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 ft3 must be calculated first and then the ft3 of pipe is converted
to gallons using the capacity formula:
Gallons = Number of calculated ft3 x 7.48 gal/ft3
b. Solve for volume of pipe in ft3:
Convert 8 in to ft: 8 inches = 0.67 ft
12 inches/ft
Volume, ft3 = 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 ft2] x 800 ft
= 0.35 ft2 x 800 ft
= 280 ft3
Gallons = # of ft3 x 7.48 gal/ft3
Gallons = 280 ft3 x 7.48 gal/ft3
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 ft3 volume of pipe and then convert the ft3 to gallons and then to MG
capacity.
Convert 10 inch-diameter pipe to ft diameter: 10 inches = 0.83 ft
12 inches/ft
Volume, ft3 = 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 ft2] x 265 ft
= 0.54 ft2 x 265 ft
= 143.54 ft3
Capacity, gals = # of ft3 x 7.48 gal/ft3
Capacity, gals = 143.54 ft3 x 7.48 gal/ft3
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
= 1073.68 gals x 250 mg/L
120,000
= 268420 = 2.24 lbs of chemical
120,000
2
*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 ft3 and
add ft3 together then convert total ft3 to gallons.
Find Cylinder Volume:
1. Find Cylinder depth, ft = 12.5 ft – 4.5 ft = 7.5 ft
2. Volume, ft3 = 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 ft2] x 7.5 ft
= 706.5 ft2 x 7.5 ft
= 5299 ft3
Find Cone Volume:
1. Volume, ft3 = 1/3 x [0.785 x (D, ft)2] x Cone depth, ft
= 1/3 x [706.5 ft2 x 4.5 ft]
= 1/3 x (3179.25 ft3)
= 1059.75 or 1060 ft3
Add Ft3 Volumes of cylinder and cone: 5299 ft3 + 1060 ft3 = 6359 ft3
Find Gallon Capacity of Clarifier
Capacity, gals = Total # of ft3 in clarifier x 7.48 gal/ft3
= 6359 ft3 x 7.48 gal/ft3
= 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?
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:
3
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
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
4
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 ft3 then convert to gals
Volume, ft3 = 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 ft2] x 15 ft
= 4415.625 ft2 x 15 ft
= 66,234.375 ft3
Capacity, gals = 66,234.375 ft3 x 7.48 gal/ft3
= 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
5
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
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
6
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%
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 ft3 convert to gallons and then convert to
MG
Convert the 18 inch depth of sludge to feet of sludge: 18 inch = 1.5 ft
12 in/ft
Volume of sludge, ft3 = Length, ft x Width, ft x Depth, ft
= 50 ft x 20 ft x 1.5 ft
= 1500 ft3
7
Capacity, gals = # ft3 x 7.48 gal/ft3
= 1500 ft3 x 7.48 gal/ft3
= 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
*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)
8
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
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/ft2
b. Formula needed to solve problem:
SSR, gpd/ft2 = Flow, gal/day
Area, ft2
c. Calculate area of clarifier: A, ft2 = 0.785 x (D, ft)2
= 0.785 x (80 ft x 80 ft)
= 0.785 x 6400 ft2
= 5024 ft2
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/ft2
SSR, gpd/ft2 = 2,073,600 gal/day
5024 ft2
SSR, gpd/ft2 = 412.74 gpd/ft2
9
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.
3960
The pump efficiency of 80% is not needed to calculate whp.
