Sludge Volume from Suspended Solids
Hardsludgesolids
A wastewater plant removes 220 mg/L of suspended solids from a flow of 4,000 m³/day. The sludge is withdrawn at 4% solids with a specific gravity of 1.02. The volume of sludge produced is most nearly:
- A8.8 m³/day
- B21.6 m³/day
- C44 m³/day
- D220 m³/day
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Answer: B. 21.6 m³/day
First the dry solids mass: (4,000 m³/day)(220 g/m³)/1,000 = 880 kg/day. The sludge volume then follows from the solids concentration and density: V = 880/[(0.04)(1.02)(1,000)] ≈ 21.6 m³/day. Dividing by the solids fraction alone, without the specific gravity, gives 22.0 and is close enough to pass unnoticed — but omitting the 0.04 entirely gives a wildly low answer.
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