Steady-State and Transient Stability
The same swing equation answers two quite different questions depending on how hard the system is pushed — a small nudge is settled by the sign of two coefficients, a short circuit only by the whole nonlinear excursion — and the machinery for each is worked out here.
- Why the size of the disturbance, not its cause, divides stability into steady-state, dynamic and transient problems, and why each needs a different tool.
- How the linearised swing equation of Chapter 27 yields \(\omega_n\), \(\zeta\) and the characteristic roots, and how each root tells a different story about the machine.
- Why the steady-state stability limit is \(P_{max}=|E||V|/X\) at \(\delta=90^\circ\), and how to compute it for a real machine and load.
- Which internal voltage and which reactance to use — \(E_q\) behind \(X_d\), \(E'\) behind \(X_d'\), or a regulated terminal voltage — and how much difference the choice makes.
- How an automatic voltage regulator raises the steady-state limit past \(90^\circ\) yet can introduce negative damping, and what a power system stabiliser does about it.
- What decides the transient-stability verdict: fault type, fault location, clearing time, inertia, transfer reactance and pre-fault loading.
- The practical measures that improve stability, and where each one acts in the swing equation.
One Equation, Two Questions
Chapter 27 finished with a single nonlinear differential equation and two ways of attacking it: linearise about the operating point, or integrate step by step. The choice is not a matter of taste. It is dictated by how far the rotor angle moves, and that in turn is dictated by the size of the disturbance.
A power system is disturbed continuously. Loads switch on and off, tap changers step, wind output fluctuates, and the aggregate of all this is a permanent low-level jitter in every rotor angle on the network. None of it moves \(\delta\) more than a degree or two. For disturbances of that size the sine in \(P_e = P_{max}\sin\delta\) may be replaced by its tangent at the operating point without appreciable error, the equation becomes linear, and the whole apparatus of linear system theory applies. The question it answers is: does this operating point hold? That is steady-state stability.
A power system is also occasionally struck hard. A three-phase fault removes a transmission path entirely for a hundred milliseconds or more; the rotor angle swings through sixty or eighty degrees. No linearisation survives that, because the very feature that decides the outcome — the fact that \(P_e\) turns over and starts falling past \(90^\circ\) — is the part a linear model throws away. The question now is: does the machine survive this particular event? That is transient stability.
A system can be steady-state stable and transiently unstable — indeed almost every system is, for a severe enough fault. The reverse is impossible: a system that cannot hold its operating point against a nudge will not survive a fault either.
Between the two sits a third classification with a longer history than clear boundaries. Dynamic stability, now more often called small-signal stability, is also a small-disturbance problem, but it retains the machine's control equipment — exciter, voltage regulator, governor — in the model rather than treating the field flux as frozen. The distinction matters because those controllers act on the same one-to-two-second time scale as the electromechanical swing, and their effect on damping can be either sign. Section 28-5 pursues this.
| Steady-state | Dynamic (small-signal) | Transient | |
|---|---|---|---|
| Disturbance | Infinitesimal, gradual load change | Infinitesimal, but repeated | Large — fault, line trip, load rejection |
| Model | Constant flux linkage, no controls | Machine plus exciter, AVR, governor | Classical model, network switched |
| Equation | Linearised, \(D\) often ignored | Linearised, \(D\) essential | Full nonlinear swing equation |
| Decided by | Sign of \(P_s = P_{max}\cos\delta_0\) | Sign of the real part of the roots | Whether \(\delta\) turns before \(\delta_{max}\) |
| Answer takes the form | A maximum transmissible power | A damping ratio or growth rate | A critical clearing time |
| Time scale | Tens of seconds | 1–20 s | 0.3–2 s, first swing |
The Linearised Swing Equation
Take the swing equation with damping from Chapter 27 and perturb the operating point. Write \(\delta = \delta_0 + \Delta\delta\), where \(\delta_0\) satisfies \(P_m = P_{max}\sin\delta_0\), and expand the sine:
The constant term cancels against \(P_m\), leaving only the increment, and the swing equation becomes an ordinary linear second-order equation in \(\Delta\delta\):
A mass \(2H/\omega_s\), a dashpot \(D/\omega_s\), and a spring \(P_s\). Every conclusion of this chapter's first half is read off this one equation, and the reader who has met the standard second-order system in Chapter 8 of the control course already knows most of what follows.
