconds, minutes, hours or days. Detention timeexpress the flow. For example, if the flow is in gal/min, is used in conjunction with various treatment processes,the detention time will be in days. If the detention time is including sedimentation and coagulation-flocculation.in the wrong time units, simply convert to the appropriate Generally, in practice, detention time is associatedunits. with the amount of time required for a tank to empty. The range of detention time varies with the process. For exam- EXAMPLE 4.51 ple, in a tank used for sedimentation, detention time is commonly measured in minutes. Problem: The calculation methods used to determine detention The reservoir for the community is 110,000 gal. The well time are illustrated in the following sections. will produce 60 gal/min. What is the detention time in the reservoir in hours? 4.8.1.1 Detention Time in Days Solution: The general hydraulic detention time calculation is: 110, 000 gal Tank Volume DT = = 1834 min HDT = (4.15) 60 gal min Flow Rate 1834 min or = 30.6 h This general formula is then modified based upon the 60 min h information provided or available and the normal range of detention times for the unit being evaluated. Equation 4.28 EXAMPLE 4.52 shows another form of the general equation: Problem: Find the detention time in a 55,000-gal reservoir if the HDT (d ) = ( ) Tank Volume ft 3 ¥ 7.48 gal ft 3 (4.16) flow rate is 75 gal/min. Flow (gal d ) © 2003 by CRC Press LLC
  • EXAMPLE 4.54 Note: The tank volume and the flow rate must be in the same dimensions before calculating Problem: the hydraulic detention time. An anaerobic digester has a volume of 2,200,000 gal. What is the detention time in days when the influent flow rate is 0.06 MGD? 4.9 CHEMICAL DOSAGE CALCULATIONS Chemicals are used extensively in water and wastewater Solution: treatment plant operations. Water and wastewater treat- ment plant operators add chemicals to various unit pro- 2, 200, 000 gal cesses for slime-growth control, corrosion control, odor DT (d ) = control, grease removal, BOD reduction, pH control, 0.06 MGD ¥ 1, 000, 000 gal MG sludge-bulking control, ammonia oxidation, bacterial = 37 d reduction, fluoridation, and other reasons. In order to apply any chemical dose correctly it is4.8.1.2 Detention Time in Hours important to be able to make certain dosage calculations. One of the most frequently used calculations in water and wastewater mathematics is the conversion of milligrams HDT ( h ) = per liter (mg/L) concentration to pounds per day (lb/d) or ( ) Tank Volume ft 3 ¥ 7.48 gal ft 3 ¥ 24 h d (4.17) pounds (lb) dosage or loading. The general types of mg/L to lb/d or lb calculations are for chemical dosage, BOD, Flow (gal d ) chemical oxygen demand, or SS loading/removal, pounds of solids under aeration and WAS pumping rate. These EXAMPLE 4.55 calculations are usually made using either of the following equation 4.31 or 4.32. Problem: A settling tank has a volume of 40,000 ft.3 What is the mg L ¥ MGD flow ¥ 8.34 lb gal = lb d (4.19) detention time in hours when the flow is 4.35 MGD? mg L ¥ MGD volume ¥ 8.34 lb gal = lb (4.20) Solution: 40, 000 ft 3 ¥ 7.48 gal ft 3 ¥ 24 h d Note: If mg/L concentration represents a concentra- DT ( h ) = tion in a flow, then MGD flow is used as the 4.35 MGD ¥ 1, 000, 000 gal MG second factor. However, if the concentration = 1.7 h pertains to a tank or pipeline volume, then MG volume is used as the second factor.4.8.1.3 Detention Time in Minutes 4.9.1 CHLORINE DOSAGEHDT (min) = (4.18) Chlorine is a powerful oxidizer commonly used in water treatment for purification and in wastewater treatment for ( ) Tank Volume ft 3 ¥ 7.48 gal ft 3 ¥ 1440 min d disinfection, odor control, bulking control, and other Flow (gal d ) applications. When chlorine is added to a unit process, we want to ensure that a measured amount is added. In describing the amount of chemical added or EXAMPLE 4.56 required two ways are used: (1) mg/L, and (2) lb/d. Problem: In the conversion from mg/L (or ppm) concentration to lbs/day, we use equation 4.33. A grit channel has a volume of 1240 ft.3 What is the detention time in minutes when the flow rate is 4.1 MGD? mg L ¥ MGD ¥ 8.34 = lb d (4.21) Solution: Note: In previous years it was normal practice to use 1240 ft ¥ 7.48 gal ft ¥ 1440 min d 3 3 the expression parts per million (ppm) as an DT ( min ) = expression of concentration, since 1 mg/L = 4,1000, 000 gal d 1 ppm. However, current practice is to use mg/L = 3.26 minutes as the preferred expression of concentration. © 2003 by CRC Press LLC
  • EXAMPLE 4.57 and the U.S. Environmental Protection Association many facilities are deciding to substitute the hazardous chemical Problem: chlorine with nonhazardous hypochlorite. Obviously, the Determine the chlorinator setting (lb/d) needed to treat a potential liability involved with using deadly chlorine is flow of 8 MGD with a chlorine dose of 6 mg/L. also a factor involved in the decision to substitute it with a less toxic chemical substance. Solution: For whatever reason the wastewater treatment plant decides to substitute chlorine for hypochlorite, there are mg L ¥ MGD ¥ 8.34 = lb d differences between the two chemicals of which the waste- water operator needs to be aware. 6 mg L ¥ 8 MGD ¥ 8.34 lb gal = lb d Chlorine is a hazardous material. Chlorine gas is used = 400 lb d in wastewater treatment applications at 100% available chlorine. This is an important consideration to keep in EXAMPLE 4.58 mind when making or setting chlorine feed rates. For example, if the chlorine demand and residual requires Problem: 100-lb/d chlorine, the chlorinator setting would be just that — 100-lb/24 h. What should the chlorinator setting be (lb/d) to treat a flow of 3 MGD if the chlorine demand is 12 mg/L and a Hypochlorite is less hazardous than chlorine; it is sim- chlorine residual of 2 mg/L is desired? ilar to strong bleach and comes in two forms: dry HTH and liquid sodium hypochlorite. HTH contains about 65%Note: The chlorine demand is the amount of chlorine available chlorine; sodium hypochlorite contains about used in reacting with various components of the 12 to 15% available chlorine (in industrial strengths). wastewater, such as harmful organisms and Note: Because either type of hypochlorite is not 100% other organic and inorganic substances. When pure chlorine, more lb/day must be fed into the the chlorine demand has been satisfied, these system to obtain the same amount of chlorine reactions stop. for disinfection. This is an important economic consideration for those facilities thinking about mg L ¥ MGD ¥ 8.34 = lb d substituting hypochlorite for chlorine. Some studies indicate that such a substitution can Solution: increase operating costs, overall, by up to 3 times the cost of using chlorine. In order to find the unknown value (lb/d), we must first determine chlorine dose. To calculate the lb/d hypochlorite required a two-step calculation is used: Cl Dose ( mg L ) = Cl Demand ( mg L ) + Step 1: Calculate the lb/d chlorine required using the mg/L Cl Residual ( mg L ) to lb/d equation: = 12 mg L + 2 mg L mg L ¥ MGD ¥ 8.34 = lb d (4.22) = 14 mg L Step 2: Calculate the lb/d hypochlorite required: Then we can make the mg/L to lb/d calculation: Cl (lb d ) ¥ 100 = Hypochlorite (lb d ) (4.23) 12 mg L ¥ 3 MGD ¥ 8.34 lb gal = 300 lb d % Available4.9.2 HYPOCHLORITE DOSAGE EXAMPLE 4.59At many wastewater facilities sodium hypochlorite or cal- Problem:cium hypochlorite are used instead of chlorine. The reasons A total chlorine dosage of 10 mg/L is required to treat afor substituting hypochlorite for chlorine vary. However, particular wastewater. If the flow is 1.4 MGD and thewith the passage of stricter hazardous chemicals regulations hypochlorite has 65% available chlorine how many lb/dunder the Occupational Safety and Health Administration of hypochlorite will be required? © 2003 by CRC Press LLC
  • Solution: ( ) v 10 6 gal = Cl concentration ( mg L ) ¥ 8.34 Step 1: Calculate the lb/day chlorine required using the mg/L to lb/d equation: = lb Cl Step 2: Substitute the numbers into the equation: mg L ¥ MGD ¥ 8.34 = lb d 10 mg L ¥ 1.4 MGD ¥ 8.34 lb gal = 117 lb d 5 ¥ 10 6 gal ¥ 2 mg L ¥ 8.34 = 83 lb Cl Step 2: Calculate the lb/d hypochlorite required. Since only 65% of the hypochlorite is chlorine, more than 117 4.10 PERCENT REMOVAL lb/d will be required: Percent removal is used throughout the wastewater treat- ment process to express or evaluate the performance of 117 lb d Cl the plant and individual treatment unit processes. The ¥ 100 = 180 lb d hypochlorite 65% Available Cl results can be used to determine if the plant is performing as expected or in troubleshooting unit operations by com- EXAMPLE 4.60 paring the results with those listed in the plant’s operations and maintenance manual. It can be used with either con- Problem: centration or quantities (see Equations 4.24 and 4.25). For concentrations use: A wastewater flow of 840,000 gal/d requires a chlorine dose of 20 mg/L. If sodium hypochlorite (15% available chlorine) is to be used, how many lbs/day of sodium % Removal = [Influent Conc. - Eff . Conc.] ¥ 100 (4.24) hypochlorite are required? How many gal/day of sodium Influent Conc. hypochlorite is this? For quantities use: Solution: Step 1: Calculate the lb/day chlorine required: % Removal = [Influent Quantity - Eff . Quantity] ¥ 100 Influent Quantity (4.25) mg L ¥ MGD ¥ 8.34 = lb d 20 mg L ¥ Note: The calculation used for determining the per- 0.84 MGD ¥ 8.34 lb gal formance (percent removal) for a digester is different from that used for performance (percent = 140 lb d Cl removal) for other processes such as some pro- cess residuals or biosolids treatment processes. Step 2: Calculate the lb/day sodium hypochlorite: Ensure the right formula is selected. 140 lb d Cl EXAMPLE 4.62 ¥ 100 = 933 lb d hypochlorite 15% Available Cl Problem: The plant influent contains 259 mg/L BOD and the plant Step 3: Calculate the gal/d sodium hypochlorite: effluent contains 17 mg/L BOD. What is the percentage of BOD removal? 933 lb d = 112 gal d sodium hypochlorite 8.34 lb gal Solution: % Removal = [259 mg L - 17 mg L ] ¥ 100 = 93.4% EXAMPLE 4.61 259 mg L Problem: How many pounds of chlorine gas are necessary to 4.11 POPULATION EQUIVALENT OR UNIT 5,000,000 gal of wastewater at a dosage of 2 mg/L? LOADING FACTOR When it is impossible to conduct a wastewater characteriza- Solution: tion study and other data are unavailable, population equiv- Step 1: Calculate the pounds of chlorine required: alent (PE) or unit per capita loading factors are used to© 2003 by CRC Press LLC
  • estimate the total waste loadings to be treated. If the BOD Solution:contribution of a discharger is known, the loading placedupon the wastewater treatment system in terms of equivalent Weight (lb gal) = 1.4515 ¥ 8.34 lb galnumber of people can be determined. The BOD contributionof a person is normally assumed to be 0.17 lb BOD/d. = 12.1 lb BOD Contribution (lb d ) 4.13 PERCENT VOLATILE MATTER PE ( people) = (4.26) 0.17 lb BOD d person REDUCTION IN SLUDGE The calculation used to determine percent volatile matter EXAMPLE 4.63 (VM) reduction is complicated because of the changes Problem: occurring during sludge digestion. A new industry wishes to connect to the city’s collection system. The industrial discharge will contain an average %VM ( reduction) = BOD concentration of 349 mg/L and the average daily flow will be 50,000 gal/d. What is the population equiv- [%VM in ] - %VM out ¥ 100 (4.28) alent of the industrial discharge? [%VM - (%VM in in ¥ %VM out )] Solution: EXAMPLE 4.65 Step 1: Convert flow rate to MGD: Problem: 50, 000 gal d Using the digester data provided below, determine the Flow = = 0.050 MGD 1, 000, 000 gal MG percent volatile matter reduction for the digester. Step 2: Calculate the population equivalent: Data: PE ( people) Raw Sludge Volatile Matter = 72% Digested Sludge Volatile Matter = 51% 349 mg L ¥ 0.050 MGD ¥ 8.34 lb mg L MG = 0.17 lb BOD d person [0.72 - 0.51] ¥ 100 %VM ( reduction ) = = 59% = 856 people d [0.72 - (0.72 ¥ 0.51)]4.12 SPECIFIC GRAVITY 4.14 HORSEPOWERSpecific gravity (sp gr) is the ratio of the density of a In water and wastewater treatment operations, horsepowersubstance to that of a standard material under standard (hp) is a common expression for power. One horsepowerconditions of temperature and pressure. The standard is equal to 33,000 foot pounds (ft-lb) of work/min. Thismaterial for gases is air, and for liquids and solids, it is value is determined, for example, for selecting a pump orwater. Specific gravity can be used to calculate the weight combination of pumps to ensure adequate pumping capac-of a gallon of liquid chemical. ity. Pumping capacity depends upon the flow rate desired and the feet of head against which the pump must pump (also known as effective height). Chemical (lb gal) = H 2O (lb gal) ¥ (4.27) Calculations of horsepower are made in conjunction Chemicals sp gr with many treatment plant operations. The basic concept from which the horsepower calculation is derived is the EXAMPLE 4.64 concept of work. Work involves the operation of a force (lb) over a Problem: specific distance (ft). The amount of work accomplished The label states of the chemical states that the contents is measured in foot-pounds: of the bottle have a specific gravity of 1.4515. What is the weight of 1 gal of solution? ft ¥ lb = ft-lb (4.29) © 2003 by CRC Press LLC
  • The rate of doing work (power) involves a time factor. EXAMPLE 4.67Originally, the rate of doing work or power compared the Problem:power of a horse to that of the steam engine. The rate atwhich a horse could work was determined to be about Under the specified conditions, the pump efficiency is 73%. If the required water horsepower is 40 hp, what is550 ft-lb/sec (or expressed as 33,000 ft-lb/min). This rate the required brake horsepower?has become the definition of the standard unit called horse-power (see Equation 4.44). 40 Whp Bhp = = 55 Bhp 0.73 Power (ft-lb min) Horsepower ( hp) = (4.30) 33, 000 ft-lb min hp 4.14.3 MOTOR HORSEPOWER Motor horsepower (Mhp) is the horsepower the motor As mentioned, in water and wastewater operations, must generate to produce the desired brake and waterthe major use of horsepower calculation is in pumping horsepower.operations. When used for this purpose, the horsepowercalculation can be modified as shown in Section 4.14.1. Bhp Mhp = (4.33) Motor % Efficiency4.14.1 WATER HORSEPOWER EXAMPLE 4.68The amount of power required to move a given volumeof water a specified total head is known as water horse- Problem:power (Whp). The motor is 93% efficient. What is the required motor horsepower when the required brake horsepower is 49 Bhp?Whp = (4.31) Solution: Pump Rate (gal min) ¥ Total Head (ft ) ¥ 8.34 lb gal 49 Bhp 33, 000 ft-lb min hp Mhp = = 53 Mhp 0.93 EXAMPLE 4.66 4.15 ELECTRICAL POWER Problem: In this age of energy conservation, water and wastewater A pump must deliver 1210 gal/min to total head of 130 ft. operators (especially senior operators) are often required What is the required water horsepower? to make electrical power calculations, especially regarding electrical energy required/consumed during a period. To Solution: accomplish this, horsepower is converted to electrical energy (kilowatts), and then multiplied by the hours of 1210 gal min ¥ 130 ft ¥ 8.34 lb gal operation to obtain kilowatt-hours (kW-h). Whp = 33, 000 ft-lb min hp = 40 Whp kW - h = hp ¥ 0.746 kW hp ¥ (4.34) Operating Time ( h )4.14.2 BRAKE HORSEPOWER EXAMPLE 4.69Brake horsepower (Bhp) refers to the horsepower suppliedto the pump from the motor. As power moves through the Problem:pump, additional horsepower is lost from slippage and A 60-hp motor operates at full load 12 h/d, 7 d/week.friction of the shaft and other factors; thus, pump efficien- How many kilowatts of energy does it consume per day?cies range from about 50 to 85% and must be taken intoaccount. Solution: kW-h day = 60 hp ¥ 0.746 kW hp ¥ 12 h d Whp Bhp = (4.32) Pump % Efficiency = 537 kW-h day © 2003 by CRC Press LLC
  • Note: Given the cost per kilowatt-hour, the operator 4.16.2 CALCULATING SOLUTION STRENGTH (or anyone else) may calculate the cost of power for any given period of operation. Use the following procedure to calculate solution strength. EXAMPLE 4.72 Cost = Power Required d ¥ kw-h d ¥ (4.35) Problem: d Period ¥ Cost kW-h Eight pounds of alum are added to 115 lb of H2O. What is the solution strength? EXAMPLE 4.70 Problem: Solution: A 60-hp motor requires 458 kW-h/d. The pump is in 8 70% = ¥ 100 = 6.5% Solution operation every day. The current cost of electricity is (8 + 115) $0.0328/kW-h. What is the yearly electrical cost for this pump? We use this same concept in determining other solu- tion strengths. Solution: EXAMPLE 4.73 Cost ($) = 458 kW-d ¥ 365 days year ¥ Problem: $0.0328 kW-h = $5483.18 Approximately 25 lb of alum are added to 90 lb of H2O. What is the solution strength?4.16 CHEMICAL COAGULATION AND SEDIMENTATION Solution:Chemical coagulation consists of treating the water with 25 ¥ 100 = 22% Solutioncertain chemicals to bring nonsettleable particles together (25 + 30)into larger heavier masses of solid material (called floc)that are then relatively easy to remove. In the previous examples, we added pounds of chem- icals to pounds of water. Recall that 1 gal of H2O = 8.34 lb.4.16.1 CALCULATING FEED RATE By multiplying the number of gallons by the 8.34 factor, we can find pounds.The following equation is used to calculate the feed rateof chemicals used in coagulation. EXAMPLE 4.74 Chemical Feed Rate (lb d ) = Dose ( mg L ) ¥ (4.36) Problem: Approximately 40 lb of soda ash are added to 65 gal of Flow (MGD) ¥ 8.34 H2O. What is the solution strength? EXAMPLE 4.71 Solution: Problem: The units must be consistent, so convert gallons of H2O to pounds of H2O: A water treatment plant operates at a rate of 5 MGD. The dosage of alum is 40 ppm (or mg/L). How many pounds of alum are used a day? 65 gal ¥ 8.34 lb gal = 542.7 lb H 2 O 40 Solution: ¥ 100 = 6.9% Solution (542.7 + 40) Chemical Feed Rate = Dose ( mg L ) ¥ Flow (MGD) ¥ (8.34) 4.17 FILTRATION = 40 mg L ¥ 5 MGD ¥ (8.34) In waterworks operation (and to an increasing degree in wastewater treatment), the rate of flow through filters is = 1668 lb d of alum an important operational parameter. While flow rate can © 2003 by CRC Press LLC
  • be controlled by various means or may proceed at a vari- 4.17.2 FILTER BACKWASHable declining rate, the important point is that with flowsuspended matter continuously builds up within the filter In filter backwashing, one of the most important operationalbed, affecting the rate of filtration. parameters to be determined is the amount of water in gallons required for each backwash. This amount depends4.17.1 CALCULATING THE RATE OF FILTRATION on the design of the filter and the quality of the water being filtered. The actual washing typically lasts 5 to 10 min and EXAMPLE 4.75 uses amounts of 1 to 5% of the flow produced. Problem: EXAMPLE 4.76 A filter box is 20 ¥ 30 ft (also the sand area). If the influent Problem: value is shut, the water drops 3 in./min. What is the rate of filtration in MGD? A filter has the following dimensions: Solution: L = 30 ft Given: W = 20 ft Depth of filter media = 24 in. Filter box = 20 ¥ 30 ft Water drops = 3 in./min Assuming a backwash rate of 15 gal/ft2/min per minute is recommended, and 10 minutes of backwash is required, Find the volume of water passing through the filter calculate the amount of water in gallons required for each backwash. v=A¥H Solution: A=W¥L Given:Note: The best way to perform calculations of this L = 30 ft type is systematic, breaking down the problem into what is given and what is to be found. W = 20 ft Depth of filter media = 24 in. Step 1: Calculate the area; convert 3 in. to feet, and divide 3 by 12 to find feet: Rate = 15 gal ft 2 min A = 20 ft ¥ 30 ft = 600 ft 2 10-min backwash time 3.0 12 = 0.25 ft Find the amount of water in gallons required v = 600 ft ¥ 0.25 ft 2 Step 1: Calculate the area of the filter: = 150 ft of H 2 O passing through the filter 3 30 ft ¥ 20 ft = 600 ft2 in 1 min Step 2: Calculate the gallons of H2O used per square foot Step 2: Convert ft3 to gal: of filter: 150 ft 3 ¥ 7.48 gal ft 3 = 1122 gal min 15 gal ft 2 min ¥ 10 min = 15 gal ft 2 Step 3: The problem asks for the rate of filtration in MGD. Step 3: Calculate the gallons required: To find MGD, multiply the number of gallons per minute by the number of minutes per day. 150 gal ft 2 ¥ 600 ft 2 = 1122 gal min ¥ 1440 min d = 1.62 MGD 90, 000 gal required for backwash © 2003 by CRC Press LLC
  • 4.18 PRACTICAL WATER DISTRIBUTION Solution: SYSTEM CALCULATIONS Change MGD to ft3/sec, inches to ft; solve for velocity using Equation 4.25.After water is adequately treated, it must be conveyed ordistributed to the customer for domestic, commercial,industrial, and fire-fighting applications. Water distribu- Q=A ¥ Vtion systems should be capable of meeting the demandsplaced on them at all times and at satisfactory pressures. 2 ¥ 1.55 = 0.785 ¥ 0.832 VWaterworks operators responsible for water distribution 3.1 = 0.785 ¥ 0.69 Vmust be able to perform basic calculations for both prac-tical and licensure purposes; such calculations deal with 5.7 = Vwater velocity, rate of water flow, water storage tanks, andwater disinfection. V = 5.7 ft sec4.18.1 WATER FLOW VELOCITY EXAMPLE 4.78The velocity of a particle (any particle) is the speed at Problem:which it is moving. Velocity is expressed by indicating the A 24-in. diameter pipe carries water at a velocity of 140length of travel and how long it takes to cover the distance. ft/min. What is the flow rate in gal/min)?Velocity can be expressed in almost any distance and timeunits. Solution: Distance Traveled Change ft/min to ft/sec and inches to ft; solve for flow. V= (4.37) t Q=A ¥ V Note that water flow that enters the pipe (any pipe) isthe same flow that exits the pipe (under steady flow con- = 0.785 ¥ 2 2 ¥ 2.3ditions). Water flow is continuous. Water is incompress- = 7.2 ft 3 secible; it cannot accumulate inside. The flow at any givenpoint is the same flow at any other given point in the = 7.2 ft 3 sec ¥ 7.48 ft 3 ¥ 60 minpipeline. Therefore, a given flow volume may not change (it = 3, 231 gal minshould not), but the velocity of the water may change. Atany given flow, velocity is dependent upon the cross- EXAMPLE 4.79sectional area of the pipe or conduit. Velocity (the speedat which the flow is traveling) is an important parameter. Problem: When dealing with velocity of flow, another most If water travels 700 ft in 5 minutes, what is the velocity?basic hydraulic equation is: Solution: Q=A¥V Distance Traveled v= where t Q = Flow 700 ft A = Area (cross-sectional area of conduit — = 5 min [0.785 ¥ Diameter2]) V = Velocity = 140 ft min EXAMPLE 4.77 EXAMPLE 4.80 Problem: Problem: A flow of 2 MGD occurs in a 10-in. diameter conduit. Flow in a 6-in. pipe is 400 gal/min. What is the average What is the water velocity? velocity? © 2003 by CRC Press LLC
  • Solution: A=W ¥D Step 1: Calculate the area. Convert 6 in. to feet by dividing = 2.0 ft ¥ 1.2 ft by 12: 6/12 = 0.5 ft = 2.4 ft 2 A = 0.785 ¥ Diameter 2 Step 3: Calculate the velocity: = 0.785 ¥ 0.52 = 0.785 ¥ 0.25 V (ft sec) = ( Q ft 3 sec ) = 0.196 ft 2 ( rounded) ( ) A ft 2 110 ft 3 sec Step 2: Calculate the flow: = 2.4 ft 2 Q (gal min ) ¥ ft 3 = 4.6 ft sec ( rounded ) ( ) Q ft 3 sec = 7.48 gal ¥ 1 min 60 sec 4.18.2 STORAGE TANK CALCULATIONS 400 gal min ¥ ft 3 = 7.48 gal ¥ 1 min 60 sec Water is stored at a waterworks operation to provide allow- ance for differences in water production rates and high- 400 ft 3 lift pump discharge to the distribution system. Water = 448.3 sec within the distribution system may be stored in elevated = 0.89 ft 3 sec tanks, standpipes, covered reservoirs, and/or underground basins. Step 3: Calculate the velocity: The waterworks operator should be familiar with the basic storage tank calculation illustrated in the following V (ft sec) = ( Q ft 3 sec ) example. A ft( )2 EXAMPLE 4.82 3 0.89 ft sec = Problem: 0.196 ft 2 A cylindrical tank is 120 feet high and 25 feet in diameter. = 4.5 ft sec How many gallons of water will it contain? EXAMPLE 4.81 Solution: Problem: Given: Flow in a 2.0-ft wide rectangular channel is 1.2 ft deep and measures 11.0 ft3/sec. What is the average velocity? Height = 120 ft Diameter = 25 ft Solution: Given: Find the total gallons of water contained in the tank. Q = Rate of flow = 11.0 ft3/sec Step 1: Find the volume in ft3: A = Area in ft2 2.0 ft wide 1.2 ft deep v = 0.785 ¥ Diameter 2 ¥ H = 0.785 ¥ (25 ft ) ¥ 120 ft 2 Find the average velocity. Step 1: Transpose Q = VA to V = Q/A = 0.785 ¥ 625 ft 2 ¥ 120 ft Step 2: Calculate the area: = 58.875 ft 3© 2003 by CRC Press LLC
  • Step 2: Find the number of gallons of water the cylindrical Chemical Weight (lb) = tank will contain: Chemical Dose ( mg L ) ¥ = 58.875 ft 3 ¥ 7.48 gal ft 3 Water Volume (MG ) ¥ 8.34 = 440, 385 gal Step 3: Calculate the amount of available chlorine:4.18.3 DISTRIBUTION SYSTEM DISINFECTION CALCULATIONS Cl = 50 mg L ¥ 0.367 MG ¥ 8.34Before being placed in service, all facilities and appurte- = 153 lb (available)nances associated with the treatment and distribution ofwater must be disinfected. This is because water maybecome tainted anywhere in the system; delivering a clean, Note: The fundamental concept to keep in mind whenpathogen-free product to the customer is what water treat- computing hypochlorite calculations is thatment operation is all about. once we determine how many pounds of chlo- In the examples that follow, we demonstrate how to rine will be required for disinfection, we willperform the necessary calculations for this procedure. always need more pounds hypochlorite as com- pared to elemental chlorine. EXAMPLE 4.83 Step 4: Calculate the amount of HTH required: Problem: A waterworks has a tank containing water that needs to Available Chlorine Hypochlorite = be disinfected using HTH 70% available chlorine. The Chlorine Fraction tank is 100 ft high and 25 ft in diameter. The dose to use is 50 ppm. How many pounds of HTH are needed? 153 lb = 0.7 Solution: = 218.6 lb HTH required ( rounded ) Given: H = 100 ft EXAMPLE 4.84 Diameter = 25 ft Problem: Cl dose = 50 ppm Available Cl = 70% When treating 4000 ft of 8-in. water line by applying enough chlorine for 80-ppm dosage, how many pounds Find the pounds of HTH of hypochlorite of 70% available chlorine are required? Step 1: Find the volume of the tank: Solution: v = 3.14 ¥ r 2 ¥ H Given: Diameter 25 ft r= = = 12.5 ft Length = 4000 ft 2 2 Available Cl = 70% v = 3.14 ¥ 12.52 ¥ 100 Diameter = 8 in. Cl dose = 80 ppm = 3.14 ¥ 156.25 ¥ 100 Find the pounds of hypochlorite required. = 3.14 (15, 625) Step 1: Find the volume of the pipe (change 8 in. to ft by = 49, 062.5 f 3 dividing by 12): Step 2: Convert cubic feet to million gallons (MG). 8 in. Diameter = = 0.66 ft = 0.70 ft ( rounded ) 7.48 gal MG 12 in. ft 49, 062.5 ft 3 ¥ = ft 3 1, 000, 000 gal Diameter 0.70 ft r= = = 0.35 ft = 0.367 MG 2 2 © 2003 by CRC Press LLC
