Review of Exam Wastewater Math Problems (Grades 2-4)

Review of Exam Wastewater Math Problems (Grades 2-4) 1. A pump runs continuously for 8 hrs at full speed and delivers 9,350 gallons, what is the capacity (pumping rate) of the pump in gallons per minute? Pumping Rate, gpm = Volume, gals Time, mins a. Change hrs to mins: 8 hrs x 60 mins = 480 mins hr b. Solve for gpm discharging pumping rate using the above formula: Pumping Rate, gpm = 9350 gal 480 mins Pumping Rate, gpm = 19.5 gpm 2. Approximately how many gallons of wastewater would 800 feet of 8-inch pipe hold? a. x (the unknown) = number of gallons pipe will hold The pipe s volume in ft 3 must be calculated first and then the ft 3 of pipe is converted to gallons using the capacity formula: b. Solve for volume of pipe in ft 3 : Gallons = Number of calculated ft 3 x 7.48 gal/ft 3 Convert 8 in to ft: 8 inches = 0.67 ft 12 inches/ft Volume, ft 3 = 0.785 x (D, ft) 2 x length of pipe, ft = 0.785 x (0.67 ft) 2 x 800 ft = [0.785 x (0.67 ft x 0.67 ft)] x 800 ft = [0.785 x 0.45 ft 2 ] x 800 ft = 0.35 ft 2 x 800 ft = 280 ft 3 Gallons = # of ft 3 x 7.48 gal/ft 3 Gallons = 280 ft 3 x 7.48 gal/ft 3 Gallons = 2094.4 gals! 1

3. A sewer pipe is 265 feet long and has a diameter of 10-inches. The pipe is to be treated with a root-killing chemical containing a 250-mg/L concentration. How many pounds of chemical are needed? a. x (unknown) = lbs. of chemical needed to kill the roots in the pipe The lbs formula is the main formula needed to solve this problem Lbs. Chemical = (pipe capacity, MG) x (Chemical conc. mg/l) x 8.43 lbs/gal b. First the ft 3 volume of pipe and then convert the ft 3 to gallons and then to MG capacity. Convert 10 inch-diameter pipe to ft diameter: 10 inches = 0.83 ft 12 inches/ft Volume, ft 3 = 0.785 x (D, ft) 2 x length of pipe, ft = [0.785 x (0.83 ft) 2 ] x 265 ft = [0.785 x (0.83 ft x 0.83 ft)] x 265 ft = [0.785 x 0.69 ft 2 ] x 265 ft = 0.54 ft 2 x 265 ft = 143.54 ft 3 Capacity, gals = # of ft 3 x 7.48 gal/ft 3 Capacity, gals = 143.54 ft 3 x 7.48 gal/ft 3 Capacity, gals = 1073.68 gals MG = 1073.68 gals = 0.001074 MG 1,000,000 gals/mg c. Solve for lbs of chemical using the lbs. formula: lbs. Chemical = 0.001074 MG x 250 mg/l x 8.34 lbs/gal lbs. Chemical = 2.24 lbs of chemical d. There is another formula that can be used to solve for lbs of chemical Lbs. Chemical = gallons treated x chemical concentration, mg/l 120,000! 2

= 1073.68 gals x 250 mg/l 120,000 = 268420 = 2.24 lbs of chemical 120,000 4. Find the gallon capacity of a secondary (also can be called the final clarifier) that has a diameter of 30 feet, a water depth of 12 feet. The cone portion of the tank is 4.5 feet deep. a. x(unknown) = gallon capacity of tank b. Find the volumes of both the cylinder part and cone part of clarifier in ft 3 and add ft 3 together then convert total ft 3 to gallons. Find Cylinder Volume: 1. Find Cylinder depth, ft = 12.5 ft 4.5 ft = 7.5 ft 2. Volume, ft 3 = 0.785 x (D, ft. 2 x depth of cylinder, ft = 0.785 x (30 ft) 2 x 7.5 ft = [0.785 x (30 ft x 30 ft)] x 7.5 ft = [0.785 x 900 ft 2 ] x 7.5 ft = 706.5 ft 2 x 7.5 ft = 5299 ft 3 Find Cone Volume: 1. Volume, ft 3 = 1/3 x [0.785 x (D, ft) 2 ] x Cone depth, ft = 1/3 x [706.5 ft 2 x 4.5 ft] = 1/3 x (3179.25 ft 3 ) = 1059.75 or 1060 ft 3 Add Ft 3 Volumes of cylinder and cone: 5299 ft 3 + 1060 ft 3 = 6359 ft 3 Find Gallon Capacity of Clarifier Capacity, gals = Total # of ft 3 in clarifier x 7.48 gal/ft 3 = 6359 ft 3 x 7.48 gal/ft 3 = 47,565 gals 5. How many pounds of chlorine gas are necessary to treat 8,000,000 gallons of wastewater at a dosage of 4 mg/l?! 3

a. x = lbs. of Chlorine needed to treat wastewater The lbs formula is the main formula needed to solve this problem Lbs. Chlorine = (Vol. of wastewater, MG) x (Chlorine dose, mg/l) x 8.43 lbs/gal Convert gallons of wastewater to MG: 8,000,000 gals = 8.0 MG 1,000,000 gal/mg Continued on next page: b. Find lbs of chlorine needed Lbs. Chlorine = (Vol. of wastewater, MG) x (Chlorine dose, mg/l) x 8.43 lbs/gal = 8.0 MG x 4 mg/l x 8.34 lbs/gal = 266.9 or 267 lbs of chlorine 6. A WWTP uses 1-ton cylinders of chlorine for disinfection. The average daily chlorine demand is 8 mg/l requiring an average daily dosage of 11 mg/l. How many cylinders will the plant need for the month of May? The average plant flow for the month of May is 12 MGD. Data needed to solve this problem: ADF in May = 12 MGD Cl dose = 11 mg/l Days in May = 31 days Ton cylinder = 2,000 lbs a. x(unknown) = Number of ton chlorine cylinders used in May b. Find lbs Cl/day used = 12 MGD x 11 mg/l x 8.34 lbs/gal = 1100.88 c. Find lbs. Chlorine used in May = 1100.88 lbs Cl/day x 31 days/may = 34,127.28 lbs used/may d. #Ton Cl cylinders/may = 34,127.28 lbs used/may 2000 lbs/cylinder = 17.06 or 18 cylinders used in May 7. How may pound of chlorine has to be added per day to maintain a chlorine residual of 0.5 mg/l with a chlorine demand of 19.5 mg/l and a plant flow of 5 MGD? a. X(unknown) = Cl dose, lbs/day! 4

