Design of the Aeration Basin Based on Solids Retention Time - sewer construction:

You are provided the following information about a municipal wastewater treatment plant. That plant will use the classically activated sludge process. Population = 150,000 people, flow rate of 33.75 × 10⁶ l/day (equals 225 l/person/day) and influent BOD5 concentration of 444 mg/l (note this is high strength wastewater). Suppose that the regulatory agency enforces an effluent standard of BOD5 = 20 mg/l and suspended solids standard of 20 mg/l in the treated wastewater. A wastewater sample is collected from the biological reactor and is found to contain a suspended solids concentration of 4,300 mg/l. A suspended solids concentration within the secondary sludge is 15,000 mg/l and the concentration in the secondary sludge is 5,000 mg/l. The concentration of suspended solids in the plant influent is 200 mg/l and that which leaves the primary clarifier is 100 mg/l. A microorganism within the activated sludge procedure can convert 100 grams of BOD5 into 55 grams of biomass. They have a maximum growth rate of 0.1/day, a first-order death rate constant of 0.05/day, and they reach 1/2 of their maximum growth rate when the BOD5 concentration is 10 mg/l. The mean cell retention time of the solids is 4 days and sludge is processed on the belt filter press every 5 days.