Wind Turbine Power Output

A rotor cannot take all the power in the wind, so the resource equation gains one factor: $P = \tfrac{1}{2}\rho A v^3 C_p$, where $C_p$ is the power coefficient — the fraction of the kinetic flux through the swept disc that actually reaches the shaft. Everything else here is the wind; $C_p$ is the machine.

Read $C_p$ against 0.5926, not against 1. Betz showed in 1920 that no open-stream device can exceed 16/27, so a rotor at $C_p = 0.45$ is capturing about three-quarters of everything physics allows. A reader who calls that "45% efficient, so 55% wasted" has misread the problem: the missing 55% is not a loss the engineer failed to prevent, it is air that must keep moving to make room for the air behind it. The cube law still governs, and it is why hub-height wind speed dominates every other input on this page.

$C_p$ is not a property of a turbine the way mass is. It is a function of tip-speed ratio and blade pitch, and it has one sharp peak. That is the entire argument for variable-speed operation: holding the rotor at its best tip-speed ratio as the wind changes keeps $C_p$ near its maximum across the working range, and is worth several percent of annual energy over a fixed-speed machine. Above rated wind speed, the controller deliberately pitches the blades out of the flow to hold the generator at its limit, so $C_p$ falls continuously while the wind keeps rising.

Three things separate this figure from a meter reading. Real turbines cut in near 3–4 m/s, because below that the rotor cannot overcome its own losses; they hold flat at rated power above roughly 12–15 m/s; and they cut out entirely near 25 m/s to protect themselves, so the cube does not continue upward forever. Beneath all that sit gearbox, generator and converter losses of a further 5–10%. And note which area is meant: the swept disc $\pi d^2/4$, not the blade surface, which on a slender three-bladed rotor is only a few percent of the circle.