Solved Problems · Set 33

Alternator Fundamentals and the EMF Equation

Part 5 · Synchronous Machines — one equation, \(E = 4.44\,k_w f \Phi T_{ph}\), and the bookkeeping around it: poles into frequency, slots and conductors into turns per phase, phase into line.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 33 — Alternator Fundamentals and the EMF Equation

An alternator generates because a rotating field sweeps past stationary conductors, and the voltage it produces is settled by five numbers: the frequency, the flux per pole, the series turns per phase, and the two winding factors that account for the coils not all being in step. This set does nothing but assemble those numbers correctly and substitute them.

The arithmetic is easy and the traps are all in the bookkeeping. Conductors per slot must be halved to give turns and turns per coil must not; the equation returns a phase voltage that is not the terminal voltage unless the machine is in delta; the slot angle is electrical, not mechanical; and a harmonic carries its own frequency, its own flux and its own pair of winding factors. Each problem here is built around one of those distinctions.

Part 5 · Alternators · 6 solved problems

i Method Recap

  • Poles and speed fix the frequency. A rotor with \(P\) poles turning at \(N\) rev/min sweeps \(P/2\) cycles per revolution:

    \[ f = \frac{PN}{120} \qquad\Longleftrightarrow\qquad N_s = \frac{120f}{P} \]

  • The e.m.f. equation is one line and never changes. Per phase, with \(T_{ph}\) series turns per phase and \(\Phi\) the flux per pole:

    \[ E_{ph} = 4.44\,k_p k_d\, f\, \Phi\, T_{ph} = 4.44\,k_w f \Phi T_{ph} \]

    The 4.44 is \(\pi\sqrt{2}\) — the form factor 1.11 of a sine wave times the 4 of Faraday's law for a full cycle.

  • Everything hard is the bookkeeping into \(T_{ph}\). From conductors per slot \(Z_s\) in \(S\) slots, or from turns per coil in a double-layer winding:

    \[ T_{ph} = \frac{S Z_s}{3 \times 2} \qquad\text{or}\qquad T_{ph} = \frac{S \times \text{turns/coil}}{3} \]

    A double-layer winding has as many coils as slots. Two conductors make one turn — halve once, and only once.

  • The distribution factor accounts for the coils of a phase belt lying in different slots, so their e.m.f.s add as phasors rather than arithmetically. With \(m\) slots per pole per phase and slot angle \(\beta\):

    \[ \beta = \frac{180^\circ P}{S} = \frac{180^\circ}{\text{slots/pole}}, \qquad m = \frac{S}{3P}, \qquad k_d = \frac{\sin(m\beta/2)}{m\sin(\beta/2)} \]

    The angles are in electrical degrees. Feeding radians to a calculator left in degree mode is the single most common numerical error in this set.

  • The pitch factor accounts for a coil spanning less than a pole pitch, so its two sides are not exactly in opposition. If the coil falls short by \(\alpha\) electrical degrees, \(k_p = \cos(\alpha/2)\), and \(k_p = 1\) for a full-pitched coil.

  • Phase is not line. The e.m.f. equation always gives the phase value. Then

    \[ E_L = \sqrt{3}\,E_{ph}\ \text{(star)}, \qquad E_L = E_{ph}\ \text{(delta)} \]

  • Harmonics ride on the same equation with their own frequency, flux and winding factors, and combine in r.m.s.:

    \[ E_{ph} = \sqrt{E_1^2 + E_3^2 + E_5^2 + \cdots}, \qquad \Phi_n = \frac{1}{n}\left(\frac{B_n}{B_1}\right)\Phi_1 \]

    Triplen harmonics are in phase in all three windings, so they cancel in every line voltage of a star or delta machine.

VideoWalkthrough
Problem 1CoreFrequency and Line E.M.F.

