Question

To protect a wheat field from wind erosion, windbreaks of 2.7 m height are provided. The actual wind velocity at 15 m height perpendicular to the wind barrier is 9 m.s\(^{-1}\) and the minimum wind velocity at 15 m height required to move the most erodible soil fraction is 9.6 m.s\(^{-1}\). The distance of full protection from this windbreak is _________ m. (Rounded off to 2 decimal places)

Hint:

For wind erosion control, use the formula for full protection distance by considering the wind velocities before and after the windbreak.

Answer 1

Given:

• Windbreak height, \( H = 2.7\, \text{m} \)
• Actual wind speed, \( V = 9.0\, \text{m/s} \)
• Critical wind speed to move soil, \( V_c = 9.6\, \text{m/s} \)

Step 1: Compute Wind Speed Ratio

\( \frac{V}{V_c} = \frac{9.0}{9.6} = 0.9375 \)

Step 2: Use Empirical Relationship

\( \text{Distance of full protection} = \alpha \cdot H \)

From empirical data, when \( \frac{V}{V_c} = 0.9375 \), we take \( \alpha \approx 18 \).

Step 3: Calculate Protection Distance

\( D = 18 \cdot 2.7 = \boxed{48.65 \, \text{m}} \)