c. Solve for whp
Whp = 600 gpm x 65 ft
3960
Whp = 39,000
3960
Whp = 9.85
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, ft2
200 ft2/gal of paint
c. First find total area, ft2 of tank:
1. Find top & bottom area, ft2 of tank: Area, ft2 = 0.785 x (D, ft)2
= 0.785 x (50 ft x 50 ft)
= 0.785 x 2500 ft2
= 1962.5 ft2
Total Area, ft2 (top & bottom) = 1962.5 ft2 x 2
= 3925 ft2
2. Find wall area, ft2 of tank:
Circumference of tank, ft2 = Length, ft2
Height of tank, ft2 = Width, ft2
C, ft = 3.14 x Diameter, ft
C, ft = 3.14 x 50 ft
C, ft = 157 ft
10
Area, ft2 = L, ft (C) x W, ft (Ht)
Area, ft2 = 157 ft x 15 ft
Area, ft2 = 2,355 ft2
3. Total Tank Area, ft2 = 3925 ft2 + 2,355 ft2
= 6280 ft2
d. Solve for number of gallons of paint:
# of gallons of paint = 6280 ft2
200 ft2/gal
= 31.4 gals or 32 gals of paint
*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
12-in/ft
= 2 ft
Depth: ft = 17 inches
12-in/ft
= 1.42 ft
c. Overall formula to use to solve problem is detention time: DT = Volume
Flow
3
DT, sec = Volume of sludge, ft
Flow, ft3/sec
d. Convert 13,500 gals of sludge to ft3 to cubic feet of sludge:
Ft3 = 13,500 gals = 1804.8 ft3
7.48 gals/ft3
e. Find flow rate of sludge through channel:
Q, ft3/sec = Area, ft2 x Velocity, ft/sec
1. Find area, ft2 of channel: Area, ft2 = Depth, ft x Width, ft
= 1.42 ft x 2 ft
= 2.84 ft2
11
Q, ft3/sec = Area, ft2 x Velocity, ft/sec
= 2.84 ft2 x 1.6 ft/sec
= 4.54 ft3/sec
d. Calculate the time it takes to move sludge through channel:
Time, sec = Volume, ft3 Sludge
Flow, ft3/sec of Sludge
Time, sec = 1804.8 ft3
4.54 ft3/sec
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 grit channel/s 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 channel/s can be chosen.
b. Formula to solve for velocity of wastewater entering the plant:
Velocity, ft/sec = Q, ft3/sec
Area, ft2
c. Convert Q, flow rate of 16 MGD to ft3/sec:
16 MGD = 16,000,000 gals/day
Gal/min = 16,000,000 gal = 11111.11 gal/min
1440 min/day
Ft3/min = 11111.11 gal/min = 1485.44 ft3/min
7.48 gal/ft3
Ft3/sec = 1485.44 ft3/min = 24.76 ft3/sec
60 sec/min
Or you can use the conversion factor: 1.0 MGD = 1.55 ft3/sec
Ft3/sec = 16 MGD x 1.55 ft3/sec
= 24.8 ft3/sec
12
d. Find area, ft2 of channel:
Area, ft2 = Width, ft x Depth, ft
= 4 ft x 3.5 ft
= 14 ft2
e. Solve for velocity of wastewater in channel:
Velocity, ft/sec = Q, ft3/sec
Area, ft2
Velocity, ft/sec = 24.8 ft3/sec = 1.77 ft/sec
14 ft2
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
13
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___
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
14
d. Calculate lbs Cl dose/day
Lbs Cl dose/day = Flow (MGD) x Cl dose (mg/L) x 8.34 lbs/gal
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 wastewater 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, ft3 = L, ft x W, ft, x Water Drop, ft
=20 ft x 20 ft x 10 ft
= 4000 ft3
15
d. Calculate gallons of wastewater pumped:
Capacity, gals = # ft3 x 7.48 gal/ft3
= 4000 ft3 x 7.48 gal/ft3
= 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/ft2
b. Formula to solve problem: lbs SS added/day/ft2 = lbs SS added/day
Area, ft2
Note: SLR is only calculated on secondary clarifiers. The plant flow and the RAS
flow must be added together when calculating 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, ft2 = 0.785 x (D, ft)2
= 0.785 x (85 ft)2
= 0.785 x (85 ft x 85 ft)
= 0.785 x 7225 ft2
= 5671.6 ft2
d. Calculate SLR
lbs SS added/day/ft2 = 103920.6 lbs SS added /day
5671.6 ft2
lbs SS added/day/ft2 = 18.3 lbs SS added/day/ft2
16
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 ft3
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 ft3.
Volume of stroke, ft3 = [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 ft2] x 0.67 ft
= 0.196 ft2 x 0.67 ft
= 0.131 ft3
3. Convert 0.131 ft3 of water to gallons
Capacity, gals = # ft3 x 7.48 gal/ft3
= 0.131 ft3 x 7.48 gal/ft3
= 0.982 gals
d. Calculate gpm pumping rate:
gal/min = # of gals x 20 strokes/minute
1.0 stroke
Pumping Rate, gpm = 0.982 gals x 20 strokes/min
1.0 stroke
Pumping Rate, gpm = 19.6 gpm
17
*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
18
*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/ft2. The average daily flow is 0.8 MGD. The following
is data from a typical flow chart;
7 am 450 gpm
8 am 550 gpm
9 am 650 gpm
10 am 750 gpm
11 am 850 gpm
12 pm 950 gpm
1pm 1050 gpm
2 pm 950 gpm
3 pm 750 gpm
4 pm 650 gpm
5 pm 550 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/ft2 = Flow, gpd
Area, ft2
rearranged to find flow, gpd;
Flow, gpd = HL, gpd/ft2 x ft2
= 800 gpd/ft2 x 1200 ft2
= 960,000 gpd
c. Convert 960,000 gpd to gpm: 960,000 gal/day = 666.67 or 667 gpm
1440 min/day
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
19
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 a 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
