Discussion
A small horizontal-axis wind turbine (HAWT) is being designed to deliver 2 kW shaft power at a steady wind speed of 7 m.s−1. The HAWT operates with an aerodynamic power coefficient of 0.36, drivetrain efficiency of 0.90, and air density of 1.225 kg.m−3. Neglecting other losses, the required rotor diameter is X m. When the wind speed drops to 6 m.s−1, the shaft power is Y kW. Identify the nearest correct option for X and Y combination
(Take π = 3.14)
Answer: A
The nearest combination is X = 6.12 m and Y = 1.28 kW.
Shaft power is the power in the wind through the swept area, times the power coefficient and the drivetrain efficiency:
\[ \begin{aligned} P &= \tfrac{1}{2}\rho A V^{3} C_{p}\eta \\ 2000 &= 0.5(1.225)A(343)(0.36)(0.90) \\ A &= \frac{2000}{68.07} = 29.38\ \text{m}^{2} \\ D &= \sqrt{\frac{4A}{\pi}} = \sqrt{37.43} = 6.12\ \text{m} \end{aligned} \]
With the same rotor, power varies as the cube of wind speed:
\[ \begin{aligned} Y &= 2\left(\frac{6}{7}\right)^{3} \\ &= 2 \times 0.630 = 1.26\ \text{kW} \end{aligned} \]
That is 1.26 kW, and the option set offers 1.28 alongside the correct diameter, so it is the nearest correct combination, as the question asks. X = 3.06 halves the diameter (it is the radius), X = 6.12 with Y = 2.00 ignores the fall in wind speed, and X = 6.75 does not come from the power equation.
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