Power System Stability

Start with the mechanical law. For a rotating body of moment of inertia \(J\) at mechanical angular position \(\theta_m\):

\(T_m\) is the mechanical torque from the turbine, \(T_e\) the electromagnetic torque opposing it, and \(T_a\) the accelerating torque. Nothing here is peculiar to electrical machines — it is the rotational form of \(F = ma\).

Measure the angle relative to a synchronously rotating reference, because \(\theta_m\) itself grows without limit and tells us nothing:

The synchronous term differentiates away twice, so the equation is unchanged in form but \(\delta_m\) is now a small, bounded quantity — the rotor's departure from where it would be if nothing had happened. This is the single step that makes stability analysis tractable.

Convert torque to power by multiplying through by \(\omega_m\):

Strictly \(\omega_m\) varies during the transient, but it never departs from synchronous speed by more than a per cent or two even in a severe swing, so replacing it by the constant \(\omega_{sm}\) is an excellent approximation and the standard one.

Define the inertia constant \(M = J\omega_{sm}\), the angular momentum at synchronous speed:

Units of \(M\) are joule-seconds per radian, or equivalently MJ·s/rad when the powers are in MW.

Change to electrical angle. For a machine with \(p\) poles, electrical and mechanical angles are related by \(\delta = (p/2)\delta_m\):

Electrical angle is the useful variable because the power-angle relation \(P_e = P_{\max}\sin\delta\) is written in it. From here on \(M\) is understood to be per electrical radian and the pole factor is absorbed.

The final form, written per unit on the machine's own rating with the \(H\) constant defined in Problem 2:

Or compactly \(M\ddot\delta = P_a\) with \(M = 2H/\omega_s = H/(\pi f)\). Every stability calculation in this chapter is an application of this one second-order equation.

Its character. Written as a system of two first-order equations:

This is the equation of a pendulum — a mass in a sinusoidal potential well, driven by a constant torque. Every property of transient stability has a pendulum analogue: a stable equilibrium, an unstable one, a separatrix, and a critical energy above which the pendulum goes over the top instead of swinging back.

What has been assumed, and it should be stated plainly:

All four are good for the first swing, which lasts roughly half a second and is where most instability occurs. For multi-swing studies over several seconds the exciter, governor and flux decay must be modelled, and the classical model is abandoned.

The definition. \(H\) is the stored kinetic energy at synchronous speed divided by the machine's rating:

The units deserve a moment. MJ/MVA is MJ per MW, which is seconds — \(H\) is the time the machine could supply its rated output from its stored kinetic energy alone. That physical reading is why \(H\), not \(J\), is the constant engineers quote.

The relation to \(M\):

For the machine of this chapter, \(H = 5.0\) at 50 Hz gives \(M = 5.0/157.08 = 0.031831\) pu·s²/electrical radian. Note that \(M\) is frequency-dependent while \(H\) is not — one more reason \(H\) is the quantity tabulated.

Typical values, which are worth knowing because they vary by a factor of five across machine types:

The last row is the defining problem of modern power systems: converter-interfaced generation contributes no natural inertia at all, so as it displaces synchronous plant the system's total \(H\) falls and every swing becomes faster.

Combining two machines that swing together — for example two units on a common bus. Kinetic energies add:

So on a common base \(S_{\text{base}}\), the equivalent inertia is \(H_{\text{eq}} = (H_1S_1+H_2S_2)/S_{\text{base}}\). Two 100 MVA machines with \(H = 5\) each become a single 200 MVA machine with \(H = 5\), or equivalently \(H = 10\) on a 100 MVA base.

Combining two machines that swing against each other is different, and the formula surprises people:

The reduced-mass formula from two-body mechanics, and for exactly the same reason. Note the consequence: a small machine swinging against a very large one has \(M_{\text{eq}} \approx M_{\text{small}}\) — which is precisely the infinite-bus idealisation used in this chapter.

Changing base. \(H\) is quoted on the machine's own rating, so a system study must convert:

A 250 MVA machine with \(H = 4.0\) on its own base becomes \(H = 4.0\times250/100 = 10.0\) on a 100 MVA system base. Forgetting this conversion is the commonest single error in a first stability study, and it changes every swing time by the square root of the ratio.

