Question

2. [20] 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 BODs 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 m 3/s soluble BODs-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 day1, yield coefficient-0.60 mgVSS mg1 BODs removed, maximum specific growth rate constant- 2.50 day1, 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 (m'/d) for the new activated sludge plant (given density of air is 1.185 kg/m3)

Answer 1

Sol:-- How= 0. 2.000 ms/s D Compude tle volune of aesation touk Flous of Se unXo ewe goD 020 X86 400/80 6880 x ) O, 00024