Wind, like any moving body, has kinetic energy of movement. Therefore, it can be used as an energy source for water.

The amount of motion energy $\mathit{A}$ is equal to half the product of the mass of the body moving at its speed $\mathit{V}$ squared:

$\mathit{A}\mathit{=}\frac{\mathit{m}{\mathit{V}}^{\mathit{2}}}{\mathit{2}}$

The amount of mass $\mathit{m}$ wind that is coming on any surface $\mathit{F}$ in one second, is determined by the product of air density $\mathit{\rho}$ to the surface $\mathit{F}$, through which the wind passes and its speed $\mathit{V}$:

$\mathit{m}\mathit{=}\mathit{\rho}\mathit{F}\mathit{V}$

Substituting this value of mass $\mathit{m}$ in determining the amount of movement $\mathit{A}$, which is described above, note that the wind energy is equal:

$\mathit{A}\mathit{=}\frac{\mathit{\rho}\mathit{F}{\mathit{V}}^{\mathit{3}}}{\mathit{2}}$

Considering the propeller-type wind wheel, we note that its working plane is described by the blades, there is a circle. The formula of the area of a circle is known, and has the following form:

$\mathit{F}\mathit{=}\frac{\mathit{\pi}{\mathit{D}}^{\mathit{2}}}{\mathit{4}}$

Substituting the area value $\mathit{F}$ its expression, and substituting this expression in determining the energy of motion $\mathit{A}$, get the energy of the air flow in one second:

$\mathit{A}\mathit{=}\frac{\mathit{\rho}\mathit{\pi}{\mathit{D}}^{\mathit{2}}{\mathit{V}}^{\mathit{3}}}{\mathit{8}}$

However, only a fraction of this energy is converted into mechanical work. Professor G. Kh. Sabinin in "The Ideal Windmill Theory" proved that the maximum coefficient of use of wind energy (CWE) $\mathit{\xi}$, cannot be more than 0.687 or 68.7%. Putting this energy conversion factor into work and remembering that the amount of work done in one second of time is power $\mathit{N}$, we will get:

$\mathit{N}\mathit{=}\mathit{\xi}\mathit{A}$

By replacing the expression $\mathit{A}$ its value, we get the power $\mathit{N}$ wind wheels, depending on its diameter $\mathit{D}$:

$\mathit{N}\mathit{=}\frac{\mathit{\xi}\mathit{\rho}\mathit{\pi}{\mathit{D}}^{\mathit{2}}{\mathit{V}}^{\mathit{3}}}{\mathit{8}}$

But, as a rule, in wind turbines the calculated value is the diameter at a given power, which you want to get.

Based on the above, we can determine the required diameter of the windmill $\mathit{D}$, for a given power $\mathit{N}$:

$\mathit{D}\mathit{=}\sqrt{\frac{\mathit{8}\mathit{N}}{\mathit{\xi}\mathit{\rho}\mathit{\pi}{\mathit{V}}^{\mathit{3}}}}$

It is also proved that the power of the wind wheel does not depend on the number of blades. All aspects related to the number of blades, their shape and other nuances in CWE $\mathit{\xi}$.

Equally important is the calculation of the electrical wind turbine - power loss during torque transmission. ${\mathit{\eta}}_{\mathit{m}\mathit{e}\mathit{h}}$ generator, as well as the coefficient of performance of the generator itself ${\mathit{\eta}}_{\mathit{e}\mathit{l}}$.

In this case, the final expression to determine the required diameter of the windmill would look like:

$\mathit{D}\mathit{=}\sqrt{\frac{\mathit{8}\mathit{N}}{\mathit{\xi}\mathit{\rho}\mathit{\pi}{\mathit{V}}^{\mathit{3}}{\mathit{\eta}}_{\mathit{m}\mathit{e}\mathit{h}}{\mathit{\eta}}_{\mathit{e}\mathit{l}}}}$

When installing a generator on one axis with a wind wheel (without multiplier), only losses in the generator are taken into account ${\mathit{\eta}}_{\mathit{e}\mathit{l}}$. If with the help of a windmill only mechanical work will be carried out, then only the transmission efficiency is taken into account ${\mathit{\eta}}_{\mathit{m}\mathit{e}\mathit{h}}$.

And also, you can always use the program **"Aerodinama"**, which takes into account all the nuances of the aerodynamic calculation of the propeller-type wind turbine.