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 = [tex]\frac{Q*In (\frac{R_1}{R_2}) }{2 \pi B(H_2-H_1)}[/tex]
K = [tex]\frac{2000*In (\frac{160}{453}) }{2 \pi *33(250-249)}[/tex]
K = [tex]\frac{2081.43662}{207.3451151}[/tex]
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 = [tex]\frac{Q* In(\frac{R_2}{R_1} )}{2 \pi B (H_2-H_1)} =\frac{Q* In(\frac{R_2}{R_3} )}{2 \pi B (H_2-H_3)}[/tex]
[tex]\frac{ In(\frac{453}{160} )}{(250-249)} =\frac{ In(\frac{453}{100} )}{ (250-H_3)}[/tex]
H₃ = 248.548 m
Drawdown (s) = H₂ - H₃
s = (250 - 248.548)m
s = 1.452 m