A 3-phase, 16-pole alternator has a star-connected winding with 144 stator slots and 10 conductors per slot. The flux per pole is 0.03 Wb, sinusoidally distributed, and the machine is driven at 375 rpm. The coils are full-pitched. Determine

  1. the frequency of the generated e.m.f.;
  2. the e.m.f. generated per phase;
  3. the line e.m.f. at the terminals.
Solution

Frequency from poles and speed. Nothing about the winding is needed for this — only how fast the poles pass a conductor:

\[ f = \frac{PN}{120} = \frac{16 \times 375}{120} = 50\ \text{Hz} \]

A 16-pole machine on a 50 Hz system is locked to 375 rpm and to no other speed.

Reduce the slot geometry to \(\beta\) and \(m\). There are 144 slots spread over 16 poles, and the three phases share them equally:

\[ \text{slots/pole} = \frac{144}{16} = 9, \qquad \beta = \frac{180^\circ}{9} = 20^\circ, \qquad m = \frac{144}{16 \times 3} = 3 \]

The two winding factors. The coils are full-pitched, so the pitch factor is unity; the three slots of a phase belt are 20° apart, so their e.m.f.s add as phasors:

\[ k_p = 1, \qquad k_d = \frac{\sin(3 \times 20^\circ/2)}{3\sin(20^\circ/2)} = \frac{\sin 30^\circ}{3\sin 10^\circ} = \frac{0.5}{0.5209} = 0.960 \]

Distributing the winding costs 4 % of the fundamental e.m.f. and buys a far cleaner wave shape — a bargain that every practical machine takes.

Conductors to turns per phase. The 144 slots hold \(144 \times 10 = 1440\) conductors in all; one third belong to each phase, and two conductors make one turn:

\[ Z_{ph} = \frac{144 \times 10}{3} = 480, \qquad T_{ph} = \frac{Z_{ph}}{2} = 240 \]

Apply the e.m.f. equation with the phase quantities just assembled:

\[ \begin{aligned} E_{ph} &= 4.44\,k_p k_d\, f\, \Phi\, T_{ph} \\ &= 4.44 \times 1 \times 0.960 \times 50 \times 0.03 \times 240 \\ &= 1534\ \text{V} \end{aligned} \]

Phase to line. The winding is star-connected, so the terminal voltage is \(\sqrt{3}\) times the phase e.m.f.:

\[ E_L = \sqrt{3}\,E_{ph} = 1.732 \times 1534 = 2657\ \text{V} \]
The e.m.f. equation itself is trivial; the four numbers fed into it are where the marks are. Frequency comes from poles and speed, \(k_d\) from slots per pole per phase, \(T_{ph}\) from a conductor count halved exactly once, and the line value from the connection. Write those four down before touching the 4.44.
Answer(a)\(f = 50\ \text{Hz}\)   (b)\(E_{ph} = 1534\ \text{V}\)   (c)\(E_L = 2657\ \text{V}\)
Problem 2Exam levelChorded Double-Layer Winding

Find the no-load phase and line voltage of a star-connected, 3-phase, 6-pole alternator which runs at 1200 rpm and has a sinusoidally distributed flux per pole of 0.1 Wb. Its stator carries 54 slots with a double-layer winding; each coil has 8 turns and every coil is chorded by one slot.

Solution

Frequency. Six poles at 1200 rpm is not a 50 Hz machine:

\[ f = \frac{PN}{120} = \frac{6 \times 1200}{120} = 60\ \text{Hz} \]

Slot angle and phase spread.

\[ \text{slots/pole} = \frac{54}{6} = 9, \qquad \beta = \frac{180^\circ}{9} = 20^\circ, \qquad m = \frac{54}{6 \times 3} = 3 \]
\[ k_d = \frac{\sin 30^\circ}{3 \sin 10^\circ} = 0.960 \]

The chording. One slot is worth \(\beta = 20^\circ\) electrical, so the coil falls short of a full pole pitch by that angle:

\[ \alpha = 20^\circ, \qquad k_p = \cos\frac{\alpha}{2} = \cos 10^\circ = 0.985 \]

A 1.5 % loss of fundamental e.m.f. In return the fifth and seventh harmonics are cut, the end-windings shorten and copper is saved — which is why nearly every alternator is short-pitched.

