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Suppose two wells are located 6 miles apart, and are drilled into a aquifer of average sand composition (i.e., hydraulic conductivity = 10 m/day). The water level in one of the wells is 40 feet, and the water level in the other well is 75 feet. If the aquifer is 3 miles wide and 50 feet deep, which of the following numbers is closest to the aquifer's actual discharge? (1 m = 3.28 feet)

Mathematics
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Q= KA(H/L) Groundwater discharge can be calculated with the following equation:  in which Q is an aquifer’s discharge, K is the aquifer’s hydraulic conductivity, A is the cross-sectional area of the aquifer, H is the difference in elevation of the water table between two wells, and L is the horizontal distance between the two wells. (Note that H/L is the gradient, or slope, of the water table.) 150 m3/day 800 m3/day 1,200 m3/day 8,000 m3/day
H/L appears to be 35/(6*5280) feet
and I guess they want you to use a rectangle to approximate the CSA?

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Other answers:

yes 79,200ft
ft^2
\[Q=\frac{ (32.808ft)(79200ft ^{2})(35ft }{ 31680ft }\]
then i converted to meters
\[Q=\frac{ (9.997)(24140.16)(10.668) }{ 9656.06 }\]
will ft^2 convert to meters^2?
You seem to have an error there...
Better to convert the height and width to meters first... then find the CSA in meters^2
working
15.24 m * 4828 m
73,578.72m^2
yes
((9.997)(73578.72)(10.668))/(9656.06)
working
812.65m^3/day

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