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:
\[ PE \ = \ KE_m \times m_{ air } \]
Expert Answer
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:
\[ PE \ = \ KE_m \times \rho_{ air } \times \dfrac{ 1 }{ 4 } \pi D^2 \times v \]
\[ \Rightarrow PE \ = \ ( 2 ) \times ( 1.25 ) \times \dfrac{ 1 }{ 4 } \pi ( 10 )^2 \times ( 2 ) \]
\[ \Rightarrow PE \ = \ 392.7 \ W \]