b. Find Cl dose in mg/l: Cl dose, mg/l = Cl demand, mg/l + Cl residual, mg/l = 19.5 mg/l + 0.5 mg/l = 20.0 mg/l c. Find lbs/day of Chlorine dose Lbs. Cl dose/day = Flow (MGD) x Chlorine Dose (mg/l) x 8.34 lbs/gal = 5 MGD x 20.0 mg/l x 8.34 lbs/gal = 834 lbs Cl dose/day 8. Calculate the weir-loading rate for a primary clarifier that has a total weir length of 520 feet and receives a flow of 5 MGD. a. x(unknown) = Weir Loading rate also called weir overflow rate in gpd/ft First, Convert 5 MGD to gpd: 5 MGD x 1,000,000 gal/day = 5,000,000 gpd MGD b. Formula: WLR, gpd/ft = Flow, gpd Weir Length, ft. WLR, gpd/ft = 5,000,000 gpd 520 ft = 9615.4 gpd/ft 9. A primary clarifier has a diameter of 75 ft with a depth of 15 ft and a daily flow of 5.0 MGD that passes through the tank. What is the detention time (in hours) of the tank? a. x(unknown) = Detention time, hrs b. Formula: DT, hrs = Clarifier Vol., gals x 24 hr/day Flow, gal/day c. First, find clarifier volume in ft 3 then convert to gals Volume, ft 3 = 0.785 x (D, ft.) 2 x depth of clarifier, ft = 0.785 (75 ft.) 2 x 15 ft = [0.785 x (75 ft x 75 ft)] x 15 ft = [0.785 x 5625 ft 2 ] x 15 ft! 5

= 4415.625 ft 2 x 15 ft = 66,234.375 ft 3 Capacity, gals = 66,234.375 ft 3 x 7.48 gal/ft 3 = 495,433.13 gals d. Convert 5 MGD to gal/day: 5 MGD x 1,000,000 gal/day = 5,000,000 gpd MGD e. Find detention time of clarifier: DT, hrs = 495,433.13 gals x 24 hr/day 5,000,000 gal/day =.099 day x 24 hr/day = 2.4 hrs 10. A pump delivers 250,000 gallons per day at a static head of 310 feet. Calculate the pressure equivalent to this head expressed as pounds per square inch (psi)? a. x(unknown) = psi of static head b. Conversion Formula: #psi = Number of feet 2.31 ft/psi c. Solve for number of psi # psi = 310 ft 2.31 ft/psi = 134.2 psi 11. Given the following information calculate the BOD value of this sample. Initial Sample DO = 8.2 mg/l Final Sample DO = 4.9 mg/l Total Sample Volume = 300 ml Amt. of Sample Used = 10 ml a. x(unknown) = BOD, mg/l of sample b. Formula: BOD, mg/l = Initial BOD, mg/l Final BOD, mg/l Amt. of Sample, ml 300 ml! 6

BOD, mg/l = 8.2 mg/l 4.9 mg/l 10 ml_ 300 ml = 3.3 mg/l 0.03 = 110 mg/l BOD 12. Given the following information determine the percent volatile suspended solids of this sample. Wt. of dish = 21.03 grams Wt. of dish + wet sample = 54.12 grams Wt. of dish + dry sample = 22.60 grams Wt. of dish + ash = 21.75 grams a. x (unknown) = % Volatile Solids b. Formulas: Dry Solids wt. grams = (Wt. of dish, g + Dry sample, g) Wt. of Dish, g Ash wt. = (wt. of dish + ash wt.) Wt. of Dish, g Volatile Solids, % = Dry solids wt. g Ash wt. g x 100 Dry solids wt. g Volatile Solids, % = (22.60 g 21.03 g) (21.75 g 21.03 g) x 100 22.60 g 21.03 g = 1.57 0.72 x 100 1.57 = 0.85 x 100 1.57 = 0.54 x 100 = 54%! 7

13. A drying bed dimensions are 50 feet in length and 20 feet in width. A bed of sludge has been drawn to a depth of 18 inches. The sludge contains 3.8% solids. How many pounds of solids are in the bed? a. x(unknown) = lbs of solids in drying bed b. Formula to solve for lbs of solids: Lbs of solids = MG (capacity of drying bed) x mg/l (conc. of solids in sludge) x 8.34 lbs/gal c. Find Volume of sludge in drying bed in ft 3 converts to gallons and then convert to MG Convert the 18 inch depth of sludge to feet of sludge: 18 inches = 1.5 ft 12 in/ft Volume of sludge, ft 3 = Length, ft x Width, ft x Depth, ft = 50 ft x 20 ft x 1.5 ft = 1500 ft 3 Capacity, gals = # ft 3 x 7.48 gal/ft 3 = 1500 ft 3 x 7.48 gal/ft 3 = 11220 gals MG = 11220 gals = 0.0112 MG 1,000,000 gals/mg d. Convert 3.8% solids to mg/l Conversion factor: 1% = 10,000 mg/l 3.8% x 10,000 mg/l = 38,000 mg/l 1% c. Calculate lbs of solids in drying bed: Lbs of solids = MG (capacity of drying bed) x mg/l (conc. of solids in sludge) x 8.34 lbs/gal Lbs of Solids = 0.0112 MG x 38,000 mg/l x 8.34 lbs/gal Lbs of Solids = 3555.8! 8

14. Given the following information, determine the excess solids in pounds that should be wasted from the aerated sludge system. Target F:M = 0.6 MLVSS = 3250 mg/l BOD loading = 20,500 lb/day Vol. of aeration basin = 2.0 MG a. X(unknown) = Lbs of solids (MLSS) must be wasted to achieve the target F:M ratio. b. Formulas Needed to Solve this Problem: 1. Lbs. MLVSS to waste = lbs MLVSS in Aerator lbs. MLVSS required in basin to meet target F:M 2. Lbs. MLVSS in Aerator = Aerator Vol. (MG) x MLVSS conc. mg/l x 8.34 lbs/gal 3. Rearrange F:M formula to calculate the lbs of MLVSS required in basin to meet the target F:M. F:M = lbs. BOD/day can be rearranged to the following formula: lbs MLVSS Lbs. MLVSS = lbs. BOD/day F:M (lbs BOD/day/lb MLVSS) c. Find lbs MLVSS already in basin: lbs MLVSS = 2.0 MG x 3250 mg/l x 8.34 lbs/gal = 54,210 lbs MLVSS d. Calculate lbs of MLVSS required in basin to meet target F:M using the rearranged F:M formula: lbs MLVSS = 20,500 lbs BOD/day 0.6 lbs BOD/day/lb MLVSS Lbs. MLVSS = 34,166.67 lbs MLVSS e. Calculate the lbs MLVSS to be wasted to meet target F:M lbs MLVSS to waste = 54,210 lbs MLVSS 34,167 lbs MLVSS = 20,043 This type of problem is on a Grade 4 and possibly on a Grade 3 exam! 9

