Mathematics Manual for WATER AND WASTEWATER TREATMENT PLANT OPERATORS © 2004 by CRC Press LLC Mathematics Manual for WATER AND WASTEWATER TREATMENT PLANT OPERATORS Frank R Spellman CRC PR E S S Boca Raton London New York Washington, D.C © 2004 by CRC Press LLC L1675_C00.fm Page iv Wednesday, February 4, 2004 8:03 AM Library of Congress Cataloging-in-Publication Data Spellman, Frank R Mathematics manual for water and wastewater treatment plant operators / Frank R Spellman p cm Includes index ISBN 1-56670-675-0 (alk paper) Water—PuriÞcation—Mathematics Water quality management—Mathematics Water—PuriÞcation—Problems, exercises, etc Water quality management—Problems, exercises, etc Sewage—PuriÞcation—Mathematics Sewage disposal—Mathematics Sewage—PuriÞcation—Problems, exercises, etc Sewage disposal—Problems, exercises, etc I Title TD430.S64 2004 628.1¢01¢51—dc22 2003065830 This book contains information obtained from authentic and highly regarded sources Reprinted material is quoted with permission, and sources are indicated A wide variety of references are listed Reasonable efforts have been made to publish reliable data and information, but the author and the publisher 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 SpeciÞc permission 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 © 2004 by CRC Press LLC No claim to original U.S Government works International Standard Book Number 1-56670-675-0 Library of Congress Card Number 2003065830 Printed in the United States of America Printed on acid-free paper © 2004 by CRC Press LLC L1675_C00.fm Page v Wednesday, February 4, 2004 8:03 AM PREFACE To properly operate a waterworks or wastewater treatment plant and to pass the examination for a waterworks/wastewater operator’s license, it is necessary to know how to perform certain calculations In reality, most of the calculations that operators at the lower level of licensure need to know how to perform are not difÞcult, but all operators need a basic understanding of arithmetic and problem-solving techniques to be able to solve the problems they typically encounter How about waterworks/wastewater treatment plant operators at higher levels of licensure — they also need to be well versed in mathematical operations? The short answer is absolutely The long answer is that anyone who works in water or wastewater treatment and who expects to have a successful career that includes advancement to the highest levels of licensure or certiÞcation (usually prerequisites for advancement to higher management levels) must have knowledge of math at both the basic or fundamental level and at the advanced practical level It is simply not possible to succeed in this Þeld without the ability to perform mathematical operations Keep in mind that mathematics is a universal language Mathematical symbols have the same meaning to people speaking many different languages throughout the world The key to using mathematics is learning the language, symbols, deÞnitions, and terms of mathematics that allow us to grasp the concepts necessary to solve equations In Mathematics Manual for Water/Wastewater Treatment Plant Operators, we begin by introducing and reviewing concepts critical to the qualiÞed operators at the fundamental or entry level; however, this does not mean that these are the only math concepts that a competent operator must know to solve routine operation and maintenance problems After covering the basics, therefore, the text progressively advances, step-by-step, to higher more practical applications of mathematical calculations — that is, the math operations that operators at the highest level of licensure would be expected to know how to perform The basic level reviews fractions and decimals, rounding numbers, determining the correct number of signiÞcant digits, raising numbers to powers, averages, proportions, conversion factors, calculating ßow and detention times, and determining the areas and volumes of different shapes This review also explains how to keep track of units of measurement (inches, feet, gallons, etc.) during calculations and demonstrates how to solve real-life problems that require calculations After building a strong foundation based on theoretical math concepts (the basic tools of mathematics, such