Rpm = 120 x Hz/n_poles. Rearrange terms and solve.

V/400 = 37/50. Ditto.

allyrat; i would buy a 6 pole motor, 1000 rpm minus slip, raise it to 57 HZ & 456 volts perry

Some motors may be suitable for "overspeed" applications, but more conservative and safer (and probably cheaper) would be a 4-pole (1500 synchronous rpm) motor at 37 Hz, just as in the original assumptions.

The poles are fixed based on the motor design, name plate details will give you the information on poles . Generally stator poles and number of rotor poles are same.

f= 120n/p

or p= 120N/F,

slip= Ns-N/Ns

Frotor= slip * Fstator,

Terminal voltage is increased as slip increases.

Hai, The correction in the formula. N= (120*F) / P Where N= Speed, F= Frequency, P= Poles

Let me try to be as simple as i can .

A three phase 400v, 50 Hz induction motor is to have its speed controlled by a variable frequency drive unit to 1100rpm. The frequency required is found to be 37Hz, and the motor terminal voltage is reduced at the same ratio. For the conditions above, calculate the number of poles in the motor and the motor terminal voltage.

BASIC

Any motor as you know has number of pole pairs (the poles always exist in N-S pairs).

If these poles are fed with AC current, say 50 Hz then each pole will be N-50 times/ second and S- 50 times/sec. Suppose there are only 2 poles (1 pole pair) along a cylinder then if there is another magnet inside, its N gets pulled by the S of this pole and vice-versa. Hence this will rotate 50 times inside the other magnet per second. Or 50x60=3000 times per minute (3000RPM)

Suppose there are 4 poles on the cylinder instead of 2. Now the S still attracts the N of the inner magnet, but now it rotates only by 90 degrees (each pole is equally spaced- ie at 360/4=90 degrees interval) instead of earlier 180 degrees.

This happens in half the time period, since in the other-half, it is again back so it will now rotate at half the earlier speed.

Had there been 6 poles, it would have rotated at 1/3^rd speed.

Hence the speed of the AC, motor ( we call it the synchronous speed since it is in synchronism with the supply frequency) is

= 3000/ no of pairs of poles

= 60* frequency of supply / no of pairs of pole

= 2*60*frequency of supply/ no of poles.

We normally talk in terms of poles (eg 2 pole machine)

Any Induction motor will not work at its synchronous speed but just at a tiny bit less called slip.

The speed of IM = slip x synchronous speed

Eg for a 2 pole IM, if slip is 0.90,

The speed = 0.90*3000 = 2700 RPM

So the first part is clear?

You have 37 Hz supply , poles unknown

Speed = 1100 = slip (since it is IM) x 120 * 37 (frequency) / no of poles

No of poles = slip x 120 x 37 /1100 = slip x 4.036

Obviously the slip will be quite small, and the number of poles have to be even integer (remember it is always in N-S pairs)

So poles = 4

Slip = 0.9918 (4/4.036)

Part-2

The voltage is reduced by the same- In motors under variable frequency, usually

V/f is kept constant – this is directly proportional to the magnetic flux density and hence have to be maintained so that neither the core saturates nor the efficiency is compromised.

So with V/f = constant, you have

400/50 = V / 37

V = 37*400/50 = 296 V

Thank you so much... that makes a lot of sense and I realy appreicate your time and effort. I have another 10 of these questions all very similar but with different values or figures given.. I am now going to have a go...

Thanks Again

allyrat