Why Is Wind Energy Velocity Cubed Instead of Squared?

Question

Formula for everything seems to be squared, for example, kinetic energy $KE = \frac{1}{2}mv^2$
But for wind energy, this formula is cubed, e.g. $\frac{1}{2}\rho A v^3$
I couldn't find any results for the reasoning behind this.

Answer 1

Because that's a formula for wind power, not wind energy. Mass flux per unit time multiplied by energy per unit mass.

For energy E, time t, constant power P, constant density, constant area, constant velocity:

Mass flux per unit time

$\dfrac{dm}{dt} = \dfrac{\rho Adx}{dt} = \rho Av$

$P = \dfrac{dE}{dt} = \dfrac{d(.5m v^2)}{dt} = .5 \rho Av^3$

Answer 2

As we know, the kinetic energy of any given mass is $\frac{1}{2}mv^2$. The reason why there's an extra factor of $v$ in wind energy is because the amount of mass of air that passes over a given area is dependent on the velocity, since the faster the wind is moving, the larger the volume of air that will intersect the surface. Thus we get energy proportional to $v^2$ for any given molecule of air, but since the number of molecules contacting the surface per unit time depends on $v$, the power (energy per unit time) is proportional to $v^3$.

The $m$ in $\frac{1}{2}mv^2$ essentially becomes $t\rho Av$ where $t$ is time, so we get $KE = \frac{1}{2}t\rho A v^3$. This checks out because a velocity times a time is a length, times an area is a volume, times a density is a mass.

We divide both sides by $t$ to get $\frac{KE}{t} = \frac{1}{2}\rho A v^3$, and $\frac{KE}{t}$ (energy per unit time) is just power ($P$).