Waste Water Engineering

A conventional activated-sludge wastewater treatment plant without primary clarification operates under the following conditions:

Design flow = 2.14 MGD;

Influent BOD₅ = 185 mg/L;

Influent SS = 212 mg/L

Aeration basin: 4 units; 40 ft width, 40 ft length and 15.5 ft depth

MLVSS = 2,600 mg/L and it is 80% of MLSS

Waste flow rate = 39,000 gal/day

Concentration of microorganisms in returned activated sludge = 8,600 mg VSS/L

Effluent BOD₅ = 15 mg/L

and Effluent TSS = 15 mg/L

Calculate:

A) Aeration period

B) BOD loading

C) Suspended Solids removal efficiency

D) BOD removal efficiency

E) F/M ration

F) Sludge Age

G) Return activated sludge rate

H) Will this process remove organic nitrogen?