Advanced Waterworks Mathematics, 2019a

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Calculating the length of the weir is required in order to calculate the WOR. Sometimes the weir can be a circular structure requiring the circumference to be calculated in order to find the actual length. Other times it is a linear structure, in which case the length would be known. Example: What is the weir overflow rate through a 12.85 MGD treatment plant if the weir is 90 feet long? (Express your answer in MGD/ft and gpm/ft). The values provided in the problem statement are in MGD and feet. Therefore, to calculate the weir overflow rate in MGD/ft, substitute the values for flow and weir length into the equation. Weir Overflow Rate (MGD/ft) = 12.85 MGD 90 ft 0.142778 MGD/ft You can approach calculating the solution in gpm/ft in two ways. Option #1: One way is to convert the MGD/ft solution previously calculated to gpm/ft. 0.142778 MG 1,000,000 gal 1 day 1 hour D 1 MG 24 hour 60 min = ft 99.15 gpm ft Option #2: You can convert the treatment plant flow rate provided in the problem statement from MGD to gpm and then calculate the weir overflow rate. 12.85 MG 1,000,000 gal 1 day 1 hour = 8,923.611 gpm D 1 MG 24 hour 60 min Weir Overflow Rate (gpm/ft) = 8,923.611 gpm 99.15 gpm 90 ft ft Note that the final answer is the same using either option. Example: A drainage channel has a 32-foot weir and a weir overflow rate of 14.5 gpm/ft. What is the daily flow expressed in MGD? Substitute the values provided in the question into the Weir Overflow Rate formula and solve for the unknown flow rate. 118 | A dvanced Waterworks Mathematics
Weir Overflow Rate (gpm/ft) = ? gpm 32 ft 14.5 gpm/ft Rearrange the terms and solve for the flow rate in gpm. 14.5 gpm ? gpm = 32 ft = 464 gpm ft Now you can convert the gpm to MGD. 464 gal 1 MG 24 hour 60 min = 0.66816 MGD = 0.67 MGD min 1,000,000 gal 1 day 1 hour Example: What is the length of a weir if the daily flow is 4.3 MG and the weir overflow rate is 52 gpm/ft? The units of the values provided in the problem statement, do not match. You have MGD for the daily flow and gpm/ft for the weir overflow rate. Therefore, in order to solve for length, either MGD must be converted to gpm or gpm/ft must be converted to MGD/ft. Both will provide you with the same final answer. Converting 4.3 MGD to gpm. 4.3 MG 1,000,000 gal 1 day 1 hour = 2,986.1 gpm D 1 MG 24 hour 60 min Now you can substitute the values into the Weir Overflow Rate formula and solve for the weir length. Weir Overflow Rate (gpm/ft) = 2,986.1 gpm ? ft 52 gpm/ft Rearrange the terms and solve for the weir length in feet. Weir Length (ft) = 2,986.1 gpm 52 gpm/ft 57.425 ft = 57.43 ft Converting 52 gpm/ft to MGD/ft. 52 gal 1 MG 24 hour 60 min min 1,000,000 gal 1 day 1 hour = ft 0.07488 MGD ft 119 | A dvanced Waterworks Mathematics
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1 | A dvanced Waterworks Mathematic
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Acknowledgements College of the Can
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Unit 10 201 10.1 Horsepower and Eff
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Multiple units may need to be conve
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Practice Problems 1.1a Show how eac
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Word problems may involve unit conv
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6. A pipe flows at a rate of 6.3 cf
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6. A well flows at a rate of 1,250
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Well 2: Total Flow in MGD 1,000 gal
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Practice Problems 1.2 1. A water ut
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3. A water utility has 12,300 servi
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The shape of a tank is often a cyli
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Circles A circle is a round figure
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Example: What is the area of an 813
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Trapezoids Trapezoids are four-side
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Figure 2.9 9 Circumference of a Cir
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The formula for the area of a spher
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Example: What is the entire interio
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Practice Problems 2.1 1. What is th
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8. What is the entire interior surf
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5. A standpipe needs to be painted.
