Question

The drainage coefficient of an area is 3.6 cm/day. What would be the discharge in \(\text{m}^3\text{/s}\) from an area of 60 ha?

• A. \(25\text{ m}^3\text{/s}\)
• B. \(0.25\text{ m}^3\text{/s}\)
• C. \(15\text{ m}^3\text{/s}\)
• D. \(0.025\text{ m}^3\text{/s}\)

Hint:

Fast shortcut: $Q = \frac{60 \times 10^4 \times 0.036}{86400} = \frac{21600}{86400} = 0.25\text{ m}^3\text{/s}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Drainage Coefficient (DC) is the design depth of excess water to be removed from a drainage basin in a 24-hour period.
Key Formula or Approach:
\[ Q = \frac{A \times \text{DC}}{86,400} \]
where \(A\) is in \(\text{m}^2\) and \(\text{DC}\) is in meters/day.

Step 2: Detailed Explanation:
Given parameters:
- Drainage coefficient: \(\text{DC} = 3.6\text{ cm/day} = 0.036\text{ m/day}\)
- Agricultural area: \(A = 60\text{ ha} = 60 \times 10,000\text{ m}^2 = 600,000\text{ m}^2\)

Step 1: Calculate total daily drainage volume:
\[ V = A \times \text{DC} = 600,000\text{ m}^2 \times 0.036\text{ m} = 21,600\text{ m}^3\text{/day} \]
Convert daily discharge into cubic meters per second (\(1\text{ day} = 86,400\text{ seconds}\)):
\[ Q = \frac{21,600\text{ m}^3}{86,400\text{ s}} = \frac{216}{864} = \frac{1}{4} = 0.25\text{ m}^3\text{/s} \]

Step 3: Final Answer:
Therefore, the drainage discharge is \(0.25\text{ m}^3\text{/s}\), corresponding to option (B).