as fractions, decimals, percents, areas, volumes), we move on to applied math — basic math concepts applied when solving practical water/wastewater operational problems Even though considerable crossover of basic math operations used by both waterworks and wastewater operators occurs, this book separates applied math problems for wastewater and water to aid operators dealing with speciÞc unit processes unique to either waterworks or wastewater operations The text is divided into Þve parts Part I covers basic math concepts used in both water and wastewater treatment Part II covers advanced math concepts for waterworks operators Part III covers advanced math concepts for wastewater operators Part IV covers fundamental laboratory calculations used in both water and wastewater treatment operations Part V presents a comprehensive workbook of more than 1400 practical math problems that highlight the type of math exam questions operators can expect to see on state licensure examinations What makes Mathematics Manual for Water/Wastewater Treatment Plant Operators different from other math books available? Consider the following: © 2004 by CRC Press LLC L1675_C00.fm Page vi Wednesday, February 4, 2004 8:03 AM • • • • • The author has worked in and around water/wastewater treatment and taught water/wastewater math for several years The author has sat at the table of licensure examination preparation boards to review, edit, and write state licensure exams This step-by-step training manual provides concise, practical instruction in the math skills that operators must have to pass certiÞcation tests The text is completely self-contained in one complete volume The advantage should be obvious — one text that combines math basics and advanced operator math concepts eliminates shufßing from one volume to another to Þnd the solution to a simple or more complex problem The text is user friendly; no matter the difÞculty of the problem to be solved, each operation is explained in straightforward, plain English Moreover, numerous example problems (several hundred) are presented to enhance the learning process To assure correlation to modern practice and design, the text provides illustrative problems dealing with commonly encountered waterworks/wastewater treatment operations and associated parameters and covers typical math concepts for waterworks/wastewater treatment unit process operations found in today’s waterworks/wastewater treatment facilities ߜ Note: The symbol ߜ displayed in various locations throughout this manual indicates or emphasizes an important point or points to study carefully This text is accessible to those who have little or no experience in treatment plant math operations Readers who work through the text systematically will be surprised at how easily they can acquire an understanding of water/wastewater math concepts, thus adding another critical component to their professional knowledge A Þnal point before beginning our discussion of math concepts: It can be said with some accuracy and certainty that without the ability to work basic math problems (i.e., those typical to water/wastewater treatment) candidates for licensure will Þnd any attempts to successfully pass licensure exams a much more difÞcult proposition Frank R Spellman Norfolk, Virginia © 2004 by CRC Press LLC L1675_C00.fm Page vii Wednesday, February 4, 2004 8:03 AM Table of Contents PART I Basic Math Concepts Chapter Introduction Math Terminology and DeÞnitions Calculation Steps Key Words Calculators Chapter Sequence of Operations Sequence of Operations — Rules Sequence of Operations — Examples Chapter Fractions, Decimals, and Percent Fractions Decimals Percent Chapter Rounding and SigniÞcant Digits Rounding Numbers Determining SigniÞcant Figures Chapter Powers of Ten and Exponents Rules Examples Chapter Averages (Arithmetic Mean) and Median Averages Median Chapter Solving for the Unknown Equations Axioms Solving Equations Setting Up Equations Chapter Ratio and Proportion Ratio Proportion Working with Ratio and Proportion © 2004 by CRC Press LLC L1675_C00.fm Page viii Wednesday, February 4, 2004 8:03 AM Chapter Dimensional Analysis Dimensional Analysis in Problem Solving Basic Operation Basic Operation Basic Operation