18 Induction Motor Slip and Speed

In this short video clip, I'm going to run through the calculations and the examples which are covered on your handouts regarding induction motor speed. Um So, straight into it. There are a couple of different calculations we can do on synchronous speed. And synchronous speed is the speed of the magnetic field. So, if we

have a 50 Hz supply, the magnetic field is rotating at 50 uh revs per second or 3,000 rpm. And that's the fastest possible speed that the rotor could ever turn at. So, that is synchronous speed. So, couple of quick calculations. We can work it out in revs per second. Lower case n is revs per second, and

it's f over p. K. Do a couple of quick examples on your handout. This is example one and two. So, example one. Uh ns is f over p. The frequency is 50, and in the question, it is a two-pole machine. P is pairs of poles. There is one pair. That would be a trip up if this question

occurs in an exam. So, 50 revs per second. Nice and simple. Let's do example two. Uh and this is a four-pole machine on 50 Hz. So, synchronous speed ns equals f over p. Put the numbers in. The frequency is 50. There are four poles. That is two pairs. And that comes to 25.

Ah. Revs per second. Sorry about that. Okay, so that was all about being in revs per second. Let's move on and do the second half of this particular page on your handout and let's look at RPM. The formula has moved from lower case to upper case and it's 60 F over P. The reason for the 60 is quite

simple. There are 60 seconds in a minute and we're now doing revs per minute. So, if I look at example three, we have a two-pole machine on 50 hertz and we have NS equals 60 F over P. 60 times the frequency over the pairs of poles.