Turns per phase — the step that decides the answer. A double-layer winding has one coil per slot, so there are 54 coils of 8 turns each:

\[ T_{\text{total}} = 54 \times 8 = 432, \qquad T_{ph} = \frac{432}{3} = 144\ \text{turns/phase} \]

Cross-check through conductors: two coil sides per slot, 8 conductors per side, gives 16 conductors per slot; \(Z = 54 \times 16 = 864\), \(Z_{ph} = 288\), \(T_{ph} = 288/2 = 144\). The two routes agree, and taking the second route without first counting 16 conductors per slot is what halves the answer wrongly.

The phase e.m.f.:

\[ \begin{aligned} E_{ph} &= 4.44\,k_p k_d\, f\, \Phi\, T_{ph} \\ &= 4.44 \times 0.985 \times 0.960 \times 60 \times 0.1 \times 144 \\ &= 3626\ \text{V} \end{aligned} \]

The line voltage, the winding being in star:

\[ E_L = \sqrt{3} \times 3626 = 6280\ \text{V} \approx 6.28\ \text{kV} \]

A sensible rating: 6.6 kV class machines of this size are standard, and the no-load voltage sitting slightly under the nominal figure is exactly what one expects before excitation is trimmed.

Turns per coil and conductors per slot are different quantities and the factor of two between them is not interchangeable. When the data says turns per coil, multiply by the number of coils and divide by 3 — there is no further halving. Halve only when the data counts conductors.
Answer\(f = 60\ \text{Hz},\quad E_{ph} = 3626\ \text{V},\quad E_L = 6280\ \text{V}\)
Problem 3ChallengeThird-Harmonic E.M.F.

Calculate the r.m.s. value of the induced e.m.f. per phase of a 10-pole, 3-phase, 50 Hz alternator with 2 slots per pole per phase and 4 conductors per slot in two layers. The coil span is 150° electrical. The flux per pole has a fundamental component of 0.12 Wb together with a third-harmonic component whose peak density is 20 % of the fundamental. Find also the line voltage.

Solution

Assemble the winding data. With \(m = 2\) slots per pole per phase there are 6 slots per pole, so

\[ \text{slots/pole} = 3m = 6, \qquad \beta = \frac{180^\circ}{6} = 30^\circ \]
\[ Z_{ph} = P \times m \times Z_s = 10 \times 2 \times 4 = 80, \qquad T_{ph} = \frac{80}{2} = 40 \]

Each phase owns \(10 \times 2 = 20\) slots and each slot holds 4 conductors.

Fundamental winding factors. A 150° span falls 30° short of the pole pitch:

\[ \alpha = 180^\circ - 150^\circ = 30^\circ, \qquad k_{p1} = \cos 15^\circ = 0.966 \]
\[ k_{d1} = \frac{\sin(2 \times 30^\circ/2)}{2\sin(30^\circ/2)} = \frac{\sin 30^\circ}{2\sin 15^\circ} = 0.966 \]

Fundamental e.m.f.:

\[ E_1 = 4.44 \times 0.966 \times 0.966 \times 50 \times 0.12 \times 40 = 994\ \text{V} \]

The third harmonic sees different winding factors. Every electrical angle is multiplied by the harmonic order \(n = 3\):

\[ k_{p3} = \cos\frac{3\alpha}{2} = \cos 45^\circ = 0.707 \]
\[ k_{d3} = \frac{\sin(3m\beta/2)}{m\sin(3\beta/2)} = \frac{\sin 90^\circ}{2\sin 45^\circ} = 0.707 \]

The same geometry that barely touches the fundamental removes 30 % of the third harmonic twice over — this is precisely what chording and distribution are for.

