Production of the Rotating Magnetic Field

Stationary coils, wound and supplied for two-phase or three-phase respectively, produce a uniformly rotating or revolving magnetic flux of constant value.

This chapter proves both assertions, first for the simpler two-phase case and then for the three-phase machine that matters. The result underpins everything that follows: the induction motor of Part 4 and the synchronous machine of Part 5 both depend on it entirely.

Consider two windings \(P\) and \(Q\) placed at \(90^{\circ}\) with respect to each other. Exciting them with a two-phase supply, the flux produced by each is taken as purely sinusoidal:

The direction of the flux along each axis is taken as positive one way; the opposite side indicates negative values. Here \(\theta = \omega t\) is the electrical angle, advancing steadily with time.

Resolve both fluxes onto a fixed pair of axes. Take \(x\) along the axis of \(P\) and \(y\) along the axis of \(Q\).

The Pythagorean identity does all the work. Because the two fluxes are in quadrature in space and also in quadrature in time, their squares always sum to the same value — which is precisely the condition for a rotating rather than a pulsating field.

Now the case that matters. Three windings are placed at \(120^{\circ}\) to one another in space, and carry currents displaced by \(120^{\circ}\) in time:

Note carefully that the displacement is twofold: \(120^{\circ}\) in space between the winding axes, and \(120^{\circ}\) in time between the currents. Both are essential — remove either and the field pulsates instead of rotating.

As before, a table of instants is only suggestive. The general result follows from two trigonometric identities.

Notice that the two components have exactly the form found in the two-phase case, merely scaled by 1.5. Once the three fluxes are resolved onto fixed axes, the three-phase system becomes a two-phase system of larger amplitude — which is the mathematical content of the Scott connection met in Chapter 55, seen from the other direction.

The same argument works for any number of phases, provided the windings are equally spaced in space and the currents equally spaced in time.

The proof showed that the field's direction advances at the same rate as \(\theta\). But \(\theta\) is the electrical angle, and only for a two-pole machine is that the same as the mechanical angle.

A two-pole machine has \(p = 1\), so its field makes one revolution per supply cycle — 3000 rev/min at 50 Hz. A four-pole machine needs two cycles per revolution, and so runs at half that speed, and so on.

The proof gave the direction of the resultant as \(\alpha = \theta - 90^{\circ}\), which advances from the axis of \(A\) towards the axis of \(B\), then towards \(C\).

Swapping all three leads simply relabels the phases and restores the original sequence, so nothing changes — the point already made in Chapter 58 and worth verifying against this proof: replacing \(\theta\) by \(-\theta\) in the two component equations gives \(\alpha = -\theta - 90^{\circ}\), a field of the same magnitude turning the other way.

With the field established, the argument of Chapter 58 can now be stated in full.

1. The three-phase supply given to the stator winding produces a magnetic flux of constant magnitude but rotating at synchronous speed.
2. The flux passes through the air gap, sweeps past the rotor surface, and so cuts the rotor conductors, which as yet are stationary.
3. Owing to the relative speed between the rotating flux and the stationary conductors, an EMF is induced in the latter, according to Faraday's laws of electromagnetic induction. Its magnitude is proportional to the relative velocity, and its direction is given by Fleming's right-hand rule.
4. Since the rotor bars form a closed circuit, rotor current is produced. Its direction, by Lenz's law, is such as to oppose the very cause producing it.
5. That cause is the relative velocity between the stator's rotating flux and the rotor conductors.
6. Hence, to reduce the relative speed, the rotor starts running in the same direction as the flux and tries to catch up with it.

• Stationary coils supplied with polyphase currents produce a flux of constant magnitude rotating uniformly. Both parts of that claim need proof.

• Two phases: resolving gives \(\Phi_x = \Phi_m\sin\theta\) and \(\Phi_y = -\Phi_m\cos\theta\), so \(\left|\Phi\right| = \Phi_m\) for every \(\theta\) and \(\alpha = \theta - 90^{\circ}\).

• Three phases: the sum-to-product identities give \(\Phi_x = 1.5\Phi_m\sin\theta\) and \(\Phi_y = -1.5\Phi_m\cos\theta\), hence \(\left|\Phi\right| = 1.5\Phi_m\), constant, rotating uniformly.

• Once resolved, the three-phase system becomes a two-phase system of 1.5 times the amplitude.

