Question

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Question 2: (8 Marks) At the Whitelee windfarm situated south of Glasgow, the blades of a wind turbine accelerate from rest up to a maximum rotation rate of Wmax = 17 rpm in a time of At1 = 150 s. The increase in angular velocity of the blades is uniform, and after reaching the maximum angular speed, the turbine rotates at a constant rate. It is required to determine: a) the average angular acceleration for the time interval Ati; (2 Marks) b) how many revolutions will be done in 3.5 minutes after starting from rest? (4 Marks) c) the linear speed of the tip of the blade, 1 minutes after starting from rest, if the diameter of the blades is Dolades = 91.44 m ? (2 Marks)

Answer 1

$\omega _{max}=17rpm=17X\frac{2\pi }{60}=1.78 rad / s$

tiܠ 1505

a)   Average angular accelaration  $\alpha$=  $\omega _{max}$ / $\Delta t$ ₁= 1.78 / 150 =0.012 rad/s²

b) we need to find number of revlutions N=?? but   $\theta =2\pi N$ then

N 27

To find  $\theta$ formula is $\theta =\omega _{0}t+(1/2)\alpha t^{2}$

$\omega _{0}=0$ t=3.5 minutes =3.5x60=210s   $\alpha =0.012rad/s^{2}$

$\theta =$(1/2)x0.012x(210)²=264.6rad

$\theta =2\pi N$

$N= \frac{\theta}{2\pi }=264.6/(2X3.14)=42 revolutions$

c)  linear speed  $V=r\omega$

t=1minute D_blades=91.44m radius r= D/2= 91.44/2=45.72m

  
w=wo + at
but   $\omega _{0}=0$ then   $\omega =\alpha t$ and  t=1minute

$\omega =0.012X60=0.72 rad /s$

V=r$\omega$

V=45.72x0.72=32.4 m/s