Question

The water requirement of a crop is 280 mm and the electrical conductivity (EC) of irrigation and percolation water below the root zone respectively are 2 and 8 ds/m respectively. How much irrigation water is required to meet both the crop water requirement and the leaching requirement?

• A. 350 mm
• B. 373.3 mm
• C. 280 mm
• D. 1120 mm

Hint:

USSL Leaching Formula: $D_{iw} = \frac{ET}{1 - LR} = \frac{280}{1 - (2/8)} = \frac{280}{0.75} = 373.3\text{ mm}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Leaching Requirement (LR) is the fraction of total applied irrigation water that must pass beyond the root zone to prevent root zone salt accumulation.
Key Formula or Approach:
\[ \text{LR} = \frac{\text{EC}_{iw}}{\text{EC}_{dw}} \]
\[ D_{iw} = \frac{\text{ET}}{1 - \text{LR}} \quad \text{(USSL Steady-State Leaching Equation)} \]

Step 2: Detailed Explanation:
Given parameters:
- Crop evapotranspiration requirement: \(\text{ET} = 280\text{ mm}\)
- Electrical conductivity of irrigation water: \(\text{EC}_{iw} = 2\text{ dS/m}\)
- Electrical conductivity of drainage percolation water: \(\text{EC}_{dw} = 8\text{ dS/m}\)

Step 1: Compute Leaching Requirement fraction (LR):
\[ \text{LR} = \frac{\text{EC}_{iw}}{\text{EC}_{dw}} = \frac{2}{8} = 0.25 \]
Calculate total gross irrigation depth (\(D_{iw}\)):
\[ D_{iw} = \frac{\text{ET}}{1 - \text{LR}} = \frac{280\text{ mm}}{1 - 0.25} = \frac{280}{0.75} = \frac{280 \times 4}{3} = \frac{1120}{3} = 373.33\text{ mm} \approx 373.3\text{ mm} \]

Step 3: Final Answer:
Therefore, total irrigation water required is 373.3 mm, matching option (B).