Calculate wind turbine power output from rotor size, wind speed, air density, and efficiency, plus annual kWh and the rotor size needed for a target power.
Wind Turbine Power Output Formula
- P = electrical power output (W)
- rho = air density (kg/m3), 1.225 at sea level and 15 C
- A = swept area of the rotor (m2)
- Cp = rotor power coefficient as a fraction; the ideal Betz ceiling is 16/27 (about 0.592593). Enter Cp as a percentage in the calculator
- eta = combined drivetrain and generator efficiency as a fraction; the calculator’s 90% default is illustrative
- v = wind speed (m/s)
- L = blade length, the rotor radius of a horizontal axis turbine (m)
- D = rotor diameter (m); H = rotor height of a vertical axis turbine (m)
- E = annual energy scenario (kWh); F = annual average output as a fraction of the entered-speed reference output (enter percent in the calculator); 8760 = hours in a 365-day year
The power output mode applies the first formula directly: it computes the swept area from your rotor dimensions, multiplies by the wind power density 0.5 * rho * v3, then applies the power coefficient and drivetrain efficiency to give the electrical output in watts or kilowatts. The annual energy mode extends that result with the third formula, multiplying the electrical power by the 8,760 hours in a year and your annual-average fraction of that reference output to calculate a kWh-per-year scenario and, if you enter an electricity price, the yearly dollar value. The rotor sizing mode rearranges the first formula to A = P / (0.5 * rho * Cp * eta * v3), which returns the swept area, rotor diameter, and blade length needed to hit a target power at a given wind speed. Except when solving for air density, select standard density, custom density, or an estimate from elevation and temperature. The estimate retains p = 101325 * exp(-elevation/8434) Pa and rho = p / (287.05 * temperature in kelvins), a constant-scale-height pressure approximation rather than a standard-atmosphere model; it excludes humidity and weather-pressure variation. Its supported range is -500 to 11,000 m and -80 to 60 C. Blank elevation and temperature use 0 m and 15 C. Unit changes convert entered quantities.
Wind Power Density and Elevation Reference Tables
The table below shows how much power the wind carries per square meter of swept area at sea level air density, and the idealized electrical output using illustrative Cp of 35% and 90% drivetrain efficiency. Because power scales with the cube of wind speed, small speed differences change the result dramatically.
| Wind speed | Equivalent | Available power (W/m2) | Illustrative electrical output (W/m2) |
|---|---|---|---|
| 3 m/s | 6.7 mph | 16.5 | 5.2 |
| 5 m/s | 11.2 mph | 76.6 | 24.1 |
| 7 m/s | 15.7 mph | 210.1 | 66.2 |
| 9 m/s | 20.1 mph | 446.5 | 140.7 |
| 10 m/s | 22.4 mph | 612.5 | 192.9 |
| 12 m/s | 26.8 mph | 1,058.4 | 333.4 |
| 15 m/s | 33.6 mph | 2,067.2 | 651.2 |
| 20 m/s | 44.7 mph | 4,900.0 | 1,543.5 |
| 25 m/s | 55.9 mph | 9,570.3 | 3,014.6 |
Air gets thinner with altitude, and thinner air carries less energy at the same wind speed. The next table shows standard-atmosphere air density by elevation and the power you give up compared with an identical turbine at sea level. These are rounded standard-atmosphere reference values, separate from the calculator’s constant-scale-height pressure approximation with an independently entered temperature. Use a measured custom density when available.
| Elevation | Air density (kg/m3) | Power vs sea level |
|---|---|---|
| 0 m (sea level) | 1.225 | 100% |
| 500 m (1,640 ft) | 1.167 | -4.7% |
| 1,000 m (3,281 ft) | 1.112 | -9.2% |
| 1,500 m (4,921 ft) | 1.058 | -13.6% |
| 2,000 m (6,562 ft) | 1.007 | -17.8% |
| 2,500 m (8,202 ft) | 0.957 | -21.9% |
| 3,000 m (9,843 ft) | 0.909 | -25.8% |
Wind Turbine Power Output Example Problems
Example 1: power output of a small HAWT. A horizontal axis turbine has a blade length of 2 m and sees a steady 10 m/s wind at sea level (rho = 1.225 kg/m3). Assume Cp = 35% and drivetrain efficiency = 90%. First find the swept area: A = pi * 22 = 12.57 m2. The power available in the wind is 0.5 * 1.225 * 12.57 * 103 = 7,697 W. Applying the power coefficient gives 7,697 * 0.35 = 2,694 W of aerodynamic power, and the drivetrain leaves 2,694 * 0.90 = 2,425 W, so the turbine delivers about 2.42 kW.
Example 2: sizing a rotor for a target power. You want 5 kW of electrical output at 8 m/s from a site at 1,500 m elevation and 10 C. The estimated air density is about 1.044 kg/m3, so the electrical power density is 0.5 * 1.044 * 83 * 0.40 * 0.90 = 96.2 W/m2 using Cp = 40%. The required swept area is 5,000 / 96.2 = 52 m2, which corresponds to a rotor diameter of √(4 * 52 / pi) = 8.1 m, or a blade length of about 4.1 m.
Wind Turbine Power Output Frequently Asked Questions
Why does wind turbine power depend on the cube of wind speed? The kinetic energy of moving air grows with the square of its speed, and the amount of air passing through the rotor each second grows linearly with speed. Multiplying the two gives a cubic relationship. The practical consequence is that a 10% increase in steady wind speed raises idealized power by about 33%, and doubling steady speed multiplies it by eight while Cp and efficiency are held fixed. Cubing an average speed does not establish average power; real turbines also have rated limits and control behavior.
What is the Betz limit and what Cp should I use? The ideal Betz ceiling is 16/27, about 59.2593% (often rounded to 59.3%). It is a theoretical rotor limit, not achievable electrical efficiency. Use the manufacturer’s Cp at the relevant operating point. The 35% Cp and 90% drivetrain defaults are illustrative, not equipment specifications.
Why is my real annual energy lower than the power output suggests? The power formula describes one steady wind speed. A conventional capacity factor is annual energy divided by rated power times annual hours. This calculator instead multiplies the electrical output at your entered speed by 8,760 hours and your selected annual-average share of that reference output. Enter 30% only for an explicit scenario where annual average output is 30% of that reference; it is not automatically a rated-power capacity factor. Site forecasts require a wind distribution, manufacturer power curve and losses. Optional electricity value uses USD per kWh and is not profit.
