MMF of Distributed Windings and Winding Factors

Two preliminary facts about the windings themselves, which distinguish AC machines from the DC machines of Part 2:

• DC machines have closed-circuit windings — the winding forms a continuous closed path round the armature.

• Alternator windings are open: there is no closed path for the currents in the winding itself. One end is joined to neutral and the other brought out, giving a star connection.

Five quantities describe the geometry of any distributed winding, and every formula in this chapter is built from them.

Two sides of any coil should be under two adjacent poles, that is, the coil span should equal one pole pitch. For maximum EMF the two sides of a coil should be one pole pitch — 180° electrical — apart, so that their EMFs add directly rather than partially cancelling.

The winding must be arranged in the armature slots so that it produces a sinusoidal EMF. That requirement, not merely the maximum EMF, is what drives the design decisions of the rest of this chapter.

A winding may be arranged in one layer or two, and this is the first choice a designer makes.

Single-Layer Winding

The complete slot contains only one coil side of a coil.

Example — single-layer, one-turn, full-pitch, 4-pole, 12-slot machine. Here there are 3 slots per pole, so 1 slot per pole per phase, and the pole pitch is 3 slots. For maximum EMF the two sides of a coil should be one pole pitch (180° electrical) apart, which gives:

• \(R\)-phase: slot 1 \(\Rightarrow\) 4, 7 \(\Rightarrow\) 10

• \(Y\)-phase: slot 3, that is 120° afterwards — since 3 slots = 180°, 2 slots = 120°

• \(B\)-phase: slot 5 \(\Rightarrow\) 8, 11 \(\Rightarrow\) 2

The ends of the windings are joined to form a \(Y\)-connection.

Concentric or Chain Winding

• Number of slots \(= 2 \times\) number of coils \(=\) number of coil sides.

• The polar group of each phase is 360° electrical apart.

• It is necessary to use two different shapes of coil to avoid fouling of the end connections.

• Since polar groups of each phase are 360° apart, all such groups are connected in the same direction.

• Disadvantage: short-pitched coils cannot be used.

Double-Layer Winding

• Either wave-wound or lap-wound. This is the simplest and most commonly used arrangement.

• The number of slots is a multiple of the number of poles and phases. For a 4-pole, 3-phase machine this permits 12, 24, 36, 48 and so on — multiples of \(4\times3 = 12\).

• Number of slots = number of coils, all of the same shape. Each slot therefore contains two coil sides, one at the bottom and one at the top, and the coils overlap each other.

Example — 4-pole, 24-slot machine. The pole pitch is \(24/4 = 6\) slots. For maximum EMF the coils should be full-pitched, meaning one side of a coil sits in slot 1 and the other in slot 7, the two slots being one pole pitch — 180° — apart. Each phase then has \(24/3 = 8\) coils.

So far full-pitched coils have been assumed, in which one pole pitch equals 180°. In a 24-slot, 4-pole machine a full-pitched coil has its sides in slots 1 and 7.

A short-pitched or fractional-pitched coil is instead placed in slots 1 and 6, so that the coil span is \(5/6\) of a pole pitch. It falls short by \(1/6\) of a pole pitch, that is

The pitch factor — also called the coil-span factor or chording factor — is defined as

Let \(E_s\) be the induced EMF in each side of the coil. For a full-pitched coil the two sides are exactly in phase, so the total induced EMF is simply \(2E_s\).

If the coil is short-pitched by 30°, the two coil-side EMFs are 30° apart and their resultant is

The harmonic pay-off. The \(n\)th space harmonic sees \(n\) times the electrical angle, so its pitch factor is

and this is where short-pitching earns its keep. A harmonic can be eliminated entirely by choosing \(\alpha\) so that \(n\alpha/2 = 90^{\circ}\).

In each phase the coils are not concentrated or bunched in one slot, but are distributed over a number of slots to form polar groups under each pole.

• The coils of a phase are displaced from each other by a certain angle.

• As a result the EMFs induced in the coil sides constituting a polar group are not in phase, but differ by the angular displacement of the slots.

Take a three-phase, four-pole machine with 36 slots. This gives 9 slots per pole, hence 3 slots per pole per phase, and the angular displacement between any two adjacent slots is

If the three coils were bunched in one slot, the total EMF induced in the three coil sides would be their arithmetic sum, \(3E_s\), where \(E_s\) is the EMF induced in one coil side. Since the coils are distributed, the individual EMFs have a phase difference of 20° with each other, and the resultant is their vector sum:

The General Case

Let

\[\begin{array}{cl} \beta = & \text{angular displacement between slots} \\ n = & \text{no. of slots/pole} \\ m = & \text{no. of slots/pole/phase} \\ m\beta = & \text{phase spread angle} \end{array}\]

The \(m\) coil EMFs, each displaced from the next by \(\beta\), form a regular polygon whose vertices lie on a circle of radius \(r\). Each coil EMF is a chord subtending \(\beta\) at the centre, and the resultant is a chord subtending \(m\beta\):

\[\begin{aligned} K_d &= \frac{\text{vector sum of coil emfs}}{\text{arithmetic sum of coil emfs}} = \frac{E_r}{mE_s} \\[6pt] &= \frac{2r\sin\left(m\beta/2\right)}{m\times 2r\sin\left(\beta/2\right)} \\[6pt] K_d &= \frac{\sin\left(m\beta/2\right)}{m\sin\left(\beta/2\right)} \end{aligned}\]

The radius \(r\) cancels, so the result depends only on how many coils there are and how far apart they sit.

