Question

A sewage treatment plant (capacity 10 MLD) treats wastewater using an aerobic biological process. Inlet BOD = 100 mg/L, outlet BOD = 30 mg/L. Microbial yield = 0.5 g VSS/g BOD removed. At 80% plant capacity, the sludge production rate (in kg/day, rounded off to one decimal place) is __________.

Hint:

Sludge = (BOD removed) × (yield coefficient).

Answer 1

Correct Answer: 280

Plant capacity used:
\[ 0.80 \times 10\,\text{MLD} = 8\,\text{MLD}. \]
Flow rate in L/day:
\[ 8\ \text{MLD} = 8\times10^6\ \text{L/day}. \]
BOD removed:
\[ (100 - 30)\ \text{mg/L} = 70\ \text{mg/L}. \]
Mass of BOD removed per day:
\[ 70\ \text{mg/L} \times 8\times10^6\ \text{L/day} = 5.6\times10^8\ \text{mg/day}. \]
Convert mg → kg:
\[ 5.6\times10^8\ \text{mg} = 560\ \text{kg}. \]
Biomass yield: 0.5 g/g = 0.5 kg/kg.
Sludge production:
\[ \text{Sludge} = 0.5 \times 560 = 280\ \text{kg/day}. \]
\[ \boxed{280.0\ \text{kg/day}} \]