Question Bank • QB_IDE_ADV_053 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Soil-Water-Plant Relationship

Subtopic: Measurement of infiltration, soil moisture and irrigation water infiltration

Question Statement:

Philip's two-term infiltration equation expresses cumulative infiltration depth as $I = S \cdot t^{1/2} + A \cdot t$, where sorptivity $S = 24.0\text{ mm/h}^{1/2}$ and transmissivity factor $A = 5.0\text{ mm/h}$. The infiltration rate $i = \frac{dI}{dt}$ at elapsed time $t = 4.0\text{ hours}$ in millimeters per hour ($\text{mm/h}$) is ________ (round off to one decimal place).

Answer Key:

11.0

Step-by-Step Solution & Explanation

**Method 1: Derivative of Philip's Equation** 1. Infiltration rate $i(t)$: $i(t) = \frac{dI}{dt} = \frac{1}{2} S \cdot t^{-1/2} + A$ 2. Substituting given parameters at $t = 4.0\text{ h}$: - $S = 24.0\text{ mm/h}^{1/2}$ - $A = 5.0\text{ mm/h}$ - $t^{-1/2} = \frac{1}{\sqrt{4.0}} = \frac{1}{2.0} = 0.50\text{ h}^{-1/2}$ $i(4.0) = \frac{1}{2} \times 24.0 \times 0.50 + 5.0 = 12.0 \times 0.50 + 5.0 = 6.0 + 5.0 = 11.0\text{ mm/h}$ **Method 2: Two-Component Physical Separation** - Sorptive (matric suction) component: $i_{\text{sorptive}} = \frac{24.0}{2 \sqrt{4}} = 6.0\text{ mm/h}$ - Gravity transmission component: $i_{\text{gravity}} = A = 5.0\text{ mm/h}$ - Total rate $= 6.0 + 5.0 = 11.0\text{ mm/h}$
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