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.

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