Slip, Rotor Frequency and Rotor EMF

• Slip is a ratio and has no units.

• In motoring, \(N_s \gt N_r\) always.

• If \(N_s = N_r\) then the relative speed is zero \(\Rightarrow\) no rotor EMF or current \(\Rightarrow\) no torque.

• At standstill or blocked rotor, \(N_r = 0\) so \(s = 1\).

• At synchronous speed, \(N_s = N_r\) so \(s = 0\).

Typical full-load slips are 2 to 6 %, smaller in large machines. The whole useful operating range of an induction motor therefore lies in the first few percent of the slip axis, with the remaining 95 % of it reserved for starting and abnormal conditions.

• When the rotor is stationary, the frequency of the rotor current is the same as the supply frequency.

• When the rotor starts revolving, the frequency depends upon the relative speed, or on the slip speed.

Let the frequency of the rotor current at any slip speed be \(f_r\). Then the slip speed must correspond to that frequency in the same way that the synchronous speed corresponds to the supply frequency:

The rotor is a winding cut by a flux, so its EMF obeys the same equation as any other — the transformer EMF equation of Chapter 41:

The rotor winding has leakage inductance \(L_2\), fixed by its geometry. Its reactance, however, depends on the frequency at which it is worked:

This single dependence explains the shape of every characteristic in Part 4: the low starting torque despite huge starting current, the existence of a maximum torque at an intermediate slip, and the almost-linear torque-slip relation near full load.

With the three quantities established, the rotor circuit per phase follows at once.

Rotor currents have a frequency \(f_r = sf\), and when flowing through the individual phases of the rotor winding they give rise to rotor magnetic fields. Those individual fields combine, exactly as in Chapter 59, into a single rotating field.

A cage rotor deserves a note here. It has no defined phase windings, yet the same result holds: the induced bar currents arrange themselves into whatever pole pattern the stator field demands, and their combined field rotates at \(sN_s\) relative to the rotor just as a wound rotor's would.

The rotor field travels at \(sN_s\) relative to the rotor, but the rotor itself is running at speed \(N\) with respect to space. Adding the two gives the speed of the rotor field in space.

The slip has cancelled completely. Whether the machine is at standstill, at half speed or at full load, the two fields keep station with one another — and that is what makes a steady, non-pulsating torque possible. Two fields at a fixed relative angle produce a constant torque; two fields in relative motion would produce a torque that alternated and averaged to nothing.

Torque depends on the flux, on the rotor current, and on the phase angle between them — only the component of rotor current in phase with the EMF produces torque.

Since the standstill rotor EMF is proportional to the flux, \(E_{20} \propto \Phi\), the constant can absorb the proportionality:

Substituting the expressions of Section 60-5 turns this into a formula in slip alone, which is the business of Chapter 61:

• Slip \(s = (N_s - N)/N_s\) is dimensionless. At standstill \(s = 1\); at synchronous speed \(s = 0\) and there is no torque.

• Rotor frequency \(f_r = sf\), obtained by applying \(N = 120f/P\) to the slip speed and to the synchronous speed and dividing. The pole number cancels.

• Rotor EMF \(E_2 = sE_{20}\), since flux and turns are fixed and only frequency varies.

• Rotor reactance \(X_2 = sX_{20}\) — the crucial relation, because \(R_2\) does not vary with slip.

• Hence \(Z_2 = \sqrt{R_2^{2} + (sX_{20})^{2}}\), \(I_2 = sE_{20}/Z_2\) and \(\cos\phi_2 = R_2/Z_2\).

• As slip rises the current rises much less than proportionally (5.19× for a 25× EMF change in Example 60.2) while the power factor collapses from 0.982 to 0.204.

• The rotor field travels at \(sN_s\) relative to the rotor, hence at \(sN_s + N = N_s\) in space.

• Both fields therefore rotate at \(N_s\) and are stationary with respect to each other, whatever the slip — which is what permits a steady torque.

• \(T \propto \Phi I_2\cos\phi_2 = K_1E_{20}I_2\cos\phi_2\): a poor rotor power factor wastes current.

• The product is greatest when \(sX_{20} = R_2\), where \(\cos\phi_2 = 1/\sqrt2\) and \(\phi_2 = 45^{\circ}\).

• Taking rotor reactance as constant. It is \(sX_{20}\), and the whole character of the machine follows from that.

• Making rotor resistance depend on slip. It does not — resistance is independent of frequency.

• Using \(E_2\) for the standstill value. Here \(E_{20}\) is at standstill and \(E_2 = sE_{20}\) is the running value.

• Expecting rotor current to rise in proportion to slip. The impedance rises too, so it flattens off.

• Judging torque by rotor current alone. The current is largest at standstill, where the torque is nearly least.

• Using \(E_2\) rather than \(E_{20}\) in the torque product. It is the flux — hence the standstill EMF — that appears there.