Put it in standard form by dividing through by the coefficient of the second derivative:
The characteristic equation \(s^2 + 2\zeta\omega_n s + \omega_n^2 = 0\) has roots \(s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}\), and the whole of steady-state stability is contained in where those roots sit. Four cases exhaust the possibilities, and each is a physically distinct machine behaviour.
| Condition | Roots | Behaviour of \(\Delta\delta\) | Verdict |
|---|---|---|---|
| \(P_s>0\), \(D>0\) | Complex, negative real part | Damped oscillation about \(\delta_0\) | Stable |
| \(P_s>0\), \(D=0\) | Pure imaginary, \(\pm j\omega_n\) | Sustained oscillation | Marginal |
| \(P_s>0\), \(D<0\) | Complex, positive real part | Growing oscillation | Dynamically unstable |
| \(P_s<0\) | Real, one positive | Monotonic runaway | Steady-state unstable |
The Steady-State Stability Limit
Imagine loading a generator gradually. The governor admits more steam, \(P_m\) rises in small increments, and after each increment the rotor settles at a slightly larger angle where \(P_{max}\sin\delta\) again equals \(P_m\). Provided the field current is held fixed, \(|E|\) does not change and the machine simply climbs its own power-angle curve.
The climb cannot continue indefinitely. At \(\delta = 90^\circ\) the curve reaches its crest, \(P_s = P_{max}\cos 90^\circ = 0\), and the spring holding the rotor to the network has gone slack. Any further increment of mechanical power cannot be met electrically, the rotor accelerates, \(\delta\) advances past \(90^\circ\), \(P_e\) falls, and the machine runs away. The crest of the power-angle curve is therefore not merely the largest power the machine can transmit; it is the boundary of stable operation.
\(X\) is the total reactance from the internal voltage \(E\) to the receiving-end voltage \(V\), including the machine, its transformer and the line. Operation with \(\delta>90^\circ\) is impossible with fixed excitation, since \(P_s<0\) there. The limit is a property of the machine and the network: the same generator has a different limit on a different line.
Two refinements matter in practice. First, no system is operated at its steady-state limit. A margin is specified, usually as
with \(P_0\) the scheduled output. A margin of \(30\%\) to \(50\%\) is typical, and it exists precisely because the limit above assumes an infinitesimal disturbance, whereas real disturbances are finite.
Second, for a salient-pole machine the reluctance term of Chapter 26 shifts the crest. With \(P_e = \frac{|E||V|}{X_d}\sin\delta + \frac{|V|^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\), the synchronising coefficient is
Setting \(P_s=0\) gives a critical angle below \(90^\circ\) — typically \(70^\circ\) to \(80^\circ\) — but the maximum power itself is higher than the cylindrical-rotor value, because the reluctance term adds to the fundamental over most of the range. Saliency helps the limit and hurts the angle at which it occurs.
Which Reactance, Which Voltage?
The formula \(P_{SSSL}=|E||V|/X\) is unambiguous only once \(E\) and \(X\) are pinned down, and the correct pair depends on what is assumed to remain constant while the load is increased. This is the single point at which steady-state stability calculations go wrong most often.
If the machine has no voltage regulator, the field current is fixed by the operator and the flux is set by it. The voltage that stays constant is the one behind the synchronous reactance, \(E_q\) behind \(X_d\), and the calculation proceeds in two steps: find \(E_q\) from the present operating condition, then compute the limit with the same \(E_q\).
