An activated sludge aeration basin is a rectangular tank 120 ft long × 60 ft wide with a water depth (SWD) of 15 ft. The basin has 2 ft of freeboard (do not include freeboard in liquid volume). Average flow to the aeration basin is 2.4 MGD. Current MLSS is 2,800 mg/L and SVI is 140 mL/g.

Using 1 ft³ = 7.48 gal and 1 day = 24 hr, what is the theoretical detention time in the aeration basin (hours)?

Question

An activated sludge aeration basin is a rectangular tank 120 ft long × 60 ft wide with a water depth (SWD) of 15 ft. The basin has 2 ft of freeboard (do not include freeboard in liquid volume). Average flow to the aeration basin is 2.4 MGD. Current MLSS is 2,800 mg/L and SVI is 140 mL/g.

Using 1 ft³ = 7.48 gal and 1 day = 24 hr, what is the theoretical detention time in the aeration basin (hours)?

Accepted Answer

A) 8.1 hr — Correct. Basin volume = 120 ft × 60 ft × 15 ft = 108,000 ft³, then convert: 108,000 ft³ × 7.48 gal/ft³ = 807,840 gal. Detention time = 807,840 gal ÷ 2,400,000 gal/day = 0.3366 day, and 0.3366 day × 24 hr/day = 8.08 hr ≈ 8.1 hr. Sanity check: a ~0.8 MG basin at ~2.4 MGD should be about 0.33 day (~8 hr).

Answer

B) 1.1 hr — Wrong because it skips the ft³ → gal conversion and effectively treats 108,000 ft³ as 108,000 gal (or 0.108 MG). That underestimates volume by a factor of 7.48, giving about 1.1 hr instead of about 8 hr.

Answer

C) 0.34 hr — Wrong because it converts gallons to days but then reports the result as hours without multiplying by 24 hr/day. 0.3366 day is about 8.1 hr, not 0.34 hr.

Answer

D) 9.2 hr — Wrong because it includes the 2 ft freeboard in the liquid depth (using 17 ft instead of 15 ft). That inflates the volume and detention time to about 9.2 hr, but freeboard is not part of the wastewater volume.

Detention Time Calculation with Extraneous Data

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