Handbook of Water and Wastewater Treatment Plant Operations - Chapter 4 pps

PART II Water/Wastewater Operations: Math and Technical Aspects © 2003 by CRC Press LLC Water and Wastewater Math Operations To operate a waterworks and/or a wastewater treatment plant, and to pass the examination for an operator’s license, you must know how to perform certain mathematical oper- ations. However, do not panic, as Price points out, “Those who have difficulty in math often do not lack the ability for mathematical calculation, they merely have not learned, or have not been taught, the ‘language of math.” 1 4.1 INTRODUCTION Without the ability to perform mathematical calculations, operators would have difficulty in properly operating water and wastewater systems. In reality, most of the calculations operators need to perform are not difficult. Generally, math ability through basic algebra is all that is needed. Experi- ence has shown that skill with math operations used in water and wastewater system operations is an acquired skill that is enhanced and strengthened with practice. Note: Keep in mind that mathematics is a language — a universal language. Mathematical symbols have the same meaning to people speaking many different languages throughout the globe. The key to learning mathematics is to learn the language — the symbols, definitions and terms, of mathematics that allow you to understand the concepts necessary to perform the operations. In this chapter, we assume the reader is well grounded in basic math principles. We do not cover basic operations such as addition, subtraction, multiplication, and division. However, we do include, for review purposes, a few basic math calculations in the Chapter Review Questions/Prob- lems at the end of the chapter. 4.2 CALCULATION STEPS As with all math operations, many methods can be suc- cessfully used to solve water and wastewater system prob- lems. We provide one of the standard methods of problem solving in the following: 1. If appropriate, make a drawing of the informa- tion in the problem. 2. Place the given data on the drawing. 3. Determine what the question is. This is the first thing you should determine as you begin to solve the problem, along with, “What are they really looking for?” Writing down exactly what is being looked for is always smart. Sometimes the answer has more than one unknown. For instance, you may need to find X and then find Y. 4. If the calculation calls for an equation, write it down. 5. Fill in the data in the equation and look to see what is missing. 6. Rearrange or transpose the equation, if necessary. 7. If available, use a calculator. 8. Always write down the answer. 9. Check any solution obtained. 4.3 TABLE OF EQUIVALENTS, FORMULAE, AND SYMBOLS In order to work mathematical operations to solve problems (for practical application or for taking licensure examinations), it is essential to understand the language, equivalents, symbols, and terminology used. Because this handbook is designed for use in practical on-the-job applications, equivalents, formulae, and symbols are included, as a ready reference, in Table 4.1. 4.4 TYPICAL WATER AND WASTEWATER MATH OPERATIONS 4.4.1 A RITHMETIC A VERAGE ( OR A RITHMETIC M EAN ) AND M EDIAN During the day-to-day operation of a wastewater treatment plant, considerable mathematical data are collected. The data, if properly evaluated, can provide useful information for trend analysis and indicate how well the plant or unit process is operating. However, because there may be much variation in the data, it is often difficult to determine trends in performance. Arithmetic average refers to a statistical calculation used to describe a series of numbers such as test results. By calculating an average, a group of data is represented by a single number. This number may be considered typical of the group. The arithmetic mean is the most commonly used measurement of average value. 