An activated sludge process with an aeration basin volume of 0.3 MG is treating wastewater with...

An activated sludge process with an aeration basin volume of 0.3 MG is treating wastewater with the average daily flow of 2 MGD. The raw sewage entering the treatment plant has an average BOD5 of 400 mg/L. The primary treatment removes 25% of BOD5 and the subsequent activated sludge process is designed to remove 90% of BOD5.

Given:

Plant effluent BOD5 concentration = 30 mg/L

Biomass concentration in the aeration tank = 2,000 mg/L

Biomass concentration in the plant effluent = 20 mg/L

Biomass concentration in the recycle (RAS) = 8,000 mg/L

Flow rate of waste sludge (WAS) = 0.025 MGD

Endogenous decay rate (kd) = 0.01 day-1

a) Calculate the MCRT of the process.

b) Determine the yield coefficient, Y.

c) If µnet, nitrifiers = 0.2/day, would you expect to get nitrification in this system? Briefly describe why or why not and show any supporting calculations.

Expert Solution

Ally Wells answered 11 months ago

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