Question

The largest wind turbines have rotor diameters \(^{40}\) around \(150 \mathrm{~m}\). Using a sensible efficiency of \(50 \%\), what power does such a jumbo turbine deliver at a maximum design wind speed of \(13 \mathrm{~m} / \mathrm{s} ?\)

Short answer

Answer: The power delivered by the wind turbine at a wind speed of 13 m/s and an efficiency of 50% is approximately 6214 kW.

Step-by-step solution

Write the formula for wind power

The formula for wind power is given by: P = (1/2) * ρ * A * v^3 * η, where P is the power delivered by the turbine, ρ is the air density, A is the area swept by the rotor, v is the wind speed, and η is the efficiency of the turbine. We are given η = 0.5 (50%), and we will estimate ρ = 1.225 kg/m^3 as the air density at sea level in a standard atmosphere.

Calculate the area swept by the rotor

The area swept by the rotor, A, can be calculated using the formula: A = π * (D/2)^2, where D is the diameter of the rotor. In this case, D = 150 m, so: A = π * (150/2)^2 = 17671.5 m^2.

Use the given wind speed

We are given the maximum design wind speed as v = 13 m/s.

Calculate the power delivered by the turbine

Now, we can use the wind power formula to calculate the power delivered by the turbine: P = (1/2) * 1.225 kg/m^3 * 17671.5 m^2 * (13 m/s)^3 * 0.5 P ≈ 6214 kW So, at a wind speed of 13 m/s and an efficiency of 50%, such a large wind turbine delivers a power of approximately 6214 kW.

Key concepts

Wind Power Formula

To calculate the power generated by a wind turbine, we use the wind power formula: \[ P = \frac{1}{2} \times \rho \times A \times v^3 \times \eta \]Where:

• \( P \) is the power produced by the wind turbine.
• \( \rho \) represents the air density, typically around 1.225 kg/m³ at sea level.
• \( A \) is the rotor swept area.
• \( v \) is the wind speed, given in meters per second.
• \( \eta \) is the turbine's efficiency, often provided as a percentage.

The formula quantifies how much kinetic energy in the wind is converted into electrical energy by the turbine. Maximizing each variable can improve the power output from a wind turbine.

Turbine Efficiency

Turbine efficiency, denoted as \( \eta \), is a measure of how well a turbine converts wind's kinetic energy into electrical energy. The maximum theoretical efficiency limit is known as the Betz limit and is approximately 59.3%. However, practical turbines operate below this limit.
In this scenario, the turbine efficiency is provided as 50%. This is a reasonable estimate, acknowledging some energy losses due to mechanical, electrical, and aerodynamic factors.
The efficiency parameter is crucial because it directly influences the power output. A higher efficiency means less wind energy is wasted, making the turbine more effective.

Rotor Diameter

A wind turbine's rotor diameter significantly impacts its power output. It determines the area over which the turbine can capture wind energy. The larger the diameter, the greater the area, thus capturing more wind.To calculate the swept area \( A \), used in the power formula, use:\[ A = \pi \times \left(\frac{D}{2}\right)^2 \]Where \( D \) is the rotor diameter. For a diameter of 150 meters, the calculation gives a large swept area of approximately 17,671.5 m².
A larger rotor allows the turbine to harness more wind energy, essential for generating higher power, especially at lower wind speeds.

Air Density

Air density, represented as \( \rho \), plays a vital role in wind power calculations. It influences how much kinetic energy is available in the wind for the turbine to convert into electricity.
Air density depends on atmospheric conditions, primarily temperature and altitude. Standard sea level conditions give an average air density of 1.225 kg/m³. This value is often used in calculations unless specific local atmospheric data indicates otherwise.
Higher air densities mean more energy potential in the wind. Conversely, if air density decreases—say at higher altitudes—less energy is available to be captured by the turbine.