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Wind power varies as cube of velocity

Short answer

Power in wind is P = 1/2 ρ A v³, so output is proportional to the cube of velocity. Other factors include swept area, air density, power coefficient (Betz limit), hub height, wind speed distribution, turbine cut-in/rated/cut-out speeds, mechanical/electrical losses, availability, wake effects, and grid constraints.

Show that the power output of wind power plant is cube of the velocity of air. What are the other factors that may guide the possible output of wind power project?

Power varies as cube of wind velocity

Consider air of density ρ\rho moving at speed vv through the swept area AA of the rotor.

Mass of air passing per second:

m˙=ρAv\dot{m} = \rho A v

Kinetic energy carried per second (power):

P=12m˙v2=12(ρAv)v2=12ρAv3P = \frac{1}{2}\dot{m}v^2 = \frac{1}{2}(\rho A v)v^2 = \frac{1}{2}\rho A v^3

The turbine extracts a fraction CpC_p (power coefficient):

Pout=12ρAv3Cp  ⇒  Pout∝v3P_{out} = \frac{1}{2}\rho A v^3 C_p \;\Rightarrow\; P_{out} \propto v^3

So doubling wind speed gives 23=82^3 = 8 times the power, and 10% higher speed gives 1.13=1.331.1^3 = 1.33 times the power.

Other factors that guide the output of a wind project

  1. Swept area / blade length – A=πr2A = \pi r^2, so power ∝r2\propto r^2.
  2. Air density – lower at high altitude and high temperature; at about 3000 m density is roughly 25–30% less than at sea level.
  3. Power coefficient – blade design, tip-speed ratio, pitch control; limited by Betz limit (0.593).
  4. Hub height – wind speed increases with height above ground (wind shear).
  5. Wind speed distribution – mean speed, variability and duration (Weibull distribution), turbulence.
  6. Cut-in, rated and cut-out speeds of the turbine.
  7. Mechanical and electrical efficiency – gearbox, generator, converter, transformer losses.
  8. Availability and wake losses – downtime for maintenance; turbines in a wind farm shading each other.
  9. Grid availability and curtailment.

✓ Answer checked by Claude Opus 5.5 (high effort)

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