Question

Surface drainage coefficient of a watershed is 1.44 mm/day. What should be the capacity of the surface drain at the outlet of the watershed, if its area is 500 ha ?

• A. $3 \text{ m}^3\text{/min}$
• B. $10 \text{ m}^3\text{/min}$
• C. $5 \text{ m}^3\text{/min}$
• D. $50 \text{ m}^3\text{/min}$

Hint:

To perform this conversion quickly, remember the conversion factor:
$1 \text{ ha-mm/day} = 0.0694 \text{ m}^3\text{/min}$.
Here, $\text{Total ha-mm} = 500 \times 1.44 = 720 \text{ ha-mm/day}$.
$Q = 720 \times 0.0694 \approx 5 \text{ m}^3\text{/min}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The drainage coefficient ($D$) is the depth of water that must be drained from a given watershed area in a 24-hour period.
To find the required drain capacity (discharge $Q$), we multiply the drainage coefficient by the watershed area and convert the units to $\text{m}^3/\text{min}$.
Key Formula or Approach:
The formula for peak drainage discharge is: \[ Q = \text{Area} \times \text{Drainage Coefficient} \] Convert units:
- $1 \text{ ha} = 10,000 \text{ m}^2$
- $1 \text{ mm} = 10^{-3} \text{ m}$
- $1 \text{ day} = 1440 \text{ minutes}$

Step 2: Detailed Explanation:
Given values from the problem:
- Drainage coefficient ($D$) = $1.44 \text{ mm/day} = 1.44 \times 10^{-3} \text{ m/day}$
- Watershed Area ($A$) = $500 \text{ ha} = 500 \times 10,000 = 5 \times 10^6 \text{ m}^2$
First, calculate the total daily discharge volume ($Q_{\text{day}}$): \[ Q_{\text{day}} = A \times D \] \[ Q_{\text{day}} = \left( 5 \times 10^6 \text{ m}^2 \right) \times \left( 1.44 \times 10^{-3} \text{ m/day} \right) = 7,200 \text{ m}^3\text{/day} \] Second, convert this daily discharge rate to cubic meters per minute ($\text{m}^3/\text{min}$): \[ \text{Total minutes in a day} = 24 \times 60 = 1440 \text{ minutes} \] \[ Q_{\text{min}} = \frac{7,200 \text{ m}^3}{1440 \text{ min}} = 5 \text{ m}^3\text{/min} \] Thus, the required capacity of the surface drain is $5 \text{ m}^3\text{/min}$.

Step 3: Final Answer:
The required drain capacity is $5 \text{ m}^3\text{/min}$, which corresponds to Option (C).