A drainage system for a 40 acre field is designed to remove 3/8 inch of standing water from the entire field in 24 hours.
What flow rate (gpm) must the drainage system achieve to meet the design specification?
volume of water 40 acre * 3/8 " = 40 acre * 1/16 ft = 40*43560/16 ft^3 = 108900 ft^3
1 gal = 231 in^3 = 231/1728 = .13368 ft^3
so, our field is covered by 814632 gal of water
to drain that in 24 hours requires a rate of 814632/24 gal/hr = 33943 gal/hr = 565.72 gal/min
oops
3/8 " is not 1/16 ft.
make that fix and adjust the final figure by the right factor.
To find the flow rate in gallons per minute (gpm) that the drainage system must achieve, we need to convert the given specifications into a consistent unit and then calculate the flow rate.
First, let's convert the 40-acre field area into square feet, as most drainage systems are designed in terms of square footage:
1 acre = 43,560 square feet
So, 40 acres = 40 * 43,560 = 1,742,400 square feet.
Next, we need to convert the depth of standing water to the same unit. Since 1 inch is equivalent to 0.0833 feet, the standing water depth of 3/8 inch is:
(3/8) * 0.0833 = 0.03125 feet.
Now, we can calculate the volume of water that needs to be removed from the field:
Volume = Area * Depth
= 1,742,400 square feet * 0.03125 feet
= 54,450 cubic feet.
To convert cubic feet to gallons, we need to know that 1 cubic foot is equal to 7.48052 gallons. So, the volume in gallons is:
54,450 cubic feet * 7.48052 gallons/cubic foot = 406,581.366 gallons.
Since we want to remove this volume of water in 24 hours (1 day), the flow rate required would be:
Flow rate = Volume / Time
= 406,581.366 gallons / 24 hours
= 16,941.724 gallons per hour.
To convert the flow rate to gallons per minute, divide this by 60:
Flow rate = 16,941.724 gallons per hour / 60
= 282.362 gpm (rounded to three decimal places).
Therefore, the drainage system must achieve a flow rate of approximately 282.362 gpm to meet the design specification.