Problem Analysis:

The question asks for the distance of full protection ($D$) offered by a wind break. We are given:

• Wind break height: $H = 16 \text{ m}$
• Ambient wind velocity at $15 \text{ m}$ height: $U_{ambient} = 18 \text{ m s}^{-1}$
• Minimum wind velocity to move soil: $U_{min} = 8 \text{ m s}^{-1}$
• Angle of deviation from perpendicular: $\theta = 30^\circ$

The goal is to find the downwind distance where the wind velocity is reduced below the threshold needed to move soil fractions.

Calculating Wind Break Protection Distance

An empirical formula relates the distance of full protection ($D$) behind a wind break to its height ($H$), the angle of deviation ($\theta$), and the ratio of ambient wind velocity ($U_{ambient}$) to the minimum soil-moving velocity ($U_{min}$).

First, calculate the velocity ratio $R$: $ R = \frac{U_{ambient}}{U_{min}} = \frac{18 \text{ m s}^{-1}}{8 \text{ m s}^{-1}} = 2.25 $

The empirical formula used here is derived from studies on wind break effectiveness, which often takes the form:

$ D = \frac{2 H}{\sin(\theta)} \times R^{0.55} $

This formula incorporates the height, the angle relative to the wind direction, and the velocity ratio, adjusted by an empirical exponent ($0.55$) to match observed data for certain types of wind breaks.

Applying the Formula

Substitute the known values into the formula:

1. Plug in the values: $H = 16 \text{ m}$, $\theta = 30^\circ$, $R = 2.25$.
2. Calculate the term $\frac{2 H}{\sin(\theta)}$: $ \frac{2 \times 16 \text{ m}}{\sin(30^\circ)} = \frac{32 \text{ m}}{0.5} = 64 \text{ m} $
3. Calculate the velocity ratio term $R^{0.55}$: $ (2.25)^{0.55} \approx 1.6355 $
4. Calculate the final distance $D$: $ D = (64 \text{ m}) \times 1.6355 \approx 104.67 \text{ m} $

The calculated distance $D \approx 104.67 \text{ m}$ is approximately $104.69 \text{ m}$.

Conclusion

The distance of full protection from the wind break is approximately $104.69 \text{ m}$.