The current between the two sources:

Resistance is neglected throughout. On a transmission network \(X/R\) is 5–15, and within a machine it is far higher, so this is a very good approximation — and it is also the conservative one, since resistance provides damping.

The complex power leaving the sending source:

Taking the real part and using \(\angle\delta = \cos\delta+j\sin\delta\):

The result:

Real power depends on the angle between the sources; reactive power on the difference in their magnitudes. That separation is the single most useful fact in power system operation, and it is exact only when \(R = 0\).

Reading the three quantities in a stability context:

The whole of transient stability lives in the third row. A fault, a breaker opening, an auto-reclose — each changes \(X\) discontinuously, and therefore \(P_{\max}\) discontinuously, while \(\delta\) and \(\dot\delta\) must remain continuous because the rotor has mass.

Why \(\delta\) cannot jump is worth dwelling on, because it is what makes the analysis possible:

A finite torque cannot change a finite angular momentum instantaneously. So at each switching instant the operating point moves vertically from one power-angle curve to another at fixed \(\delta\) — which is exactly the geometry the equal-area criterion exploits.

The salient-pole refinement, for completeness:

The second term is reluctance power, arising because a salient rotor has a preferred orientation even with no field current. It peaks at \(45°\) and typically adds 10–20% to \(P_{\max}\), shifting the peak of the total curve below \(90°\). Round-rotor machines have \(X_d = X_q\) and the term vanishes; this chapter uses the round-rotor form throughout.

The prefault transfer reactance, with both circuits in service:

The two circuits are in parallel, so the line contributes half of one circuit's reactance. Note that the machine's own \(x'_d\) is the single largest element — half the total — which is typical and is why generator reactance dominates stability.

The prefault power-angle curve:

A peak of 200 MW against a rating of 100 MVA — the machine has ample steady-state margin, as a well-designed unit should.

The initial operating point:

Exactly 30°, which will keep the arithmetic clean throughout. A rotor angle of 30° at full output is comfortable; operating angles above 45° are usually a sign that the system is stressed.

Check the initial condition against the reactive flow, to confirm \(|E'| = 1.20\) is consistent:

A small lagging reactive export — a plausible operating condition, not an artificial one. The machine is delivering 100 MW and 6.5 MVAr.

The post-fault curve, after one circuit has been tripped:

Losing one of two circuits raises the transfer reactance by a third and cuts the peak power by a quarter. The new equilibrium angle would be \(\arcsin(1.0/1.5) = 41.81°\) — the machine must sit further out to push the same power through a weaker network.

The during-fault curve depends on where the fault is, and is computed in Problem 11. For a three-phase fault at the midpoint of one circuit:

The three curves — 2.000, 0.750, 1.500 — are the whole input to every equal-area calculation that follows.

The inertia constant in working form:

Which sets the timescale of everything. A useful feel for the number: an unbalance of 1.0 pu accelerates the rotor at \(1/M = 31.4\) rad/s², so it covers 30° (0.524 rad) in \(\sqrt{2\times0.524/31.4} = 0.183\) s. First swings last a fraction of a second, which is why breakers must clear in a few cycles.

Summary of the study system:

Every subsequent problem in this set refers to this table.

Perturb the equilibrium. Let \(\delta = \delta_0 + \Delta\delta\) with \(\Delta\delta\) small, and linearise:

The first term is the equilibrium power, which cancels \(P_m\). The coefficient of \(\Delta\delta\) is what matters.

The synchronising power coefficient:

It is a stiffness: the restoring power produced per radian of angular displacement. The linearised swing equation becomes \(M\Delta\ddot\delta + P_s\Delta\delta = 0\), the equation of a spring-mass system.

The stability condition follows immediately:

Beyond 90° an increase in angle reduces the transferred power, so the rotor accelerates further — a positive feedback. The machine loses synchronism without any disturbance beyond the infinitesimal.

The steady-state stability limit is therefore the peak of the curve:

With a modern fast exciter the effective \(E'\) is regulated upward as the angle grows, and the practical limit is higher — this is the dynamic stability limit. The classical figure below is the conservative one.

