The two sedimentation tanks at a surface water treatment plant are each 40' x 90' long and 10'6" deep; one tank is out of service and the plant is treating 2.37 MGD. What is the detention time in the sedimentation stage, in hours?

• A. 2.9 hours
• B. 1.5 hours
• C. 3.6 hours
• D. 4.2 hours

Answer: (A)

Detention time in the sedimentation stage is the volume available for flow divided by the plant flow, using only the active tank(s). Since one sedimentation tank is out of service, use the volume of a single tank: 40 ft × 90 ft × 10.5 ft = 37,800 ft³.

Convert the plant flow to cubic feet per day. 2.37 million gallons per day equals 2.37 × 10^6 gallons/day. Using 1 ft³ = 7.4805 gallons, the flow is 2.37 × 10^6 / 7.4805 ≈ 316,824 ft³/day.

Detention time = volume / flow = 37,800 ft³ / 316,824 ft³/day ≈ 0.119 days. Convert to hours: 0.119 × 24 ≈ 2.86 hours, which rounds to about 2.9 hours.