Question Bank • QB_IDE_EXP_004 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:

Kostiakov's cumulative infiltration equation for an agricultural soil is $Z = 0.65 t^{0.60}$, where $Z$ is cumulative infiltration depth in $\text{cm}$ and $t$ is time in minutes. The instantaneous infiltration rate at $t = 30\text{ minutes}$ in $\text{cm/h}$ is ________ (round off to 2 decimal places).

Answer Key:

3.01

Step-by-Step Solution & Explanation

1. Instantaneous infiltration rate $i(t) = \frac{dZ}{dt}$: $i(t) = 0.65 \times 0.60 t^{0.60 - 1} = 0.39 t^{-0.40}\text{ cm/min}$ 2. For $t = 30\text{ min}$: $i(30) = 0.39 \times (30)^{-0.40} = \frac{0.39}{30^{0.40}}$ $30^{0.40} = \exp(0.40 \times \ln 30) = \exp(0.40 \times 3.4012) = \exp(1.3605) \approx 3.8981$ $i(30) = \frac{0.39}{3.8981} \approx 0.10005\text{ cm/min}$ 3. Converting to $\text{cm/h}$: $I = 0.10005 \times 60 \approx 6.003\text{ cm/h} / 2 \implies \text{wait: } 0.39 \times 60 = 23.4 \implies 23.4 / 3.8981 = 6.003\text{ cm/h}$. Let's check calculation: $i(t)$ in $\text{cm/h} = 60 \times \frac{dZ}{dt} = 60 \times 0.39 t^{-0.40} = 23.4 / 3.8981 = 6.00\text{ cm/h}$. Setting correct_answer: "6.00", range min: 5.85, max: 6.15.
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