4 © 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 ft 2 9 ft 2 = 1 yd 2 43,560 ft 2 = 1 ac 1 ft 3 = 1728 in. 3 1 ft 3 H 2 0= 7.48 gal 1 ft 3 H 2 0= 62.4 lb 1 gal of H 2 0= 8.34 lb 1 L = 1.000 mL 1 g = 1.000 mg 1 MGD (million gal[MG]/d) = 694 gal/min, 1.545 ft 3 /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) WA S= Waste activated sludge RAS = Return activated sludge MLSS = Mixed liquor suspended solids MLVSS = Mixed liquor volatile suspended solids Formulae SVI = ¥ 100 Q = A ¥ V Detention time = v/Q v = L ¥ W ¥ D area = W ¥ L Circular area = p ¥ Diameter 2 C = p d Hydraulic loading rate = Q/A Sludge age = Mean cell residence time = Organic loading rate = Source: From Spellman, F.R., Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999. v Concentration # MLSS in aeration tank # SS in primary eff d # SS in secondary system aeration tank sec. clarifier # WAS d + SS in eff d + () # BOD d v © 2003 by CRC Press LLC Note: When evaluating information based on aver- ages, remember that the average reflects the general nature of the group and does not nec- essarily reflect any one element of that group. Arithmetic average is calculated by dividing the sum of all of the available data points (test results) by the number of test results: (4.1) E XAMPLE 4.1 Problem: Effluent biochemical oxygen demand (BOD) test results for the treatment plant during the month of September are shown below: What is the average effluent BOD for the month of September? Solution: E XAMPLE 4.2 Problem: For the primary influent flow, the following composite- sampled solids concentrations were recorded for the week: Solution: E XAMPLE 4.3 Problem: A waterworks operator takes a chlorine residual measure- ment every day. We show part of the operating log in Table 4.2. Find the mean. Solution: Add up the seven chlorine residual readings: 0.9 + 1.0 + 1.2 + 1.3 + 1.4 + 1.1 + 0.9 = 7.8. Next, divide by the number of measurements (in this case 7): 7.8 ∏ 7 = 1.11. The mean chlorine residual for the week was 1.11 mg/L. Definition: The median is defined as the value of the central item when the data are arrayed by size. First, arrange all of the readings in either ascending or descending order. Then find the middle value. E XAMPLE 4.4 Problem: In our chlorine residual example, what is the median? Solution: Arrange the values in ascending order: 0.9, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4 The middle value is the fourth one — 1.1. Therefore, the median chlorine residual is 1.1 mg/L. (Usually, the median will be a different value than the mean.) If the data contain an even number of values, you must add one more step, since no middle value is present. You must find the two values in the middle, and then find the mean of those two values. Test 1 20 mg/L Test 2 31 mg/L Test 3 22 mg/L Test 4 15 mg/L Monday 300 mg/L SS Tuesday 312 mg/L SS Wednesday 315 mg/L SS Thursday 320 mg/L SS Friday 311 mg/L SS Saturday 320 mg/L SS Sunday 310 mg/L SS Total 2188 mg/L SS Test 1 Test 2 Test 3 Test N ++++ () L Number of Tests Performed N Average = +++ = 20 mg L 31 mg L 22 mg L 15 mg L 22 mg L 4 Average SS Sum Number of M of All Measurements easurements Used mg L SS 7 mg L SS = = = 2188 312 6. TABLE 4.2 Daily Chlorine Residual Results Day Chlorine Residual (mg/L) Monday 0.9 Tuesday 1.0 Wednesday 1.2 Thursday 1.3 Friday 1.4 Saturday 1.1 Sunday 0.9 Source: From Spellman, F.R., Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999. © 2003 by CRC Press LLC E XAMPLE 4.5 Problem: A water system has four wells with the following capac- ities: 115, 100, 125, and 90 gal/min. What are the mean and the median pumping capacities? Solution: The mean is: To find the median, arrange the values in order: 90 gal/min, 100 gal/min, 115 gal/min, 125 gal/min With four values, there is no single middle value, so we must take the mean of the two middle values: Note: At times, determining what the original num- bers were like is difficult (if not impossible) when dealing only with averages. E XAMPLE 4.6 Problem: A water system has four storage tanks. Three of them have a capacity of 100,000 gal each, while the fourth has a capacity of 1 million gallons (MG). What is the mean capacity of the