Chapter 10 Units of Measurement, Conversions, and Electrical Calculations Units of Measurement Conversion Factors Weight, Concentration, and Flow Conversions Typical Water/Wastewater Conversion Examples Temperature Conversions Population Equivalent (PE) or Unit Loading Factor SpeciÞc Gravity and Density Flow Detention Time Chemical Addition Conversions Horsepower and Energy Costs Electrical Power Electrical Calculations Ohm’s Law Electric Power Electric Energy Series D-C Circuit Characteristics Parallel D-C Circuits Chapter 11 Measurements: Circumference, Area, and Volume Perimeter and Circumference Perimeter Circumference Area Area of a Rectangle Area of a Circle Surface Area of a Circular or Cylindrical Tank Volume Volume of a Rectangular Basin Volume of Round Pipe and Round Surface Areas Volume of a Cone and Sphere Volume of a Circular or Cylindrical Tank Chapter 12 Force, Pressure, and Head Calculations Force and Pressure Head Static Head Friction Head Velocity Head Total Dynamic Head (Total System Head) © 2004 by CRC Press LLC L1675_C00.fm Page ix Wednesday, February 4, 2004 8:03 AM Pressure/Head Head/Pressure Force, Pressure, and Head Example Problems Chapter 13 Mass Balance and Measuring Plant Performance Mass Balance for Settling Tanks Mass Balance Using BOD Removal Measuring Plant Performance Plant Performance/EfÞciency Unit Process Performance/EfÞciency Percent Volatile Matter Reduction in Sludge PART II Applied Math Concepts: Water Treatment Chapter 14 Pumping Calculations Pumping Basic Water Hydraulics Calculations Weight of Water Weight of Water Related to the Weight of Air Gauge Pressure Water in Motion Pipe Friction Basic Pumping Calculations Pumping Rates Calculating Head Loss Calculating Horsepower and EfÞciency SpeciÞc Speed Positive Displacement Pumps Volume of Biosolids Pumped (Capacity) Chapter 15 Water Source and Storage Calculations Water Sources Water Source Calculations Well Drawdown Well Yield SpeciÞc Yield Well-Casing Disinfection Deep-Well Turbine Pump Calculations Vertical Turbine Pump Calculations Water Storage Water Storage Calculations Copper Sulfate Dosing Chapter 16 Coagulation and Flocculation Calculations Coagulation Flocculation © 2004 by CRC Press LLC L1675_C00.fm Page x Wednesday, February 4, 2004 8:03 AM Coagulation and Flocculation Calculations Chamber and Basin Volume Calculations Detention Time Determining Dry Chemical Feeder Setting (lb/d) Determining Chemical Solution Feeder Setting (gpd) Determining Chemical Solution Feeder Setting (mL/min) Determining Percent of Solutions Determining Percent Strength of Liquid Solutions Determining Percent Strength of Mixed Solutions Dry Chemical Feeder Calibration Solution Chemical Feeder Calibration Determining Chemical Usage Chapter 17 Sedimentation Calculations Sedimentation Tank Volume Calculations Calculating Tank Volume Detention Time Surface Overßow Rate Mean Flow Velocity Weir Loading Rate (Weir Overßow Rate) Percent Settled Biosolids Determining Lime Dosage (mg/L) Determining Lime Dosage (lb/day) Determining Lime Dosage (g/min) Chapter 18 Filtration Calculations Water Filtration Flow Rate through a Filter (gpm) Filtration Rate Unit Filter Run Volume (UFRV) Backwash Rate Backwash Rise Rate Volume of Backwash Water Required (gal) Required Depth of Backwash Water Tank (ft) Backwash Pumping Rate (gpm) Percent Product Water Used for Backwatering Percent Mud Ball Volume Chapter 19 Water Chlorination Calculations Chlorine Disinfection Determining Chlorine Dosage (Feed Rate) Calculating Chlorine Dose, Demand and Residual Breakpoint Chlorination Calculations Calculating Dry Hypochlorite Feed Rate Calculating Hypochlorite Solution Feed Rate © 2004 by CRC Press LLC L1675_C00.fm Page xi Wednesday, February 4, 2004 8:03 AM Calculating Percent Strength of Solutions Calculating Percent Strength Using Dry Hypochlorite Calculating Percent Strength Using Liquid Hypochlorite Chemical Use Calculations Chapter 20 Fluoridation Water Fluoridation Fluoride Compounds Sodium Fluoride Sodium Fluorosilicate Fluorosilicic Acid Optimal Fluoride Levels Fluoridation Process Calculations Percent Fluoride Ion in a Compound Fluoride Feed Rate Fluoride Feed Rates for Saturator Calculated Dosages Calculated Dosage Problems Chapter 21 Water