Dear reader, due to some heavy work.we couldn’t touch with u last few months, OK.

Now we see as an with small example!

**Aeration tank:**

Influent BOD: 250mg/l(after 30% reduction in the Physio-chemical treatment followed by primary clarifier

Process selected: Activated sludge process

**Design Condition:**

F/M (Food/Microbes) =0.15

MLSS (Mixed Liquor Suspended Solids)=3000 mg/l

**Calculation of Aeration Tank Volume**:

F/M: So (?-MLVSS)

F/M: Food to Microbes ratio in d

^{-1}So: influent substrate concentration in mg/l

MLVSS: Mixed Liquor Volatile Suspended solids in mg/l =0.8 x MLSS

? – 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,

0.15=250/(v/1400 x 0.8 x 3000)

V=972 KL, selected volume =1400 KL

I have just read your topic on aeration volume calculation today. It’s simple and very interesting to me but I would like to know where did you get the number 1400 in your formula or equation? 0.15= 250/(v/1400×0.8×3000). Thank you.