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The 500-bed Lotta Hart Hospital has a small activated sludge plant to treat its wastewater. The average daily hospital discharge is 1,500 L per day per bed, and the average soluble BOD5 after primary settling is 500 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide by 10.0 m long by 4.5 m deep. The plant operating parameters are as follow:
MLVSS = 2000 mg/L, MLSS = 1.2 x MLVSS, and return sludge concentration = 12,000 mg/L (VSS).
Determine:
(a) Aeration Period in hrs.
(b) F/M ratio.

Answer :

mubarakadedeji7

Answer:

1. t= 0.3days = 7.2hrs

2. F/M = 0.69kg BOD / kg MLSS . day

Explanation:

1. Aeration time or hydralic retention time = Volume / Flow - rate

                                         t = Va / Q

where Va= aeration tank volume

           Q=  flow rate

   Va= 10 x 10 x 4.5 =450

   Q= 1500

     t = 450/ 1500= 0.3days = 7.2hrs

2. Food to microorganism ratio = Mass of BOD applied per day / Mass of suspended solids in aeration tank

F / M = ( Qo )( BODo ) / ( Va ) ( MLSS )

Where BOD5= 500 mg/L

            Va= 450

           MLVSS= 2000

            MLSS= 1.2 X MLVSS= 1.2 X 2000= 2400

       F / M =  \frac{1500 × 500}{450 × 2400}

               = \frac{750000}{1080000}

               = 0.69kg BOD / kg MLSS . day

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