15. A primary clarifier has a diameter of 80 feet with an incoming flow of 1440 gpm. Calculate the surface-settling rate (hydraulic loading) on the clarifier. a. X(unknown) = SSR, gpd/ft 2 b. Formula needed to solve problem: SSR, gpd/ft 2 = Flow, gal/day Area, ft 2 c. Calculate area of clarifier: A, ft 2 = 0.785 x (D, ft) 2 d. Convert 1440 gpm to gal/day gal/day = 1440 gal/min x 1440 min/day gal/day = 2,073,600 gal/day e. Solve for SSR, gpd/ft 2 SSR, gpd/ft 2 = 2,073,600 gal/day 5024 ft 2 = 0.785 x (80 ft x 80 ft) = 0.785 x 6400 ft 2 = 5024 ft 2 SSR, gpd/ft 2 = 412.74 gpd/ft 2 16. Determine the water horsepower delivered by a pump if the flow is 600 gpm and the TDH is 65 feet and the pump s efficiency is 80%. a. x(unknown) = whp delivered by pump b. Formula to solve problem: Whp = Flow, gpm x head, ft x 0.0% 3960 Convert pump efficiency of 80% to its mathematical equivalent: 80% = 80/100 = 0.80 c. Solve for whp Whp = 600 gpm x 65 ft 3960 Whp = 39,000 3960 Whp = 9.85! 10

17. A water tank needs to be painted. The tank is 15 feet tall and has a diameter of 50 feet. The bottom and top of the tank must also be painted. A gallon of paint will cover 200 sq. ft./gal. How many gallons of paint must be bought to paint the tank? a. x(unknown) = Number of gallons of paint needed to paint the tank b. Formula to solve problem: # of gallons paint = Total Tank Area, ft 2 c. First find total area, ft 2 of tank: 200 ft 2 /gal of paint 1. Find top & bottom area, ft 2 of tank: Area, ft 2 = 0.785 x (D, ft) 2 Total Area, ft 2 (top & bottom) = 1962.5 ft 2 x 2 = 3925 ft 2 2. Find wall area, ft 2 of tank: Circumference of tank, ft 2 = Length, ft 2 Height of tank, ft 2 = Width, ft 2 C, ft = 3.14 x Diameter, ft C, ft = 3.14 x 50 ft C, ft = 157 ft Area, ft 2 = L, ft (C) x W, ft (Ht) Area, ft 2 = 157 ft x 15 ft Area, ft 2 = 2,355 ft 2 3. Total Tank Area, ft 2 = 3925 ft 2 + 2,355 ft 2 = 6280 ft 2 d. Solve for number of gallons of paint: # of gallons of paint = 6280 ft 2 200 ft 2 /gal = 31.4 gals or 32 gals of paint = 0.785 x (50 ft x 50 ft) = 0.785 x 2500 ft 2 = 1962.5 ft 2! 11

18. A channel has the dimensions of 24 inches in width and 17 inches in depth. 13,500 gallons of sludge is wasted through this channel at a velocity of 1.6 ft/sec to maintain the desired mean cell residence time. The waste activated sludge flow meter is out of service, how long does it take to waste this amount of sludge? a. X(unknown) = Time, minutes it takes to waste 13,500 gallons of sludge b. Convert all inch dimensions in problem to feet: Width: ft = 24 inches Depth: ft = 17 inches 12-in/ft 12-in/ft = 2 ft = 1.42 ft c. Overall formula to use to solve problem is detention time: DT = Volume Flow DT, sec = Volume of sludge, ft 3 Flow, ft 3 /sec d. Convert 13,500 gals of sludge to ft 3 to cubic feet of sludge: Ft 3 = 13,500 gals = 1804.8 ft 3 7.48 gals/ft 3 e. Find flow rate of sludge through channel: Q, ft 3 /sec = Area, ft 2 x Velocity, ft/sec 1. Find area, ft 2 of channel: Area, ft 2 = Depth, ft x Width, ft = 1.42 ft x 2 ft = 2.84 ft 2 Q, ft 3 /sec = Area, ft 2 x Velocity, ft/sec = 2.84 ft 2 x 1.6 ft/sec = 4.54 ft 3 /sec d. Calculate the time it takes to move sludge through channel: Time, sec = Volume, ft 3 Sludge Flow, ft 3 /sec of Sludge Time, sec = 1804.8 ft 3 4.54 ft 3 /sec! 12

Time, sec = 397.5 sec Time, min = 397.5 sec = 6.6 minutes to waste 13,500 gals of sludge 60 sec/min 19. A plant has 5 grit channels. They are 20 ft long, 4 ft wide and 42 inches deep. The plant flow is 16 MGD. How many grit channels are needed; 2, 3, 4 or 5? a. x(unknown) = # of grit channels needed To find the number of grit channels needed, the velocity of the wastewater entering the plant must be calculated, then based on the information that grit settles out at an average velocity of 1.0 ft/sec the number of grit channels can be chosen. b. Formula to solve for velocity of wastewater entering the plant: Velocity, ft/sec = Q, ft 3 /sec Area, ft 2 c. Convert Q, flow rate of 16 MGD to ft 3 /sec: 16 MGD = 16,000,000 gals/day Gal/min = 16,000,000 gal = 11111.11 gal/min 1440 min/day Ft 3 /min = 11111.11 gal/min = 1485.44 ft 3 /min 7.48 gal/ft 3 Ft3/sec = 1485.44 ft 3 /min = 24.76 ft 3 /sec 60 sec/min Or you can use the conversion factor: 1.0 MGD = 1.55 ft 3 /sec Ft 3 /sec = 16 MGD x 1.55 ft 3 /sec = 24.8 ft 3 /sec d. Find area, ft 2 of channel: Area, ft 2 = Width, ft x Depth, ft = 4 ft x 3.5 ft = 14 ft 2 e. Solve for velocity of wastewater in channel:! 13