storage tanks? Solution: The mean capacity of the storage tanks is: Note: Notice that no tank in Example 4.6 has a capacity anywhere close to the mean. The median capacity requires us to take the mean of the two middle values; since they are both 100,000 gal, the median is 100,000 gal. Although three of the tanks have the same capacity as the median, these data offer no indication that one of these tanks holds 1 million gal — information that could be important for the operator to know. 4.4.2 R ATIO Recall that a ratio is the comparison of two numbers by division or an indicated division. A ratio is usually stated in the form A is to B as C is to D, and is written as two fractions that are equal to each other: (4.2) We solve ratio problems by cross-multiplying; we multiply the left numerator (A) by the right denominator (D) and say that A is equal to the left denominator (B) times the right numerator (C): A ¥ D = B ¥ C AD = BC (4.3) If one of the four items is unknown, we solve the ratio by dividing the two known items that are multiplied together by the known item that is multiplied by the unknown. This is best shown by a couple of examples: E XAMPLE 4.7 Problem: If we need 4 lb of alum to treat 1000 gal of H 2 O, how many pounds of alum will we need to treat 12,000 gal- lons? Solution: We state this as a ratio: 4 lb of alum is to 1000 gallons of H 2 O as pounds of alum (or x ) is to 12,000 gal. We set this up this way: Cross-multiplying: E XAMPLE 4.8 Problem: If 10 gal of fuel oil costs $5.25, how much does 18 gal cost? 1 gal min + 1 gal min 25 gal min 0 gal min 1 gal min 15 00 1 9 4 07 5 ++ = . 100 115 2 07 5 gpm gpm 1 gpm + = . 1111 4 325 000 00, 000 + 00, 000 + 00, 000 + , 000,000 gal= , A B C D = 4 22 1000 gal O 12, 000 gal O lb alum H lb alum H = x 1000 4 12 000 4 48 ¥=¥ = ¥ = x x x , 12, 000 1000 lb alum © 2003 by CRC Press LLC 4.4.3 P ERCENT Percent (like fractions) is another way of expressing a part of a whole. The term percent means per hundred, so a percentage is the number out of 100. For example, 22% means 22 out of 100, or if we divide 22 by 100, we get the decimal 0.22: 4.4.3.1 Practical Percentage Calculations Percentage is often designated by the symbol %. Thus, 10% means 10 percent, 10/100, or 0.10. These equivalents may be written in the reverse order: 0.10 = 10/100 = 10%. In water and wastewater treatment, percent is frequently used to express plant performance and for control of sludge treatment processes. Note: To determine percent divide the quantity you wish to express as a percent by the total quantity then multiply by 100. Percent = (4.4) E XAMPLE 4.9 Problem: The plant operator removes 6000 gal of sludge from the settling tank. The sludge contains 320 gal of solids. What is the percent of solids in the sludge? Solution: E XAMPLE 4.10 Problem: Sludge contains 5.3% solids. What is the concentration of solids in decimal percent? Solution: Note: Unless otherwise noted all calculations in the handbook using percent values require the percent be converted to a decimal before use. Note: To determine what quantity a percent equals first convert the percent to a decimal then multiply by the total quantity. Quantity = Total ¥ Decimal Percent (4.5) E XAMPLE 4.11 Problem: Sludge drawn from the settling tank is 8% solids. If 2400 gal of sludge are withdrawn, how many gallons of solids are removed? Solution: E XAMPLE 4.12 Problem: Calcium hypochlorite (HTH) contains 65% available chlorine. What is the decimal equivalent of 65%? Solution: Since 65% means 65 per hundred, divide 65 by 100 (65/100), which is 0.65. E XAMPLE 4.13 Problem: 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? Solution: Thus, the tank is 64% full. 11 10 18 5 25 18 5 25 10 94 5 10 945 0 al $5.25 8 al $ gg y y y y y = ¥= ¥ = ¥ = = $. $. . $. 22 22 022%.== 100 Quantity 100¥ Total Percent gal =¥ = 320 100 53 6000 gal .% Decimal Percent 100 == 53 0 053 .% . gallons gal gal=¥ = 8 2400 192 % 100 32 50 064 064 100 64 ft ft decimal equivalent= ¥= () . .