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Sagot :
Answer:the aquifer’s hydraulic conductivity is K = 10.039 m/day
transmissivity T = 331.287 m/day
the drawdown 100 m away from the well is s = 1.452 m
Explanation:Given that :
The constant pumps rate Q = 2000 m³/day
R₁ =160 m → H₁ = 249 m
R₂ = 453 m → H₂ = 250 m
The confined aquifer B is 33 m thick
The hydraulic conductivity K =
K =
K =
K = 10.039 m/day
Transmissivity T = K × B
T = 10.039×33
T = 331.287 m/day
TO find the drawdown 100 m away from the well; we have:
K =
H₃ = 248.548 m
Drawdown (s) = H₂ - H₃
s = (250 - 248.548)m
s = 1.452 m
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