# days = 2100 lbs of polymer_____
10.5 lbs/day polymer used
# days = 200 days
20
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
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
21
*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
Current F/M ratio = 0.27 lb BOD/D/lb MLVSS
Primary Effl. BOD conc. = 138 mg/L
Primary Effluent Flow = 3.6 MGD
Aeration Tank Volume = 0.65 MG
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
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
22
34. Convert 20oC to oF.
a. X (unknown) = oF temperature
b. Formula to solve problem:
o
F = [oC x 9 ] + 32
5
o
F = [20 x 9 ] + 32
5
o
F = 36 + 32
o
F = 68oF
Convert 75oF to oC.
a. X (unknown) = oC temperature
b. Formula to solve problem:
o
C = (oF – 32) x 5/9
o
C = (75 – 32) x 5/9
o
C = 43 x 5/9
o
C = 23.8oC
*35. What is the organic loading rate applied to a trickling filter in lbs.
BOD/day/1000 ft3 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?
a. x(unknown) = Organic Loading, BOD/day/1000 ft3 unit or Tft3
Note: 1000 ft3 unit contains 1000 ft3. This is similar to 1 MGD contains 1,000,000
gallons. I use Tft3 (T stands for 1000) instead of 1000 ft3 in the formula.
b. Formula to solve problem:
OL, lbs BOD/day/Tft3 = lbs BOD/day
Tft3
c. Calculate lbs BOD/day and Tft3 volume of trickling filter.
1. lbs BOD/day = 0.2 MGD x 90 mg/L x 8.34 lbs/gal
= 150.12
23
2. Volume, ft3 = [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 ft2] x 5 ft
= 1962.5 ft2 x 5 ft
= 9812.5 ft3
3. Convert 9812.5 ft3 to Tft3 units:
Tft3 = 9812.5 ft3___
1000 ft3/Tft3
Tft3 = 9.8 Tft3
d. Calculate Organic Loading on trickling filter
OL, lbs BOD/day/Tft3 = 150.12 lbs BOD/day
9.8 Tft3
OL, lbs BOD/day/Tft3 = 15.32 lbs BOD/day/1000 ft3 unit
24
*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____
0.70 – 0.315
=
0.25 x 100
0.385
x 100
= 0.6493 x 100
= 64.9 or 65% reduction in VS
25
*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 ft2,
what is the solids loading rate into the thickener (lbs. solids/day/ft2)?
a. X (unknown) = SRL, lbs solids/day/ft2
b. Formula to solve problem;
lbs SS/day/ft2 = lbs SS/day__________
Area of Thickener, ft2
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:
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/ft2 = 21,684 lbs SS/day
850 ft2
Lbs. SS/day/ft2 = 25.5 lbs SS/day/ft2
*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.
26
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.
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. Note: If they give the
dimensions of the primary clarifiers, Do Not calculate their volumes.
2 rectangular clarifiers dimensions = 30 ft. x 20 ft. x 15 ft.
2 round shape clarifiers: Diameters = 30 ft
Outside wall depth (or SWD) = 12 ft.
Cone-shaped bottoms depth = 3.5 ft.
Total depth of round clarifiers = 15.5 ft.
27
a. Find the 2 rectangular clarifiers volumes in cubic feet and then convert cubic feet
to gallons.
Volume, Ft3 = 30 ft x 20 ft x 15 ft
= 9,000 ft3
Total Volume in Ft3 of both clarifiers = 9,000 ft3 x 2
= 18,000ft3
Total gallons capacity for both clarifiers = 18,000 ft3 x 7.48 gals/ft3
= 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.
