Question

The blades of a wind turbine are 30 m in length and rotate at a maximum rotation rate of 20 rev/min. (a) If the blades are 6000 kg each and the rotor assembly has three blades, calculate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes?

Answer 1

Explanation:

It is given that,

Length of the blades, L = 30 m

Maximum rotation of the blades, $\omega=20\ rev/min=2.09\ rad/s$

(a) Mass of the blades, m = 6000 kg

Let L is the angular momentum of the turbine at this rotation rate. Its formula is given by :

$L=I\omega$

I is the moment of inertia of the rod, $I=\dfrac{ml^2}{3}$

$L=\dfrac{ml^2}{3}\omega$                    

For three blades,

$L=3\dfrac{ml^2}{3}\omega$                

$L=ml^2\omega$                    

$L=6000\times 30^2\times 2.09$          

$L=1.12\times 10^7\ kg-m^2/s$

(b) Let $\tau$is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes or 300 seconds.

$\tau=I\times \alpha$

For three blades,

$\tau=3\dfrac{ml^2}{3}\times \alpha$    

$\tau=ml^2\times \dfrac{\omega}{t}$

$\tau=6000\times 30^2\times \dfrac{2.09}{300}$          

$\tau=37620\ N-m$

Hence, this is the required solution.