  • v = 3.14 ¥ r 2 ¥ H Solution: lb = 2580 mg L ¥ 0.90 MG ¥ 8.34 lb MG mg L = 3.14 ¥ 0.352 ¥ 4000 ft = 19, 366 lb = 3.14 (0.1225) ¥ 4000 ft = 1538.6 ft 3 4.19.1.2 Concentration (Milligrams per Liter) to Pounds/Day Step 2: Convert ft3 to MG: lb d = Concentration (mg L ) ¥ Q (MGD) ¥ 3 (4.39) 7.48 gal ft MG 1538.6 ft 3 ¥ = 8.34 lb MG mg L ft 3 1, 000, 000 gal = 0.0115 MG EXAMPLE 4.86 Problem: Step 3: Calculate the amount of available chlorine: Given: Cl = 80 mg L ¥ 0.0115 MG ¥ 8.34 Effluent BOD = 23 mg/L Effluent flow = 4.85 MGD = 7.67 lb (available) Find the concentration in lb/d. Step 4: Calculate the amount of hypochlorite required: Solution: 7.67 lb Cl = 11 lb of hypochlorite ( rounded ) lb d = 23 mg L ¥ 4.85 MGD ¥ 8.34 lb MG mg L 0.7 = 930 lb d4.19 COMPLEX CONVERSIONS 4.19.1.3 Concentration (Milligrams per Liter)Water and wastewater operators use complex conversions, to Kilograms per Dayfor example, in converting laboratory test results to otherunits of measure, which can be used to adjust or control kg d = Concentration (mg L ) ¥ Q (MGD) ¥the treatment process. Conversions such as these require (4.40)the use of several measurements (i.e., concentration, flow 3.785 lb MG mg Lrate, tank volume, etc.) and an appropriate conversionfactor. The most widely used of these conversions are EXAMPLE 4.87discussed in the following sections. Problem:4.19.1 CONCENTRATION TO QUANTITY Given:4.19.1.1 Concentration (Milligrams per Liter) Effluent TSS = 29 mg/L Effluent flow = 11.5 MGD to Pounds Find the concentration in kg/d. lb = Concentration (mg L ) ¥ Tank Volume (MG) ¥ Solution: 8.34 lb MG mg L (4.38) kg d = 29 mg L ¥ 11.5 MGD ¥ 3.785 lb MG mg L EXAMPLE 4.85 = 1263 kg d Problem: 4.19.1.4 Concentration (milligrams/kilogram) Given: to pounds/ton MLSS = 2580 mg/L Aeration tank volume = 0.90 MG lb ton = Concentration (mg KG) ¥ (4.41) Find the concentration in pounds. 0.002 lb ton mg kg © 2003 by CRC Press LLC
  • EXAMPLE 4.88 4.19.2.3 Kilograms per Day to Concentration Problem: (Milligrams per Liter) Biosolids contain 0.97 mg/kg of lead. How many pounds (4.44) Quantity ( kg d ) Concentration (mg L ) = of lead are being applied per acre if the current application rate is 5 dry tons of solids per acre? Q (MG) ¥ 3.785 kg mg L MG Solution: 4.19.3 QUANTITY TO VOLUME OR FLOW RATE lb ac = 0.97 mg kg ¥ 5 tons ac ¥ 4.19.3.1 Pounds to Tank Volume (Million Gallons) 0.002 lb ton mg kg (4.45) Quantity (lb) = 0.0097 lb ac v (MG) = Concentration (mg L ) ¥ 8.34 lb mg L MG4.19.2 QUANTITY TO CONCENTRATION 4.19.3.2 Pounds per Day to Flow4.19.2.1 Pounds to Concentration (Million Gallons per Day) (Milligrams per Liter) (4.46) Quantity (lb d ) (4.42) Q (MGD) = Quantity (lb) Concentration (mg L ) ¥ 8.34 lb mg L MG Concentration (mg L ) = v (MG) ¥ 8.34 lb mg L MG EXAMPLE 4.91 EXAMPLE 4.89 Problem: Problem: You must remove 8485 lb of solids from the activated The aeration tank contains 73,529 lb of solids. The volume sludge process. The waste activated sludge solids concen- of the tank is 3.20 MG. What is the concentration of solids tration is 5636 mg/L. How many MG must be removed? in the aeration tank in mg/L? Solution: Solution: 8485 (lb d ) 73, 529 lbs Q (MGD) = Concentration ( mg L ) = 5636 mg L ¥ 8.34 lb mg L MG 3.20 MG ¥ 8.34 lb mg L MG = 0.181 MGD = 2755.1 (2755) mg L 4.19.3.3 Kilograms per Day to Flow4.19.2.2 Pounds per Day to Concentration (Million Gallons per Day) (Milligrams per Liter) (4.47) Quantity ( kg d ) (4.43) Quantity (lb) Q (MG) = Concentration (mg L ) = Concentration (mg L ) ¥ 3.785 kg MG mg L Q (MG) ¥ 8.34 lb mg L MG EXAMPLE 4.90 4.20 CHAPTER REVIEW QUESTIONS Problem: AND PROBLEMS What is the chlorine dose in milligrams/liter when 490 BASIC MATH QUESTIONS lb/d of chlorine is added to an effluent flow of 11.0 MGD? 4.1. Fractions are used to express a portion of a Solution: _____________. 4.2. In 1.2, 1.6, 1.9, 1.8, 1.0, 1.5, what is the 490 lb d mean? Dose ( mg L ) = 11.0 MG ¥ 8.34 lb mg L MG 4.3. [(25 – 4 – 6) ∏ (3 ¥ 5)] + 4 ¥ 3 = 4.4. 2/3 is equal to how many ninths (x/9): = 5.34 mg L 4.5. 3/4 ¥ 5/6 = © 2003 by CRC Press LLC
  • 4.6. 3/7 ∏ 2/3 = 4.25. The digester has a diameter of 60 ft and is 4.7. What is the fraction equivalent of 0.625? 26 ft deep. If the operator pumps 5200 gal- 4.8. What is the decimal equivalent of 3/4? lons of residuals (sludge) to the digester per 4.9. Write 10,000,000 as powers of ten. day, what is the HDT in the digester in days? 4.10. What is the area of a rectangle 9 ¥ 30 ft? 4.26. If 4000 gal of solids are removed from the 4.11. What is the volume of a tank 25 ¥ 60 ¥ 8 ft primary settling tank, how many pounds of deep? solids are removed? 4.27. The plant influent contains 240 mg/L BOD, 4.12. A pipe has a diameter of 8 in. Water is flow- the primary effluent contains 180 mg/L ing through it at 4 ft/min. How much water BOD, and the final effluent contains 22 mg/L is passing through in 1 min? In 1 h? BOD. What is the percent removal for the 4.13. Find the volume of a fuel tank 5 ft in diam- primary treatment process and for the entire eter and 12 ft long. plant? 4.14. A reservoir is 40 ft deep. What will the pres- 4.28. The plant influent includes an industrial flow sure be at the bottom of the reservoir? that contains 335 mg/L BOD. The industrial 4.15. The term __________ base is used to iden- flow is 0.68 MGD. What is the population tify the bottom leg of a triangle. equivalent of the industrial discharge in 4.16. The distance from the center of a circle to people? the edge is the ___________. 4.29. The label on the chemical container states that 4.17. The distance around a circular object is the specific gravity is 1.1435. What is the called ______________ while the distance weight of 1 gal of the chemical solution? around areas other than circles is the 4.30. The influent flow rate is 5.0 MG, and the ___________. current return activated sludge flow rate is 4.18. What do the key words: of, and, per, and less 2.0 MGD. If the SSV60 is 36%, what should than mean in solving math problems? be the return rate? 4.19. When no grouping is given, in what order 4.31. The operator determines that 12,000 lb of are problems worked? activated sludge must be removed from the process. The waste activated sludge solidsTYPICAL UNIT PROCESS CALCULATIONS concentration currently is 5500 mg/L. What is the required waste activated sludge flow 4.20. The sludge contains 6.55% solids. If 9000 rate in MGD and gal/min? gal of sludge are removed from the primary 4.32. The plant effluent currently requires a chlo- settling tank, how many pounds of solids are rine dose of 7.0 mg/L to produce the required removed? 1.1-mg/L chlorine residual in the chlorine 4.21. The operator wishes to remove 3440 lb/d of contact tank. What is the chlorine demand in solids from the activated sludge process. The milligrams per liter? waste activated sludge concentration is 4.33. The ferric chloride solution (stock solution) 3224 mg/L. What is the required flow rate in delivered to the plant contains 42.0% ferric MGD? chloride. Testing indicates that the optimum 4.22. The plant influent includes an industrial flow concentration for the working solution used that contains 240 mg/L BOD. The industrial in the treatment of the wastewater is 3.50%. flow is 0.90 MGD. What is the population How many gallons of stock ferric chloride equivalent for the industrial contribution in solution must be used to prepare 6000 gal of people per day? working solution? 4.23. Determine the per capita characteristics of 4.34. Sludge is added to a 560,000-gal digester at BOD and SS, if garbage grinders are installed the rate of 12,500 gal/d. What is the sludge in a community. Assume that the average per retention time? capita flow is 120 gal/d and that the typical 4.35. What is the surface area of a circular tank average per capita contributions for domestic with a 25 ft diameter? wastewater with ground kitchen wastes are 4.36. What is the volume of a round tank 10 ft BOD: 0.29 lb/capita/d; SS: is 0.22 lb/capita/d. deep, with a 35-ft diameter? 4.24. The label of hypochlorite solution states that 4.37. If the same water is passing through a 200-gal the specific gravity of the solution is 1.1347. tank, what is the detention time? What is the weight of 1 gal of the hypochlorite 4.38. Find the volume of a chlorine cylinder 25 in. solution? in diameter and 44 in. tall. © 2003 by CRC Press LLC
  • 4.39. The average daily winter demand of a com- the water drops 4.0 in./min. What is the rate munity is 14,000 gal. If the summer demand of filtration in MGD? is estimated to be 73% greater than the win- 4.47. A pump must pump 1800 gal/min against a ter demand, what is the estimated summer total head of 30 ft. What horsepower is demand? required for this work?Note: Demand (when related to use) is the amount of 4.48. The operator withdraws 5450 gal of solids water used in a period. The term refers to the from the digester. How many pounds of sol- “demand” put onto the system to meet the needs ids have been removed? of customers. 4.49. A total of 40 hp is required for a particular pumping application. If the pump efficiency 4.40. A reservoir is 50 ft deep. What will be the is 80%, what is the brake horsepower pressure at the bottom of the reservoir? 4.41. Find the flow in gal/min when the flow is required? 1.5 ft3/sec. 4.50. The plant effluent contains 38 mg/L solids. 4.42. Find the flow in a 5-in. pipe when the veloc- The effluent flow rate is 3.89 MGD. How ity is 1.3 ft/sec. many pounds per day of solids are dis- 4.43. The sedimentation basin of a waterworks charged? contains 6575 gallons. What is the detention 4.51. The label of hypochlorite solution states that time if the flow is 160 gal/min? the specific gravity of the solution is 1.1540. 4.44. A sedimentation tank handles a flow of What is the weight of a gallon of the 7.5 MGD. The tank is 70 ¥ 20 ¥ 15 ft and hypochlorite solution? rectangular. What is the detention time? 4.45. A circular clarifier handles a flow of 0.85 MGD. The clarifier has a 20 ft radius REFERENCE and a depth of 12 ft. Find the detention time. 4.46. A filter box is 40 ¥ 20 ft, which also includes 1. Price, J.M, Applied Math for Wastewater Plant Opera- the sand area. If the influent valve is shut, tors, Technomic Publ., Lancaster, PA, 1991, p. vii. © 2003 by CRC Press LLC
  • Water Hydraulics 5 Beginning students of water hydraulics and its principles choice is oil. Some common examples of hydraulic fluid often come to the subject matter with certain misgivings. power systems include automobile braking and power For example, water/wastewater operators quickly learn on steering systems, hydraulic elevators, and hydraulic jacks the job that their primary operational/maintenance con- or lifts. Probably the most familiar hydraulic fluid power cerns involves a daily routine of monitoring, sampling, systems in water and wastewater operations are used on laboratory testing, operation and process maintenance. dump trucks, front-end loaders, graders, and earth-moving How does water hydraulics relate to daily operations? The hydraulic functions of the treatment process have already and excavation equipment. In this text, we are concerned been designed into the plant. Why learn water hydraulics with liquid water. at all? Many find the study of water hydraulics difficult and puzzling (especially those related questions on the licen- Simply put, while having hydraulic control of the plant is sure examinations), but we know it is not mysterious or obviously essential to the treatment process, maintaining difficult. It is the function or output of practical applica- and ensuring continued hydraulic control is also essential. tions of the basic principles of water physics. No water/wastewater facility (and/or distribution collec- Because water and wastewater treatment is based on tion system) can operate without proper hydraulic control. the principles of water hydraulics, concise, real-world The operator must know what hydraulic control is and training is necessary for operators who must operate the what it entails to know how to ensure proper hydraulic control. Moreover, in order to understand the basics of plant and for those sitting for state licensure/certification piping and pumping systems, water/wastewater mainte- examinations. nance operators must have a fundamental knowledge of basic water hydraulics.1 5.2 BASIC CONCEPTSNote: The practice and study of water hydraulics is not new. Even in medieval times, water hydrau- Air Pressure ( @ Sea Level) = 14.7 psi lics was not new. “Medieval Europe had inher- ited a highly developed range of Roman The relationship shown above is important because our hydraulic components.”2 The basic conveyance study of hydraulics begins with air. A blanket of air, many technology, based on low-pressure systems of miles thick surrounds the earth. The weight of this blanket pipe and channels, was already established. In on a given square inch of the earth’s surface will vary studying modern water hydraulics, it is impor- according to the thickness of the atmospheric blanket tant to remember that the science of water above that point. As shown above, at sea level, the pressure hydraulics is the direct result of two immediate exerted is 14.7 pounds per square inch (psi). On a moun- and enduring problems: “The acquisition of taintop, air pressure decreases because the blanket is not fresh water and access to continuous strip of as thick. land with a suitable gradient between the source and the destination.”3 1 ft3 H2O = 62.4 lb5.1 WHAT IS WATER HYDRAULICS? The relationship shown above is also important: both cubic feet and pounds are used to describe a volume ofThe word hydraulic is derived from the Greek words hydro water. There is a defined relationship between these two(meaning water) and aulis (meaning pipe). Originally, the methods of measurement. The specific weight of water isterm hydraulics referred only to the study of water at rest defined relative to a cubic foot. One cubic foot of waterand in motion (flow of water in pipes or channels). Today weighs 62.4 lb. This relationship is true only at a temper-it is taken to mean the flow of any liquid in a system. ature of 4°C and at a pressure of 1 atm (known as standard What is a liquid? In terms of hydraulics, a liquid can temperature and pressure (STP) — 14.7 psi at sea levelbe either oil or water. In fluid power systems used in containing 7.48 gal). The weight varies so little that, formodern industrial equipment, the hydraulic liquid of practical purposes, this weight is used from a temperature © 2003 by CRC Press LLC
  • 0°C to 100°C. One cubic inch of water weighs 0.0362 lb. 5.2.1 STEVIN’S LAWWater 1 ft deep will exert a pressure of 0.43 psi on thebottom area (12 in. ¥ 0.0362 lb/in.3). A column of water Stevin’s law deals with water at rest. Specifically, the lawtwo feet high exerts 0.86 psi, a column 10 ft high exerts states: “The pressure at any point in a fluid at rest depends4.3 psi, and a column 55 ft high exerts on the distance measured vertically to the free surface and the density of the fluid.” Stated as a formula, this becomes 55 ft ¥ 0.43 psi ft = 23.65 psi p=w¥h (5.1) A column of water 2.31 ft high will exert 1.0 psi. To whereproduce a pressure of 50 psi requires a water column p = pressure in pounds per square foot (lb/ft2) w = density in pounds per cubic foot (lb/ft3) 50 psi ¥ 2.31 ft psi = 115.5 ft h = vertical distance in feet The important points being made here are: EXAMPLE 5.2 Problem: 1. 1 ft3 H2O = 62.4 lb (see Figure 4.11) 2. A column of water 2.31 ft high will exert What is the pressure at a point 18 ft below the surface of 1.0 psi. a reservoir? Another relationship is also important: Solution: Note: To calculate this, we must know that the density 1 gal H2O = 8.34 lb of the water (w) is 62.4 lb/ft3. At STP, 1 ft3 H2O contains 7.48 gal. With these two p=w ¥ hrelationships, we can determine the weight of 1 gal H2O.This is accomplished by = 62.4 lb ft 3 ¥ 18 ft 62.4 lb = 1123 lb ft 2 wt 1 gal H2O = = 8.34 lb gal 7.48 gal Water and wastewater operators generally measure Thus, pressure in pounds per square inch rather than pounds per square foot; to convert, divide by 144 in.2/ft2 (12 in. ¥ 1 gal H2O = 8.34 lb 12 in. = 144 in.2):Note: Further, this information allows cubic feet to be 1123 lb ft 2 converted to gallons by simply multiplying the P= = 7.8 psi ( rounded ) 144 in.2 ft 2 number of cubic feet by 7.48 gal/ft.3 EXAMPLE 5.1 5.3 PROPERTIES OF WATER Problem: Table 5.1 shows the relationship between temperature, specific weight, and the density of water. Find the number of gallons in a reservoir that has a volume of 855.5 ft.3 5.3.1 DENSITY AND SPECIFIC GRAVITY Solution: When we say that iron is heavier than aluminum, we say that iron has greater density than aluminum. In practice, 855.5 ft 3 ¥ 7.48 gal ft 3 = 6399 gal ( rounded ) what we are really saying is that a given volume of iron is heavier than the same volume of aluminum.Note: As mentioned in Chapter 4, the term head is Note: What is density? Density is the mass per unit used to designate water pressure in terms of the volume of a substance. height of a column of water in feet. For exam- ple, a 10-ft column of water exerts 4.3 psi. This Suppose you had a tub of lard and a large box of cold can be called 4.3-psi pressure or 10 ft of head. cereal, each having a mass of 600 g. The density of the © 2003 by CRC Press LLC
  • changes. This occurs because substances usually increase TABLE 5.1 in volume (size — they expand) as they become warmer. Water Properties (Temperature, Specific Because of this expansion with warming, the same weight Weight, and Density) is spread over a larger volume, so the density is lower when a substance is warm than when it is cold. Temperature Specific Weight Density (∞F) (lb/ft3) (slugs/ft3) Note: What is specific gravity? Specific gravity is the weight (or density) of a substance compared to 32 62.4 1.94 the weight (or density) of an equal volume of 40 62.4 1.94 water. [Note: The specific gravity of water is 1]. 50 62.4 1.94 60 62.4 1.94 This relationship is easily seen when 1 ft3 H2O, which 70 62.3 1.94 weighs 62.4 lb as shown earlier, is compared to 1 ft3 of 80 62.2 1.93 aluminum, which weights 178 lb. Aluminum is 2.7 times 90 62.1 1.93 as heavy as water. 100 62.0 1.93 It is not that difficult to find the specific gravity (sp gr) 110 61.9 1.92 120 61.7 1.92 of a piece of metal. All you have to do is to weigh the 130 61.5 1.91 metal in air, then weigh it under water. Its loss of weight 140 61.4 1.91 is the weight of an equal volume of water. To find the 150 61.2 1.90 specific gravity, divide the weight of the metal by its loss 160 61.0 1.90 of weight in water. 170 60.8 1.89 180 60.6 1.88 Wt of Substance 190 60.4 1.88 sp gr = (5.3) 200 60.1 1.87 Wt of Equal Volume of Water 210 59.8 1.86 EXAMPLE 5.3 Source: From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001. Problem: Suppose a piece of metal weighs 150 lb in air and 85 lbcereal would be much less than the density of the lard underwater. What is the specific gravity?because the cereal occupies a much larger volume thanthe lard occupies. Solution: The density of an object can be calculated by usingthe formula: Step 1: Calculate the loss of weight in water: Mass 150 lb – 85 lb = 65 lb loss of weight in H2O Density = (5.2) v Step 2: Calculate the specific gravity In water and wastewater treatment, perhaps the mostcommon measures of density are pounds per cubic foot 150 sp gr = = 2.3(lb/ft3) and pounds per gallon (lb/gal). 65 Note: In a calculation of specific gravity, it is essential 1 ft 3 H 2O = 62.4 lb — Density = 62.4 lb ft 3 that the densities be expressed in the same units. 1 gal H 2O = 8.34 lb — Density = 8.34 lb gal As stated earlier, the specific gravity of water is 1.00. This is the standard — the reference to which all other liquid The density of a dry material, such as cereal, lime, or solid substances are compared. Specifically, any objectsoda, and sand, is usually expressed in pounds per cubic that has a specific gravity greater than 1.0 will sink in waterfoot. The density of a liquid, such as liquid alum, liquid (rocks, steel, iron, grit, floc, sludge). Substances with achlorine, or water, can be expressed either as pounds per specific gravity of less than 1.0 will float (wood, scum,cubic foot or as pounds per gallon. The density of a gas, gasoline). Considering the total weight and volume of a ship,such as chlorine gas, methane, carbon dioxide, or air, is its specific gravity is less than one; therefore, it can float.usually expressed in pounds per cubic foot. The most common use of specific gravity in water and As shown in Table 5.1, the density of a substance like wastewater treatment operations is in gallons-to-poundswater changes slightly as the temperature of the substance conversions. In many cases, the liquids being handled have © 2003 by CRC Press LLC
  • a specific gravity of 1.00 or very nearly 1.00 (between Earlier we pointed out that pounds per square inch or0.98 and 1.02), so 1.00 may be used in the calculations pounds per square foot are common expressions of pres-without introducing significant error. However, in calcu- sure. The pressure on the bottom of the cube is 62.4 lb/ft2lations involving a liquid with a specific gravity of less (see Figure 4.11). It is normal to express pressure inthan 0.98 or greater than 1.02, the conversions from gallons pounds per square inch. This is easily accomplished byto pounds must consider specific gravity. The technique determining the weight of 1 in.2 of a cube 1 ft high. If weis illustrated in the following example. have a cube that is 12 inches on each side, the number of square inches on the bottom surface of the cube is 12 in. ¥ EXAMPLE 5.4 12 in. = 144 in.2 Dividing the weight by the number of square inches determines the weight on each square inch. Problem: There are 1455 gal of a certain liquid in a basin. If the 62.4 lb ft specific gravity of the liquid is 0.94, how many pounds psi = = 0.433 psi ft 144 in.2 of liquid are in the basin? This is the weight of a column of water one-inch Solution: square and 1 ft tall. If the column of water were 2 ft tall, Normally, for a conversion from gallons to pounds, we the pressure would be 2 ft ¥ 0.433 psi/ft = 0.866 psi. would use the factor 8.34 lb/gal (the density of water) if the Note: 1 ft H2O = 0.433 psi substance’s specific gravity were between 0.98 and 1.02. However, in this instance the substance has a specific gravity With the above information, feet of head can be con- outside this range, so the 8.34 factor must be adjusted. verted to pounds per square inch by multiplying the feet of head times 0.433 psi/ft. Step 1: Multiply 8.34 lb/gal by the specific gravity to obtain the adjusted factor: EXAMPLE 5.5 8.34 lb gal ¥ 0.94 = 7.84 lb gal ( rounded ) Problem: A tank is mounted at a height of 90 ft. Find the pressure Step 2: Convert 1455 gal to lb using the corrected factor: at the bottom of the tank. 1455 gal ¥ 7.84 lb gal = 11, 407 lb ( rounded ) Solution: 90 ft ¥ 0.433 psi ft = 39 psi ( rounded )5.4 FORCE AND PRESSURE Note: To convert pounds per square inch to feet, youWater exerts force and pressure against the walls of its would divide the pounds per square inch bycontainer, whether it is stored in a tank or flowing in a 0.433 psi/ft.pipeline. There is a difference between force and pressure,though they are closely related. Force and pressure are EXAMPLE 5.6defined below. Force is the push or pull influence that causes motion. Problem:In the English system, force and weight are often used in Find the height of water in a tank if the pressure at thethe same way. The weight of 1 ft3 H2O is 62.4 lb. The bottom of the tank is 22 psi.force exerted on the bottom of a 1-ft cube is 62.4 lb (seeFigure 4.11). If we stack two cubes on top of one another, Solution:the force on the bottom will be 124.8 lb. Pressure is a force per unit of area. In equation form, 22 psithis can be expressed as: H ( ft ) = = 51 ft ( rounded ) 0.433 psi ft F Important Point: One of the problems encountered in P= (5.4) A a hydraulic system is storing the liquid. Unlike air, which is readily compressible and is capa-where ble of being stored in large quantities in rela- P = pressure tively small containers, a liquid such as water F = force cannot be compressed. Therefore, it is not pos- A = area over which the force is distributed sible to store a large amount of water in a small © 2003 by CRC Press LLC
