anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
H/L appears to be 35/(6*5280) feet
anonymous
  • anonymous
and I guess they want you to use a rectangle to approximate the CSA?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.