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Then, calculate the volume of the t
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2 1 2 Trapezoidal Prism = H L 8.5
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Exercise 2.2 1. What is the volume
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2.3 AREA AND VOLUME WORD PROBLEMS W
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Figure 2.20 19 Step 4: Put the info
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Step 3: Make a sketch to clarify th
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Figure 2.22 21 When looking at a pr
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Calculate the volume of the cylinde
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4. A water tank truck delivered 30
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Exercise 2.3 1. A water utility ope
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6. A private contractor needs water
Page 67 and 68: 2.4 FLOW RATE Flow rate is the meas
Page 69 and 70: Key Terms circumference - the dist
Page 71 and 72: 4. What is the diameter of a pipe t
Page 73 and 74: 4. What is the diameter of a pipe t
Page 75 and 76: Substance Specific Gravity Weight C
Page 77 and 78: 5. The specific gravity of 25% Alum
Page 79 and 80: 5. The specific gravity of 25% Alum
Page 81 and 82: To convert between a percentage and
Page 83 and 84: And now we are moving to the row fo
Page 85 and 86: Practice Problems 3.2 1. An 87.5% c
Page 87 and 88: Exercise 3.2 Solve the following pr
Page 89 and 90: 3.3 MIXING AND DILUTING SOLUTIONS T
Page 91 and 92: Total Volume of the new solution is
Page 93 and 94: 3. A chlorine storage tank that is
Page 95 and 96: 3. A chlorine storage tank that is
Page 97 and 98: The following charts or Pie Wheels
Page 99 and 100: At only a 10% concentration, you wi
Page 101 and 102: Now the question is asking how many
Page 103 and 104: Key Terms chlorine demand - the
Page 105 and 106: 3. How many pounds of 24.5% sodium
Page 107 and 108: 7. A water treatment operator adjus
Page 109 and 110: 11. An operator added 422 gallons o
Page 111 and 112: 3. How many pounds of 12.5% sodium
Page 113 and 114: 7. A water treatment operator adjus
Page 115 and 116: 11. An operator added 275 gallons o
Page 117: Weirs can either be sharp-crested o
Page 121 and 122: Key Terms weir - an overflow stru
Page 123 and 124: 5. A treatment plant processes 8.4
Page 125 and 126: 5. A treatment plant processes 15 M
Page 127 and 128: UNIT 6 6.1 DETENTION TIME Detention
Page 129 and 130: D = t Volume Flow gallons million
Page 131 and 132: Rearranging the terms in the detent
Page 133 and 134: 3. A water utility engineer is desi
Page 135 and 136: 7. A 70-foot diameter, 35-foot-deep
Page 137 and 138: 3. A water utility engineer is desi
Page 139 and 140: 7. The chlorine residual decay rate
Page 141 and 142: 11. A circular clarifier processes
Page 143 and 144: Although filtration rates are commo
Page 145 and 146: Key Terms contact time - detenti
Page 147 and 148: 4. A water treatment plant processe
Page 149 and 150: Exercise 6.2 1. A water treatment p
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Page 153 and 154: Pin It! Misconception Alert Sometim
Page 155 and 156: Next, there are two crucial terms t
Page 157 and 158: Finding the Correct CT Table Typica
Page 159 and 160: and resulting disinfection inactiva
Page 161 and 162: plant maintains a chloraminated res
Page 163 and 164: Practice Problems 7.1 1. What is th
Page 165 and 166: 5. A conventional water treatment p
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4. A conventional water treatment p
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6. A direct filtration water treatm
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UNIT 8 8.1 PRESSURE Pressure is the
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From this example, it is clear that
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5. Two houses are served by a nearb
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Exercise 8.1 Solve the following pr
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8.2 HEAD LOSS As water travels thro
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Friction will need to be added at t
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3. A well located at 200 feet above
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Exercise 8.2 Solve the following pr
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UNIT 9 9.1 WELL YIELD, SPECIFIC CAP
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Cone of Depression The triangular s
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Practice Problems 9.1 1. A well has
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7. A well has a specific capacity o
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4. A well has an hour meter attache
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UNIT 10 10.1 HORSEPOWER AND EFFICIE
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The above equation is used to calcu
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Practice Problems 10.1 1. What is t
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Exercise 10.1 Calculate the require
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10.2 HEAD LOSS AND HORSEPOWER As di
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Practice Problems 10.2 1. A well is
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Exercise 10.2 Calculate the followi
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10.3 CALCULATING POWER COSTS It is
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93.25 kW $ 0.088 8.33 hr $ 68.35
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Practice Problems 10.3 1. A well fl
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4. A well draws water from an aquif
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7. Complete the table below based o
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Exercise 10.3 Solve the following p
