induction motor
A 4 pole 50 Hz induction motor is running at 1470 rpm. What is the slip value?
Solution – My attempt:
Slip S = \$\frac{N_s – N_r}{N_s}\$
Also \$N_s = \frac{120f}{P} = 1500\$
So we have,
S = .02
I am looking for shortcut to find slip? As in objective exams we have lesser time to solve questions.
So, just an induction motor employed as a generator? Yes there IS an ac magnetization on the stator winding. Spin the shaft, and a sine wave appears. An induction generator is an electromechanical sinewave oscillator. Small residual polarization of iron parts gets it started, and it builds up as a mechanically-driven RLC resonance between the capacitors and the generator inductance (but operating way off resonance, of course.)
In that case the "synchronous" speed would be the frequency of the AC signal measured at the stator coil (or at cap bank terminals,) same as when running in motor-mode. The slip is then taking place between this coil frequency versus generator rotor RPM. Just put the AC frequency in terms of RPM = 2/#poles x 60 x HzFreq
So, if a 4-pole induction motor (as a generator) with a particular capacitor value puts out 70Hz across the cap bank, the b-field inside the motor is rotating at 2/4*60*70 = 2100 RPM. If the actual shaft RPM is 2200, then slip factor is (2100 - 2200)/2200 = -0.045 I put it as negative slip, since it's opposite of the grid-driven slip of an induction motor.
I haven't messed with one of these beasts myself, so take this all with a grain of salt.
Classic diy page: QSL ham radio site
A variable frequency drive (VFD) will generally try to maintain a constant magnetizing current in the motor with any frequency and load. The results will vary depending on the VFD design and set-up. That will result in an operating power factor that is generally similar to the pf vs load plot provided by the motor manufacturer for rated voltage and frequency sine wave operation.
To calculate output power, motor efficiency must also be considered. That will vary with speed.
You can not accurately determine the output mechanical power from the motor using the method that you propose. If the VFD provides a torque display, that is likely to be more accurate than any determination made by measuring motor current. A power determination using DC bus current of the VFD might also be more accurate. There are clamp-on wattmeters on the market that can accurately measure VFD output power. Of course, that still leaves the problem of estimating motor efficiency. I have some empirical data that might help, but I don't have time to post it right now.
Here is the data that I mentioned above. It is calculated from motor design data, not empirical test data.
I believe this data was calculated in the 1970's by Alberto Abbondanti of Westinghouse, but I can't find where it was originally published. I received it in class notes for a tutorial.
Best Answer
The slip is the fractional difference between synchronous and actual speed.
You have done exactly the minimum number of calculations required to get that from the information given. Starting from this information, there is no shorter cut.
It might be slightly quicker if you precompute the synchronous frequencies before the exams and remember them, so in a 50Hz system, 2 and 4 pole motors have synch speeds of 3k and 1.5krpm, and they're 3.6k and 1.8k in 60Hz land. But that won't work with questions that ask you about other frequencies!
It will be yet quicker if you precompute the difference for 1% slip for all the different frequencies and pole numbers. So with 50Hz/4 pole with 1500 synch frequency, 1% is 15rpm. Then in the exam, 1470rpm is obviously 30rpm short, so it's 2%. But that won't work with questions that ask you to show your working!
Given that you know how to calculate it, and can, I would not suggest you burden your memory with precomputed single-purpose equations, but instead practice on speeding up your general arithmetic skills.