• Thinking the rotor field travels at \(N\). It travels at \(sN_s\) relative to the rotor and at \(N_s\) in space.

• Supposing the two fields move relative to each other. They never do, at any slip.

• Believing a cage rotor cannot produce a rotating field because it has no phase windings. The bar currents arrange themselves accordingly.

• Forgetting to cut out the external rotor resistance once a slip-ring motor is running.

Find the slip first; then \(f_r\), \(E_2\) and \(X_2\) all follow by multiplying by \(s\). Only \(R_2\) stays put.

1. P60.1 A 6-pole, 50 Hz motor runs at 940 rev/min. Find the slip, slip speed and rotor frequency.

   Show answer

   \[N_s = \frac{120(50)}{6} = 1000~\mathrm{rev/min}\]

   \[N_s - N = 60~\mathrm{rev/min}, \qquad s = \frac{60}{1000} = 0.0600\]

   \[f_r = (0.06)(50) = 3.00~\mathrm{Hz}\]

2. P60.2 A motor has \(E_{20} = 100\) V, \(R_2 = 0.30~\Omega\) and \(X_{20} = 1.00~\Omega\) per phase. Find \(E_2\), \(X_2\), \(Z_2\), \(I_2\) and \(\cos\phi_2\) at 5 % slip.

   Show answer

   \[E_2 = (0.05)(100) = 5.00~\mathrm{V}, \qquad X_2 = (0.05)(1.00) = 0.0500~\Omega\]

   \[Z_2 = \sqrt{0.09 + 0.0025} = \sqrt{0.0925} = 0.30414~\Omega\]

   \[I_2 = \frac{5.00}{0.30414} = 16.440~\mathrm{A}, \qquad \cos\phi_2 = \frac{0.30}{0.30414} = 0.98639\]

3. P60.3 For P60.2, repeat at standstill and find the ratio of standstill to 5 % slip current.

   Show answer

   \[Z_2 = \sqrt{0.09 + 1.00} = \sqrt{1.09} = 1.04403~\Omega\]

   \[I_2 = \frac{100}{1.04403} = 95.783~\mathrm{A}, \qquad \cos\phi_2 = \frac{0.30}{1.04403} = 0.28735\]

   \[\frac{95.783}{16.440} = 5.826\]

   The EMF rose 20-fold but the current only 5.83-fold.

4. P60.4 For P60.2, find the slip at which the torque is maximum and the rotor power factor there.

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   \[s_{max} = \frac{R_2}{X_{20}} = \frac{0.30}{1.00} = 0.300\]

   There \(X_2 = 0.30 = R_2\), so

   \[Z_2 = \sqrt{2}\,(0.30) = 0.42426~\Omega, \qquad \cos\phi_2 = \frac{1}{\sqrt2} = 0.70711, \qquad \phi_2 = 45^{\circ}\]

5. P60.5 An 8-pole, 50 Hz motor runs at 720 rev/min. Find the speed of the rotor field relative to the rotor and relative to space.

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   \[N_s = \frac{120(50)}{8} = 750, \qquad s = \frac{750-720}{750} = 0.0400\]

   \[\text{relative to rotor} = sN_s = (0.04)(750) = 30~\mathrm{rev/min}\]

   \[\text{in space} = 30 + 720 = 750 = N_s \quad\checkmark\]

6. P60.6 A slip-ring motor has \(R_2 = 0.20~\Omega\) and \(X_{20} = 0.80~\Omega\). What external resistance per phase gives maximum torque at starting?

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   Maximum torque at start requires \(s_{max} = 1\), that is \(R_2 + R_{ext} = X_{20}\):

   \[R_{ext} = 0.80 - 0.20 = 0.60~\Omega\ \text{per phase}\]

7. P60.7 Derive \(f_r = sf\), \(E_2 = sE_{20}\) and \(X_2 = sX_{20}\), and explain why the third is the most consequential.

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   Frequency. Applying \(N = 120f/P\) to the slip speed and to the synchronous speed:

   \[N_s - N = \frac{120f_r}{P}, \qquad N_s = \frac{120f}{P} \quad\Longrightarrow\quad \frac{f_r}{f} = \frac{N_s-N}{N_s} = s\]

   so \(f_r = sf\). The pole number cancels.

   EMF. From \(E = 4.44k_wf\Phi N_2\), the flux and turns are fixed by the machine and only the frequency varies, so \(E_2/E_{20} = f_r/f = s\) and \(E_2 = sE_{20}\).

   Reactance. The leakage inductance \(L_2\) is fixed by geometry, but reactance depends on frequency:

   \[X_2 = 2\pi f_rL_2 = 2\pi\left(sf\right)L_2 = s\left(2\pi fL_2\right) = sX_{20}\]

   Why the third matters most. The rotor resistance is independent of frequency and so does not vary with slip. The rotor impedance therefore changes its character as the machine speeds up — dominated by reactance at standstill, almost purely resistive at low slip.

