Stability of Synchronous Machines

What the steady-state model predicts. With rated voltage behind \(x_d = 1.80\) pu:

Less than rated current. Taken literally this says a bolted three-phase fault at the machine terminals is less onerous than normal operation — which is plainly absurd, and every measurement contradicts it.

What is actually measured. An oscillogram of a sudden three-phase short circuit on this machine shows:

A factor of eight between the first cycle and the final value. The synchronous reactance is right, eventually — but it takes several seconds to become right, and every protection and stability question is decided long before then.

The physical reason is Lenz's law applied to closed rotor windings:

The armature's sudden current tries to drive flux through the rotor. Every closed winding on the rotor responds with an induced current that cancels that flux, and while those currents persist the flux is forced into the leakage paths — a much lower-reluctance route than through the rotor iron, hence a much lower reactance.

The magnetic circuit view. Three different flux paths, three different reactances:

The ordering \(x''_d < x'_d < x_d\) is universal and is a statement about reluctance: the more paths that are blocked, the more flux is forced into air, and air has low permeability, so the inductance is low.

Which reactance for which question:

Sets 18 and 22–23 used \(x''_d\) throughout for fault calculations; Set 24 used \(x'_d\) for the swing. Both were correct choices, and this problem is why.

The order-of-magnitude sanity check. Real machines satisfy:

For this machine \(1.80/0.30 = 6.0\) and \(0.30/0.22 = 1.36\) — both squarely typical. A data set violating these ratios almost always contains a transcription error, and checking them takes five seconds.

The three windings on the direct axis: armature (\(a\)), field (\(f\)) and damper (\(D\)), all coupled through the common mutual reactance \(x_{ad}\):

\(x_l\) is the armature leakage reactance — flux that links the stator winding but never crosses the air gap — and it sets the floor below which no reactance of the machine can fall.

With the field winding closed, it appears in parallel with the magnetising branch as seen from the armature:

where \(x_f\) is the field's own leakage. The parallel combination is smaller than \(x_{ad}\), so \(x'_d < x_d\) necessarily. This is the shorted-secondary transformer result, and the machine is exactly that: a transformer whose secondary happens to rotate.

With the damper closed as well, a third branch joins the parallel group:

Smaller again. The nesting \(x_l < x''_d < x'_d < x_d\) follows immediately, and the lower bound is \(x_l\) — approached only if the rotor windings were superconducting.

Numerically for the study machine with \(x_l = 0.15\):

Note how much of \(x''_d\) is pure leakage: 0.15 of 0.22, or 68%. The subtransient reactance is dominated by a quantity that has nothing to do with the rotor at all, which is exactly why it is so nearly independent of rotor position.

Why \(x''_d \approx x''_q \approx x_2\). Negative-sequence current produces an mmf rotating backwards at synchronous speed, so relative to the rotor it sweeps past at twice synchronous speed:

The wave alternately faces the \(d\) and \(q\) axes, so the machine presents the average of the two subtransient reactances. And because it sweeps at 100 Hz, the damper and field currents it induces never have time to decay — the machine is permanently in its subtransient state as far as negative sequence is concerned.

Checking against the study machine's data:

Which retroactively justifies Set 22's assumption that \(x_2 = x''_d\) for a round-rotor machine. For a salient-pole machine \(x''_q\) exceeds \(x''_d\) by 20–40% and the averaging matters.

The zero-sequence reactance completes the set and is different in kind:

Zero-sequence current produces no rotating mmf at all — the three phase mmfs are displaced 120° in space and in phase, so they cancel. Only the leakage flux remains, and even that is partly cancelled between slots, so \(x_0\) is below the leakage reactance. This is why a solidly earthed generator neutral gives such a severe earth fault (Set 23, Practice 8).

The open-circuit constants are the natural \(L/R\) of each rotor winding with the armature open:

The field winding is designed for low loss and is physically large, so \(T'_{d0}\) is measured in seconds. The damper is a set of short bars with substantial resistance, so \(T''_{d0}\) is measured in tens of milliseconds. The ratio here is 267.

With the armature short-circuited the rotor windings see the armature's low reactance reflected in parallel, which reduces their effective inductance and therefore their time constant:

Each ratio is the reactance seen by that winding with the next-faster effect already accounted for. Both ratios are less than one, so short-circuit constants are always shorter than open-circuit ones.

Numerically:

The field's transient decays six times faster into a short circuit than it would on open circuit — a factor of \(x_d/x'_d\). This is the same effect that makes a shorted transformer secondary decay faster than an open one.

The armature time constant governs the dc offset, and is set by the stator's own \(L/R\):

\(x_2\) appears rather than \(x''_d\) because the dc component in the stator is stationary in space, so relative to the rotor it rotates backwards at synchronous speed — the negative-sequence condition again. In cycles, \(T_a = 11.7\) cycles at 50 Hz.