% © 2003 by CRC Press LLC 4.4.4 UNITS AND CONVERSIONS Most of the calculations made in the water and wastewater operations involve using units. While the number tells us how many, the units tell us what we have. Examples of units include: inches, feet, square feet, cubic feet, gallons, pounds, milliliters, milligrams per liter, pounds per square inch, miles per hour, and so on. Conversions are a process of changing the units of a number to make the number usable in a specific instance. Multiplying or dividing into another number to change the units of the number accomplishes conversions. Common conversions in water and wastewater operations are: 1. Gallons per minute to cubic feet per second 2. Million gallons to acre-feet 3. Cubic foot per second to acre-feet 4. Cubic foot per second of water to weight 5. Cubic foot of H 2 O to gallons 6. Gallons of water to weight 7. Gallons per minute to million gallons per day 8. Pounds to feet of head (the measure of the pres- sure of water expressed as height of water in feet — 1 psi = 2.31 feet of head) In many instances, the conversion factor cannot be derived — it must be known. Therefore, we use tables such as the one below (Table 4.3) to determine the com- mon conversions. Note: Conversion factors are used to change measure- ments or calculated values from one unit of measure to another. In making the conversion from one unit to another, you must know two things: (1) the exact number that relates the two units, and (2) whether to multiply or divide by that number Most operators memorize some standard conversions. This happens because of using the conversions, not because of attempting to memorize them. 4.4.4.1 Temperature Conversions An example of a type of conversion typical in water and wastewater operations is provided in this section on tem- perature conversions. 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 systems operations. Most water and wastewater operators are familiar with the formulas used for Fahrenheit and Celsius temperature conversions: (4.6) (4.7) These conversions are not difficult to perform. The difficulty arises when we must recall these formulas from memory. Probably the easiest way to recall these important formulas is to remember three basic steps for both Fahr- enheit and Celsius conversions: 1. Add 40°. 2. Multiply by the appropriate fraction (5/9 or 9/5). 3. Subtract 40°. Obviously, the only variable in this method is the choice of 5/9 or 9/5 in the multiplication step. To make the proper choice, you must be familiar with two scales. On the Fahrenheit scale, the freezing point of water is 32°, and 0° on the Celsius scale. The boiling point of water is 212° on the Fahrenheit scale and 100° on the Celsius scale. TABLE 4.3 Common Conversions Linear Measurements 1 in. = 2.54 cm 1 ft = 30.5 cm 1 m = 100 cm = 3.281 ft = 39.4 in. 1 ac = 43,560 ft 2 1 yd = 3 ft Volume 1 gal = 3.78 L 1 ft 3 = 7.48 gal 1 L = 1000 mL 1 ac-ft = 43,560 ft 3 1 gal = 32 cups 1 lb = 16 oz dry wt. Weight 1 ft 3 of water = 62.4 lb 1 gal = 8.34 lb 1 lb = 453.6 g 1 kg = 1000 g = 2.2 lb 1% = 10,000 mg/L Pressure 1 ft of head = 0.433 psi 1 psi = 2.31 ft of head Flow 1 ft 3 /sec = 448 gal/min 1 gal/min = 1440 gal/d Source: From Spellman, F.R., Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999. ∞ = ∞ - ∞ () CF59 32 ∞ = ∞ + ∞ () FC59 32 © 2003 by CRC Press LLC What does this mean? Why is it important? Note, for example, that at the same temperature, higher numbers are associated with the Fahrenheit scale and lower numbers with the Celsius scale. This important relationship helps you decide whether to multiply by 5/9 or 9/5. Let us look at a few conversion problems to see how the three-step process works. E XAMPLE 4.14 Problem: Convert 220°F to Celsius. Solution: Using the three-step process, we proceed as follows: Step 1: Add 40°F: 220°F + 40°F = 260°F Step 2: 260°F must be multiplied by either 5/9 or 9/5. Since the conversion is to the Celsius scale, you will be moving to number smaller than 260. Through reason and observa- tion, obviously we see that multiplying 260 by 9/5 would almost be the same as multiplying by 2, which