The output power is 100hp and slip is 0.02 and efficiency of 90%,4 pole induction motor
Want to find the shaft torque(output torque).
So I guess to find the torque we use the equation
P(out) = T(out)/w , but i'm not sure w is same with input(?) w.
here is my tried solution
Td = p(out)/w
S= ns(1-s) , 0.02 /(1-0.02) = ns = 0.0204
And find the n
S= ns-n/ns , n= 0.01992
Pg = Tw,
T(out)= 74.6kw/[(2pi/60)n]= 35761
can i use the same w for output power torque? and why?
Also, please check the solution.
Best Answer
Torque is given by:
$$ \tau = \frac{P_{shaft}}{\omega} $$
Where \$P_{shaft}\$ is the output power of the motor (shaft power) and \$\omega\$ is the angular frequency of rotation.
The power supply in my country is 50 Hz, so a 4-pole motor has a synchronous (zero slip) speed of 1,500 RPM. Converting to angular frequency (radians per second), we get
$$ 1,500\ RPM \times \frac{2\pi\ rad.s^{-1}}{60\ RPM} = 157.1\ rad.s^{-1} $$
Slip is 2%, so the motor is running at 98% of synchronous speed :
$$ \omega = 157.1\ rad.s^{-1} \times 0.98 = 153.9\ rad.s^{-1} $$
The output power of your motor is 100 HP. In metric units that is
$$ P_{shaft} = 100\ HP \times \frac{746\ W}{1\ HP} = 74,600\ W $$
Giving
$$ \tau = \frac{74,600\ W}{153.9\ rad.s^{-1}} = 484.6\ N.m $$
Note I have written down the units of every figure I have used. This makes it much easier to spot mistakes.