Question

A 60-cm diameter well is located in an aquifer with a transmissivity of 125 m2/day and a storativity of 5 x10-2. A fault, which acts as a barrier to flow, is located 300 m from the well pumping at 2700 m3/d. At the end of 1 year of pumping, what is the drawdown at the midpoint between the fault and the well

Answer 1

Answer:

The drawdown at the midpoint is $K = 29.11 \ m$

Explanation:

From the question we are told that

  The  transmissivity  is  $T = 125 \ m^3/day$

   The diameter is  $d = 60 \ cm = 0.60 \ m$

    The storativity is  $S = 5 *10^{-2}$

    The distance  of the vault from the pumping well is  $d = 300 \ m$

   The pumping rate is  $R = 2700 \ m^3 / day$

Generally the radius is mathematically represented as

         $r = \frac{ d}{2}$

=>       $r = \frac{ 0.60 }{2}$

=>       $r = 0.30 \ m$

Generally the time is  1 year =  365  days as stated in the question

Generally the drawdown at the midpoint between the fault and the well is mathematically represented as

      $K = \frac{R }{ 4 \pi * T } * ln ((2.2459)*\frac{T * t }{r * S} )$

=>    $K = \frac{2700 }{ 4* 3.142 * 125 } * ln ((2.2459 ) * \frac{125 * 365}{03^2 * 5 *10^{-2}} )$

=>    $K = 29.11 \ m$