  • Liquid level T FIGURE 5.1 Hydrostatic pressure. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) tank — 62.4 lb of water occupies a volume of 1 ft3, regardless of the pressure applied to it. Flow5.4.1 HYDROSTATIC PRESSURE ThrustFigure 5.1 shows a number of differently shaped, con- 90°nected, open containers of water. Note that the water levelis the same in each container, regardless of the shape orsize of the container. This occurs because pressure isdeveloped, within water (or any other liquid), by the Flowweight of the water above. If the water level in any onecontainer were to be momentarily higher than that in any FIGURE 5.2 Shows direction of thrust in a pipe in a trenchof the other containers, the higher pressure at the bottom (viewed from above). (From Spellman, F.R. and Drinan, J.,of this container would cause some water to flow into the Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)container having the lower liquid level. In addition, thepressure of the water at any level (such as Line T) is the When a rapidly closing valve suddenly stops watersame in each of the containers. Pressure increases because flowing in a pipeline, pressure energy is transferred to theof the weight of the water. The further down from the valve and pipe wall. Shockwaves are set up within thesurface, the more pressure is created. This illustrates that system. Waves of pressure move in a horizontal yo-yothe weight, not the volume, of water contained in a vessel fashion — back and forth — against any solid obstaclesdetermines the pressure at the bottom of the vessel. in the system. Neither the water nor the pipe will compress Nathanson (1997) points out some very important to absorb the shock, which may result in damage to pipes,principles that always apply for hydrostatic pressure. valves, and shaking of loose fittings. Another effect of water under pressure is called thrust. 1. The pressure depends only on the depth of Thrust is the force that water exerts on a pipeline as it water above the point in question (not on the rounds a bend. As shown in Figure 5.2, thrust usually acts water surface area). perpendicular (at 90°) to the inside surface its pushes 2. The pressure increases in direct proportion to against. As stated, it affects bends, but also reducers, dead the depth. ends, and tees. Uncontrolled, the thrust can cause move- 3. The pressure in a continuous volume of water ment in the fitting or pipeline, which will lead to separa- is the same at all points that are at the same tion of the pipe coupling away from both sections of depth. pipeline, or at some other nearby coupling upstream or 4. The pressure at any point in the water acts in downstream of the fitting. all directions at the same depth. There are two types of devices commonly used to control thrust in larger pipelines: thrust blocks and thrust5.4.2 EFFECTS OF WATER UNDER PRESSURE5 anchors. A thrust block is a mass of concrete cast in placeWater under pressure and in motion can exert tremendous onto the pipe and around the outside bend of the turn. Anforces inside a pipeline. One of these forces, called example is shown in Figure 5.3. These are used for pipeshydraulic shock or water hammer, is the momentary with tees or elbows that turn left or right or slant upward.increase in pressure that occurs when there is a sudden The thrust is transferred to the soil through the largerchange of direction or velocity of the water. bearing surface of the block. © 2003 by CRC Press LLC
  • 14.7 psi of atmospheric pressure, but subtracting this Top view 14.7 psi leaves a gauge pressure of 0 psi. This shows that the water would rise 0 feet above the reservoir surface. If the gauge pressure in a water main were 120 psi, the water would rise in a tube connected to the main: 120 psi ¥ 2.31 ft psi = 277 ft ( rounded ) Thrust The total head includes the vertical distance the liquid must be lifted (static head), the loss to friction (friction head), and the energy required to maintain the desired FIGURE 5.3 Thrust block. (From Spellman, F.R. and Drinan, velocity (velocity head). J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) Total Head = Static Head + Friction Head + Couplings (5.5) Velocity Head 5.5.1 STATIC HEAD Static head is the actual vertical distance the liquid must be lifted. Static Head = Discharge Elevation ¥ Thrust Shackle direction (5.6) rods Supply Elevation FIGURE 5.4 Thrust anchor. (From Spellman, F.R. and Dri- EXAMPLE 5.7 nan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) Problem: A thrust anchor is a massive block of concrete, often The supply tank is located at elevation 118 ft. The dis- charge point is at elevation 215 ft. What is the static heada cube, cast in place below the fitting to be anchored (see in feet?Figure 5.4). As shown in Figure 5.4, imbedded steelshackle rods anchor the fitting to the concrete block, effec-tively resisting upward thrusts. Solution: The size and shape of a thrust control device depends Static Head (ft) = 215 ft ¥ 118 ft = 97 fton pipe size, type of fitting, water pressure, water hammer,and soil type. 5.5.2 FRICTION HEAD Friction head is the equivalent distance of the energy that5.5 HEAD must be supplied to overcome friction. Engineering refer- ences include tables showing the equivalent vertical dis-Head is defined as the vertical distance the water or waste- tance for various sizes and types of pipes, fittings, andwater must be lifted from the supply tank to the discharge, valves. The total friction head is the sum of the equivalentor as the height a column of water would rise due to the vertical distances for each component.pressure at its base. A perfect vacuum plus atmosphericpressure of 14.7 psi would lift the water 34 ft. If the topof the sealed tube is opened to the atmosphere and the Friction Head (ft ) = Energy Losses Due (5.7)reservoir is enclosed, the pressure in the reservoir is to Frictionincreased; the water will rise in the tube. Because atmo-spheric pressure is essentially universal, we usually ignore 5.5.3 VELOCITY HEADthe first 14.7-psi of actual pressure measurements, andmeasure only the difference between the water pressure Velocity head is the equivalent distance of the energyand the atmospheric pressure; we call this gauge pressure. consumed in achieving and maintaining the desired veloc-For example, water in an open reservoir is subjected to the ity in the system. © 2003 by CRC Press LLC
  • Velocity Head (ft ) = Energy Losses to 5.6 FLOW/DISCHARGE RATE: WATER (5.8) IN MOTION Maintain Velocity The study of fluid flow is much more complicated than that of fluids at rest, but it is important to have an under-5.5.4 TOTAL DYNAMIC HEAD (TOTAL SYSTEM HEAD) standing of these principles. This is because the water in a waterworks and distribution system and in a wastewater Total Dynamic Head = Static Head + Friction Head + treatment plant and collection system is nearly always in motion. Velocity Head (5.9) Discharge (or flow) is the quantity of water passing a given point in a pipe or channel during a given period.5.5.5 PRESSURE/HEAD This is stated another way for open channels: the flow rate through an open channel is directly related to the velocityThe pressure exerted by water and wastewater is directly of the liquid and the cross-sectional area of the liquid inproportional to its depth or head in the pipe, tank, or the channel.channel. If the pressure is known, the equivalent head canbe calculated. Q=A¥V (5.12) where Head (ft ) = Pressure ( psi) ¥ 2.31 ft psi (5.10) Q = flow (discharge in cubic feet per second [ft3/sec]) EXAMPLE 5.8 A = cross-sectional area of the pipe or channel (ft2) Problem: V = water velocity in feet per second (ft/sec) The pressure gauge on the discharge line from the influent EXAMPLE 5.10 pump reads 72.3 psi. What is the equivalent head in feet? Problem: Solution: The channel is 6 ft wide and the water depth is 3 ft. The velocity in the channel is 4 ft/sec. What is the discharge Head (ft ) = 72.3 psi ¥ 2.31 ft psi = 167 ft or flow rate in ft3/sec?5.5.6 HEAD/PRESSURE Solution:If the head is known, the equivalent pressure can be cal- ( ) Q ft 3 sec = A ¥ Vculated using the following equation: = 6 ft ¥ 3 ft ¥ 4 ft sec = 72 ft 3 sec Head (ft ) Pressure ( psi) = (5.11) Discharge or flow can be recorded as gallons per day 2.31 ft psi (gal/d), gallons per minute (gal/min), or cubic feet (ft3/sec). Flows treated by many waterworks or wastewater EXAMPLE 5.9 treatment plants are large, and often referred to in million gallons per day (MGD). The discharge or flow rate can Problem: be converted from cubic feet per second to other units The tank is 22 ft deep. What is the pressure in psi at the such as gallons per minute or million gallons per day by bottom of the tank when it is filled with water? using appropriate conversion factors. Solution: EXAMPLE 5.11 Problem: 22 ft Pressure ( psi ) = 2.31 ft psi A pipe 12 in. in diameter has water flowing through it at 10 ft/sec. What is the discharge in (a) ft3/sec, (b) gal/min, = 9.52 psi ( rounded ) and (c) MGD? © 2003 by CRC Press LLC
  • Solution: Before we can use the basic formula (Equation 5.13), we must determine the area of the pipe. The formula for the area of a circle is B D2 A=p¥ = p ¥ r2 4 FIGURE 5.5 Laminar (streamline) flow. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lan- where caster, PA, 2001.) D = diameter of the circle in feet r = radius of the circle in feet p = the constant value 3.14159 (or simply 3.14) Therefore, the area of the pipe is: D2 (1 ft )2 A=p¥ = 3.14 ¥ = 0.785 ft 2 Streamline Turbulent 4 4 Now we can determine the discharge in ft3/sec (part [a]): Q = A ¥ V = 0.785 ft 2 ¥ 10 ft sec = 7.85 ft 3 sec For part (b), we need to know that 1 ft3/sec is 449 gal/min, FIGURE 5.6 Turbulent flow. (From Spellman, F.R. and Dri- so 7.85 ft3/sec ¥ 449 gal/min/ft3/sec = 3525 gal/min nan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, (rounded). 2001.) Finally, for part (c), 1 MGD is 1.55 ft3/sec, so: any other point (if water does not leave or enter the pipe or channel). That means that under the assumption of steady state flow, the flow that enters the pipe or channel 7.85 ft 3 sec = 5.06 MGD is the same flow that exits the pipe or channel. In equation 1.55 ft 3 sec MGD form, this becomesNote: Flow may be laminar (streamline — see Q1 = Q2 or A1 ¥ V1 = A2 ¥ V2 (5.13) Figure 5.5) or turbulent (see Figure 5.6). Lam- inar flow occurs at extremely low velocities. Note: In regards to the area/velocity relationship, The water moves in straight parallel lines, Equation 5.13 also makes clear that for a given called streamlines, or laminae, that slide upon flow rate the velocity of the liquid varies indi- each other as they travel, rather than mixing up. rectly with changes in the cross-sectional area Normal pipe flow is turbulent flow that occurs of the channel or pipe. This principle provides because of friction encountered on the inside of the basis for many of the flow measurement the pipe. The outside layers of flow are thrown devices used in open channels (weirs, flumes, into the inner layers; the result is that all the and nozzles). layers mix and are moving in different direc- tions and at different velocities. However, the direction of flow is forward. EXAMPLE 5.12Note: Flow may be steady or unsteady. For our pur- Problem: poses, we consider steady state flow only; most A pipe 12 in. in diameter is connected to a 6-in. diameter of the hydraulic calculations in this manual pipe. The velocity of the water in the 12-in. pipe is 3 ft/sec. assume steady state flow. What is the velocity in the 6-in. pipe?5.6.1 AREA/VELOCITY Solution:The law of continuity states that the discharge at each Using the equation A1 ¥ V1 = A2 ¥ V2, we need to deter-point in a pipe or channel is the same as the discharge at mine the area of each pipe: © 2003 by CRC Press LLC
  • For 12-in. pipe: treatment systems involve water in motion — in pipes under pressure or in open channels under the force of D2 gravity. The volume of water flowing past any given point A=p¥ in the pipe or channel per unit time is called the flow rate 4 or discharge, or just flow. (1 ft )2 In regards to flow, continuity of flow and the continu- = 3.14 ¥ ity equation have been discussed (i.e., Equation 5.15). 4 Along with the continuity of flow principle and continuity = 0.785 ft 2 equation, the law of conservation of energy, piezometric surface, and Bernoulli’s theorem (or principle) are also For 6-in. pipe: important to our study of water hydraulics. 0.52 5.7.1 LAW OF CONSERVATION OF ENERGY A = 3.14 ¥ 4 Many of the principles of physics are important to the 2 study of hydraulics. When applied to problems involving = 0.196 ft the flow of water, few of the principles of physical science are more important and useful to us than the law of con- The continuity equation now becomes: servation of energy. Simply, the law of conservation of energy states that energy can neither be created nor 0.785 ft2 ¥ 3ft/sec = 0.196 ft2 ¥ V2 destroyed, but it can be converted from one form to another. In a given closed system, the total energy is Solving for V2 : constant. 0.785 ft 2 ¥ 3 ft sec 5.7.2 ENERGY HEAD V2 = = 12 ft sec 0.196 ft 2 Two types of energy, kinetic and potential, and three forms5.6.2 PRESSURE/VELOCITY of mechanical energy exist in hydraulic systems: potential energy due to elevation, potential energy due to pressure,In a closed pipe flowing full (under pressure), the pressure and kinetic energy due to velocity. Energy has the unitsis indirectly related to the velocity of the liquid. This prin- of foot pounds (ft-lb). It is convenient to express hydraulicciple, when combined with the principle discussed in the energy in terms of energy head, in feet of water. This isprevious section, forms the basis for several flow measure- equivalent to foot-pounds per pound of water (ft-lb/lbment devices (venturi meters and rotameters) as well as the H2O = ft H2O).injector used for dissolving chlorine into water, and chlo-rine, sulfur dioxide and/or other chemicals into wastewater. 5.7.3 PIEZOMETRIC SURFACE7 Velocity1 ¥ Pressure1 = Velocity2 ¥ Pressure2 (5.14) As mentioned earlier, we have seen that when a vertical tube, open at the top, is installed onto a vessel of water, the water will rise in the tube to the water level in theor tank. The water level to which the water rises in a tube is V1 ¥ P1 = V2 ¥ P2 the piezometric surface. The piezometric surface is an imaginary surface that coincides with the level of the water to which water in a system would rise in a piezometer (an5.7 PIEZOMETRIC SURFACE AND instrument used to measure pressure). BERNOULLI’S THEOREM The surface of water that is in contact with the atmo- sphere is known as free water surface. Many important They will take your hand and lead you to the pearls of hydraulic measurements are based on the difference in the desert, those secret wells swallowed by oyster crags height between the free water surface and some point in of wadi, underground caverns that bubble rusty salt water the water system. The piezometric surface is used to locate you would sell your own mothers to drink.6 this free water surface in a vessel, where it cannot be observed directly.To keep the systems in your plant operating properly and To understand how a piezometer actually measuresefficiently, you must understand the basics of hydraulics — pressure, consider the following example.the laws of force, motion, and others. As stated previously, If a clear, see-through pipe is connected to the side ofmost applications of hydraulics in water and wastewater a clear glass or plastic vessel, the water will rise in the © 2003 by CRC Press LLC
  • Open end Pressure applied Free water Piezometric surface surface Piezometric surface Piezometer FIGURE 5.7 A container not under pressure where the pie- zometric surface is the same as the free water surface in the vessel. (From Spellman, F.R. and Drinan, J., Water Hydrau- lics, Technomic Publ., Lancaster, PA, 2001.) FIGURE 5.8 A container under pressure where the piezo- metric surface is above the level of the water in the tank.pipe to indicate the level of the water in the vessel. Such (From Spellman, F.R. and Drinan, J., Water Hydraulics, Tech-a see-through pipe, the piezometer, allows you to see the nomic Publ., Lancaster, PA, 2001.)level of the top of the water in the pipe; this is the piezo-metric surface. situation is quite different when water is flowing. Con- In practice, a piezometer is connected to the side of a sider, for example, an elevated storage tank feeding atank or pipeline. If the water-containing vessel is not under distribution system pipeline. When the system is at rest,pressure (as is the case in Figure 5.7), the piezometric all valves closed, all the piezometric surfaces are the samesurface will be the same as the free water surface in the height as the free water surface in storage. On the othervessel, just as it would if a drinking straw (the piezometer) hand, when the valves are opened and the water begins towere left standing a glass of water. flow, the piezometric surface changes. This is an important When pressurized in a tank and pipeline system, as point because as water continues to flow down a pipeline,they often are, the pressure will cause the piezometric less pressure is exerted. This happens because some pres-surface to rise above the level of the water in the tank. The sure is lost (used up) keeping the water moving over thegreater the pressure, the higher the piezometric surface (see interior surface of the pipe (friction). The pressure that isFigure 5.8). An increased pressure in a water pipeline sys- lost is called head loss.tem is usually obtained by elevating the water tank.Note: In practice, piezometers are not installed on 5.7.3.1 Head Loss water towers, because water towers are hundreds Head loss is best explained by example. Figure 5.9 shows of feet high, or on pipelines. Instead, pressure an elevated storage tank feeding a distribution system gauges are used that record pressure in feet of pipeline. When the valve is closed (Figure 5.9A), all the water or in pounds per square inch. piezometric surfaces are the same height as the free water Water only rises to the water level of the main body surface in storage. When the valve opens and water beginsof water when it is at rest (static or standing water). The to flow (Figure 5.9B), the piezometric surfaces drop. The Piezometric surface Piezometric surface 1 2 3 1 2 3 HGL HGL Closed valve Open valve (A) (B) FIGURE 5.9 Shows head loss and piezometric surface changes when water is flowing. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • further along the pipeline, the lower the piezometric sur-face, because some of the pressure is used up keeping thewater moving over the rough interior surface of the pipe.Thus, pressure is lost and is no longer available to pushwater up in a piezometer; this is the head loss.5.7.3.2 Hydraulic Grade LineWhen the valve is opened as in Figure 5.9, flow begins FIGURE 5.10 Demonstrates Bernoulli’s principle. (Fromwith a corresponding energy loss due to friction. The Spellman, F.R. and Drinan, J., Water Hydraulics, Technomicpressures along the pipeline can measure this loss. In Publ., Lancaster, PA, 2001.)Figure 5.9B, the difference in pressure heads betweensections 1, 2, and 3 can be seen in the piezometer tubes Note: The basic principle explained above ignoresattached to the pipe. A line connecting the water surface friction losses from point to point in a fluidin the tank with the water levels at sections 1, 2, and 3 system employing steady state flow.shows the pattern of continuous pressure loss along thepipeline. This is called the hydraulic grade line (HGL) or 5.7.4.1 Bernoulli’s Equationhydraulic gradient of the system. (It is important to point In a hydraulic system, total energy head is equal toout that in a static water system, the HGL is always hor- the sum of three individual energy heads. This can beizontal. The HGL is a very useful graphical aid when expressed asanalyzing pipe flow problems.)Note: During the early design phase of a treatment Total Head = Elevation Head + Pressure Head + plant, it is important to establish the hydraulic Velocity Head grade line across the plant because both the where proper selection of the plant site elevation and Elevation head = pressure due to the elevation of the the suitability of the site depend on this consid- water eration. Typically, most conventional water Pressure head = the height of a column of water that treatment plants required 16 to 17 ft of head a given hydrostatic pressure in a loss across the plant. system could supportKey Point: Changes in the piezometric surface occur Velocity head = energy present due to the velocity when water is flowing. of the water This can be expressed mathematically as5.7.4 BERNOULLI’S THEOREM8Swiss physicist and mathematician Samuel Bernoulli P V2 E =z+ + (5.15)developed the calculation for the total energy relationship w 2gfrom point to point in a steady state fluid system in the1700s. Before discussing Bernoulli’s energy equation, it whereis important to understand the basic principle behind Ber- E = total energy headnoulli’s equation. z = height of the water above a reference plane (ft) Water (and any other hydraulic fluid) in a hydraulic P = pressure (psi)system possesses two types of energy — kinetic and w = unit weight of water (62.4 lb/ft3)potential. Kinetic energy is present when the water is in V = flow velocity (ft/sec)motion. The faster the water moves, the more kinetic g = acceleration due to gravity (32.2 ft/sec2)energy is used. Potential energy is a result of the waterpressure. The total energy of the water is the sum of the Consider the constriction in section of pipe shown inkinetic and potential energy. Bernoulli’s principle states Figure 5.11. We know, based on the law of energy con-that the total energy of the water (fluid) always remains servation, that the total energy head at section A, E1, mustconstant. Therefore, when the water flow in a system equal the total energy head at section B, E2, and usingincreases, the pressure must decrease. When water starts Equation 5.16, we get Bernoulli’s equation.to flow in a hydraulic system, the pressure drops. When 2the flow stops, the pressure rises again. The pressure gauges PA VA P V2 zA = + = zB + B + B (5.16)shown in Figure 5.10 indicate this balance more clearly. w 2g w 2g © 2003 by CRC Press LLC
  • Total energy line v A2/2g v B2 Pressure 2g drop PA w PB E1 w E2 A Q B Constriction zA zB Reference plane FIGURE 5.11 Shows the result of the law of conservation. Since the velocity and kinetic energy of the water flowing in the constricted section must increase, the potential energy may decrease. This is observed as a pressure drop in the constriction. (Adapted from Nathanson, J.A., Basic Environmental Technology: Water Supply, Waste Management, and Pollution Control, 2nd ed. Prentice Hall, Upper Saddle River, NJ: 1997. p. 29.) The pipeline system shown in Figure 5.11 is horizon- Solution:tal. Therefore, we can simplify Bernoulli’s equationbecause zA = zB. Step 1: Compute the flow area at each section, as follows: Because they are equal, the elevation heads cancel outfrom both sides, leaving: p ¥ (0.666 ft ) 2 AA = = 0.349 ft 2 ( rounded ) 4 PA VA 2 P V2 + = B+ B (5.17) p ¥ (0.333 ft ) 2 w 2g w 2g AB = = 0.087 ft 2 4 As water passes through the constricted section of thepipe (section B), we know from continuity of flow that Step 2: From Q = A ¥ V or V = Q/A, we get:the velocity at section B must be greater than the velocityat section A, because of the smaller flow area at section 3.0 ft 3 sec VA = = 8.6 ft sec ( rounded )B. This means that the velocity head in the system 0.349 ft 2increases as the water flows into the constricted section.However, the total energy must remain constant. For this 3.0 ft 3 secto occur, the pressure head, and therefore the pressure, VB = = 34.5 ft sec ( rounded ) 0.087 ft 2must drop. In effect, pressure energy is converted intokinetic energy in the constriction. Step 3: Applying Equation 5.18, we get: The fact that the pressure in the narrower pipe section(constriction) is less than the pressure in the bigger section 100 ¥ 144 8.6 2 PB ¥ 144 34.52seems to defy common sense. However, it does follow + = + 62.4 2 ¥ 32.2 62.4 2 ¥ 32.2logically from continuity of flow and conservation ofenergy. The fact that there is a pressure difference allows Note: The pressures are multiplied by 144 in2/ft2 tomeasurement of flow rate in the closed pipe. convert from psi to lb/ft2 to be consistent with the units for w; the energy head terms are in EXAMPLE 5.13 feet of head. Problem: Continuing, we get In Figure 5.11, the diameter at Section A is 8 in. and at section B, it is 4 in. The flow rate through the pipe is 3.0 231 + 1.15 = 2.3PB + 18.5 ft3/sec and the pressure at Section A is 100 psi. What is the pressure in the constriction at Section B? and © 2003 by CRC Press LLC
  • TABLE 5.2Pump Applications in Water and Wastewater Systems11Application Function Pump TypeLow service To lift water from the source to treatment processes, or from storage to filter-backwashing Centrifugal systemHigh service To discharge water under pressure to distribution system; to pump collected or intercepted Centrifugal water/wastewater and pump to treatment facilityBooster To increase pressure in the distribution/collection system or to supply elevated storage tanks CentrifugalWell To lift water from shallow or deep wells and discharge it to the treatment plant, storage Centrifugal or jet facility, or distribution systemChemical feed To add chemical solutions at desired dosages for treatment processes Positive displacementSampling To pump water/wastewater from sampling points to the laboratory or automatic analyzers Positive displacement or centrifugalSludge/biosolids To pump sludge or biosolids from sedimentation facilities to further treatment or disposal Positive displacement or centrifugalSource: From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001. 232.2 - 18.5 213.7 Several methods are available for transporting water, PB = = = 93 psi ( rounded ) wastewater, and chemicals for treatment between process 2.3 2.3 equipment:5.8 HYDRAULIC MACHINES (PUMPS) 1. Centrifugal force inducing fluid motion 2. Volumetric displacement of fluids, either mechan- Only the sail can contend with the pump for the title of ically, or with other fluids the earliest invention for the conversion of natural energy 3. Transfer of momentum from another fluid to useful work, and it is doubtful that the sail takes prece- 4. Mechanical impulse dence. Since the sail cannot, in any event, be classified as 5. Gravity induction a machine, the pump stands essentially unchallenged as the earliest form of machine that substituted natural energy Depending on the facility and unit processes contained for muscular effort in the fulfillment of man’s needs.9 within, all of the methods above may be important to the maintenance operator.Conveying water and wastewater to and from processequipment is an integral part of the water and wastewater 5.8.1 PUMPING HYDRAULICS12industry that requires energy consumption. The amount ofenergy required depends on the height to which the water During operation, water enters a pump on the suction side,or wastewater is raised, the length and diameter of the where the pressure is lower. Since the function of the pumpconveying conduits, the rate of flow, and the water or is to add pressure to the system, discharge pressure willwastewater’s physical properties (in particular, viscosity always be higher. In pump systems, an important conceptand density). In some applications, external energy for to keep in mind is measurements are taken from the pointtransferring water or wastewater is not required. For exam- of reference to the centerline of the pump (horizontal lineple, when water or wastewater flows to a lower elevation drawn through center of pump).under the influence of gravity, a partial transformation of In order to understand pump operation, or pumpingthe water or wastewater’s potential energy into kinetic hydraulics, we need to be familiar with certain basic termsenergy occurs. However, when conveying water or waste- and then relate these terms pictorially (as we do inwater through horizontal conduits, especially to higher Figure 5.12) to illustrate how water is pumped from oneelevations within a system, mechanical devices such as point to another.pumps are employed. Requirements vary from small unitsused to pump only a few gallons per minute to large units 1. Static head — The distance between the suctioncapable of handling several hundred cubic feet per sec- and discharge water levels when the pump isond.10 Table 5.2 lists pump applications in water and shut off. We indicate static head conditions withwastewater treatment operations. the letter Z (see Figure 5.12). 2. Suction lift — The distance between the suctionNote: In determining the amount of pressure or force water level and the center of the pump impeller. a pump must provide to move the water or This term is only used when the pump is in a wastewater, the term pump head was estab- suction lift condition; the pump must have the lished. energy to provide this lift. A pump is said to be © 2003 by CRC Press LLC
