Q.3. The effluent from a primary clarifier has the following characteristics: Fl
ID: 951115 • Letter: Q
Question
Q.3. The effluent from a primary clarifier has the following characteristics: Flow = 0.029m3/s; BOD5 = 240 mg/l; Growth constants: Ks = 100mg/l; m = 10 d-1; kd = 0.025 d-1; Y = 0.8 mg VSS/mg BOD5 removed; the design MLVSS is 3000 mg/l This effluent is further treated in an activated sludge system in order to meet an effluent standard of 25 mg/l BOD5 and 30 mg/l suspended solids (SS). Assuming that BOD5 of the SS may be estimated as equal to 70% of the SS concentration, estimate the required volume of the aeration tank.
Explanation / Answer
Solution:
MLVSS: Mixed Liquor Volatile Suspended solids in mg/l =0.7 x MLSS = 3000 mg/lt
Hydraulic retention time =V/Q
Where V=Aeration tank volume in cu.m
Q=Effluent flow in cu.m per day
Hence, substituting these values,
Here taking basis of 1400 lt for volume selection, so we get
0.15=240/(v/1400 * 0.7* 3000)
V=968 KL, selected volume =1400 KL
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