Armature Reaction and Compensation

Chapter 29 used both consequences repeatedly — the drooping internal characteristic came from the first, and Chapter 23's assumption that the commutating coil sits at zero EMF is invalidated by the second. This chapter supplies the mechanism and the arithmetic.

• The flux is distributed symmetrically with respect to the polar axis.

• The Magnetic Neutral Axis (MNA) is the axis along which no EMF is produced in the armature conductors, because they move parallel to the lines of flux.

• The MNA is \(\perp\) to the flux passing through the armature.

• The MNA is also called the axis of commutation, because the reversal of current in armature conductors takes place across this axis.

• This is the flux set up by \(I_a\) alone, the field coils being unexcited.

• The mmfs of the armature conductors combine to send flux downwards through the armature, shown by the vector \(OF_A\).

The direction is worth noting: with the brushes on the geometrical neutral axis, \(OF_A\) lies along the brush axis, which is perpendicular to the main polar axis. The armature mmf therefore acts on the quadrature axis — the q-axis of Chapter 22 — while the field mmf acts on the direct axis.

Under actual conditions both mmfs — main and armature — exist simultaneously.

• The flux through the armature is no longer uniform and symmetrical but distorted.

• The flux is crowded at the trailing pole tips but weakened or thinned out at the leading pole tips.

The pole tip which is first met during rotation by armature conductors is known as the leading pole tip and the other as the trailing pole tip.

• The new position of the MNA is always \(\perp\) to the resultant mmf \(OF\).

• With the shift of the MNA by \(\theta\), the brush also shifts, to lie in the new position of the MNA.

• Due to the brush shift — the forward lead — a redistribution of \(I_a\) occurs: conductors which were earlier under the influence of an \(N\)-pole come under the influence of an \(S\)-pole, and vice versa.

Let

This conversion is the commonest source of error in armature-reaction problems, and the factor can be large: in an eight-pole machine, \(\theta_m = \theta_e/4\).

The cross-magnetising conductors are all the rest — those lying between \(\angle AOD\) and \(BOC\), outside the demagnetising sectors.

The total armature ampere-conductors per pole, accounting for both cross-magnetising and demagnetising effects, is therefore \(ZI/P\), or \(ZI/2P\) ampere-turns per pole.

For neutralising the demagnetising effect, an extra number of turns may be put on each pole.

The two denominators differ because the extra turns are added to the winding that already exists. In a shunt machine that winding carries the small field current \(I_{sh}\), so many extra turns are needed; in a series machine it carries the full armature current, so very few are.

• It is embedded in pole-shoe slots, connected in series with the armature, such that the current in it flows in the opposite direction to that flowing in the armature conductors directly below the pole shoes.

• It is used for large DC machines subjected to large fluctuations in load — for example rolling-mill motors and turbo-generators.

• Its purpose is to neutralise the cross-magnetising effect of armature reaction.

• Armature reaction is the effect of the armature current's magnetic field on the distribution of flux under the main poles. It demagnetises — reducing the generated voltage — and cross-magnetises — distorting the field and causing sparking.

• With the main field alone, flux is symmetrical about the polar axis and the MNA — the axis of commutation, where no EMF is induced — coincides with the GNA.

• The armature mmf \(OF_A\) acts along the brush axis, on the quadrature axis.

• With both fields, \(\overrightarrow{OF} = \overrightarrow{OF_M} + \overrightarrow{OF_A}\), the flux is crowded at the trailing pole tip and thinned at the leading tip, and the net flux falls because the crowded tip saturates.

• The MNA shifts to lie \(\perp\) to \(OF\), and the brushes must follow it. This tilts \(OF_A\) by \(\theta\) and resolves it into \(OF_d\) and \(OF_c\).

• \(AT_d = ZI\theta_m/360\) and \(AT_c = ZI\left(1/2P - \theta_m/360\right)\), with \(I = I_a/A\). Their sum is always \(ZI/2P\).

