Set 38 — Synchronising and Parallel Operation
An alternator running alone sets its own voltage and its own frequency. Put it on a bus alongside others and it can set neither: the bus decides both, and the machine must be brought into agreement with them before the switch is closed. Those four conditions of agreement, and the penalty for each one violated, are where this set begins.
After that the questions are about stiffness and about sharing. A displaced rotor produces a resultant e.m.f. across its own reactance and a restoring power follows — the synchronising power, quoted per mechanical degree, which measures how firmly the machine is held in step. Load sharing is a separate matter and a simpler one: kilowatts are divided by the governor droop lines, kilovars by the excitation, and the two never interfere. Almost every problem in this set is solved by balancing \(P\) and \(Q\) separately and recombining them only at the last line.
Four conditions must hold before the switch is closed. The incoming machine's terminal voltage must equal the bus voltage in magnitude, in frequency, in phase and in phase sequence. Only the last is a wiring matter settled once; the other three are set by the operator every time.
Each condition, violated, produces its own symptom. Unequal magnitude drives a purely reactive circulating current; unequal phase drives a real power surge; unequal frequency makes the phase error sweep continuously, so the machine slips poles; reversed sequence is a short circuit through two phases and must never be attempted.
A magnitude error alone gives a circulating current and no power. With two machines of equal reactance and no load,
\[ I_{circ} = \frac{E_A - E_B}{Z_{sA}+Z_{sB}} \approx \frac{\Delta E}{2X_s} \]Because \(\Delta E\) is in phase with both e.m.f.s and the impedance is nearly pure reactance, this current is at right angles to both. It carries kVAR from the more strongly excited machine to the other and transfers no kilowatts at all.
A phase error produces a restoring power. If the rotor is displaced by \(\delta\) electrical radians the resultant e.m.f. across the synchronising path is
\[ E_r = 2E\sin\frac{\delta}{2} \approx E\delta, \qquad I_{sy} = \frac{E_r}{X_{\text{eff}}} \]Take \(X_{\text{eff}} = X_s\) against an infinite bus and \(X_{\text{eff}} = 2X_s\) between two identical machines.
Synchronising power and torque follow directly:
\[ P_{sy} = 3EI_{sy} \approx \frac{3E^2\delta}{X_{\text{eff}}}, \qquad T_{sy} = \frac{P_{sy}}{\omega_s}, \qquad \omega_s = \frac{2\pi N_s}{60} \]Mechanical degrees convert to electrical degrees through the pole count: \(\delta_e = (P/2)\,\delta_m\).
On load the synchronising coefficient is smaller. Differentiating the power-angle characteristic,
\[ P = \frac{3EV}{X_s}\sin\delta \;\Longrightarrow\; \frac{dP}{d\delta} = \frac{3EV}{X_s}\cos\delta \]It is largest at no load and falls to zero at \(\delta = 90^\circ\), which is precisely why that angle is the stability limit.
Real power is shared by the governors, not the excitation. Each machine follows a straight droop line from its no-load frequency \(f_0\) down to its full-load frequency, and the machines settle at the one common frequency at which their outputs add up to the load:
\[ P_k = \frac{f_{0k}-f}{\text{droop}_k}, \qquad \sum_k P_k = P_{\text{load}} \]Reactive power is shared by the excitation, not the governors. Raising a machine's field makes it take more kVAR and leaves its kilowatts untouched; the other machines shed exactly the kVAR it picks up. In every load-sharing problem, kW and kVAR are balanced separately and then recombined:
\[ \sum P_k = P_{\text{load}}, \qquad \sum Q_k = Q_{\text{load}}, \qquad S_k = \sqrt{P_k^2+Q_k^2} \]
Two identical 10 MVA, 11 kV, 50 Hz, 4-pole, star-connected alternators, each of synchronous reactance 12 Ω per phase with negligible resistance, are running in parallel on no load. Through an oversight one machine's excitation is left set for 11.5 kV while the other is correct at 11 kV.