Softening Water Hardness Calculating Calcium Hardness as CaCO3 Calculating Magnesium Hardness as CaCO3 Calculating Total Hardness Calculating Carbonate and Noncarbonate Hardness Alkalinity Determination Determining Bicarbonate, Carbonate, and Hydroxide Alkalinity Lime Dosage Calculation for Removal of Carbonate Hardness Calculation for Removal of Noncarbonate Hardness Recarbonation Calculation Calculating Feed Rates Ion Exchange Capacity Water Treatment Capacity Treatment Time Calculation (Until Regeneration Required) Salt and Brine Required for Regeneration PART III Wastewater Math Concepts Chapter 22 Preliminary Treatment Calculations Screening Screening Removal Calculations Screening Pit Capacity Calculations Grit Removal Grit Removal Calculations Grit Channel Velocity Calculation © 2004 by CRC Press LLC L1675_C11.fm Page 102 Saturday, January 31, 2004 5:25 PM Mathematics for Water/Wastewater Treatment Plant Operators 102 r = 10 ft H = 25 ft FIGURE 11.8 Circular or cylindrical water tank Example 11.13 Problem What is the volume (in cubic feet) of a gas storage container that is spherical and has a diameter of 60 ft? Solution Volume (ft3) = 0.524 ¥ 60 ft ¥ 60 ft ¥ 60 ft = 113,184 ft3 VOLUME OF A CIRCULAR OR CYLINDRICAL TANK Circular process and various water/chemical storage tanks are commonly found in water/wastewater treatment A circular tank consists of a circular floor surface with a cylinder rising above it (see Figure 11.8) The volume of a circular tank is calculated by multiplying the surface area times the height of the tank walls Example 11.14 Problem A tank is 20 feet in diameter and 25 feet deep How many gallons of water will it hold? ߜ Hint: In this type of problem, calculate the surface area first, multiply by the height, and then convert to gallons Solution r = D ÷ = 20 ft ÷ = 10 ft A = p ¥ r = p ¥ 10 ft ¥ 10 ft A = 314 ft2 © 2004 by CRC Press LLC L1675_C11.fm Page 103 Saturday, January 31, 2004 5:25 PM Measurements: Circumference, Area, and Volume V=A¥H V = 314 ft2 ¥ 25 ft = 7850 ft3 V = 7850 ft3 ¥ 7.5 gal/ft3 = 58,875 gal © 2004 by CRC Press LLC 103 L1675_C12.fm Page 105 Saturday, January 31, 2004 5:28 PM Pressure, and 12 Force,Calculations Head TOPICS • • • Force and Pressure Head • Static Head • Friction Head • Velocity Head • Total Dynamic Head • Pressure/Head • Head/Pressure Force, Pressure, and Head Example Problems Before we study calculations involving force, pressure, and head, we must first define some terms: • • • Force — Push exerted by water on any confined surface; force can be expressed in pounds, tons, grams, or kilograms Pressure — Force per unit area; the most common way of expressing pressure is in pounds per square inch (psi) Head — Vertical distance or height of water above a reference point; head is usually expressed in feet In the case of water, head and pressure are related FORCE AND PRESSURE Figure 12.1 helps to illustrate these terms A cubical container measuring foot on each side can hold cubic foot of water It is a basic fact of science that cubic foot of water weighs 62.4 pounds and contains 7.48 gallons The force acting on the bottom of the container would be 62.4 pounds per square foot The area of the bottom in square inches is: ft2 = 12 in ¥ 12 in = 144 in2 Therefore, the pressure in pounds per square inch (psi) is: 62.4 lb ft 62.4 lb ft = = 0.433 lb in.2 ( psi) ft 144 in.2 ft If we use the bottom of the container as our reference point, the head would be foot From this, we can see that foot of head is equal to 0.433 psi — an important parameter to remember Figure 12.2 illustrates some other important relationships between pressure and head 105 © 2004 by CRC Press LLC L1675_C12.fm Page 106 Saturday, January 31, 2004 5:28 PM Mathematics for Water/Wastewater Treatment Plant Operators 106 ft ft ft FIGURE 12.1 One cubic foot of water weights 62.4 lbs lb of water 2.31 ft 0.433 lb of water ft sq in AREA ft water = 0.433 psi sq in AREA psi = 2.31 ft water FIGURE 12.2 Shows the relationship between pressure and head ߜ Important Point: Force acts in a particular direction; water in a tank exerts force down on the bottom and out of the sides Pressure, however, acts in all directions A marble at a water depth of foot would experience 0.433 psi of pressure acting