The three timescales in perspective:

The separation is what makes the classical model work. The subtransient is over in a cycle, so it can be ignored in a stability study; the transient lasts 1.33 s, comfortably longer than the 0.5 s first swing, so \(E'\) can be treated as constant over it. That is the whole justification for Set 24.

How well does "constant \(E'\)" hold? Over a 0.5 s first swing:

So the transient component has fallen by 31% by the end of the swing — not negligible. The classical model is nonetheless conservative in practice, because the AVR is simultaneously boosting the field to counteract exactly this decay, and the two errors largely cancel. That accidental cancellation is why the classical model survived for fifty years, and why it must be checked rather than trusted when the AVR is slow or out of service.

The last row matters for a different reason. \(T'_{d0} = 8\) s is the field's own response time, and it is what any exciter must fight:

With \(K_A = 200\) the effective field time constant falls to 0.04 s, which is what makes a modern exciter fast enough to act within a single swing. It is also, as Problem 15 shows, what makes it capable of producing negative damping.

The envelope equation, which is three exponentials in the reciprocals of the reactances:

It is written in reciprocals because currents, not reactances, superpose. Each bracket is the extra current that a particular rotor winding is responsible for, and it decays with that winding's time constant.

The three amplitudes:

The three components must add to the subtransient current at \(t = 0\), which is the structural check on the decomposition.

Evaluated through the transient, with \(T'_d = 1.333\) s and \(T''_d = 0.022\) s:

With \(I_{\text{base}} = 100\times10^6/(\sqrt3\times220\times10^3) = 262.4\) A. The current falls by a factor of eight over five seconds.

The subtransient is essentially over in one cycle. At \(t = 0.020\) s:

40% of the subtransient contribution remains after one cycle, 16% after two, 3% after four. This is why the subtransient reactance is used for the first cycle momentary duty and the transient reactance for anything slower — and why a breaker that parts its contacts at 3 cycles sees a current governed almost entirely by \(x'_d\), not \(x''_d\).

The three protection-relevant numbers:

The last row is the awkward one. A generator's sustained short-circuit current is below its rated current, so a plain overcurrent relay set above load will never operate on a sustained terminal fault. This is why generator protection uses voltage-restrained or voltage-controlled overcurrent, or differential protection, and never plain overcurrent alone.

The effect of prefault load, since the calculation above assumed no-load:

A machine at full load with 1.0 pu terminal voltage has \(|E''| \approx 1.05\)–\(1.10\), raising every figure above by 5–10%. The convention in fault studies — used throughout Sets 18 and 22 — is to take \(E'' = 1.0\) and accept the small error, or to use a flat 1.05 prefault voltage as a blanket allowance.

Step 1 — remove the dc offset. Take the half-difference of the upper and lower envelopes at each instant:

The three phases have different dc offsets — their sum is zero — so this must be done phase by phase. Working with a phase whose fault incidence angle gave little offset makes the job easier.

Step 2 — read the steady value from the tail of the trace, several seconds in:

Straightforward in principle. In practice the machine is decelerating (the prime mover cannot hold speed into a short circuit) and the field may be forced by the AVR, so a true steady state is rarely reached and \(x_d\) is normally measured by a separate slip test instead.

Step 3 — plot \((I - I_\infty)\) on semi-log paper. After the first few cycles only the transient term survives, and it plots as a straight line:

The slope gives \(T'_d = \mathbf{1.333\ s}\) and the intercept gives the transient amplitude. This is the classic graphical extraction, and it works because the two time constants differ by a factor of 60 — the two straight-line regions do not overlap.

Step 4 — extrapolate that line back to \(t = 0\):

The extrapolation is essential: the current at \(t = 0\) is not the transient value, because the subtransient is superimposed on it. Reading the trace at \(t = 0\) and calling it \(x'_d\) is the classic beginner's error, and it gives 0.220 instead of 0.300.

Step 5 — subtract the extrapolated transient line and plot the remainder:

This second peeling is where the accuracy is lost. Only two or three cycles of data contain the subtransient, and they are the cycles most contaminated by dc offset and instrument transient.

The five practical difficulties, in order of severity:

The fourth row is the one most often forgotten, and it biases the answer in the dangerous direction — a shorter apparent \(T'_d\) makes the machine look worse for stability than it is.

The modern alternative is the standstill frequency response test (SSFR), in which the stationary machine is excited over 0.001–100 Hz and the transfer functions are fitted:

SSFR also identifies second-order rotor models, which a short-circuit test cannot, and is now the standard method for machines being modelled for detailed stability studies. But the short-circuit test remains the acceptance test, because it is the one that answers the question a protection engineer actually asks.

The origin is continuity of current in an inductance. Before the fault the current in a phase is \(i_0\); the instant after, the ac component demands \(I''\cos(\theta)\) at that same instant:

where \(\theta\) is the point on the voltage wave at which the fault occurred. The dc term is whatever is needed to stop the current jumping — it has no independent existence, and no source drives it.