Velocity, ft/sec = Q, ft 3 /sec Area, ft 2 Velocity, ft/sec = 24.8 ft 3 /sec = 1.77 ft/sec 14 ft 2 Operator would choose 2 grit channels based on average grit channel velocity of 1.0 ft/sec 20. A WWTP has an average daily flow of 2.6 MGD with a daily dosage rate of 4 mg/l and a chlorine residual of 0.5 mg/l. The plant has 9 full cylinders. When the plant is down to 3 full cylinders, chlorine is reordered. How many days will the chlorine last before it has to be reordered? Assume a one-day delivery time. a. x(unknown) = Number of days before Chlorine has to be reordered Have to reorder chlorine when 3 full cylinders are left. 9 cylinders 3 cylinders = 6 cylinders are only used for chlorine dose b. Formula to solve problem: # of days of Chlorine feed = Total lbs of Cl in 6 cylinders Lbs. Cl used/day 1. Calculate lbs Cl/day feed Lbs Cl used/day = 2.6 MGD x 4.0 mg/l x 8.34 lbs/gal = 86.74 2. Calculate total amount of Chlorine in 6 one-ton cylinders 6 cylinders x 2000 lbs/cylinder = 12,000 lbs of chlorine c. Calculate the number of days of Chlorine will last before reorder # of days = 12,000 lbs Chlorine = 138.3 days (137 days with a one day delivery) 86.74 lbs Cl used/day 21. Estimate the velocity of wastewater flowing through a grit channel if a float travels 45 feet in one minute and 10 seconds. a. x(unknown) = Velocity of flow, ft/sec b. Formula to solve problem: Velocity, ft/sec = Distance traveled, ft! 14

Total Travel Time, sec 1. Convert time from minutes to seconds: 1 minute = 60 seconds 60 seconds + 10 seconds = 70 seconds Velocity, ft/sec = 45 ft_ = 0.64 ft/sec 70 sec 22. A tank is 15 feet in height and has a 30 ft diameter. What is the force at the bottom of the tank in psi? a. x(unknown) = PSI at bottom of tank b. Formula to solve problem: Conversion factor: 1.0 psi = 2.31 ft # of psi = 15 ft 2.31 ft/psi = 6.49 psi 23. How many pounds of chlorine are added to a 4 MGD plant flow with a chlorine demand of 19.5 mg/l is needed to maintain a chlorine residual of 0.3 mg/l? a. x(unknown) = lbs Cl used/day b. Formulas to use to solve problem: Lbs Cl dose/day = Flow (MGD) x Cl dose (mg/l) x 8.34 lbs/gal Cl dose, mg/l = Cl demand, mg/l + Cl residual, mg/l c. First, calculate mg/l of Chlorine dose: Cl dose, mg/l = Cl demand, mg/l + Cl residual, mg/l = 19.5 mg/l + 0.3 mg/l = 19.8 mg/l d. Calculate lbs Cl dose/day Lbs Cl dose/day = Flow (MGD) x Cl dose (mg/l) x 8.34 lbs/gal! 15

Lbs Cl dose/day = 4 MGD x 19.8 mg/l x 8.34 lbs/gal = 660.53 24. Digested sludge is being drawn from the secondary digester at a rate of 1000 gal/day with 3% solids content. How many gallons would have to be drawn at 6% solids to remain the same + 30 gallons either way? 1000 gals, 500 gals, 750 gals, or 1500 gals. a. x(unknown) = gallons of 6% sludge withdrawn from digester b. Formula to solve problem: Volume1 x Concentration1 = Volume2 x Concentration2 Volume1 = 1,000 gals Concentration1 = 3% Volume2 = X (the unknown) Concentration2 = 6% Rearrange above formula to solve for Volume2 (X): V2 = V1 x C1 C2 V2 = 1,000 gals x 3% 6% V2 = 3,000 gals = 500 gals 6 25. A wet well has a width of 20 ft, a length of 20 ft and a depth of 25 ft. The pump kicks on at 20 ft and stops when the water level reaches 10 ft. The pump ran for 5 minutes. How many gallons were pumped out of the well? a. x(unknown) = gallons of wastewater pumped from well b. Find ft of water pumped from well in 5 minutes: 20 ft 10 ft = 10 ft of water pumped c. Calculate Volume of wastewater pumped from well: Volume Wastewater, ft 3 = L, ft x W, ft, x Water Drop, ft =20 ft x 20 ft x 10 ft = 4000 ft 3! 16

d. Calculate gallons of wastewater pumped: Capacity, gals = # ft 3 x 7.48 gal/ft 3 = 4000 ft 3 x 7.48 gal/ft 3 = 29,920 gals 26. A secondary clarifier 85 feet in diameter receives a primary effluent flow of 2.8 MGD and a RAS flow of 0.75 MGD. If the MLSS concentration is 3510 mg/l, what is the solids loading rate on the clarifier? a. x(unknown) = Solids loading rate on clarifier expressed as lbs SS added/day/ft 2 b. Formula to solve problem: lbs SS added/day/ft 2 = lbs SS added/day Area, ft 2 Note: SLR is only calculated on secondary clarifiers. The plant flow and the RAS flow must be added together when calculation lbs SS added/day. NOTE: If the plant flow is only given in the problem, it is assumed that the RAS flow is part of the given total flow entering the secondary clarifier. When the flows are separated out in the problem, they must be added together 1. Add the two flows that enter the sec. clarifier: 2.8 MGD + 0.75 MGD = 3.55 MGD flow into sec. clarifier 2. Calculate lbs SS added /day Lbs SS added/day = 3.55 MGD x 3510 mg/l x 8.34 lbs/gal = 103,920.6 c. Calculate area of sec. clarifier: Area, ft 2 = 0.785 x (D, ft) 2 = 0.785 x (85 ft) 2 = 0.785 x (85 ft x 85 ft) = 0.785 x 7225 ft 2 = 5671.6 ft 2 d. Calculate SLR lbs SS added/day/ft 2 = 103920.6 lbs SS added /day 5671.6 ft 2! 17

lbs SS added/day/ft 2 = 18.3 lbs SS added/day/ft 2 27. A single-piston reciprocating pump has a 6-inch diameter piston with an 8-inch length of stroke. If it makes 20 discharges strokes per minute, calculate the pumping rate in gallons per minute. a. x(unknown) = pumping rate, gpm b. Formula to solve problem: gal/min = # of gals x 20 strokes/minute 1.0 stroke c. Calculate volume of water discharge from cylinder per stroke from cylinder in ft 3 and then convert to gallons. 1. Convert all inch measurements to feet. Depth of stroke = 8 inches/stroke = 0.67 ft/stroke 12 in/ft Diameter of cylinder = 6 inches = 0.5 ft 12 in/ft 2. Calculate volume of water discharged from cylinder per stroke in ft 3. Volume of stroke, ft 3 = [0.785 x (D, ft) 2 ] x Depth of stroke, ft = [0.785 x (0.5 ft) 2 ] x 0.76 ft = [0.785 x (0.5 ft x 0.5 ft)] x 0.67 ft = [0.785 x 0.25 ft 2 ] x 0.67 ft = 0.196 ft 2 x 0.67 ft = 0.131 ft 3 3. Convert 0.131 ft 3 of water to gallons Capacity, gals = # ft 3 x 7.48 gal/ft 3 = 0.131 ft 3 x 7.48 gal/ft 3 = 0.982 gals d. Calculate gpm pumping rate: gal/min = # of gals x 20 strokes/minute 1.0 stroke! 18

