Answer:
a) The slip for the given conditions is 0.2074 or 20.74% and the developed torque is 7.14 Nm.
b) The new slip after reducing the torque to 4 Nm is 0.1162 or 11.62% and the new motor speed is 1193.13 rpm.
Explanation:
In order to find the slip we can use the definition as the relative difference between Synchronous speed and the rotor speed and that the developed torque is the ratio between output power and rotor angular velocity.
Slip.
We can find the synchronous speed u sing the following formula
Where f stands for the frequency and p the number of poles, so we have
Replacing on the slip formula
we get
Thus the slip for the given conditions is 0.2074 or 20.74%.
Developed torque.
We can find the angular velocity in radians per second
Thus we can replace on the torque formula
We get
The developed torque is 7.14 Nm.
Slip for the reduced torque.
The torque is proportional to the slip, so we can write
Thus solving for the new slip we have:
Replacing the values obtained on the previous part we have
So the new slip after reducing the torque to 4 Nm is 0.1162 or 11.62%
Motor speed.
We can use the slip definition
Solving for the motor speed we have
Replacing values we have
The new motor speed is 1193.13 rpm.