- Find the circulating current and state what it does.
- State, for each of the four synchronising conditions in turn, what happens when it alone is violated.
Reduce to one phase. Both e.m.f.s are in phase — only the magnitudes differ — so the difference is a scalar:
The current it drives flows round the local loop formed by the two armatures, through both synchronous reactances in series:
Rated current is \(10\times10^6/(\sqrt3\times11000) = 524.9\) A, so this is 2.3 % of rating — a nuisance rather than a hazard, but it heats both machines for no return.
What the current does. With the resistance neglected the impedance is \(j24\ \Omega\), so \(I_{circ}\) lags \(\Delta E\) by exactly 90° — and \(\Delta E\) is in phase with both e.m.f.s. The power exchanged is therefore
The over-excited machine delivers \(3\times6639.5\times12.03 = 239.6\) kVAR lagging and the under-excited one absorbs it. No kilowatts move, no prime mover is loaded, and no governor notices.
The four conditions, taken one at a time.
| Condition | If violated alone | Consequence |
|---|---|---|
| Equal voltage magnitude | A steady \(\Delta E\) in phase with the e.m.f.s | Purely reactive circulating current, extra \(I^2R\) loss, no power transfer — 12.03 A here |
| Equal phase | A resultant \(2E\sin(\delta/2)\) in quadrature with the e.m.f.s | A real power surge that accelerates one rotor and retards the other until they align — Problem 4 |
| Equal frequency | The phase error sweeps through 360° continuously | The synchronising current beats between zero and \(2E/X\); the machine slips poles and may never lock in |
| Same phase sequence | Two of the three phases are permanently opposed | Effectively a line-to-line short circuit; violent current and mechanical shock. Never a matter of timing — it is checked once, at commissioning |
Why the first three are checked with lamps or a synchroscope, and the fourth is not. Magnitude, phase and frequency are all under continuous operator control — a rheostat and a throttle — and the classic three-lamp "two bright, one dark" method displays all three at once: the lamps go dark together only when the phase error passes through zero, and the rate at which they beat measures the frequency error. Phase sequence, by contrast, is fixed by the cabling; it is proved once with a sequence indicator and never rechecked.
One of the 10 MVA, 11 kV, 50 Hz, 4-pole alternators of Problem 1, with \(X_s = 12\ \Omega\) per phase, runs on no load in parallel with an infinite bus, its excitation adjusted so that its e.m.f. equals the bus voltage. Its rotor is then displaced by one mechanical degree from the position it would occupy in perfect step. Calculate
- the synchronising current;
- the synchronising power;
- the synchronising torque.
Convert the mechanical displacement to electrical degrees first. This is the step that is forgotten most often. A 4-pole machine turns one electrical revolution for every half mechanical revolution:
The resultant e.m.f. across the synchronising path. Two equal phasors \(\delta_e\) apart subtract to a phasor of magnitude
The small-angle form \(E\delta_e = 6350.9\times0.03491 = 221.7\) V agrees to four figures, as it will for any displacement worth calling small.
Only the machine's own reactance stands in the way, because the bus is infinite and contributes none:
This current is almost exactly in phase with \(E\), not in quadrature with it. \(E_r\) is perpendicular to the bisector of the two e.m.f.s, and dividing by \(jX_s\) turns it through a further 90°, leaving it along the e.m.f. So it carries real power:
The torque follows from the synchronous speed, which for a 4-pole 50 Hz machine is 1500 rev/min:
Read the sign physically. If the rotor is ahead, the machine delivers 352 kW to the bus, which it can only do by decelerating; if it is behind, it draws 352 kW from the bus and accelerates. Either way the power flows in the direction that removes the displacement. That is what makes the machine self-synchronising, and it is entirely due to \(E_r\) being in quadrature with the e.m.f.s.
