# Induction motor simple power and torque

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.

• You have a number of errors here due to using the wrong numbers in the wrong places. Try writing down the units of each and every quantity you are using, i.e. watts, newton-metres, RPM. (Hint: Torque is power divided by angular frequency.) – Li-aung Yip Oct 28 '13 at 2:53
• You also appear to be confusing angular velocity, $\omega$ (measured in radians/sec) with input power ($P_{input}$ measured in watts, W.) – Li-aung Yip Oct 28 '13 at 2:58

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.

• When you find the Developed torque, Used the original w=157.1? – kimchi Oct 28 '13 at 4:06
• @sayuri: I'm not going to do your homework for you. I suggest looking up the "power flow diagram" for an induction motor. Mulukutla S. Sarma's Electric Machines (WM. C. Brown Publishers, 1985) talks about this in section §7.4 on Polyphase Induction Machine Performance. – Li-aung Yip Oct 28 '13 at 4:19

So the key information:

Power = 100HP = 74,570W Slip = 2% Efficiency = 90% Pole = 4

mains freq = 60 (determined from OP initial calc). sync speed = 1800rpm (main_freq/(pole/2))*60

Key equations:

P=T $\omega$ P=Power, T= Torque, $\omega$ = rotor speed s = (fs - fr)/fs (s=slip, fs = slip speed, fr = sync speed)

so the slip speed = 1764rpm.

$Via P = T\omega => T = P/\omega => T = 74570/((1764/60)*2*pi) => T = 403Nm$

• Naib: you may want to look up this guide on how to use math markup in answers, so instead of writing P=Tw you can write $P=\tau\times\omega$. – Li-aung Yip Oct 28 '13 at 14:00
• Naib: Remember that the dollar signs need backslashes - $...$, not $...$. As in $x^2 + y^2 = z^2$. – Li-aung Yip Oct 28 '13 at 14:15
• thanks, the issue is it is not being rendered in chrome (neither is your post below specific for 50Hz) ... – JonRB Oct 28 '13 at 15:09