Hence the two extremes. Starting from no load (\(i_0 = 0\)):

Since the circuit is almost purely inductive, current lags voltage by nearly 90°. A fault initiated at the voltage zero therefore demands a current at its peak — and the dc term must supply the whole of it.

The three phases cannot all be worst. Since the three dc offsets are determined by three phase currents 120° apart, and those sum to zero:

so at most one phase can carry the full offset, and the other two carry roughly half of it in the opposite sense. The worst case is examined per phase, and the standard design assumption is that one phase gets the maximum.

The decay is governed by the armature time constant found in Problem 3:

Note this decays through the stator resistance, not the rotor's, and is therefore governed by a completely different physical mechanism from the ac decay. A low-loss machine has a longer dc time constant, which is one of the few respects in which efficiency is a disadvantage.

The peak asymmetrical current, occurring about half a cycle after the fault:

Almost exactly \(2\sqrt2\) times the subtransient rms current — the classical "doubling effect". The theoretical maximum, with no decay at all, is \(2\sqrt2 I'' = 12.857\) pu, so the machine's own resistance has removed only 2%.

What this number is used for:

The second row is why busbar bracing is designed on the asymmetrical peak: the mechanical force goes as the square of current, so a doubling of current is a quadrupling of force. The third row is the reason protection engineers care about \(T_a\) at all — a long dc offset drives current transformers into saturation and can delay differential protection by several cycles.

Asymmetry factor as a function of time, which is what the standards tabulate:

Using \(I_{\text{rms,asym}} = I_{\text{rms,sym}}\sqrt{1+2e^{-2t/T_a}}\). At \(t=0\) the factor is \(\sqrt3 = 1.732\), the familiar maximum, and by the time a 3-cycle breaker parts its contacts it has fallen to 1.48.

The statement. The flux linkage of a closed winding of finite resistance cannot change instantaneously, and decays thereafter with the winding's own time constant.

A finite \(r\) and finite \(i\) give a finite \(d\lambda/dt\), so \(\lambda\) is continuous. For a superconducting winding \(\lambda\) would be constant forever.

The proof of continuity is the standard integration argument. Over an interval \([0^-,0^+]\):

The integrand is bounded, the interval has zero measure, so the integral vanishes and \(\lambda(0^+) = \lambda(0^-)\). Note carefully that \(\lambda\) is continuous, not the current — the field current jumps at the instant of a fault precisely so that \(\lambda_f\) need not.

Applied to the field winding. Its flux linkage is:

If \(i_d\) jumps (as it does when a fault suddenly demands armature current), then \(i_f\) must jump in the opposite sense to hold \(\lambda_f\) fixed. That induced field current is what supplies the extra short-circuit current of Problem 4, and it decays with \(T'_d\).

Define \(E'_q\) as the flux linkage in voltage units:

This is the entire content of "constant \(E'\)". It is not an approximation about the machine's behaviour; it is an exact consequence of the flux-linkage theorem at the instant of the disturbance. The approximation is only in extending it forward in time.

How far forward can it be extended? The governing equation is:

With \(T'_{d0} = 8.0\) s, \(E'_q\) moves slowly. Over a 0.5 s first swing it can change by at most a few per cent from the \(E_{fd}-E'_q\) term — provided the exciter does not act, and provided \(i_d\) does not change much, which it does.

Quantifying the error for the study machine during a first swing. Take \(i_d\) rising from 0.97 to 1.30 pu as the rotor swings out, with the exciter on manual:

A 2.6% fall in \(E'\) means a 2.6% fall in every power-angle curve — small, and in the conservative direction. This is the quantitative justification for Set 24, and it is a good one.

Where the assumption fails, and it does fail:

The first two rows tend to cancel, which is why the classical model has survived. The third row is where it must be abandoned outright: over ten seconds \(E'_q\) has decayed most of the way towards its steady-state value and the machine's power-angle curve is nothing like the one assumed.

Three state variables, hence "third order":

The classical model has the first two only; the one-axis model adds the third. Damper windings are omitted, which is legitimate because \(T''_{d0} = 0.03\) s is far shorter than any timescale of interest in a stability study.

The three differential equations:

The first two are Set 24's swing equation split into first-order form; the third is new and is what makes the model a machine model rather than a mass model.

The algebraic equations that close the system, for a machine on an infinite bus through \(X_e\):

The second term in \(P_e\) is the transient saliency term, present because \(x_q \ne x'_d\) even in a round-rotor machine — indeed especially in one, since \(x_q = 1.70\) while \(x'_d = 0.30\).

Expanding \(P_e\) in terms of \(\delta\) alone gives the familiar form:

A fundamental term and a second-harmonic term, exactly as in the salient-pole steady-state equation of Problem 9 — but with \(x'_d\) in place of \(x_d\). Because \(x_q \gg x'_d\), the second term is negative here, unlike the steady-state case.

