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Wind turbine output: 6.5 m blade, 9 and 18 m/s

Short answer

A = π(6.5)² = 132.73 m². At 9 m/s: P = 0.5×1.225×132.73×9³×0.35 ≈ 20.74 kW. At 18 m/s: P = 0.5×1.225×132.73×18³×0.35 ≈ 165.95 kW (8× due to v³).

Assume that we live in an area slightly above sea level that has an air density of 1.225 kg/m³ and we have installed a 35% efficient wind turbine which has a rotor blade radius of 6.5 m. Calculate the output power from the turbine at a wind speed of 9 m/s and again at double the velocity of 18 m/s.

Turbine output power:

P=12ρAv3η,A=πr2P = \frac{1}{2}\rho A v^3 \eta, \qquad A = \pi r^2

Given

ρ=1.225\rho = 1.225 kg/m³, r=6.5r = 6.5 m, η=0.35\eta = 0.35, v1=9v_1 = 9 m/s, v2=18v_2 = 18 m/s.

Swept area

A=π×6.52=132.73 m2A = \pi \times 6.5^2 = 132.73\ \text{m}^2

At v1=9v_1 = 9 m/s

Pw1=0.5×1.225×132.73×93=0.5×1.225×132.73×729=59,267 WP1=0.35×59,267=20,743 W\begin{aligned} P_{w1} &= 0.5 \times 1.225 \times 132.73 \times 9^3 \\ &= 0.5 \times 1.225 \times 132.73 \times 729 = 59{,}267\ \text{W} \\ P_1 &= 0.35 \times 59{,}267 = 20{,}743\ \text{W} \end{aligned}

At v2=18v_2 = 18 m/s

Pw2=0.5×1.225×132.73×183=0.5×1.225×132.73×5832=474,133 WP2=0.35×474,133=165,947 W\begin{aligned} P_{w2} &= 0.5 \times 1.225 \times 132.73 \times 18^3 \\ &= 0.5 \times 1.225 \times 132.73 \times 5832 = 474{,}133\ \text{W} \\ P_2 &= 0.35 \times 474{,}133 = 165{,}947\ \text{W} \end{aligned}

Check

P2P1=(189)3=8,8×20.743=165.95 kW\frac{P_2}{P_1} = \left(\frac{18}{9}\right)^3 = 8, \qquad 8 \times 20.743 = 165.95\ \text{kW}

Answer: Output ≈20.74\approx 20.74 kW at 9 m/s and ≈165.95\approx 165.95 kW at 18 m/s. Doubling the wind speed gives 8 times the power because P∝v3P \propto v^3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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