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The question i'm having trouble with is:

Calculate the speed of the rotor of a 6 pole induction motor with a slip of 2.2%. The motor is connected to a 230V single-phase supply.

I am familiar with the table below and a certain formulae but i'm doing something incorrect and am confused on how to tackle this question.

Any formulaes or help would be appriciated as I am really stuck

enter image description here

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  • \$\begingroup\$ "i'm doing something incorrect " Start with showing us what you do, even if you know it is incorrect. Homework question with no attempt at a solution will be closed. See also here \$\endgroup\$
    – Oldfart
    Nov 2, 2019 at 10:07
  • \$\begingroup\$ Which "certain formulae" have you tried? Do you know the supply frequency? \$\endgroup\$
    – user80875
    Nov 2, 2019 at 13:53
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    \$\begingroup\$ I'm voting to close this question as off-topic because this is a homework or study problem with no demonstration of attempt to solve? \$\endgroup\$
    – user80875
    Nov 2, 2019 at 13:54
  • \$\begingroup\$ If the voltage is 230V what is the likely supply frequency? Having got that go to your course documentation and look up the formula for slip.; \$\endgroup\$
    – RoyC
    Nov 2, 2019 at 14:40
  • \$\begingroup\$ the calculation im doing is 120-50hz/6 = 1000 \$\endgroup\$
    – JakeeYm
    Nov 3, 2019 at 0:40

1 Answer 1

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This is pretty straight forward. The slip= (ws-wr)/ws, where ws= 1000 rpm at 50 Hz and 1200 at 60 Hz , wr is the speed of the motor ,

wr at 50 Hz = 1000(1-2.2%) = 978 rpm, wr at 60 Hz = 1200(1-2.2%) = 1173.6 rpm

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