Numerically for the study machine with \(X_e = 0.30\), \(E'_q = 1.20\), \(V = 1.0\):

The transient saliency term is large — 29% of the fundamental amplitude — and negative. It suppresses the curve below 90° and lifts it above, moving the peak from 90° to 113.5°.

How the classical model is recovered. Two approximations, applied in order:

The second assumption is well justified; the first is not. Setting \(x_q = x'_d\) is a substantial distortion of a round-rotor machine, and it is made purely because the resulting single sine is analytically tractable and the equal-area criterion depends on it.

Does the distortion matter? Compare the two curves at the operating point:

Large differences — but note that the rotor angle in the one-axis model is measured to the q-axis, not to the classical model's fictitious \(E'\) phasor, so the two are not describing the same operating point at the same \(\delta\). Once both are matched to the same terminal conditions (Problem 11) the discrepancy in power at the operating point disappears, and what remains is a difference in the shape of the curve — typically 5–10% in the critical clearing angle.

Resolve the current onto the two axes. With the q-axis at angle \(\delta\) ahead of the infinite bus:

The d-axis current is driven by the difference in the two voltages' q-axis components; the q-axis current by the infinite bus's d-axis component. Two independent circuits, which is precisely what the two-axis decomposition achieves.

The power is the sum of the two axes' contributions:

Using \(2\sin\delta\cos\delta = \sin2\delta\):

The first term is the familiar excitation power; the second is reluctance power, and it exists because a salient rotor prefers to align its pole axis with the stator field. It requires no field current at all.

Numerically:

The reluctance term is 16% of the fundamental amplitude — typical for a salient-pole machine, and not negligible.

The maximum, found by differentiating:

Two things to note. The peak is 1.1258 pu at 74.3° — 4.5% higher than the round-rotor value of 1.0769, and reached at an angle well below 90°. Saliency both raises the limit and moves it inward.

The consequence for stability is a stiffer machine at every operating angle:

At \(\delta = 30°\) this gives \(0.9326+0.1709 = 1.1035\) pu/rad against 0.9326 without saliency — 18% stiffer. Since the oscillation frequency goes as \(\sqrt{P_s}\), a salient-pole machine oscillates 8.8% faster than a round-rotor machine of the same inertia and rating.

The most striking consequence: a salient machine can carry load with no field current at all.

17% of rating, purely from reluctance. This is the operating principle of the reluctance motor, and it is why a salient-pole machine that loses its excitation may hold synchronism at light load while a round-rotor machine (for which \(x_d \approx x_q\), so the reluctance term nearly vanishes) certainly will not — as Problem 19 examines.

Typical saliency ratios, for reference:

The study machine of the rest of this set has \(x_q/x_d = 1.70/1.80 = 0.944\) — round rotor, so its steady-state reluctance term is only 0.012 pu and is legitimately neglected. Its transient saliency, however, is severe (Problem 8), because \(x_q/x'_d = 5.7\).

The difficulty. To resolve \(\bar I\) into \(i_d\) and \(i_q\) you must know the rotor angle \(\delta\). But to compute \(\delta\) you appear to need \(i_d\) and \(i_q\). The q-axis locator breaks the circle.

It is not obvious that any such shortcut should exist, and the reason it does is a small algebraic accident worth seeing.

Write the terminal voltage in components:

These are the two-axis machine equations in the steady state, with \(E\) the internal EMF on the q-axis.

Now add \((r_a+jx_q)\bar I\) to \(\bar V_t\) and look at the components:

The d-axis component cancels identically. That is the accident: adding \(jx_q\bar I\) exactly removes the d-axis part of \(\bar V_t\), whatever the operating point.

Hence the locator:

Its direction is exact and its magnitude is not the internal EMF. Using \(|\bar E_q|\) as the excitation voltage is a common and consequential error — for the study machine it gives 2.255 instead of 2.648, a 15% understatement.

The construction, step by step:

Steps 4 and 5 differ only in which reactance multiplies \(i_d\), which is the whole content of the three-reactance picture: same current, different flux path, different EMF.

The resolution convention. With the q-axis at angle \(\delta\), any phasor \(\bar X\) resolves as:

The sign on \(X_d\) is the one convention that varies between textbooks and is the source of most sign errors in this subject. The check is simple: a purely lagging current in an over-excited generator must give \(i_d > 0\), since it demagnetises.

Why the diagram is worth constructing rather than avoided:

Every stability simulation begins by solving a load flow and then running this construction at each machine. Get it wrong and the run starts from a state that is not an equilibrium, and the machine oscillates for reasons that have nothing to do with the disturbance being studied.

Find the reactive output that gives \(E'_q = 1.20\). Iterating the construction of Problem 10 on \(Q\):

A modest lagging export, which is what an over-excited machine on a stiff system does. The value is found by a one-dimensional search because \(E'_q\) is not an explicit function of \(Q\).

