Derivation of Betz limit for wind turbine

Assumptions

Ideal rotor as an actuator disc, steady, incompressible, frictionless flow, uniform velocity over the disc and no rotation of the wake. Let $v_1$ be the upstream (free) wind speed, $v$ the speed through the disc of area $A$, and $v_2$ the speed far downstream.

  v1 ->  \   |  /  -> v2 (slow)
 --------\--|--/--------
 ->      | disc |  v
 --------/--|--\--------

Power from momentum

Mass flow through the disc: $\dot m = \rho A v$.

Force on the disc (change of momentum): $F = \dot m (v_1 - v_2)$.

Power extracted: $P = Fv = \rho A v^2 (v_1 - v_2)$.

Power also equals the loss of kinetic energy of the stream:

$$P = \tfrac{1}{2}\dot m (v_1^2 - v_2^2) = \tfrac{1}{2}\rho A v (v_1 - v_2)(v_1 + v_2)$$

Equating the two expressions for $P$:

$$v = \frac{v_1 + v_2}{2}$$

So the wind speed at the disc is the mean of the upstream and downstream speeds.

Power coefficient

Substituting $v$:

$$P = \tfrac{1}{4}\rho A (v_1 + v_2)(v_1^2 - v_2^2)$$

Let $a = v_2/v_1$. Then

$$P = \tfrac{1}{2}\rho A v_1^3 \cdot \tfrac{1}{2}(1+a)(1-a^2)$$

The available power in the wind is $P_0 = \tfrac{1}{2}\rho A v_1^3$, so

$$C_p = \frac{P}{P_0} = \tfrac{1}{2}(1+a)(1-a^2)$$

Maximum

$$\frac{dC_p}{da} = \tfrac{1}{2}\left(1 - 2a - 3a^2\right) = 0 \Rightarrow 3a^2 + 2a - 1 = 0$$

giving $a = \tfrac{1}{3}$ (the other root is $-1$, not physical). Then

$$C_{p,max} = \tfrac{1}{2}\left(\tfrac{4}{3}\right)\left(\tfrac{8}{9}\right) = \frac{16}{27} = 0.593$$

Meaning

No wind turbine can capture more than 59.3 % of the kinetic energy of the wind passing through its swept area, because the air must keep moving behind the rotor. Practical large turbines reach $C_p$ of 0.40–0.50 due to blade drag, tip losses and wake rotation; the maximum for a Savonius rotor is about 0.15–0.20.