Third-harmonic frequency and flux. The harmonic field has three times as many poles, so its pole pitch is one third of the fundamental's and the flux under one harmonic pole is reduced accordingly:

\[ f_3 = 3f = 150\ \text{Hz}, \qquad \Phi_3 = \frac{1}{3} \times 0.20 \times \Phi_1 = \frac{0.20 \times 0.12}{3} = 0.008\ \text{Wb} \]

The \(1/3\) is geometry, not a fudge: a 20 % harmonic flux density integrated over a third of the area gives a 6.7 % harmonic flux.

Third-harmonic e.m.f., from the same equation with harmonic values throughout:

\[ E_3 = 4.44 \times 0.707 \times 0.707 \times 150 \times 0.008 \times 40 = 106.6\ \text{V} \]

The tripled frequency almost cancels the reduced flux, which is why harmonic e.m.f.s are never as small as the harmonic fluxes suggest.

Combine in r.m.s. Components of different frequency are orthogonal, so they add in quadrature and never linearly:

\[ E_{ph} = \sqrt{E_1^2 + E_3^2} = \sqrt{994^2 + 106.6^2} = 1000\ \text{V} \]

An 11 % harmonic raises the r.m.s. by only 0.6 % — a quadrature sum is very forgiving of small terms.

The line voltage contains no third harmonic at all. The third harmonic e.m.f.s of the three phases are displaced by \(3 \times 120^\circ = 360^\circ\), that is, they are in phase; in any line-to-line difference they cancel exactly. The same holds for the ninth, fifteenth and every other triplen:

\[ E_L = \sqrt{3}\,E_1 = 1.732 \times 994 = 1722\ \text{V} \]
The third harmonic exists in the phase winding and vanishes between the lines. A voltmeter across a phase of a star-connected machine reads 1000 V; one across two lines reads \(\sqrt3 \times 994 = 1722\) V, not \(\sqrt3 \times 1000 = 1732\) V. The 10 V difference is the harmonic that the connection threw away.
Answer\(E_1 = 994\ \text{V},\ E_3 = 106.6\ \text{V},\ E_{ph} = 1000\ \text{V},\ E_L = 1722\ \text{V}\)
Problem 4CorePoles, Speed and Frequency

A hydro-electric station couples its alternator directly to a water turbine whose best efficiency lies near 150 rpm. The station feeds a 50 Hz network.

  1. How many poles must the alternator have, and at what speed does it then run?
  2. A steam turbo-alternator on the same network runs at 3000 rpm. How many poles has it?
  3. While still disconnected from the network the hydro machine's governor lets it settle at 144 rpm. What frequency does it generate?
  4. Could the turbine designer have specified 160 rpm instead?
Solution

One relation governs all four parts. A pair of poles passing a conductor produces one cycle, so in one revolution a \(P\)-pole rotor produces \(P/2\) cycles:

\[ f = \frac{P}{2}\cdot\frac{N}{60} = \frac{PN}{120} \]

Read it in whichever direction the question needs — for \(P\), for \(N\) or for \(f\). There is no other equation in this problem.

(a) The pole number for 150 rpm at 50 Hz:

\[ P = \frac{120f}{N} = \frac{120 \times 50}{150} = 40\ \text{poles} \]

Forty poles is an even integer, so the requirement is met exactly and the machine runs at precisely 150 rpm. Slow speed means a large pole number, which is why hydro alternators are wide, flat, salient-pole machines while turbo-alternators are long and thin.

(b) The turbo-alternator:

\[ P = \frac{120 \times 50}{3000} = 2\ \text{poles} \]

3000 rpm is the fastest a 50 Hz machine can turn, because two poles is the fewest it can have.

(c) Off-speed running. The pole number is fixed by construction, so the frequency follows the speed directly:

\[ f = \frac{PN}{120} = \frac{40 \times 144}{120} = 48\ \text{Hz} \]

A 4 % speed error is a 4 % frequency error. Once the machine is synchronised to the network the speed is no longer free — the network holds it at 150 rpm and the governor setting then decides power, not speed.

