Set 35 — Armature Reaction and Synchronous Impedance
Open-circuited, an alternator produces the e.m.f. that Set 33 computed. Load it, and the armature current builds a rotating magnetic field of its own that turns in step with the poles and adds to theirs. The flux the conductors actually cut is no longer the flux the field winding established, and the terminal voltage parts company with the generated e.m.f. That is armature reaction, and this set measures it.
The size of the effect depends only on the armature current; its direction depends only on the power factor. Rather than carry a variable flux through every calculation, engineering practice converts the whole business into a fictitious reactance and adds it to the real leakage reactance to give the synchronous reactance \(X_s\). The problems here obtain that reactance from open- and short-circuit test data by interpolation, distinguish its saturated and unsaturated values, express it as a short-circuit ratio, and finally take it apart again with a zero-power-factor test to see how little of it is genuine leakage.
Armature reaction is the stator's own rotating field. Load the machine and the three phase currents produce a m.m.f. wave that turns at exactly synchronous speed, so it is stationary with respect to the poles. Its peak value per pole depends on the armature current alone:
\[ \mathcal{F}_a = \frac{3}{2}\cdot\frac{4}{\pi}\cdot\frac{k_w T_{ph}\sqrt2\,I_a}{P} = 2.7\,\frac{k_w T_{ph} I_a}{P}\ \text{A-turns/pole} \]The \(3/2\) is the three-phase resultant, the \(4/\pi\) the fundamental of a rectangular m.m.f. step, and the \(\sqrt2\) the peak of the current. Together they give 2.7.
The load power factor decides the direction, never the size. With \(\phi\) the angle of \(\mathbf{I}_a\) behind the terminal voltage, the armature m.m.f. resolves about the pole axis as
\[ \mathcal{F}_{\text{cross}} = \mathcal{F}_a\cos\phi, \qquad \mathcal{F}_{\text{direct}} = \mathcal{F}_a\sin\phi \]Direct component demagnetising for a lagging load, magnetising for a leading one, and zero at unity power factor where the effect is purely cross-magnetising.
Its effect is lumped into a reactance. The e.m.f. the armature reaction destroys behaves exactly like a voltage drop proportional to \(I_a\) and in quadrature with it, so it is written as a fictitious reactance and added to the true leakage reactance:
\[ X_s = X_L + X_a, \qquad Z_s = R_a + jX_s \]Two tests measure \(Z_s\) between them. The open-circuit characteristic (OCC) gives the e.m.f. against field current with the terminals open; the short-circuit characteristic (SCC) gives the armature current with the terminals shorted. Read both at the same field current:
\[ Z_s = \frac{E_{oc}\ \text{per phase}}{I_{sc}}, \qquad X_s = \sqrt{Z_s^2 - R_a^2} \]The answer depends on where you read it. Taken on the straight part of the OCC — the air-gap line — the result is the unsaturated synchronous reactance; taken at the field current that produces rated open-circuit voltage it is the saturated value, and it is smaller because the OCC has bent over while the SCC has not.
The short-circuit ratio is the same information in one number:
\[ \text{SCR} = \frac{I_f\ \text{for rated open-circuit voltage}}{I_f\ \text{for rated short-circuit current}} = \frac{1}{X_s\ \text{(saturated, per unit)}} \]The identity is exact whenever the SCC is straight, which it always is. A large SCR means a large air gap, a heavy field winding, small regulation and good stability.
The Potier (zero-power-factor) construction splits \(X_s\) back into its parts. Two points on the zero-power-factor characteristic at rated current — the short-circuit point and the full-voltage point — fix a right-angled triangle whose vertical side is the leakage drop \(I_aX_L\) and whose horizontal side is the field current that produces it on the air-gap line.
Per unit keeps the reading honest. On the machine's own base,
\[ Z_{base} = \frac{V_{ph,\text{rated}}}{I_{a,\text{rated}}} \]A synchronous reactance is typically 0.8–1.5 per unit and a leakage reactance 0.1–0.2, so the great majority of \(X_s\) is armature reaction and not leakage at all.
A 1000 kVA, 3300 V, 50 Hz, 6-pole, star-connected alternator has 72 stator slots carrying a double-layer winding of 8 conductors per slot. Each coil spans 10 slots. Determine
- the winding factor;
- the series turns per phase;
- the peak armature-reaction m.m.f. per pole at rated armature current.
