Principle of Operation

Chapter 57 built the machine and named it a rotating transformer. This chapter follows what actually happens when the supply is switched on, step by step.

1. The three-phase stator winding carries three currents displaced by \(120^{\circ}\) in time.
2. Because the three windings are also displaced by \(120^{\circ}\) in space, they together produce a rotating magnetic field of constant magnitude, turning at synchronous speed \(N_s = 120f/P\). Chapter 59 proves this.
3. The rotor is initially at rest, so there is relative motion between the field and the rotor conductors.
4. The conductors are therefore cut by the flux, and an EMF is induced in each — dynamically induced EMF, Chapter 10.
5. The rotor circuit is closed — permanently, in a cage — so current flows.
6. Those current-carrying conductors lie in a magnetic field, so each experiences a force, and the forces around the periphery add to a torque.

Take one rotor bar of length \(l\), lying in a flux density \(B\), with the field sweeping past it at a relative speed \(v\).

The rotor current is limited by the rotor impedance \(Z_2\), which is itself a function of slip because the rotor reactance depends on rotor frequency. That single complication is what gives the induction motor its characteristic torque-slip curve, and Chapters 60 to 62 develop it.

The formulas give the magnitude of the torque, but not its direction. For that, Lenz's law of Chapter 10 is enough — and it settles the matter immediately.

This is a particularly clean use of Lenz's law, because the rotor is mechanically free to respond. In a transformer the induced current opposes the change but nothing moves; here the opposition expresses itself as motion, and that motion is the useful output of the machine.

The rotor chases the field, but it can never catch it. The reason is the chain of Section 58-1 read backwards.

The result is a genuinely stable equilibrium, not a precarious one. If the rotor speeds up, the torque falls and pulls it back; if it slows, the torque rises and pushes it forward. Section 58-6 follows this through.

The shortfall in speed is given a name and a symbol, and it becomes the fundamental variable of the whole of Part 4.

Two quantities follow at once, and Chapter 60 derives them properly:

The machine regulates itself, and the mechanism is worth following in sequence.

1. A load torque is applied to the shaft, exceeding the torque the motor is currently developing.
2. The rotor slows down.
3. Slowing increases the relative motion, so the slip rises.
4. Higher slip means more rotor EMF, hence more rotor current, hence more torque.
5. The rotor settles at the speed where the developed torque again equals the load torque.

The comparison with a DC shunt motor made in Chapter 57 now makes sense: both drop a few percent in speed from no load to full load, and for the same underlying reason — the machine must develop a larger internal quantity (armature current there, slip here) to produce more torque.

The rotor follows the field, so the direction of rotation is decided entirely by the direction in which the field turns — which is decided by the phase sequence of the supply.

Chapter 57 called the induction motor a rotating transformer. The analogy is exact in some respects and misleading in others, and it is worth being precise about which.

Chapter 64 will resolve the second difference elegantly. By dividing the rotor resistance by the slip, the whole rotor circuit can be referred to the stator at supply frequency, and the induction motor's equivalent circuit becomes the transformer circuit of Chapter 44 with one term changed.

Nothing in the argument so far restricted the rotor to speeds between zero and synchronous. Allowing any speed gives three distinct modes, and the slip alone identifies which.

• The chain is: stator current → rotating field → relative motion → rotor EMF → rotor current → force → torque. Every link is a result from Part 1.

• Torque comes from \(e = Blv\), then \(i = e/Z_2\), then \(F = Bil\). \(B\) appears twice, so torque depends on the square of the flux and hence of the voltage.

• Lenz's law fixes the direction: the induced effects oppose the relative motion, and the only way the rotor can reduce it is to chase the field.

• At \(N = N_s\) there is no relative motion, so no EMF, no current and no torque. The machine must always slip — hence asynchronous.

• \(s = (N_s - N)/N_s\), \(N = (1-s)N_s\), \(f_r = sf\), \(E_2 = sE_{20}\).

• Full torque comes from a very small relative motion — under 1 m/s at the rotor surface in Example 58.2, a factor \(1/s\) below the field's own speed.

• Under load the machine self-regulates: more load → lower speed → more slip → more current → more torque, until balance is reached. For small \(s\), \(T \propto s\).

