Wind turbine output: 12 m blade, 18 m/s

The output of a wind turbine is the power in the wind multiplied by the efficiency (power coefficient):

$$P = \frac{1}{2}\rho A v^3 \eta$$

Given

• $\rho = 1.225$ kg/m³
• $r = 12$ m
• $v = 18$ m/s
• $\eta = 0.45$

Step 1: Swept area

$$A = \pi r^2 = \pi \times 12^2 = 452.39\ \text{m}^2$$

Step 2: Power available in the wind

$$\begin{aligned} P_w &= \frac{1}{2}\rho A v^3 \\ &= 0.5 \times 1.225 \times 452.39 \times 18^3 \\ &= 0.5 \times 1.225 \times 452.39 \times 5832 \\ &= 1{,}615{,}980\ \text{W} \approx 1.616\ \text{MW} \end{aligned}$$

Step 3: Output power

$$P = \eta P_w = 0.45 \times 1{,}615{,}980 = 727{,}191\ \text{W}$$

Answer: Output power $\approx 727.2$ kW (about 0.727 MW).

Note: 45% is below the Betz limit of 59.3%, so it is a realistic value.