Question Bank • QB_IDE_ADV_080 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Agricultural Drainage

Subtopic: Planning, design and layout of surface and sub-surface drainage systems

Question Statement:

In an unsteady state subsurface drainage scenario governed by the Glover-Dumm equation, the mid-spacing water table height above drain level decays as $h_t = \frac{4}{\pi} h_0 e^{-\alpha t}$, where drainage factor $\alpha = \frac{\pi^2 K d}{\mu L^2}$. An initial water table rise of $h_0 = 1.20\text{ m}$ must be lowered to $h_t = 0.60\text{ m}$ within $t = 2.0\text{ days}$. The soil has hydraulic conductivity $K = 1.00\text{ m/day}$, equivalent depth $d = 2.00\text{ m}$, and drainable porosity (specific yield) $\mu = 0.050$. Taking $\pi = 3.1416$, the required drain spacing $L$ in meters is ________ (round off to one decimal place).

Answer Key:

29.1

Step-by-Step Solution & Explanation

**Method 1: Glover-Dumm Transient Water Table Recession** 1. Water table drawdown ratio: $\frac{h_t}{h_0} = \frac{0.60\text{ m}}{1.20\text{ m}} = 0.50$ 2. Glover-Dumm first-term equation: $h_t = \frac{4}{\pi} h_0 e^{-\alpha t} \implies e^{-\alpha t} = \frac{\pi}{4} \left( \frac{h_t}{h_0} \right) = \frac{3.1416}{4} \times 0.50 = 0.39270$ 3. Solving for $\alpha$: $-\alpha t = \ln(0.39270) = -0.93472$ $\alpha = \frac{0.93472}{t} = \frac{0.93472}{2.0\text{ days}} = 0.46736\text{ day}^{-1}$ 4. Drain spacing $L$ relation: $\alpha = \frac{\pi^2 K d}{\mu L^2} \implies L^2 = \frac{\pi^2 K d}{\mu \alpha}$ $L^2 = \frac{(3.1416)^2 \times 1.00\text{ m/day} \times 2.00\text{ m}}{0.050 \times 0.46736\text{ day}^{-1}} = \frac{9.8696 \times 2.00}{0.023368} = \frac{19.7392}{0.023368} = 844.71\text{ m}^2$ $L = \sqrt{844.71} = 29.064\text{ m} \approx 29.1\text{ m}$ **Method 2: Dimensionless Time Factor Check** $u = \alpha t = 0.935 \implies L = \sqrt{\frac{\pi^2 K d t}{\mu u}} = \sqrt{\frac{9.8696 \times 1 \times 2 \times 2}{0.05 \times 0.935}} = \sqrt{844.5} = 29.06\text{ m} \approx 29.1\text{ m}$
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