  • 2 Velocity Head V /2g = 1 ft Headloss HL = 19 ft THD = 100 ft Total Dynamic Head Static Head Z = 80 ft Suction Lift FIGURE 5.12 Components of total dynamic head. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) in a suction lift condition any time the center presented and defined, and they are related pictorially in (eye) of the impeller is above the water being Figure 5.13. Also discussed are wet wells, which are impor- pumped (see Figure 5.12). tant, both in water and wastewater operations. 3. Suction head — A pump is said to be in a suction head condition any time the center (eye) 5.9.1 WELL HYDRAULICS of the impeller is below the water level being pumped. Specifically, suction head is the dis- 1. Static water level — The water level in a well tance between the suction water level and the when no water is being taken from the ground- center of the pump impeller when the pump is water source (i.e., the water level when the in a suction head condition (see Figure 5.12). pump is off; see Figure 5.13). Static water level is normally measured as the distance from the 4. Velocity head — The amount of energy ground surface to the water surface. This is an required to bring water or wastewater from important parameter because it is used to mea- standstill to its velocity. For a given quantity of sure changes in the water table. flow, the velocity head will vary indirectly with 2. Pumping water level — The water level when the pipe diameter. Velocity head is often shown the pump is off. When water is pumped out of mathematically as V2/2g (see Figure 5.12). a well, the water level usually drops below the 5. Total dynamic head — The total energy needed level in the surrounding aquifer and eventually to move water from the centerline of a pump stabilizes at a lower level; this is the pumping (eye of the first impeller of a lineshaft turbine) level (see Figure 5.13). to some given elevation or to develop some 3. Drawdown — the difference, or the drop, given pressure. This includes the static head, between the static water level and the pumping velocity head and the head loss due to friction water level, measured in feet. Simply, it is the (see Figure 5.12). distance the water level drops once pumping begins (see Figure 5.13).5.9 WELL AND WET WELL HYDRAULICS 4. Cone of depression — In unconfined aquifers, there is a flow of water in the aquifer from allWhen the source of water for a water distribution system directions toward the well during pumping. Theis from a groundwater supply, knowledge of well hydraulics free water surface in the aquifer then takes theis important to the operator. Basic well hydraulics terms are shape of an inverted cone or curved funnel line. © 2003 by CRC Press LLC
  • Discharge Ground surface Pump Static water level Cone of depression Drawdown Pump water level Zone of influence FIGURE 5.13 Hydraulic characteristics of a well. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) The curve of the line extends from the pumping For large wells, it may be measured in cubic water level to the static water level at the outside feet per second (cubic meters per second). edge of the zone (or radius) of influence (see 2. Specific capacity is the pumping rate per foot Figure 5.13). of drawdown (gallon per minute per foot), orNote: The shape and size of the cone of depression is Well Yield dependent on the relationship between the Specific Capacity = (5.18) Drawdown pumping rate and the rate at which water can move toward the well. If the rate is high, the EXAMPLE 5.14 cone will be shallow and its growth will stabi- lize. If the rate is low, the cone will be sharp Problem: and continue to grow in size. If the well yield is 300 gal/min and the drawdown is measured to be 20 ft, what is the specific capacity? 5. Zone (or radius) of influence — The distance between the pump shaft and the outermost area affected by drawdown (see Figure 5.13). The Solution: distance depends on the porosity of the soil and 300 other factors. This parameter becomes impor- Specific Capacity = 20 tant in well fields with many pumps. If wells are set too close together, the zones of influence = 15 gal min ft of drawdown will overlap, increasing the drawdown in all wells. Obviously, pumps should be spaced apart Specific capacity is one of the most important con- to prevent this from happening. cepts in well operation and testing. The calculation should be made frequently in the monitoring of well operation. A sudden drop in specific capacity indicates problemsNote: Two important parameters not shown in such as pump malfunction, screen plugging, or other prob- Figure 5.13 are well yield and specific capacity. lems that can be serious. Such problems should be iden- tified and corrected as soon as possible. 1. Well yield is the rate of water withdrawal that a well can supply over a long period. Alterna- 5.9.2 WET WELL HYDRAULICS tively, this is simply the maximum pumping rate that can be achieved without increasing the Water pumped from a wet well by a pump set above the drawdown. The yield of small wells is usually water surface exhibits the same phenomena as the ground- measured in gallons per minute (liters per water well. In operation, a slight depression of the water minute) or gallons per hour (liters per hour). surface forms right at the intake line (drawdown), but in © 2003 by CRC Press LLC
  •  this case it is minimal because there is free water at the pipes flow completely full (as in water lines), we alsopump entrance at all times (at least there should be). The address pipes that flow partially full (wastewater lines,most important consideration in wet well operations is to normally treated as open channels) in this section.ensure that the suction line is submerged far enough below The solution of practical pipe flow problems resultingthe surface, so that air entrained by the active movement from application of the energy principle, the equation ofof the water at this section is not able to enter the pump. continuity, and the principle and equation of water resis- Because water or wastewater flow is not always con- tance are also discussed. Resistance to flow in pipes is notstant or at the same level, variable speed pumps are com- only the result of long reaches of pipe but is also offeredmonly used in wet well operations, or several pumps are by pipe fittings, such as bends and valves, that dissipateinstalled for single or combined operation. In many cases, energy by producing relatively large-scale turbulence.pumping is accomplished in an on/off mode. Control ofpump operation is in response to water level in the well. 5.10.2 PIPE AND OPEN FLOW BASICSLevel control devices, such as mercury switches, are usedto sense a high and low level in the well and transmit the In order to gain understanding of what friction head losssignal to pumps for action. is all about, it is necessary to review a few terms presented earlier in the text and to introduce some new terms perti- nent to the subject.135.10 FRICTION HEAD LOSSMaterials or substances capable of flowing cannot flow 1. Laminar flow — Laminar flow is ideal flow;freely. Nothing flows without encountering some type of that is, water particles moving along straight,resistance. Consider electricity, the flow of free electrons parallel paths, in layers or streamlines. More-in a conductor. Whatever type of conductor used (i.e., over, in laminar flow there is no turbulence incopper, aluminum, silver, etc.) offers some resistance. In the water and no friction loss. This is not typicalhydraulics, the flow of water or wastewater is analogous of normal pipe flow because the water velocityto the flow of electricity. Within a pipe or open channel, is too great, but is typical of groundwater flow.for instance, flowing water, like electron flow in a con- 2. Turbulent flow — Characterized as normal forductor, encounters resistance. However, resistance to the a typical water system, turbulent flow occursflow of water is generally termed friction loss (or more when water particles move in a haphazard fash-appropriately, head loss). ion and continually cross each other in all direc- tions resulting in pressure losses along a length5.10.1 FLOW IN PIPELINES of pipe. 3. Hydraulic grade line (HGL) — Recall that theThe problem of waste and wastewater flow in pipelines — hydraulic grade line (HGL) (shown inthe prediction of flow rate through pipes of given charac- Figure 5.14) is a line connecting two points toteristics, the calculation of energy conversions therein, and which the liquid would rise at various placesso forth — is encountered in many applications of water along any pipe or open channel if piezometersand wastewater operations and practice. Although the sub- were inserted in the liquid. It is a measure of theject of pipe flow embraces only those problems in which pressure head available at these various points. V 12 hL V 12 hL 2g Energy grade line 2g Energy grade line V 22 V 22 Water surface Hydraulic grade line 2g 2g y1 y1 V1 Piezometers V2 y2 y2 V1 V2 z1 z1 Channel bottom Center line of pipe z2 z2 Datum 1 Pipe flow 2 1 Open channel flow 2 FIGURE 5.14 Comparison of pipe flow and open-channel flow. (Adapted from Metcalf & Eddy. Wastewater Engineering: Collection and Pumping of Wastewater, Tchobanoglous, G. (Ed.), McGraw-Hill, New York, 1981, p. 11.) © 2003 by CRC Press LLC
  • Note: When water flows in an open channel, the HGL ally varied or rapidly varied (i.e., when the coincides with the profile of the water surface. depth of flow changes abruptly) as shown in Figure 5.15. 4. Energy grade line — the total energy of flow 9. Slope (gradient) — The head loss per foot of in any section with reference to some datum channel. (i.e., a reference line, surface or point) is the sum of the elevation head, z, the pressure head, y, and the velocity head, V2/2g. Figure 5.14 5.10.3 MAJOR HEAD LOSS shows the energy grade line or energy gradient, Major head loss consists of pressure decreases along the which represents the energy from section to length of pipe caused by friction created as water encoun- section. In the absence of frictional losses, the ters the surfaces of the pipe. It typically accounts for most energy grade line remains horizontal, although of the pressure drop in a pressurized or dynamic water the relative distribution of energy may vary system. between the elevation, pressure, and velocity heads. In all real systems, however, losses of 5.10.3.1 Components of Major Head Loss energy occur because of resistance to flow, and the resulting energy grade line is sloped (i.e., The components that contribute to major head loss: rough- the energy grade line is the slope of the specific ness, length, diameter, and velocity. energy line). 5.10.3.1.1 Roughness 5. Specific energy (E) — sometimes called spe- cific head, is the sum of the pressure head, y, Even when new, the interior surfaces of pipes are rough. and the velocity head, V2/2g. The specific The roughness varies depending on pipe material, corro- energy concept is especially useful in analyzing sion (tuberculation and pitting), and age. Because normal flow in open channels. flow in a water pipe is turbulent, the turbulence increases with pipe roughness, which in turn causes pressure to drop 6. Steady flow — Occurs when the discharge or over the length of the pipe. rate of flow at any cross section is constant. 7. Uniform and nonuniform flow — Uniform flow 5.10.3.1.2 Pipe Length occurs when the depth, cross-sectional area, and With every foot of pipe length, friction losses occur. The other elements of flow are substantially con- longer the pipe, the more head loss. Friction loss because stant from section to section. Nonuniform flow of pipe length must be factored into head loss calculations. occurs when the slope, cross-sectional area, and velocity change from section to section. The 5.10.3.1.3 Pipe Diameter flow through a venturi section used for measur- Generally, small diameter pipes have more head loss than ing flow is a good example. large diameter pipes. This is the case because in large 8. Varied flow — Flow in a channel is considered diameter pipes less of the water actually touches the inte- varied if the depth of flow changes along the rior surfaces of the pipe (encountering less friction) than length of the channel. The flow may be gradu- in a small diameter pipe. RVF GVF RVF GVF RVF GVF RVF Sluice gate Hydraulic jump Flow over a weir Hydraulic drop RVF – Rapidly Varied Flow GVF – Gradually Varied Flow FIGURE 5.15 Varied flow. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • 5.10.3.1.4 Water VelocityTurbulence in a water pipe is directly proportional to the TABLE 5.3speed (or velocity) of the flow. Thus, the velocity head C FACTORS15also contributes to head loss. Type of Pipe C FactorNote: For the same diameter pipe, when flow Asbestos cement 140 increases, head loss increases. Brass 140 Brick sewer 1005.10.3.2 Calculating Major Head Loss Cast iron 10 years old 110Henry Darcy, Julies Weisbach, and others developed the 20 years old 90first practical equation used to determine pipe friction in Ductile iron (cement lined) 140about 1850. The equation or formula now known as the Concrete or concrete linedDarcy-Weisbach equation for circular pipes is: Smooth, steel forms 140 Wooden forms 120 LV 2 Rough 110 hf = f (5.19) Copper 140 D2 g Fire hose (rubber lined) 135 In terms of the flow rate Q, the equation becomes: Galvanized iron 120 Glass 140 Lead 130 8fLQ 2 hf = (5.20) Masonry conduit 130 p 2 gD5 Plastic 150 Steelwhere Coal-tar enamel lined 150 hf = head loss (ft) New unlined 140 f = coefficient of friction Riveted 110 L = length of pipe (ft) Tin 130 V = mean velocity (ft/sec) Vitrified 120 Wood stave 120 D = diameter of pipe (ft) g = acceleration due to gravity (32.2 ft/sec2) Source: From Spellman, F.R. and Drinan, J., Water Q = flow rate (ft3/sec) Hydraulics, Technomic Publ., Lancaster, PA, 2001. The Darcy-Weisbach formula as such was meant to 5.10.3.2.1 C Factorapply to the flow of any fluid. Into this friction factor was The C factor, as used in the Hazen-Williams formula,incorporated the degree of roughness and an element designates the coefficient of roughness. C does not varycalled the Reynold’s number, which was based on the appreciably with velocity, and by comparing pipe typesviscosity of the fluid and the degree of turbulence of flow. and ages, it includes only the concept of roughness, ignor- The Darcy-Weisbach formula is used primarily for ing fluid viscosity and Reynold’s number.determining head loss calculations in pipes. For making Based on experience (experimentation), acceptedthis determination in open channels, the Manning equation tables of C factors have been established for pipe (seewas developed during the later part of the 19th century. Table 5.3). Generally, C factor decreases by one with eachLater, this equation was used for both open channels and year of pipe age. Flow for a newly designed system isclosed conduits. often calculated with a C factor of 100, based on averaging In the early 1900s, a more practical equation, the it over the life of the pipe system.Hazen-Williams equation, was developed for use in mak-ing calculations related to water pipes and wastewater Note: A high C factor means a smooth pipe. A low Cforce mains: factor means a rough pipe. Q = 0.435 ¥ CD2.63 ¥ S0.54 (5.21) Note: An alternate to calculating the Hazen-Williams formula, called an alignment chart, has becomewhere quite popular for fieldwork. The alignment Q = flow rate (ft3/sec) chart can be used with reasonable accuracy. C = coefficient of roughness (C decreases with roughness) 5.10.3.2.2 Slope D = hydraulic radius r (ft) Slope is defined as the head loss per foot. In open chan- S = slope of energy grade line (ft/ft) nels, where the water flows by gravity, slope is the amount © 2003 by CRC Press LLC
  • of incline of the pipe and is calculated as feet of drop per Of course, pipes must be constructed to withstand thefoot of pipe length (ft/ft). Slope is designed to be just expected conditions of exposure, and pipe configurationenough to overcome frictional losses, so that the velocity systems for water distribution and/or wastewater collectionremains constant, the water keeps flowing, and solids will and interceptor systems must be properly designed andnot settle in the conduit. In piped systems, where pressure installed in terms of water hydraulics. Because the waterloss for every foot of pipe is experienced, slope is not and wastewater operator should have a basic knowledgeprovided by slanting the pipe, but instead by pressure of water hydraulics related to commonly used standardadded to overcome friction. piping configurations, piping basics are briefly discussed in this section.5.10.4 MINOR HEAD LOSS 5.11.1 PIPING NETWORKSIn addition to the head loss caused by friction betweenthe fluid and the pipe wall, losses also are caused by It would be far less costly and make for more efficientturbulence created by obstructions (i.e., valves and fittings operation if municipal water and wastewater systems wereof all types) in the line, changes in direction, and changes built with separate single pipe networks extending fromin flow area. treatment plant to user’s residence, or from user’s sink or bathtub drain to the local wastewater treatment plant.Note: In practice, if minor head loss is less than 5% Unfortunately, this ideal single-pipe scenario is not prac- of the total head loss, it is usually ignored. tical for real world applications. Instead of a single piping system, a network of pipes is laid under the streets. Each5.11 BASIC PIPING HYDRAULICS of these piping networks is composed of different mate- rials that vary (sometimes considerably) in diameter, Water, regardless of the source, is conveyed to the water- length, and age. These networks range in complexity to works for treatment and distributed to the users. Convey- varying degrees, and each of these joined-together pipes ance from the source to the point of treatment occurs by aqueducts, pipelines, or open channels, but the treated contribute energy losses to the system. water is normally distributed in pressurized closed con- duits. After use, whatever the purpose, the water becomes 5.11.1.1 Energy Losses in Pipe Networks wastewater, which must be disposed of somehow, but almost always ends up being conveyed back to a treatment Water and wastewater flow networks may consist of pipes facility before being outfalled to some water body, to arranged in series, parallel, or some complicated combina- begin the cycle again. tion. In any case, an evaluation of friction losses for the flows is based on energy conservation principles applied to We call this an urban water cycle, because it provides a the flow junction points. Methods of computation depend human-generated imitation of the natural water cycle. on the particular piping configuration. In general, however, Unlike the natural water cycle, however, without pipes, they involve establishing a sufficient number of simulta- the cycle would be nonexistent or, at the very least, neous equations or employing a friction loss formula where short-circuited. the friction coefficient depends only on the roughness of the pipe (e.g., Hazen-Williams equation). (Note: Demon-For use as water mains in a distribution system, pipes must strating the procedure for making these complex computa-be strong and durable in order to resist applied forces and tions is beyond the scope of this text. We only present thecorrosion. The pipe is subjected to internal pressure from operator “need to know” aspects of complex or compoundthe water and to external pressure from the weight of the piping systems in this text.)backfill (soil) and vehicles above it. The pipe may also haveto withstand water hammer. Damage due to corrosion or 5.11.1.2 Pipes in Seriesrusting may also occur internally because of the water qual-ity or externally because of the nature of the soil conditions. When two pipes of different sizes or roughnesses are Pipes used in a wastewater system must be strong and connected in series (see Figure 5.16), head loss for a givendurable to resist the abrasive and corrosive properties of discharge, or discharge for a given head loss, may bethe wastewater. Like water pipes, wastewater pipes must calculated by applying the appropriate equation betweenalso be able to withstand stresses caused by the soil back- the bonding points, taking into account all losses in thefill material and the effect of vehicles passing above the interval. Thus, head losses are cumulative.pipeline. Series pipes may be treated as a single pipe of constant Joints between wastewater collection/interceptor pipe diameter to simplify the calculation of friction losses. Thesections should be flexible, but tight enough to prevent approach involves determining an equivalent length of aexcessive leakage, either of sewage out of the pipe or constant diameter pipe that has the same friction loss andgroundwater into the pipe. discharge characteristics as the actual series pipe system. © 2003 by CRC Press LLC
  • 1 2 3 2 FIGURE 5.16 Pipes in series. (From Spellman, F.R. and 3 Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)In addition, application of the continuity equation to the FIGURE 5.17 Pipe in parallel. (From Spellman, F.R. andsolution allows the head loss to be expressed in terms of Drinan, J., Water Hydraulics, Technomic Publ., Lancaster,only one pipe size. PA, 2001.)Note: In addition to the head loss caused by friction 5.11.1.3 Pipes in Parallel between the water and the pipe wall, losses also are caused by minor losses: obstructions in the Two or more pipes connected (as in Figure 5.17) so that line, changes in directions, and changes in flow flow is first divided among the pipes and is then rejoined area. In practice, the method of equivalent comprise a parallel pipe system. A parallel pipe system is length is often used to determine these losses. a common method for increasing the capacity of an exist- The method of equivalent length uses a table to ing line. Determining flows in pipes arranged in parallel convert each valve or fitting into an equivalent are also made by application of energy conservation prin- length of straight pipe. ciples — specifically, energy losses through all pipes con- In making calculations involving pipes in series, necting common junction points must be equal. Each legremember these two important basic operational tenets: of the parallel network is treated as a series piping system and converted to a single equivalent length pipe. The fric- 1. The same flow passes through all pipes con- tion losses through the equivalent length parallel pipes are nected in series. then considered equal and the respective flows determined 2. The total head loss is the sum of the head losses by proportional distribution. of all of the component pipes. Note: Computations used to determine friction losses In some operations involving series networks where in parallel combinations may be accomplishedthe flow is given and the total head loss is unknown, we using a simultaneous solution approach for acan use the Hazen-Williams equation to solve for the slope parallel system that has only two branches.and the head loss of each pipe as if they were separate However, if the parallel system has three orpipes. Adding up the head losses to get the total head loss more branches, a modified procedure using theis then a simple matter. Hazen-Williams loss formula is easier.16 Other series network calculations may not be as sim-ple to solve using the Hazen-Williams equation. For exam-ple, one problem we may be faced with is what diameter 5.12 OPEN-CHANNEL FLOWto use with varying sized pipes connected together in aseries combination. Moreover, head loss is applied to both Water is transported over long distances through aque-pipes (and other multiples), and it is not known how much ducts to locations where it is to be used and/or treated.loss originates from each one. This makes determining Selection of an aqueduct type rests on such factors asslope difficult, but not impossible. topography, head availability, climate, construction prac- In such cases the equivalent pipe theory, as mentioned tices, economics, and water quality protection. Along withearlier, can be used. Again, one single equivalent pipe is pipes and tunnels, aqueducts may also include or be solelycreated which will carry the correct flow. This is practical composed of open channels.17because the head loss through it is the same as that in theactual system. The equivalent pipe can have any C factorand diameter, just as long as those same dimensions are In this section, we deal with water passage in open chan-maintained all the way through to the end. Keep in mind nels, which allow part of the water to be exposed to thethat the equivalent pipe must have the correct length, so atmosphere. This type of channel — an open-flowthat it will allow the correct flow through, which yields channel — includes natural waterways, canals, culverts,the correct head loss (the given head loss).16 flumes, and pipes flowing under the influence of gravity. © 2003 by CRC Press LLC