Pumping Rate, gpm = 0.982 gals x 20 strokes/min 1.0 stroke Pumping Rate, gpm = 19.6 gpm 28. A plant permit requires that all plant effluent be dechlorinated to 0 mg/l using sulfur dioxide. Given the following current operating conditions, what is the minimum amount of sulfur required to dechlorinate the effluent? (Grade 4) Effluent Flow 10 MGD Chlorine Demand 4.5 mg/l Chlorine Dosage 6.0 mg/l Sulfur dioxide-to-chlorine ratio required 1.0:1.0 a. x = Lbs. SO2 required to dechlorinate b. Calculate the chlorine residual. Cl Residual, mg/l = Chlorine Dose, mg/l Chlorine Demand, mg/l Cl Residual, mg/l = 6.0 mg/l 4.5 mg/l Cl Residual, mg/l = 1.5 mg/l c. Calculate the lbs./day of Chlorine residual Lbs./day chlorine residual = 10 mg/l x 1.5 mg/l x 8.34 lbs./gal Lbs./day chlorine residual = 125.1 1 lb. of SO2 per 1 lb. Chlorine Residual Lbs. of SO2 needed = 125.1! 19

29. A rectangular clarifier is 20 ft wide, 60 ft long, 10 ft deep; it was designed for a surface-loading rate of 800 gpd/ft 2. The average daily flow is 0.8 MGD. The following is data from a typical flow chart; 7 am 450 gpm 1pm 1050 gpm 8 am 550 gpm 2 pm 950 gpm 9 am 650 gpm 3 pm 750 gpm 10 am 750 gpm 4 pm 650 gpm 11 am 850 gpm 5 pm 550 gpm 12 pm 950 gpm 6 pm 450 gpm Based on the above daily surface-loading rate, the design capacity was exceeded by approximately how many hours each day? Answer 6 hrs a. x(unknown) = How many hours was the designed SLR on clarifier exceeded that day. b. Formula to solve problem: Rearrange hydraulic loading formula to find designed flow in gpd then convert flow to gpm. HL, gpd/ft 2 = Flow, gpd rearranged to find flow, gpd; Area, ft 2 Flow, gpd = HL, gpd/ft 2 x ft 2 = 800 gpd/ft 2 x 1200 ft 2 = 960,000 gpd c. Convert 960,000 gpd to gpm: 960,000 gal/day = 666.67 or 667 gpm 1440 min/day! 20

d. Go back to gpm data in problem and determine how many hours 667 gpm were exceeded during the time period given. There are two time periods when the 667 gpm was in between the flows; 9 am 10:00 am and 3 pm 4 pm, you must take an average to come up with the last 6th hour. 9 am 10 am: 667 650 = 17 gpm 3 pm 4 pm: 750 667 = 83 gpm 17 gpm + 83 gpm = 100 gpm/2 = 50 gpm Since flows increase by 100 gpm per hour there is an average 50 gpm between the two time periods then there is a half an hour in morning 667 is exceeded and an half an hour in the afternoon that is exceeded and they must be added together to make the 6 hour. 30. A wastewater treatment plant used a polymer for solids dewatering centrifuges. The current polymer inventory is 3100 lbs. Standard procedure requires the polymer to be reordered when 1,000 lbs remain in inventory. Given the following data, when would the polymer have to be reordered? Feed Solids to centrifuges at 1400 lbs/day Polymer dosage = 15 lbs/ton of solids a. x(unknown) = Number of days polymer will last before reordering b. 3100 lbs 1,000 lbs = 2100 lbs of polymer used before reordering c. Calculate the polymer usage in lbs/day (set-up as a proportion) 15 lbs polymer = X, lbs/day polymer used 2000 lbs solids 1400 lbs solids/day 2000X = 15 x 1400 2000X = 21,000 X = 21,000 2000 X = 10.5 lbs/day of polymer used d. Calculate number of days polymer will last before reordering! 21

# days = 2100 lbs of polymer 10.5 lbs/day polymer used # days = 200 days 31. The cost of chlorine is $0.08 per pound, the average Chlorine dose is 5 mg/l with a chlorine residual of 1.5 mg/l with an average flow of 2.0 MGD and the effluent BOD concentration is 10 mg/l. How many pounds of chlorine are required per year? a. X(unknown) = lbs/year of chlorine used b. Formulas to solve problem: Lbs. Cl/day = Flow (MGD) x Cl concentration (mg/l) x 8.34 lbs/gal Lbs. Cl/day = lbs Cl/day x 365 days/yr c. First calculate lbs Cl/day and lbs Cl/year used: 1. Lbs. Cl/day = 2.0 MGD x 5 mg/l x 8.34 lbs/gal = 83.4 2. Lbs. Cl/year = 83.4 lbs. Cl x 365 days day year Lbs. Cl used/year = 30,441 32. A WWTP has two 1-ton cylinders of chlorine on hand. One cylinder contains 450 lbs and the other is full. The chlorine demand is 2.9 mg/l and the chlorine residual is 0.65 mg/l with an average flow of 2.6 MGD. How many days until the chlorine has to be reordered? a. x(unknown) = Number of days left until Chlorine has to be reordered. b. Calculate Chlorine on hand: 2000 lbs + 450 lbs = 2450 lbs of Chlorine! 22

c. Calculate Chlorine dose, mg/l; lbs Cl used/day; Number of days of Cl on hand before reordering 1. Cl dose, mg/l = Cl Demand, mg/l + CL Residual, mg/l = 2.9 mg/l + 0.65 mg/l = 3.55 mg/l 2. lbs. Cl used/day = 2.6 MGD x 3.55 mg/l x 8.34 lbs/gal = 76.98 3. # of days of Cl on = 2450 lbs of Cl hand before reorder 76.98 lbs Cl used/day = 31.83 or 32 days 33. Given the following data, how much MLVSS must be wasted to achieve the desired F/M ratio? Desired F/M ratio = 0.33 lb BOD/D/lb MLVSS Primary Effluent Flow = 3.6 MGD Current F/M ratio = 0.27 lb BOD/D/lb MLVSS Aeration Tank Volume = 0.65 MG Primary Effl. BOD conc. = 138 mg/l MLVSS conc. = 2840 mg/l a. X (unknown) = Lbs of solids (MLSS) must be wasted to achieve the target F:M ratio. b. Formulas Needed to Solve this Problem: 1. Lbs. MLVSS to waste = lbs MLVSS in Aerator lbs. MLVSS required in basin to meet desired F:M 2. Lbs. MLVSS in Aerator = Aerator Vol. (MG) x MLVSS conc. mg/l x 8.34 lbs/gal 3. Rearrange F:M formula to calculate the lbs of MLVSS required in basin to meet the target F:M. F:M = lbs. BOD/day can be rearranged to the following formula: lbs MLVSS Lbs. MLVSS = lbs BOD/day F:M (lbs BOD/day/lb MLVSS) c. Find lbs MLVSS already in basin: Lbs. MLVSS = 0.65 MG x 2840 mg/l x 8.34 lbs/gal = 15,395.64 lbs MLVSS! 23

