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## ludwig457 Group Title 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) one year ago one year ago

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1. ludwig457 Group Title

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

2. Algebraic! Group Title

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

3. Algebraic! Group Title

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

4. ludwig457 Group Title

yes 79,200ft

5. ludwig457 Group Title

ft^2

6. ludwig457 Group Title

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

7. ludwig457 Group Title

then i converted to meters

8. ludwig457 Group Title

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

9. ludwig457 Group Title

will ft^2 convert to meters^2?

10. Algebraic! Group Title

You seem to have an error there...

11. Algebraic! Group Title

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

12. ludwig457 Group Title

working

13. Algebraic! Group Title

15.24 m * 4828 m

14. ludwig457 Group Title

73,578.72m^2

15. Algebraic! Group Title

yes

16. ludwig457 Group Title

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

17. ludwig457 Group Title

working

18. ludwig457 Group Title

812.65m^3/day