• Electrical degrees must be converted: \(\theta_m = \theta_e/(P/2)\).

• Extra field turns \(AT_d/I_{sh}\) (shunt) or \(AT_d/I_a\) (series) neutralise the demagnetising effect, but not the distortion.

• Compensating windings in the pole shoes, in series with the armature and carrying current the opposite way, neutralise the cross-magnetising effect. They need \(Z_c = Z_a/A\) conductors, or \(0.7\times ZI/2P\) ampere-turns per pole.

• Using \(I_a\) where \(I\) is meant. The formulas need the current in one conductor, which is \(I_a/A\).

• Failing to convert electrical to mechanical degrees. The error is a factor of \(P/2\) — three times over in a six-pole machine.

• Halving only the first term when converting cross-magnetising amp-conductors to amp-turns. Both terms must be halved.

• Skipping the sum check. \(AT_d + AT_c\) must equal \(ZI/2P\); if it does not, one formula is wrong.

• Omitting the current from the compensating ampere-turns. \(0.7\times Z/2P\) is a turn count; the ampere-turns need \(\times I\).

• Confusing MNA with GNA. They coincide only on no load.

• Thinking distortion alone costs no flux. It does, because the crowded tip saturates and cannot gain what the weakened tip loses.

• Expecting extra field turns to stop sparking. They restore the flux but leave the cross-magnetising distortion untouched.

• Expecting a compensating winding to cover the whole armature. Only the conductors under the pole arc — about 70 % — can be compensated.

• Confusing leading and trailing pole tips. The leading tip is met first by the armature conductors, and it is the one that is weakened.

Always compute \(I = I_a/A\) first, convert the angle if necessary, and finish by checking \(AT_d + AT_c = ZI/2P\) — which confirms the formulas, though not the angle.

1. P30.1 A 4-pole lap generator has 500 conductors and an armature current of 80 A, with a brush lead of 8 mechanical degrees. Find \(AT_d\) and \(AT_c\).

   Show answer

   \[I = \frac{80}{4} = 20~\mathrm{A}, \qquad ZI = (500)(20) = 10\,000\]

   \[AT_d = (10\,000)\frac{8}{360} = 222.2~\mathrm{AT/pole}\]

   \[AT_c = (10\,000)\left(\frac{1}{8} - \frac{8}{360}\right) = (10\,000)(0.125 - 0.02222) = 1027.8~\mathrm{AT/pole}\]

   Check: \(222.2 + 1027.8 = 1250 = 10\,000/8\) \(\checkmark\)

2. P30.2 An 8-pole wave generator has 720 conductors and 200 A armature current, brush lead 5 mechanical degrees. Find both components.

   Show answer

   \[I = \frac{200}{2} = 100~\mathrm{A}, \qquad ZI = 72\,000\]

   \[AT_d = (72\,000)\frac{5}{360} = 1000~\mathrm{AT/pole}\]

   \[AT_c = (72\,000)\left(\frac{1}{16} - \frac{5}{360}\right) = (72\,000)(0.0625 - 0.013889) = 3500~\mathrm{AT/pole}\]

   Check: \(1000 + 3500 = 4500 = 72\,000/16\) \(\checkmark\)

3. P30.3 Repeat P30.2 with the brush lead given as 20 electrical degrees.

   Show answer

   \[\theta_m = \frac{20}{8/2} = \frac{20}{4} = 5^{\circ}\ \text{mechanical}\]

   which is exactly the case of P30.2, so \(AT_d = 1000\) and \(AT_c = 3500\) AT/pole.

   Had the 20° been used directly, \(AT_d\) would have been 4000 AT — four times too large — and \(AT_c\) only 500 AT.

   Note that the sum check would not have caught this. The identity \(AT_d + AT_c = ZI/2P\) holds for any value of \(\theta_m\) whatever, so it validates the formulas but not the input angle. What should raise suspicion is the physics: the pole pitch of an 8-pole machine is \(360/8 = 45\) mechanical degrees, so a "lead" of 20° would be nearly half a pole pitch — absurd for a brush shift, which is normally a few degrees at most.