As with the pitch factor, the \(n\)th harmonic sees \(n\) times the slot angle:

The two effects are independent and multiply:

Typical values of \(k_w\) for the fundamental lie between 0.90 and 0.96 in three-phase machines. The number looks like a small correction and is easy to forget, but omitting it overstates the EMF by 5 to 10 % — enough to make a machine unusable.

So far the winding factors have been derived from EMF. The same factors govern the mmf the winding produces, and this is what connects the chapter to the torque results of Chapter 18.

A single full-pitch concentrated coil of \(N\) turns carrying current \(i\) produces a rectangular mmf wave of amplitude \(Ni/2\) around the air gap. Distributing the winding over several slots replaces the rectangle by a staircase, which is a much closer approximation to a sinusoid.

Extracting the fundamental of the stepped wave gives the result used throughout machine analysis:

This is the \(F_1\) and \(F_2\) that appeared in the torque expression \(T = -kF_1F_2\sin\delta\) of Section 18-7. The winding factor therefore reduces the torque in exactly the same proportion as the EMF, which is why a single number suffices for both.

• Windings are distributed over several slots and usually short-pitched, to improve cooling, improve the waveform and even out the iron loading.

• Key geometry: \(n = S/P\) slots per pole, \(\beta = 180^{\circ}/n\) slot angle, \(m = n/q\) slots per pole per phase, phase spread \(m\beta\).

• DC machines have closed windings; alternator windings are open, with one end at neutral and the other brought out.

• Double-layer windings dominate: identical coils, one per slot, two coil sides per slot, and short-pitching is possible. Concentric windings cannot be short-pitched.

• A coil short-pitched by \(\alpha\) has pitch factor \(k_c = \cos(\alpha/2)\), and for the \(n\)th harmonic \(k_{cn} = \cos(n\alpha/2)\).

• The distribution factor is \(K_d = \sin(m\beta/2)/\left[m\sin(\beta/2)\right]\), also called the breadth or spread factor.

• The winding factor \(k_w = k_cK_d\) corrects the EMF equation to \(E_{\text{ph}} = 4.44k_wfN_{\text{ph}}\Phi\). Typical values are 0.90 to 0.96.

• The same \(k_w\) multiplies the mmf: \(F_1 = (4/\pi)(k_wN_{\text{ph}}/P)i\), so torque is reduced in the same proportion as EMF.

• The 3 to 7 % of fundamental given up buys a 70 to 80 % reduction in the fifth and seventh harmonics — a bargain by any measure.

• Confusing electrical and mechanical angles. The slot angle \(\beta = 180^{\circ}/n\) is electrical. In a four-pole machine one mechanical degree is two electrical degrees.

• Using the full chording angle in the cosine. It is \(\cos(\alpha/2)\), not \(\cos\alpha\) — half the angle, because the resultant bisects the two coil-side phasors.

• Forgetting the \(m\) in the denominator of \(K_d\). Without it the formula gives a number far greater than one.

• Using degrees in a calculator set to radians. A perennial source of impossible winding factors.

• Adding \(k_c\) and \(K_d\) instead of multiplying. The two effects are independent, so their factors multiply.

• Omitting \(k_w\) from the EMF equation. This overstates the voltage by 5 to 10 %.

• Confusing turns per coil with turns per phase. \(N_{\text{ph}}\) is turns in series per phase — coils per phase times turns per coil, divided by any parallel paths.

• Counting coils wrongly in a double-layer winding. There is one coil per slot, not one per pair of slots.

• Assuming a negative harmonic pitch factor is an error. It merely indicates a phase reversal; the magnitude is what matters.

• Trying to short-pitch a concentric winding. Its geometry does not allow it — that is the disadvantage of the type.

Always compute \(n\), \(\beta\) and \(m\) first — every other quantity follows from them.

1. P19.1 A three-phase, 6-pole alternator has 54 slots. Find \(n\), \(\beta\), \(m\) and the phase spread.

   Show answer

   \[n = \frac{54}{6} = 9, \qquad \beta = \frac{180^{\circ}}{9} = 20^{\circ}, \qquad m = \frac{9}{3} = 3\]

   \[\text{phase spread} = m\beta = 60^{\circ}\]

   Identical to the 36-slot, 4-pole machine of Example 19.1 — the winding factors depend on \(n\) and \(m\), not on the absolute slot count.