Current and terminal voltage:

Terminal voltage 1.128 pu — high, but this machine is heavily excited. The terminal apparent power is \(1.000+j0.6176\), so the machine itself delivers 61.8 MVAr while only 29.2 MVAr reaches the bus; the difference is absorbed by \(X_e\).

Locate the q-axis:

The rotor angle relative to the infinite bus. Note it is far larger than the classical model's 30° (Set 24), because it is measured to the q-axis, not to a fictitious \(E'\) phasor behind \(x'_d\).

Resolve onto the two axes using \(X_q = \mathrm{Re}(\bar Xe^{-j\delta})\), \(X_d = -\mathrm{Im}(\bar Xe^{-j\delta})\):

Check: \(\sqrt{0.96522^2+0.39196^2} = 1.04176 = |\bar I|\) ✓ and \(\sqrt{0.66633^2+0.91044^2} = 1.12822 = |\bar V_t|\) ✓. Note \(i_d\) is much the larger component — the machine is heavily demagnetising, which is what an over-excited generator's armature reaction does.

The two internal EMFs:

The target is met exactly, confirming the search. Note the enormous difference: 2.648 against 1.200, a factor of 2.2. That gap is entirely the armature reaction drop \((x_d-x'_d)i_d = 1.448\) pu, and it is why the steady-state and transient stability limits differ so dramatically.

Check the power by a route that has not been used:

Exactly 1.000 to five decimal places, computed from dq components alone. The d-axis carries 64% of the power here, which surprises people — but there is nothing special about the q-axis as far as power is concerned.

The complete operating point:

This is the initial condition for every dynamic simulation of this machine, and every problem from here to the end of the set uses it.

The field current requirement. \(E = 2.648\) pu means:

The machine needs 2.65 times its no-load field current. Since exciters are typically rated for 2.5–3.0 times no-load field current continuously, with a ceiling of 1.5–2 times that for short periods, this machine is near its continuous excitation limit — an operating point that a real over-excitation limiter would be watching closely.

Constant field current — the machine with its exciter on manual, or with no AVR at all. The EMF behind \(x_d\) is fixed:

A margin of only 26% over the operating point of 1.000. This is the true classical steady-state stability limit, and it is uncomfortably tight — which is exactly why every machine built since about 1950 has an automatic voltage regulator.

Constant field flux linkage — the assumption of Set 24, valid for a first swing:

Exactly the value used throughout Set 24, which is not a coincidence — the operating point was chosen to reproduce it. A margin of 100%.

An ideal voltage regulator holding \(|\bar V_t|\) constant. The machine's own reactance disappears from the problem entirely:

A margin of 276%. The limiting angle is now the angle across \(X_e\) alone reaching 90°, because as far as the network is concerned the machine has become an ideal voltage source at its own terminals.

The comparison:

A factor of three between the extremes, from the same machine at the same operating point. The models are not approximations of one another; they answer different questions.

Why the spread is so large. Follow what each model assumes about the field:

The three cases are "the curve falls", "the curve stays", "the curve rises" — and the corresponding effective reactance is \(x_d\), \(x'_d\), or zero. Everything else follows.

Which model for which question:

The third row is a genuine operational concern: a machine running comfortably at 1.00 pu with its AVR in service has only 26% margin if the AVR reverts to manual, and system operators do reduce output when that happens.

The ideal-AVR figure is optimistic, and must be qualified:

The last row is the important one and is the subject of Problems 14–16. A fast AVR raises the steady-state limit dramatically while simultaneously reducing damping — so the machine may be able to sit at 3 pu in principle and still oscillate itself out of step in practice.

The elements, in signal order:

The plant is the slowest element by a factor of eight, which is what makes high gain both possible and necessary.

The closed loop, simplified to the two dominant elements:

Substituting the first into the second, and noting that \(|V_t|\) depends on \(E'_q\) with some sensitivity \(K_6 = \partial|V_t|/\partial E'_q\):

The effective time constant:

High gain does not make the field respond faster in any physical sense; it makes the closed loop's dominant pole move out along the negative real axis, which amounts to the same thing for control purposes.

Numerically for the study machine, taking \(K_6 \approx 0.3\) (a typical value at this loading):

Compare the 0.5 s first swing. With \(K_A = 10\) the exciter is far too slow to influence it. With \(K_A = 200\) the exciter acts four times within one swing, and the constant-\(E'\) assumption of Set 24 becomes conservative — the field is being driven up while the rotor swings out.

Steady-state voltage regulation is the other benefit of gain:

Against 100% with no regulator. Grid codes typically require terminal voltage held within 0.5% of setpoint, which sets a floor on \(K_A\) of a few hundred.

Ceiling voltage is the second design parameter, and it governs transient rather than steady behaviour:

During a fault the terminal voltage collapses, the error is enormous, and the regulator saturates at its ceiling. What matters then is not gain but how much voltage the exciter can actually deliver — and a static exciter fed from the generator's own terminals has a problem here, because its supply collapses with the fault. Compound-source and independent-supply arrangements exist precisely to fix that.

