Consider the design of a wastewater treatment plant (WWTP) for acommunity with average
daily and peak hourly wastewater design flows of 7570 m3/day and18925 m3/day. The raw
sewage has an average of 240 mg/L BOD5 and 270 mg/L of suspendedsolids.
(a) Assume that the primary sedimentation process removes 55% ofthe suspended solids and
35% of the BOD5 of the raw sewage. Determine the SS and BOD5concentrations in the
primary sedimentation effluent flow. (1 mark)
The primary effluent is treated by two parallel trains of thecomplete-mix activated sludge
process. Assume with the average flow conditions and theperformance of primary
sedimentation achieved as above. Given the following parametersrequired for the activated
sludge process:
· Effluent BOD5 (S): 8 mg/L
· Observed biomass yield (Y): 0.55 kg biomass / kg BOD
· Endogenous decay rate (b) = 0.04 /day
· Solids retention time: 8 days
· MLVSS concentration in the aeration tank: 3000 mg/L
· Waste and recycle solids concentration: 12,000 mg/L
By using the formulas shown in appendix, determine thefollowing:
(b) Aeration tank volume (m3).
(c) Mass flow rate of wasted sludge (kg/day). (1 mark)
(d) Volumetric flow rate of wasted sludge (m3/day). (1 mark)
(e) Return (recycle) flow rate (m3/day).
(f) Volumetric BOD loading to the aeration tank (kg BOD/m3.day). (1mark)
(g) Food to microorganisms (F/M) ratio for the aeration tank (kgBOD day/kg MLVSS). (1 mark)
(h) Design hydraulic retention time (hr). (1 mark)
