Soil and Water Conservation Engineering · Soil and Water Erosion · 2 marks · Numerical answer
Windbreaks 2.7 m high protect a wheat field. Wind perpendicular to the barrier is 9 m/s at 15 m height; the minimum 15 m wind speed that moves the most erodible soil fraction is 9.6 m/s. Find distance of full protection in metres, to two decimals.
Using the empirical windbreak relation $d=17h(V_m/V)\cos\alpha$,the wind is perpendicular to the barrier so $\alpha=0$ and $\cos\alpha=1$. Substitution gives $d=17(2.7)(9.6/9)=\mathbf{48.96}$ m. This is the empirical protection-distance index intended by the key. Since the supplied actual wind is already below the stated erosion threshold, the physical need for a finite protected distance is not established by these data alone.
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