• In general \(\left|\Phi\right| = (m/2)\Phi_m\). A single phase gives a pulsating field, not a rotating one, and hence no starting torque.

• The displacement must be twofold — \(120^{\circ}\) in space and \(120^{\circ}\) in time. Remove either and the field pulsates.

• One supply cycle advances the field by one pole pair, so \(n_s = f/p = 2f/P\) rev/s and \(N_s = 120f/P\) rev/min.

• The field runs in the order of the phase sequence; interchanging any two leads reverses it.

• The rotor follows because Lenz's law makes the induced effects oppose the relative motion, and the only way to reduce it is to chase the field.

• The rotor EMF frequency equals the supply frequency only at standstill; in general it is \(sf\).

• Proving the result at a few instants and stopping. A table is suggestive; only the general algebra establishes it.

• Using one symbol for both the time angle and the space angle. They are different quantities, and conflating them makes the cosine-rule approach unreadable.

• Expecting the resultant to be \(\Phi_m\) in a three-phase machine. It is \(1.5\Phi_m\), larger than any single phase ever reaches.

• Forgetting that the displacement must be twofold. Three windings on the same axis, or three in space fed in phase, both give pulsation.

• Placing two-phase windings at \(180^{\circ}\). They must be in quadrature; \(180^{\circ}\) is a single axis.

• Confusing electrical and mechanical angle. They are equal only for a two-pole machine.

• Thinking more poles give a faster field. They give a slower one, since \(N_s = 120f/P\).

• Reversing by swapping all three leads. That restores the original sequence.

• Saying rotor EMF is always at supply frequency. True only at standstill; in general \(f_r = sf\).

• Expecting a single-phase motor to start on its own. Its field pulsates, so the starting torque is zero.

For any resultant, resolve each phase flux along its own axis into fixed \(x\) and \(y\) components and add. The method never fails and needs no special formula.

1. P59.1 Find the three-phase resultant at \(\theta = 120^{\circ}\), taking \(\Phi_m = 1\).

   Show answer

   \[\Phi_A = \sin120^{\circ} = 0.8660, \quad \Phi_B = \sin0^{\circ} = 0, \quad \Phi_C = \sin\left(-120^{\circ}\right) = -0.8660\]

   \[\Phi_x = 0.8660 - 0 - 0.5\left(-0.8660\right) = 1.2990\]

   \[\Phi_y = 0.86603\left(0 - \left(-0.8660\right)\right) = 0.7500\]

   \[\left|\Phi\right| = \sqrt{1.6874 + 0.5625} = 1.5000, \qquad \alpha = +30^{\circ} = \theta - 90^{\circ} \quad\checkmark\]

2. P59.2 Verify the general formula against P59.1.

   Show answer

   \[\Phi_x = 1.5\sin120^{\circ} = (1.5)(0.8660) = 1.2990 \quad\checkmark\]

   \[\Phi_y = -1.5\cos120^{\circ} = -(1.5)\left(-0.5\right) = 0.7500 \quad\checkmark\]

   Both agree exactly with the component sums.

3. P59.3 A 12-pole machine is supplied at 60 Hz. Find the field speed and the mechanical angle swept per supply cycle.

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   \[p = 6, \qquad n_s = \frac{60}{6} = 10.00~\mathrm{rev/s}\]

   \[N_s = \frac{120(60)}{12} = 600~\mathrm{rev/min}\]

   \[\text{mechanical angle per cycle} = \frac{360^{\circ}}{6} = 60^{\circ}\]

4. P59.4 A machine's field must rotate at 900 rev/min from a 60 Hz supply. How many poles?

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   \[P = \frac{120f}{N_s} = \frac{120(60)}{900} = 8\ \text{poles}\]

5. P59.5 A 6-phase winding is supplied with balanced six-phase currents of peak flux \(\Phi_m\) each. Find the resultant.

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   \[\left|\Phi\right| = \frac{m}{2}\Phi_m = \frac{6}{2}\Phi_m = 3.0\,\Phi_m\]

   Twice the three-phase value, since twice as many windings contribute.

6. P59.6 A 4-pole, 50 Hz motor has an air-gap flux density of 0.90 T, rotor diameter 0.25 m and core length 0.30 m. Find the EMF induced in one bar at standstill and at 5 % slip.