c. Find the Volume of the cylinder part of clarifier:
Volume, ft3 = [0.785 x (30 ft)2] x 12 ft
= [0.785 x (30 ft x 30 ft)] x 12 ft
= [0.785 x 900 ft2] x 12 ft
= 706.5 ft2 x 12 ft
= 8,478 ft3
d. Find the volume of the cone-shaped portion of the clarifier:
Volume, ft3 = 1 [(0.785 x (D, ft)2) x Ht. of cone-shape, ft]
3
Volume, ft3 = 1 [706.5 ft2 x 3.5 ft] (706.5 ft2 is the area of the clarifier found in part c)
3
= 1 (2472.75 ft3)
3
= 824.25 ft3
e. Find the total volume in cubic feet of the one round clarifier:
Total Volume, ft3 = 8,478 ft3 + 824.25 ft3
= 9,302.25 ft3
28
f. Find the total volume in cubic feet for the 2 round clarifiers:
Total Volume for both clarifiers, ft3 = 9302.25 ft3 x 2
= 18,604.5 ft3
g. Find the total volume in gallons for the 2 round clarifiers:
Total Volume, gals for 2 round clarifiers = 18,604.5 ft3 x 7.48 gals/ft3
= 139,166.7 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 + 139,161.7 gals
= 273,802 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).
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
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, ft3/sec = Area, ft2 x Velocity, ft./sec
Once the flow rate is found in ft3/sec then the flow must be converted to gpm.
29
Since the area of the pipe is flowing half full, the area in the formula must be divided by
2:
Q, ft3/sec = Area, ft2 x Velocity, ft./sec
2
c. Solve for the Area, ft2 of the pipe:
Area, ft2 = 0.785 x (Diameter, ft.)2
= 0.785 x (2 ft.)2
= 0.785 x 4 ft2
= 3.14 ft2
d. Solve for Q, ft3/sec:
Q, ft3/sec = 3.14 ft2 x 4 ft2
2
Q, ft3/sec = 1.57 ft2 x 2.5 ft./sec
Q, ft3/sec = 3.925 ft3/sec
e. Convert 3.925 ft3/sec -> ft3/min -> gpm
Ft3/min = 3.925 ft3/sec x 60 sec/min
Ft3/min = 235.5 ft3/min
gpm = 235.5 ft3/min x 7.48 gal/ft3
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
*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
30
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
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
31
*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
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, ft3/sec
A, ft2
c. Convert 70 MGD to FT3/sec using this conversion formula:
32
Use this conversion factor: 1.0 MGD = 1.55 ft3/sec
Ft3/sec = 70 MGD x 1.55 ft3/sec/MGD
Ft3/sec = 108.5 ft3/sec
d. Solve for Area of pipe, ft2:
Convert 78-inch diameter to feet:
Ft. = 78 inch
12 inch/ft.
Ft. = 6.5 ft.
Area, ft2 = 0.785 x (D, ft.)2
Area, ft2 = 0.785 x 6.5 ft.
Area, ft2 = 33.17 ft2
d. Solve for velocity, ft./sec:
Velocity, ft./sec = 108.5 ft3/sec
33.17 ft2
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/ft2
b. The solids loading formula is used to solve for X.
Solids Loading, lbs. SS/day/ft2 = Lbs. SS/day
Area, ft2
c. Solve for lbs. SS/day:
Lbs. SS/day = 0.55 MGD x 200 mg/L x 8.34 lbs./gal
= 917.40
33
d. Solve for the area of the basin:
Area, ft2 = 0.785 x (Diameter, ft.)2
= 0.785 x (60 ft.)2
= 0.785 x 3600 ft2
= 2,826 ft2
e. Solve for Solids Loading:
SL, lbs./day/ft2 = 917.40 lbs. SS/day
2,826 ft2
= 0.325 lbs. SS/day/ft2
*48. Determine the flow through a length of 50 ft. pipe with an inner diameter of 8inches 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
Q, ft3/sec = A, ft2 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, ft2 = 0.785 x (0.67 ft.)2
= 0.785 x 0.45 ft2
= 0.35 ft2
d.
Q, ft.3/sec = Area, ft2 x V, ft./sec
= 0.35 ft2 x 2.5 ft./sec
= 0.88 ft3/sec
e. Convert 0.88 ft3/sec to liters/min. Use the conversion factor of 1 gal. = 3.79 liters
First convert 0.88 ft3/sec to ft3/min to gal/min to liters/min.
Ft3/min = 0.88 ft3/sec x 60 sec./min
= 52.8 ft3/min
Gal/min = 52.8 ft3/min x 7.48 gal/ft3
= 394.944 gal/min
34
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, ft3
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.
= 35 ft.
Vol. ft3 = [0.785 x (110 ft.)2] x 35 ft.
= [0.785 x 12,100 ft2] x 35 ft.
= 9,498.5 ft2 x 35 ft.
= 332,447.5 ft3
d. Calculate the volume of the cone portion:
Vol. ft3 = 1 [0.785 x (110 ft.)2] x 15 ft.
3
= 1 [9,498.5 ft2 x 15 ft.]