(d) Why 160 rpm is not available.

\[ P = \frac{120 \times 50}{160} = 37.5 \]

Poles come in north–south pairs, so \(P\) must be an even integer. Only the discrete speeds below exist at 50 Hz, and the turbine must be designed to one of them:

Poles \(P\)2461016363840
\(N_s\) (rpm)300015001000600375166.7157.9150

The nearest available speeds to 160 rpm are 157.9 rpm (38 poles) and 166.7 rpm (36 poles).

Speed is not a free variable in a synchronous machine. Fix the frequency and the pole number, and the speed is determined to the last decimal place; that rigidity is what "synchronous" means, and it is the whole reason a synchronous machine can be paralleled with a network at all.
Answer(a)40 poles at 150 rpm   (b)2 poles   (c)\(f = 48\ \text{Hz}\)   (d)No — \(P = 37.5\) is not an even integer
Problem 5CoreFlux Per Pole Required

A 3-phase, 4-pole, 50 Hz alternator has 36 stator slots with 8 conductors per slot, all coils full-pitched. Determine the flux per pole needed to generate a no-load terminal voltage of 400 V when the stator is

  1. star-connected;
  2. delta-connected.

State also the speed at which the machine must be driven.

Solution

Driving speed, fixed by the pole number and the required frequency:

\[ N_s = \frac{120f}{P} = \frac{120 \times 50}{4} = 1500\ \text{rpm} \]

Winding constants. These depend on the slots alone and are the same for either connection:

\[ \text{slots/pole} = \frac{36}{4} = 9, \qquad \beta = 20^\circ, \qquad m = \frac{36}{4 \times 3} = 3 \]
\[ k_p = 1, \qquad k_d = \frac{\sin 30^\circ}{3\sin 10^\circ} = 0.960, \qquad k_w = 0.960 \]

Turns per phase. The slot data counts conductors, so halve once:

\[ Z_{ph} = \frac{36 \times 8}{3} = 96, \qquad T_{ph} = \frac{96}{2} = 48 \]

Invert the e.m.f. equation. Everything except the flux is now known, so collect the constant once and use it twice:

\[ \frac{E_{ph}}{\Phi} = 4.44\,k_w f\, T_{ph} = 4.44 \times 0.960 \times 50 \times 48 = 10228\ \text{V/Wb} \]
\[ \Phi = \frac{E_{ph}}{10228} \]

(a) Star connection. A 400 V terminal voltage is a line value, so each phase need only generate \(400/\sqrt3\):

\[ E_{ph} = \frac{400}{\sqrt3} = 231.0\ \text{V}, \qquad \Phi = \frac{231.0}{10228} = 0.0226\ \text{Wb} = 22.6\ \text{mWb} \]

(b) Delta connection. Now the phase winding stands across the full 400 V:

\[ E_{ph} = 400\ \text{V}, \qquad \Phi = \frac{400}{10228} = 0.0391\ \text{Wb} = 39.1\ \text{mWb} \]

The ratio is exactly \(\sqrt3 = 1.732\). The delta machine needs 73 % more flux — and therefore a much heavier magnetic circuit and a larger field current — to reach the same terminal voltage. This is one reason alternators are almost always star-connected.

