Where the wind power equation comes from
Wind carries kinetic energy, and a rotor intercepts a stream of it. Air crossing a disc of area A at speed v delivers a mass flow of ρAv kilograms per second, and each kilogram carries ½v² joules. Multiply and the power crossing the disc is ½ρAv³. That is the wind power density times the swept area, and it is the entire supply available before the turbine takes any of it.
Two consequences dominate every design decision in wind energy. First, power goes as the cube of speed: a site with 20% more wind has 73% more power in it, and a site with double the wind has eight times. Second, power goes as the square of rotor diameter, because area does. Those two exponents explain why turbines have grown enormous and why siting matters more than hardware. A taller tower buys speed, and speed is cubed.
What the rotor actually takes is Cp, the power coefficient. It cannot be one. Extracting all the kinetic energy would leave the air motionless behind the disc, which would stop the flow from continuing through it. Albert Betz worked out in 1919 the maximum possible: 16/27, about 59.3%, achieved when the rotor slows the wind to two-thirds of its upstream speed. Real rotors peak somewhere below that and fall away either side of their design tip-speed ratio.
Each term, and why air density is not 1.225
Air density. The value 1.225 kg/m³ is the International Standard Atmosphere at sea level and 15 °C, and it is wrong nearly everywhere. Density follows the ideal gas law, ρ = p/(RT) with R = 287.05 J/(kg·K) for dry air, and pressure falls with elevation on the standard lapse, p = 101325(1 − 2.25577×10⁻⁵h)^5.25588. At 1,500 m on a standard day the density is 1.058 kg/m³ — 13.6% below sea level, so 13.6% less power from the same wind through the same rotor. Temperature matters too: at fixed pressure, air at −10 °C is about 9.5% denser than air at 15 °C.
Swept area. A = πD²/4. Use the full rotor diameter including hub, since that is the disc the momentum theory refers to. The swept-area calculator covers this step alone if that is all you need.
Wind speed. This must be the speed at hub height. Anemometer data is usually recorded at 10 m, and wind shear means hub-height speed is meaningfully higher. Applying a 10 m speed to a 60 m hub understates power badly, and the cube law amplifies the error: a speed 15% low gives 0.85³ = 61.4% of the true power, a 39% shortfall.
Cp and η. Keep them separate. Cp is aerodynamic and is bounded by Betz; η is electromechanical and is bounded by 1. Manufacturers sometimes publish a single overall coefficient that folds both together, and comparing that to a rotor-only Cp makes one machine look worse than it is. The percentage-of-Betz output on this page uses Cp alone, which is the only fair comparison against the 0.5926 ceiling.
Worked example: a 10 m rotor in a 10 m/s wind
A small commercial turbine has a 10 m rotor, a power coefficient of 0.35 and a drivetrain that is 90% efficient. The site is at sea level, the air is at 15 °C, and the wind at hub height is 10 m/s.
- Swept area. A = π × 10² ÷ 4 = 78.54 m².
- Air pressure. At sea level, p = 101,325 Pa.
- Air density. ρ = 101,325 ÷ (287.05 × 288.15) = 101,325 ÷ 82,714 = 1.2250 kg/m³.
- Wind power density. ½ × 1.2250 × 10³ = 0.6125 × 1,000 = 612.5 W/m².
- Power crossing the disc. 612.5 × 78.54 = 48,106 W.
- Shaft power. 48,106 × 0.35 = 16,837 W.
- Electrical power. 16,837 × 0.90 = 15,153 W = 15.15 kW.
- Share of Betz. 0.35 ÷ 0.5926 = 59.1% of the theoretical rotor maximum.
Now move the same machine to 2,000 m elevation at 5 °C. Pressure falls to about 79,500 Pa and density to roughly 0.996 kg/m³, so every figure from step 4 onward scales by 0.996 ÷ 1.225 = 0.813. Output drops to about 12.3 kW on identical hardware in identical wind. Nothing about the turbine changed; there is simply less air.
And if the wind rises from 10 to 12 m/s, output goes to 15.15 × (12/10)³ = 15.15 × 1.728 = 26.2 kW — except that a 25 kW machine would be pitching to hold 25 kW by then. That flat top is the defining feature of a real power curve.
Reading the result honestly
The instantaneous figure is a snapshot, and a turbine spends almost none of its life at any single wind speed. What matters commercially is the integral of the power curve against the site's wind distribution — the annual energy production, and from it the capacity factor. A machine rated at 25 kW will not produce 25 kW × 8,760 hours; a good site delivers somewhere in the region of a quarter to a third of that.