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4. A well draws water from an aquif
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7. Complete the table below based o
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UNIT 11 11.1 PER CAPITA WATER USE A
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That’s a lot of gallons. This is
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Green Building Code) has resulted i
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7. A water district has a goal of 1
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4. How much water would a family of
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UNIT 12 12.1 BLENDING AND DILUTING
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To determine the percentage of Sour
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Practice Problems 12.1 1. A well ha
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5. Two wells need to achieve a dail
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3. A well with a PCE level of 7.5 u
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UNIT 13 13.1 SCADA AND THE USE OF M
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ottom to what is referred to as the
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mA (reading) - mA (offset) = percen
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4. A water tank is 52 ft tall and h
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8. A chemical injection system is m
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4. A water tank is 45 ft tall and h
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8. A chemical injection system is m
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mean extra funds to pay for securit
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Per the table, the total annual bud
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Pumps and Motors: $72,000 x% $6,2
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Variable Costs Reason 273 | A dvanc
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Practice Problems 14.1 1. A utility
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3. A utility has annual operating e
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7. A small water company has a tota
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2. A pump that has been in operatio
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5. A 250 hp well operates 9 hours a
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PRACTICE PROBLEM SOLUTIONS Practice
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3.1 cf 3,600 sec 10 hr 7.48 gal 30
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1,346,400 gal 1 AF 365 day day 3
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Practice Problems 2.3 1. A water ut
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3 22 ft 217,800.13 ft = depth ft
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Q (cfs) 3.18 cfs A (ft ) = = = 1.17
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5. The specific gravity of 25% Alum
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To convert ppb to ppm, you divide p
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Practice Problems 3.3 1. How many g
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Practice Problems 4.1 1. How many g
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6.35 MG 8.34 lbs 2,914.5 lbs ppm
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8. A water utility produced 6,000 A
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72,327.816 lbs 72,327.816 lbs 72,32
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Weir Length (ft) = 4,791.67 gpm 41
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quality objectives of a certain tre
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12 MG 1,000,000 gal 1 day 1 hour
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8. A circular clarifier processes 9
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4. A water treatment plant processe
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4,700,000 gal 1 day ? min = 1,554.
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4. A conventional water treatment p
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Now look at Table C-7 Inactivation
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500 mg/L min CT is required. Locati
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Location and Type of Disinfection T
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2 2 D = 0.33121019 ft 12 in D = 0.5
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5. A water utility has two differen
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gpm Specific Capacity = ft 717.5 gp
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3. A well with pumps located 46 ft
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(630 gpm)(130 ft) 65 hp = (3,960)(?
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electrical rate of $0.111 per kW-Hr
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(flow rate in gallons per minute)(t
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2 flushes per family member 6 famil
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Practice Problems 12.1 1. A well ha
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-137.5 mg/L ? mg/L = = 275 mg/L - (
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Practice Problems 13.1 1. A 4-20 mA
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7. A water utility uses a 4-20 mA s
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NEW PUMP: (450 gpm)(65 ft) Motor hp
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Total operational cost per year. $2
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CT TABLES APPENDIX 367 | A dvanced
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Advanced Waterworks Mathematics, 2019a

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