**A 40 cm diameter well fully penetrates an unconfined aquifer whose bottom is 75 m below the undisturbed ground water table. When pumped at a steady rate of 1.40 $m^3/min$. The drawdown observed in observation wells at radial distance of 4 m and 14 m are 4 m and 2 m respectively. Determine the drawdown in the well.**

$$H=75$$ $$Q=1.40\ m^3/min=\frac{1.40}{60}=0.0233\ m^3/s$$ S1 = 4m h1 = H – s1 = 75 – 4 = 71 at r1 = 4

S2 = 2m h2 = H – s2 = 75 – 2 = 73 at r2 = 14

$$Q=\frac{1.36k(H^2-h_1^2)}{log\frac{R}{r_1}}=\frac{1.36k(H^2-h^2)}{log\frac{R}{r_2}}$$ $$\frac{(75^2-71^2)}{log\frac{R}{r_1}}=\frac{(75^2-73^2)}{log\frac{R}{r_2}}$$

$$\log\frac{R}{14}=0.50684\log_{10}\left(\frac{R}{4}\right)$$ $$\log\frac{R}{14}=\log_{10}\left(\frac{R}{4}\right)^{0.50684}$$ $$R^{0.50684}=6.9343$$ $$R=50.72$$ Coefficient of permeability $$Q=\frac{1.36k(H^2-h^2)}{\log\left(\frac{R}{r}\right)}$$ $$0.0233=\frac{1.36k(75^2-71^2)}{\log\left(\frac{50.72}{4}\right)}$$ $$k=3.236\times10^{-5}$$ Drawdown in the well (h) of diameter 40 mm $$Q=\frac{1.36k(H^2-h^2)}{\log\left(\frac{R}{r}\right)}$$ $$0.0233=\frac{1.36\times3.236\times10^{-5}(75^2-h^2)}{\log\left(\frac{50.72}{0.2}\right)}\ \ \ \ \ \ \ r=0.4/2$$ $$h=66.10$$ Drawdown at the well s = H – h = 75 – 66.10 = 8.9 m

The drawdown in the well = 8.9 m