The two identical machines of Problem 1 now run in parallel with each other alone — there is no infinite bus — on no load and correctly excited. One rotor is displaced by one mechanical degree relative to the other. Find the synchronising current, power and torque, and explain why they differ from Problem 2. Then find the synchronising power coefficient when the machine of Problem 2 is instead running on the infinite bus at a load angle of 30°.
The resultant e.m.f. is unchanged. One mechanical degree is still two electrical degrees, and the two e.m.f.s are still equal in magnitude:
What has changed is the path. The current must now flow through both armatures in series, because neither machine is an infinite bus and each presents its own reactance:
Hence exactly half the synchronising power and torque:
The advanced machine delivers this power and the retarded one receives it; both torques act to close the gap, so the pair is restored twice as gently as a single machine on a stiff bus.
An infinite bus is the limiting case of a very large machine. Writing the general result for two machines of reactances \(X_A\) and \(X_B\),
An infinite bus has zero reactance, infinite inertia and a fixed frequency. A real system is stiffer than one small machine but softer than an ideal bus, and the true answer for this pair lies between 176 kW and 352 kW per mechanical degree.
The loaded case. Everything above assumed no load, where \(\delta = 0\). On load the synchronising power is the slope of the power-angle characteristic at the working point:
At no load the same coefficient is \(10.083\times10^6\times1 = 176.0\) kW per electrical degree — the figure found above. Loading the machine to 30° has cost 13 % of its stiffness.
Where the stiffness goes to zero. The coefficient \(3EV\cos\delta/X_s\) vanishes at \(\delta = 90^\circ\). Beyond that angle a further advance of the rotor reduces the power delivered, so the machine cannot recover and falls out of step. That is the whole content of the steady-state stability limit, and Set 40 returns to it.
The 10 MVA, 11 kV, 4-pole machine of Problem 2 is connected to the infinite bus by mistake at the moment its e.m.f. is 30 electrical degrees behind the bus voltage, the magnitudes being correctly matched. Its synchronous reactance is 12 Ω per phase and its subtransient reactance \(X_d'' = 0.15\) per unit. Determine
- the resultant e.m.f. across the machine at the instant of closure;
- the initial surge current, and the same figure computed carelessly from \(X_s\);
- the power the bus immediately forces into the machine, and what it does mechanically.
The resultant e.m.f. Two equal phasors 30° apart:
Over half the phase voltage itself. The small-angle approximation is useless here — \(E\delta\) would give 3325 V, and at larger angles the error grows fast.
The current at the instant of closure is governed by the subtransient reactance, not the synchronous one. Flux cannot change instantaneously in the field and damper circuits, and while it is held constant the machine presents a much smaller reactance:
Compare the careless calculation. Using \(X_s = 12\ \Omega\) instead:
Under-stating the surge by a factor of 6.6 and making a dangerous event look harmless. \(X_s\) is a steady-state parameter; it applies only after several seconds, by which time the machine has either pulled into step or tripped.
The power exchanged. Once the initial d.c. and subtransient components have decayed, the quasi-steady power at 30° is given by the power-angle characteristic:
Half the machine's entire rating, appearing on the shaft in a fraction of a second, and in the direction that accelerates the lagging rotor forward.
What happens next. The rotor accelerates toward alignment, overshoots because it arrives with kinetic energy, swings past zero, is decelerated by the reversed synchronising power, and oscillates about the equilibrium angle until the damper winding absorbs the swing. The electrical transient lasts a cycle or two; the mechanical swing lasts of the order of a second. The damage, when there is damage, is mechanical — a torsional shock through the coupling and the shaft, and forces on the end windings proportional to \(I''^2\), here twelve times the rated value.
A note on mixing models. Parts (b) and (c) use two different reactances for two different instants, and that is deliberate rather than sloppy: \(X_d''\) describes the machine during the first cycle, while flux linkages are still trapped, and \(X_s\) describes it once they have redistributed. A single reactance cannot do both jobs, and choosing the wrong one is the commonest error in fault and synchronising calculations alike.
