Diameter of rotor: 30m
Generator rotation speed rating : 1500 rpm
Gear box ratio: 1:48
a) Convert the rpm to rad/s to find the angular velocity (w) of the high speed shaft,
w_o (high speed shaft)=1500/60 X 2À=157 rad/s…………………………….(1)
The angular velocity of the low speed shaft is 1/48 of that of the high speed shaft due to the gearbox,
Therfore
w (low speed shaft)=157/48=3.2 rad/s……………………………………….(2)
We can now find the velocity (v) at the tips of the blades by the equation.
V=wr
V=3.2 X 15………………………………………………………………………(3)
V=48 m/s
If the wind speed now increases and the velocity of the blade tips increases by 10 m/s in 5 seconds, the following results will occur;
New Velocity=48 10=58m/s
V=wr……………………………………………………………………..(4)
58=w X 15
w( low speed shaft)= 3.9 rad/s
High speed shaft=low speed shaft X Gear box ratio…….……………..(5)
w=3.9 X 48
w (new angular velocity high speed shaft)=187.2 rad/s
The new RPM of the high speed shaft will be:
187.2/2À X 60=1788 an increase of 288 RPM........................................
The angular acceleration of the high speed shaft is found by the formula;
w=w_o αt………………………………………………(7)
Where (t) = time
187.2=157 α 5
α= (187.2-157)/5
Angular acceleration (α) = 6 rad/s2
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If the rotor length was now halved to a diameter of 15 metres;
From equation (2)
V=wr
V=3.2 X 7.5
V=24 m/s
If the wind increased at the same rate, therefore the rotor tip speed increases by 10m/s in 5 seconds, then;
New Velocity=24 10=34 m/s
V=wr
34=w X 7.5
w( low speed shaft)= 4.5 rad/s
High speed shaft=low speed shaft X Gear box ratio
w=4.5 X 48
w (high speed shaft)=216 rad/s
The new RPM of the high speed shaft will be:
216/2À X 60=2063 rpm
an increase of 563 rpm.
The new angular acceleration is as follows
From equation (7)
α= (216-157)/5
Angular acceleration (α) = 11.8 rad/s2
IS THIS CORRECT AND WHY WILL THE ACCELERATION INCREASE?
PLEASE HELP
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