 # At a certain location wind is blowing steadily at 12 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60m diameter blades at that location. Take the air density to be 1.25kg/m^3. This question aims to develop an understanding of the power generation capacity of a wind turbine generator.

A wind turbine is a mechanical device that converts the mechanical energy (kinetic energy to be precise) of the wind into electrical energy.

The energy generation potential of a wind turbine depends upon the energy per unit mass $KE_m$ of the air and mass flow rate of the air $m_{ air }$. The mathematical formula is as follows:

$PE \ = \ KE_m \times m_{ air }$

Given:

$\text{ Speed } \ = \ v \ = \ 10 \ m/s$

$\text{ Diameter } \ = \ D \ = \ 60 \ m$

$\text{ Density of Air } = \ \rho_{ air } \ = \ 1.25 \ kg/m^3$

Part (a) – Kinetic energy per unit mass is given by:

$KE_m \ = \ KE \times \dfrac{ 1 }{ m }$

$KE_m \ = \ \dfrac{ 1 }{ 2 } m v^2 \times \dfrac{ 1 }{ m }$

$\Rightarrow KE_m \ = \ \dfrac{ 1 }{ 2 } v^2$

Substituting values:

$KE_m \ = \ \dfrac{ 1 }{ 2 } ( 12 )^2$

$\Rightarrow KE_m \ = \ 72 \ J$

Part (b) – The energy generation potential of the wind turbine is given by:

$PE \ = \ KE_m \times m_{ air }$

Where $m_{ air }$ is the mass flow rate of air passing through the wind turbine blades which is given by the following formula:

$m_{ air } \ = \ \rho_{ air } \times A_{ turbine } \times v$

Since $A_{ turbine } \ = \ \dfrac{ 1 }{ 4 } \pi D^2$, the above equation becomes:

$m_{ air } \ = \ \rho_{ air } \times \dfrac{ 1 }{ 4 } \pi D^2 \times v$

Substituting this value in the $PE$ equation:

$PE \ = \ KE_m \times \rho_{ air } \times \dfrac{ 1 }{ 4 } \pi D^2 \times v$

Substituting values into this equation:

$PE \ = \ ( 72 ) \times ( 1.25 ) \times \dfrac{ 1 }{ 4 } \pi ( 60 )^2 \times ( 12 )$

$\Rightarrow PE \ = \ 3053635.2 \ W$

$\Rightarrow PE \ = \ 3053.64 \ kW$

## Numerical Result

$KE_m \ = \ 72 \ J$

$PE \ = \ 3053.64 \ kW$

## Example

Calculate the energy generation potential of a wind turbine with a blade diameter of 10 m at a wind speed of 2 m/s.

Here:

$KE_m \ = \ \dfrac{ 1 }{ 2 } v^2$

$\Rightarrow KE_m \ = \ \dfrac{ 1 }{ 2 } ( 2 )^2$

$\Rightarrow KE_m \ = \ 2 \ J$

And:

$PE \ = \ KE_m \times \rho_{ air } \times \dfrac{ 1 }{ 4 } \pi D^2 \times v$

$\Rightarrow PE \ = \ ( 2 ) \times ( 1.25 ) \times \dfrac{ 1 }{ 4 } \pi ( 10 )^2 \times ( 2 )$

$\Rightarrow PE \ = \ 392.7 \ W$