The two families:

The second row's weakness is real and is why compound-source excitation exists: a machine whose exciter dies during the very fault it needs to ride through is worse than one with a modest but reliable rotating exciter.

The decomposition. Any small perturbation of electrical torque can be written as a sum of a component in phase with angle and one in phase with speed:

This is not an approximation — any complex number can be resolved onto two non-parallel directions, and since \(\Delta\omega = j\omega\,\Delta\delta\) for a sinusoidal oscillation, "in phase with speed" means "in quadrature with angle". The decomposition is exact at any one frequency.

What each component does:

The two failure modes look completely different on a recorder. A negative \(K_S\) gives a single monotonic run-away in under a second; a negative \(K_D\) gives an oscillation that grows over ten or twenty seconds. Confusing them leads to fixing the wrong thing.

The linearised system:

The factors of \(\omega_s\) appear because \(K_D\) is conventionally defined per unit speed in per unit, while \(\dot\delta\) is in electrical radians per second.

For the study machine with \(H = 5.0\), \(K_S = 1.7321\) pu/rad (the Set 24 value at the operating point), \(M = 2H/\omega_s = 0.031831\):

The four cases:

Grid codes typically require \(\zeta > 0.03\) as a minimum and \(\zeta > 0.05\) as a planning target. \(K_D = 10\) is a representative value for a machine with damper windings and a well-tuned stabiliser.

The damped frequency is barely affected:

A 0.2% reduction. At these damping levels the oscillation frequency is set entirely by \(K_S\) and \(H\), and damping shows only in the decay envelope. This is why the frequency of a recorded oscillation identifies \(K_S\) and the envelope identifies \(K_D\) — two independent measurements from one trace.

Where damping comes from:

The fifth row is the reason the sixth exists. Every natural source of damping is positive and modest; the AVR is the only large term and it is negative under exactly the conditions — heavy loading, weak system — where damping is most needed.

Follow the signal round the loop during an oscillation in which the rotor is swinging outward:

The AVR is doing exactly what it was designed to do — hold the voltage — and the harm comes entirely from the phase lag in step 4. Nothing is malfunctioning.

The phase lag, quantified. The field path contributes a lag of:

With \(\omega_n = 7.377\) rad/s, \(T_{\text{eff}} = 0.131\) s, \(T_A = 0.05\) s, \(T_R = 0.02\) s: \(\phi = 44.0°+20.2°+8.4° = 72.7°\). Any lag beyond 90° would put the torque squarely in the negative-damping quadrant; at 72.7° a substantial fraction of it already is.

The classical statement in terms of the Heffron–Phillips constants is that the AVR path contributes a torque:

The sign of the whole expression is carried by \(K_5\), the sensitivity of terminal voltage to rotor angle. When \(K_5 < 0\) — which is the usual case at heavy loading through a large external reactance — this term produces negative damping in direct proportion to \(K_A\).

When is \(K_5\) negative? Physically, \(K_5 = \partial|V_t|/\partial\delta\) at constant \(E'_q\):

Every condition in the third row is one under which the system needs damping most: a heavily loaded machine on a weak tie is precisely the configuration prone to inter-area oscillation. The AVR's contribution turns negative exactly where it hurts.

The consequences, for the study machine with a net \(K_D = -5\):

A 1° oscillation grows to 2° in 2.8 s, 4° in 5.5 s, 32° in 14 s and out of step in about 20 s. Slow enough that an operator watching a recorder would see it develop — and far too fast for anyone to do anything about it.

Growth over time:

Note the oscillation frequency is still 1.174 Hz throughout: the trace is a clean growing sinusoid, not a runaway. That signature — a sustained or slowly growing oscillation at 0.5–2 Hz, with no obvious triggering event — is the fingerprint of negative damping, and it has caused real system separations.

The historical significance. This effect was not anticipated:

The classical theory said high gain was unambiguously good, because it raised the steady-state stability limit threefold (Problem 12). Only when machines began oscillating themselves out of step at loadings well below that limit was the damping question separated from the synchronising question — the work of deMello and Concordia in 1969, which produced both the analysis above and its remedy.

The objective. Inject an extra signal into the AVR summing junction that produces a torque in phase with speed:

This is a purely additive fix: the AVR's voltage regulation is untouched, and the stabiliser contributes nothing in the steady state. It is a damping device, not a voltage device.

Choice of input signal:

The fourth is the modern standard (IEEE type PSS2B), constructed as \(\int(P_m-P_e)dt/2H\) using a filtered speed signal to synthesise \(P_m\). It gives a clean speed-equivalent signal without the torsional content that made early speed-input stabilisers excite shaft resonances.

The phase compensation required. The stabiliser's signal must travel through the exciter and field to reach the torque, picking up the same lag computed in Problem 15:

So the stabiliser must supply about \(+73°\) of phase lead at the local mode frequency, to arrive in phase with speed rather than 73° behind it.