d. Calculate lbs BOD/day entering the aeration basin: Lbs. BOD/day = 3.6 MGD x 138 mg/l x 8.34 = 4143.3 e. Calculate lbs of MLVSS required in basin to meet desired F:M Lbs. MLVSS = 4143.3 lbs BOD/day 0.33 lbs BOD/day/lb MLVSS (use desired F:M value) Lbs. MLVSS = 12,555.49 lbs MLVSS f. Calculate the lbs MLVSS to be wasted to meet desired F:M Lbs. MLVSS to waste = 15,395.64 lbs MLVSS 12,555.49 lbs MLVSS = 2840.15 34. Convert 20 o C to o F. a. X (unknown) = o F temperature b. Formula to solve problem: o F = [ o C x 9 ] + 32 5 o F = [20 x 9 ] + 32 5 o F = 36 + 32 o F = 68 o F Convert 75 o F to o C. a. X (unknown) = o C temperature b. Formula to solve problem: o C = ( o F 32) x 5/9 o C = (75 32) x 5/9 o C = 43 x 5/9 o C = 23.8 o C 35. What is the organic loading rate applied to a trickling filter in lbs. BOD/ day/1000 ft 3 units for a filter with a diameter of 50 ft, a depth of 5 ft., a flow of 0.2 MGD, and a filter influent BOD of 90 mg/l?! 24

a. x(unknown) = Organic Loading, BOD/day/1000 ft 3 unit or Tft 3 Note: 1000 ft 3 unit contains 1000 ft 3. This is similar to 1 MGD contains 1,000,000 gallons. I use Tft 3 (T stands for 1000) instead of 1000 ft 3 in the formula. b. Formula to solve problem: OL, lbs BOD/day/Tft 3 = lbs BOD/day Tft 3 c. Calculate lbs BOD/day and Tft 3 volume of trickling filter. 1. lbs BOD/day = 0.2 MGD x 90 mg/l x 8.34 lbs/gal = 150.12 2. Volume, ft 3 = [0.785 x (D, ft) 2 ] x Depth, ft = [0.785 x (50 ft) 2 ] x 5 ft = [0.785 x (50 ft x 50 ft)] x 5 ft = [0.785 x 2500 ft 2 ] x 5 ft = 1962.5 ft 2 x 5 ft = 9812.5 ft 3 3. Convert 9812.5 ft 3 to Tft 3 units: Tft 3 = 9812.5 ft 3 1000 ft 3 /Tft 3 Tft 3 = 9.8 Tft 3 d. Calculate Organic Loading on trickling filter OL, lbs BOD/day/Tft 3 = 150.12 lbs BOD/day 9.8 Tft 3 OL, lbs BOD/day/Tft 3 = 15.32 lbs BOD/day/1000 ft 3 unit! 25

36. Calculate the reduction in volatile solids if the percent volatile solids entering the digester are 70% and the percent leaving is 45%. a. x(unknown) = % reduction of solids in the digester b. Formula to solve problem: % Reduction VS = In Out x 100 In (In x Out) Note: Decimal form of the percentages must be used with this formula. If you use the whole numbers given in the problem, the answer will be a very tiny negative number! c. Calculate for % Reduction of Volatile Solids % Reduction VS = 0.70 0.45 x 100 0.70 (0.70 x 0.45) = 0.25 x 100 0.70 0.315! 26

= 0.25 x 100 0.385 = 0.6493 x 100 = 64.9 or 65% reduction in VS 37. 65,000 gal/day of sludge from a primary clarifier is pumped to a gravity thickener. If the solids content of the sludge is 4% and the surface area of the thickener is 850 ft 2, what is the solids loading rate into the thickener (lbs. solids/day/ft 2 )? a. X (unknown) = SRL, lbs solids/day/ft 2 b. Formula to solve problem; lbs SS/day/ft 2 = lbs SS/day Area of Thickener, ft 2 c. Convert 65,000 gal/day sludge to lbs/day of sludge (note: if sludge weight per gallon is not given in the problem use 8.43 lbs/gal) lbs/day of sludge = 65,000 gal/day x 8.34 lbs/gal = 542,100 lbs/day sludge d. Calculate solids portion in the sludge:! 27

Part (solids) = Whole (sludge) x O.O%(decimal percent of solids in sludge) = 542,100 lbs/day sludge x 0.04 = 21,684 lbs solids/day Note: 4% = 0.04 4 = 0.04 100 e. Calculate Solids loading rate on thickener: Lbs. SS/day/ft 2 = 21,684 lbs SS/day 850 ft 2 Lbs. SS/day/ft 2 = 25.5 lbs SS/day/ft 2 38. Calculate the RAS flow back into the aeration basin based on the MLSS to RAS ratio. Influent flow = 5.0 MGD MLSS conc. = 2500 mg/l RAS conc. = 8,000 mg/l Primary Clarif. Effl. SS conc. = 100 mg/l a. X (unknown) = RAS flow (mgd or gpm) needed to maintain 2500 mg/l solids in basin NOTE: Most pumping rates are given in the unit gpm instead of MGD. Look at the answers units to see what flow unit is given with the four choices given. b. Formula to solve problem: You must first find the flow in MGD (with this ratio formula) before it can be converted to gpm.! 28

R, MGD = (MLSS mg/l) x (Q, MGD) (RAS SS, mg/l) (MLSS, mg/l) Note: Plant Flow = Q R, MGD = 2500 mg/l x 5 MGD 8,000 mg/l 2500 mg/l = 12,500 mg/l. MGD 5500 mg/l (The dot means the mg/l and MGD units have been joined together as one unit, thus the mg/l unit can be cancelled out in the next step) = 12,500 MGD (mg/l unit is cancelled out, leaving only the flow unit, MGD) 5,500 = 2.27 MGD RAS Flow c. Convert 5 MGD to gpm: 2.27 MGD = 2,270,000 gal/day Gal/min = 2,270,000 gal/day 1440 min./day Gal/min = 1,576.40 gpm 39. Find the total volume in gallons of the final clarifiers. 2 rectangular clarifiers dimensions = 60 ft x 15 ft x 10 ft 2 round shape clarifiers: Diameters = 30 ft Clarifiers Ht = 12 ft Cone-shaped bottoms Ht = 3.5 ft a. Find the 2 rectangular clarifiers volumes in cubic feet and then convert cubic feet to gallons. Volume, Ft 3 = 60 ft x 15 ft x 10 ft = 9,000 ft 3 Total Volume in Ft 3 of both clarifiers = 9,000 ft 3 x 2! 29

