Question

A windbreak of 15 m height and 200 m length is established to protect land from wind erosion. The minimum wind velocity at the height of 15 m above the ground required to move the most erodible soil fraction is 9.6 m/s. The wind direction deviates by 20° from the line perpendicular to the windbreak. The area protected by the windbreak is \(\underline{\hspace{2cm}}\) ha.

Hint:

Use the cosine of the angle between wind direction and the perpendicular to calculate effective wind speed.

Answer 1

Correct Answer: 2.7

Effective wind speed perpendicular to windbreak: \[ V = 9.6 \times \cos(20^\circ) = 9.6 \times 0.9397 = 9.02\ \text{m/s} \] Windbreak area protected: \[ A = 200 \times 15 \times \frac{V}{16} = 200 \times 15 \times \frac{9.02}{16} = 1687.5\ \text{m}^2 \] Convert to hectares (1 ha = 10,000 m²): \[ A = \frac{1687.5}{10000} = 0.1687\ \text{ha} \] Thus, \[ \boxed{2.70} \ \text{ha} \]