For the study system, intact:

A 1° displacement produces 3.0 MW of restoring power. Comfortable.

With one circuit out:

The stiffness falls by 35% and the margin halves. The system is still stable, but every subsequent disturbance will produce a larger and slower swing.

The two limits compared:

The last row is the crucial one: during the fault \(P_{\max} = 0.750 < P_m = 1.000\), so no equilibrium exists at all. The rotor must accelerate for as long as the fault is on the system, whatever the angle. That is why fault clearing time, and not steady-state margin, determines transient stability.

The linearised swing equation from Problem 5:

Simple harmonic motion, with \(M\) playing the part of mass and \(P_s\) that of spring constant. The solution is a pure oscillation because damping has been neglected.

The natural angular frequency:

The last form is the one usually quoted, and shows the two design levers explicitly: stiffness in the numerator, inertia in the denominator.

Numerically for the study system:

Just over one hertz. This is the characteristic local mode frequency of a single machine against a strong system, and real plant oscillations of this kind are observed at 0.8–2.0 Hz almost universally.

The two standard oscillation modes in a real network, for context:

Inter-area modes are slower because the equivalent inertia is much larger and the tie reactance between areas is high, giving small \(P_s\). They are also the hardest to damp, and are the reason power system stabilisers exist.

Dependence on loading. As the machine is loaded up, \(\delta_0\) grows and the stiffness falls:

The frequency falls to zero at the steady-state stability limit — the oscillation becomes infinitely slow and then, beyond 90°, ceases to be an oscillation at all and becomes an exponential runaway. Watching the local-mode frequency fall is one practical way of detecting an approach to the limit.

Dependence on inertia. Since \(\omega_n \propto 1/\sqrt{H}\):

Halving the inertia raises the frequency by \(\sqrt2\). As synchronous plant is displaced by converter-interfaced generation, system inertia falls and every electromechanical mode speeds up — which shortens the time available for protection and control to act.

Adding damping, which the classical model omits. With a damping power \(P_D = D\,\Delta\dot\delta\):

Damper windings, load frequency-sensitivity and — above all — a properly tuned power system stabiliser provide \(D\). A well-damped local mode has \(\zeta = 0.05\)–\(0.15\); below 0.03 the mode is considered inadequately damped. Note that a fast, high-gain exciter without a stabiliser can make \(D\) negative, which is the historical reason stabilisers were invented.

Multiply the swing equation by \(2\dot\delta\). This is the standard trick for getting a first integral of an undamped second-order equation:

The left side is the exact derivative of \((d\delta/dt)^2\) — which is why the factor of two was chosen.

Integrate once:

using \(\dot\delta = 0\) at \(\delta = \delta_0\), since the machine starts in equilibrium. This is an energy statement: the left side is proportional to the rotor's kinetic energy relative to synchronous speed, the right to the net work done on it.

The criterion. The rotor returns to synchronism if and only if \(\dot\delta\) comes back to zero at some angle \(\delta_{\max}\) before the unstable equilibrium is passed:

Splitting the integral at the clearing angle \(\delta_c\), where \(P_m > P_e\) before and \(P_e > P_m\) after:

In the familiar two-area form:

Areas on a power–angle diagram have units of energy per unit inertia. The criterion says: the energy the rotor gains while the fault is on must be given back before the angle reaches the point of no return.

The point of no return is the unstable equilibrium of the post-fault system:

Beyond this angle the post-fault curve has fallen below \(P_m\) again, so the rotor re-accelerates and never returns. For the study system, \(\delta_{\max} = 180° - \arcsin(1.0/1.5) = 138.19°\).

What the criterion tells you:

The fourth row is the practical limitation: a breaker is specified in cycles, not degrees, so every equal-area result must be converted to time by Problem 13's method. The last row is the fundamental one — the criterion depends on having a single first integral, which exists only for one machine against an infinite bus, or two machines reduced to one.

The pendulum reading, which makes the criterion obvious rather than clever:

A pendulum pushed from rest goes over the top if the work done on it exceeds the potential barrier. That is all the equal-area criterion says — and the reason it is exact, not approximate, is that the undamped swing equation conserves this energy exactly.

