You must calculate the depth of water in a stormwater pond resulting from a two
ID: 1919909 • Letter: Y
Question
You must calculate the depth of water in a stormwater pond resulting from a two hour rainfall event. the rainfall occurs over a 24-acre watershed at an average rate of 1 inch/hour. the runoff coefficient is 0.5. the average area of the pond is 2 acres and the depth at the beginning of the rainfall is 5 ft. the pond water exfiltates at a rate of 4 in./hr through the pond side walls ( area = 2100 ft^2) when the water is above 5 ft. what is the pond depth above the 5 ft. - level 8.2 hours after the rainfall ends?
Explanation / Answer
Total Volume of Water = Area * Rainfall = 24 acre * 1 inch/hr * 2 hrs
Convert everything into ft
24 acre = 1045440
1 inch = 0.08333 ft
Total Volume = 1045440 *0.08333*2 = 174240 ft^3
Runoff/ hr = 174240*0.5/2 = 43560 ft^3/hr
Total inflow volume = 43560*2 hrs = 87120 ft^3
Now Time = 2 + 8.2 =10.2 hrs
Outflow through wall = 4in/hr * 2100 ft^2 = 4/12 ft/hr *2100 ft^2 = 700 ft^3/hr
Total outflow volume = 700*10.2 = 7140 ft^3
Area of pond = 2 acres = 87120 ft^2
depth = (Inflow - Outflow)/Area of pond = (87120-7140)/ 87120 = 0.918 ft
answer is 0.918 ft
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