Sync Machine Short Circuit

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A bolted three-phase fault on the terminals of an unloaded machine — the test that puts the numbers on a generator's datasheet.

A 100 MVA, 13.8 kV round-rotor machine running open-circuited at rated voltage. At $t=0.5$ s a bolted three-phase fault closes across all three terminals and stays closed. Field voltage is fixed at 1.0 pu, shaft torque is zero, and the rotor is held at synchronous speed for the whole run.

Before the fault the machine is doing nothing interesting, and that is the point: efd0 = 1.0 pu puts the terminals at 11,267 V peak line-to-neutral, which is 0.9999 of rated, and the stator current is zero to eight decimal places. This is the open-circuit condition the standard test starts from. The rotor is held at synchronous speed for the whole run (rotor_lock with unlock_time past the end of the window) so the analysis below can treat $\omega$ as constant, exactly as the textbook derivation does. With tm0 = 0 a free rotor would coast down during the fault and drag the whole waveform off frequency.

The first few cycles

Stator currents through the first 110 ms; the dashed lines mark the rated peak of $\pm 5.92$ kA. Each phase carries the same AC component but its own DC offset, fixed by the requirement that current be continuous at the instant of the fault. Phase $a$ happens to catch nearly the full offset and reaches 54.5 kA, 9.2 times rated.

The rated peak current of this machine is $\sqrt{2}\times 100\text{MVA}/(\sqrt{3}\times 13.8\text{kV})=\mathrm{5,917}$ A, and phase $a$ reaches 54,546 A 8.5 ms after the fault — 9.22 pu. The ceiling is 10 pu: the AC component starts at $E/X_d^{″}=1/0.2=5$ pu and the worst-case DC offset can equal it, doubling the first peak. Phase $a$ gets 4.99 pu of offset out of a possible 5, so it very nearly hits the ceiling, while phases $b$ and $c$ split the remainder — their offsets must sum to zero with $a$'s, so no fault can give all three phases a full offset.

The offsets are not chosen; they are forced. Stator flux linkage cannot change instantaneously, so the current in each phase must be continuous through the fault instant, and the only way a suddenly-imposed AC waveform can start from the pre-fault value (zero, here) is to arrive with a DC component that cancels it. That component then decays with the armature time constant, and the measured decay is 0.210 s. The textbook value is $X_2/(\omega R_a)=0.2/(377\times 0.002)=0.265$ s — but the fault path is part of the loop, and 1 mΩ on a 1.904 Ω base is another 0.000525 pu of resistance, which brings the prediction to 0.210 s. The measurement is reading the whole loop, not just the machine.

The air-gap torque is worth a glance too. It peaks at 4.85 pu 4.4 ms in and swings to −3.72 pu half a cycle later, an oscillation at system frequency driven by the DC offset beating against the rotor field. That torque reversal is what shaft designers care about, and it is the reason a multi-mass shaft model exists in the component at all.

Reading the parameters off the envelope

Strip the DC offset out and what is left is the AC envelope, which contains the whole datasheet. Kundur's decomposition of the symmetrical current is

$$I_{\mathrm{ac}}(t)=E[(\frac{1}{X_d^{″}}-\frac{1}{X_d^{\prime }})e^{-t/T_d^{″}}+(\frac{1}{X_d^{\prime }}-\frac{1}{X_d})e^{-t/T_d^{\prime }}+\frac{1}{X_d}],$$

three terms with three very different time scales, which is why plotting the envelope minus its final value on a log axis turns it into two straight lines.

Left: the AC envelope of $i_a$ with the steady-state component removed, on a log axis, where the transient term is a straight line of slope $-1/T_d^{\prime }$. Right: what remains of the first 100 ms once that fitted line is subtracted, one point per half cycle — a second straight line of slope $-1/T_d^{″}$. The two intercepts at $t=0$ give $X_d^{\prime }$ and $X_d^{″}$.

Fitting the tail from 120 ms out gives a decay of 1.338 s and, extrapolated back to $t=0$, an amplitude of 2.81 pu above a steady value of 0.5556 pu. Subtract that fitted tail from the first 50 ms and the residue is a second exponential decaying in 20.1 ms from 1.69 pu. Add the pieces up and invert:

From the waveform	Recovered	On the nameplate
$1/I_{\mathrm{\infty }}$	$X_d=1.800$	1.8 pu
$1/(I_{\mathrm{\infty }}+2.81)$	$X_d^{\prime }=0.297$	0.3 pu
$1/(I_{\mathrm{\infty }}+2.81+1.69)$	$X_d^{″}=0.198$	0.2 pu
tail slope	$T_d^{\prime }=1.338$ s	$T_{d0}^{\prime }{X}_d^{\prime }/X_d=1.333$ s
residue slope	$T_d^{″}=20.1$ ms	$T_{d0}^{″}{X}_d^{″}/X_d^{\prime }=20.0$ ms

Every one lands within about 1 % of where it should. That agreement is not a coincidence of tuning — it is the same rotor-circuit model read forwards (parameters in, waveform out) and backwards (waveform in, parameters out), which is precisely the claim the standard-parameter formulation makes.

The physical story behind the three time scales is short: immediately after the fault the damper windings hold their flux, so the armature sees the smallest reactance the machine has; the dampers are resistive and give up in tens of milliseconds, leaving the field to hold flux on its own for a second or so; when the field's induced current has decayed too, all that remains is the synchronous reactance and the fault current settles at $E/X_d=0.556$ pu, or 2,324 A RMS — barely half of rated. An unregulated generator cannot hold up its own fault current, which is why real machines need an exciter that boosts the field the moment the terminal voltage collapses.

Things to try

Lengthen the run past 8 s and the current keeps creeping down: at the end of this window the envelope is still 0.567 pu against an asymptote of 0.5556, because $5.6T_d^{\prime }$ have passed and $e^{-5.6}$ of 2.81 pu is still 0.010 pu. Set fault_type = AG for a single-line-to-ground fault and the answer now depends on X0 as well, through the sequence network. Set rotor_lock = Off (free from t=0) and the rotor coasts down under zero mechanical torque, sliding the waveform off 60 Hz and breaking the constant-speed assumption the envelope analysis rests on — useful to see once, so that the assumption stops being invisible. Or wire an exciter to the machine's Efd input, as the generator controls sample does, and watch the settled fault current climb well above 0.556 pu as the regulator drives the field into ceiling.