Wastewater Mathematics Flashcards
7 cards from real Wastewater Operator Certification practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Wastewater Mathematics flashcards as text
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?
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.
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?
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.
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?
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.
A pump is delivering 600 GPM against a total dynamic head (TDH) of 50 feet. What is the water horsepower (WHP)?
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.
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.
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.
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?
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.
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?
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.