Use the power density output as a site screen before you consider any hardware. It is a property of the wind alone, independent of what you put in it, which makes it the honest way to compare locations. Because it goes as v³, a modest difference in mean speed between two candidate sites translates into a large difference in resource — and because the mean of the cubes exceeds the cube of the mean, the annual power density at a site is always higher than what its mean speed alone would suggest.
Treat the percentage-of-Betz figure as a plausibility check on your Cp. A rotor operating above 100% of the Betz limit is not a discovery; it is an error in the inputs, most often a Cp that already includes the drivetrain, or a swept area entered as radius rather than diameter.
The annual energy estimate on this page assumes a Rayleigh distribution, which is the Weibull distribution with shape factor k = 2. It is a reasonable default for many temperate open sites and a poor one for sites with strong seasonal or diurnal structure. It also omits cut-in and cut-out behaviour, availability, wake losses and electrical losses, all of which reduce real output. Use it for scale, not for a bankable number.
Power against wind speed for a 10 m rotor
| Wind speed (m/s) | Wind speed (mph) | Power density (W/m²) | Output (kW) |
|---|---|---|---|
| 2 | 4.5 | 4.9 | 0.121 |
| 4 | 8.9 | 39.2 | 0.970 |
| 6 | 13.4 | 132.3 | 3.273 |
| 8 | 17.9 | 313.6 | 7.759 |
| 10 | 22.4 | 612.5 | 15.153 |
| 12 | 26.8 | 1,058.4 | 26.185 |
| 14 | 31.3 | 1,680.7 | 41.581 |
| 16 | 35.8 | 2,508.8 | 62.068 |
| 18 | 40.3 | 3,572.1 | 88.374 |
| 20 | 44.7 | 4,900.0 | 121.226 |
Every row is the wind power equation evaluated at that speed. Doubling the speed from 10 to 20 m/s multiplies output by exactly 8, which you can confirm against the last two rows: 121.226 ÷ 15.153 = 8.00.
What this calculation leaves out
- Cut-in and cut-out. Below roughly 3 m/s a turbine cannot overcome its own friction; above roughly 25 m/s it shuts down to protect itself. The equation happily returns a number in both regimes.
- Cp is not constant. It varies with tip-speed ratio and therefore with wind speed. A single value is a design-point approximation; a real power curve is measured, not derived.
- Turbulence and yaw error. A rotor misaligned by angle θ sees roughly cos³θ of the available power, so a 15° yaw error costs about 10%.
- Wake losses. Turbines downstream of others in an array see slowed, more turbulent air. Array losses of several percent to well over ten percent are ordinary and are handled at wind-farm level, not here.
- Blade soiling, icing and blade erosion. All reduce Cp in service, and none appear in a nameplate figure.
- Humidity. Moist air is slightly less dense than dry air at the same pressure and temperature. The effect is under a percent in most conditions and is ignored here.
Betz's law is a limit on the rotor, not on the machine
The 16/27 figure comes from one-dimensional momentum theory applied to an actuator disc in an unbounded, incompressible, steady flow. It bounds the fraction of the kinetic energy flux through the disc that any such rotor can convert to shaft power. It does not bound the drivetrain, it does not apply to ducted or diffuser-augmented rotors in the same form, and it is not a statement about efficiency in the thermodynamic sense. A well-designed modern rotor reaching Cp near 0.48 is capturing about 81% of what Betz allows, which is close to the practical ceiling once tip losses and finite blade count are accounted for.
From instantaneous power to a project
Instantaneous power is the first calculation in wind energy and the least useful one on its own. The sequence that follows it runs: measure or model the wind distribution at hub height, integrate it against the manufacturer's measured power curve to get gross annual energy, deduct wake, availability, electrical and environmental losses to get net energy, and divide by rated power times 8,760 to get the capacity factor. From there the levelized cost of energy follows in the same way it does for any generation asset.
For small and micro wind, two practical notes. Rotor diameter dominates everything — doubling diameter quadruples output, while doubling Cp is impossible. And roof and low-tower mounting is almost always a mistake, because buildings and trees produce exactly the turbulent, slowed air the cube law punishes hardest. If the mean speed at your proposed hub height is below about 4.5 m/s, the arithmetic on this page will tell you what you need to know before you spend anything.
Where a stream is available instead of open wind, the same energy-flux logic gives a much steadier resource, because water is roughly 800 times denser than air — see the micro-hydro calculator for that version of the problem.