= 18,000ft 3 Total gallons capacity for both clarifiers = 18,000 ft 3 x 7.48 gals/ft 3 = 134,640 gals b. Find the 2 circular clarifiers volumes in cubic feet and then convert cubic feet to gallons. The clarifiers have two shapes: the cylinder part and the cone-shape bottom part. The volumes of each of these shapes must be calculated separately and then added together to obtain the total volume of the round clarifier in cubic feet. Since the overall height of the clarifier (including the cone-shaped bottom) is 12 ft, the cone shaped height must be subtracted from the overall height of 12 ft to obtain the height of the cylinder shape part of the clarifier. 12 ft 3.5 ft = 8.5 ft (This is the height of the cylinder portion of the clarifier) c. Find the Volume of the cylinder part of clarifier: Volume, ft 3 = [0.785 x (30 ft) 2 ] x 8.5 ft = [0.785 x (30 ft x 30 ft)] x 8.5 ft = [0.785 x 900 ft 2 ] x 8.5 ft = 706.5 ft 2 x 8.5 ft = 6005.25 ft 3 d. Find the volume of the cone-shaped portion of the clarifier: Volume, ft 3 = 1 [(0.785 x (D, ft) 2 ) x Ht. of cone-shape, ft] 3 Volume, ft 3 = 1 [706.5 ft 2 x 3.5 ft] (706.5 ft 2 is the area of the clarifier found in part c) 3 = 1 (2472.75 ft 3 ) 3 = 824.25 ft 3 e. Find the total volume in cubic feet of the one round clarifier:! 30

Total Volume, ft 3 = 6005.25 ft 3 + 824.25 ft 3 = 6829.50 ft 3 f. Find the total volume in cubic feet for the 2 round clarifiers: Total Volume for both clarifiers, ft 3 = 6829.50 ft 3 x 2 = 13,659 ft 3 g. Find the total volume in gallons for the 2 round clarifiers: Total Volume, gals for 2 round clarifiers = 13,659 ft 3 x 7.48 gals/ft 3 = 102, 169.32 gals h. Add the total gallons from the rectangular clarifiers with the total gallons of the round clarifiers: Total gals for all final clarifiers = 134,640 gals + 102,169.32 gals = 236,809.32 gals NOTE: If the problem gives you dimensions for primary clarifier/s (also called primary sedimentation basins), they must not be used in the calculation. The problem asks for the total volume in gallons for the final clarifier/s (final sedimentation basins). The catch in this problem is to recognize that the total round clarifier height must be divided into two heights; the cylinder portion height and the cone-shaped height to obtain the total volume (cubic feet) of the round clarifier. 40. A pump pulls 3 amps on a 15-amp circuit breaker, what size fuse is used? Rule: Only pull 80% of the size of the circuit breaker. a. Calculate the number of amps #amps = 15 amps x.80 # amps = 12 amps! 31

41. If the velocity in a 24-inch pipe flowing half-full is 2.5 fps, what is the flow rate in gpm? (Grade 3) a. X = Flow, gpm b. The formula needed to solve the problem: Q, ft 3 /sec = Area, ft 2 x Velocity, ft./sec Once the flow rate is found in ft 3 /sec then the flow must be converted to gpm. Since the area of the pipe is flowing half full, the area in the formula must be divided by 2: Q, ft 3 /sec = Area, ft 2 x Velocity, ft./sec 2 c. Solve for the Area, ft 2 of the pipe: Area, ft 2 = 0.785 x (Diameter, ft.) 2 = 0.785 x (2 ft.) 2 = 0.785 x 4 ft 2 = 3.14 ft 2 d. Solve for Q, ft 3 /sec: Q, ft 3 /sec = 3.14 ft 2 x 4 ft 2 2 Q, ft 3 /sec = 1.57 ft 2 x 2.5 ft./sec Q, ft 3 /sec = 3.925 ft 3 /sec e. Convert 3.925 ft 3 /sec -> ft 3 /min -> gpm Ft 3 /min = 3.925 ft 3 /sec x 60 sec/min Ft 3 /min = 235.5 ft 3 /min gpm = 235.5 ft 3 /min x 7.48 gal/ft 3 gpm = 1761.54 gpm 42. EPA defines metals as the sum of the concentrations of copper, nickel, total chromium, and lead. An analysis of industrial wastewater for an electroplating facility produced the following results: (Grade 4) Pollutant Concentration Cadmium (T) 0.6 mg/l! 32

*Chromium (T) 2.1 mg/l *Copper (T) 0.9 mg/l *Lead (T) 0.3 mg/l *Nickel (T) 1.2 mg/l Zinc (T) 1.4 mg/l Cyanide (T) 0.4 mg/l The total metal concentration is: a. 3.9 mg/l b. 4.5 mg/l c. 6.5 mg/l d. 6.9 mg/l x = Sum of metals, mg/l a. Calculate the sum of the concentrations of chromium, copper, lead, and nickel in mg/ L. Total metal, mg/l = (Cu), 0.9 mg/l + (Ni), 1.2 mg/l + (Cr), 2.1 mg/l + (Pb), 0.3 mg/l Total metal concentrations, mg/l = 4.5 mg/l 43. Determine the MCRT for an activated sludge treatment plant in days using the following information: Plant flow 4,000,000 gallons MLVSS 1,600 mg/l Aeration tank volume 0.8 MG Plant effluent SS 16 mg/l WAS 7,600 mg/l WAS flow 0.03 MGD a. 9.2 days b. 8.1 days c. 5.3 days d. 4.3 days a. x = MCRT, days b. MCRT, days = Aeration Tank, TSS, lbs. TSS Wasted, lbs./day + Effluent TSS, Lbs./day c. Convert 4,000,000 gpd -> MGD MGD = 4,000,000 gal/day 1,000,000 gal/mg MGD = 4.0 MGD! 33

