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

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Civil Engineering

An activated sludge process with an aeration basin volume of 0.3MG is treating wastewater with the average daily flow of 2 MGD. Theraw sewage entering the treatment plant has an average BOD5 of 400mg/L. The primary treatment removes 25% of BOD5 and the subsequentactivated 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 getnitrification in this system? Briefly describe why or why not andshow any supporting calculations.

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