Rated armature current first, because the armature-reaction m.m.f. is proportional to it and to nothing else:
The winding factors, by the method of Set 34. With 72 slots and 6 poles there are 12 slots per pole, so
The coil is short by two slots, a pole pitch being 12 slots, so the chording angle is \(\alpha = 2\beta = 30^\circ\):
Turns per phase. Two conductors make one turn, and the 72 slots are shared by three phases:
The m.m.f. of one phase at the instant its current is at its peak \(\sqrt2 I_a\). A distributed winding of \(k_wT_{ph}\) effective turns per phase spread over \(P\) poles produces a rectangular m.m.f. staircase whose fundamental has amplitude
Three phases give one and a half times that, because the resultant of three pulsating waves displaced by 120° in space and in time is a travelling wave of amplitude \(1.5\mathcal{F}_1\):
The constant 2.7 is \(1.5\times(4/\pi)\times\sqrt2 = 2.7009\). It is worth remembering as a number rather than re-deriving it each time.
The alternator of Problem 1 has a field winding of 280 turns per pole. Working at rated armature current, resolve the armature-reaction m.m.f. into its cross-magnetising and direct-axis components for power factors of unity, 0.8 lagging, zero lagging, 0.8 leading and zero leading. Express each direct-axis component as the equivalent change in field current, and state what the field regulator must do in each case to hold the terminal voltage.
Fix the geometry before touching any numbers. The armature m.m.f. wave is locked to the armature current: its axis lies \(90^\circ\) electrical behind the axis of the generated e.m.f. when the current is in phase with that e.m.f. The pole axis (the field m.m.f. axis) is \(90^\circ\) ahead of the e.m.f. axis. So at unity power factor the two m.m.f.s are in space quadrature, and as the current falls back by \(\phi\) the armature wave rotates \(\phi\) further into direct opposition with the poles.
Hence the resolution, with \(\mathcal{F}_a = 6993\) A-t/pole fixed by Problem 1:
The field-current equivalent of a direct-axis m.m.f. follows from the field turns per pole:
Tabulating all five conditions at the same 175 A:
| Power factor | \(\phi\) | Cross-magnetising (A-t/pole) | Direct-axis (A-t/pole) | Effect on poles | \(\Delta I_f\) needed |
|---|---|---|---|---|---|
| 1.0 | 0° | 6993 | 0 | Purely distorting | 0 |
| 0.8 lagging | 36.87° | 5595 | 4196 | Demagnetising | +14.99 A |
| 0 lagging | 90° | 0 | 6993 | Wholly demagnetising | +24.98 A |
| 0.8 leading | −36.87° | 5595 | 4196 | Magnetising | −14.99 A |
| 0 leading | −90° | 0 | 6993 | Wholly magnetising | −24.98 A |
The column of cross-magnetising values is \(6993\cos\phi\) and the direct column \(6993\sin\phi\); the two always combine to the same 6993.
Reading the table. At unity power factor the armature merely leans the flux wave sideways: one pole tip is strengthened and the other weakened, the net flux per pole barely changes, and the regulator has almost nothing to do. At 0.8 lagging the field current must be raised by nearly 15 A — well over half the 28 A that a short circuit demands — simply to stand still. At zero leading power factor the armature does the field winding's job for it, and the excitation must be reduced by 25 A or the terminal voltage will rise above rating with no load being usefully served.
Why cross-magnetisation is not harmless. Although it does not change the average flux per pole to first order, it shifts the flux axis and drives one pole tip further into saturation. Saturation removes more flux from the strengthened tip than the weakened tip gains, so a small net demagnetisation survives even at unity power factor. It is second-order and is neglected in every calculation on this page.
Open-circuit and short-circuit tests on the 1000 kVA, 3300 V, star-connected alternator of Problem 1 gave the readings below. The open-circuit e.m.f. is a line value; the short-circuit reading is the armature current per phase. The d.c. armature resistance corrected to operating temperature is 0.2 Ω per phase.
| Field current \(I_f\) (A) | 0 | 8 | 16 | 24 | 32 | 40 | 48 | 56 |
|---|---|---|---|---|---|---|---|---|
| O.C. line e.m.f. (V) | 0 | 920 | 1840 | 2600 | 3150 | 3550 | 3800 | 3960 |
| S.C. armature current (A) | 0 | 50 | 100 | 150 | 200 | 250 | 300 | 350 |
Determine
- the unsaturated synchronous impedance and reactance;
- the saturated values, taken at rated open-circuit voltage;
- both reactances in per unit on the machine's own rating.
