Problem 2 Initial water levels with respect to an arbitrary

Problem 2. Initial water levels (with respect to an arbitrary datum) and the levels after 8 hours and 22 minutes of pumping are tabulated below for six observation wells in the vicinity of the pumped well, in an unconfined aquifer. The average pumping rate for the test period is 0.0312 m\' sec. The aquifer is a mixture of clayey sand and gravel. The effective radius of the pumped well is 0.3 m. Also assume that the drawdown everywhere is small relative to the initial saturated thickness, making it permissible to use the equations for confined flow to analvze the flow in this aquifer. 1 2 3 4 5 7 Well number Distance from pumped well (m) 4.6 10.4 19.8 34.5 65.8 91.8 115.2 Initial water level (m) 96.49 96.52 96.53 96.5596.56 96.5796.517 water level after 8 hr 22 min 95.14 95.58 95.8896.14 96.3696.43 96.56 Estimate the volume of the aquifer that had its water pumped out. Calculate the specific yield using the data provided

Solution

Given

Average pumping rate Q = 0.0312m³/s^-1

Radius = 0.3m

Time t= 8hr 22min

From the given table

H₁ = h₂ - h₁ =96.49-95.14 =1.36m

H₂ = h₂ - h₁ =96.52-95.58 =0.94m

H₃ = h₂ - h₁ =96.53-95.88 =0.65m

H₄ = h₂ - h₁ =96.55-96.14 =0.41m

H₅= h₂ - h₁ =96.56-96.36 =0.2m

H₆ = h₂ - h₁ =96.57-96.43 =0.14m

H₇ = h₂ - h₁ =96.57-96.56 =0.01m

H_net = 1.36+0.94+0.65+0.41+0.2+0.14+0.01

        = 3.71m

1) Q=v/t

0.0312 = Vol_pumped                

              8hr 22min

Volume pumped out =939.744m3

2)Specific Yield = Q t    

                        7.84V

                            = 0.0312 x 8hr 22 min  

                                  7.48 x 939.744

            Specific Yield    = 0.133