  • 5.12.1 CHARACTERISTICS OF OPEN- 5.12.1.4 Parameters Used in Open-Channel Flow CHANNEL FLOW18 The three primary parameters used in open-channel flowBasic hydraulic principles apply in open-channel flow are: hydraulic radius, hydraulic depth, and slope, S.(with water depth constant) although there is no pressure 5.12.1.4.1 Hydraulic Radiusto act as the driving force. Velocity head is the only naturalenergy this water possesses, and at normal water veloci- The hydraulic radius is the ratio of area in flow to wettedties, this is a small value (V2/2g). perimeter. Several parameters can be (and often are) used todescribe open-channel flow. However, we begin our dis- A rH = (5.22)cussion with a few characteristics, including laminar or Pturbulent; uniform or varied; and subcritical, critical, orsupercritical. where rH = hydraulic radius5.12.1.1 Laminar and Turbulent Flow A = the cross sectional area of the waterLaminar and turbulent flow in open channels is analogous P = wetted perimeterto that in closed pressurized conduits (i.e., pipes). It isimportant to point out that flow in open channels is usually Why is the hydraulic radius important?turbulent. In addition, there is no important circumstance Probably the best way in which to answer this questionin which laminar flow occurs in open channels in either is by illustration. Consider, for example, that in open chan-water or wastewater unit processes or structures. nels it is of primary importance to maintain the proper velocity. This is the case because if velocity is not main-5.12.1.2 Uniform and Varied Flow tained then flow stops (theoretically). In order to maintain velocity at a constant level, the channel slope must beFlow can be a function of time and location. If the flow adequate to overcome friction losses. As with other flows,quantity is invariant, it is said to be steady. Uniform flow calculation of head loss at a given flow is necessary, andis flow in which the depth, width, and velocity remain the Hazen-Williams equation is useful (Equation 5.22).constant along a channel. This means that if the flow cross Keep in mind that the concept of slope has not changed.section does not depend on the location along the channel, The difference? We are now measuring, or calculating for,the flow is said to be uniform. Varied or nonuniform flow the physical slope of a channel (ft/ft), equivalent to headinvolves a change in depth, width, and velocity, with a loss.change in one producing a change in the others. Most The preceding seems logical, but there is a problem.circumstances of open-channel flow in water and waste- The problem is with the diameter. In conduits that are notwater systems involve varied flow. The concept of uniform circular (grit chambers, contact basins, streams and riv-flow is valuable, however, in that it defines a limit that the ers), or in pipes only partially full (drains, wastewatervaried flow may be considered to be approaching in many gravity mains, sewers, etc.) where the cross-sectional areacases. of the water is not circular, there is no diameter.Note: Uniform channel construction does not ensure If there is no diameter, then what do we do? uniform flow. Because there is no diameter in a situation where the cross-sectional area of the water is not circular, we must5.12.1.3 Critical Flow use another parameter to designate the size of the cross section, and the amount of it that contacts the sides of theCritical flow (i.e., flow at the critical depth and velocity) conduit. This is where the hydraulic radius (rH) comes in.defines a state of flow between two flow regimes. Critical The hydraulic radius is a measure of the efficiency withflow coincides with minimum specific energy for a given which the conduit can transmit water. Its value dependsdischarge and maximum discharge for a given specific on pipe size and amount of fullness. Simply, we use theenergy. Critical flow occurs in flow measurement devices hydraulic radius to measure how much of the water is inat or near free discharges, and establishes controls in open- contact with the sides of the channel, or how much of thechannel flow. Critical flow occurs frequently in water and water is not in contact with the sides (see Figure 5.18).wastewater systems and is very important in their opera-tion and design. Note: For a circular channel flowing either full or half- full, the hydraulic radius is D/4. Hydraulic radiiNote: Critical flow minimizes the specific energy and of other channel shapes are easily calculated maximizes discharge. from the basic definition. © 2003 by CRC Press LLC
  • the hydraulic radius. For use in open channels, Manning’s formula has become most commonly used: 1.5 Q= ¥ A ¥ r .66 ¥ S.5 (5.25) n where Wetted perimeter Wetted area Q = channel discharge capacity (ft3/sec) 1.5 = constant n = channel roughness coefficient FIGURE 5.18 Hydraulic radius. (From Spellman, F.R. and A = cross-sectional flow area (ft2) Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, r = hydraulic radius of the channel (ft) PA, 2001.) S = slope of the channel bottom, dimensionless5.12.1.4.2 Hydraulic DepthThe hydraulic depth is the ratio of area in flow to the width The hydraulic radius of a channel is defined as the ratioof the channel at the fluid surface. [Note that another name of the flow area to the wetted perimeter P. In formula form,for hydraulic depth is the hydraulic mean depth or hydrau- r = A/P. The new component is n (the roughness coefficient)lic radius]. and depends on the material and age for a pipe or lined channel and on topographic features for a natural stream- A bed. It approximates roughness in open channels and can dH = (5.23) range from a value of 0.01 for a smooth clay pipe to 0.1 w for a small natural stream. The value of n commonlywhere assumed for concrete pipes or lined channels is 0.013. As dH = hydraulic depth the channels get smoother, n values decrease (see Table 5.4). A = area in flow The following example illustrates the application of w = width of the channel at the fluid surface Manning’s formula for a channel with a rectangular cross section.5.12.1.4.3 SlopeThe slope, S, in open channel equations is the slope of EXAMPLE 5.15the energy line. If the flow is uniform, the slope of the Problem:energy line will parallel the water surface and channelbottom. In general, the slope can be calculated from Ber- A rectangular drainage channel is 3 ft wide and is linednoulli’s equation as the energy loss per unit length of with concrete, as illustrated in Figure 5.19. The bottomchannel. of the channel drops in elevation at a rate of 0.5/100 ft. What is the discharge in the channel when the depth of Dh water is 2 ft? S= (5.24) Dl Solution:5.12.2 OPEN-CHANNEL FLOW CALCULATIONS Assume n = 0.013As mentioned, the calculation for head loss at a given flowis typically accomplished by using Hazen-Williams equa- Referring to Figure 5.19, we see that the cross-sectionaltion. In addition, in open-channel flow problems where flow area A = 3 ft ¥ 2 ft = 6 ft2, and the wetted perimeteralthough the concept of slope has not changed, the prob- P = 2 ft + 3 ft + 2 ft = 7 ft. The hydraulic radius R =lem arises with the diameter. Again, in pipes only partially A/P = 6 ft2/7 ft = 0.86 ft. The slope, S = 0.5/100 = 0.005.full where the cross-sectional area of the water is notcircular, there is no diameter. Thus, the hydraulic radius Applying Manning’s formula, we get:is used for these noncircular areas. In the original version of the Hazen-Williams Equa- 2.0 Q= ¥ 6 ¥ 0.86 ¥ 0.005 .66 .5tion, the hydraulic radius was incorporated. Moreover, 0.013similar versions developed by Antoine Chezy (pronounced Q = 59 ft sec 3“Shay-zee”) and Robert Manning, and others incorporated © 2003 by CRC Press LLC
  • TABLE 5.4 Manning Roughness Coefficient (n) Type of Conduit n Type of Conduit n Pipe Cast iron, coated 0.012–0.014 Cast iron, uncoated 0.013–0.015 Wrought iron, galvanized 0.015–0.017 Wrought iron, black 0.012–0.015 Steel, riveted and spiral 0.015–0.017 Corrugated 0.021–0.026 Wood stave 0.012–0.013 Cement surface 0.010–0.013 Concrete 0.012–0.017 Vitrified 0.013–0.015 Clay, drainage tile 0.012–0.014 Lined Channels Metal, smooth semicircular 0.011–0.015 Metal, corrugated 0.023–0.025 Wood, planed 0.010–0.015 Wood, unplaned 0.011–0.015 Cement lined 0.010–0.013 Concrete 0.014–0.016 Cement rubble 0.017–0.030 Grass N/R–0.020a Unlined Channels Earth: straight and uniform 0.017–0.025 Earth: dredged 0.025–0.033 Earth: winding 0.023–0.030 Earth: stony 0.025–0.040 Rock: smooth and uniform 0.025–0.035 Rock: jagged and irregular 0.035–0.045 a N/R = No result Source: From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001. Free water surface Water surface = HGL hL Wetted perimeter 2.0 ft Q Channel bottom Slope = hL/L L 3.0 ft FIGURE 5.20 Steady uniform open-channel flow — whereFIGURE 5.19 For Example 5.15. (From Spellman, F.R. and Dri- the slope of the water surface (or HGL) is equal to the slopenan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) of the channel bottom. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)5.12.3 OPEN-CHANNEL FLOW: THE BOTTOM LINE The key to solving storm water and/or sanitary sewerTo this point, we have set the stage for explaining (in the routine problems is a condition known as steady uniformsimplest possible way) what open-channel flow is and flow; that is, we assume steady uniform flow. Steady flowwhat it is all about. Now that we have explained the means that the discharge is constant with time. Uniformnecessary foundational material and important concepts, flow means that the slope of the water surface and thewe are ready to explain open-channel flow in a manner cross-sectional flow area are also constant. It is commonwhereby it will be easily understood. practice to call a length of channel, pipeline, or stream that We stated that when water flows in a pipe or channel has a relatively constant slope and cross section a reach.19with a free surface exposed to the atmosphere, it is calledopen-channel flow. We also know that gravity provides The slope of the water surface, under steady uniformthe motive force, the constant push, while friction resists flow conditions, is the same as the slope of the channelthe motion and causes energy expenditure. River and bottom. The HGL lies along the water surface and, as instream flow is open-channel flow. Flow in sanitary sewers pressure flow in pipes, the HGL slopes downward in theand storm water drains are open-channel flow, except in direction of flow. Energy loss is evident as the water sur-force mains where the water is pumped under pressure. face elevation drops. Figure 5.20 illustrates a typical © 2003 by CRC Press LLC
  • Ground surface Pipe crown Air Buried pipe Pipe invert Stream partial flow (A) (B) FIGURE 5.21 Shows open-channel flow, whether in a surface stream or in an underground pipe. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)profile view of uniform steady flow. The slope of the water 3. The detention times through the treatment pro-surface represents the rate of energy loss. cesses must be calculated. This is particularly applicable to surface water plants that mustNote: Rate of energy loss (see Figure 5.20) may be meet contact ¥ time (C ¥ T) values required by expressed as the ratio of the drop in elevation the Surface Water Treatment Rule. of the surface in the reach to the length of the reach. 4. Flow measurement allows operators to maintain a record of water furnished to the distribution Figure 5.21 shows typical cross sections of open-chan- system for periodic comparison with the totalnel flow. In Figure 5.21A, the pipe is only partially filled water metered to customers. This provides awith water and there is a free surface at atmospheric pres- measure of “water accounted for,” or converselysure. This is still open-channel flow, although the pipe is (as pointed out earlier by Hauser), the amounta closed underground conduit. Remember, the important of water wasted, leaked, or otherwise not paidpoint is that gravity — not a pump — is moving the water. for (i.e., lost water). 5. Flow measurement allows operators to deter-5.13 FLOW MEASUREMENT mine the efficiency of pumps. (Note: Pumps are covered in detail in Chapter 7). Pumps that are not delivering their designed flow rate are prob- While it is clear that maintaining water/wastewater flow is at the heart of any treatment process, clearly, it is the ably not operating at maximum efficiency, and measurement of flow that is essential to ensuring the so power is being wasted. proper operation of a water/wastewater treatment system. 6. For well systems, it is very important to main- Few knowledgeable operators would argue with this state- tain records of the volume of water pumped and ment. Hauser (1996) asks: “Why measure flow?” Then the hours of operation for each well. The peri- she explains: “The most vital activities in the operation odic computation of well pumping rates can of water and wastewater treatment plants are dependent identify problems such as worn pump impellers on a knowledge of how much water is being processed.”20 and blocked well screens. 7. Reports that must be furnished to the state byIn the statement above, Hauser makes clear that flow mea- most water systems must include records of rawsurement is not only important, but also routine, in water and finished water pumpage.and wastewater operations. Routine, yes, but also the mostimportant variable measured in a treatment plant. Hauser 8. Wastewater generated by a treatment systemalso pointed out that there are several reasons to measure must also be measured and recorded.flow in a treatment plant. The American Water Works 9. Individual meters are often required for theAssociation21 lists several additional reasons to measure proper operation of individual pieces of equip-flow: ment. For example, the makeup water to a flu- oride saturator is always metered to assist in 1. The flow rate through the treatment processes tracking the fluoride feed rate. needs to be controlled so that it matches distri- bution system use. Note: Simply put, measurement of flow is essential 2. It is important to determine the proper feed rate for operation, process control, and recordkeep- of chemicals added in the processes. ing of water and wastewater treatment plants. © 2003 by CRC Press LLC
  • All of the uses addressed create the need for a number mine process variables within the treatment plant, inof flow-measuring devices, often with different capabili- wastewater collection, and in potable water distribution.ties. In this section, we discuss many of the major flow Typically, flow rate meters used are differential pressuremeasuring devices currently used in water and wastewater meters, magnetic meters, and ultrasonic meters. Flow rateoperations. meters are designed for metering flow in closed pipe or open-channel flow.5.13.1 FLOW MEASUREMENT: THE Flow amount is measured in either gallons or in cubic OLD-FASHIONED WAY feet. Typically, a totalizer, which sums up the gallons or cubic feet that pass through the meter, is used. Most ser-An approximate but very simple method to determine vice meters are of this type. They are used in private,open-channel flow has been used for many years. The commercial, and industrial activities where the totalprocedure involves measuring the velocity of a floating amount of flow measured is used in determining customerobject moving in a straight uniform reach of the channel billing. In wastewater treatment, where sampling opera-or stream. If the cross-sectional dimensions of the channel tions are important, automatic composite samplingare known and the depth of flow is measured, then flow units — flow proportioned to grab a sample every so manyarea can be computed. From the relationship Q = A ¥ V, gallons — are used. Totalizer meters can be the velocitythe discharge Q can be estimated. (propeller or turbine), positive displacement, or compound In preliminary fieldwork, this simple procedure is use- types. In addition, weirs and flumes are used extensivelyful in obtaining a ballpark estimate for the flow rate, but for measuring flow in wastewater treatment plants becauseis not suitable for routine measurements. they are not affected (to a degree) by dirty water or floating solids. EXAMPLE 5.16 Problem: 5.13.3 FLOW MEASURING DEVICES A floating object is placed on the surface of water flowing in a drainage ditch and is observed to travel a distance of In recent decades, flow measurement technology has 20 m downstream in 30 sec. The ditch is 2 m wide and evolved rapidly from the old fashioned way of measuring the average depth of flow is estimated to be 0.5 m. Esti- flow we discussed in Section 5.13.1, to the use of simple mate the discharge under these conditions. practical measuring devices, and finally to much more sophisticated devices. Physical phenomena discovered Solution: centuries ago have been the starting point for many of the viable flowmeter designs used today. Moreover, the recent The flow velocity is computed as distance over time: technology explosion has enabled flowmeters to handle many more applications than could have been imagined D centuries ago. V= t Before selecting a particular type of flow measurement 20 m device, Kawamura22 recommends consideration of several = questions. 30 s = 0.67 m sec 1. Is liquid or gas flow being measured? 2. Is the flow occurring in a pipe or in an open The channel area is: channel? A = 2 m ¥ 0.5 m = 1.0 m2 3. What is the magnitude of the flow rate? 4. What is the range of flow variation? The discharge is: 5. Is the liquid being measured clean, or does it contain suspended solids or air bubbles? Q = A ¥ V = 1.0 m2 ¥ 0.66 m2 = 0.66 m3/sec 6. What is the accuracy requirement? 7. What is the allowable head loss by the flowmeter?5.13.2 BASIS OF TRADITIONAL FLOW MEASUREMENT 8. Is the flow corrosive?Flow measurement can be based on flow rate, or flow 9. What types of flowmeters are available to theamount. Flow rate is measured in gallons per minute, region?million gallons per day, or cubic feet per second. Water 10. What types of postinstallation service are avail-and wastewater operations need flow rate meters to deter- able to the area? © 2003 by CRC Press LLC
  • 5.13.3.1 Differential Pressure Flowmeters23 5.13.3.1.2 Types of Differential Pressure FlowmetersFor many years differential pressure flowmeters have beenthe most widely applied flow-measuring device for water Differential pressure flowmeters operate on the principleflow in pipes that require accurate measurement at reason- of developing a differential pressure across a restrictionable cost. The differential pressure type of flowmeter makes that can be related to the fluid flow rate.up the largest segment of the total flow measurement Note: Optimum measurement accuracy is maintaineddevices currently being used. Differential pressure-produc- when the flowmeter is calibrated, the flowmetering meters currently on the market are the venturi, Dall is installed in accordance with standards andtype, Hershel venturi, universal venturi, and venturi inserts. codes of practice, and the transmitting instru- The differential pressure-producing device has a flow ments are periodically calibrated.restriction in the line that causes a differential pressure orhead to be developed between the two measurement loca- The most commonly used differential pressure flow-tions. Differential pressure flowmeters are also known as meter types used in water and wastewater treatment are:head meters, and, of all the head meters, the orifice flow-meter is the most widely applied device. 1. Orifice The advantages of differential pressure flowmeters 2. Venturiinclude: 3. Nozzle 4. Pitot-static tube 1. Simple construction 5.13.3.1.2.1 Orifice 2. Relatively inexpensive 3. No moving parts The most commonly applied orifice is a thin, concentric, and flat metal plate with an opening in the plate (see 4. Transmitting instruments are external Figure 5.22) that is installed perpendicular to the flowing 5. Low maintenance stream in a circular conduit or pipe. Typically, a sharp- 6. Wide application of flowing fluid that is suitable edged hole is bored in the center of the orifice plate. As for measuring both gas and liquid flow the flowing water passes through the orifice, the restriction 7. Ease of instrument and range selection causes an increase in velocity. A concurrent decrease in 8. Extensive product experience and performance pressure occurs as potential energy (static pressure) is database converted into kinetic energy (velocity). As the water leaves the orifice, its velocity decreases and its pressure The disadvantages include: increases as kinetic energy is converted back into potential energy according to the laws of conservation of energy. 1. Flow rate is a nonlinear function of the differ- However, there is always some permanent pressure loss ential pressure. due to friction, and the loss is a function of the ratio of 2. There is a low flow rate range with normal the diameter of the orifice bore (d) to the pipe diameter instrumentation. (D). For dirty water applications (i.e., wastewater), a con-5.13.3.1.1 Operating Principle centric orifice plate will eventually have impaired perfor-Differential pressure flowmeters operate on the principle mance due to dirt buildup at the plate. Instead, eccentricof measuring pressure at two points in the flow, whichprovides an indication of the rate of flow that is passingby. The difference in pressures between the two measure- Upstream Downstreamment locations of the flowmeter is the result of the change Face A Face B Angle of bevelin flow velocities. Simply, there is a set relationshipbetween the flow rate and volume, so the meter instrumen-tation automatically translates the differential pressure into Flow D Axial centerline da volume of flow. The volume of flow rate through the Downstream edgecross-sectional area is given by Dam height Q = A ¥ V(average) Upstream edgewhere Q = the volumetric flow rate FIGURE 5.22 Orifice plate. (From Spellman, F.R. and Dri- A = flow in the cross-sectional area nan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, V = the average fluid velocity 2001.) © 2003 by CRC Press LLC
  • Flow D d FIGURE 5.24 Long radius flow nozzle. (From Spellman, Concentric Eccentric Segmental F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lan- caster, PA, 2001.) FIGURE 5.23 Types of orifice plates. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lan- and abrasive fluids better than an orifice. Note that for the caster, PA, 2001.) same line size and flow rate, the differential pressure at the nozzle is lower (head loss ranges from 10 to 20% ofor segmental orifice plates (see Figure 5.23) are often the differential) than the differential pressure for an orifice;used. Measurements are typically less accurate than those hence, the total pressure loss is lower than an orifice’s.obtained from the concentric orifice plate. Eccentric or Nozzles are primarily used in steam service because ofsegmental orifices are rarely applied in current practice. their rigidity, which makes them dimensionally more sta- The orifice differential pressure flowmeter is the low- ble at high temperatures and velocities than orifices.est cost differential flowmeter, is easy to install, and hasno moving parts. However, it also has high permanent Note: A useful characteristic of nozzles it that theyhead loss (ranging from 40 to 90%) higher pumping costs, reach a critical flow condition — a point atan accuracy of ±2% for a flow range of 4:1, and is affected which further reduction in downstream pressurewith wear or damage. does not produce a greater velocity through the nozzle. When operated in this mode, nozzlesNote: Orifice meters are not recommended for perma- are very predictable and repeatable. nent installation to measure wastewater flow; solids in the water easily catch on the orifice, 5.13.3.1.2.4 Pitot Tube throwing off accuracy. For installation, it is nec- A Pitot tube is a point velocity-measuring device (see essary to have 10 diameters of straight pipe Figure 5.25). It has an impact port; as fluid hits the port, ahead of the orifice meter to create a smooth its velocity is reduced to zero and kinetic energy (velocity) flow pattern, and 5 diameters of straight pipe is converted to potential energy (pressure head). The pres- on the discharge side. sure at the impact port is the sum of the static pressure and the velocity head. The pressure at the impact port is5.13.3.1.2.2 Venturi also known as stagnation pressure or total pressure. TheA venturi is a restriction with a relatively long passage pressure difference between the impact pressure and thewith smooth entry and exit (see Figure 5.24). It has long static pressure measured at the same point is the velocitylife expectancy, simplicity of construction, and relatively head. The flow rate is the product of the measured velocityhigh-pressure recovery (i.e., produces less permanent and the cross-sectional area at the point of measurement.pressure loss than a similar sized orifice). Despite these Note that the Pitot tube has negligible permanent pressureadvantages, the venturi is more expensive, is not linear drop in the line, but the impact port must be located inwith flow rate, and is the largest and heaviest differentialpressure flowmeter. It is often used in wastewater flows Total pressure tapsince the smooth entry allows foreign material to be sweptthrough instead of building up as it would in front of an Static pressure portorifice. The accuracy of this type flowmeter is ±1% for aflow range of 10:1. The head loss across a venturi flow-meter is relatively small. It ranges from 3 to 10% of thedifferential, depending on the ratio of the throat diameterto the inlet diameter (also known as beta ratio).5.13.3.1.2.3 NozzleFlow nozzles (flow tubes) have a smooth entry and sharpexit (see Figure 5.24). For the same differential pressure,the permanent pressure loss of a nozzle is of the same FIGURE 5.25 Pitot tube. (From Spellman, F.R. and Drinan,order as that of an orifice, but it can handle wastewater J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Voltage Electromagnet Flow FIGURE 5.26 Magnetic flowmeter. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)the pipe where the measured velocity is equal to the aver- 1. Its metered liquid must be conductive (but youage velocity of the flowing water through the cross section. would not use this type meter on clean fluids anyway).5.13.3.2 Magnetic Flowmeters24 2. It is bulky, expensive in smaller sizes, and may require periodic calibration to correct driftingMagnetic flowmeters are relatively new to the water and of the signal.wastewater industry. They are volumetric flow devicesdesigned to measure the flow of electrically conductive The combination of the magnetic flowmeter and theliquids in a closed pipe. They measure the flow rate based transmitter is considered as a system. A typical system,on the voltage created between two electrodes (in accor- schematically illustrated in Figure 5.27, shows a transmitterdance with Faraday’s law of electromagnetic induction) mounted remote from the magnetic flowmeter. Some sys-as the water passes through an electromagnetic field (see tems are available with transmitters mounted integral to theFigure 5.26). Induced voltage is proportional to flow rate. magnetic flowmeter. Each device is individually calibratedVoltage depends on magnetic field strength (constant), during the manufacturing process, and the accuracy state-distance between electrodes (constant), and velocity of ment of the magnetic flowmeter includes both pieces offlowing water (variable). equipment. One is not sold or used without the other. Properties of the magnetic flowmeter include: It is also interesting to note that since 1983 almost every manufacturer now offers the microprocessor-based 1. Minimal head loss (no obstruction with line size transmitter. meter) Regarding minimum piping straight run requirements, 2. No effect on flow profile magnetic flowmeters are quite forgiving of piping config- 3. Suitablity for size range between 0.1 in. to 120 in. uration. The downstream side of the magnetic flowmeter 4. An accuracy rating of from 0.5 to 2% of flow is much less critical than the upstream side. Essentially, rate 5. Ability to measure forward or reverse flow. Meter Signal The advantages of magnetic flowmeters include: Conditioner 1. No obstruction to flow mA DC 2. Minimal head loss 3. Wide range of sizes V AC 4. Bidirectional flow measurement 5. Variations in density, viscosity, pressure, and temperature yield negligible effect. 6. Wastewater use 7. No moving parts FIGURE 5.27 Magnetic flowmeter system. (From Spell- man, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., The disadvantages include: Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • all that is required of the downstream side is that sufficient The advantages of time-of-flight ultrasonic flowmetersbackpressure is provided to keep the magnetic flowmeter include:full of liquid during flow measurement. Two diametersdownstream should be acceptable.25 1. No obstruction to flow 2. Minimal head lossNote: Magnetic flowmeters are designed to measure 3. Clamp-ons conductive liquids only. If air or gas is mixed with a. Can be portable the liquid, the output becomes unpredictable. b. No interruption of flow 4. No moving parts5.13.3.3 Ultrasonic Flowmeters 5. Linear over wide rangeUltrasonic flowmeters use an electronic transducer to send 6. Wide range of pipe sizesa beam of ultrasonic sound waves through the water to 7. Bidirectional flow measurementanother transducer on the opposite side of the unit. Thevelocity of the sound beam varies with the liquid flow The disadvantages include:rate, so the beam can be electronically translated to indi-cate flow volume. The accuracy is ±1% for a flow velocity 1. Sensitivity to solids or bubble contentranging from 1 to 25 ft/s, but the meter reading is greatly a. Interference with sound pulsesaffected by a change in the fluid composition 2. Sensitivity to flow disturbances Two types of ultrasonic flowmeters are in general use 3. Alignment of transducers is criticalfor closed pipe flow measurements. The first (time of flight 4. Clamp-on — pipe walls must freely pass ultra-or transit time) usually uses pulse transmission and is for sonic pulsesclean liquids, while the second (Doppler) usually usescontinuous wave transmission and is for dirty liquids. 5.13.3.3.2 Doppler Type Ultrasonic Flowmeters5.13.3.3.1 Time of Flight Ultrasonic Flowmeters26 Doppler ultrasonic flowmeters make use of the Doppler frequency shift caused by sound scattered or reflectedTime-of-flight flowmeters make use of the difference in from moving particles in the flow path. Doppler metersthe time for a sonic pulse to travel a fixed distance, first are not considered to be as accurate as time of flightin the direction of flow and then against the flow. This is flowmeters. However, they are very convenient to use andaccomplished by opposing transceivers positioned on generally more popular and less expensive than time ofdiagonal path across meter spool as shown in Figure 5.28. flight flowmeters.Each transmits and receives ultrasonic pulses with flow In operation, a propagated ultrasonic beam is inter-and against flow. The fluid velocity is directly proportional rupted by particles in moving fluid and reflected toward ato time difference of pulse travel. receiver. The difference of propagated and reflected fre- The time of flight ultrasonic flowmeter operates with quencies is directly proportional to fluid flow rate.minimal head loss, has an accuracy range of 1 to 2.5% full Ultrasonic Doppler flowmeters feature minimal headscale, and can be mounted as integral spool piece transduc- loss with an accuracy of 2% to 5% full scale. They areers or as externally mountable clamp-ons. They can mea- either of the integral spool piece transducer type or exter-sure flow accurately when properly installed and applied. nally mountable clamp-ons. The advantages of the Doppler ultrasonic flowmeter Transceiver includes: 1. No obstruction to flow Flow 2. Minimal head loss 3. Clamp-ons a. Can be portable b. No interruption of flow Transceiver 4. No moving parts 5. Linear over wide range FIGURE 5.28 Time-of-flight ultrasonic flowmeter. (From 6. Wide range of pipe sizes Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic 7. Low installation and operating costs Publ., Lancaster, PA, 2001.) 8. Bidirectional flow measurement © 2003 by CRC Press LLC
  • Transducer Transducer Transducer Full penetration Partial penetration Poor penetration FIGURE 5.29 Particle concentration effect. The more particles there are the more error. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) Register Shaft bearing Turbine wheel Flow Flow (A) (B) Bearing support arm FIGURE 5.30 (A) Propeller meter; (B) turbine meter. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) The disadvantages include: The advantages of the velocity meter include: 1. Minimum concentration and size of solids or 1. High accuracy bubbles for reliable operation required (see 2. Corrosion-resistant materials Figure 5.29) 3. Long-term stability 2. Minimum speed to maintain suspension 4. Liquid or gas operation required 5. Wide operating range 3. Clamp-on type limited to sonically conductive 6. Low pressure drop pipe 7. Wide temperature and pressure limits 8. High shock capability5.13.3.4 Velocity Flowmeters27 9. Wide variety of electronics availableVelocity or turbine flowmeters use a propeller or turbineto measure the velocity of the flow passing the device (see As shown in Figure 5.30, a turbine flowmeter consistsFigure 5.30). The velocity is then translated into a volu- of a rotor mounted on a bearing and shaft in a housing.metric amount by the meter register. Sizes exist from a The fluid to be measured is passed through the housing,variety of manufacturers to cover the flow range from causing the rotor to spin with a rotational speed propor-0.001 gal/min to over 25,000 gal/min for liquid service. tional to the velocity of the flowing fluid within the meter.End connections are available to meet the various piping A device to measure the speed of the rotor is employedsystems. The flowmeters are typically manufactured of to make the actual flow measurement. The sensor can bestainless steel but are also available in a wide variety of a mechanically gear-driven shaft to a meter or an elec-materials, including plastic. Velocity meters are applicable tronic sensor that detects the passage of each rotor bladeto all clean fluids. Velocity meters are particularly well generating a pulse. The rotational speed of the sensor shaftsuited for measuring intermediate flow rates on clean and the frequency of the pulse are proportional to thewater. volumetric flow rate through the meter. © 2003 by CRC Press LLC
  • Piston Sliding vane Oval Tri-rotor Bi-rotor Disc FIGURE 5.31 Six common positive displacement meter principles. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)5.13.3.5 Positive-Displacement Flowmeters28 cision metering element and is made up of the measuring chamber and the displacement mechanism. The most com-Positive-displacement flowmeters are most commonly used mon displacement mechanisms include the oscillating pis-for customer metering; they have long been used to measure ton, sliding vane, oval gear, tri-rotor, bi-rotor, and nutatingliquid products. These meters are very reliable and accurate disc types (see Figure 5.31). The counter drivetrain is usedfor low flow rates because they measure the exact quantity to transmit the internal motion of the measuring unit intoof water passing through them. Positive-displacement flow- a usable output signal. Many positive-displacement flow-meters are frequently used for measuring small flows in a meters use a mechanical gear train that requires a rotarytreatment plant because of their accuracy. Repair or replace- shaft seal or packing gland where the shaft penetrates thement is easy since they are so common in the distribution external housing.system. The positive-displacement flowmeter can offer excellent In essence, a positive-displacement flowmeter is a accuracy, repeatability, and reliability in many applications.hydraulic motor with high volumetric efficiency that It has satisfied many needs in the past and should play aabsorbs a small amount of energy from the flowing stream. vital role in serving the future needs as required.This energy is used to overcome internal friction in drivingthe flowmeter and its accessories and is reflected as a 5.13.4 OPEN-CHANNEL FLOW MEASUREMENT29pressure drop across the flowmeter. Pressure drop isregarded as unavoidable but must be minimized. It is the The majority of industrial liquid flows are carried in closedpressure drop across the internals of a positive displace- conduits that flow completely full and under pressure. How-ment flowmeter that actually creates a hydraulically unbal- ever, this is not the case for high volume flows of liquids inanced rotor, which causes rotation. waterworks, sanitary, and stormwater systems that are com- A positive-displacement flowmeter continuously monly carried in open channels. Low system heads and highdivides the flowing stream into known volumetric seg- volumetric flow rates characterize flow in open-channels.ments, isolates the segments momentarily, and returns The most commonly used method of measuring thethem to the flowing stream while counting the number of rate of flow in open-channel flow configurations is that ofdisplacements. A positive-displacement flowmeter can be hydraulic structures. In this method, flow in an open chan-broken down into three basic components: the external nel is measured by inserting a hydraulic structure into thehousing, the measuring unit, and the counter drive train. channel, which changes the level of liquid in or near the The external housing is the pressure vessel that contains structure. By selecting the shape and dimensions of thethe product being measured. The measuring unit is a pre- hydraulic structure, the rate of flow through or over the © 2003 by CRC Press LLC
  • Draw down Head Nappe Weir crest Air Flow Flow Stream or channel bed FIGURE 5.32 Side view of a weir. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)restriction will be related to the liquid level in a known dam or obstruction placed in the channel so that watermanner. Thus, the flow rate through the open channel can backs up behind it and then flows over it. The sharp crestbe derived from a single measurement of the liquid level or edge allows the water to spring clear of the weir platein or near the structure. and to fall freely in the form of a nappe. The hydraulic structures used in measuring flow in As Nathanson30 points out, when the nappe dischargesopen channels are known as primary measuring devices. freely into the air, there is a hydraulic relationship betweenThey may be divided into two broad categories, weirs and the height and depth of water flowing over the weir crestflumes and are covered in the following subsections. and the flow rate. This height, the vertical distance between the crest and the water surface, is called the head5.13.4.1 Weirs on the weir; it can be measured directly with a meter or yardstick or automatically by float-operated recordingThe weir is a widely used device to measure open-channel devices. Two common weirs, rectangular and triangular,flow. As can be seen in Figure 5.32, a weir is simply a are shown in Figure 5.33. (A) Rectangle weir Head Weir crest Crest length (B) Triangular weir V-notch angle Head Weir crest FIGURE 5.33 (A) Rectangular weir; (B) triangular V-notch weir. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Rectangular weirs are commonly used for large flows Q = 2.5 ¥ h2.5(see Figure 5.33A). The formula used to make rectangular 2.67 = 2.5 ¥ h2.5weir computations is: h = 1.03 (calculator: 1.06 INV yx2.5 = 1.026 or 1.03) 0.27 ft (original depth) + 1.03 (head on weir) = 1.3 ft Q = 3.33 ¥ L ¥ h1.5 (5.26) It is important to point out that weirs, aside from beingwhere operated within their flow limits, must also be operated Q = flow within the available system head. In addition, the opera- L = width of weir tion of the weir is sensitive to the approach velocity of h = head on weir (measured from edge of weir the water, often necessitating a stilling basin or pound in contact with the water, up to the water upstream of the weir. Weirs are not suitable for water that surface). carries excessive solid materials or silt, which deposit in the approach channel behind the weir and destroy the EXAMPLE 5.17 conditions required for accurate discharge measurements. Problem: Note: Accurate flow rate measurements with a weir cannot be expected unless the proper conditions A 4-ft high weir extends 15 ft across a rectangular channel and dimensions are maintained. in which there are 80 ft3/sec flowing. What is the depth just upstream from the weir? 5.13.4.2 Flumes Solution: A flume is a specially shaped constricted section in an Q = 3.33 ¥ L ¥ h1.5 open channel (similar to the venturi tube in a pressure 80 = 3.33 ¥ 15 h1.5 conduit). The special shape of the flume (see Figure 5.34) h = 1.4 ft (w/calculator: 1.6 INV yx1.5 = 1.36 or 1.4) restricts the channel area and/or changes the channel 4 ft height of weir + 1.4 ft head of water = 5.4 ft depth slope, resulting in an increased velocity and a change in the level of the liquid flowing through the flume. The flume Triangular weirs, also called V-notch weirs, can have restricts the flow, and then expands it in a definite fashion.notch angles ranging from 22.5∞ to 90∞, but right angle The flow rate through the flume may be determined bynotches are the most common (see Figure 5.33B). measuring the head on the flume at a single point, usually The formula used to make V-notch (90∞) weir calcu- at some distance downstream from the inlet.lations is Flumes can be categorized as belonging to one of three general families, depending upon the state of flow induced Q = 2.5 ¥ h2.5 — subcritical, critical, or supercritical. Typically, flumes that induce a critical or supercritical state of flow are mostwhere commonly used. This is because when critical or super- Q = flow critical flow occurs in a channel, one head measurement h = head on weir (measured from bottom of notch can indicate the discharge rate if it is made far enough to water surface) upstream so that the flow depth is not affected by the drawdown of the water surface as it achieves or passes EXAMPLE 5.18 through a critical state of flow. For critical or supercritical states of flow, a definitive head-discharge relationship can Problem: be established and measured based on a single head read- What should be the minimum weir height for measuring ing. Thus, most commonly encountered flumes are a flow of 1200 gal/min with a 90° V-notch weir, if the designed to pass the flow from subcritical through critical flow is mow moving at 4 ft/sec in a 2.5 ft wide rectangular or near the point of measurement. channel? The most common flume used for a permanent waste- water flowmetering installation is called the Parshall Solution: flume, shown in Figure 5.34. 1200 gpm Formulas for flow through Parshall flumes differ, = 2.67 ft 3 sec dependent on throat width. The formula below can be used 60 sec min ¥ 7.48 gal ft 3 for widths of 1 to 8 feet, and applies to a medium range Q=A¥V of flows. 2.67 = 2.5 ¥ d ¥ 4 d = 0.27 ft Q = 4 ¥ W ¥ Ha1.52 ¥ W .026 (5.27) © 2003 by CRC Press LLC
  • Converging inlet Diverging outlet Throat Flow Top view Stilling well for measuring hear FIGURE 5.34 Parshall flume. (From Spellman, F.R. and Drinan, J., Water Hydraulics, Technomic Publ., Lancaster, PA, 2001.)where 5.10. The pressure on a surface is 35 psi gauge. If Q = flow the surface area is 1.6 ft2, what is the force Ha = depth in stilling well upstream (lb) exerted on the surface? W = width of throat 5.11. Bernoulli’s principle states that the total energy of a hydraulic fluid is __________ _________.Note: Parshall flumes are low maintenance items. 5.12. What is pressure head? 5.13. What is a hydraulic grade line?5.14 CHAPTER REVIEW QUESTIONS 5.14. A flow of 1500 gal/min takes place in a 12-in. AND PROBLEMS pipe. Calculate the velocity head. 5.1. What should be the minimum weir height for 5.15. Water flows at 5.00 mL/sec in a 4-in. line measuring a flow of 900 gal/min with a 90° under a pressure of 110 psi. What is the V-notch weir, if the flow is now moving at pressure head (ft H2O)? 3 ft/sec in a 2-ft wide rectangular channel? 5.16. In Question 5.15, what is the velocity head 5.2. A 90° V-notch weir is to be installed in a in the line? 30-in. diameter sewer to measure 600 5.17. What is velocity head in a 6-in. pipe con- gal/min. What head should be expected? nected to a 1-ft pipe, if the flow in the larger 5.3. For dirty water operations, a ___________ pipe is 1.46 ft3/sec? or _________ orifice plate should be used. 5.18. What is velocity head? 5.4. A _________ has a smooth entry and sharp 5.19. What is suction lift? exit. 5.20. Explain energy grade line. 5.5. A _____________ sends a beam of ultra- sonic sound waves through the water to another transducer on the opposite side of REFERENCES the unit. 5.6. Find the number of gallons in a storage tank 1. Spellman, F.R. and Drinan, J., Water Hydraulics, CRC that has a volume of 660 ft.3 Press, Boca Raton, FL (originally published by Tech- nomic Publishing, Lancaster, PA). 2001, p. 5. 5.7. Suppose a rock weighs 160 lb in air and 125 lb 2. Magnusson, R.J., Water Technology in the Middle Ages, underwater. What is the specific gravity? The John Hopkins University Press, Baltimore, MD, 5.8. A 110-ft diameter cylindrical tank contains 2001, p. xi. 1.6 MG H2O. What is the water depth? 3. Magnusson, R.J., Water Technology in the Middle Ages, 5.9. The pressure in a pipeline is 6400 psf. What The John Hopkins University Press, Baltimore, MD, is the head on the pipe? 2001, p. 36. © 2003 by CRC Press LLC
  • 4. Nathanson, J.A., Basic Environmental Technology: 18. Adapted from McGhee, T.J., Water Supply and Sewer- Water Supply, Waste Management, and Pollution Con- age, 2nd ed., McGraw-Hill, New York, 1991, p. 45. trol, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 19. Nathanson, J.A., Basic Environmental Technology: 1997, pp. 21–22. Water Supply, Waste Management, and Pollution Con- 5. This section adapted from information contained in trol, 2nd ed., Prentice Hall, Upper Saddle River, NJ, Hauser, B.A., Hydraulics for Operators, Lewis Publish- 1997, p.34. ers, Boca Raton, FL, 1993, pp. 16–18; Basic Science 20. Adapted from Hauser, B.A., Practical Hydraulics Hand- Concepts and Applications: Principles and practices of book, 2nd ed., Lewis Publishers, Boca Raton, FL, 1996, Water Supply Operations, 2nd ed., American Water p. 91. Works Association, Denver, 1995, pp. 351–353. 21. From Water Treatment: Principles and Practices of 6. Holman, S., A Stolen Tongue, Anchor Press, Doubleday, Water Supply Operations, 2nd ed., American Water New York, 1998, p. 245. Works Association, Denver, 1995, pp. 449–450. 7. Spellman, F.R., The Science of Water: Concepts & 22. Kawamura, S., Integrated Design and Operation of Applications, Technomic Publ., Lancaster, PA, 1998, pp. Water Treatment Facilities, 2nd ed., John Wiley & Sons, 92–93. New York, 2000. 8. Adapted from Nathanson, J.A., Basic Environmental 23. Adapted from Husain, Z.D. and Sergesketter, M.J., Dif- Technology: Water Supply, Waste Management, and Pol- ferential pressure flow meters, in Flow Measurement, lution Control, 2nd ed., Prentice Hall, Upper Saddle Spitzer, D.W., Ed., Instrument Society of America, River, NJ, 1997, pp. 29–30. Research Triangle Park, NC, 1991, pp. 119–160. 9. Krutzsch, W.C., Introduction and classification of 24. Adapted from Flow Instrumentation: A Practical Work- pumps, in Pump Handbook, Karassik, I.J. et al., shop on Making Them Work, The Water & Wastewater McGraw-Hill, Inc., New York, 1976, p. 1-1. Instrumentation Testing Association and United States 10. Adapted from Cheremisinoff, N.P. and Cheremisinoff, Environmental Protection Agency, Section A, Sacra- P.N., Pumps/Compressors/Fans: Pocket Handbook, mento, CA, May 16–17, 1991. Technomic Publ., Lancaster, PA, 1989, p. 3. 25. Mills, R.C., Magnetic flow meters, in Flow Measurement, 11. Adapted from Water Transmission and Distribution, 2nd Spitzer, D.W., Ed., Instrument Society of America, ed., American Water Works Association, Denver, 1996, Research Triangle Park, NC, 1991, pp. 175–219. p. 358. 26. Adapted from Brown, A.E., Ultrasonic flow meters, in 12. Adapted from Arasmith, S., Introduction to Small Water Flow Measurement, Spitzer, D.W., Ed., Instrument Soci- Systems, ACR Publications, Inc., Albany, OR, 1993, pp. ety of America, Research Triangle Park, NC, 1991, pp. 59–61. 415–432. 13. A more complete listing of hydraulic terms may be 27. Adapted from Oliver, P.D., Turbine flow meters, in Flow found in Lindeburg, M.R., Civil Engineering Reference Measurement, Spitzer, D.W., Ed., Instrument Society of Manual, 4th ed., Professional Publications, Inc., San America, Research Triangle Park, NC, 1991, pp. Carlos, CA, 1986, pp. 5-2–5-3. 373–414. 14. Metcalf & Eddy, Wastewater Engineering: Collection 28. Barnes, R.G., Positive displacement flow meters for liq- and Pumping of Wastewater, McGraw-Hill, New York, uid measurement, in Flow Measurement, Spitzer, D.W., 1981, p. 11. Ed., Instrument Society of America, Research Triangle 15. Adapted from Lindeburg, M.R., Civil Engineering Ref- Park, NC, 1991, pp. 315–322. erence Manual, 4th ed., Professional Publications, Inc., 29. Adapted from Grant, D.M., Open channel flow measure- San Carlos, CA, 1986, p. 3-20. ment, in Flow Measurement, Spitzer, D.W., Ed., Instru- 16. Lindeburg, M.R., Civil Engineering Reference Manual, ment Society of America, Research Triangle Park, NC, 4th ed., Professional Publications, Inc., San Carlos, CA, 1991, pp. 252–290. 1986, p. 3-26. 30. Nathanson, J.A., Basic Environmental Technology: 17. Viessman, W., Jr., and Hammer, M.J., Water Supply and Water Supply, Waste Management, and Pollution Con- Pollution Control, 6th ed., Addison-Wesley, Menlo Park, trol, 2nd ed., Prentice Hall, Upper Saddle River, NJ, CA, 1998, p. 119. 1997, p. 39.© 2003 by CRC Press LLC
  • Fundamentals of Electricity 6 We believe that electricity exists, because the electric Recognizing electrical equipment is easy because we company keeps sending us bills for it, but we cannot figure use so much of it. If we ask typical operators where such out how it travels inside wires. equipment is located in their plant site, they know, because they probably operate these devices or their ancillaries. If Dave Barry we asked these same operators what a particular electrical device does, they could probably tell us. If we were to ask When Gladstone was British Prime Minister he visited if their plant electrical equipment was important to plant Michael Faraday’s laboratory and asked if some esoteric operations, the chorus would resound, “absolutely.” substance called “Electricity” would ever have practical significance. There is another question that does not always result in such a resounding note of assurance. If we asked these “One day, sir, you will tax it.” was his answer.* same operators to explain to us in very basic terms how electricity works to make their plant equipment operate,*Quoted in Science, 1994. As Michael Saunders points out, this cannot the answers we probably would receive would be varied,be correct because Faraday died in 1867 and Gladstone became prime jumbled, disjointed, and probably not all that accurate.minister in 1868. A more plausible prime minister would be Peel as Even on a more basic level, how many operators would beelectricity was discovered in 1831. Equally well it may be an urban able to accurately answer the question, what is electricity?legend. Probably very few operators would be able to answer this. Why do so many operators in both water and waste-6.1 ELECTRICITY: WHAT IS IT? water know so little about electricity? Part of the answer resides in the fact that operators are expected to know soWater and wastewater operators generally have little dif- much (and they are — and do), but are given so littleficulty in recognizing electrical equipment. Electrical opportunity to be properly trained.equipment is everywhere and is easy to spot. For example,typical plant sites are outfitted with electrical equipment We all know that experience is the great trainer. Asthat an example, let us look at what an operator assigned to change the bearings on a 5-hp 3-phase motor would need 1. Generates electricity (a generator — or emer- to know to accomplish this task. (Note: Remember, it is gency generator) not uncommon for water and wastewater operators to 2. Stores electricity (batteries) maintain as well as operate plant equipment.) The operator would have to know: 3. Changes electricity from one form to another (transformers) 4. Transports or transmits and distributes electric- 1. How to deenergize the equipment (i.e., proper ity throughout the plant site (wiring distribution lockout or tagout procedures) systems) 2. Once deenergized, how to properly disassemble 5. Measure electricity (meters) the motor coupling from the device it operates 6. Converts electricity into other forms of energy (e.g., a motor coupling from a pump shaft) and (rotating shafts — mechanical energy, heat the proper tools to use energy, light energy, chemical energy, or radio 3. Once uncoupled, how to know how to properly energy) disassemble the motor end-bells (preferably 7. Protects other electrical equipment (fuses, circuit without damaging the rotor shaft) breakers, or relays) 4. Once disassembled, how to recognize if the 8. Operates and controls other electrical equip- bearings are really in need of replacement ment (motor controllers) (though once removed from the end-bells, the 9. Converts some condition or occurrence into an bearings are typically replaced) electric signal (sensors) 10. Converts some measured variable to a representa- Questions the operator would need answered include tive electrical signal (transducers or transmitters). the following: © 2003 by CRC Press LLC
