EMF Equation of a DC Generator

Chapter 21 obtained this result while establishing the working principle. It is restated here because Chapters 28 to 39 use it constantly, and because the derivation repays a second reading now that the winding is understood.

Let

For a given machine \(Z\), \(P\) and \(A\) are fixed once the armature is wound. Only \(\Phi\) and \(N\) can change while it runs, so the equation collapses to

and in terms of angular velocity, substituting \(N = 60\omega/2\pi\):

Every line passes through the origin, because a stationary armature generates nothing whatever the flux. The lines are straight only while the flux is held constant; if the flux itself is produced by a field current, saturation bends the relationship between \(E_g\) and \(I_f\) — which is the magnetisation curve of Chapter 28.

The constant \(k_a = ZP/2\pi A\) does more than shorten the EMF equation. It also relates torque to armature current, and the reason is simple energy conservation.

The power converted from mechanical to electrical form is \(E_gI_a\), and it equals the mechanical power \(T\omega\):

In SI units this is exact: a machine with \(k_a\Phi = 2.0\) generates 2.0 V per rad/s and produces 2.0 N·m per ampere. Chapter 34 derives the torque equation independently and arrives at the same place, but the argument above shows it could not have come out otherwise.

Since \(A\) is the only term the winding decides, its effect follows at once:

The product of the two ratios is always unity, so the power is identical — the result established in Chapter 25. Note the first row: in a two-pole machine lap and wave are the same winding, since \(A = 2\) either way.

Chapter 21 insisted that the EMF inside the armature is alternating. Its frequency follows from the same reasoning as the EMF itself: a conductor completes one full cycle each time it passes a pair of poles.

This is not an academic quantity. The armature iron is magnetised alternately north and south at this frequency, so from Chapters 6 and 7:

and both rise as the machine is speeded up. This is why a DC machine's iron loss is speed-dependent while its copper loss is not — a distinction that becomes important in the efficiency testing of Chapter 33.

In practice the EMF equation is used backwards. The voltage, speed and flux are given by the specification; the designer must find a conductor count and then a physically buildable arrangement of slots and coils.

1. Find \(Z\) from \(Z = 60AE_g/\Phi PN\).
2. Choose the slot number \(S\) so that \(Z/S\) is an even integer — the conductors per slot must divide into two layers.
3. Deduce the turns per coil, \(T = Z/2S\) for a double-layer winding with one coil per slot.
4. Check the volts per segment, \(EP/C\) for lap with \(C = S\), against the 15 V limit of Chapter 22.
5. Check the winding is possible — parity of \(Y_B\) and \(Y_F\) for lap, the integer condition for wave (Chapter 25).

The EMF equation gives what the armature generates. What appears at the terminals is less, by the drops of Chapter 26.

The chain of reasoning for a complete problem therefore runs: flux and speed give \(E_g\); \(E_g\) and the drops give \(V\); \(V\) and the load give the currents — and in a shunt machine that last step feeds back into the second, which is why Chapter 26's Example 26.2 needed a simultaneous solution.

• The EMF equation is \(E_g = \Phi PNZ/60A\), from flux cut per revolution \(\Phi P\), time per revolution \(60/N\), and \(Z/A\) conductors in series per path.

• Compactly, \(E_g = k\Phi N\) with \(k = ZP/60A\), or \(E_g = k_a\Phi\omega\) with \(k_a = ZP/2\pi A\). Always, \(E_g \propto \Phi N\).

• The same constant gives the torque: \(T = k_a\Phi I_a\), which follows from \(T\omega = E_gI_a\).

• Because \(A\) is in the denominator, more paths mean less voltage, and \(E_{\text{wave}}/E_{\text{lap}} = P/2\).

• The armature frequency is \(f = PN/120\). Hysteresis loss varies as \(f\) and eddy loss as \(f^{2}\), so iron loss is speed-dependent.

• Design procedure: find \(Z\); choose \(S\) so \(Z/S\) is even; deduce \(T = Z/2S\); check \(EP/C \le 15\) V; check the winding parity or integer condition.

• The volts-per-segment check usually forces a higher slot count than the EMF equation alone would suggest.

• Terminal voltage follows by subtracting the armature, series-field and brush drops appropriate to the machine type.

• Putting \(A\) in the numerator. Parallel paths divide the conductors, so \(A\) belongs in the denominator.

• Confusing \(P\) with \(A\). They cancel for a lap winding, giving \(E = \Phi NZ/60\), but not for wave.

• Leaving the flux in milliweber. It must be in weber before substitution.

• Using \(\omega\) in the \(60\) form or \(N\) in the \(2\pi\) form. The constants differ by \(2\pi/60\).

• Treating the torque constant as a different number from the EMF constant. In SI units they are identical.

• Using \(f = PN/60\). One cycle takes a pair of poles, so the divisor is 120.

• Accepting the first slot count that gives the right \(Z\). The volts-per-segment limit must also be met.