The lead-lag blocks. Each first-order block gives at most about 55° of usable lead, so two are needed:

Two identical blocks, each supplying \(73/2 = 36.4°\) at \(\omega_n\).

Setting \(T_1\) and \(T_2\). For maximum lead at \(\omega_n\), centre the block geometrically:

Both squarely in the range published for real stabilisers (\(T_1\) 0.1–0.5 s, \(T_2\) 0.02–0.10 s), which is a reassuring check on the design.

The washout filter is what makes the device safe:

A high-pass filter. It blocks any steady or slowly varying component of the input, so the stabiliser cannot bias the terminal voltage during a sustained frequency excursion — which would otherwise be a serious side effect during system islanding. At 1.174 Hz its phase contribution is \(\arctan(1/(10\times7.377)) = 0.8°\), negligible, which is why \(T_w\) is chosen large.

Setting the gain. To restore \(K_D\) from \(-5\) to a target of \(+10\):

In practice \(K_{\text{PSS}}\) is set by field test: raise the gain until the machine becomes unstable in a different mode — usually an exciter or intra-plant mode at 3–5 Hz — then back off to one-third of that value. The one-third rule is empirical and near-universal.

The result:

The middle row is the machine that Problem 12 said had a steady-state stability limit of 3.76 pu. It cannot in fact be operated anywhere near that, because it oscillates itself apart. The stabiliser is what converts the paper limit into a usable one.

What a stabiliser cannot do:

The second row is the practical difficulty of multi-machine tuning: a stabiliser compensated for 1.2 Hz supplies the wrong phase at 0.4 Hz, so a unit tuned for its local mode may worsen an inter-area mode. Modern practice uses two lead-lag stages tuned at different frequencies, or coordinated system-wide optimisation.

The mechanism is induction-machine action. When the rotor swings, it is momentarily not at synchronous speed:

The stator's rotating field therefore sweeps past the damper bars at slip speed, induces currents in them, and those currents produce a torque opposing the relative motion — exactly as in an induction motor.

The torque opposes the slip, which is what makes it damping rather than synchronising:

It is proportional to speed deviation, not to angle, so it contributes purely to \(K_D\) and nothing to \(K_S\). This is the cleanest source of damping in the whole machine, and unlike the AVR its sign cannot change.

The induction-machine torque-slip curve gives the magnitude:

For small slip the term \(r_D/s\) dominates the denominator and \(T_D \approx V^2s/(\omega_sr_D)\) — linear in slip, which is what makes the constant-\(K_D\) model valid.

The slip at maximum torque. From the damper's open-circuit time constant:

A large slip. Compare the actual slip during a 1° amplitude oscillation at 1.174 Hz: \(s = \omega_n\Delta\delta/\omega_s = 7.377\times0.01745/314.16 = 0.041\%\). The machine operates on the far linear tail of the torque-slip curve, three orders of magnitude below the peak — which is precisely why \(K_D\) can be treated as a constant.

Why the linearity is fortunate:

Damping torque therefore grows in proportion to the disturbance over the entire range of interest, which is exactly the behaviour a linear damping model assumes. Only during a pole slip, where \(s\) can reach several per cent, does the curvature begin to matter.

The other functions of damper windings, several of which matter more than damping:

The second row is the one that dominates the design. Without a low-resistance damper cage, negative-sequence current — from an unbalanced fault or unbalanced load — induces double-frequency currents that flow in the rotor surface and forging, causing severe local heating. The \(I_2^2t\) withstand limits (typically 30 s at \(I_2 = 0.10\), or 10 s at 0.30) are set by exactly this.

Construction, and the round-rotor case:

A turbogenerator has no explicit damper winding at all, yet it has a well-defined \(x''_d\) and substantial damping — the solid steel forging is itself a short-circuited conductor, and the slot wedges provide additional paths. Because the eddy currents flow in a distributed conductor, its behaviour is really a continuum rather than a single circuit, which is why detailed models use two or three rotor circuits per axis rather than one.

The electrical resonance. A series-compensated line is an \(LC\) circuit, and its natural frequency is:

where \(k\) is the compensation level. The total inductive reactance seen by the resonance includes the machine's own \(x''\) and the transformer, so \(X_L\) is larger than the line's alone, and the practical frequency is somewhat lower than this simple formula gives.

Numerically, with the study machine (\(x''_d = 0.22\)), transformer 0.10 and line 0.40 pu, compensated 40%:

Subsynchronous, as the name promises. Note the machine's own reactance is nearly a third of \(X_L\), so the resonant frequency is a property of the machine-plus-network, not of the line alone.

The complementary frequency, which is what the rotor feels:

A stator current at 23.6 Hz produces a flux wave rotating at 23.6 Hz, and the rotor turning at 50 Hz sees it at the difference — 26.4 Hz. If the shaft has a torsional natural mode at or near 26.4 Hz, that mode is driven directly.

