Wind turbine output: 6 m blade, 8 and 16 m/s

Turbine output power:

$$P = \frac{1}{2}\rho A v^3 \eta, \qquad A = \pi r^2$$

Given

$\rho = 1.225$ kg/m³, $r = 6$ m, $\eta = 0.40$, $v_1 = 8$ m/s, $v_2 = 16$ m/s.

Swept area

$$A = \pi \times 6^2 = 113.10\ \text{m}^2$$

At $v_1 = 8$ m/s

$$\begin{aligned} P_{w1} &= 0.5 \times 1.225 \times 113.10 \times 8^3 \\ &= 0.5 \times 1.225 \times 113.10 \times 512 = 35{,}467\ \text{W} \\ P_1 &= 0.40 \times 35{,}467 = 14{,}187\ \text{W} \end{aligned}$$

At $v_2 = 16$ m/s

$$\begin{aligned} P_{w2} &= 0.5 \times 1.225 \times 113.10 \times 16^3 \\ &= 0.5 \times 1.225 \times 113.10 \times 4096 = 283{,}739\ \text{W} \\ P_2 &= 0.40 \times 283{,}739 = 113{,}495\ \text{W} \end{aligned}$$

Check

$$\frac{P_2}{P_1} = \left(\frac{16}{8}\right)^3 = 8, \qquad 8 \times 14.187 = 113.50\ \text{kW}$$

Answer: Output $\approx 14.19$ kW at 8 m/s and $\approx 113.50$ kW at 16 m/s; doubling the wind speed raises the power 8 times.