inward on all sides Using the preceding information, we can develop Equation 12.1 and Equation 12.2 for calculating pressure and head: Pressure (psi) = 0.433 ¥ head (ft) Head (ft) = 2.31 ¥ pressure (psi) (12.1) (12.2) HEAD As mentioned, head is the vertical distance the water must be lifted from the supply tank or unit process to the discharge 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 velocity (velocity head) Total head = static head + friction head + velocity head (12.3) STATIC HEAD Static head is the actual vertical distance the liquid must be lifted: Static head = discharge elevation – supply elevation © 2004 by CRC Press LLC (12.4) L1675_C12.fm Page 107 Saturday, January 31, 2004 5:28 PM Force, Pressure, and Head Calculations 107 Example 12.1 Problem A supply tank is located at an elevation of 108 ft The discharge point is at an elevation of 205 ft What is the static head in feet? Solution Static head (ft) = 205 ft – 108 ft = 97 ft FRICTION HEAD Friction head is the equivalent distance of the energy that must be supplied to overcome friction Engineering references include tables showing the equivalent vertical distance for various sizes and types of pipes, fittings, and valves The total friction head is the sum of the equivalent vertical distances for each component: Friction head (ft) = energy losses due to friction (12.5) VELOCITY HEAD Velocity head is the equivalent distance of the energy consumed in achieving and maintaining the desired velocity in the system: Velocity head (ft) = energy losses to maintain velocity (12.6) TOTAL DYNAMIC HEAD (TOTAL SYSTEM HEAD) Total head = static head + friction head + velocity head (12.7) PRESSURE/HEAD The pressure exerted by water/wastewater is directly proportional to its depth or head in the pipe, tank, or channel If the pressure is known, the equivalent head can be calculated Head (ft) = pressure (psi) ¥ 2.31 ft/psi (12.8) Example 12.2 Problem The pressure gauge on the discharge line form the influent pump reads 75.3 psi What is the equivalent head in feet? Solution Head (ft) = 75.3 ¥ 2.31 ft/psi = 173.9 ft HEAD/PRESSURE If the head is known, the equivalent pressure can be calculated by: Pressure ( psi) = © 2004 by CRC Press LLC head (ft ) 2.31 ft psi (12.9) L1675_C12.fm Page 108 Saturday, January 31, 2004 5:28 PM Mathematics for Water/Wastewater Treatment Plant Operators 108 Example 12.3 Problem A tank is 15 feet deep What is the pressure (in pounds per square inch) at the bottom of the tank when it is filled with wastewater? Solution Pressure ( psi) = 15 ft 2.31 ft psi = 6.49 psi Before we look at a few example problems dealing with force, pressure, and head, it is important to list the key points related to force, pressure, and head: • • • By definition, water weighs 62.4 pounds per cubic foot The surface of any one side of a cube contains 144 square inches (12 in ¥ 12 in = 144 in.2); therefore, the cube contains 144 columns of water that are foot tall and inch square The weight of each of these pieces can be determined by dividing the weight of the water in the cube by the number of square inches: Weight = • 62.4 lb = 0.433 lb in.2 or 0.433 psi 144 in.2 Because this is the weight of one column of water foot tall, the true expression would be 0.433 psi per foot of head, or 0.433 psi/ft ߜ Key Point: foot of head = 0.433 psi In addition to remembering the important parameter of foot of head = 0.433 psi, it is important to understand the relationship between pressure and feet of head — in other words, how many feet of head psi represents This is determined by dividing by 0.433: Feet of head = ft = 2.31 ft psi 0.433 psi If a pressure gauge read 12 psi, the height of the water that would represent this pressure would be 12 psi ¥ 2.31 ft/psi = 27.7 feet ߜ Key Point: Both the above conversions are commonly used in water/wastewater treatment calculations; however, the most accurate conversion is ft = 0.433 psi This is the conversion we use throughout this text FORCE, PRESSURE, AND HEAD EXAMPLE PROBLEMS Example 12.4 Problem Convert 40 psi to feet head © 2004 by CRC Press LLC L1675_C12.fm Page 109 Saturday, January 31, 2004 5:28 PM Force, Pressure, and Head Calculations 109 Solution psi ft ¥ = 92.4 ft 0.433 psi Example 12.5 Problem Convert 40 