If the machine has a fast automatic voltage regulator, the field current is not fixed at all. The regulator raises it as the load angle grows, holding the terminal voltage — or, in the idealised limit, the voltage behind the transient reactance. Since \(X_d'\) is three or four times smaller than \(X_d\), the effective transfer reactance collapses and the limit rises dramatically. This is why the classical model of Chapter 26 uses \(E'\) behind \(X_d'\).
The idealised extreme is a regulator so fast and so powerful that it holds the terminal voltage \(V_t\) rigidly. The machine then behaves as a constant-voltage source at its own terminals, the transfer reactance is only that of the external network, and the limit rises further still.
| Assumption | Constant voltage | Transfer reactance | Limit | Comment |
|---|---|---|---|---|
| No regulator, fixed field | \(E_q\) behind \(X_d\) | \(X_d + X_e\) | \(|E_q||V|/(X_d+X_e)\) | The classical steady-state limit |
| Fast regulator | \(E'\) behind \(X_d'\) | \(X_d' + X_e\) | \(|E'||V|/(X_d'+X_e)\) | Used for transient studies too |
| Ideal regulator | \(V_t\) at the terminals | \(X_e\) | \(|V_t||V|/X_e\) | Upper bound; the network alone limits |
Dynamic Stability and the Excitation System
The idealisation of a regulator that instantaneously holds a voltage constant is convenient for computing a limit and misleading about everything else. A real excitation system is a feedback loop with gain, time constants and its own dynamics, and it interacts with the electromechanical oscillation of Section 28-2 in a way that took the industry two decades to understand.
The mechanism is this. When the rotor swings forward, the terminal voltage dips, because more current is drawn through the machine and line reactance. The regulator sees the dip and boosts the field. The boost raises \(|E'|\) and hence \(P_e\), which opposes the forward swing — so far so good, and this is the synchronising action that raises the steady-state limit. But the field circuit has a time constant \(T_{d0}'\) of several seconds, so the boost arrives late. By the time the extra flux appears, the rotor may already be swinging back, and the extra power then assists the backward swing instead of opposing it. A torque in phase with the angle is synchronising; a torque in phase with the speed is damping; and a delayed synchronising action delivers a component that is in antiphase with the speed. That component is negative damping.
The result is the third row of the table in Section 28-2: complex roots with a positive real part, and an oscillation at around \(1\) Hz that grows instead of decaying. The system is steady-state stable — \(P_s\) is comfortably positive — yet dynamically unstable.
The remedy is not to slow the regulator down, because its speed is what buys transient stability. The remedy is to add a second input to the exciter that is deliberately proportional to speed deviation, so that its contribution to torque is in phase with speed and therefore purely damping. That auxiliary controller is the power system stabiliser, and it is developed in Chapter 33 alongside the rest of the excitation and frequency control apparatus. Its signal is usually derived from shaft speed, from electrical power, or from a combination that rejects the slow components a governor should handle.
The lesson generalises beyond excitation. Any controller acting on the swing time scale — a governor, an HVDC converter control, a static var compensator — modifies \(D\) as well as \(P_s\), and the sign of its contribution depends on its phase lag at the oscillation frequency. Designing for damping is therefore an exercise in phase compensation at around one hertz, which is exactly the problem the frequency-response methods of a control course are built for.
Transient Stability: The First Swing
Now abandon the linearisation entirely. A three-phase fault appears on a line; the breakers clear it in a few cycles; the question is whether the machine holds. Chapter 26 supplied the three power-angle curves and Chapter 27 supplied the equation of motion. Assembling them gives the sequence of events, and the sequence is always the same.
Before the fault, the machine sits at \(\delta_0 = \sin^{-1}(P_m/P_{max,1})\) on the tall pre-fault curve, in equilibrium.
At fault inception, the transfer reactance jumps and \(P_{max}\) collapses to \(P_{max,2}\). The rotor cannot move instantaneously, so \(\delta\) is unchanged and the operating point drops vertically onto the low curve. Mechanical input now exceeds electrical output by a large margin, and the rotor accelerates.