The situation. Only one power-angle curve is involved, \(P_e = 2.000\sin\delta\), but the horizontal line \(P_m\) jumps upward:

The rotor cannot move instantaneously, so it finds itself delivering 1.000 pu while receiving \(P_n\). It accelerates.

The new equilibrium and the limit angle:

\(\delta_1\) is where the rotor would settle if it could get there smoothly; \(\delta_2\) is the unstable equilibrium beyond which it cannot recover. Note the rotor does not stop at \(\delta_1\) — it arrives there at full speed and overshoots.

The equal-area condition for the critical step, taking \(\delta_{\max} = \delta_2\):

One transcendental equation in \(P_n\), since \(\delta_2\) depends on it. Solve numerically.

Iterating:

The critical new load is 1.7393 pu, so the largest survivable step is \(1.7393-1.000 = \) 0.7393 pu — a 74% increase in one instant.

Compare the steady-state limit. The machine could carry 2.000 pu if brought there gradually, but only 1.7393 pu if the load is applied in a single step:

The difference is entirely the overshoot. Energy gained accelerating from 30° to 60.42° must be given back between 60.42° and 119.58°, and if the step is larger there is not enough area left.

A smaller step, to see the swing. For \(P_n = 1.500\):

The machine settles at \(\delta_1 = 48.59°\) eventually, but on the way it swings out to 69.88° — overshooting the new equilibrium by 21.29°, slightly more than the 18.59° by which the equilibrium itself moved. With no damping it would oscillate between 30° and 69.88° forever; real damping brings it to rest at 48.59° in a few seconds.

Why this matters practically. Sudden load increases of this size do occur:

And the governor cannot help on this timescale: a steam governor takes several seconds to respond, while the first swing is over in half a second. On the first swing the machine is on its own.

The event. A single instantaneous change of curve, with no during-fault period at all:

The operating point drops vertically from 1.000 pu to \(1.500\sin30° = 0.750\) pu. The machine is now delivering less than it receives, so it accelerates.

The immediate accelerating power:

A modest acceleration, and it falls as the rotor advances and \(P_e\) rises.

The new equilibrium:

The rotor must move out by 11.81° to restore balance — and will overshoot.

Find the maximum swing by equating the areas on the single post-fault curve:

Solving: \(\delta_{\max} = 0.94950\) rad \(= \) 54.40°.

The verdict:

The rotor swings out to 54.4°, comes back, and with damping settles at 41.81°. The peak overshoot beyond the new equilibrium is \(54.40-41.81 = 12.59°\), slightly more than the 11.81° by which the equilibrium itself moved — the classic near-doubling of a step response in a lightly damped second-order system.

A margin measure. The available decelerating area beyond the swing:

Twenty-one times more area available than was used. Losing a circuit cleanly is not remotely a stability threat at this loading — it is the fault that precedes the switching that causes the problem, and that is what the rest of this set addresses.

How heavily could the machine be loaded and still survive a clean circuit loss? Set \(\delta_{\max} = \delta_2\):

Solving gives \(P_m = 1.381\) pu. So the machine can carry up to 138 MW and still survive a clean switching event — but this is also, as Problem 18 shows, the absolute ceiling for surviving any fault, even one cleared instantaneously.

The during-fault curve. A solid three-phase fault at the HV bus holds that bus at zero volts:

No power whatever can reach the infinite bus, because everything the machine produces is short-circuited to earth at the sending end. This is the most severe possible fault for stability — nothing else drives \(P_e\) lower than zero.

Consequently the acceleration is constant:

A constant, so the swing is a simple parabola in time — which is the reason this case, alone among the fault locations, has an exact closed-form clearing time.

The equal-area condition. With \(P_2 = 0\) the general formula simplifies:

The \(P_2\cos\delta_0\) term vanishes and \(P_3-P_2 = P_3\).

Substituting \(P_m = 1.000\), \(P_3 = 1.500\), \(\delta_0 = 30.00° = 0.52360\) rad, \(\delta_{\max} = 138.19° = 2.41187\) rad:

The rotor may advance by 29.1° before the breaker must open. Any further and the remaining decelerating area is insufficient.

