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The Rotor of a 4 pole, 50hz slip ring induction motor has a resistance of 5ohm per phase and inductance of 150mH per phase. Calculate the slip speed at which max torque occurs.

I have used Ns = 120f/p to find the synchronous speed, being 1500RPM, but I cannot calculate the Slip speed without knowing the rated RPM of the motor.

Is there a way to calculate this using the given information?

UPDATE: I believe this is the answer.

ns = 120f/p
= 120x50/4
= 1500 rpm

fr = X/2piL
= 5/(2pi x 0.150)
= 5.3hz (or rpm?)

Slip% = fr/f
= 5.3/50
= 0.106

Slip speed = slip% x ns
= 0.106 x 1500
= 159rpm at 10.6% slip (0.106x100)

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  • \$\begingroup\$ The proper formula for synchronous speed is Ns=120f/p. \$\endgroup\$ – Pat Jun 6 at 12:20
  • \$\begingroup\$ Yes, sorry I must have typed it wrong. \$\endgroup\$ – Dexterdave Jun 7 at 9:32
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Homework, I guess, a theoretical one which assumes ideal stator and no substantial saturation nor stray inductance effects.

The general induction motor theory tells: In that ideal case max. torque occurs at slip=R/X where R is one phase resistance (=load + rotor wire) and X=one phase rotor reactance. Actually also negative slip = -(R/X) is valid for max torque; that has meaning for generators.

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  • \$\begingroup\$ I think I have it now. = R/X = 5/0.150 = 33.33 Rated RPM = 1500-33.33 = 1466.67 S% = ns - n/ ns x100 S% = (1500-1466.67)/1500 x100 S% = 2.2% Does this look correct? \$\endgroup\$ – Dexterdave Jun 7 at 9:48
  • \$\begingroup\$ @Dexterdave Your comment was only "I think I have it now" + nothing else, when I wrote OK. Your calculation is crap. X=6,28fL, that's not 0.15 Slip as number is between 0 and 1, slip=1 means stopped, 0 means = sync speed. \$\endgroup\$ – user287001 Jun 7 at 10:03
  • \$\begingroup\$ @Dexterdave Im sorry, but I must tell you seem to use numbers very liberally, you insert them to nice looking formulas randomly without thinking their meanings. That's a big obstacle in front of succesful calculations and engineering competence. Start by fixing this. A good beginning is to know that it's useless to add or subtract quantities which have different dimensions. You subtracted from 1500 RPM a bare number 33.33. That was the 2nd error, your 33.33 was already nonsense. \$\endgroup\$ – user287001 Jun 7 at 10:18
  • \$\begingroup\$ Yes, Im new to the site and accidentally posted before I had finished my reply when I pressed enter. Sorry about that. Im not an engineering student, but an apprentice electrician, so these calculations are all a bit above my head. Thanks for trying to help anyway. \$\endgroup\$ – Dexterdave Jun 7 at 10:36
  • \$\begingroup\$ @Dexterdave then you have something that many of us have never got: Practical skills starting from how to use common tools and equipment + knowing common parts and materials. Do not give up! Find a local tutor who shows to you some math things which are needed in calculations around the electricity. \$\endgroup\$ – user287001 Jun 7 at 10:42

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