Food-to-Microorganism Ratio
Mediumactivated sludgef/m ratio
An aeration basin of 2,000 m³ holds an MLSS of 3,000 mg/L and receives 5,000 m³/day of wastewater with a BOD of 250 mg/L. The food-to-microorganism ratio is most nearly:
- A0.08 per day
- B0.21 per day
- C0.42 per day
- D0.83 per day
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Answer: B. 0.21 per day
The F/M ratio compares the daily BOD load to the mass of microorganisms: F/M = QS₀/(VX) = (5,000)(250)/[(2,000)(3,000)] = 1,250,000/6,000,000 ≈ 0.21 per day. Conventional activated sludge typically runs between 0.2 and 0.5, so this basin sits at the low-rate end.
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