Rotor Swept Area

The swept area is the circle the blade tips trace, $A = \pi d^2/4$, and every wind power equation on this site takes it. The two mistakes are both easy to make and both large: this is the diameter, not the radius, so substituting a 45 m blade length where a 90 m diameter belongs divides the answer by four; and it is the swept disc, not the blade planform area, which on a modern three-bladed rotor covers only a few percent of the circle.

That second point deserves a moment, because it is where the physics is. A rotor is not a sail. It does not need to cover the disc to work on the air passing through it — each blade sheds a vortex system that influences the whole flow, so three slender blades can decelerate the entire streamtube. Adding blades adds solidity and torque but not, past a point, power; it merely lowers the tip-speed ratio at which the rotor performs best.

Because area goes as the square of the diameter while power goes directly as area, a rotor 40% wider makes twice the power in the same wind. Tower, nacelle, foundation, grid connection, road access and the maintenance crew's time do not double alongside it, and that asymmetry is the entire economic story of the last forty years of wind energy. Chasing the last few points of $C_p$ cannot compete, because $C_p$ is already close to a ceiling that cannot move.

Two practical footnotes. Hub height rises with diameter for clearance reasons, so a bigger rotor also reaches faster wind — the two gains compound, and the cube law amplifies the second. And the limits on rotor growth are not aerodynamic: blade root bending moment, transport of a single-piece blade down a public road, and tip speed, which is capped near 80 m/s onshore because aerodynamic noise rises roughly with the fifth power of it.