Question

An irrigation stream of 30 l/s is diverted to a check basin of size \(12\text{ m} \times 10\text{ m}\). The water holding capacity of the soil is 15 %. The average soil moisture content in the root zone prior to applying water is 7 %. How long should the irrigation stream be applied in the basin to replenish the root zone moisture to its field capacity, assuming no loss due to deep percolation?
[The depth of the crop root zone may be assumed as 1.0 m. The apparent specific gravity of the root zone soil is 1.50]

• A. 6 minutes
• B. 8 minutes
• C. 10 minutes
• D. 12 minutes

Hint:

Fast shortcut: $t = \frac{\text{Area} \times A_s \times D \times \Delta M \times 10}{Q} = \frac{120 \times 1.5 \times 1 \times 0.08 \times 1000}{30} = \frac{14400}{30} = 480\text{ s} = 8\text{ min}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The depth and volume of irrigation water required to replenish a depleted root zone to field capacity depends on the moisture deficit percentage, soil apparent specific gravity, root zone depth, and basin surface area.
Key Formula or Approach:
\[ d = \frac{A_s \cdot D \cdot (\text{FC} - \text{M}_i)}{100} \]
\[ V = A \times d, \quad t = \frac{V}{Q} \]

Step 2: Detailed Explanation:
Given parameters:
- Check basin area: \(A = 12\text{ m} \times 10\text{ m} = 120\text{ m}^2\)
- Field capacity: \(\text{FC} = 15\%\)
- Initial moisture content: \(M_i = 7\%\)
- Moisture deficit: \(\Delta M = 15\% - 7\% = 8\%\)
- Apparent specific gravity (bulk density water density): \(A_s = 1.50\)
- Root zone depth: \(D = 1.0\text{ m}\)
- Inflow stream discharge: \(Q = 30\text{ L/s} = 0.030\text{ m}^3\text{/s}\)

Step 1: Calculate net depth of water required:
\[ d = \frac{1.50 \times 1.0\text{ m} \times (15 - 7)}{100} = \frac{1.50 \times 8}{100} = 0.12\text{ m} = 12\text{ cm} \]
Calculate total volume of water required:
\[ V = A \times d = 120\text{ m}^2 \times 0.12\text{ m} = 14.4\text{ m}^3 = 14,400\text{ litres} \]

Step 2: Compute application time duration:
\[ t = \frac{14,400\text{ litres}}{30\text{ litres/second}} = 480\text{ seconds} \]
Converting to minutes:
\[ t = \frac{480}{60} = 8\text{ minutes} \]

Step 3: Final Answer:
Therefore, the irrigation stream should be applied for 8 minutes, corresponding to option (B).