Question

One million liters per day (MLD) of wastewater with a soluble BOD of 200 mg/L is treated in an activated sludge process. The BOD of treated wastewater is 20 mg/L. The observed yield coefficient of the biological system is 0.35.
The daily biomass generation in the system is _____________kg (an integer value).

Answer 1

Correct Answer: 63

Step 1: Calculate the daily BOD removed. \[ \text{BOD removed} = (200 \text{ mg/L} - 20 \text{ mg/L}) \times 1,000,000 \text{ L} = 180,000,000 \text{ mg/day} = 180,000 \text{ g/day} \] Step 2: Calculate the daily biomass generation using the yield coefficient. \[ \text{Biomass generated} = 180,000 \text{ g/day} \times 0.35 = 63,000 \text{ g/day} = 63 \text{ kg/day} \]