Rated quantities and the base impedance, which every later step is measured against:
Locate the air-gap line. The first two open-circuit readings both give \(920/8 = 1840/16 = 115\) V per field ampere, so the OCC is straight up to at least 16 A and the air-gap line is
At 24 A the measured 2600 V is already below the 2760 V the line predicts, so saturation has begun by then.
Unsaturated synchronous impedance, read at \(I_f = 16\) A where the machine is still linear. The short-circuit test is always linear, because the flux on short circuit is tiny:
Subtracting \(R_a\) changes the reactance in the fourth significant figure. For any machine of this size \(X_s \approx Z_s\).
The field current for rated open-circuit voltage, by linear interpolation between the 32 A and 40 A readings:
The short-circuit current at that same field current, interpolated on the SCC:
Equivalently \(I_{sc} = 6.25\,I_f\), the SCC being a straight line through the origin.
Saturated synchronous impedance:
In per unit, and side by side:
| Read at | \(I_f\) | \(E_{oc}\) per phase | \(I_{sc}\) | \(Z_s\) | \(X_s\) (p.u.) |
|---|---|---|---|---|---|
| Air-gap line | 16 A | 1062.3 V | 100 A | 10.62 Ω | 0.976 |
| Rated O.C. voltage | 35 A | 1905.3 V | 218.75 A | 8.71 Ω | 0.800 |
The unsaturated value is 22 % larger. Both come from the same two curves; only the reading point differs.
Why the saturated value is the smaller one. Between 16 A and 35 A the OCC has bent over — the numerator of \(Z_s\) has grown by less than the field current — while the SCC has continued in a straight line and the denominator has grown in proportion. The ratio must therefore fall. A machine driven harder into saturation shows a smaller \(X_s\) and behaves better on load than the unsaturated figure suggests, which is exactly why the e.m.f. method of Set 36 over-estimates regulation.
For the machine and test data of Problem 3, determine
- the short-circuit ratio;
- the relation between the short-circuit ratio and the per-unit saturated synchronous reactance, verified numerically;
- the sustained current the machine would deliver into a three-phase terminal short circuit if the excitation were left at the value that gives rated voltage on open circuit.
The two field currents the definition needs are both already in hand. Rated open-circuit voltage of 3300 V needs \(I_f = 35.0\) A (Problem 3), and rated short-circuit current of 175 A needs, by interpolation on the SCC,
Hence the short-circuit ratio:
Why it is the reciprocal of the per-unit reactance. Let the SCC have slope \(k\) amperes of armature current per ampere of field current. Then the short-circuit current at the excitation \(I_{f,oc}\) that gives rated voltage is \(kI_{f,oc}\), and
The slope \(k\) cancels, so the identity holds for any machine whose SCC is straight — which is all of them.
The sustained short-circuit current at rated excitation. Leave the field at 35 A and short the terminals; the SCC gives the answer directly:
So the short-circuit ratio is literally what its name says: the per-unit current a machine sustains on a short circuit at rated excitation.
What the number is worth in design terms. A machine with a large SCR has a long air gap, needs more field ampere-turns and therefore more field copper and a larger frame — it is expensive. In return the armature reaction is a smaller fraction of the field m.m.f., the regulation is smaller, the machine is stiffer against pull-out, and the charging current of a long line it may have to energise is easier to handle. Turbo-alternators are built with SCR between 0.5 and 0.8, hydro machines between 1.0 and 1.5. At 1.25 this machine sits at the generous end.
A zero-power-factor lagging test on the machine of Problem 3 holds the armature current at its rated 175 A and the terminal voltage at rated 3300 V with a field current of 60 A. Using the open-circuit and short-circuit data already tabulated, determine
- the leakage (Potier) reactance per phase;
- the armature-reaction m.m.f. at rated current, in field amperes and in ampere-turns per pole;
- the split of the saturated synchronous reactance into its leakage and armature-reaction parts.
Two points on the zero-power-factor characteristic are enough. At rated current the ZPF curve passes through the short-circuit point \((I_f, V) = (28.0,\ 0)\) — a short circuit is a zero-power-factor load — and through the given full-voltage point \((60.0,\ 3300)\). The whole construction rests on the fact that the ZPF curve is the OCC shifted bodily to the right and downwards by a fixed triangle.
