Question Bank • QB_IDE_ADV_091 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 well taps an unconfined aquifer with a horizontal impermeable bed. The initial saturated thickness of the aquifer before pumping is $H = 30.0\text{ m}$, and hydraulic conductivity is $K = 12.0\text{ m/day}$. During steady pumping, water table elevations observed in two observation wells are $h_1 = 27.0\text{ m}$ at radial distance $r_1 = 20.0\text{ m}$ and $h_2 = 29.0\text{ m}$ at $r_2 = 80.0\text{ m}$. Using Dupuit's steady-state unconfined flow equation, the well pumping discharge $Q$ in cubic meters per day ($\text{m}^3/\text{day}$) is ________ (round off to nearest integer). Take $\pi = 3.1416$.

Answer Key:

3046

Step-by-Step Solution & Explanation

**Method 1: Dupuit Unconfined Aquifer Formulation** $Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$ 1. Parameters: - $K = 12.0\text{ m/day}$ - $h_1 = 27.0\text{ m} \implies h_1^2 = 729.0\text{ m}^2$ - $h_2 = 29.0\text{ m} \implies h_2^2 = 841.0\text{ m}^2$ - $h_2^2 - h_1^2 = 841.0 - 729.0 = 112.0\text{ m}^2$ - $\frac{r_2}{r_1} = \frac{80.0}{20.0} = 4.0$ - $\ln(4.0) = 1.386294$ 2. Evaluating discharge $Q$: $Q = \frac{3.1416 \times 12.0 \times 112.0}{1.386294} = \frac{4222.31}{1.386294} = 3045.75\text{ m}^3/\text{day} \approx 3046\text{ m}^3/\text{day}$ **Method 2: Equivalent Thiem Average Saturated Thickness** $\bar{h} = \frac{h_1 + h_2}{2} = \frac{27 + 29}{2} = 28.0\text{ m}$ $s_1 - s_2 = (30 - 27) - (30 - 29) = 3.0 - 1.0 = 2.0\text{ m}$ $Q = \frac{2 \pi K \bar{h} (s_1 - s_2)}{\ln(r_2/r_1)} = \frac{2 \times 3.1416 \times 12 \times 28 \times 2}{1.3863} = \frac{4222.3}{1.3863} = 3045.7\text{ m}^3/\text{day}$
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