Question

A completely mixed activated sludge process is designed to treat 20,000 m3 /d of domestic wastewater having a BOD5 concentration of 250 mg/L following primary treatment. The NPDES permit requirement for effluent BOD5 and TSS is 20 mg/L for each. The biokinetic parameters have been established as: yield = 0.6 , k = 5/d, Ks = 60 mg/L, and kd = 0.06/d. The MLVSS concentration in the aeration tank is to be maintained at 3000 mg/L and the ratio of VSS/TSS is 0.75. Assume that the particulate BOD5 oxygen demand is approximately 95% of the effluent VSS. The secondary clarifier concentrates the settled solids to 10,000 mg/L.

Determine: (a) the effluent soluble BOD5 (sBOD5) necessary to meet the NPDES effluent total BOD5 (tBOD5) requirement of 20 mg/L. (b) the solid residence time required (c ) the volume of the aeration basin in m3 (d) the mass of sludge produced (kg/d) (e) the organic loading rate on a mass basis (F/M loading rate) (f) the organic loading rate on a volumetric basis (g) the volumetric sludge wasting rate (m3 /d) from the underflow of the secondary clarifier (compare the approximation approach in Mihelcic to the more precise approach in the FE Handbook. (h) If the sludge wastage rate is increased in the plant, will the solids retention time go up, go down, or remain the same?

Solve all parts if possible?

Answer 1