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

Expert verified
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

01

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.
02

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.
03

Use the given wind speed

We are given the maximum design wind speed as v = 13 m/s.
04

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.

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