A sedimentation basin has a flow of 0.5 million gallons per day and surface area 2,000 ft². What is the loading rate in gpd/ft²?

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Prepare for the ADEQ Water Treatment Grade 4 Test. Study with flashcards and multiple-choice questions, with each question offering hints and explanations. Ace your exam!

Multiple Choice

A sedimentation basin has a flow of 0.5 million gallons per day and surface area 2,000 ft². What is the loading rate in gpd/ft²?

Explanation:
Loading rate tells you how much water passes through each square foot of surface per day. To find it, divide the total daily flow by the surface area. Here, 0.5 million gallons per day equals 500,000 gpd, and the basin area is 2,000 ft². So 500,000 ÷ 2,000 = 250 gpd per ft². The loading rate is 250 gpd/ft².

Loading rate tells you how much water passes through each square foot of surface per day. To find it, divide the total daily flow by the surface area. Here, 0.5 million gallons per day equals 500,000 gpd, and the basin area is 2,000 ft². So 500,000 ÷ 2,000 = 250 gpd per ft². The loading rate is 250 gpd/ft².

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