Question

A tubewell operating 10 hours per day discharges water at a rate of \(25\ \text{m}^3.\text{h}^{-1}\). The average irrigation depth is 80 mm, and the irrigation interval is 20 days. If the intensity of irrigation is 80%, the culturable command area (in hectares) is ________. (Rounded off to two decimal places)

Hint:

Start by finding the total volume the tubewell delivers over one irrigation interval.

Answer 1

Correct Answer: 7.81

Step 1: Set up one combined formula for the command area.
The area a tubewell can command is found from $\text{CCA} = \dfrac{Q \times t \times T}{d \times I}$, where $Q$ is the hourly discharge, $t$ is the daily running hours, $T$ is the irrigation interval in days, $d$ is the irrigation depth, and $I$ is the irrigation intensity as a fraction.

Step 2: Put in the given numbers with consistent units.
Take $Q = 25\ \text{m}^3/\text{h}$, $t = 10\ \text{h}$, $T = 20$ days, $d = 0.08\ \text{m}$ (80 mm), and $I = 0.80$.
$\text{CCA} = \dfrac{25 \times 10 \times 20}{0.08 \times 0.80}\ \text{m}^2 = \dfrac{5000}{0.064}\ \text{m}^2 = 78125\ \text{m}^2$.

Step 3: Convert square metres to hectares.
Since $1\ \text{ha} = 10000\ \text{m}^2$, $\text{CCA} = \dfrac{78125}{10000} = 7.8125\ \text{ha}$.

Final Answer:
This rounds to 7.81 hectares, matching the step by step check. \[ \boxed{CCA \approx 7.81\ \text{ha}} \]