Induction motor slip calculator

Synchronous speed, slip and rotor frequency of an induction motor from the supply frequency, its poles and its shaft speed.

Motor
Hz

The number of poles, not pole pairs.

rpm

Measured, or from the nameplate at full load.

Result

Slip
2.778%
Rotor frequency
1.667Hz
Synchronous speed, ns
1,800rpm
Slip speed
50rpm
Slip, per unit
0.02778

Motoring range: shaft speed from standstill (slip 1) to synchronous speed (slip 0).

What it calculates

The synchronous speed of an induction motor's stator field, how far the shaft lags behind it, and the frequency of the currents that lag induces in the rotor. The stator field turns at the synchronous speed, set by the supply frequency f and the number of poles P:

ns=120 fPn_s = \dfrac{120\,f}{P}

The rotor turns a little slower, at n. Slip is that shortfall as a fraction of synchronous speed, and the rotor bars see the field pass at the slip frequency:

s=ns−nnss = \dfrac{n_s - n}{n_s}fr=s ff_r = s\,f

Worked example

A 4-pole motor on a 60 Hz supply runs at 1,750 rpm. The calculator opens on this case:

ns=120×604=1,800 rpms=1,800−1,7501,800=0.02778fr=0.02778×60=1.667 Hz\begin{aligned} n_s &= \dfrac{120 \times 60}{4} = 1{,}800\ \text{rpm} \\[4pt] s &= \dfrac{1{,}800 - 1{,}750}{1{,}800} = 0.02778 \\[4pt] f_r &= 0.02778 \times 60 = 1.667\ \text{Hz} \end{aligned}

The same 4-pole motor on a 50 Hz supply, running at 1,440 rpm:

ns=120×504=1,500 rpmn_s = \dfrac{120 \times 50}{4} = 1{,}500\ \text{rpm}s=1,500−1,4401,500=0.04s = \dfrac{1{,}500 - 1{,}440}{1{,}500} = 0.04fr=0.04×50=2.0 Hzf_r = 0.04 \times 50 = 2.0\ \text{Hz}

Reading the result

  • At standstill the slip is 1 and the rotor frequency equals the supply frequency. That is the locked-rotor condition, when the motor draws its largest current. The locked-rotor amps calculator finds that current from the nameplate.
  • At full load, slip is typically a few percent, so the rotor frequency is only a few hertz and the rotor's own reactance is small. Torque then rises almost in proportion to slip.
  • The air-gap power splits by slip: s times the air-gap power is lost as heat in the rotor, and (1 − s) times it becomes mechanical power.
  • On a variable-frequency drive, the synchronous speed moves with the drive's output frequency. Enter that frequency as the supply frequency.

Assumptions

  • The speed is steady, so the slip is constant.
  • The motor is motoring: the shaft speed is between standstill and synchronous speed. Above synchronous speed an induction machine generates, with negative slip; this calculator does not cover that range.
  • Poles means the total number of poles, not pole pairs, as the 120 in the formula assumes.

Questions

How do I find the number of poles from the nameplate speed?
Take the synchronous speed just above the nameplate speed. On a 60 Hz supply the synchronous speeds are 3,600, 1,800, 1,200 and 900 rpm for 2, 4, 6 and 8 poles, so a 1,760 rpm motor has 4 poles and an 880 rpm motor has 8. Full-load slip is only a few percent, so the nearest speed above is the right one.
Why is the rotor frequency so low while the motor runs?
The rotor bars see the stator field sweep past them only at the slip speed, the difference between synchronous speed and shaft speed. At a few percent slip on a 60 Hz supply that is one or two hertz. At standstill the slip is 1 and the rotor frequency equals the supply frequency.
What is the difference between slip speed and slip?
Slip speed is the difference in rpm, n_s − n. Slip is that difference as a fraction of synchronous speed, (n_s − n) / n_s, often quoted in percent. Formulas such as f_r = s × f, and rotor copper loss = s × air-gap power, use the fraction.

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