The e.m.f. equation knows nothing about the connection. It always returns the voltage of one phase winding. Deciding whether the number quoted in the question is a phase value or a line value, and dividing by \(\sqrt3\) at the right moment, is a separate step that the formula will not do for you.
Answer\(N_s = 1500\ \text{rpm}\); (a)\(\Phi = 22.6\ \text{mWb}\)   (b)\(\Phi = 39.1\ \text{mWb}\)
Problem 6Exam levelSlots and Conductors to Turns

A 3-phase, 12-pole, star-connected alternator runs at 500 rpm. Its stator has 180 slots with 8 conductors per slot arranged in two layers, and every coil spans 13 slots. The flux per pole is 0.025 Wb. Determine

  1. the frequency;
  2. the distribution factor;
  3. the pitch factor;
  4. the series turns per phase;
  5. the phase and line e.m.f.
Solution

(a) Frequency:

\[ f = \frac{PN}{120} = \frac{12 \times 500}{120} = 50\ \text{Hz} \]

Slot geometry. Both winding factors come from these three numbers:

\[ \text{slots/pole} = \frac{180}{12} = 15, \qquad \beta = \frac{180^\circ}{15} = 12^\circ, \qquad m = \frac{180}{12 \times 3} = 5 \]

(b) Distribution factor. Five slots per phase belt, 12° apart, spreading the belt over the standard 60°:

\[ k_d = \frac{\sin(m\beta/2)}{m\sin(\beta/2)} = \frac{\sin(5 \times 6^\circ)}{5\sin 6^\circ} = \frac{\sin 30^\circ}{5 \times 0.10453} = \frac{0.5}{0.52264} = 0.9567 \]

Check the spread: \(m\beta = 5 \times 12^\circ = 60^\circ\), as it must be for a three-phase 60° phase-belt winding whatever the slot count.

(c) Pitch factor. A full pitch is 15 slots; the coil spans 13, falling short by 2 slots:

\[ \alpha = 2 \times 12^\circ = 24^\circ, \qquad k_p = \cos\frac{24^\circ}{2} = \cos 12^\circ = 0.9781 \]
\[ k_w = k_p k_d = 0.9781 \times 0.9567 = 0.9358 \]

(d) Turns per phase. The data gives conductors, so halve once at the end:

\[ Z_{ph} = \frac{180 \times 8}{3} = 480, \qquad T_{ph} = \frac{480}{2} = 240 \]

Cross-check by coils, as in Problem 2: two layers means 4 conductors per coil side, so each coil has 4 turns; 180 coils give 720 turns, and 720/3 = 240 turns per phase. The two counts agree.

(e) The e.m.f.:

\[ \begin{aligned} E_{ph} &= 4.44\,k_w f \Phi T_{ph} \\ &= 4.44 \times 0.9358 \times 50 \times 0.025 \times 240 \\ &= 1246\ \text{V} \end{aligned} \]
\[ E_L = \sqrt{3} \times 1246 = 2159\ \text{V} \]

A 12-pole, 500 rpm, 2.2 kV machine — a small hydro or diesel set, and entirely realistic.

The winding factor is a single number that never strays far from 0.93. Distribution costs about 4 %, a modest chording another 2 %, and every practical three-phase alternator lands between 0.90 and 0.96. If your \(k_w\) comes out at 0.6 or at 1.2, the error is in \(\beta\) or in a calculator left in radian mode — check before going on.
Answer\(f = 50\ \text{Hz},\ k_d = 0.957,\ k_p = 0.978,\ T_{ph} = 240,\ E_{ph} = 1246\ \text{V},\ E_L = 2159\ \text{V}\)
Formulas