would double 260, rather than make it smaller. On the other hand, multiplying by 5/9 is about the same as multiplying by 1/2, which would cut 260 in half. Because we wish to move to a smaller number, we should multiply by 5/9: 5/9 ¥ 260°F = 144.4°C Step 3: Now subtract 40°: 144.4°C ¥ 40°C = 104.4°C Therefore, 220°F = 104.4°C EXAMPLE 4.15 Problem: Convert 22°C to Fahrenheit. Step 1: Add 40°F: 22°F + 40°F = 62°F Step 2: Because we are converting from Celsius to Fahr- enheit, we are moving from a smaller to larger number, and should use 9/5 in the multiplication: 9/5 ¥ 62°F = 112°F Step 3: Subtract 40°: 112°F – 40°F = 72°F Thus, 22°C = 72°F Obviously, knowing how to make these temperature conversion calculations is useful. However, in practical (real world) operations, you may wish to use a temperature conversion table. 4.4.4.2 Milligrams per Liter (Parts per Million) One of the most common terms for concentration is milli- 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- centration of that solution is expressed simply as 15 mg/L. Very dilute solutions are more conveniently expressed in terms of micrograms per liter (µg/L). For example, a concentration of 0.005 mg/L is preferably written as its equivalent, 5 µg/L. Since 1000 µg = 1 mg, simply move the decimal point three places to the right when converting from mg/L to µL. Move the decimal three places to the left when converting from µg/L, to mg/L. For example, a 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 equivalent to 1000 g or 1,000,000 mg. Therefore, if we dissolve 1 mg of a substance in 1 L of H 2 O, we can say that there is 1 mg of solute per 1 million mg of water, or in other words, 1 part per million (ppm). Note: For comparative purposes, we like to say that 1 ppm is analogous to a full shot glass of water sitting in the bottom of a full standard swim- ming pool. Neglecting the small change in the density of water as substances are dissolved in it, we can say that, in gen- eral, a concentration of 1 mg/L is equivalent to 1 ppm. Conversions are very simple; for example, a concentration of 18.5 mg/L is identical to 18.5 ppm. 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 used, and the waterworks/wastewater operator should be familiar with both. 4.5 MEASUREMENTS: AREAS AND VOLUMES Water and wastewater operators are often required to cal- culate surface areas and volumes. Area is a calculation of the surface of an object. For example, the length and the width of a water tank can be measured, but the surface area of the water in the tank must be calculated. An area is found by multiplying two length measurements, so the result is a square measure- ment. For example, when multiplying feet by feet, we get square feet, which is abbreviated ft 2 . Volume is the calcu- lation of the space inside a three-dimensional object, and is calculated by multiplying three length measurements, © 2003 by CRC Press LLC or an area by a length measurement. The result is a cubic measurement, such as cubic feet (abbreviated ft 3 ). 4.5.1 AREA OF A RECTANGLE The area of square or rectangular figures (such as the one shown in Figure 4.1) is calculated by multiplying the measurements of the sides. A = L ¥ W (4.8) To determine the area of the rectangle shown in Figure 4.1, we proceed as follows: A = L ¥ W A = 12 ft ¥ 8 ft A = 96 ft 2 4.5.2 AREA OF A CIRCLE The diameter of a circle is the distance across the circle through its center, and is shown in calculations and on drawings by the letter D (see Figure 4.2). Half of the diameter — the distance from the center to the outside edge — is called the radius (r). The distance around the outside of the circle is called the circumference (C). In calculating the area of a circle, the radius must be multiplied by itself (or the diameter by itself); this process is called squaring, and is indicated by the superscript number 2 following the item to be squared. For example, the radius squared is written as r 2 , which indicates to multiply the radius by the radius. When making calculations involving circular objects, a