feet to psi Solution 40 ft 0.433 psi ¥ = 17.32 psi 1 ft As the above examples demonstrate, when converting psi to feet, we divide by 0.433; to convert feet to psi, we multiply by 0.433 The above process can be most helpful in clearing up any confusion about whether to multiply or divide We have another way, however — one that may be more beneficial and easier for many operators to use Notice that the relationship between psi and feet is almost two to one It takes slightly more than feet to make one psi; therefore, when looking at a problem where the data are in pressure, the result should be in feet, and the answer should be at least twice as large as the starting number For instance, if the pressure is 25 psi, we intuitively know that the head is over 50 feet; therefore, we must divide by 0.433 to obtain the correct answer Example 12.6 Problem Convert a pressure of 45 pounds per square inch to feet of head Solution 45 ft psi ¥ = 104 ft 0.433 psi 15 psi ft ¥ = 34.6 ft 0.433 psi Example 12.7 Problem Convert 15 psi to feet Solution Example 12.8 Problem Between the top of a reservoir and the watering point, the elevation is 125 feet What will the static pressure be at the watering point? Solution 125 © 2004 by CRC Press LLC ft psi ¥ = 54.1 ft 0.433 psi L1675_C12.fm Page 110 Saturday, January 31, 2004 5:28 PM Mathematics for Water/Wastewater Treatment Plant Operators 110 Example 12.9 Problem Find the pressure (in psi) in a tank 12 feet deep at a point feet below the water surface Solution Pressure (psi) = 0.433 ¥ ft = 2.17 psi Example 12.10 Problem A pressure gauge at the bottom of a tank reads 12.2 psi How deep is the water in the tank? Solution Head (ft) = 2.31 ¥ 12.2 psi = 28.2 ft Example 12.11 Problem What is the pressure (static pressure) miles beneath the ocean surface? Solution Change miles to ft, then to psi 5280 ft mile ¥ = 21,120 ft 21,120 ft = 9143 psi 2.31 ft psi Example 12.12 Problem A 150-ft-diameter cylindrical tank contains 2.0 MG water What is the water depth? At what pressure would a gauge at the bottom read in psi? Solution • Step 1: Change MG to cu ft: 2, 000, 000 gal = 267, 380 cu ft 7.48 • Step 2: Using volume, solve for depth: Volume (V ) = 785 ¥ D2 ¥ depth V = 267,380 cu ft = 785 ¥ (150)2 ¥ depth Depth = 15.1 ft © 2004 by CRC Press LLC L1675_C12.fm Page 111 Saturday, January 31, 2004 5:28 PM Force, Pressure, and Head Calculations 111 Example 12.13 Problem The pressure in a pipe is 70 psi What is the pressure in feet of water? What is the pressure in psf? Solution • Step 1: Convert pressure to feet of water: 70 psi ¥ 2.31 ft/psi = 161.7 ft of water • Step 2: Convert psi to psf: 70 psi ¥ 144 sq in./sq ft = 10,080 psf Example 12.14 Problem The pressure in a pipeline is 6476 psf What is the head on the pipe? Solution Head on pipe = ft of pressure Pressure = weight ¥ height 6476 psf = 62.4 lb/cu ft Ơ height Height = 104 ft â 2004 by CRC Press LLC L1675_C13.fm Page 113 Saturday, January 31, 2004 5:29 PM Balance and 13 Mass Performance Measuring Plant TOPICS • • • Mass Balance for Settling Tanks Mass Balance Using BOD Removal Measuring Plant Balance • Plant Performance/Efficiency • Unit Process Performance/Efficiency • Percent Volatile Matter Reduction in Biosolids The simplest way to express the fundamental engineering principle of mass balance is to say, “Everything has to go somewhere.” More precisely, the law of conservation of mass says that when chemical reactions take place, matter is neither created nor destroyed This important concept allows us to track materials (that is, pollutants, microorganisms, chemicals, and other materials) from one place to another The concept of mass balance plays an important role in treatment plant operations (especially wastewater treatment), where we assume a balance exists between the material entering and leaving the treatment plant or a treatment process: “What comes in must equal what goes out.” This concept is very helpful in evaluating biological systems, sampling and testing procedures, and many other unit processes within the treatment system In the following sections, we illustrate how the mass balance concept is used to determine the quantity of solids entering and leaving settling tanks and mass balance using biological oxygen demand (BOD) removal MASS BALANCE FOR SETTLING TANKS The mass balance for the settling tank calculates the quantity of solids entering and leaving the unit ߜ Key Point: The two numbers — in (influent) and out (effluent) — must be within 10 to 15% of each other to be considered acceptable Larger discrepancies may indicate sampling errors or increasing solids levels in the unit or undetected solids discharge in the tank effluent To get a better feel for how the mass balance for settling tanks procedure is formatted for actual use, consider Figure 13.1 and the accompanying steps provided below We will use an example computation to demonstrate how mass balance is actually used in wastewater operations: • • • • Step Step Step Step 1: 2: 3: 4: Solids in = pounds of influent suspended solids Pounds of effluent suspended solids Biosolids solids out = pounds of biosolids solids pumped per day Solids in — (solids out + biosolids solids pumped) Example 13.1 Problem A settling tank receives a daily flow of 4.20 MGD The influent contains 252 mg/L suspended solids, and the unit effluent contains 140 mg/L suspended solids The biosolids pump operates 10 © 2004 by CRC Press LLC L1675_C13.fm Page 114 Saturday, January 31, 2004 5:29 PM TSS x Flow x 8.34 TSS x Flow x 8.34 (Influent) (Effluent) Biosolids lb/day FIGURE 13.1 Mass balance for settling tanks min/h and removes biosolids at the rate of 40 gpm The biosolids content is 4.2% solids Determine if the mass balance for solids removal is within the acceptable 10 to 15% range Solution • • • • Step Step Step Step 1: 2: 3: 4: Solids in = 252 mg/L ¥ 4.20 MGD ¥ 8.34 = 8827 lb/d Solids out = 140 mg/L ¥ 4.20 MGD ¥ 8.34 = 4904 lb/d Biosolids solids = 10 min/hr ¥ 24 hr/day ¥ 40 gpm ¥ 8.34 ¥ 0.042 = 3363 lb/d Balance = 8827 lb/day – (4904 lb/day + 3363 lb/day) = 560 lb, or 6.3% MASS BALANCE USING BOD REMOVAL The amount of BOD removed by a treatment process is directly related to the quantity of solids the process will generate Because the actual amount of solids generated will vary with operational conditions and design, exact figures must be determined on a case-by-case basis; however, research has produced general conversion rates for many of the common treatment processes These values are given in Table 13.1 and can be used if plant-specific information is unavailable Using these factors, the mass balance procedure determines the amount of solids the process is anticipated to produce This quantity is compared with actual biosolids production to determine the accuracy of the sampling and/or the potential for solids buildup in the system or unrecorded solids discharges (see Figure 13.2) TABLE 13.1 General Conversion Rates Process Type Primary treatment Trickling filters Rotating biological contactors Activated biosolids with primary Activated biosolids without primary Conventional Extended air Contact stabilization Step feed Oxidation ditch © 2004 by CRC Press LLC Conversion Factor (lb Solids/lb BOD Removal) 1.7 1.0 1.0 0.7 0.85 0.65 1.0 0.85 0.65 L1675_C13.fm Page 115 Saturday, January 31, 2004 5:29 PM BOD In BOD Out Solids Out Solids Removed FIGURE 13.2 Comparison of BOD in and BOD out • • • • • • Step Step Step Step Step Step 1: 2: 3: 4: 5: 6: BODin = influent BOD ¥ flow ¥ 8.34 BODout = effluent BOD ¥ flow ¥ 8.34 BOD pounds removed = BODin – BODout Solids generated (lb) = BOD removed (lb ¥ factor) Solids removed = sludge pumped (gpd) ¥ % solids ¥ 8.34 Effluent solids (mg/L) ¥ flow (MGD) ¥ 8.34 Example 13.2 Problem A conventional activated biosolids system with primary treatment is operating at the levels listed below Does the mass balance for the activated biosolids system indicate a problem? Plant influent BOD Primary effluent BOD Activated biosolids system effluent BOD Activated biosolids system effluent total suspended solids (TSS) Plant flow Waste concentration Waste flow 250 mg/L 166 mg/L 25 mg/L 19 mg/L 11.40 MGD 6795 mg/L 0.15 MGD Solution • • • • • • BODin = 166 mg/L ¥ 11.40 MGD ¥ 8.34 = 15,783 lb/day BODout = 25 mg/L ¥ 11.40 MGD ¥ 8.34 = 2377 lb/day BOD removed = 15,783 lb/d – 2377 lb/d = 13,406 lb/day Solids produced = 13,406 lb/day ¥ 0.7 lb solids/lb BOD = 9384 lb solids/day Solids removed = 6795 mg/L ¥ 0.15 MGD ¥ 8.34 = 8501 lb/day Difference = 9384 lb/day – 8501 