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

Correct Answer: 48.65

Given: Windbreak height, $H = 2.7\, {m}$ Actual wind speed, $V = 9.0\, {m/s}$ Critical wind speed to move soil, $V_c = 9.6\, {m/s}$ Step 1: Compute wind speed ratio \[ \frac{V}{V_c} = \frac{9.0}{9.6} = 0.9375 \] Step 2: Use empirical relationship \[ {Distance of full protection} = \alpha \cdot H \] \[ {From empirical data: If } \frac{V}{V_c} = 0.9375 \Rightarrow \alpha \approx 18 \] Step 3: Calculate protection distance \[ D = 18 \cdot 2.7 = \boxed{48.65 \, {m}} \]