Two alternators run in parallel supplying a common load of 900 kW. Their governors have straight-line droop characteristics:
| Machine | Rating | No-load frequency | Frequency at full load |
|---|---|---|---|
| A | 600 kW | 50.5 Hz | 48.5 Hz |
| B | 400 kW | 50.0 Hz | 48.0 Hz |
Determine
- the load carried by each machine and the common frequency;
- the largest total load the pair can take without overloading either machine;
- the change of governor setting that would make them share in proportion to their ratings.
Write each droop line as power against frequency. Both machines fall 2 Hz from no load to full load, so
Both machines run at one and the same frequency, because they are electrically locked together. That single shared variable is what makes the problem solvable.
Impose the load. The two outputs must add to 900 kW:
Back-substitute:
Machine A is at 100 % of its rating while machine B is at only 75 % of its own. The higher no-load setting has made A take more than its share.
The limit on total load is therefore reached exactly here. Any further load lowers \(f\) below 48.5 Hz and pushes A past 600 kW while B still has 100 kW of spare capacity:
Ten per cent of the installed capacity is unusable purely because of a half-hertz difference in governor setting.
Sharing in proportion to rating means 540 kW and 360 kW. From A's own droop line, the frequency at which it delivers 540 kW is
B's no-load setting must be raised from 50.0 Hz to 50.5 Hz — that is, until the two no-load frequencies are equal.
The general rule that this exposes. Both machines already have the same per-unit droop: 2 Hz over full load, which is 4 % of 50 Hz for each. With equal per-unit droop, equal no-load frequency is exactly the condition for sharing in proportion to rating, whatever the total load:
Two alternators in parallel on a 6.6 kV bus supply a total load of 1000 kW at 0.8 power factor lagging. Machine A's governor is set so that it delivers 600 kW, and its excitation is set so that it works at 0.9 power factor lagging.
- Find the output, current and power factor of machine B.
- Machine A's excitation is now reduced until A works at unity power factor, its governor untouched. Find the new condition of machine B.
Balance kilowatts and kilovars separately. This is the whole technique; the phasor diagram is never needed. The load first:
Machine A, from its two settings:
Machine B takes whatever is left, in both currencies:
The currents, at the common 6.6 kV bus:
The load current is \(1250\times10^3/(\sqrt3\times6600) = 109.3\) A, and \(58.3+53.3 = 111.6\) A. The difference is not an error: the two machine currents are at different angles and must be added as phasors, which returns 109.3 A exactly.
Now weaken A's field. The governor is untouched, so \(P_A\) stays at 600 kW and \(P_B\) at 400 kW — excitation cannot move a kilowatt. What changes is the reactive split: at unity power factor \(Q_A = 0\), so B must carry the lot:
Tabulating the change makes the rule unmistakable:
| Quantity | Before | After A is weakened | Change |
|---|---|---|---|
| \(P_A\) | 600 kW | 600 kW | none |
| \(P_B\) | 400 kW | 400 kW | none |
| \(Q_A\) | 290.6 kVAR | 0 | −290.6 |
| \(Q_B\) | 459.4 kVAR | 750 kVAR | +290.6 |
| \(S_B\) | 609.1 kVA | 850.0 kVA | +40 % |
| \(\cos\phi_B\) | 0.657 lag | 0.471 lag | much worse |
Every kilovar A stops supplying appears in B, one for one. B's kilowatts have not moved, but its kVA has risen 40 % and its armature current with it — which is why a machine can be overloaded by someone else's excitation setting.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Circulating current | \(I_{circ} = \Delta E/(X_{sA}+X_{sB})\) | Purely reactive, no power — Problem 1 |
| Mechanical to electrical angle | \(\delta_e = (P/2)\,\delta_m\) | The most-forgotten step — Problem 2 |
