A steady discharge of the well is 18.72 $m^3/hr$ the drawdown at a distance of $R_1$ equal to 15 m from the centre of the test when is found to be 1.8 m what will be the drawn down at a distance of $R_2$ = 50m if the permeability of the soil is 3.8*$10^{-4}$ m/s. Also estimate approximate Drone down at the test will also.

q = 18.72 $m^3/hr = \frac{18.72}{60*60} = 5.2. * 10^{-8} m^3/s$

1)

$R_1 = 15m$, $h_1 = 1.8m$, $k = 3.8*10^{-4} m^3/s$, $R_2 = 50m$

$k = \frac{2.303*5.2*10^{-3}*\log_{10}\frac{50}{15}}{\pi[h_2^2(1.8)^2]}$

$h_2 = 2.91m$