The clearing time, from the parabola:

Which is 9.0 cycles at 50 Hz. Modern transmission breakers clear in 2–3 cycles with a protection operating time of 1–2 cycles, so total clearing of 4–5 cycles is achievable and the system has roughly a factor-of-two margin.

Verifying the areas at \(\delta_c = \delta_{cc}\), as an independent check:

Equal to five decimal places, computed by a completely different route from the formula. The critical clearing angle is confirmed.

Why other fault locations need numerical integration. With \(P_2 \ne 0\) the swing equation during the fault is:

whose solution involves elliptic integrals — the pendulum equation again. The angle is still available in closed form from the equal-area criterion, because that only requires integrating \(P\,d\delta\); the time requires integrating \(d\delta/\dot\delta\), which is where the difficulty lies. Problem 13 handles it numerically.

The faulted network. Label the machine node \(E'\), the high-voltage bus \(A\), the fault point \(F\) and the infinite bus \(B\):

Both circuits are still connected — the breakers have not yet operated. The fault node \(F\) is held at zero volts, and it is that constraint which throttles the power transfer.

Apply the star-delta transformation to the star centred on node \(A\), whose three arms go to \(E'\) (0.40), \(F\) (0.20) and \(B\) (0.40):

The delta arms are \(\Sigma\) divided by the opposite star arm — the standard formula, and the whole calculation is three divisions.

The three delta branches:

Note which arm produced which: \(X_{E'B}\) is divided by \(X_{AF} = 0.20\), the arm not connecting \(E'\) or \(B\). That is why a fault close to the bus (small \(X_{AF}\)) gives a large \(X_{E'B}\) and therefore a severe fault.

Now discard two of them. Node \(F\) is earthed, so \(X_{E'F}\) and \(X_{FB}\) are shunt paths to earth:

Shunt branches carry current away but do not affect the transfer between two ideal voltage sources. This is the whole point of the star-delta step: it isolates the transfer path from the shunt paths so the earthed node can simply be deleted.

The during-fault curve:

The transfer capability has fallen to 37.5% of its prefault value. Critically, \(0.750 < P_m = 1.000\), so no equilibrium exists during the fault at any angle — the rotor accelerates continuously until the fault is cleared.

The three curves together, which is the complete input to Problem 12:

The pattern \(P_1 > P_3 > P_2\) is universal for a line fault: clearing the fault by tripping the faulted circuit improves matters even though it weakens the network, because removing a short circuit helps far more than removing a circuit hurts.

A useful shorthand for a fault anywhere along a circuit at fractional distance \(\alpha\) from the sending bus, with \(a = x'_d+x_T\):

Checking at \(\alpha = 0.5\): \(0.40+0.40+0.40/0.5 = 0.80+0.80 = 1.60\) ✓. As \(\alpha\to0\) the last term diverges — a fault at the sending bus gives infinite transfer reactance and zero power, exactly as Problem 10 assumed.

Derive the general formula rather than quoting it. Equating the two areas with three different curves:

\(P_2\) applies before clearing, \(P_3\) after. Both integrals are elementary.

Evaluate both sides:

Collect the \(\cos\delta_c\) terms on one side and everything else on the other.

The result:

Angles in radians in the first term, since it came from integrating \(d\delta\). Forgetting that is the single commonest arithmetic error in this subject.

Assemble the numbers:

\(\delta_{\max} = \pi - \arcsin(1.000/1.500) = \pi - 0.72973 = 2.41186\) rad.

Substitute:

The rotor may advance 50.74° from its initial 30° before the breaker must open — far more room than the 29.1° available for a sending-end fault, because the mid-line fault still lets 0.750 pu through.

Verify by computing the areas separately. First the accelerating area, from 0.52360 to 1.40914 rad:

Positive, as it must be — the rotor gains energy.

Then the decelerating area, from 1.40914 to 2.41186 rad:

Equal to five decimal places, computed by an entirely independent route ✓. The critical clearing angle is confirmed.

What happens on either side of it:

Note that at the critical angle the rotor swings all the way to the unstable equilibrium and, in the undamped model, sits there. In reality any damping brings it back and any perturbation pushes it over — which is why practical clearing times carry a margin of at least 30% on the critical value.

