GATE 2021 ES – Question 42

Water and Wastewater Treatment and Management · Sewage Collection and Treatment · 2 marks · Multiple choice

An STP treats20000 m³/day with 200 mg/L suspended solids. Primary clarification removes60% into sludge containing5 wt% solids, density1000 kg/m³. Find sludge volume in m³/day.

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Correct answer: (A) 48

Explanation

Feed suspended solids=20000(.200)=4000 kg/day. Captured solids=.60(4000)=2400 kg/day. Wet sludge mass=2400/.05=48000 kg/day. Volume=48000/1000=**48 m³/day, A**. Five percent is a mass fraction of sludge, not the removal efficiency.

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