| Resultant e.m.f. | \(E_r = 2E\sin(\delta_e/2)\) | Exact; \(\approx E\delta_e\) only for small \(\delta\) |
| Synchronising current | \(I_{sy} = E_r/X_{\text{eff}}\) | \(X_{\text{eff}} = X_s\) (bus) or \(2X_s\) (two machines) |
| Synchronising power | \(P_{sy} = 3EI_{sy} \approx 3E^2\delta_e/X_{\text{eff}}\) | Per mechanical degree — Problems 2, 3 |
| Synchronising torque | \(T_{sy} = P_{sy}/\omega_s\), \(\omega_s = 2\pi N_s/60\) | \(N_s = 120f/P\) |
| Power-angle relation | \(P = (3EV/X_s)\sin\delta\) | Cylindrical rotor, \(R_a\) neglected |
| Synchronising coefficient | \(dP/d\delta = (3EV/X_s)\cos\delta\) | Falls to zero at 90° — Problem 3 |
| Subtransient surge | \(I'' = E_r/X_d''\) | First cycle only, not \(X_s\) — Problem 4 |
| Base impedance | \(Z_{base} = V_L^2/S\) | Converts per-unit reactances to ohms |
| Droop line | \(P_k = (f_{0k}-f)/\text{droop}_k\) | Straight line, one common \(f\) — Problem 5 |
| Load balance | \(\sum P_k = P_{\text{load}}\), \(\sum Q_k = Q_{\text{load}}\) | Solve separately, recombine last — Problem 6 |
| Reactive power | \(Q = P\tan\phi\), \(S = \sqrt{P^2+Q^2}\) | Sign positive for lagging |
| Machine current | \(I = S/(\sqrt3 V_L)\) | Machine currents add as phasors, not scalars |
Common Mistakes
Using the mechanical angle in an electrical formula. One mechanical degree on a 4-pole machine is two electrical degrees, and on a 12-pole machine six — Problem 2.
Using one \(X_s\) when two machines are in parallel. Neither is an infinite bus; the current sees both reactances in series and the synchronising power halves — Problem 3.
Writing \(E_r = E\delta\) at a large angle. The exact form \(2E\sin(\delta/2)\) is no harder and is right everywhere — Problem 4.
Computing a switching surge from \(X_s\). The synchronous reactance applies seconds later; the first-cycle current is set by \(X_d''\) and is several times larger — Problem 4.
Believing a circulating current transfers power. With a magnitude error alone it is at 90° to both e.m.f.s and moves only kVAR — Problem 1.
Expecting excitation to change load sharing. It changes the kVAR split and the bus voltage; the kilowatts are the governors' business alone — Problems 5 and 6.
Adding two machine currents arithmetically. 58.3 A and 53.3 A make 109.3 A, not 111.6 A, because they differ in angle — Problem 6.
Solving a droop problem for power before frequency. The frequency is the shared unknown; find it first, then read each machine's output from its own line — Problem 5.
Assuming equal droop alone gives proportional sharing. The no-load settings must agree too, or one machine reaches full load while the other still has capacity — Problem 5.
Treating phase sequence as something to check at each synchronising. It is a wiring property, verified once; the three-lamp method checks magnitude, phase and frequency, not sequence — Problem 1.
Once a machine is on the bus, its terminal voltage and its frequency stop being its own. What is left under the operator's hand is the load angle, set by the throttle, and the excitation, which decides only how much reactive power the machine carries. Every problem on this page has been one of those two knobs turned, and the arithmetic has been the same each time: a resultant e.m.f. divided by a reactance, or a pair of power balances solved separately.
Now reverse the power flow. Drive the same machine from the bus instead of into it, and the rotor falls behind the field rather than leading it. Nothing in the phasor equation changes except a sign, but the machine becomes something rather unusual — a constant-speed motor whose power factor is set by a rheostat, and which can be made to draw leading current at will.
Next: Set 39 — Synchronous Motor Performance and V-Curves, where the excitation is swept at constant load and the armature current traces the curve that gives the machine its second career as a power-factor corrector.