During the fault, \(\delta\) increases along the low curve. The accelerating power stays positive throughout, because \(P_{max,2}\) is generally smaller than \(P_m\) — for a fault at the sending bus it is zero and the acceleration is uniform, the parabolic case of Chapter 27.
At clearing, the faulted circuit is isolated and the operating point jumps vertically onto the post-fault curve at whatever angle \(\delta_c\) the rotor has reached. If it lands above the \(P_m\) line, the accelerating power is now negative and the rotor begins to decelerate — but it is still turning faster than synchronous speed, so \(\delta\) continues to increase while the excess kinetic energy is given back.
After clearing, one of two things happens. Either the rotor's excess speed reaches zero at some angle \(\delta_{max}\) short of the post-fault unstable equilibrium \(\delta_u = 180^\circ - \sin^{-1}(P_m/P_{max,3})\), in which case the machine swings back and is stable; or \(\delta\) passes \(\delta_u\) with speed still to spare, \(P_e\) falls below \(P_m\) again, and the machine pulls out of step.
Two conditions are hidden in this. There must exist a post-fault equilibrium at all, which requires \(P_{max,3} > P_m\); and the rotor must arrive at \(\delta_u\) with no kinetic energy left. Chapter 29 turns the second condition into an area balance and thereby obtains the critical clearing angle in closed form.
The kinetic-energy language used above is not decoration. Multiply the swing equation by \(d\delta/dt\) and integrate once, and the result is an energy statement: the accelerating power integrated over angle is the kinetic energy stored in the rotor's excess speed. The rotor gains energy while \(P_m>P_e\) and gives it back while \(P_e>P_m\), and stability is the question of whether there is enough angle left over which to give it back. That observation is the equal-area criterion in embryo, and it is the reason damping must be excluded from the model — a dissipative term would break the energy bookkeeping the criterion depends on.
What the Verdict Depends On
Six quantities decide the outcome, and each acts on a different part of the picture. Knowing which lever moves which term is what separates an engineer from a formula.
Clearing time. The dominant factor by a wide margin. A longer fault means more angle accumulated at constant acceleration and more kinetic energy stored, and both grow as \(t_c^2\). Halving the clearing time roughly quarters the energy that must subsequently be dissipated.
Fault type. Chapter 24 ordered the faults by severity of current; stability orders them by severity of power loss, and the order is the same. A three-phase fault shorts the positive-sequence network directly and kills the transfer completely at the fault point. An unsymmetrical fault appears in the positive-sequence network as a shunt impedance — \(Z_2\) for a line-to-line fault, \(Z_2+Z_0\) for a single line-to-ground, \(Z_2\parallel Z_0\) for a double line-to-ground — so some power continues to flow throughout. The single line-to-ground fault is the mildest and by far the commonest; the three-phase fault is the harshest and rarest, and it is the one systems are designed against.
Fault location. A fault at the sending bus gives \(P_{max,2}=0\) and is the worst case. The further along the line the fault, the larger the residual transfer, and the less severe the disturbance.
Inertia. \(H\) appears in the denominator of the acceleration. More inertia means a slower swing and more time for the breakers, but the benefit goes as \(\sqrt H\) in clearing time, not linearly. Inverter-based generation with no rotating mass, discussed in Chapter 39, removes inertia from the system entirely, and that is one of the central problems of the modern grid.
Transfer reactance. It sets all three values of \(P_{max}\). Anything that reduces \(X\) — a second circuit, a higher voltage, series compensation — raises the post-fault curve and increases the angle over which the rotor can decelerate.
