Question Bank • QB_IDE_ADV_065 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Irrigation Water Conveyance and Application Methods

Subtopic: Surface, sprinkler and micro irrigation methods

Question Statement:

A border strip is irrigated with an inflow stream of $Q = 0.030\text{ m}^3/\text{s}$ ($108.0\text{ m}^3/\text{h}$). The soil has an average infiltration capacity of $I = 20.0\text{ mm/h} = 0.020\text{ m/h}$, and the average depth of surface flow is $y = 75.0\text{ mm} = 0.075\text{ m}$. Using the Lewis-Milne equation, $t = 2.303 \frac{y}{I} \log_{10}\left(\frac{Q}{Q - I A}\right)$, the time $t$ required for water to cover a border strip area of $A = 3000.0\text{ m}^2$ in hours is ________ (round off to two decimal places).

Answer Key:

3.04

Step-by-Step Solution & Explanation

**Method 1: Lewis-Milne Border Advance Formulation** 1. Infiltration discharge over target area ($I A$): $I A = 0.020\text{ m/h} \times 3000.0\text{ m}^2 = 60.0\text{ m}^3/\text{h}$ 2. Net inflow parameter: $Q - I A = 108.0 - 60.0 = 48.0\text{ m}^3/\text{h}$ $\frac{Q}{Q - I A} = \frac{108.0}{48.0} = 2.25$ $\log_{10}(2.25) = 0.35218$ 3. Depth-to-infiltration coefficient: $2.303 \times \frac{y}{I} = 2.303 \times \frac{0.075}{0.020} = 2.303 \times 3.75 = 8.63625\text{ hours}$ 4. Time of advance $t$: $t = 8.63625 \times 0.35218 = 3.0415\text{ hours} \approx 3.04\text{ hours}$ **Method 2: Natural Logarithm Evaluation** $t = \frac{y}{I} \ln\left(\frac{Q}{Q - I A}\right) = 3.75 \times \ln(2.25) = 3.75 \times 0.81093 = 3.041\text{ hours}$
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