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An aquifer test with a constant pumping rate of 200 gpm was conducted in a confined aquifer with a saturated thickness of 12
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Answer #1

Ans) Given,

Pumping rate (Q) = 200 gpm

Thickness of aquifer (D)= 12 m

Distance between observation well and pumping well (r) = 250 ft

Step 1) Find to from the plot

From the above pot we can see that at time = 0.30 min , drawdown = 0

therefore, to = 0.30 min

Step 2) Find \Delta Z for values t1 = 10 min and t2 = 100 min

Z(drawdown) at t = 10 min , S = 0.70 m

At t = 100 min , S = 1.10 m

\DeltaZ=1.10 - 0.70

\DeltaZ = 0.40 m

or 1.31 ft

Step 3) Compute Transmissivity (T),

T = 264 Q / \Delta Z

  T = 264 x 200 / 1.31

  T = 40305.34 gpd / ft

or 500.5 m2/day

Step 4) Compute storativity (S)

S = Tt0 / 4790r2

S = 40305.34 x 0.30 /(4790 x 2502)

S = 4.04 x 10-5

Step 5) Compute Hydraulic conductivity (K)

T = K x D

500.5 = K x 12

K = 41.708 m/day

or K = 4.83 x 10-4 m/s

Step 6) Compute Specific storage (Ss)

S = Ss x D

4.04 x 10-5 = Ss x 12

Ss = 3.36 x 10-6 m-1

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