A windbreak, 15 m in height and 200 m in length, is established to protect the land from wind erosion in an arid area. The minimum wind velocity at the height of 15 m above the ground required to move the most erodible soil fraction is $9.6\ m\cdot s^{-1}$. If 5-year return period wind velocity at 15 m height is $16\ m\cdot s^{-1}$ and the wind direction deviates $20^\circ$ from the line perpendicular to the windbreak, the area protected by the windbreak in ha is ________________.
[round off to 2 decimal places]
Windbreak Protected Area Calculation
This solution details the calculation of the protected area provided by a windbreak, considering factors like its dimensions and wind direction deviation.
Effective Windbreak Length Calculation
The effective length ($L_{eff}$) of the windbreak is its actual length adjusted for the angle of the wind relative to it.
- Given Windbreak Length, $L = 200\ m$.
- Given Wind Direction Deviation, $\theta = 20^\circ$.
- The formula for effective length is $L_{eff} = L \cos(\theta)$.
- Substituting the values: $L_{eff} = 200\ m \times \cos(20^\circ)$.
- Using the value $\cos(20^\circ) \approx 0.9397$, the effective length is $L_{eff} \approx 200 \times 0.9397 = 187.94\ m$.
Downwind Protected Distance Estimation
The downwind protected distance ($D$) is the distance behind the windbreak where wind velocity is reduced. This is often estimated empirically based on the windbreak's height ($H$).
- Given Windbreak Height, $H = 15\ m$.
- A common empirical estimation for the protected distance is $D \approx 10H$.
- Calculating the distance: $D = 10 \times 15\ m = 150\ m$.
Protected Area Calculation
The protected area ($A$) is calculated by multiplying the effective length by the downwind protected distance.
- The formula is $A = D \times L_{eff}$.
- Using the calculated values: $A \approx 150\ m \times 187.94\ m$.
- This gives an area of $A \approx 28191\ m^2$.
Area Conversion to Hectares
The question asks for the area in hectares (ha). Conversion is necessary.
- Conversion factor: $1\ ha = 10000\ m^2$.
- To convert the area to hectares, divide by 10000: $A_{ha} = A / 10000$.
- Calculation: $A_{ha} \approx 28191\ m^2 / 10000\ m^2/ha$.
- Result: $A_{ha} \approx 2.8191\ ha$.
Final Result Rounding
Rounding the calculated area to two decimal places as required:
- The protected area is approximately $2.82\ ha$.
Important Questions from Wind Erosion
- A $16 \text{ m}$ high wind break is constructed to protect soil from wind erosion due to wind velocity of $18$ $\text{m s}^{-1}$ at $15 \text{ m}$ height. The minimum wind velocity at $15 \text{ m}$ height capable of moving the soil fraction is $8 \text{ m s}^{-1}$. The angle of deviation of prevailing wind direction from the perpendicular to the wind break is $30^\circ$. The distance of full protection from the wind break in $\text{m}$ is
In wind erosion, the maximum portion of soil is transported by the process of ____________.
- In wind erosion process, the rate of soil movement (S) depends on wind speed (v), threshold limit of wind speed required to move the soil ($v_m$) and average soil particle size (d). This interrelationship is expressed as
- 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)
- In wind erosion, the amount of soil transported varies