Key Formulas

QuantityRelationNotes
Frequency\(f = PN/120\)Problems 1, 2, 4, 6
Synchronous speed\(N_s = 120f/P\)\(P\) even; discrete speeds only — Problem 4
E.m.f. per phase\(E_{ph} = 4.44\,k_p k_d f \Phi T_{ph}\)The whole set in one line
Same, by conductors\(E_{ph} = 2.22\,k_w f \Phi Z_{ph}\)Because \(T_{ph} = Z_{ph}/2\)
Slot angle\(\beta = 180^\circ P/S = 180^\circ/(\text{slots per pole})\)Electrical degrees
Slots per pole per phase\(m = S/(3P)\)\(m\beta = 60^\circ\) always — Problem 6
Distribution factor\(k_d = \dfrac{\sin(m\beta/2)}{m\sin(\beta/2)}\)Typically 0.955–0.96
Pitch factor\(k_p = \cos(\alpha/2)\)\(\alpha\) = shortfall in electrical degrees
Turns per phase, conductor data\(T_{ph} = S Z_s/6\)Problems 1, 3, 5, 6
Turns per phase, coil data\(T_{ph} = S \times (\text{turns/coil})/3\)Double layer: coils = slots — Problem 2
Line voltage\(E_L = \sqrt3 E_{ph}\) star, \(E_L = E_{ph}\) deltaProblem 5 contrasts the two
Harmonic winding factors\(k_{pn} = \cos(n\alpha/2)\), \(k_{dn} = \dfrac{\sin(nm\beta/2)}{m\sin(n\beta/2)}\)Every angle scaled by \(n\) — Problem 3
Harmonic flux per pole\(\Phi_n = (1/n)(B_n/B_1)\Phi_1\)Harmonic pole pitch is \(1/n\)
Resultant phase e.m.f.\(E_{ph} = \sqrt{E_1^2+E_3^2+E_5^2+\cdots}\)Quadrature sum — Problem 3
Line e.m.f. with triplens\(E_L = \sqrt3\sqrt{E_1^2+E_5^2+E_7^2+\cdots}\)3rd, 9th, 15th cancel in the line
Pitfalls

Common Mistakes

  1. Halving turns that were never conductors. When the data gives turns per coil, the coil count already carries the factor of two; dividing by 2 again halves the answer — Problem 2.

  2. Forgetting that a double-layer winding has two coil sides per slot. Eight turns per coil means sixteen conductors per slot, not eight — Problem 2.

  3. Reporting the phase e.m.f. as the terminal voltage. The equation gives one phase; the star factor \(\sqrt3\) is a separate, deliberate step — Problems 1 and 5.

  4. Dividing a required line voltage by \(\sqrt3\) for a delta machine. In delta the phase winding sees the whole line voltage, so the flux needed is \(\sqrt3\) times greater — Problem 5.

  5. Computing \(k_d\) with the calculator in radian mode. \(\beta\) is in electrical degrees; a radian slip turns 0.96 into something absurd — Problems 1 and 6.

  6. Using the mechanical angle for \(\beta\). The slot angle is \(180^\circ P/S\) electrical, which is \(P/2\) times the mechanical slot pitch — Problem 6.

  7. Measuring the coil shortfall in slots and calling it degrees. Two slots short is \(2\beta = 24^\circ\), not 2° — Problem 6.

  8. Leaving the harmonic frequency at \(f\). The third harmonic is generated at \(3f\), and the tripled frequency largely offsets the reduced flux — Problem 3.

  9. Taking \(\Phi_3\) as 20 % of \(\Phi_1\). A 20 % harmonic density over a third of a pole pitch gives one third of that as harmonic flux — Problem 3.

  10. Adding harmonic e.m.f.s arithmetically, or leaving the triplen in the line voltage. Different frequencies add in quadrature, and triplens vanish between lines — Problem 3.

→ Looking Ahead

Every number on this page came from the same line, \(E_{ph} = 4.44\,k_w f \Phi T_{ph}\), and the effort went into the four quantities feeding it rather than into the equation itself. The frequency came from the pole count, the turns from a careful reading of the slot data, and the terminal voltage from the connection. The winding factor \(k_w\) was quoted here and used without much comment.

That is the piece still owing. \(k_p\) and \(k_d\) are not arbitrary correction factors: they are what a designer chooses when deciding how many slots to use and how far to chord the coils, and the choice is made to kill particular harmonics rather than to maximise the fundamental. The next set treats them as design variables in their own right.

Next: Set 34 — Winding Factors and Coil Span, where the coil pitch is chosen to eliminate a named harmonic and fractional-slot windings are reduced to an equivalent integral-slot one.