Question

Consider the design of a wastewater treatment plant (WWTP) for a community with average
daily and peak hourly wastewater design flows of 7570 m3/day and 18925 m3/day. The raw
sewage has an average of 240 mg/L BOD5 and 270 mg/L of suspended solids.

(a) Assume that the primary sedimentation process removes 55% of the suspended solids and
35% of the BOD5 of the raw sewage. Determine the SS and BOD5 concentrations in the
primary sedimentation effluent flow. (1 mark)

The primary effluent is treated by two parallel trains of the complete-mix activated sludge
process. Assume with the average flow conditions and the performance of primary
sedimentation achieved as above. Given the following parameters required 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 the following:
(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). (1 mark)
(g) Food to microorganisms (F/M) ratio for the aeration tank (kg BOD day/kg MLVSS). (1 mark)
(h) Design hydraulic retention time (hr). (1 mark)

Answer 1