Lesson Nine: Calculating Detention Time

Detention time is the amount of time that water remains in a basin as the water travels from the entrance point to the exit point of the basin. All detention time calcuations on introductory-level water treatment examinations will involve water that is passing through flocculation basins or sedimentation basins. The answers to detention time problems will be expressed in minutes, hours, or days.
An interesting and easily understandable way to view detention time is to view it as the time needed to completely fill a basin if the basin is completely empty. Two numerical figures are always needed to calculate detention time. The first figure is the total number of gallons that the basin holds. The second figure is the flow of water through the basin, expressed as gpm, gph, or gpd.
To calculate detention time, simply divide the number of gallons that the basin holds by the flow through the basin in either gpm, gph, or gpd. Below are several good examples of detention time problems.

1. If a basin holds 50,000 gallons and the flow through the basin is 50,000 gph, what is the detention time through the basin as expressed in hours?

Answer: 50,000 gallons divided by 50,000 gph = 1 hour

2. If a basin holds 75,000 gallons and the flow through the basin is 2 MGD, what is the detention time through the basin expressed in hours?

Answer: The first step is to convert 2 MDG to gph. To do this, divide 2,000,000 by 24 to obtain 83,333 gph. Next, divide 75,000 by 83,333. The answer is .9 hours or 9/10 of one hour. Thus, water is detained in this basin for .9 hours.

3. If a basin holds 75,000 gallons and the flow through the basin is .85 MGD, how many hours does it take for water to flow through the basin?

Answer: First, divide 850,000 by 24 to obtain 35,417 gph. Next, divide 75,000 by 35,417. The answer is 2.1 hours.

4. A water treatment plant has a flocculation basin that holds 240,000 gallons. This plant is located in a state that requires a minimum flocculation time of thirty minutes. A water treatment plant operator wishes to increase the plant flow from 9.2 MGD to 10.8 MGD. At a flow of 10.8 MGD, what will be the detention time through the flocculation basins in minutes?

Answer: Divide 10,800,000 by 1440 (minutes in a day) to obtain a flow of 7500 gpm. Next, divide the 240,000 gallons that the flocculation basin holds by the flow of 7500 gpm to obtain a detention time of 32 minutes. Yes, this plant operator can legally increase the plant flow to 10.8 MGD.

5. A sedimentation basin is 150 feet wide, 120 feet long, and 45 feet deep. What is the basin detention time in minutes if the plant flow is 24.7 MDG?

Answer: Convert the area of the basin into gallons. To do this, multiply 150 x 120 x 45 to obtain 810,000 cubic feet. Then, multiply 810,000 by 7.48 gallons per cubic foot to obtain a basin volume of 6,058,800 gallons. Next, divide 24,700,000 gallons by 1440 minutes in a day to obtain a flow of 17,153 gpm. Lastly, divide 6,058,800 by 17,153 to obtain a detention time of 353 minutes. (If the problem had asked for detention time in hours, 353 minutes divided by 60 equals 5.88 hours.)