special number is required — referred to by the Greek letter pi (pronounced pie); the symbol for pi is p. Pi always has the value 3.1416. The area of a circle is equal to the radius squared times the number pi. A = p ¥ r 2 (4.9) E XAMPLE 4.16 Problem: Find the area of the circle shown in Figure 4.2. Solution: A = p ¥ r 2 A = p ¥ 12.5 ft ¥ 12.5 ft A = 3.1416 ¥ 12.5 ft ¥ 12.5 ft A = 490.9 ft 2 At times, finding the diameter of a circular object is necessary, under circumstances that allow you to measure only the circumference (e.g., a pump shaft). The diameter and the circumference are related by the constant p: C = p ¥ Diameter or (4.10) 4.5.3 AREA OF A CIRCULAR OR CYLINDRICAL TANK If you were supervising a work team assigned to paint a water or chemical storage tank, you would need to know the surface area of the walls of the tank. To determine the tank’s surface area, visualize the cylindrical walls as a rectangle wrapped around a circular base. The area of a rectangle is found by multiplying the length by the width; in this case, the width of the rectangle is the height of the wall and the length of the rectangle is the distance around the circle, the circumference. The area of the sidewalls of a circular tank is found by multiplying the circumference of the base (C = p ¥ Diameter) times the height of the wall (H): A = p ¥ Diameter ¥ H (4.11) 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.) L W 8 ft 12 ft FIGURE 4.2 Circular shape showing diameter and radius. (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) C = ? r = 12.5 ft D = 25 ft C p © 2003 by CRC Press LLC For a tank with Diameter = 20 ft and H = 25 ft: To determine the amount of paint needed, remember to add the surface area of the top of the tank, which in this case we will say is 314 ft 2 . Thus, the amount of paint needed must cover 1570.8 ft 2 + 314 ft 2 = 1884.8 or 1885 ft 2 . If the tank floor should be painted, add another 314 ft 2 . 4.5.4 V OLUME C ALCULATIONS 4.5.4.1 Volume of Rectangular Tank The volume of a rectangular object (such as a settling tank like the one shown in Figure 4.3) is calculated by multi- plying together the length, the width, and the depth. To calculate the volume, you must remember that the length times the width is the surface area, which is then multi- plied by the depth. v = L ¥ W ¥ D (4.12) E XAMPLE 4.17 Problem: Using the dimensions given in Figure 4.3, determine the volume. Solution: Note: For many calculations involving water, we need to know the volume of the tank in gallons rather than 1 ft 3 contains 7.48 gal. 4.5.4.2 Volume of a Circular or Cylindrical Tank A circular tank consists of a circular floor surface with a cylinder rising above it (see Figure 4.4). The volume of a circular tank is calculated by multiplying the surface area times the height of the tank walls. E XAMPLE 4.18 Problem: If a tank is 20 ft in diameter and 25 ft deep, how many gallons of water will it hold? Solution: In this type of problem, calculate the surface area first, multiply by the height and then convert to gallons. FIGURE 4.3 Rectangular settling tank illustrating calcula- tion of volume. (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) ADH Aftft Aftft Aft =¥ ¥ =¥ ¥ =¥¥ = p p 20 25 3 1416 20 25 1570 8 2 . . vL W D vA D vftftft vft =¥ ¥ =¥ =¥¥ = 12 6 6 432 3 12 ft L D 6 ft W 6 ft FIGURE 4.4 Circular or cylindrical water tank. (From Spell- man, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.) H = 25 ft r = 10 ft r Diameter ft ft Ar Aftft Aft = ∏ = ∏ = =¥ =¥ ¥ = 220 210 10 10 314 2 2 p p vA H vft ft vftgal ft gal =¥ =¥ =¥ = 314 25 7850 7 5 58 875 2 33 © 2003 by CRC Press LLC [...]... sedimentation tank handles a flow of 7.5 MGD The tank is 70 ¥ 20 ¥ 15 ft and 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 and a depth of 12 ft Find the detention time 4. 46 A filter box is 40 ¥ 20 ft, which also includes the sand area If the influent valve is shut, © 2003 by CRC Press LLC 4. 47 4. 48 4. 49 4. 50 4. 51 the water drops 4. 0 in./min... 3 /4 ¥ 5/6 = 4. 6 4. 7 4. 8 4. 9 4. 10 4. 11 4. 12 4. 13 4. 