This solution outlines the steps to calculate the groundwater drawdown at a specific distance from a pumping well in a confined aquifer using steady-state flow principles.
For steady-state radial flow to a fully penetrating well in a confined aquifer, the drawdown (s) at any radius (r) is given by:
$ s = \frac{Q}{2 \pi K b} \ln\left(\frac{R}{r}\right) $
Where:
The term $\frac{Q}{2 \pi K b}$ is a constant for a given pumping condition and aquifer, often denoted as C.
$ s = C \ln\left(\frac{R}{r}\right) $
$ s_w = C \ln\left(\frac{R}{r_w}\right) $
$ 3 = C \ln\left(\frac{3000}{0.15}\right) $
$ 3 = C \ln(20000) $
Therefore, the constant C is:
$ C = \frac{3}{\ln(20000)} $
$ s = C \ln\left(\frac{R}{r}\right) $
Substitute the value of C and the given R and r:
$ s = \left(\frac{3}{\ln(20000)}\right) \ln\left(\frac{3000}{300}\right) $
$ s = \frac{3}{\ln(20000)} \ln(10) $
Calculate the natural logarithms:
$ \ln(20000) \approx 9.9035 $
$ \ln(10) \approx 2.3026 $
Now, compute s:
$ s \approx 3 \times \frac{2.3026}{9.9035} $
$ s \approx 3 \times 0.2325 $
$ s \approx 0.6975 \text{ m} $
The calculated drawdown of approximately 0.6975 m falls within the range of 0.68 m to 0.71 m.
Water from a confined aquifer having transmissivity of 1000 $m^2 day^{-1}$ is pumped through a fully penetrating well of 300 mm diameter at a rate of 1200 $m^3 day^{-1}$. If the radius of influence is 400 m, the drawdown in the well under steady-state flow condition in meter is __________ [round off to two decimal places]
(Take $\pi = 3.14$)