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GATE 2021 ES – Question 60

Water and Wastewater Treatment and Management · Sewage Collection and Treatment · 2 marks · Numerical answer

A 10 MLD activated-sludge plant receives primary-treated sewage BOD₅140 mg/L into a 1500 m³ aeration tank. MLVSS3000 mg/L;effluent BOD₅5 mg/L. Find F/M in day⁻¹,to two decimals.

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Correct answer: .25 to .35

Explanation

Food loading uses influent substrate:Q S₀=10000(.140)=1400 kg BOD/day. Biomass inventory=VX=1500(3)=4500 kg VSS. F/M=1400/4500=.31111 day⁻¹,or **.31 day⁻¹**. Subtracting effluent BOD would give a removal loading, not the conventional influent F/M.

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