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I have a task to find a number of poles and rated slip of induction motor if frequency is 50 HZ and full load speed is 2950 rpm. However synchronous speed is not given. Is it correct to put full load speed value into P = f x 120 / n, where P is poles number and just round the result?

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Since this seems to be study related. I won't spell out the whole answer for you but give you some guidance.

I have a task to find a number of poles and rated slip of induction motor if frequency is 50 Hz and full load speed is 2950 rpm.

If there was no slip and the minimum number of poles what would the motor speed be in revolutions per second? In RPM?

However synchronous speed is not given.

You can assume that it's the nearest no-slip value for 2, 4, 6, etc., poles above the rated speed. The difference should be < 10%.

Is it correct to put full load speed value into P = f x 2 / n, where P is poles number and just round the result?

I'd be more inclined to round the RPM to the nearest no-slip value and then do the calculation but either would work.

Put your calculations in a comment and I'll mark it for you!

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  • \$\begingroup\$ So number of poles: P = 50x120/2950=2.03 =~ 2 poles. After I can find synchronous speed: n = 120x50/2=3000 rpm. And finally slip: s = (3000-2950)/3000 x 100% = 1.7% \$\endgroup\$ – ArtyomShep Jan 5 at 18:42
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    \$\begingroup\$ That looks good to me. Just a note: watch out if you're ever working with variable frequency drives (VFDs). Some will have slip compensation so that when you ask for 50 Hz they'll run that 2-pole motor at 3000 RPM meaning that the output is somewhat higher than 50 Hz. In other words, they compensate for the slip and give the requested speed. \$\endgroup\$ – Transistor Jan 5 at 18:49

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