Question

An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answer 1

Volume of aeration tank = 1.29 x 10^4 m³

Step-by-step explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³