Calculations!:Aeration Tank Volume, HRT, MLSS Values.

Calculations!:Aeration Tank Volume, HRT, MLSS Values.

I am working in a STP & ETP plant. I need to know how to calculate: 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 Flo3, 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 optiimise 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 Prof. R. V. Saraf Director Viraj Envirozing India Pvt. Ltd. 21 Radhakrishna Near SBI, Paud Road, Pune 38 watersgs@vsnl.net 13 Aug, 2008   |  Taral Kumar Dear Jeyaroopa, Though the calculation is not that simple, I will give you a thumb rule. Multiply BOD with Quantity of effluent and divide it by 500 to get aeration tank volume. For example, for 10 cum/day with 1200 BOD, aeration tank volume shall be 10 x 1200 / 500 = 24 cum. That is 2.5 days storage nearly. But that is because the BOD is excessively high. With best regards, Taral Kumar Executive Director Akar Impex Pvt. Ltd. Noida, Uttar Pradesh http://www.indiawaterportal.org/Network/askq/kb/?View=entry&EntryID=343