• Interchanging any two supply leads reverses the phase sequence, the field and the rotor. Doing it while running gives plugging, with \(s \gt 1\) and a very severe rotor duty.

• The transformer analogy holds for induction and ampere-turn balance, but the secondary quantities all vary with slip and the air gap makes the magnetising current 30 to 50 % of rated.

• Three modes: motoring \(0 \lt s \lt 1\), generating \(s \lt 0\), braking \(s \gt 1\).

• Saying the rotor is "dragged round by magnetic attraction". There is no attraction between poles here — the torque comes from \(F = Bil\) on induced currents.

• Expecting the rotor to reach synchronous speed given enough time. At that speed the torque is exactly zero.

• Using the synchronous speed in \(e = Blv\). The EMF depends on the relative speed, which is \(sN_s\).

• Thinking the rotor EMF is constant. It falls to a few percent of its standstill value in normal running.

• Assuming rotor iron loss matters at full load. The rotor frequency is only about 2 Hz.

• Reversing a three-phase motor by swapping all three leads. That restores the original sequence and changes nothing.

• Treating plugging as harmless. The slip approaches 2, so rotor EMF is nearly twice the standstill value and the rotor absorbs power from both sides.

• Forgetting that torque depends on \(B^{2}\). A 10 % fall in supply voltage costs about 19 % of the torque.

• Applying \(T \propto s\) at large slip. It holds only while the rotor reactance is small compared with the resistance.

• Believing negative slip is impossible. It simply means the machine is generating.

In every problem find \(N_s\) first, then the slip; every rotor quantity follows from it.

1. P58.1 A 6-pole, 50 Hz motor runs at 960 rev/min. Its standstill rotor EMF per phase is 150 V. Find the slip speed, slip, rotor frequency and rotor EMF.

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

   \[\text{slip speed} = 1000 - 960 = 40~\mathrm{rev/min}, \qquad s = \frac{40}{1000} = 0.0400\]

   \[f_r = (0.04)(50) = 2.00~\mathrm{Hz}, \qquad E_2 = (0.04)(150) = 6.00~\mathrm{V}\]

2. P58.2 For P58.1, the rotor diameter is 0.30 m. Find the field surface speed and the relative surface speed.

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   \[c = \pi(0.30) = 0.94248~\mathrm{m}\]

   \[v_s = \left(\frac{1000}{60}\right)(0.94248) = 15.708~\mathrm{m/s}\]

   \[v_r = \left(\frac{40}{60}\right)(0.94248) = 0.6283~\mathrm{m/s}\]

   The ratio is again \(1/s = 25\).

3. P58.3 A 4-pole, 50 Hz motor has a full-load slip of 3 %. The load torque is increased by 50 %. Estimate the new speed, assuming \(T \propto s\).

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   \[s_2 = (1.5)(0.03) = 0.045\]

   \[N = (1 - 0.045)(1500) = 1432.5~\mathrm{rev/min}\]

   A drop of 22.5 rev/min from the original 1455 rev/min.

4. P58.4 An 8-pole, 50 Hz motor running at 720 rev/min has two supply leads interchanged. Find the slip and rotor frequency immediately afterwards.

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   \[N_s = \frac{120(50)}{8} = 750~\mathrm{rev/min}, \ \text{now reversed to } -750\]

   \[s = \frac{-750 - 720}{-750} = \frac{-1470}{-750} = 1.96\]

   \[f_r = (1.96)(50) = 98.0~\mathrm{Hz}\]

   The same slip as Example 58.4, since the original slip was 4 % in both cases.

5. P58.5 A 4-pole, 50 Hz induction machine is driven at 1575 rev/min. Find the slip and state the mode of operation.

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   \[s = \frac{1500 - 1575}{1500} = -0.0500\]

   Negative slip, so the machine is generating: it is being driven above synchronous speed and delivers electrical power to the supply.

   \[f_r = (0.05)(50) = 2.50~\mathrm{Hz}\]

6. P58.6 The supply voltage to an induction motor falls by 10 %. By what percentage does the developed torque fall, at a given slip?