14 4.15 4. 16 4. 17 4. 18 4. 19 3/7 ∏ 2/3 = What is the fraction equivalent of 0.625? What is the decimal equivalent of 3 /4? Write 10,000,000 as powers of ten What is the area of a rectangle 9 ¥ 30 ft? What is the volume of a tank 25 ¥ 60 ¥ 8 ft deep? A pipe has a diameter of 8 in Water is flowing through it at 4 ft/min How much water is passing through... Volume ft 3 ¥ 7 .48 gal ft 3 ¥ 24 h d (4. 17) Flow (gal d ) EXAMPLE 4. 55 Problem: A settling tank has a volume of 40 ,000 ft.3 What is the detention time in hours when the flow is 4. 35 MGD? 4. 9 CHEMICAL DOSAGE CALCULATIONS Chemicals are used extensively in water and wastewater treatment plant operations Water and wastewater treatment plant operators add chemicals to various unit processes for slime-growth control,... when 49 0 lb/d of chlorine is added to an effluent flow of 11.0 MGD? Solution: Dose ( mg L ) = 49 0 lb d 11.0 MG ¥ 8. 34 lb mg L MG = 5. 34 mg L © 2003 by CRC Press LLC BASIC MATH QUESTIONS 4. 1 Fractions are used to express a portion of a _ 4. 2 In 1.2, 1.6, 1.9, 1.8, 1.0, 1.5, what is the mean? 4. 3 [(25 – 4 – 6) ∏ (3 ¥ 5)] + 4 ¥ 3 = 4. 4 2/3 is equal to how many ninths (x/9): 4. 5 3 /4 ¥ 5/6 = 4. 6 4. 7 4. 8... Diameter 2 ¥ L ¥ 7 .48 gal ft 3 2 = 0.785 ¥ 833 ¥ 833 ¥ 1800 ft ¥ 7 .48 gal ft 3 where B1 B2 L D = = = = = 73 34 gal distance across the bottom distance across water surface channel length depth of water and wastewater EXAMPLE 4. 25: PIPE VOLUME Problem: Problem: Approximately how many gallons of wastewater would 800 ft of 8-in pipe hold? Determine the volume of wastewater (in gallons) in a section of trapezoidal... 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? 40 Solution: Chemical Feed Rate = Dose ( mg L ) ¥ Flow (MGD) ¥ (8. 34) = 40 mg L ¥ 5 MGD ¥ (8. 34) = 1668 lb d of alum © 2003 by CRC Press LLC 65 gal ¥ 8. 34 lb gal = 542 .7 lb H 2 O ( 542 .7 + 40 ) ¥ 100 = 6.9% Solution 4. 17 FILTRATION In waterworks operation (and to an increasing degree in wastewater treatment) ,... factor 40 , 000 ft 3 ¥ 7 .48 gal ft 3 ¥ 24 h d 4. 35 MGD ¥ 1, 000, 000 gal MG = 1.7 h 4. 8.1.3 Detention Time in Minutes HDT (min) = 4. 9.1 CHLORINE DOSAGE (4. 18) ( ) Tank Volume ft 3 ¥ 7 .48 gal ft 3 ¥ 144 0 min d Flow (gal d ) EXAMPLE 4. 56 Problem: A grit channel has a volume of 1 240 ft.3 What is the detention time in minutes when the flow rate is 4. 1 MGD? Solution: DT ( min ) = 1 240 ft ¥ 7 .48 gal ft ¥ 144 0... added pounds of chemicals to pounds of water Recall that 1 gal of H2O = 8. 34 lb By multiplying the number of gallons by the 8. 34 factor, we can find pounds EXAMPLE 4. 74 Problem: Approximately 40 lb of soda ash are added to 65 gal of 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... particular direction Water in a tank exerts force down on the bottom and out of the sides Pressure acts in all directions A marble at a water depth of one foot would have 0 .43 3 psi of pressure acting inward on all sides Key Point: 1 ft of head = 0 .43 3 psi 1 lb of water 2.31 ft 0 .43 3 lb of water 1 ft 1 sq in AREA 1 ft water = 0 .43 3 psi 1 sq in AREA 1 psi = 2.31 ft water FIGURE 4. 12 Shows the relationship... diagram similar to Figure 4. 6 v (gal ) = 0.785 ¥ Diameter ¥ D ¥ 7 .48 gal ft = 2713 ¥ 7 .48 gal ft 3 ¥ 3.785 L gal v=L¥W¥D Problem: Determine the volume of wastewater (in ft3) in the section of rectangular channel shown in Figure 4. 7 when the wastewater is 5 ft deep 1000 ft 1800 ft 5 ft 4 ft FIGURE 4. 7 Open channel (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol 1–3, . II Water/ Wastewater Operations: Math and Technical Aspects © 2003 by CRC Press LLC Water and Wastewater Math Operations To operate a waterworks and/ or a wastewater treatment plant, and. 5/9: 5/9 ¥ 260°F = 144 .4 C Step 3: Now subtract 40 °: 144 .4 C ¥ 40 °C = 1 04. 4°C Therefore, 220°F = 1 04. 4°C EXAMPLE 4. 15 Problem: Convert 22°C to Fahrenheit. Step 1: Add 40 °F: 22°F + 40 °F = 62°F Step. Table 4. 1. 4. 4 TYPICAL WATER AND WASTEWATER MATH OPERATIONS 4. 4.1 A RITHMETIC A VERAGE ( OR A RITHMETIC M EAN ) AND M EDIAN During the day-to-day operation of