Question Bank • QB_IDE_GWH_003 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Groundwater Hydrology

Subtopic: Steady and unsteady flow in confined and unconfined aquifers

Question Statement:

A fully penetrating tubewell pumps an unconfined aquifer at a constant discharge $Q = 1200\text{ m}^3\text{/day}$. At observation distances $r_1 = 15\text{ m}$ and $r_2 = 60\text{ m}$, the saturated water table heights above the horizontal impervious bed are $h_1 = 23.0\text{ m}$ and $h_2 = 24.5\text{ m}$, respectively. Using the Dupuit-Thiem formula $Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)}$, the hydraulic conductivity $K$ in $\text{m/day}$ is ________ (round off to 2 decimal places). Take $\pi = 3.1416$.

Answer Key:

7.43

Step-by-Step Solution & Explanation

From Dupuit's steady state radial flow equation for unconfined aquifers: $Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)}$ Given: • $Q = 1200\text{ m}^3\text{/day}$ • $h_2 = 24.5\text{ m} \implies h_2^2 = 600.25\text{ m}^2$ • $h_1 = 23.0\text{ m} \implies h_1^2 = 529.00\text{ m}^2$ • $h_2^2 - h_1^2 = 600.25 - 529.00 = 71.25\text{ m}^2$ • $\ln(60/15) = \ln(4) \approx 1.38629$ $1200 = \frac{3.1416 \times K \times 71.25}{1.38629} = \frac{223.839}{1.38629} K = 161.466 K$ $K = \frac{1200}{161.466} \approx 7.432\text{ m/day} \approx 7.43\text{ m/day}$
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