Question Bank • QB_IDE_ADV_073 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:

A subsurface relief pipe drainage system is installed in a homogeneous agricultural soil overlying an impermeable horizontal floor. The hydraulic conductivity of the soil is $K = 1.50\text{ m/day}$. The depth of the impermeable floor below the drain centers is $D = 6.0\text{ m}$, yielding an equivalent depth of $d = 3.20\text{ m}$. The steady design drainage recharge rate is $q = 0.0040\text{ m/day}$ ($4.0\text{ mm/day}$) and the maximum permissible water table rise midway between drains above drain level is $h = 1.00\text{ m}$. Using Hooghoudt's equation, $q = \frac{8 K d h + 4 K h^2}{L^2}$, the required drain spacing $L$ in meters is ________ (round off to two decimal places). ``` Ground Surface ───────────────────────────────────────────────────────────── | | | Water Table | | _--~~~~~~~~~--_ | | / h \ | h = 1.0 m Drain Pipe (O)─────────' / \ `─────────(O) Drain Pipe | | | d = 3.2 m (Equivalent depth) | L (Drain spacing) | | ═════════════╧═════════════════════════════════════════════════════════╧════════════ Impermeable Floor ```

Answer Key:

105.36

Step-by-Step Solution & Explanation

**Method 1: Hooghoudt Drainage Equation** $q = \frac{8 K d h + 4 K h^2}{L^2} \implies L = \sqrt{\frac{8 K d h + 4 K h^2}{q}}$ 1. Parameters: - $K = 1.50\text{ m/day}$ - $d = 3.20\text{ m}$ - $h = 1.00\text{ m}$ - $q = 0.0040\text{ m/day}$ 2. Calculating numerator terms: $\text{Term}_1 = 8 K d h = 8 \times 1.50 \times 3.20 \times 1.00 = 38.40\text{ m}^3/\text{day}$ $\text{Term}_2 = 4 K h^2 = 4 \times 1.50 \times (1.00)^2 = 6.00\text{ m}^3/\text{day}$ $\text{Numerator} = 38.40 + 6.00 = 44.40\text{ m}^3/\text{day}$ 3. Drain spacing $L$: $L^2 = \frac{44.40}{0.0040} = 11,100.0\text{ m}^2$ $L = \sqrt{11,100.0} = 105.356\text{ m} \approx 105.36\text{ m}$ **Method 2: Flow Zone Contribution Check** - Semi-confined flow below drains: $8 K d h / q = 38.40 / 0.004 = 9600\text{ m}^2$ ($86.5\%$ of capacity) - Unconfined flow above drains: $4 K h^2 / q = 6.00 / 0.004 = 1500\text{ m}^2$ ($13.5\%$ of capacity) - $L = \sqrt{9600 + 1500} = \sqrt{11100} = 105.36\text{ m}$
← Previous (QB_IDE_ADV_001) Next (QB_IDE_ADV_080) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test