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   Torque is proportional to the square of the flux, hence of the voltage:

   \[\frac{T_2}{T_1} = \left(\frac{V_2}{V_1}\right)^{2} = (0.90)^{2} = 0.81\]

   \[\text{fall} = (1 - 0.81)\times100 = 19\,\%\]

   A 10 % voltage dip costs nearly a fifth of the torque — which is why induction motors stall on severe voltage dips.

7. P58.7 Trace the chain by which an induction motor produces torque, and identify the law behind each step.

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   1. The three-phase stator winding carries currents displaced by \(120^{\circ}\) in time, in windings displaced by \(120^{\circ}\) in space, producing a rotating field at \(N_s = 120f/P\).

   2. The rotor is slower, so there is relative motion and the rotor conductors are cut by the flux.

   3. EMF is induced: \(e = Blv\) — Faraday's law, dynamically induced EMF.

   4. The cage is short-circuited, so current flows: \(i = e/Z_2\) — Ohm's law.

   5. Each current-carrying bar in the field feels a force \(F = Bil\) — the Lorentz force law.

   6. The forces around the periphery sum to a torque \(T = \sum Fr\).

   Direction is given by Lenz's law: the induced effects oppose the relative motion, and the only way the rotor can reduce it is to move in the same direction as the field.

   Note that \(B\) enters twice, in steps 3 and 5, so \(T \propto B^{2} \propto V^{2}\).

8. P58.8 Explain why an induction motor cannot run at synchronous speed, and why the resulting equilibrium is stable.

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   Read the causation chain backwards at \(N = N_s\):

   \[\text{relative motion} = 0 \ \Longrightarrow\ e = 0 \ \Longrightarrow\ i = 0 \ \Longrightarrow\ F = 0 \ \Longrightarrow\ T = 0\]

   With zero torque the machine could not overcome even its own friction and windage, so it must slow down.

   Why the equilibrium is stable. The moment the rotor slows, relative motion reappears, EMF and current return, and torque is restored. Conversely, if the rotor speeds up the slip falls, the torque falls, and it is pulled back.

   The machine therefore settles at the exact speed where developed torque equals load torque, always a little below synchronous — a self-correcting equilibrium, not a precarious one. Hence asynchronous motor.

9. P58.9 How is the direction of rotation reversed, and what happens if this is done while the motor is running?

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   Reversal. The rotor follows the field, and the field's direction is set by the phase sequence. Interchanging any two of the three supply leads changes the sequence from \(RYB\) to \(RBY\), so the field and the rotor both reverse. No internal change to the machine is needed.

   Swapping all three leads restores the original sequence and changes nothing.

   Doing it while running — plugging. The field reverses instantly but the rotor's inertia keeps it turning the old way. The relative motion becomes the sum of the two speeds instead of the difference, so

   \[s = \frac{-N_s - N}{-N_s} \approx 2 \ \text{for a lightly slipping motor}\]

   Rotor EMF is then nearly twice its standstill value, the current is very large, and the torque reverses and brakes the machine hard.

   The rotor absorbs power from both sides at once — electrical power from the supply and mechanical energy from the decelerating load — and all of it becomes heat in the rotor. Plugging is therefore effective but thermally severe, and machines intended for it are given high-resistance rotors and generous thermal capacity.

10. P58.10 In what respects is an induction motor like a transformer, and in what respects unlike?

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    Like a transformer:

    • The rotor receives power purely by induction, never by conduction — the stator is the primary, the rotor the secondary.

    • Ampere-turn balance holds exactly as in Chapter 43.

    • The equivalent circuit has the same form: series impedances and a shunt magnetising branch.

    Unlike a transformer:

    • The secondary quantities depend on speed — \(f_r = sf\), \(E_2 = sE_{20}\), \(X_2 = sX_{20}\) — so the machine has a torque-speed characteristic rather than a fixed ratio.

    • The magnetic circuit contains an air gap, so the magnetising current is 30 to 50 % of rated instead of 2 to 6 %. This is the entire reason for the induction motor's lagging power factor.

    • The output is mechanical, so part of the air-gap power becomes shaft work rather than all of it reaching an electrical load.

    Chapter 64 reconciles the two: dividing the rotor resistance by the slip refers the rotor circuit to supply frequency, and the induction motor's equivalent circuit becomes the transformer's with one term changed.