oluble BOD5 = 80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50% of the suspended solids concentration, the microorganism concentration in the aeration tank (MLVSS) is 2000 mg/L, and the net activated sludge produced is 400.0 kg/d; also assume the following values for the growth constants: bacterial decay rate = 0.10 day-1 , yield coefficient = 0.60 mgVSS ∙ mg-1 BOD5 removed, maximum specific growth rate constant = 2.50 day-1 , and half saturation constant = 100 mg/L. (1) Estimate the volume of the aeration tank
A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BOD5 and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics: Flow = 0.2000 m3 /s soluble BOD5 = 80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50% of the suspended solids concentration, the microorganism concentration in the aeration tank (MLVSS) is 2000 mg/L, and the net activated sludge produced is 400.0 kg/d; also assume the following values for the growth constants: bacterial decay rate = 0.10 day-1 , yield coefficient = 0.60 mgVSS ∙ mg-1 BOD5 removed, maximum specific growth rate constant = 2.50 day-1 , and half saturation constant = 100 mg/L. (1) Estimate the volume of the aeration tank; (2) Compute the F/M ratio; (3) If the BOD5 is 60% of the ultimate BOD and that the oxygen transfer efficiency is 10%, estimate the volume of air to be supplied (m3 /d) for the new activated sludge plant (given density of air is 1.185 kg/m3 ).
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