  • 1. If the bearings are in need of replacement, how mysterious forces of attraction and repulsion exhibited by are they to be removed without causing damage amber when it was rubbed with a cloth. They did not to the rotor shaft? understand the nature of this force. They could not answer 2. Once removed, what bearings should be used the question, “What is electricity?” The fact is this ques- to replace the old bearings? tion still remains unanswered. Today, we often attempt to 3. When the proper bearings are identified and answer this question by describing the effect and not the obtained, how are they to be installed properly? force. That is, the standard answer given is, “the force that 4. When the bearings are replaced properly, how moves electrons” is electricity; this is about the same as is the motor to be reassembled properly? defining a sail as “that force that moves a sailboat.” 5. Once the motor is correctly put back together, At the present time, little more is known than the how is it properly aligned to the pump and then ancient Greeks knew about the fundamental nature of reconnected? electricity, but we have made tremendous strides in har- 6.. What is the test procedure to ensure that the nessing and using it. As with many other unknown (or motor has been restored properly to full oper- unexplainable) phenomena, elaborate theories concerning ational status? the nature and behavior of electricity have been advanced and have gained wide acceptance because of their apparent Every one of the steps and questions on the above truth — and because they work.procedures is important — errors at any point in the pro- Scientists have determined that electricity seems tocedure could cause damage (maybe more damage than behave in a constant and predictable manner in givenoccurred in the first place). Another question is, does the situations or when subjected to given conditions. Scien-operator really need to know electricity to perform the tists, such as Michael Faraday, George Ohm, Fredericksequence of tasks described above? Lenz, and Gustav Kirchhoff, have described the predict- The short answer is no, not exactly. Fully competent able characteristics of electricity and electric current inoperators (who received most of their training via on-the- the form of certain rules. These rules are often referred tojob experience) are usually qualified to perform the bear- as laws. Though electricity itself has never been clearlying-change-out activity on most plant motors with little defined, its predictable nature and form of energy hasdifficulty. made it one of the most widely used power sources in The long answer is yes. Consider the motor mechanic modern times.who tunes your automobile engine. Then ask yourself, is The bottom line on what you need to learn aboutit important that the auto mechanic have some understand- electricity can be summed up as follows: anyone can learning of internal combustion engines? We think it is important. about electricity by learning the rules or laws applying toYou probably do, too. We also think it is important for the the behavior of electricity; and by understanding the meth-water or wastewater operator who changes bearings on an ods of producing, controlling, and using it. Thus, thiselectrical motor to have some understanding of how the learning can be accomplished without ever having deter-electric motor operates. mined its fundamental identity. Here is another issue to look at. Have you ever taken You are probably scratching your head — puzzled.an operator’s state licensure examination? If you have, I understand the main question running through thethen you know that, typically, these examinations test the reader’s brain cells at this exact moment: “This is a chapterexaminee’s knowledge of basic electricity. (Note: This is about basic electricity and the author cannot even explainespecially the case for water operators.) Therefore, some what electricity is?”states certainly consider operator knowledge of electricity That is correct; we cannot. The point is no one canimportant. definitively define electricity. Electricity is one of those For reasons of licensure and of job competence, subject areas where the old saying, “we don’t know whatwater/wastewater operators should have some basic elec- we don’t know about it,” fits perfectly.trical knowledge. How and where can operators quickly Again, there are a few theories about electricity thatand easily learn this important information? have so far stood the test of extensive analysis and much In this chapter, we provide the how and the where — time (relatively speaking, of course). One of the oldest andhere and now. most generally accepted theories concerning electric cur- rent flow (or electricity), is known as the electron theory.6.2 NATURE OF ELECTRICITY The electron theory states that electricity or current flow is the result of the flow of free electrons in a con-The word electricity is derived from the Greek word elec- ductor. Thus, electricity is the flow of free electrons ortron (meaning amber). Amber is a translucent (semitrans- simply electron flow. In addition, this is how we defineparent) yellowish fossilized mineral resin. The ancient electricity in this text —electricity is the flow of freeGreeks used the words electric force in referring to the electrons. © 2003 by CRC Press LLC
  • Electrons are extremely tiny particles of matter. To Electrongain understanding of electrons and exactly what is meantby electron flow, it is necessary to briefly discuss thestructure of matter. Proton6.3 THE STRUCTURE OF MATTERMatter is anything that has mass and occupies space. Tostudy the fundamental structure or composition of any FIGURE 6.2 One proton and one electron = electricallytype of matter, it must be reduced to its fundamental neutral. (From Spellman, F.R. and Drinan, J., Electricity,components. All matter is made of molecules, or combi- Technomic Publ., Lancaster, PA, 2001.)nations of atoms (Greek: not able to be divided), that arebound together to produce a given substance, such as salt, Most of the weight of the atom is in the protons andglass, or water. For example, if we divide water into neutrons of the nucleus. Whirling around the nucleus issmaller and smaller drops, we would eventually arrive at one or more negatively charged electrons. Normally, therethe smallest particle that was still water. That particle is is one proton for each electron in the entire atom, so thatthe molecule, which is defined as the smallest bit of a the net positive charge of the nucleus is balanced by thesubstance that retains the characteristics of that substance. net negative charge of the electrons rotating around the nucleus (see Figure 6.2).Note: Molecules are made up of atoms, which are bound together to produce a given substance. Note: Most batteries are marked with the symbols + Atoms are composed, in various combinations, of sub- and – or even with the abbreviations POS (pos-atomic particles of electrons, protons, and neutrons. These itive) and NEG (negative). The concept of aparticles differ in weight (a proton is much heavier than positive or negative polarity and its importancethe electron) and charge. We are not concerned with the in electricity will become clear later. However,weights of particles in this text, but the charge is extremely for the moment, we need to remember that animportant in electricity. The electron is the fundamental electron has a negative charge and that a protonnegative charge (–) of electricity. Electrons revolve about has a positive charge.the nucleus or center of the atom in paths of concentric We stated earlier that in an atom the number of protonsorbits, or shells (see Figure 6.1). The proton is the funda- is usually the same as the number of electrons. This is anmental positive (+) charge of electricity. Protons are found important point because this relationship determines thein the nucleus. The number of protons within the nucleus kind of element (the atom is the smallest particle thatof any particular atom specifies the atomic number of that makes up an element; an element retains its characteristicsatom. For example, the helium atom has 2 protons in itsnucleus so the atomic number is 2. The neutron, which is when subdivided into atoms) in question. Figure 6.3 showsthe fundamental neutral charge of electricity, is also found a simplified drawing of several atoms of different materi-in the nucleus. als based on the conception of electrons orbiting about the nucleus. For example, hydrogen has a nucleus consisting of one proton, around which rotates one electron. The helium atom has a nucleus containing two protons and two neutrons, with two electrons encircling the nucleus. Electrons Nucleus Both of these elements are electrically neutral (or bal- anced) because each has an equal number of electrons and protons. Since the negative (–) charge of each electron is equal in magnitude to the positive (+) charge of each proton, the two opposite charges cancel. A balanced (neutral or stable) atom has a certain amount of energy that is equal to the sum of the energies of its electrons. Electrons, in turn, have different energies called energy levels. The energy level of an electron is proportional to its distance from the nucleus. Therefore, FIGURE 6.1 Electrons and nucleus of an atom. (From Spell- the energy levels of electrons in shells further from the man, F.R. and Drinan, J., Electricity, Technomic Publ., Lan- nucleus are higher than that of electrons in shells nearer caster, PA, 2001.) the nucleus. © 2003 by CRC Press LLC
  • electron Orbit Nucleus (2 Protons) 2P (2 Neutrons) 1P 2N Nucleus (1 Proton) Hydrogen Helium 8P 9P 10N 8N Oxygen Fluorine FIGURE 6.3 Atomic structure of elements. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) Electrons Force Current (Voltage) Flow FIGURE 6.4 Electron flow in a copper wire. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) When an electric force is applied to a conducting excess energy absorbed by an atom may become sufficientmedium, such as copper wire, electrons in the outer orbits to cause loosely bound outer electrons (as shown inof the copper atoms are forced out of orbit (i.e., liberating Figure 6.4) to leave the atom against the force that acts toor freeing electrons) and are impelled along the wire. This hold them within.electrical force, which forces electrons out of orbit, can Note: An atom that has lost or gained one or morebe produced in a number of ways, such as by moving a electrons is said to be ionized. If the atom losesconductor through a magnetic field; by friction, as when electrons it becomes positively charged and isa glass rod is rubbed with cloth (silk); or by chemical referred to as a positive ion. Conversely, if theaction, as in a battery. atom gains electrons, it becomes negatively When the electrons are forced from their orbits, they charged and is referred to as a negative ion.are called free electrons. Some of the electrons of certainmetallic atoms are so loosely bound to the nucleus thatthey are relatively free to move from atom to atom. These 6.4 CONDUCTORS, SEMICONDUCTORS,free electrons constitute the flow of an electric current in AND INSULATORSelectrical conductors. Electric current moves easily through some materials, butNote: When an electric force is applied to a copper with greater difficulty through others. Substances that per- wire, free electrons are displaced from the cop- mit the free movement of a large number of electrons are per atoms and move along the wire, producing called conductors. The most widely used electrical con- electric current as shown in Figure 6.4. ductor is copper because of its high conductivity (how If the internal energy of an atom is raised above its good a conductor the material is) and cost-effectiveness.normal state, the atom is said to be excited. Excitation Electrical energy is transferred through a copper ormay be produced by causing the atoms to collide with other metal conductor by means of the movement of freeparticles that are impelled by an electric force as shown electrons that migrate from atom to atom inside the con-in Figure 6.4. In effect, what occurs is that energy is ductor (see Figure 6.4). Each electron moves a very shorttransferred from the electric source to the atom. The distance to the neighboring atom where it replaces one or © 2003 by CRC Press LLC
  • TABLE 6.1 TABLE 6.2 Electrical Conductors Common Insulators Silver Brass Rubber Plastics Copper Iron Mica Glass Gold Tin Wax or paraffin Fiberglass Aluminum Mercury Porcelain Dry wood Zinc Bakelite Air Source: From Spellman, F.R. and Source: From Spellman, F.R. and Drinan, J., Electricity, Technomic Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001. Publ., Lancaster, PA, 2001.more electrons by forcing them out of their orbits. The raised, however, a limited number of electrons becomereplaced electrons repeat the process in other nearby atoms available for conduction.until the movement is transmitted throughout the entirelength of the conductor. A good conductor is said to havea low opposition, or resistance, to the electron (current) 6.5 STATIC ELECTRICITYflow. Electricity at rest is often referred to as static electricity.Note: If lots of electrons flow through a material with More specifically, when two bodies of matter have unequal only a small force (voltage) applied, we call charges, and are near one another, an electric force is that material a conductor. exerted between them because of their unequal charges. Because they are not in contact, their charges cannot Table 6.1 lists many of the metals commonly used as equalize. The existence of such an electric force whereelectric conductors. The best conductors appear at the top current cannot flow is static electricity.of the list, with the poorer ones shown last. Static, or electricity at rest, will flow if given theNote: The movement of each electron (e.g., in copper opportunity. An example of this phenomenon is often wire) takes a very small amount of time, almost experienced when one walks across a dry carpet and then instantly. This is an important point to keep in touches a doorknob; a slight shock is usually felt and a mind later in the book, when events in an elec- spark at the fingertips is likely noticed. In the workplace, trical circuit seem to occur simultaneously. static electricity is prevented from building up by properly bonding equipment to ground or earth. While it is true that electron motion is known to existto some extent in all matter, some substances, such as 6.5.1 CHARGED BODIESrubber, glass, and dry wood have very few free electrons.In these materials, large amounts of energy must be To fully grasp the understanding of static electricity, it isexpended in order to break the electrons loose from the necessary to know one of the fundamental laws of elec-influence of the nucleus. Substances containing very few tricity and its significance.free electrons are called insulators. Insulators are impor- The fundamental law of charged bodies states that liketant in electrical work because they prevent the current charges repel each other and unlike charges attract eachfrom being diverted from the wires. other. A positive charge and negative charge, being oppositeNote: If the voltage is large enough, even the best or unlike, tend to move toward each other, attracting each insulators will break down and allow their elec- other. In contrast, like bodies tend to repel each other. trons to flow. Electrons repel each other because of their like negative Table 6.2 lists some materials that we often use as charges, and protons repel each other because of their likeinsulators in electrical circuits. The list is in decreasing positive charges. Figure 6.5 demonstrates the law oforder of ability to withstand high voltages without con- charged bodies.ducting. It is important to point out another significant part of A material that is neither a good conductor nor a good the fundamental law of charged bodies — the force ofinsulator is called a semiconductor. Silicon and germa- attraction or repulsion existing between two magneticnium are substances that fall into this category. Because poles decreases rapidly as the poles are separated fromof their peculiar crystalline structure, these materials may each other. More specifically, the force of attraction orunder certain conditions act as conductors; under other repulsion varies directly as the product of the separate poleconditions they act as insulators. As the temperature is strengths and inversely as the square of the distance © 2003 by CRC Press LLC
  • Unlike charges attract Like charges repel (A) (B) (C) FIGURE 6.5 Reaction between two charged bodies. The opposite charge in (A) attracts. The like charges in (B) and (C) repel each other. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.)separating the magnetic poles, provided the poles are tric charge is Q, expressed in units of coulombssmall enough to be considered as points. (C). A charge of + 1 C means a body contains Let us look at an example. If we increased the distance a charge of 6.25 ¥ 1018. A charge of –1 C meansbetween 2 north poles from 2 to 4 ft, the force of repulsion a body contains a charge of 6.25 ¥ 1018 morebetween them is decreased to 1/4 of its original value. If electrons than protons.either pole strength is doubled, the distance remaining thesame, the force between the poles will be doubled. 6.5.3 ELECTROSTATIC FIELDS6.5.2 COULOMB’S LAW The fundamental characteristic of an electric charge is its ability to exert force. The space between and aroundSimply put, Coulomb’s law points out that the amount of charged bodies in which their influence is felt is called anattracting or repelling force that acts between two electri- electric field of force. The electric field is always termi-cally charged bodies in free space depends on two things: nated on material objects and extends between positive and negative charges. This region of force can consist of 1. Their charges air, glass, paper, or a vacuum, and is referred to as an 2. The distance between them electrostatic field. When two objects of opposite polarity are brought Specifically, Coulomb’s law states, “Charged bodies near each other, the electrostatic field is concentrated inattract or repel each other with a force that is directly the area between them. Lines that are referred to as elec-proportional to the product of their charges, and is trostatic lines of force generally represent the field. Theseinversely proportional to the square of the distance lines are imaginary and are used merely to represent thebetween them.” direction and strength of the field. To avoid confusion, theNote: The magnitude of electric charge a body pos- positive lines of force are always shown leaving charge, sesses is determined by the number of electrons and for a negative charge, they are shown as entering. compared with the number of protons within Figure 6.6 illustrates the use of lines to represent the field the body. The symbol for the magnitude of elec- about charged bodies. (A) (B) FIGURE 6.6 Electrostatic lines of force: (A) represents the repulsion of like-charged bodies and their associated fields; (B) represents the attraction between unlike-charged bodies and their associated fields. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Note: A charged object will retain its charge tempo- rarily if there is no immediate transfer of elec- trons to or from it. In this condition, the charge is said to be at rest. Remember, electricity at N S rest is called static electricity.6.6 MAGNETISMMost electrical equipment depends directly or indirectly FIGURE 6.8 Magnetic field of force around a bar magnet,upon magnetism. Magnetism is defined as a phenomena indicated by lines of force. (From Spellman, F.R. and Drinan,associated with magnetic fields; it has the power to attract J., Electricity, Technomic Publ., Lancaster, PA, 2001.)such substances as iron, steel, nickel, or cobalt (metalsthat are known as magnetic materials). Correspondingly,a substance is said to be a magnet if it has the propertyof magnetism. For example, a piece of iron can be mag-netized and therefore is a magnet. When magnetized, the piece of iron (note: we willassume a piece of flat bar is 6 ¥ 1 ¥ 5 in.; a bar magnet —see Figure 6.7) will have two points opposite each other,which most readily attract other pieces of iron. The pointsof maximum attraction (one on each end) are called the S Nmagnetic poles of the magnet: the north (N) pole and thesouth (S) pole. Just as like electric charges repel each otherand opposite charges attract each other, like magneticpoles repel each other and unlike poles attract each other. N SAlthough invisible to the naked eye, its force can be shownto exist by sprinkling small iron filings on a glass covering FIGURE 6.9 Horseshoe magnet. (From Spellman, F.R. anda bar magnet as shown in Figure 6.7. Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) Figure 6.8 shows how the field looks without ironfilings; it is shown as lines of force (known as magnetic the magnetic flux. These circuits are similar toflux or flux lines; the symbol for magnetic flux is the Greeklowercase letter f [phi]) in the field, repelled away from the electric circuit (an important point), whichthe north pole of the magnet and attracted to its south pole. is a complete path through which current is caused to flow under the influence of an elec-Note: A magnetic circuit is a complete path through tromotive force. which magnetic lines of force may be estab- lished under the influence of a magnetizing There are three types or groups of magnets: force. Most magnetic circuits are composed largely of magnetic materials in order to contain 1. Natural magnets — These magnets are found in the natural state in the form of a mineral (an Glass sheet iron compound) called magnetite. 2. Permanent magnets (artificial magnet) — Iron filings These magnets are hardened steel or some alloy, N S such as Alnico bars, that has been permanently magnetized. The permanent magnet most peo- Magnet ple are familiar with is the horseshoe magnet (see Figure 6.9). 3. Electromagnets (artificial magnet) — These magnets are composed of soft-iron cores aroundFIGURE 6.7 Shows the magnetic field around a bar magnet. If which are wound coils of insulated wire. Whenthe glass sheet is tapped gently, the filings will move into adefinite pattern that describes the field of force around the an electric current flows through the coil, the coremagnet. (From Spellman, F.R. and Drinan, J., Electricity, Tech- becomes magnetized. When the current ceasesnomic Publ., Lancaster, PA, 2001.) to flow, the core loses most of the magnetism. © 2003 by CRC Press LLC
  • 6.6.1 MAGNETIC MATERIALS because weak magnetic materials can be important in some applications, present studies classify materials intoNatural magnets are no longer used (they have no practical one of three groups: paramagnetic, diamagnetic, and fer-value) in electrical circuitry because more powerful and romagnetic.more conveniently shaped permanent magnets can beproduced artificially. Commercial magnets are made from 1. Paramagnetic materials — These include alumi-special steels and alloys — magnetic materials. num, platinum, manganese, and chromium — Magnetic materials are those materials that are materials that become only slightly magnetizedattracted or repelled by a magnet and that can be magne- even though they are under the influence of atized. Iron, steel, and alloy bar are the most common strong magnetic field. This slight magnetizationmagnetic materials. These materials can be magnetized by is in the same direction as the magnetizing field.inserting the material (in bar form) into a coil of insulated Relative permeability is slightly more than 1wire and passing a heavy direct current through the coil. (i.e., considered nonmagnetic materials).The same material may also be magnetized if it is stroked 2. Diamagnetic materials — These include bis-with a bar magnet. It will then have the same magnetic muth, antimony, copper, zinc, mercury, gold,property that the magnet used to induce the magnetism and silver — materials that can also be slightlyhas; there will be two poles of attraction, one at either magnetized when under the influence of a veryend. This process produces a permanent magnet by induc- strong field. Relative permeability is less thantion —the magnetism is induced in the bar by the influence 1 (i.e., considered nonmagnetic materials).of the stroking magnet. 3. Ferromagnetic materials — These include iron,Note: Permanent magnets are those of hard magnetic steel, nickel, cobalt, and commercial alloys — materials (hard steel or alloys) that retain their materials that are the most important group for magnetism when the magnetizing field is applications of electricity and electronics. Ferro- removed. A temporary magnet is one that has magnetic materials are easy to magnetize and no ability to retain a magnetized state when the have high permeability, ranging from 50 to 3000. magnetizing field is removed. 6.6.2 MAGNETIC EARTH Even though classified as permanent magnets, it isimportant to point out that hardened steel and certain The earth is a huge magnet, and surrounding earth is thealloys are relatively difficult to magnetize and are said to magnetic field produced by the earth’s magnetism. Mosthave a low permeability. This is because the magnetic lines people would have no problem understanding or at leastof force do not easily permeate, or distribute themselves, accepting this statement. If people were told that thereadily through the steel. earth’s north magnetic pole is actually its south magnetic pole and that the south magnetic pole is actually the earth’sNote: Permeability refers to the ability of a magnetic north magnetic pole, they might not accept or understand material to concentrate magnetic flux. Any this statement. However, in terms of a magnet, it is true. material that is easily magnetized has high per- meability. A measure of permeability for differ- As can be seen from Figure 6.10, the magnetic polar- ent materials in comparison with air or vacuum ities of the earth are indicated. The geographic poles are is called relative permeability, symbolized by m also shown at each end of the axis of rotation of the earth. or (mu). Clearly, as shown in Figure 6.10, the magnetic axis does not coincide with the geographic axis. Therefore, the mag- Once hard steel and other alloys are magnetized, they netic and geographic poles are not at the same place onretain a large part of their magnetic strength and are called the surface of the earth.permanent magnets. Conversely, materials that are relatively Recall that magnetic lines of force are assumed toeasy to magnetize, such as soft iron and annealed silicon emanate from the north pole of a magnet and to enter thesteel, are said to have a high permeability. Such materials south pole as closed loops. Because the earth is a magnet,retain only a small part of their magnetism after the magne- lines of force emanate from its north magnetic pole andtizing force is removed and are called temporary magnets. enter the south magnetic pole as closed loops. A compass The magnetism that remains in a temporary magnet needle aligns itself in such a way that the earth’s lines ofafter the magnetizing force is removed is called residual force enter at its south pole and leave at its north pole.magnetism. Because the north pole of the needle is defined as the end Early magnetic studies classified magnetic materials that points in a northerly direction, it follows that themerely as being magnetic and nonmagnetic, meaning magnetic pole near the north geographic pole is in realitybased on the strong magnetic properties of iron. However, a south magnetic pole and vice versa. © 2003 by CRC Press LLC