Pre-fault loading. A heavily loaded machine starts at a large \(\delta_0\), which leaves less angular room before \(\delta_u\), and has a small \(P_s\), which makes it a weak spring. Both effects push in the same direction, and this is the quantitative reason that transmission limits computed for stability fall as loading rises.
| Factor | Acts on | Direction | Chapter |
|---|---|---|---|
| Clearing time \(t_c\) | Angle and energy accumulated during fault | Shorter is better; effect goes as \(t_c^2\) | 36 |
| Fault type | \(P_{max,2}\) via the shunt at the fault point | LG mildest, LLG, LL, 3-φ harshest | 24 |
| Fault location | \(P_{max,2}\) via the star–delta transfer | Nearer the machine is worse | 26 |
| Inertia \(H\) | Acceleration \(\propto 1/H\) | More is better, as \(\sqrt H\) in \(t_{cr}\) | 27 |
| Transfer reactance \(X\) | All three \(P_{max}\) | Lower is better | 6, 14, 38 |
| Pre-fault loading \(P_m\) | \(\delta_0\) up, \(\delta_u\) down, \(P_s\) down | Lighter is better | 18 |
Improving Stability
Every practical measure for improving stability can be located in the swing equation, and organising them that way makes a long list memorable. There are only three places to intervene: reduce the accelerating power during the fault, increase the decelerating power after it, or increase the inertia that resists both.
Reduce the duration and severity of the disturbance. Fast circuit breakers — two-cycle rather than eight — and fast protective relaying are the most cost-effective stability investment available, and they are the subject of Chapters 35 and 36. Single-pole switching, in which only the faulted phase is opened for a single line-to-ground fault, keeps two phases transmitting throughout. High-speed auto-reclosing restores the healthy circuit after the arc has deionised, so that the post-fault curve becomes the pre-fault curve; since about \(80\%\) of overhead-line faults are transient, this succeeds most of the time.
Raise the post-fault power-angle curve. Anything that lowers the transfer reactance does this. Double-circuit lines and bundled conductors (Chapter 8) reduce \(X\) directly. Series capacitors cancel part of the line reactance and are the most direct method of all, with the caveat of subsynchronous resonance noted in Chapter 38. Higher transmission voltage reduces per-unit reactance for a given power. Machines with lower \(X_d'\) cost more but transmit more.
Act on the machine during the swing. A fast, high-ceiling excitation system boosts \(|E'|\) within a few hundred milliseconds of the fault, lifting the post-fault curve when it is most needed — the transient benefit that Section 28-5 warned must be paid for with a stabiliser. Fast valving momentarily closes the turbine intercept valves to cut \(P_m\), attacking the accelerating power at its source. A braking resistor switched in at the generator bus creates artificial electrical load, doing the same job from the other side. Both act within one swing.
Control the network electronically. Static var compensators hold mid-line voltage during and after the fault, keeping the effective \(P_{max}\) high. HVDC links and FACTS controllers can modulate their power in response to measured speed deviation, injecting damping directly. Chapter 38 develops these.
| Measure | Acts on | Time scale | Chapter |
|---|---|---|---|
| Fast breakers and relays | Shortens the accelerating period | Cycles | 35, 36 |
| Single-pole switching | Keeps \(P_{max,2}\) high for LG faults | Cycles | 36 |
| High-speed auto-reclosing | Restores \(P_{max,3}\) to \(P_{max,1}\) | 0.3–1 s | 36 |
| Series capacitors | Lowers \(X\), raises all \(P_{max}\) | Permanent | 38 |
| Additional circuits, higher voltage | Lowers \(X\) | Permanent | 2, 14 |
| Fast, high-ceiling excitation | Raises \(|E'|\) and \(P_{max,3}\) | 0.1–0.5 s | 33 |
| Power system stabiliser | Restores positive \(D\) | Continuous | 33 |
| Fast valving | Reduces \(P_m\) | 0.2–1 s | 33 |
| Braking resistor | Raises \(P_e\) during the swing | 0.1–0.5 s | 38 |
| SVC, HVDC modulation | Holds voltage, adds damping | Cycles onward | 38 |
| Larger \(H\), or synthetic inertia | Slows the swing | Design stage | 27, 39 |
The Modern Classification
The threefold division into steady-state, dynamic and transient stability is the classical one and it is what examinations ask for. It is organised by the size of the disturbance, and it takes for granted that the thing at risk is synchronism. Both assumptions have been overtaken by events.
