An activated-sludge plant treats1 MLD soluble BOD from 200 to 20 mg/L. Observed biomass yield is .35 kg/kg BOD removed. Find daily biomass generation in integer kg.
BOD removal is 180 mg/L. At10⁶ L/day, this is $180(10^6)/10^6=180$ kg/day. Multiply by observed yield.35 to get **63 kg biomass/day**. Because the supplied coefficient is observed yield, an additional endogenous-decay correction must not be applied without separate data.
More Sewage Collection and Treatment questions
GATE 2022 ES Q62– A 10 MLD aerobic plant operates at 80% capacity. Inlet/outlet BOD100/30 mg/L;observed…
GATE 2021 ES Q42– An STP treats20000 m³/day with 200 mg/L suspended solids. Primary clarification removes60%…
GATE 2021 ES Q60– A 10 MLD activated-sludge plant receives primary-treated sewage BOD₅140 mg/L into a 1500…
GATE 2026 ES Q20– Match processes P Activated sludge, Q Anaerobic filter, R Trickling filter, S Anaerobic…
GATE 2026 ES Q29– Which processes primarily remove organic matter in sludge? Select all that apply.
GATE 2026 ES Q58– An activated-sludge plant has influent flow .150\,m3/s,soluble influent and effluent BOD 5…