Set the triangle up. From the full-voltage point \(B(60,\,3300)\) step back horizontally by the short-circuit field current:
Point \(A\) is where the machine would sit if there were no leakage drop and no armature reaction — the two effects the triangle is about to separate.
Draw a line from \(A\) parallel to the air-gap line and find where it cuts the OCC. The air-gap line has slope 115 V per field ampere, so the line through \(A\) is \(E = 3300 + 115(I_f - 32)\). Between 24 A and 32 A the OCC is the chord
The vertical side of the triangle is the leakage drop. Dropping from \(C\) to the horizontal through \(A\):
The horizontal side is the field current that produces that drop as an air-gap e.m.f.:
Check: \(32.00 - 28.76 = 3.24\) A, as the geometry requires.
Everything left over is armature reaction. Of the 28 A the field must supply to circulate rated current on short circuit, only 3.24 A goes into producing the resultant air-gap flux; the balance is spent cancelling the armature's own m.m.f.:
Problem 1 obtained 6993 A-t/pole from the winding data alone. The two routes — one a measurement, one a design calculation — agree to within 0.9 %.
Splitting the synchronous reactance:
| Component | Value (Ω/phase) | Per unit | Share of \(X_s\) | Physical origin |
|---|---|---|---|---|
| Leakage \(X_L\) | 1.23 | 0.113 | 14 % | Slot, end-winding and tooth-tip flux that never crosses the gap |
| Armature reaction \(X_a\) | 7.48 | 0.687 | 86 % | Main-gap flux produced by the stator, opposing the field |
| Synchronous \(X_s\) | 8.71 | 0.800 | 100 % | The sum, treated as one reactance for convenience |
Take the armature-reaction m.m.f. of Problem 1 and the Potier leakage reactance of Problem 5 as given, and use them to predict the field current needed to circulate rated armature current on a sustained three-phase short circuit. Compare the prediction with the measured short-circuit characteristic, and say what the comparison tells you about the shape of that characteristic.
What the field must do on short circuit. With the terminals shorted, \(V = 0\), so the resultant air-gap e.m.f. has nothing to support except the machine's own internal impedance drop:
Barely a ninth of rated voltage. The flux in a short-circuited machine is tiny, which is exactly why the SCC never saturates.
The field current that produces it. At this flux level the machine is deep in its linear region, so the air-gap line applies:
The field current that cancels armature reaction. On short circuit the armature current lags the air-gap e.m.f. by \(\tan^{-1}(1.23/0.2) = 80.8^\circ\) — near enough 90° that the armature m.m.f. can be taken as wholly demagnetising:
Add them, because on the direct axis the two requirements are collinear:
| Contribution | Field current | Share |
|---|---|---|
| Balancing armature reaction | 24.98 A | 88 % |
| Producing the resultant air-gap flux | 3.29 A | 12 % |
| Predicted total | 28.26 A | — |
| Measured from the SCC | 28.00 A | Prediction high by 0.9 % |
What the agreement means. Nearly nine-tenths of the excitation on short circuit does nothing except cancel the stator's own m.m.f. The short-circuit characteristic is therefore a direct measurement of armature reaction, and its straightness is not a coincidence: with the working flux held at a ninth of rated value by the leakage drop alone, the iron never saturates however far the field current is pushed. The OCC bends and the SCC does not, and that single asymmetry is the source of everything on this page — the difference between saturated and unsaturated \(X_s\), the meaning of the short-circuit ratio, and the pessimism of the e.m.f. method.
