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Wind turbine power, tip speed ratio and energy

Short answer

Swept area 1256.6 m²; P_w = 0.5·1.225·1256.6·12³ = 1330 kW, P_m = 532 kW, P_e = 479 kW. Tip speed ratio λ = 3.49. At 80 m, v = 5(8)^0.14 = 6.69 m/s, power density = 183 W/m². Annual energy = 479×0.3×8760 = 1.26×10⁶ kWh.

A horizontal axis wind turbine has a rotor diameter of 40 m. At a wind speed of 12 m/s (air density 1.225 kg/m³) the power coefficient is 0.40, the mechanical-to-electrical efficiency is 0.90 and the rotor turns at 20 rpm. (a) Find the power in the wind, the mechanical power and the electrical power. (b) Find the tip speed ratio. (c) At a site the wind speed measured at 10 m is 5 m/s. Using the power law with exponent 0.14, find the wind speed at a hub height of 80 m and the wind power density there. (d) Taking the electrical power in (a) as the rated power and a capacity factor of 30 %, find the annual energy output.

(a) Powers

Swept area A=πD24=π(40)24=1256.6 m2A = \dfrac{\pi D^2}{4} = \dfrac{\pi (40)^2}{4} = 1256.6\ \text{m}^2.

Pw=12ρAv3=0.5(1.225)(1256.6)(12)3=1.330×106 W=1330 kWPm=CpPw=0.40×1330=532 kWPe=0.90×532=479 kW\begin{aligned} P_w &= \tfrac{1}{2}\rho A v^3 = 0.5(1.225)(1256.6)(12)^3 \\ &= 1.330\times10^{6}\ \text{W} = 1330\ \text{kW} \\ P_m &= C_p P_w = 0.40 \times 1330 = 532\ \text{kW} \\ P_e &= 0.90 \times 532 = 479\ \text{kW} \end{aligned}

(b) Tip speed ratio

ω=2πN60=2π(20)60=2.094 rad/s\omega = \frac{2\pi N}{60} = \frac{2\pi(20)}{60} = 2.094\ \text{rad/s} λ=ωRv=2.094×2012=3.49\lambda = \frac{\omega R}{v} = \frac{2.094 \times 20}{12} = 3.49

(c) Wind shear

vhv10=(h10)α⇒v80=5 (8)0.14=5(1.338)=6.69 m/s\frac{v_h}{v_{10}} = \left(\frac{h}{10}\right)^{\alpha} \Rightarrow v_{80} = 5\,(8)^{0.14} = 5(1.338) = 6.69\ \text{m/s}

Power density (power per unit swept area):

PA=12ρv3=0.5(1.225)(6.69)3=183 W/m2\frac{P}{A} = \tfrac{1}{2}\rho v^3 = 0.5(1.225)(6.69)^3 = 183\ \text{W/m}^2

(d) Annual energy

E=Prated×CF×8760=479×0.30×8760=1.26×106 kWhE = P_{rated}\times CF \times 8760 = 479 \times 0.30 \times 8760 = 1.26\times10^{6}\ \text{kWh}

Answer: Pw=1330P_w = 1330 kW, Pm=532P_m = 532 kW, Pe=479P_e = 479 kW; λ=3.49\lambda = 3.49; v80=6.69v_{80} = 6.69 m/s with 183 W/m²; annual energy ≈1.26\approx 1.26 GWh (1 258 000 kWh).

✓ Answer checked by Claude Opus 5.5 (high effort)

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