• Assuming iron loss is constant. It varies with speed through \(f\); only copper loss is speed-independent.

• Forgetting the wave-winding integer condition after rounding \(Z\) up from a fractional design value.

• Stopping at \(E_g\). The terminal voltage is lower by the armature, series-field and brush drops.

List \(\Phi\) (in weber), \(P\), \(N\), \(Z\) and \(A\) before substituting — and state which winding type fixes \(A\).

1. P27.1 A 4-pole lap generator has 720 conductors, 25 mWb per pole, at 900 rev/min. Find \(E_g\).

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   \[E_g = \frac{(0.025)(4)(900)(720)}{(60)(4)} = \frac{64\,800}{240} = 270~\mathrm{V}\]

2. P27.2 Repeat P27.1 for a wave winding, and state the current ratio.

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   \[E_g = \frac{64\,800}{(60)(2)} = 540~\mathrm{V}\]

   Twice the lap value, since \(P/2 = 2\). The current capability is halved, so the power is unchanged.

3. P27.3 A 6-pole wave generator must give 400 V at 800 rev/min with 30 mWb per pole. Find \(Z\).

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   \[Z = \frac{60AE_g}{\Phi PN} = \frac{(60)(2)(400)}{(0.030)(6)(800)} = \frac{48\,000}{144} = 333.3\]

   So 334 conductors, rounded up — then checked against the wave integer condition: \((334 + 2)/6 = 56\), an integer \(\checkmark\)

4. P27.4 A generator gives 200 V at 800 rev/min. Find the EMF at 1000 rev/min with the flux reduced by 15 %.

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   Since \(E \propto \Phi N\):

   \[E = 200\left(\frac{1000}{800}\right)(0.85) = 200(1.25)(0.85) = 212.5~\mathrm{V}\]

5. P27.5 An 8-pole lap machine has \(Z = 800\). Find \(k\) and \(k_a\), and the EMF at 20 mWb and 600 rev/min.

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   \[k = \frac{ZP}{60A} = \frac{(800)(8)}{(60)(8)} = 13.33, \qquad k_a = \frac{ZP}{2\pi A} = \frac{800}{2\pi} = 127.3\]

   \[E_g = k\Phi N = (13.33)(0.020)(600) = 160~\mathrm{V}\]

   Check by the other form: \(\omega = 2\pi(600)/60 = 62.83\) rad/s, and \((127.3)(0.020)(62.83) = 160\) V \(\checkmark\)

6. P27.6 For the machine of P27.5 carrying 50 A, find the torque and the converted power.

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   \[T = k_a\Phi I_a = (127.3)(0.020)(50) = 127.3~\mathrm{N\,m}\]

   \[P = E_gI_a = (160)(50) = 8000~\mathrm{W}\]

   Check: \(T = P/\omega = 8000/62.83 = 127.3\) N·m \(\checkmark\)

7. P27.7 Find the armature frequency of a 4-pole machine at 1500 rev/min, and of a 12-pole machine at 500 rev/min.

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   \[f = \frac{(4)(1500)}{120} = 50~\mathrm{Hz}, \qquad f = \frac{(12)(500)}{120} = 50~\mathrm{Hz}\]

   Both 50 Hz — the pole count and speed trade off exactly, which is why slow machines need many poles.

8. P27.8 A machine's iron loss is 800 W at 1000 rev/min, of which two-thirds is hysteresis. Find the loss at 1400 rev/min at constant flux.

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   \[P_h = 533~\mathrm{W}, \qquad P_e = 267~\mathrm{W}, \qquad r = 1.4\]

   \[P_h' = (533)(1.4) = 747~\mathrm{W}, \qquad P_e' = (267)(1.96) = 523~\mathrm{W}\]

   \[P_{\text{iron}}' = 747 + 523 = 1270~\mathrm{W}\]

   A 40 % speed rise gives a 59 % loss increase.

9. P27.9 A 4-pole lap machine needs \(Z = 480\) and generates 300 V. Find the minimum slot number satisfying the 15 V per segment limit.

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   \[\frac{EP}{C} \le 15 \quad\Longrightarrow\quad C \ge \frac{(300)(4)}{15} = 80\]

   With \(C = S\), at least 80 slots. Then conductors per slot \(= 480/80 = 6\), giving 3-turn coils.

   A choice of, say, 60 slots would give 8 conductors per slot and 4-turn coils, but \(V_{\text{seg}} = 1200/60 = 20\) V — above the limit.

10. P27.10 Explain why a DC machine's iron loss depends on speed but its copper loss does not.

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    Iron loss arises from the alternating magnetisation of the armature core at frequency \(f = PN/120\). Since \(P_h \propto f\) and \(P_e \propto f^{2}\), both rise directly with speed.

    Copper loss is \(I^{2}R\) and depends only on the current the load draws and the resistance of the winding — neither of which the speed affects.

    This is exactly why the Swinburne test of Chapter 33 can measure the two separately: run the machine on no load at rated speed and the iron loss is present while the copper loss is negligible.