Since the 1990s, following a series of collapses that were plainly not losses of synchronism, the industry has classified stability by which variable fails to hold rather than by how hard the system was pushed. Three families are recognised.
Rotor angle stability is everything this Part has discussed: the ability of synchronous machines to remain in step. It subdivides into small-disturbance (the steady-state and dynamic problems of this chapter) and large-disturbance or transient stability.
Voltage stability is the ability of the system to hold acceptable voltages at all buses after a disturbance. Its failure mode is a progressive, irreversible voltage decline driven by reactive-power shortage and by load restoration through tap changers and thermostats — a mechanism with no rotor angle in it at all. It is the subject of Chapter 34.
Frequency stability is the ability of the system to hold frequency after a severe imbalance between generation and load, typically following islanding. It is governed by the aggregate inertia and by the response of governors and load shedding, and it belongs to Chapter 33.
The classical names map onto the new scheme cleanly enough: steady-state and dynamic stability are both small-disturbance rotor angle stability, and transient stability keeps its name. What the new scheme adds is the recognition that a system can be perfectly synchronised and still collapse.
Worked Examples
Problem. A generator of synchronous reactance \(X_d = 1.20\) pu delivers \(0.60\) pu at unity power factor to an infinite bus of \(1.0\angle0^\circ\) pu through an external reactance of \(0.40\) pu. The field current is held constant. Find the internal voltage, the load angle, the steady-state stability limit, and the stability margin.
Solution. With constant field current the voltage that stays fixed is \(E_q\) behind the synchronous reactance, so the whole reactance from that node to the infinite bus is \(X = 1.20+0.40 = 1.60\) pu. Unity power factor at the infinite bus gives \(I = 0.60\angle0^\circ\) pu.
The machine is operating at \(43.8^\circ\), roughly halfway to the crest, with a \(44\%\) margin — an ordinary, comfortable condition. Note how large \(\delta_0\) already is: a synchronous reactance above one per unit means that even a modest load produces a substantial angle, which is precisely why the classical steady-state limit is so much lower than what the machine can actually be made to deliver.
Problem. The machine of Example 1 has \(X_d' = 0.30\) pu. Recompute the stability limit (a) assuming a fast regulator that holds the voltage behind \(X_d'\) constant, and (b) assuming an ideal regulator that holds the terminal voltage rigid. Compare all three.
Solution. Nothing about the operating point changes; only the node at which the constant voltage is assumed to sit.
| Assumption | Constant voltage | \(X\) (pu) | Limit (pu) | Margin on 0.6 pu |
|---|---|---|---|---|
| Fixed field | \(|E_q|=1.386\) | 1.60 | 0.866 | 44% |
| Fast regulator | \(|E'|=1.085\) | 0.70 | 1.550 | 158% |
| Ideal regulator | \(|V_t|=1.028\) | 0.40 | 2.571 | 329% |
The limit rises by a factor of \(1.79\) with a fast regulator and by \(2.97\) with an ideal one, and every bit of the gain comes from the shrinking denominator — the constant voltage itself falls as the node moves toward the terminals. This is the calculation behind the statement of Section 28-4 that a regulated machine may be operated past \(\delta=90^\circ\) as measured to the node behind \(X_d\). With the field frozen at its Example 1 value the machine simply could not deliver \(1.55\) pu at all, since the whole fixed-field curve crests at \(0.866\); under regulation the field rises as the load does, \(|E_q|\) grows with it, and the angle to that internal node passes \(90^\circ\) without any loss of synchronism, because \(E_q\) is no longer the quantity being held constant.
Problem. A \(50\) Hz machine has \(H = 4.0\) s, operates with \(P_s = 1.8\) pu per electrical radian, and has a damping coefficient \(D = 6.0\) pu. Find \(\omega_n\), \(\zeta\), the damped frequency, the decay time constant, and the response of \(\delta\) to a sudden increase of \(0.05\) pu in mechanical input.
