Wind Power Output Calculator
Turbine power from rotor size, wind speed, air density, and Cp.
What it does
Wind power grows with the disc the blades sweep and with the cube of wind speed. Doubling diameter quadruples the catch area while doubling wind speed multiplies power eight times, which is why tall towers in windy sites dominate project math.
The tool above takes rotor diameter with wind speed, air density, and the Cp efficiency in metric or imperial. It returns swept area, output power, and the Betz cap power. The default 80 m rotor at 12 m/s with Cp 0.35 gives 1.862 MW from 5026.55 m2 with a 3.155 MW cap.
The formula follows the Department of Energy small wind guidebook: power is Cp times half rho times A times V cubed with A as pi D squared over 4. Cp runs 0.25 to 0.45 in practice against the 0.59 theoretical ceiling, and ratings sit at 59 F sea level conditions.
Siting thresholds frame the rest. Energy agencies flag about 9 mph annual average for small turbines and 13 mph for utility scale, since weak winds cube into thin output. Density falls with altitude and heat, so mountain and desert sites derate against the sea level standard.
How to use
- Pick meters or feet for the rotor.
- Type rotor diameter, not blade length.
- Type average wind speed at hub height.
- Keep density at 1.225 unless altitude is known.
- Type Cp near 0.35 unless the datasheet says otherwise.
- Read output first, swept area and Betz cap beside it.
Formula
- A
- Rotor swept area from diameter D.
- rho
- Air density, 1.225 kg/m3 at sea level standard.
- Cp
- Rotor efficiency, 0.25 to 0.45 typical, 0.593 max.
Imperial diameters and speeds convert first (ft x 0.3048, mph x 0.44704), then run identical SI math, so both systems agree. Power cubes with speed: 10% more wind is about 33% more power.
Worked example
Example: 80 m rotor at 12 m/s
Metric defaults: D 80 m, v 12 m/s, rho 1.225, Cp 0.35.
- Swept area: 3.1416 x 80 x 80 / 4 = 5026.55 m2.
- Wind cube: 12 x 12 x 12 = 1728.
- Available: 0.5 x 1.225 x 5026.55 x 1728 = 5,320,099 W.
- Output: 5,320,099 x 0.35 = 1,862,035 W, shown as 1.862 MW.
- Betz cap: 5,320,099 x 0.593 = 3,154,819 W, shown as 3.155 MW.
- Tool shows 1.862 MW, 5026.55 m2, 3.155 MW.
Reference table
| Item | Value | Note |
|---|---|---|
| 10 m rotor | 78.54 m2 | Small turbine class |
| 80 m rotor | 5026.55 m2 | Default case |
| 120 m rotor | 11309.73 m2 | Utility class |
| Cp practice | 0.25 to 0.45 | Betz cap 0.593 |
| Small wind site | 9 mph up | Annual average floor |
| Utility site | 13 mph up | Annual average floor |
| Rating conditions | 59 F at sea level | DOE standard air |
Limits
- Steady wind assumed; gusts and shear ignored.
- Cp is one number; real curves vary with speed.
- Losses past the rotor, like gearbox and line, excluded.
- Annual energy needs the speed distribution, not one speed.
Common mistakes
- Typing blade length as rotor diameter, halving area math twice.
- Scaling power linearly with wind instead of cubed.
- Setting Cp past the 0.593 Betz cap.
- Using ground wind for hub height output.
- Quoting available power as delivered power.
Try it yourself
Quick checks. Answers sit tucked under each question.
Swept area of a 10 m rotor?
3.1416 x 100 / 4 = 78.54 m2.
Swept area of a 120 m rotor?
3.1416 x 14400 / 4 = 11309.73 m2.
Cp of 0.9 allowed?
No: the tool caps Cp at the 0.593 Betz limit.
Questions
What does the default turbine give?
1.862 MW at Cp 0.35 from an 80 m rotor in 12 m/s wind.
What is swept area?
The disc the blades cover: pi times diameter squared over 4.
What is the Betz limit?
The 0.593 ceiling on Cp from actuator disc physics; here 3.155 MW.
Why does wind speed dominate?
Power cubes with speed, so small wind gains multiply fast.
Does the imperial mode match?
Yes. 262.5 ft at 26.84 mph gives the same 1.862 MW.
What wind makes a site viable?
Roughly 9 mph up for small and 13 mph up for utility scale annual averages.