Free WW Wastewater Mathematics Questions and Answers — Questions and Answers
Question 1: A circular clarifier with a diameter of 80 feet treats a flow of 2.5 MGD. What is the surface overflow rate (SOR) in gpd/ft²?
- 31,250 gpd/ft²
- 497 gpd/ft² (Correct answer)
- 995 gpd/ft²
- 1,243 gpd/ft²
Correct answer: 497 gpd/ft²
First, calculate the surface area of the clarifier: Area = π * r². The radius is half the diameter (40 ft), so Area = 3.14159 * (40 ft)² = 5,026.5 ft². Next, convert the flow from MGD to gpd: 2.5 MGD * 1,000,000 gal/MG = 2,500,000 gpd. Finally, divide the flow by the area: 2,500,000 gpd / 5,026.5 ft² ≈ 497 gpd/ft².
Question 2: An aeration basin has a volume of 1.2 million gallons and the MLVSS concentration is 2,800 mg/L. If the plant influent flow is 4.0 MGD with a BOD of 220 mg/L, what is the Food to Microorganism (F/M) ratio?
- 0.15
- 0.52
- 0.26 (Correct answer)
- 1.20
Correct answer: 0.26
First, calculate the pounds of food (BOD) entering per day: 4.0 MGD * 220 mg/L * 8.34 = 7,339 lbs/day. Next, calculate the pounds of microorganisms (MLVSS) in the aeration basin: 1.2 MG * 2,800 mg/L * 8.34 = 28,022 lbs. The F/M ratio is the food divided by the microorganisms: 7,339 lbs/day / 28,022 lbs ≈ 0.26.
Question 3: What is the detention time in hours for a sedimentation tank that is 100 feet long, 30 feet wide, and has a water depth of 12 feet, if the flow rate is 3.0 MGD?
- 4.31 hours
- 1.55 hours
- 3.60 hours
- 2.15 hours (Correct answer)
Correct answer: 2.15 hours
First, calculate the volume of the tank in cubic feet: 100 ft * 30 ft * 12 ft = 36,000 ft³. Convert cubic feet to gallons: 36,000 ft³ * 7.48 gal/ft³ = 269,280 gallons. Then, calculate detention time in days: Volume (gal) / Flow (gpd) = 269,280 gal / 3,000,000 gpd = 0.0898 days. Finally, convert days to hours: 0.0898 days * 24 hr/day ≈ 2.15 hours.
Question 4: A pump is delivering 600 GPM against a total dynamic head (TDH) of 50 feet. What is the water horsepower (WHP)?
- 7.58 WHP (Correct answer)
- 12.12 WHP
- 5.05 WHP
- 9.45 WHP
Correct answer: 7.58 WHP
The formula for water horsepower is (Flow in GPM * Total Head in ft) / 3960. Plugging in the values: (600 GPM * 50 ft) / 3960 = 30,000 / 3960 = 7.58 WHP. This calculation determines the actual work being done on the water itself, before accounting for pump and motor inefficiencies.
Question 5: Calculate the Mean Cell Residence Time (MCRT) in days given the following: Aeration Tank Volume = 2.5 MG, MLSS = 3,000 mg/L, WAS Flow = 0.06 MGD, WAS Conc. = 8,000 mg/L, Effluent TSS = 12 mg/L, and Plant Flow = 5.0 MGD.
- 8.5 days
- 15.6 days
- 13.9 days (Correct answer)
- 11.2 days
Correct answer: 13.9 days
First, find the pounds of solids in the aeration system: 2.5 MG * 3,000 mg/L * 8.34 = 62,550 lbs. Next, find the pounds of solids leaving per day (wasted + effluent): (0.06 MGD * 8,000 mg/L * 8.34) + (5.0 MGD * 12 mg/L * 8.34) = 4,003 lbs/day + 500 lbs/day = 4,503 lbs/day. MCRT = Solids in System / Solids Leaving = 62,550 lbs / 4,503 lbs/day ≈ 13.9 days.
Question 6: A secondary clarifier receives a flow of 3.5 MGD containing 3,200 mg/L of MLSS. If the clarifier has a diameter of 90 feet, what is the solids loading rate (SLR) in lbs/day/ft²?
- 22.5 lbs/day/ft²
- 14.7 lbs/day/ft² (Correct answer)
- 10.1 lbs/day/ft²
- 28.9 lbs/day/ft²
Correct answer: 14.7 lbs/day/ft²
First, calculate the pounds of solids applied per day: 3.5 MGD * 3,200 mg/L * 8.34 = 93,408 lbs/day. Next, calculate the surface area of the clarifier: Area = π * (45 ft)² = 6,361.7 ft². Finally, divide the solids load by the area: 93,408 lbs/day / 6,361.7 ft² ≈ 14.7 lbs/day/ft².
Question 7: An anaerobic digester is fed raw sludge with 72% volatile solids and produces digested sludge with 55% volatile solids. What is the percent volatile solids reduction?
- 23.6%
- 61.3%
- 48.9%
- 52.5% (Correct answer)
Correct answer: 52.5%
Using the Van Kleeck formula for volatile solids reduction: Reduction % = [(VS in - VS out) / (VS in - (VS in * VS out))] * 100. Using decimal form: [(0.72 - 0.55) / (0.72 - (0.72 * 0.55))] * 100 = [0.17 / (0.72 - 0.396)] * 100 = [0.17 / 0.324] * 100 ≈ 52.5%. This calculation measures the efficiency of the digestion process in breaking down organic matter.
Free WW Wastewater Mathematics Questions and Answers
A circular clarifier with a diameter of 80 feet treats a flow of 2.5 MGD.
What is the surface overflow rate (SOR) in gpd/ft²?