Typical shaft torsional modes of a large turbine-generator, for comparison:

Mode 2 at 24–27 Hz sits directly on the 26.4 Hz complementary frequency computed above. This is not a coincidence of the example — the torsional modes of a large steam turbine-generator and the complementary frequencies of practical compensation levels occupy the same band, which is precisely why subsynchronous resonance is a real engineering problem rather than a curiosity.

The three mechanisms, which are genuinely distinct and require different remedies:

The first is purely electrical and slow. The third is a single transient event: a nearby fault excites the resonance and the resulting torque pulsation, arriving at the shaft's own natural frequency, can exceed the shaft's design torque several times over in a handful of cycles. That is what destroyed two shafts at Mohave in 1970 and 1971 — the events that established the whole subject.

Why the induction generator effect gives negative resistance:

The rotor turns faster than the subsynchronous field, so the slip is negative and the reflected rotor resistance is negative. If that negative resistance exceeds the positive resistance of the network at \(f_{er}\), the subsynchronous current grows spontaneously. The machine is acting as an induction generator at that frequency.

The countermeasures, in order of increasing cost:

The last row is the modern answer and is the main non-economic reason TCSC installations are preferred over fixed capacitors on lines near large thermal plant: a properly controlled TCSC is inherently SSR-neutral, which a fixed capacitor can never be.

Which machines are at risk:

The last row is a live subject: doubly-fed induction generators have produced subsynchronous control interaction with series-compensated lines in Texas and China, at frequencies set by the converter's control loops rather than by any shaft — the same resonance, an entirely different amplifier.

The sequence of events:

Note it is not a fast event. The 8-second field time constant means several seconds elapse between the loss of excitation and the pole slip, which is enough time for protection to act deliberately rather than instantaneously.

Can it hold synchronism? With \(E \to 0\) only reluctance power remains:

1.2% of rating. This round-rotor machine cannot hold synchronism at any meaningful load, and will slip poles. Contrast the salient-pole hydro machine of Problem 9, whose reluctance term was 0.171 pu — 14 times larger — which could hold synchronism at light load.

The reactive power drawn once slipping, which is what makes the event a system problem rather than a machine problem:

A machine that was exporting 0.62 pu of reactive power is now importing up to 1.7 pu — a swing of well over 2 pu on the system. On a weak system that alone can collapse the voltage and cascade to other units, which is why loss of field is treated as an urgent trip rather than an alarm.

The damage mechanisms, which are why the machine must come off:

The first row is the binding one for a round-rotor machine: the solid forging was never designed to carry the continuous induced current of asynchronous operation, and permanent damage follows in tens of seconds.

The protection principle. Measure impedance at the terminals. During normal generation it sits in the first quadrant; on loss of field it swings into the fourth:

This is the classic mho-characteristic loss-of-field relay (device 40), and it is one of the very few protection functions defined by a locus rather than a magnitude.

The two-zone setting for this machine:

Both circles are offset downward from the origin by half the transient reactance, and lie entirely in the negative-reactance half-plane. The offset is what makes the characteristic immune to power swings and to faults, both of which produce trajectories that stay above the origin.

Coordination with the excitation limiters, which is where this function meets machine control:

The under-excitation limiter (UEL) in the AVR must act before the relay, so that a machine drifting towards under-excitation is pulled back rather than tripped. Both must sit inside the machine's own steady-state stability limit and inside the stator end-iron heating limit. Getting this three-way coordination wrong is a common cause of spurious generator trips, and it is checked on the capability chart rather than by calculation.

The hierarchy, by number of state variables:

Each row adds a physical effect and a data requirement. The eighth-order model is the machine as Park wrote it in 1929; everything above it is a reduction obtained by neglecting a fast transient.

What each can represent:

The last two rows require the stator flux-linkage derivatives that every phasor-based model discards. That is the fundamental divide: models 1–4 above work in phasors at 50 Hz and cannot see anything at any other frequency; the full Park model works in instantaneous quantities and can.

The decision rule:

Note that the rule is set by the question, not by the availability of data. Using a sixth-order model to answer a first-swing question wastes effort and, worse, requires eight parameters that may be estimates — introducing more error than the model removes.

What the extra fidelity actually buys. For the study machine's Set 24 problem — mid-line fault, \(t_{cc}\):

Every refinement moves the answer in the optimistic direction except flux decay with the AVR out of service. The classical model is therefore conservative in normal operation — which is the real reason it has survived, and the reason a 30% margin on \(t_{cc}\) is considered adequate.

The data problem, which limits fidelity in practice more than computation does:

A sixth-order model built on \(\pm30\%\) subtransient time constants is not obviously better than a third-order model built on well-measured transient data. This is the practical argument for parsimony, and it is why the one-axis model with a good exciter representation is the workhorse of the industry.

The full chain of this chapter, read as one argument:

Every one of these is a reason the classical model of Set 24 is not the whole story, and every one is also a reason it works as well as it does.