  • electrons (current) will flow along the conductor. This flow South Magnetic Pole is from the negatively charged body to the positively charged body until the two charges are equalized and the potential difference no longer exists. North Geographic Note: The basic unit of potential difference is the volt Pole (V). The symbol for potential difference is V, indicating the ability to do the work of forcing electrons (current flow) to move. Because the Magnetic Earth volt unit is used, potential difference is called voltage. 6.7.1 THE WATER ANALOGY In attempting to train individuals in the concepts of basic electricity, especially in regards to difference of potential (voltage), current, and resistance relationships in a simple electrical circuit, it has been common practice to use what is referred to as the water analogy. We use the water analogy later to explain (in a simple, straightforward fash- South Geographic Pole ion) voltage, current, and resistance and their relationships North Magnetic Pole in more detail. For now we use the analogy to explain the basic concept of electricity: difference of potential, or FIGURE 6.10 Earth’s magnetic poles. (From Spellman, F.R. voltage. Because a difference in potential causes current and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, flow (against resistance), it is important that this concept 2001.) be understood first before the concept of current flow and resistance are explained. Consider the water tanks shown in Figure 6.11 — two6.7 DIFFERENCE IN POTENTIAL water tanks connected by a pipe and valve. At first, theBecause of the force of its electrostatic field, an electric valve is closed and all the water is in Tank A. Thus, thecharge has the ability to do the work of moving another water pressure across the valve is at its maximum. Whencharge by attraction or repulsion. The force that causes the valve is opened, the water flows through the pipe fromfree electrons to move in a conductor as an electric current A to B until the water level becomes the same in bothmay be referred to as follows: tanks. The water then stops flowing in the pipe, because there is no longer a difference in water pressure (difference 1. Electromotive force (EMF) in potential) between the two tanks. 2. Voltage Just as the flow of water through the pipe in Figure 6.11 3. Difference in potential is directly proportional to the difference in water level in the two tanks, current flow through an electric circuit is When a difference in potential exists between two directly proportional to the difference in potential acrosscharged bodies that are connected by a wire (conductor), the circuit. Tank A Tank B FIGURE 6.11 Water analogy of electric difference of potential. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Important Point: A fundamental law of current elec- 6.8 CURRENT tricity is that the current is directly proportional The movement or the flow of electrons is called current. to the applied voltage; that is, if the voltage is To produce current, the electrons must be moved by a increased, the current is increased. If the voltage potential difference. is decreased, the current is decreased. Note: The terms current, current flow, electron flow, or electron current, etc., may be used to6.7.2 PRINCIPAL METHODS OF PRODUCING VOLTAGE describe the same phenomenon.There are many ways to produce electromotive force, or Electron flow, or current, in an electric circuit is fromvoltage. Some of these methods are much more widely a region of less negative potential to a region of moreused than others. The following is a list of the six most positive potential.common methods of producing electromotive force. Note: The letter I is the basic unit that represents current measured in amperes or amps (A). The 1. Friction — Voltage produced by rubbing two measurement of 1 A of current is defined as the materials together. movement of 1 C past any point of a conductor 2. Pressure (piezoelectricity) — Voltage produced during 1 sec of time. by squeezing crystals of certain substances. Earlier we used the water analogy to help us under- 3. Heat (thermoelectricity) — Voltage produced stand potential difference. We can also use the water anal- by heating the joint (junction) where two unlike ogy to help us understand current flow through a simple metals are joined. electric circuit. 4. Light (photoelectricity) — Voltage produced by Figure 6.12 shows a water tank connected via a pipe light striking photosensitive (light sensitive) to a pump with a discharge pipe. If the water tank contains substances. an amount of water above the level of the pipe opening to the pump, the water exerts pressure (a difference in 5. Chemical action — Voltage produced by chem- potential) against the pump. When sufficient water is avail- ical reaction in a battery cell. able for pumping with the pump, water flows through the 6. Magnetism — Voltage produced in a conductor pipe against the resistance of the pump and pipe. The when the conductor moves through a magnetic analogy should be clear — in an electric circuit, if a field, or a magnetic field moves through the difference of potential exists, current will flow in the circuit. conductor in such a manner as to cut the mag- Another simple way of looking at this analogy is to netic lines of force of the field. consider Figure 6.13 where the water tank has been replaced with a generator, the pipe with a conductor In the study of basic electricity, we are most concerned (wire), and water flow with the flow of electric current.with magnetism and chemistry as a means to produce Again, the key point illustrated by Figure 6.12 andvoltage. Friction has little practical applications, though Figure 6.13 is that to produce current, the electrons must be moved by a potential difference.we discussed it earlier in static electricity. Pressure, heat, Electric current is generally classified into two generaland light do have useful applications, but we do not need types:to consider them in this text. Magnetism and chemistry,on the other hand, are the principal sources of voltage and 1. Direct current (DC)are discussed at length in this text. 2. Alternating current (AC) Water Pump Water pipe (resistance) Tank Water flow FIGURE 6.12 Water analogy: current flow. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Wire (resistance) Generator (pump) Electron flow (current) FIGURE 6.13 Simple electric circuit with current flow. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) Direct current is current that moves through a conduc- water between the tanks. This friction is similar to elec-tor or circuit in one direction only. Alternating current trical resistance. The resistance of the pipe to the flow ofperiodically reverses direction. water through it depends upon 1. The length of the pipe6.9 RESISTANCE 2. Diameter of the pipeIn Section 6.4, we discussed conductors and insulators. 3. The nature of the inside walls (rough or smooth)We pointed out that free electrons, or electric current,could move easily through a good conductor, such as Similarly, the electrical resistance of the conductorscopper, but that an insulator, such as glass, was an obstacle depends uponto current flow. In the water analogy shown in Figure 6.12and the simple electric circuit shown in Figure 6.13, either 1. The length of the wiresthe pipe or the conductor indicates resistance. 2. The diameter of the wires Every material offers some resistance, or opposition, 3. The material of the wires (copper, silver, etc.)to the flow of electric current through it. Good conductors,such as copper, silver, and aluminum, offer very little It is important to note that temperature also affects theresistance. Poor conductors, or insulators, such as glass, resistance of electrical conductors to some extent. In mostwood, and paper, offer a high resistance to current flow. conductors (copper, aluminum, etc.) the resistance increases with temperature. Carbon is an exception. InNote: The amount of current that flows in a given carbon, the resistance decreases as temperature increases. circuit depend on two factors: voltage and resis- tance. Important Note: Electricity is a study that is frequently explained in terms of opposites. The term that isNote: The letter R represents resistance. The basic unit exactly the opposite of resistance is conductance. in which resistance is measured is the ohm (W). Conductance (G) is the ability of a material to The measurement of 1 W is the resistance of a pass electrons. The unit of conductance is the circuit element, or circuit, that permits a steady Mho, which is ohm spelled backwards. The rela- current of 1 ampere (1 C/sec) to flow when a tionship that exists between resistance and con- steady EMF of 1 V is applied to the circuit. ductance is the reciprocal. A reciprocal of a num- Manufactured circuit parts containing definite ber is obtained by dividing the number into one. amounts of resistance are called resistors. If the resistance of a material is known, dividing The size and type of material of the wires in an electric its value into one will give its conductance. Sim-circuit are chosen to keep the electrical resistance as low ilarly, if the conductance is known, dividing itsas possible. In this way, current can flow easily through value into one will give its resistance.the conductors, just as water flows through the pipebetween the tanks in Figure 6.11. If the water pressure 6.10 BATTERY-SUPPLIED ELECTRICITYremains constant, the flow of water in the pipe will dependon how far the valve is opened. The smaller the opening, Battery-supplied direct current electricity has many applica-the greater the opposition (resistance) to the flow, and the tions and is widely used in water and wastewater treatmentsmaller the rate of flow will be in gallons per second. operations. Applications include providing electrical In the simple electric circuit shown in Figure 6.13, the energy in plant vehicles and emergency diesel generators;larger the diameter of the wire, the lower will be its elec- material handling equipment (forklifts); portable electrictrical resistance (opposition) to the flow of current through or electronic equipment; backup emergency power forit. In the water analogy, pipe friction opposes the flow of light-packs, hazard warning signal lights, and flashlights; © 2003 by CRC Press LLC
  • tively charge atoms. When the plates of these materials are immersed in an electrolyte, chemical action between Electron Flow the two begins. In the dry cell (see Figure 6.15), the electrodes are the carbon rod in the center and the zinc container in which the cell is assembled. Zinc Carbon The electrolyte is the solution that acts upon the elec- trodes that are placed in it. The electrolyte may be a salt, an acid, or an alkaline solution. In the simple voltaic cell and in the automobile storage battery, the electrolyte is in a liquid form, while in the dry cell (see Figure 6.15) the electrolyte is a moist paste. Electrolyte 6.10.2 PRIMARY AND SECONDARY CELLS FIGURE 6.14 Simple voltaic cell. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) Primary cells are normally those that cannot be recharged or returned to good condition after their voltage drops tooand standby power supplies or uninterruptible power sup- low. Dry cells in flashlights and transistor radios are exam-plies for computer systems. In some instances, they are ples of primary cells. Some primary cells have been devel-used as the only source of power, while in others (as oped to the state where they can be recharged.mentioned above) they are used as a secondary or standby A secondary cell is one in which the electrodes andpower supply. the electrolyte are altered by the chemical action that takes place when the cell delivers current. These cells are6.10.1 THE VOLTAIC CELL rechargeable. During recharging, the chemicals that pro- vide electric energy are restored to their original condition.The simplest cell (a device that transforms chemical energy Recharging is accomplished by forcing an electric currentinto electrical energy) is known as a voltaic (or galvanic) through them in the opposite direction to that of discharge.cell (see Figure 6.14). It consists of a piece of carbon (C) Connecting as shown in Figure 6.16 recharges a cell.and a piece of zinc (Zn) suspended in a jar that contains a Some battery chargers have a voltmeter and an ammetersolution of water (H2O) and sulfuric acid (H2SO4). that indicate the charging voltage and current.Note: A simple cell consists of two strips, or elec- The automobile storage battery is the most common trodes, placed in a container that hold the elec- example of the secondary cell. trolyte. A battery is formed when two or more cells are connected. 6.10.3 BATTERY The electrodes are the conductors by which the current As was stated previously, a battery consists of two or moreleaves or returns to the electrolyte. In the simple cell cells placed in a common container. The cells are con-described above, they are carbon and zinc strips placed in nected in series, in parallel, or in some combination ofthe electrolyte. Zinc contains an abundance of negatively series and parallel, depending upon the amount of voltagecharged atoms, while carbon has an abundance of posi- and current required of the battery. Positive terminal Negative terminal Sealer Zinc container (negative electrode) Carbon rod (positive electrode) Wet paste electrolyte FIGURE 6.15 Dry cell (cross-sectional view). (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • Note: When current flows through a cell, the zinc gradually dissolves in the solution and the acid is neutralized. 6.10.3.2 Combining Cells Cell Battery (battery) charger In many operations, battery-powered devices may require more electrical energy than one cell can provide. Various devices may require either a higher voltage or more current, FIGURE 6.16 Hookup for charging a secondary cell with a and some cases both. Under such conditions, it is necessary battery charger. (From Spellman, F.R. and Drinan, J., Elec- to combine, or interconnect, a sufficient number of cells tricity, Technomic Publ., Lancaster, PA, 2001.) to meet the higher requirements. Cells connected in series provide a higher voltage, while cells connected in parallel provide a higher current capacity. To provide adequate6.10.3.1 Battery Operation power when both voltage and current requirements areThe chemical reaction within a battery provides the voltage. greater than the capacity of one cell, a combination series-This occurs when a conductor is connected externally to parallel network of cells must be interconnected. When cells are connected in series (see Figure 6.17),the electrodes of a cell, causing electrons to flow under the the total voltage across the battery of cells is equal to theinfluence of a difference in potential across the electrodes sum of the voltage of each of the individual cells. In Figurefrom the zinc (negative) through the external conductor to 6.17, the 4 1.5-V cells in series provide a total batterythe carbon (positive), returning within the solution to the voltage of 6 V. When cells are placed in series, the positivezinc. After a short period, the zinc will begin to waste away terminal of one cell is connected to the negative terminalbecause of the acid. of the other cell. The positive electrode of the first cell and The voltage across the electrodes depends upon the negative electrode of the last cell then serve as the powermaterials from which the electrodes are made and the takeoff terminals of the battery. The current flowing throughcomposition of the solution. The difference of potential such a battery of series cells is the same as from one cellbetween the carbon and zinc electrodes in a dilute solution because the same current flows through all the series cells.of sulfuric acid and water is about 1.5 V. To obtain a greater current, a battery has cells con- nected in parallel as shown in Figure 6.18. In this parallel The current that a primary cell may deliver depends connection, all the positive electrodes are connected toupon the resistance of the entire circuit, including that of one line, and all negative electrodes are connected to thethe cell. The internal resistance of the primary cell depends other. Any point on the positive side can serve as theupon the size of the electrodes, the distance between them positive terminal of the battery, and any point on thein the solution, and the resistance of the solution. The negative side can be the negative terminal.larger the electrodes and the closer together they are in The total voltage output of a battery of three parallelsolution (without touching), the lower the internal resis- cells is the same as that for a single cell (Figure 6.18), buttance of the primary cell and the more current it is capable the available current is three times that of one cell; thatof supplying to the load. is, the current capacity has been increased. 6 volts 1.5V 1.5V 1.5V 1.5V Cell 1 Cell 2 Cell 3 Cell 4 (schematic representation) 1.5V 1.5V 1.5V 1.5V FIGURE 6.17 Cells in series. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) © 2003 by CRC Press LLC
  • 1.5V 1.5V 1.5V 1.5V Cell 1 Cell 2 Cell 3 1.5V 1.5V 1.5V (schematic representation) FIGURE 6.18 Cells in parallel. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) FIGURE 6.19 Series-parallel connected cells. (From Spellman, F.R. and Drinan, J., Electricity, Technomic Publ., Lancaster, PA, 2001.) Identical cells in parallel all supply equal parts of the is a moist paste). The dry cell battery is one of the oldestcurrent to the load. For example, of 3 different parallel and most widely used commercial types of dry cell. Thecells producing a load current of 210 mA, each cell con- carbon, in the form of a rod that is placed in the center oftributes 70 mA. the cell, is the positive terminal. The case of the cell is Figure 6.19 depicts a schematic of a series-parallel made of zinc, which is the negative terminal (seebattery network supplying power to a load requiring both Figure 6.15). Between the carbon electrode and the zinca voltage and current greater than one cell can provide. case is the electrolyte of a moist chemical paste-like mix-To provide the required increased voltage, groups of three ture. The cell is sealed to prevent the liquid in the paste1.5-V cells are connected in series. To provide the required from evaporating. The voltage of a cell of this type isincreased amperage, four series groups are connected in about 1.5 V.parallel. 6.10.4.2 Lead-Acid Battery6.10.4 TYPES OF BATTERIES The lead-acid battery is a secondary cell, commonlyIn the past 25 years, several different types of batteries termed a storage battery, that stores chemical energy untilhave been developed. In this text, we briefly discuss five it is released as electrical energy.types: the dry cell, lead-acid battery, alkaline cell, nickel-cadmium, and mercury cell. Note: The lead-acid battery differs from the primary cell type battery mainly in that it may be6.10.4.1 Dry Cell recharged, whereas most primary cells are not normally recharged. In addition, the term stor-The dry cell, or carbon-zinc cell, is so known because its age battery is somewhat deceiving because thiselectrolyte is not in a liquid state (however, the electrolyte battery does not store electrical energy, but is a © 2003 by CRC Press LLC
  • source of chemical energy that produces elec- Mercury cells also produce a constant output voltage trical energy. under different load conditions. There are two different types of mercury cells. One is As the name implies, the lead-acid battery consists of a flat cell that is shaped like a button, while the other isa number of lead-acid cells immersed in a dilute solution a cylindrical cell that looks like a standard flashlight cell.of sulfuric acid. Each cell has two groups of lead plates; The advantage of the button-type cell is that several ofone set is the positive terminal and the other is the negative them can be stacked inside one container to form a battery.terminal. Active materials within the battery (lead plates A cell produces 1.35 V.and sulfuric acid electrolyte) react chemically to producea flow of direct current whenever current consumingdevices are connected to the battery terminal posts. This 6.10.4.6 Battery Characteristicscurrent is produced by the chemical reaction between the Batteries are generally classified by their various charac-active material of the plates (electrodes) and the electro- teristics. Parameters such as internal resistance, specificlyte (sulfuric acid). gravity, capacity, and shelf life are used to classify batter- This type of cell produces slightly more than 2 V. Most ies by type.automobile batteries contain 6 cells connected in series so Regarding internal resistance, it is important to keep inthat the output voltage from the battery is slightly more mind that a battery is a DC voltage generator. As such, thethan 12 V. battery has internal resistance. In a chemical cell, the resis- Besides being rechargeable, the main advantage of the tance of the electrolyte between the electrodes is responsiblelead-acid storage battery over the dry cell battery is that for most of the cell’s internal resistance. Because any cur-the storage battery can supply current for a much longer rent in the battery must flow through the internal resistance,time than the average dry cell. this resistance is in series with the generated voltage. WithSafety Note: Whenever a lead-acid storage battery is no current, the voltage drop across the resistance is zero so charging, the chemical action produces danger- that the full-generated voltage develops across the output ous hydrogen gas; thus, the charging operation terminals. This is the open-circuit voltage, or no-load volt- should only take place in a well-ventilated area. age. If a load resistance is connected across the battery, the load resistance is in series with internal resistance. When6.10.4.3 Alkaline Cell current flows in this circuit, the internal voltage drop decreases the terminal voltage of the battery.The alkaline cell is a secondary cell that gets its name The ratio of the weight of a certain volume of liquidfrom its alkaline electrolyte — potassium hydroxide. to the weight of the same volume of water is called theAnother type battery, sometimes called the alkaline bat- specific gravity of the liquid. Pure sulfuric acid has atery, has a negative electrode of zinc and a positive elec- specific gravity of 1.835 since it weighs 1.835 times astrode of manganese dioxide. It generates 1.5 V. much as water per unit volume. The specific gravity of a mixture of sulfuric acid and water varies with the strength6.10.4.4 Nickel-Cadmium Cell of the solution from 1.000 to 1.830. The specific gravity of the electrolyte solution in aThe nickel-cadmium cell, or Ni-Cad cell, is the only dry lead-acid cell ranges from 1.210 to 1.300 for new, fullybattery that is a true storage battery with a reversible charged batteries. The higher the specific gravity, the lesschemical reaction, allowing recharging many times. In the internal resistance of the cell and the higher the possiblesecondary nickel-cadmium dry cell, the electrolyte is load current. As the cell discharges, the water formedpotassium hydroxide, the negative electrode is nickel dilutes the acid and the specific gravity graduallyhydroxide, and the positive electrode is cadmium oxide. decreases to about 1.150, at which time the cell is consid-The operating voltage is 1.25 V. Because of its rugged ered to be fully discharged.characteristics (stands up well to shock, vibration, andtemperature changes) and availability in a variety of The specific gravity of the electrolyte is measured withshapes and sizes, it is ideally suited for use in powering a hydrometer, which has a compressible rubber bulb atportable communication equipment. the top, a glass barrel, and a rubber hose at the bottom of the barrel. In taking readings with a hydrometer, the dec-6.10.4.5 Mercury Cell imal point is usually omitted. For example, a specific gravity of 1.260 is read simply as “twelve-sixty.” AThe mercury cell was developed because of space explo- hydrometer reading of 1210 to 1300 indicates full charge,ration activities — the development of small transceivers about 1250 is half-charge, and 1150 to 1200 is completeand miniaturized equipment where a power source of min- discharge.iaturized size was needed. In addition to reduced size, the The capacity of a battery is measured in ampere-hoursmercury cell has a good shelf life and is very rugged. (Ah). © 2003 by CRC Press LLC
  • Note: T