Question

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)

Answer 1

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

Answer 2

H/L appears to be 35/(6*5280) feet

Answer 3

and I guess they want you to use a rectangle to approximate the CSA?

Answer 4

yes 79,200ft

Answer 5

ft^2

Answer 6

\[Q=\frac{ (32.808ft)(79200ft ^{2})(35ft }{ 31680ft }\]

Answer 7

then i converted to meters

Answer 8

\[Q=\frac{ (9.997)(24140.16)(10.668) }{ 9656.06 }\]

Answer 9

will ft^2 convert to meters^2?

Answer 10

You seem to have an error there...

Answer 11

Better to convert the height and width to meters first... then find the CSA in meters^2

Answer 12

working

Answer 13

15.24 m * 4828 m

Answer 14

73,578.72m^2

Answer 15

yes

Answer 16

((9.997)(73578.72)(10.668))/(9656.06)

Answer 17

working

Answer 18

812.65m^3/day