An activated sludge plant is treating an average flow of 520 gpm. The BOD5 entering the aeration basins is 210 mg/L and the final effluent BOD5 is 18 mg/L. (Influent TSS is 180 mg/L.) Using the pounds formula (lbs/day = mg/L × MGD × 8.34) and 1 MGD = 694.4 gpm, what is the approximate pounds per day (lb/day) of BOD removed across the treatment process?

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A) 1,200 lb/day — Correct. First convert flow: 520 gpm ÷ 694.4 (gpm per MGD) = 0.749 MGD. BOD removed = (210 − 18) mg/L = 192 mg/L, so lbs/day removed = 192 mg/L × 0.749 MGD × 8.34 = about 1,200 lb/day. Sanity check: influent load (~1,311 lb/day) minus effluent load (~112 lb/day) ≈ 1,199 lb/day removed.

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B) 144 lb/day — This is what you get if you forget to multiply by 8.34. 192 mg/L × 0.749 MGD ≈ 144, but that value is not in lb/day without the 8.34 conversion factor.

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C) 833,000 lb/day — This happens if you mistakenly use 520 gpm as if it were 520 MGD (skipping the gpm-to-MGD conversion). That makes the load hundreds of thousands of lb/day, which is not reasonable for a 520 gpm plant.

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D) 1,310 lb/day — This is the approximate influent BOD load, not the BOD removed. It uses 210 mg/L instead of the required concentration difference (210 − 18) mg/L.

Pounds Formula: BOD Removal Calculation

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