Turbo Turbines
Problem
A landowner wants to calculate whether it would be worthwhile for her to install a triple-blade wind turbine. The turbine would face the wind, which is parallel to the ground and of speed $V$. However, a wind turbine changes the speed locally around it, so that $V_{local} = V \left(\frac{3}{4} + \frac{x^2}{4L^2}\right)$, where $x$ is the distance along the blade from the centre, and the blades
are of length $L$.
The force in the direction of the blade rotation is $F = k V_{local}$ at each position $x$ along each blade ($k$ is a coefficient determined by the shape of the blade).
The central pivot is resisted by a torque $T$. It is directly connected to a large gear, which drives a smaller gear (gearing ratio 1:50). The frictional torque that the small gear exerts on the larger one is $4T$.
Draw a diagram to accurately represent the turbine and the wind.
What is the minimum wind speed $V_{crit}$ in terms of $T$, $k$, and $L$, that will produce power?
How could you decrease this minimum wind speed, assuming the mechanical torque is fixed?
Power is generated by the small gear. It has a torque $A\omega_g$, where $\omega_g$ is its angular velocity, and $A$ is a constant. The angular velocity of the blades can be approximated by $\omega = B k V$, when $V > V_{crit}$.
You may know the formula
Power = Force $\times$ Velocity
There is a rotational analogy for many such formulae. Can you find an equation for the power produced by this turbine?
Getting Started
As in many applied mathematics problems, the first part of the problem involves making the right equation from the given facts. Such aspects are important and often a major part of the the problem solving process.
Finding $V_{crit}$
The total rotational force (moment) on each of the arms is $M = \int^L_0{F}dx$. Moment and torque are equivalent.
Rotational version of Power = Force $\times$ Velocity
For the two terms on the right hand side, use their rotational equivalents.
Student Solutions
$M_{total} = 3 \int^L_0{k V \left( \frac{3}{4} + \frac{x^2}{4L^2}\right) } dx$
$= 3 k V \left[\frac{3 x}{4} + \frac{x^3}{12 L^2} \right]^L_0$
$= 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 $V > V_{crit}$.