4. P30.4 For the machine of P30.1, find the extra shunt field turns per pole if \(I_{sh} = 1.8\) A.

   Show answer

   \[\text{extra turns/pole} = \frac{222.2}{1.8} = 123.5 \quad\Longrightarrow\quad 124~\text{turns}\]

5. P30.5 A 4-pole lap machine has 400 conductors and 160 A armature current. Find the compensating ampere-turns and conductors per pole, taking pole arc/pole pitch = 0.7.

   Show answer

   \[I = \frac{160}{4} = 40~\mathrm{A}, \qquad \frac{ZI}{2P} = \frac{(400)(40)}{8} = 2000~\mathrm{AT/pole}\]

   \[AT_{\text{comp}} = (0.7)(2000) = 1400~\mathrm{AT/pole}\]

   \[\text{turns} = \frac{1400}{160} = 8.75 \quad\Longrightarrow\quad 17.5~\text{conductors/pole}, \ \text{say 18}\]

   Cross-check: \(Z_a = (400/4)(0.7) = 70\), so \(Z_c = 70/4 = 17.5\) \(\checkmark\)

6. P30.6 A machine loses 20 V to armature reaction. The local OCC slope is 0.02 V/AT. Find \(AT_d\).

   Show answer

   \[AT_d = \frac{\Delta E}{\mathrm{d}E/\mathrm{d}(AT)} = \frac{20}{0.02} = 1000~\mathrm{AT/pole}\]

7. P30.7 Explain why distorting the flux across a pole face reduces the total flux per pole.

   Show answer

   Because the machine is worked past the knee of its magnetisation curve. The leading tip loses mmf and, being on the relatively straight part of the curve, loses flux in full proportion. The trailing tip gains the same mmf but is pushed further into saturation, so it gains much less flux than the leading tip lost.

   The two do not cancel, and the net flux per pole falls. On a perfectly linear magnetic circuit there would be no such loss — distortion would be free, and armature reaction would produce sparking but no voltage droop.

8. P30.8 Why are compensating windings fitted to rolling-mill motors but not to small general-purpose machines?

   Show answer

   A rolling-mill motor is subject to large and very rapid load fluctuations — from no load to several times full load in a fraction of a second as the billet enters the rolls.

   Without compensation, the flux shifts forward and backward with each such change, and the resulting \(\mathrm{d}\Phi/\mathrm{d}t\) induces a statically induced EMF in the armature coils whose magnitude depends on the rapidity and the amount of the change. That voltage can flash over between commutator segments.

   A small machine on a steady load has slow, modest flux shifts, so the induced voltage is negligible and the expense of pole-face slots and heavy compensating conductors is not justified.

9. P30.9 A designer proposes to cure sparking by adding extra field turns. Comment.

   Show answer

   It will not work. Extra field turns neutralise the demagnetising component, restoring the flux per pole and hence the generated voltage.

   Sparking is caused by the cross-magnetising component, which distorts the field and shifts the magnetic neutral axis away from the brushes. Extra field turns leave that entirely untouched — and by restoring the flux they may even make the distortion slightly worse.

   The correct remedies are a compensating winding for the flux under the pole shoes, and interpoles for the interpolar region — the subject of Chapter 31.

10. P30.10 Two identical machines run at the same armature current, one with its iron heavily saturated and one only lightly. Which loses more voltage to armature reaction?

    Show answer

    The lightly saturated machine loses more voltage. The demagnetising ampere-turns are the same in both, since \(AT_d\) depends only on \(Z\), \(I\) and \(\theta_m\) — not on the iron.

    But the voltage lost is \(AT_d\) multiplied by the local slope of the magnetisation curve. Well into saturation that slope is small, so the same lost ampere-turns cost fewer volts.

    Working the iron hard therefore buys some immunity to armature reaction — though at the cost of higher excitation and a poorer response to field control.