2. P19.2 Find the distribution factor for the machine of P19.1.

   Show answer

   \[K_d = \frac{\sin\left(3\times10^{\circ}\right)}{3\sin 10^{\circ}} = \frac{0.5000}{0.52094} = 0.9598\]

3. P19.3 A coil spans 7 slots in a machine with a pole pitch of 9 slots. Find the chording angle and the pitch factor.

   Show answer

   The coil falls short by 2 slots out of 9, and each slot is \(180^{\circ}/9 = 20^{\circ}\):

   \[\alpha = 2\times20^{\circ} = 40^{\circ}, \qquad k_c = \cos 20^{\circ} = 0.9397\]

4. P19.4 A three-phase, 4-pole, 24-slot alternator runs at 1500 rev/min with 0.05 Wb per pole and 10 turns per coil, double-layer, full-pitch. Find the frequency, \(k_w\) and the phase EMF.

   Show answer

   \[f = \frac{PN}{120} = \frac{(4)(1500)}{120} = 50~\mathrm{Hz}\]

   \[n = 6, \quad \beta = 30^{\circ}, \quad m = 2, \qquad K_d = \frac{\sin 30^{\circ}}{2\sin 15^{\circ}} = \frac{0.5000}{0.51764} = 0.9659\]

   Full-pitch, so \(k_c = 1\) and \(k_w = 0.9659\). With \(24/3 = 8\) coils per phase at 10 turns, \(N_{\text{ph}} = 80\):

   \[E_{\text{ph}} = (4.44)(0.9659)(50)(80)(0.05) = 857.7~\mathrm{V}\]

5. P19.5 For the machine of P19.4, the coils are now short-pitched by one slot. Find the new \(k_w\) and phase EMF.

   Show answer

   \[\alpha = \beta = 30^{\circ}, \qquad k_c = \cos 15^{\circ} = 0.9659\]

   \[k_w = (0.9659)(0.9659) = 0.9330\]

   \[E_{\text{ph}} = (4.44)(0.9330)(50)(80)(0.05) = 828.5~\mathrm{V}\]

   A 3.4 % reduction, in exchange for the harmonic suppression of Table 19.2.

6. P19.6 What chording angle eliminates the seventh harmonic, and what is the resulting fundamental pitch factor?

   Show answer

   \[\frac{7\alpha}{2} = 90^{\circ} \quad\Longrightarrow\quad \alpha = \frac{180^{\circ}}{7} = 25.7^{\circ}\]

   \[k_{c1} = \cos 12.86^{\circ} = 0.9750\]

   Only 2.5 % of the fundamental lost. However the fifth would then give \(\cos 64.3^{\circ} = 0.4339\) — considerably worse than the 0.259 of a 5/6 pitch.

7. P19.7 Find the fifth-harmonic distribution factor for a winding with \(m = 2\) and \(\beta = 30^{\circ}\).

   Show answer

   \[K_{d5} = \frac{\sin\left(5\times30^{\circ}\right)}{2\sin\left(5\times15^{\circ}\right)} = \frac{\sin 150^{\circ}}{2\sin 75^{\circ}} = \frac{0.5000}{1.9319} = 0.2588\]

   The fifth harmonic is reduced to a quarter, against a fundamental \(K_d\) of 0.9659.

8. P19.8 A three-phase machine has \(k_w = 0.94\), \(N_{\text{ph}} = 120\) and 6 poles. Find the peak fundamental mmf per pole at 15 A.

   Show answer

   \[F_1 = \frac{4}{\pi}\left(\frac{k_wN_{\text{ph}}}{P}\right)i = (1.2732)\frac{(0.94)(120)}{6}(15)\]

   \[F_1 = (1.2732)(18.80)(15) = 359.0~\mathrm{AT/pole}\]

9. P19.9 Explain why the winding factor reduces the torque by exactly the same proportion as the EMF.

   Show answer

   The winding factor measures how effectively the turns of a phase are aligned with the fundamental flux wave. A winding whose coils are spread out or chorded links less of the fundamental flux, so its EMF is reduced by \(k_w\). By the same geometry it also produces a fundamental mmf reduced by \(k_w\).

   Since torque follows the product of the two mmf waves, \(T \propto F_1F_2\sin\delta\), and \(F\) carries a factor \(k_w\), torque falls in the same ratio. This reciprocity is why one factor suffices for both calculations.

10. P19.10 Why is a small loss of fundamental EMF accepted in order to suppress harmonics?

    Show answer

    The fundamental loss is 3 to 7 %, and can be recovered simply by adding a few more turns. The harmonics, if left in place, cause:

    • extra iron loss, since hysteresis and eddy-current loss rise steeply with frequency (Chapters 6 and 7);

    • parasitic torques — crawling and cogging in induction motors;

    • noise and vibration from harmonic radial force waves;

    • rotor surface loss, since harmonic fields sweep past the rotor at high relative speed;

    • possible non-compliance with grid distortion limits.

    None of these can be fixed by adding turns. A few percent of a recoverable quantity is traded for a large reduction in an unrecoverable one — which is why every AC machine of any size is both distributed and chorded.