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:

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

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

\[ \Rightarrow PE \ = \ 392.7 \ W \]