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I would like to know how the mechanical rotor frequency [Hz] can be obtained from the electrical stator frequency [Hz]? This is no homework! I have consulted several books as well as the web but could not find an exact relationship expressed as formula. I hope somebody can give advise.

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The synchronous speed of the rotor in RPM N = 120f/P where f is the frequency of the stator current and P is the number of poles.

The operating speed No = N - Ns where Ns is the slip speed.

The frequency of the rotor current fr = Ns x P/120, so fr = 0 if the slip is zero.

If the operating speed is zero, Ns = N and the frequency of the rotor current is the same as the frequency of the stator current.

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  • \$\begingroup\$ Thank you Charles! According to my calculations, this confirms the equation given in my comment in alphasierra's answer assuming that there p stands for the number of pole pairs. \$\endgroup\$ – Maxwell1919 May 24 '17 at 1:00
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$$f = \frac {P\ N_S}{120}$$

  • \$f\$ = source frequency in Hz. Your electrical stator frequency.
  • P = # of three-phase pole pairs (2, 4, 6, etc.).
  • \$N_S\$ = synchronous speed in rpm.

Synchronous speed at 60Hz will be:

  • 3600 rpm for a 2-pole machine (2-pole/phase = 6 total poles).
  • 1800 rpm for a 4-pole machine (12 total).

Motor nameplate

If we take a motor nameplate to illustrate the mechanical.

$$f_r = \frac {P\ N}{120}= \frac {4\ poles\times 1765\ rpm}{120} = 58.83Hz$$

  • \$f_r\$ = mechanical rotor frequency in Hz.
  • N = rotor speed in rpm.

\$f\$ and \$f_r\$ are related by slip. For a motor to develop torque, \$f_r < f\$ and \$N < N_s\$. $$ s = \frac {N_S - N}{N_S} = \frac {1800\ rpm - 1765\ rpm}{1800\ rpm} = 0.01944$$

Slip at rated motor load is 1.94%. Slip and rotor speed actual depend on the load being driven with most motors having a slip between 0.5% and 5%. Heavier the load, the larger the slip.

This is a 4-pole motor because:

$$P = \frac {120\ f}{N} = \frac {120 \times 60\ Hz}{1765\ rpm} = 4.079$$

Rounding down gives us 4-poles/phase.

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    \$\begingroup\$ On question regarding the P in your equations f=(P*Ns)/120 and fr=(P*N)/120. In the first equation you say it is the "# of three-phase pole pairs" and in the second equation "4 poles ×1765rpm". So, does P now stand for the number of pole pairs (e.g. 1,2,3) or the number of poles (2,4,6)? \$\endgroup\$ – Maxwell1919 May 24 '17 at 7:33
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    \$\begingroup\$ Yes. 2 poles/phase means 6 poles for a 3 phase machine. \$\endgroup\$ – StainlessSteelRat May 24 '17 at 12:52
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It depends on the motor itself, and the load. When motoring the rotor frequency is always lower than the stator frequency. The difference between these two frequencies is called the slip. The amount of slip depends on the input power, the applied load, and the characteristics of the specific motor.

The equation is $$s=\frac{(n_s-n_r)}{n_s}$$ $$n_s=stator frequency$$$$n_r=rotor frequency$$

This video should help give a simple explanation of it.

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  • \$\begingroup\$ I think that the rotor frequency in your formula is not the mechanical rotary frequency. I would expect the number of pole pairs somewhere in this equation. \$\endgroup\$ – Maxwell1919 May 23 '17 at 23:44
  • \$\begingroup\$ Could this perhaps be the equation we are looking for: s = (fs - p*fr_mech) / fs ? \$\endgroup\$ – Maxwell1919 May 23 '17 at 23:50
  • \$\begingroup\$ It is the rotor rotational frequency. I just checked it against a textbook I have here. Not sure if I'm allowed to post screenshots of it. \$\endgroup\$ – alphasierra May 24 '17 at 1:34

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