Solution. The inertia coefficient is \(M = 2H/\omega_s = 8/314.16 = 0.025465\) s²/elec rad.
For the step response, the new steady state is where the extra mechanical power is matched by the extra electrical power the larger angle produces:
The rotor overshoots its new equilibrium by \(87\%\) — nearly doubling the final displacement on the first swing — then rings at \(1.34\) Hz with the envelope decaying with a \(2.67\) s time constant, so the disturbance is essentially over after about \(11\) s. That near-\(100\%\) overshoot is characteristic of every lightly damped electromechanical system and explains why a machine's transient excursion is always roughly twice its steady change.
Problem. The excitation system of the machine in Example 3 is retuned for higher gain, and the net damping coefficient becomes \(D = -1.0\) pu while \(P_s\) is unchanged at \(1.8\). Find the characteristic roots and describe the behaviour. How long does an oscillation take to double in amplitude?
Solution. The characteristic equation is unchanged in form; only the sign of the middle coefficient has flipped.
The synchronising coefficient is untouched, so the machine is perfectly steady-state stable in the sense of Section 28-3 — it holds its operating point against a static load change. Yet any disturbance sets off a \(1.34\) Hz oscillation whose amplitude doubles every eleven seconds; within a minute an initial one-degree swing has grown past forty degrees and the machine trips. This is dynamic instability, it is invisible to any calculation that only checks \(P_s>0\), and it is exactly what the power system stabiliser of Chapter 33 exists to prevent.
Problem. A generator with \(X_d' = 0.25\) pu and internal voltage \(|E'| = 1.25\) pu feeds an infinite bus at \(1.0\) pu through a transformer of \(0.10\) pu and a double-circuit line of \(0.40\) pu per circuit. It delivers \(P_m = 0.90\) pu. A three-phase fault occurs at the midpoint of one circuit and is cleared by opening that circuit. Find the three power-angle curves, the initial angle, the electrical output the instant the fault strikes, and the post-fault equilibrium and unstable angles.
Solution. Pre-fault: the two circuits in parallel give \(0.20\) pu.
During fault: the fault point is a third node grounded at the middle of one circuit. The three star arms meeting at the sending bus run to the machine node (\(0.25+0.10 = 0.35\)), to the infinite bus through the healthy circuit (\(0.40\)), and to the fault (\(0.20\)). Apply the star–delta transformation of Chapter 26:
The instant the fault strikes, electrical output falls from \(0.90\) to \(0.341\) pu while mechanical input stays at \(0.90\), so \(0.559\) pu of accelerating power appears at once. The rotor has \(147.32^\circ - 23.33^\circ = 123.99^\circ\) of angular room in which to store and then return that energy, and a post-fault curve whose crest of \(1.667\) pu comfortably exceeds \(P_m\) — so a post-fault equilibrium exists and the machine is stable provided the fault is cleared quickly enough. Chapter 29 converts "quickly enough" into a number.
Problem. The system of Example 5 is loaded to \(P_m = 1.80\) pu instead of \(0.90\) pu. Show that clearing the fault by opening one circuit cannot restore stability at any clearing time, and find the degree of series compensation on the remaining circuit that would restore an equilibrium.
Solution. The pre-fault operating point still exists:
But the post-fault curve peaks at \(P_{max,3}=1.667\) pu, which is less than the mechanical input of \(1.80\) pu. The equation \(1.80 = 1.667\sin\delta\) has no solution: at every angle the accelerating power is positive, the rotor never decelerates, and no clearing time however short can help. The machine loses synchronism.
Series compensation of the surviving circuit to a degree \(k\) replaces its reactance \(0.40\) by \(0.40(1-k)\). Requiring the post-fault crest to exceed the input:
Fourteen per cent compensation restores an equilibrium, but only barely — the post-fault curve would then peak at exactly the input power, with \(\delta_{eq}=90^\circ\) and \(P_s=0\), which is useless. Taking a practical \(40\%\) compensation:
Now a genuine post-fault equilibrium exists with \(63.7^\circ\) of decelerating room beyond it, and the transient stability question becomes meaningful again. The example shows the two-stage nature of every stability assessment: first check that a post-fault equilibrium exists at all, and only then ask how fast the breakers must be.