A caution about the arithmetic. Adding the two field currents as scalars assumes the armature m.m.f. lies exactly opposite the resultant m.m.f. It does not: the armature current lags the air-gap e.m.f. by 80.8° while the resultant m.m.f. leads it by 90°, so the two m.m.f.s are 170.8° apart rather than 180°. The honest vector sum is
The scalar shortcut gave 28.26 A — high by 0.1 %, far inside the uncertainty of either input. On short circuit the two m.m.f.s really are as good as collinear.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Rated armature current | \(I_a = S/(\sqrt3 V_L)\) | Power factor plays no part — Problem 1 |
| Turns per phase | \(T_{ph} = SZ_s/6\) | Slots × conductors per slot, halved and thirded |
| Winding factor | \(k_w = k_pk_d\) | Set 34 — Problem 1 |
| Armature m.m.f., one phase | \(\mathcal{F}_1 = 1.8\,k_wT_{ph}I_a/P\) | Peak of the fundamental — Problem 1 |
| Armature m.m.f., three phase | \(\mathcal{F}_a = 2.7\,k_wT_{ph}I_a/P\) | 1.5 times the single-phase peak |
| Cross-magnetising part | \(\mathcal{F}_a\cos\phi\) | Distorts, does not reduce — Problem 2 |
| Direct-axis part | \(\mathcal{F}_a\sin\phi\) | Demagnetising lagging, magnetising leading |
| Field-current equivalent | \(\Delta I_f = \mathcal{F}/N_f\) | \(N_f\) = field turns per pole — Problems 2, 5, 6 |
| Air-gap line | \(E_{oc} = (\text{slope})\times I_f\) | Fitted to the straight part of the OCC |
| Synchronous impedance | \(Z_s = E_{oc(ph)}/I_{sc}\) at equal \(I_f\) | Both readings at one excitation — Problem 3 |
| Synchronous reactance | \(X_s = \sqrt{Z_s^2-R_a^2}\) | Practically equal to \(Z_s\) |
| Base impedance | \(Z_{base} = V_{ph}/I_{a,\text{rated}}\) | Machine's own rating — Problems 3, 4 |
| Short-circuit ratio | \(\text{SCR} = I_{f(oc)}/I_{f(sc)} = 1/X_{s\,\text{p.u.}}\) | Saturated value — Problem 4 |
| Reactance split | \(X_s = X_L + X_a\) | Leakage plus armature reaction — Problem 5 |
| Potier leakage drop | Vertical side of the ZPF triangle | \(X_L = (I_aX_L)/I_a\) per phase — Problem 5 |
| Short-circuit m.m.f. balance | \(E_r = I_a\sqrt{R_a^2+X_L^2}\) | All the rest of \(I_f\) fights armature reaction — Problem 6 |
Common Mistakes
Using \(T_{ph}I_a/P\) without the 2.7. The factor collects the three-phase resultant, the fundamental of a stepped wave and the peak of the current; dropping any one of them puts the m.m.f. out by tens of per cent — Problem 1.
Believing the armature reaction is smaller at unity power factor. Its magnitude is \(2.7k_wT_{ph}I_a/P\) at every power factor; only its direction changes — Problem 2.
Reading the OCC and the SCC at different field currents. The ratio \(E_{oc}/I_{sc}\) is a synchronous impedance only when both are taken at one excitation — Problem 3.
Forgetting that the open-circuit reading is a line voltage. Divide by \(\sqrt3\) before dividing by the short-circuit current, or \(Z_s\) comes out 73 % too large — Problems 3 and 5.
Taking the unsaturated reactance from a saturated point on the OCC. The unsaturated value must be read on the air-gap line, not on the measured curve — Problem 3.
Quoting the short-circuit ratio as \(X_s\) in per unit. It is the reciprocal: SCR 1.25 means \(X_s = 0.80\) p.u., not 1.25 — Problem 4.
Interpolating on the OCC as though it were straight over a wide span. Interpolate only between adjacent tabulated points, and never extrapolate a saturated curve — Problems 3 and 5.
Treating the whole of \(X_s\) as leakage. Here leakage is 14 % of it; the other 86 % is a fiction standing in for armature reaction — Problem 5.
Stepping back from the wrong point in the Potier construction. The horizontal step is the short-circuit field current, taken leftwards from the full-voltage ZPF point — Problem 5.
Assuming the short-circuit flux is large because the current is. It is about a ninth of rated, held down by the leakage drop alone, which is why the SCC is a straight line — Problem 6.
Armature reaction has now been counted three separate ways — as ampere-turns from the winding data, as a field current from the zero-power-factor test, and as ohms of reactance from the open- and short-circuit characteristics — and the three agree to within one per cent. That agreement is what licenses the single symbol \(X_s\). From here on the stator's own field never has to be thought about again; it travels inside one reactance, added on a phasor diagram like any other.
What remains is to spend it. The gap between the generated e.m.f. and the terminal voltage is now a phasor sum with two terms, and its size at rated load — the voltage regulation — is the number a customer actually cares about. Whether \(X_s\) is read saturated or unsaturated turns out to matter a great deal there, and the choice is what distinguishes the three standard methods from one another.
Next: Set 36 — Voltage Regulation — EMF, MMF and ZPF Methods, where the same open- and short-circuit data are used to predict how far the terminal voltage will fall when the load is applied, and why the pessimistic and optimistic answers bracket the truth.