FREE WW Wastewater Mathematics 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?

Accepted Answer

497 gpd/ft²

Answer

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?

Accepted Answer

0.26

Answer

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?

Accepted Answer

2.15 hours

Answer

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)?

Accepted Answer

7.58 WHP

Answer

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.

Accepted Answer

13.9 days

Answer

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.

Wastewater Operator Certification Practice Test

Wastewater Operator Certification Practice Test

FREE WW Wastewater Mathematics Questions and Answers