Chapter Summary
Small disturbance → linearise; large disturbance → integrate the nonlinear equation.
\(\dfrac{2H}{\omega_s}\Delta\ddot\delta + \dfrac{D}{\omega_s}\Delta\dot\delta + P_s\Delta\delta = \Delta P_m\).
\(\omega_n=\sqrt{\omega_sP_s/2H}\), typically \(0.5\)–\(2\) Hz; \(\zeta\) only a few per cent.
\(P_s\) decides whether it oscillates; \(D\) decides whether the oscillation survives.
\(P_{SSSL}=|E||V|/X\) at \(\delta=90^\circ\), where \(P_s\) vanishes.
\(E_q\)/\(X_d\) with fixed field, \(E'\)/\(X_d'\) with a fast regulator, \(V_t\)/\(X_e\) ideally.
More synchronising torque, less damping torque — hence the stabiliser.
Stable if the swing turns before \(\delta_u = 180^\circ-\sin^{-1}(P_m/P_{max,3})\).
No post-fault equilibrium unless \(P_{max,3}>P_m\); no clearing time can rescue that.
Clearing time, fault type, fault location, \(H\), \(X\), and pre-fault loading.
Cut the fault short, raise the post-fault curve, or reduce \(P_m\) during the swing.
Rotor angle, voltage and frequency stability — a system can synchronise and still fall.
Practice Problems
Take \(f = 50\) Hz and work in per unit on a common base. State clearly, in every steady-state stability problem, which voltage you are holding constant and why.
- A generator of \(X_d = 1.0\) pu supplies \(0.80\) pu at \(0.95\) power factor lagging to an infinite bus at \(1.0\) pu through a total external reactance of \(0.35\) pu. With the field current fixed, find \(E_q\), the load angle, the steady-state stability limit and the percentage margin.
- Repeat Problem 1 with \(X_d' = 0.28\) pu and a fast regulator holding \(E'\) constant. By what factor does the limit increase, and what is the new load angle measured to \(E'\)?
- A machine with \(H = 3.0\) s operates at \(\delta_0 = 40^\circ\) on a curve with \(P_{max}=2.2\) pu. Find \(P_s\), \(\omega_n\) in hertz, and the period of the natural oscillation. If \(D = 4.0\) pu, find \(\zeta\) and the time for the envelope to decay to \(10\%\) of its initial value.
- For the machine of Problem 3, the load is now raised until \(\delta_0 = 80^\circ\). Recompute \(P_s\) and \(\omega_n\), and comment on what has happened to both the stiffness and the speed of response.
- An excitation system contributes \(-2.5\) pu of damping while the machine and load contribute \(+1.5\) pu. With \(H = 5\) s and \(P_s = 1.6\) pu/rad, find the characteristic roots and the time for an oscillation to grow by a factor of ten. What magnitude of stabiliser damping contribution would be needed to achieve \(\zeta = 0.05\)?
- A generator with \(|E'| = 1.20\) pu and \(X_d' = 0.20\) pu feeds an infinite bus at \(1.0\) pu through a transformer of \(0.12\) pu and two parallel lines of \(0.50\) pu each, delivering \(1.0\) pu. A three-phase fault occurs at the sending end of one line. Find \(P_{max,1}\), \(P_{max,2}\), \(P_{max,3}\), \(\delta_0\), \(\delta_{eq}\) and \(\delta_u\).
- Repeat Problem 6 for a fault at the midpoint of one line, using the star–delta transformation, and state which of the two faults is the more severe and by how much in terms of \(P_{max,2}\).
- For the system of Problem 6, find the minimum degree of series compensation on the surviving line that would allow the machine to carry \(1.5\) pu with a post-fault synchronising coefficient of at least \(0.8\) pu per electrical radian.