   That single asymmetry between \(R_2\) and \(X_2\) produces the poor starting power factor, the existence of a maximum torque at an intermediate slip, and the near-linear torque-slip relation near full load. Had both varied alike, the machine would have no characteristic shape at all.

8. P60.8 Prove that the stator and rotor fields are stationary relative to each other at every slip, and explain why it matters.

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   The rotor currents have frequency \(f_r = sf\), so the field they jointly produce rotates relative to the rotor at

   \[\frac{120f_r}{P} = \frac{120\,sf}{P} = s\,\frac{120f}{P} = sN_s\]

   The rotor itself runs at \(N\) with respect to space, so the rotor field's speed in space is

   \[sN_s + N = sN_s + N_s\left(1-s\right) = N_s\]

   The slip cancels completely. Both fields therefore rotate at \(N_s\) and are stationary with respect to one another, whatever the rotor is doing.

   Why it matters. Two fields at a fixed relative angle produce a steady torque. Two fields in relative motion would produce a torque that alternated in sign and averaged to nothing, and the machine could deliver no useful output at all.

   It also means the two fields can be superimposed to give the single resultant air-gap field that actually exists — which is what makes the equivalent circuit of Chapter 64 possible.

9. P60.9 Why does a cage motor draw about six times rated current at start yet develop only about twice rated torque?

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   Because torque depends on \(I_2\cos\phi_2\), not on \(I_2\) alone, and the two move in opposite directions as slip rises.

   The current. At standstill the rotor EMF is at its maximum \(E_{20}\), so the current is large — but the impedance is also at its maximum, since \(X_2 = sX_{20} = X_{20}\). In Example 60.2 the EMF rose 25-fold from full load to standstill while the current rose only 5.19-fold.

   The power factor. With the reactance at its largest, the rotor current lags its EMF by nearly 78°, so \(\cos\phi_2\) falls from 0.982 to 0.204 — most of that large current is doing nothing useful.

   \[\frac{I_2\cos\phi_2\ \text{at }s=1}{I_2\cos\phi_2\ \text{at }s=0.04} = \frac{(97.898)(0.20395)}{(18.856)(0.98206)} = \frac{19.966}{18.518} = 1.078\]

   So a five-fold rise in current buys only about 8 % more torque-producing current in this machine.

   The remedy is to raise \(R_2\) at starting — external resistance in a slip-ring machine, or a deep-bar or double-cage rotor — which improves \(\cos\phi_2\) and shifts maximum torque towards standstill.

10. P60.10 Explain the effect of adding rotor resistance, using the rotor power factor.

    Show answer

    Adding resistance raises \(R_2\) without altering \(X_{20}\), so the slip for maximum torque

    \[s_{max} = \frac{R_2}{X_{20}}\]

    moves towards standstill. Choosing \(R_2 = X_{20}\) puts it exactly at \(s = 1\).

    What happens to the current and power factor. In Example 60.5 raising \(R_2\) from 0.25 to 1.20 Ω changed the standstill values from

    \[I_2 = 97.9~\mathrm{A},\ \cos\phi_2 = 0.204 \quad\longrightarrow\quad I_2 = 70.7~\mathrm{A},\ \cos\phi_2 = 0.707\]

    The current fell while the torque rose to its maximum — more torque from less current, because the power factor improved by a factor of 3.5.

    Why it is removed afterwards. At full-load slip the extra resistance would dissipate a great deal of power for no benefit, so the rings are short-circuited once the machine is up to speed. Chapter 62 adds the striking companion result: the value of maximum torque is entirely unaffected by rotor resistance — only its position moves.

Every quantity the torque equation needs is now in hand. Chapter 61 substitutes them into \(T = K_1E_{20}I_2\cos\phi_2\) and obtains the torque equation in slip alone:

It also settles the constant \(k\), which turns out to be \(3/2\pi n_s\) when torque is measured in newton metres — so that the equation can be used numerically and not merely proportionally.

Chapter 62 then draws the torque-slip characteristic and establishes the two results this chapter has been pointing towards. Maximum torque occurs at \(s_{max} = R_2/X_{20}\), where the rotor reactance equals its resistance and \(\phi_2 = 45^{\circ}\). And the value of that maximum, \(T_{max} = kE_{20}^{2}/2X_{20}\), contains no \(R_2\) at all — so rotor resistance moves the peak without changing its height, which is exactly the property that makes a slip-ring machine worth its cost.

Chapter 63 accounts for the power flow and shows that the air-gap power divides in the ratio \(s : (1-s)\) between rotor copper loss and mechanical output — so that a machine running at 50 % slip wastes half of everything crossing the gap, which is why rotor-resistance speed control is so inefficient. Chapter 64 then assembles the equivalent circuit, and the whole of Part 3's transformer theory returns with the rotor resistance divided by the slip.