FAQ

1. Volume of Aeration tank , 2. HRT, 3. F/M Ratio, 4. MLSS value.
The relevant details of the ETP are as below:
Influent flow = 10 m3 / day
In. flow BOD = 1200 mg / L
In.flow COD = 2200 mg / L
TSS  = 850 mg / L
  Jeyaroopa
 
jeyaroopa79@gmail.com

 

Dear Jayroopa

You may please refer Water & Waste Water Engineering by Metcalf & Eddy.
BOD of effluent is 1200 mg/l. It will be difficult to bring down the BOD by Activated Sludge process to desire permissible limits.

Simplified method of calculation is as below.

Process Activated sludge
Flow, CUM 10
Type Extended aeration
Food to Microorganism ratio (F/M) 0.15
Total BOD load, Kg BOD*Flow/1000.= 12
Total mixed liquor suspended solids(MLSS) Total BOD/(F/M) = 80 kg
MLSS in Tank say mg/l 3000
Volume of tank, CUM Total MLSS/(MLSS/1000) = 26.6
Retention time, hr 24*Volume of Tank/Flow = 63.84

You can select F/M & MLSS values and optimize0 the volume of a tank.
The waste water has high suspended solids therefore sedimentation is must.

Maximum efficiency of Activated sludge process is 95% therefore if you need lower value of BOD in treated effluent then you have to opt two stage process.
For further information please contact