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   \[n_s = \frac{2(50)}{4} = 25~\mathrm{rev/s}, \qquad v = \pi(0.25)(25) = 19.635~\mathrm{m/s}\]

   \[e_{standstill} = (0.90)(0.30)(19.635) = 5.301~\mathrm{V}\]

   \[e_{5\%} = (0.05)(5.301) = 0.2651~\mathrm{V}\]

7. P59.7 Prove that three windings displaced by \(120^{\circ}\) in space, carrying currents displaced by \(120^{\circ}\) in time, produce a resultant of constant magnitude \(1.5\Phi_m\).

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   Resolve each phase flux along its own axis into components on fixed \(x\) and \(y\) axes, with \(x\) along the axis of phase A:

   \[\frac{\Phi_x}{\Phi_m} = \sin\theta - 0.5\sin\left(\theta-120^{\circ}\right) - 0.5\sin\left(\theta-240^{\circ}\right)\]

   Using \(\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\):

   \[\sin\left(\theta-120^{\circ}\right) + \sin\left(\theta-240^{\circ}\right) = 2\sin\left(\theta-180^{\circ}\right)\cos60^{\circ} = -\sin\theta\]

   \[\therefore\ \Phi_x = \Phi_m\left(\sin\theta + 0.5\sin\theta\right) = 1.5\Phi_m\sin\theta\]

   For the \(y\) component, using \(\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\):

   \[\Phi_y = 0.866\Phi_m\left[2\cos\left(\theta-180^{\circ}\right)\sin60^{\circ}\right] = -1.5\Phi_m\cos\theta\]

   Hence

   \[\left|\Phi\right| = 1.5\Phi_m\sqrt{\sin^{2}\theta + \cos^{2}\theta} = 1.5\Phi_m, \qquad \alpha = \theta - 90^{\circ}\]

   The magnitude is constant for every \(\theta\) and the direction advances at the same rate as \(\theta\) — a field of constant magnitude rotating uniformly, which is what was to be proved.

8. P59.8 Why does a single-phase winding fail to produce a rotating field, and what follows?

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   With one winding there is only one axis. The flux along it varies as \(\Phi_m\sin\theta\), growing, collapsing and reversing, but it can never point in any other direction.

   The field therefore pulsates along a fixed line rather than rotating. Its magnitude is not constant, and it has no direction of travel.

   The consequence: a stationary rotor placed in a pulsating field experiences equal and opposite torque impulses over each half cycle, so the net starting torque is exactly zero. The motor will not start on its own, though it will continue to run if started by other means.

   This is why every single-phase induction motor needs a starting arrangement — an auxiliary winding, a capacitor, or shaded poles — to fabricate an artificial second phase and so produce at least a crude rotating field. The whole subject matter of Part 6 follows from this single deficiency.

9. P59.9 Explain why one supply cycle does not advance the field by one revolution, and derive \(N_s\).

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   Because the winding is arranged for \(P\) poles, the magnetic pattern repeats \(p = P/2\) times around the periphery. One complete electrical cycle carries the field through one repetition of that pattern — one pole pair — not once round the machine.

   \[\theta_{mech} = \frac{\theta_{elec}}{p}\]

   So in \(f\) cycles per second the field makes

   \[n_s = \frac{f}{p} = \frac{2f}{P}\ \mathrm{rev/s} \quad\Longrightarrow\quad N_s = \frac{120f}{P}\ \mathrm{rev/min}\]

   Check: a two-pole machine has \(p = 1\), so one cycle is one revolution and \(N_s = 3000\) rev/min at 50 Hz. A four-pole machine needs two cycles per revolution and runs at half that.

   Adding poles slows the field, and at a fixed frequency the available speeds are therefore a coarse discrete set — 3000, 1500, 1000, 750 — with nothing between.

10. P59.10 Trace the direction argument from a clockwise stator field to a clockwise rotor.

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    1. Assume the stator field rotates clockwise.

    2. The rotor is initially stationary, so its motion relative to the field is anticlockwise.

    3. Applying Fleming's right-hand rule to that relative motion, the induced EMF in the rotor conductors is outwards.

    4. The cage is closed, so current flows in that direction.

    5. Applying Fleming's left-hand rule to a current-carrying conductor in the field, the force on the conductor is clockwise.

    The rotor is therefore set into rotation in the same direction as the stator flux.

    The result is exactly what Lenz's law demands, and could have been written down at once: the induced effects oppose the relative motion, and the only way the rotor can reduce that motion is to chase the field. The two rules simply confirm the sign.