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'Turbo Turbines' printed from https://nrich.maths.org/
$M_{total} = 3 \int^L_0{k V \left( \frac{3}{4} +
\frac{x^2}{4L^2}\right) } dx$
$= 3 k V \left[\frac{3 x}{4} + \frac{x^3}{12 L^2}
\right]^L_0$
$= 3 k V \left[\frac{3}{4} L + \frac{1}{12} L \right] = \frac{5}{2}
k V L $
Equating this to the resistive torque, $\frac{5}{2} k V L =
5T$
$\therefore V_{crit} = 2T/(kL)$.
Since the torque is fixed, you might decrease the minimum wind
speed by improving the geometry of the blade and thus increasing
$k$, or by increasing the blade length. Increasing the number of
blades on the turbine would also decrease $V_{crit}$.
The rotational analogy is
Power = Torque $\times$ Angular Velocity
The power produced by the generator is thus $A \omega_g^2$. The
1:50 ratio of the gearbox tells us that $\omega_g$ can be
approximated by $50 B k V$, when $V > V_{crit}$.