An activated sludge plant receives 2 MGD with influent BOD 240 mg/L. If the desired F/M ratio is 0.4 and primary clarifiers remove 30% of influent BOD, how many pounds of MLVSS should be maintained?

An activated sludge plant receives 2 MGD with influent BOD 240 mg/L. If the desired F/M ratio is 0.4 and primary clarifiers remove 30% of influent BOD, how many pounds of MLVSS should be maintained?

• A. 6500 lbs
• B. 7200 lbs
• C. 7005 lbs
• D. 6800 lbs

Answer: (C)

The key idea is that the F/M ratio equals the pounds of food (daily BOD to the aeration basin) divided by the pounds of active biomass (MLVSS) in the reactor. To meet a target F/M of 0.4, you compute how much BOD is entering the aeration basin each day, then divide by 0.4 to get the required MLVSS. First, after the primary clarifier removes 30%, the BOD concentration available to the aeration tank is 0.7 × 240 mg/L = 168 mg/L. The daily BOD load to the aeration basin is F = Q × 8.34 × BOD = 2 MGD × 8.34 × 168 ≈ 2802 lb/day. With F/M = 0.4, the required mass of MLVSS is MLVSS = F / (F/M) = 2802 / 0.4 ≈ 7005.6 lb. So about 7005 pounds of MLVSS should be maintained.

The key idea is that the F/M ratio equals the pounds of food (daily BOD to the aeration basin) divided by the pounds of active biomass (MLVSS) in the reactor. To meet a target F/M of 0.4, you compute how much BOD is entering the aeration basin each day, then divide by 0.4 to get the required MLVSS.

First, after the primary clarifier removes 30%, the BOD concentration available to the aeration tank is 0.7 × 240 mg/L = 168 mg/L. The daily BOD load to the aeration basin is F = Q × 8.34 × BOD = 2 MGD × 8.34 × 168 ≈ 2802 lb/day.

With F/M = 0.4, the required mass of MLVSS is MLVSS = F / (F/M) = 2802 / 0.4 ≈ 7005.6 lb.

So about 7005 pounds of MLVSS should be maintained.