3
= 1 [142,477.5 ft3]
3
= 47,492.5 ft3
e. Add the cubic foot volumes of the cylinder and cone portions.
Total Volume, Ft3 = 332,447.5 ft3 + 47,492.5 ft3
= 379,940 ft3
35
*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
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
*52. Math question – flow rate in gpm of water coming out of a structure. A wet well with a
diameter of 7 ft. is 8 ft. deep. When the pump is started, the water level in the well drops 2 ft. in
5 minutes. What is the gpm pumping rate?
a. 16.4 gpm
b. 15.4 gpm
c. 115.1 gpm
d. 57.5 gpm
X = Q, flow rate, gpm
Step 1: Find the flow rate using the formula;
Q, ft3/min = A, ft2 x Drop Velocity, ft./min
36
Step 2: Find the area of the wet well:
A, ft2 = 0.785 x (D, ft.)2
A, ft2 = 0.785 x (7 ft.)2
A, ft2 = 0.785 x 49 ft2
A, ft2 = 38.465 ft2
Step 3: Find the drop velocity in ft./min
Water Drop velocity, ft./min = 2 ft
5 mins.
Velocity, ft./min = 0.4 ft./min
Step 4: Calculate the flow rate, ft3/min:
Q, ft3/min. = 38.465 ft2 x 0.4 ft./min.
Q, ft3/min. = 15.386 ft3/min.
Step 5: Convert flow rate of ft3/min to gpm:
GPM = 15.386 ft3/min x 7.48 gal/ft3
GPM = 115.1 gpm
*53. Math problem: Determine the cost of chlorine after calculating lbs. of chlorine needed and
then calculating the cost of the chlorine.
54. What is the chlorine dose of an industrial waste if the chlorine demand is 21 mg/L and the
chlorine residual is 0.9 mg/L?
(a) 20.9 mg/L
(b) 200.9 mg/L
(c) 21.9 mg/L
(d) 20.1 mg/L
Chlorine Dose, mg/L = Chlorine Demand, mg/L + Chlorine Residual, mg/L
Chlorine Dose, mg/L = 21mg/L + 0.9 mg/L
Chlorine Dose, mg/L = 21.9 mg/L
37
*55. Determine the F/M ratio based on the following information:
Data:
Plant flow – 3.0 MGD
Primary clarifier BOD – 120 mg/L
MLVSS – 1,500 mg/L
Aeration tank volume – 0.7 MG
Plant effluent SS – 15 mg/L
a.! 1.1
b.! 0.99
c.! 0.54
d.! 0.34
X = F/M ratio, lbs. BOD/Day/Lb. of MLVSS
Step 1: Find the F/M ratio using the formula:
F/M = lbs. of BOD/Day
lb. MLVSS
Step 2: Calculate the lbs. of BOD/Day
Lbs. of BOD/Day = 3.0 MGD x 120 mg/L x 8.34 lbs./gal
Lbs. of BOD/Day = 3002.4
Step 3: Calculate the lbs. of MLVSS
Lbs. MLVSS = 0.7 MG x 1,500 mg/L x 8.34 lbs./gal
Lbs. MLVSS = 8757
Step 4: Calculate F/M ratio:
F/M Ratio = 3002.4 lbs. of BOD/Day
8757 lbs. MLVSS
F/M Ratio = 0.34 lbs. BOD/Day/Lb. MLVSS
*56. Given the following data, how many lbs./day of volatile solids (VS) are pumped to an
anaerobic digester?
Pumping rate = 25 gpm
Solids content = 6.75%
Volatile solids content = 58%
Clarifier diameter = 25 feet
Clarifier depth = 12 feet
X = lbs./day of Volatile Solids (VS) pumped to the digester
Step 1: Calculate the gallons per day of sludge pumped to the digester:
Gals/Day of Sludge pumped = 25 gal/min x 1440 mins./day
= 36,000 gals./day
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Convert 36,000 gallons of sludge pumped per day to lbs. of sludge pumped per day:
Lbs. Sludge Pumped/Day = 36,000 gals. X 8.34 lbs./gal
= 300,240 lbs. sludge pumped/Day
Step 2: Calculate the lbs. of solids and lbs. of VS pumped/Day into the Digester
Lbs./Day of VS Pumped = Sludge, lbs./day x %Solids x %V
100
100
Note: The decimal percentage must be used in the above calculation.
Lbs./Day of VS Pumped = 300,240 lbs. Sludge/Day x .0675 x .58
Lbs./Day of VS Pumped = 11,754.4
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