lb/day = 883 lb/day, or 9.4% These results are within the acceptable range ߜ Key Point: We have demonstrated two ways in which mass balance can be used; however, it is important to note that the mass balance concept can be used for all aspects of wastewater and solids treatment In each case, the calculations must take into account all of the sources of material entering the process and all of the methods available for removal of solids © 2004 by CRC Press LLC L1675_C13.fm Page 116 Saturday, January 31, 2004 5:29 PM MEASURING PLANT PERFORMANCE To evaluate how well a plant or unit process is performing, performance efficiency or percent (%) removal is used The results obtained can be compared with those listed in a plant’s operation and maintenance (O&M) manual to determine if the facility is performing as expected This section presents sample calculations often used to measure plant performance/efficiency The efficiency of a unit process is its effectiveness in removing various constituents from the wastewater or water Suspended solids and BOD removal are therefore the most common calculations of unit process efficiency In wastewater treatment, the efficiency of a sedimentation basin may be affected by such factors as the types of solids in the wastewater, the temperature of the wastewater, and the age of the solids Typical removal efficiencies for a primary sedimentation basin are as follows: • • • • Settleable solids Suspended solids Total solids BOD 90 40 10 20 to to to to 99% 60% 15% 50% PLANT PERFORMANCE/EFFICIENCY ߜ Key Point: The calculation used for determining the performance (percent removal) for a digester is different from that used for performance (percent removal) for other processes Care must be taken to select the right formula: % Removal = [influent concentration - effluent concentration] ¥ 100 influent concentration (13.1) Example 13.3 Problem As shown in Figure 13.3, the influent BOD5 is 247 mg/L, and the plant effluent BOD is 17 mg/L What is the percent removal? Solution % Removal = (247 mg L - 17 mg L ) ¥ 100 = 93% 247 mg L BOD 247 mg/L BOD 17 mg/L Total Entering 93% Removal FIGURE 13.3 Refers to Example 13.3 © 2004 by CRC Press LLC L1675_C13.fm Page 117 Saturday, January 31, 2004 5:29 PM BOD 235 mg/L BOD 169 mg/L % Removal = 28% FIGURE 13.4 Refers to Example 13.4 UNIT PROCESS PERFORMANCE/EFFICIENCY Equation 13.1 is used again to determine unit process efficiency The concentration entering the unit and the concentration leaving the unit (e.g., primary, secondary) are used to determine the unit performance: % Removal = [influent concentration - effluent concentration] ¥ 100 influent concentration (13.1) Example 13.4 Problem As shown in Figure 13.4, the primary influent BOD is 235 mg/L, and the primary effluent BOD is 169 mg/L What is the percent removal? Solution % Removal = (235 PERCENT VOLATILE MATTER REDUCTION mg L - 169 mg L ) ¥ 100 = 28% 235 mg L IN SLUDGE The calculation used to determine percent volatile matter (%VM) reduction is more complicated because of the changes occurring during biosolids digestion % VM reduction = (% VMin - % VMout ) ¥ 100 [% VMin - (% VMin ¥ % VMout )] Example 13.5 Problem: Using the digester data provided below, determine the percent volatile matter reduction Raw biosolids volatile matter = 74% Digested biosolids volatile matter = 54% © 2004 by CRC Press LLC L1675_C13.fm Page 118 Saturday, January 31, 2004 5:29 PM Solution % VM reduction = [ (0.74 - 0.54) ¥ 100 = 59% 0.74 - (0.74 ¥ 0.54) ] See Figure 13.5 VMin 74% VMout 54% % VM Reduction = 59% FIGURE 13.5 Refers to Example 13.5 © 2004 by CRC Press LLC ... 1 1 Powers of 10 10 0 10 1 10 2 10 3 10 4 = = = = = 10 10 0 10 00 10 ,000 L1675_C05.fm Page 22 Saturday, January 31, 2004 5 :10 PM EXAMPLES Example 5 .1 Problem How is the term 23 written in expanded form?... Cataloging-in-Publication Data Spellman, Frank R Mathematics manual for water and wastewater treatment plant operators / Frank R Spellman p cm Includes index ISBN 1- 5 667 0-6 7 5-0 (alk paper) Water? ??PuriÞcation? ?Mathematics. .. concepts used in both water and wastewater treatment Part II covers advanced math concepts for waterworks operators Part III covers advanced math concepts for wastewater operators Part IV covers fundamental