d. Aer. Tank TSS, lbs. = 0.8 MG x 1600 mg/l x 8.34 lbs./gal Aer. Tank TSS, lbs. = 10,675.2 e. Wasted Solids, lbs./day = 0.03 MGD x 7,600 mg/l x 8.34 lbs./gal Wasted Solids, lbs./day = 1,901.52 f. Effluent Lost TSS, lbs./day = 4 MGD x 16 mg/l x 8.34 lbs./gal Effluent Lost TSS, lbs./day = 533.76 g. Calculate MCRT, days: MCRT, days = 10,675.2 lbs. 1,901.52 lbs./day + 533.76 lbs./day MCRT, days = 10,675.2 lbs. 2,435.28 lbs./day MCRT, days = 4.4 days 44. An activated sludge treatment plant produces 23,000 pounds of solids per year and the cost of hauling the solids is $120 per dry ton. What is the hauling cost per year? a. $1380 b. $1440 c. $1520 d. $1660 a. x = Hauling cost per year b. Formula to use to solve problem: 1. Tons/year = Total lbs. of solids/year 2000 lbs. solids/ton 2. Hauling Cost = Tons solids/year x $120.00/ton of solids c. Solve for Tons of solids produced during the year: Tons/year = 23,000 lbs. solids/year 2000 lbs/ton Tons/year = 11.5 Tons Solids/year! 34

d. Solve for hauling cost: Hauling Cost = 11.5 Tons Solids/year x $120.00/Ton Hauling Cost = $1,380/year 45. The flow through a 78-inch diameter mag-meter is 70 MGD. What is the flow velocity? a. 2.1 ft/sec b. 3.3 ft/sec c. 21.0 ft/sec d. 33.3 ft/sec a. x = Velocity, ft/sec b. Solve for the velocity of water through the pipe in ft/sec using the following formula: Velocity, Ft./sec = Q, ft 3 /sec A, ft 2 c. Convert 70 MGD to FT 3 /sec using this conversion formula: Use this conversion factor: 1.0 MGD = 1.55 ft 3 /sec Ft 3 /sec = 70 MGD x 1.55 ft 3 /sec/mgd Ft 3 /sec = 108.5 ft 3 /sec d. Solve for Area of pipe, ft 2 : Convert 78-inch diameter to feet: Ft. = 78 inch 12 inch/ft. Ft. = 6.5 ft. Area, ft 2 = 0.785 x (D, ft.) 2 Area, ft 2 = 0.785 x 6.5 ft. Area, ft 2 = 33.17 ft 2 d. Solve for velocity, ft./sec:! 35

Velocity, ft./sec = 108.5 ft 3 /sec 33.17 ft 2 Velocity, ft./sec = 3.27 ft./sec 46. Math question gave annual cost of chlorine, gave chlorine dose, and average daily flow. Determine how much it cost to treat with chlorine for one week. 52 weeks/year 47. What is the solids loading on a sedimentation basin 60 ft. in diameter if the flow is 0.55 MGD and the solids concentration is 200 mg/l? a. X (unknown) = Solids Loading, lbs. SS/day/ft 2 b. The solids loading formula is used to solve for X. Solids Loading, lbs. SS/day/ft 2 = Lbs. SS/day Area, ft 2 c. Solve for lbs. SS/day: Lbs. SS/day = 0.55 MGD x 200 mg/l x 8.34 lbs./gal = 917.40 d. Solve for the area of the basin: Area, ft 2 = 0.785 x (Diameter, ft.) 2 = 0.785 x (60 ft.) 2 = 0.785 x 3600 ft 2 = 2,826 ft 2 e. Solve for Solids Loading: SL, lbs./day/ft 2 = 917.40 lbs. SS/day 2,826 ft 2 = 0.325 lbs. SS/day/ft 2 48. Determine the flow through a length of 50 ft. pipe with an inner diameter of 8-inches and a velocity of 2.5 ft./sec. in liters/min. a. X (unknown) = Flow (Q), Liters/mins b. The flow rate formula is used to solve for X! 36

Q, ft 3 /sec = A, ft 2 x V, ft./sec c. Solve for the area of pipe. Convert 8 inches to ft.: 8 inches = 0.67 ft. 12 inches/ft. Area, ft 2 = 0.785 x (0.67 ft.) 2 = 0.785 x 0.45 ft 2 = 0.35 ft 2 d. Q, ft. 3 /sec = Area, ft 2 x V, ft./sec = 0.35 ft 2 x 2.5 ft./sec = 0.88 ft 3 /sec e. Convert 0.88 ft 3 /sec to liters/min. Use the conversion factor of 1 gal. = 3.79 liters First convert 0.88 ft 3 /sec to ft 3 /min to gal/min to liters/min. Ft 3 /min = 0.88 ft 3 /sec x 60 sec./min = 52.8 ft 3 /min Gal/min = 52.8 ft 3 /min x 7.48 gal/ft 3 = 394.944 gal/min Liters/min = 394.944 gal/min x 3.79 liters/gal = 1,496.84 L/min 49. Determine the volume of an anaerobic digester in cubic feet. The digester has a diameter of 110 ft. with a total height of 50 ft. The cone portion of the digester s height is 15 ft. a. X (unknown) = Volume of digester, ft 3 b. The digester contains two shapes; a cylinder and a cone. The volumes for each shape must be calculated and then added together to determine the total volume of the digester. c. Calculate the volume of the cylinder portion: Find the height of the cylinder portion: Ht. of cylinder, ft. = 50 ft. 15 ft.! 37

= 35 ft. Vol. ft 3 = [0.785 x (110 ft.) 2 ] x 35 ft. = [0.785 x 12,100 ft 2 ] x 35 ft. = 9,498.5 ft 2 x 35 ft. = 332,447.5 ft 3 d. Calculate the volume of the cone portion: Vol. ft 3 = 1 [0.785 x (110 ft.) 2 ] x 15 ft. 3 = 1 [9,498.5 ft 2 x 15 ft.] 3 = 1 [142,477.5 ft 3 ] 3 = 47,492.5 ft 3 e. Add the cubic foot volumes of the cylinder and cone portions. Total Volume, Ft 3 = 332,447.5 ft 3 + 47,492.5 ft 3 = 379,940 ft 3 50. If an activated sludge plant s MLSS had a concentration of 1,200 mg/l and settled out at 600 ml/l, what would the sludge volume index (SVI) be? Grade 3 a. X (unknown) = SVI, ml/gram b. Use the SVI formula: SVI, ml/g = (SSV30, ml/l) x (1,000 mg/g) MLSS, mg/l SVI, ml/g = 600 ml/l x 1,000 mg/g 1,200 mg/l SVI, ml/g = 600,000 ml/g 1,200 SVI, ml/g = 500 ml/g! 38

51. One side of the grounds for a lift station is 325 ft. long. How many trees will be required if the trees are to be 25 ft. apart with a tree at each end? Hint; draw a diagram of the spacing of the trees. *------*------*------*------*------*------*------*------*------*------*------*------*------* 1 2 3 4 5 6 7 8 9 10 11 12 13 14 * Represents a tree Spaces between trees = 325 ft. = 13 spaces 25 ft. 14 trees are needed! 39