The problem. During the fault the swing equation is:

Nonlinear in \(\delta\). Separating variables gives \(t = \int d\delta/\dot\delta\) with \(\dot\delta = \sqrt{(2/M)\int(P_m-P_2\sin\delta)d\delta}\) — an elliptic integral, which is why no elementary closed form exists. Only the \(P_2 = 0\) case of Problem 10 collapses to a parabola.

The point-by-point method, which is how this was done before computers and is still how it is understood. Use the modified Euler scheme with step \(h\):

with \(\omega = \dot\delta\) the slip in electrical rad/s and \(P_a = P_m - P_2\sin\delta\). A step of \(h = 0.05\) s was the classical hand choice; \(h = 0.01\) s or less is used numerically.

The first few steps by hand, with \(h = 0.05\) s, \(\delta_0 = 0.52360\) rad, \(\omega_0 = 0\):

The accelerating power falls steadily as the rotor advances into a region where the during-fault curve delivers more — which is exactly why the swing is slower than the parabola of the sending-end case.

The answer:

Against 0.1798 s (9.0 cycles) for the sending-end fault. The mid-line fault gives the machine almost twice as long, and both are comfortably above the 4–5 cycles a modern protection-plus-breaker scheme achieves.

Compare with the crude estimate that treats \(P_a\) as constant at its initial value:

8% short — the estimate is conservative because it ignores the fact that \(P_a\) decreases as the rotor swings. Conservative in the safe direction, so it is a legitimate quick check, but not a substitute for integration.

Sensitivity to step size, which matters if you are doing this by hand:

Second-order accuracy — halving the step quarters the error, and the scheme errs on the short side, which is the safe direction for a breaker specification. A 0.05 s step is entirely adequate for engineering purposes, which is why the hand method survived so long. Note that the classical scheme must be restarted with an averaged \(P_a\) at each switching instant, or the discontinuity introduces a first-order error.

Putting it in context. Total fault clearing time is protection time plus breaker time:

Even the older scheme clears this fault in time. Stability becomes binding only at high loading (Problem 18), with low inertia (Problem 17), or for a sending-end fault on a heavily loaded weak system — which is precisely where transient stability studies concentrate.

The during-fault transfer reactance from the star-delta result of Problem 11, with the fault at fraction \(\alpha\) along the circuit:

As \(\alpha\) grows the denominator grows and \(X_{\text{fault}}\) falls, so more power gets through and the fault is milder.

Worked at \(\alpha = 0.25\):

A quarter of the way along the line, only 25% of the prefault transfer survives — against 37.5% at the midpoint.

The complete comparison:

Both the critical angle and the critical time increase monotonically with distance from the machine. The clearing time available more than doubles between a bus fault and a fault three-quarters of the way along.

Why the trend runs this way. Two effects reinforce each other:

A fault far from the machine is separated from it by line reactance, which acts as a buffer. A fault at the machine's own terminals is separated by nothing at all.

The design consequence is that protection must be specified for the worst location, which is always the closest one:

And this is where distance protection creates a difficulty. A zone-1 element covers 80–85% of the line and operates instantaneously; the remaining 15–20% is cleared in zone 2, typically after a 0.3–0.4 s delay. For a fault at the remote end that delay is tolerable because \(t_{cc}\) is large there — but the arrangement must be checked, and this is exactly the check.

The remote-end check. A fault at \(\alpha = 0.85\) cleared in zone 2 at 0.35 s:

The zone-2 delay is acceptable at this loading. At higher loading it would not be, which is why heavily loaded lines are given a communications-aided scheme that clears the whole line instantaneously.

The one exception to "closer is worse" deserves a note. A fault behind the machine — on its own busbar or in its unit transformer — is cleared by tripping the machine itself, so the stability question does not arise in the same form. The severity ranking above applies to faults on the transmission system that the machine must ride through.

Similarly, on a radial line with no parallel circuit, tripping the faulted line disconnects the machine entirely and stability is meaningless. Both cases are handled by different criteria.

The key idea. An unsymmetrical fault does not short the positive-sequence network to earth. It connects a shunt impedance there, whose value is built from the negative- and zero-sequence networks seen at the fault point:

These come directly from the network interconnections of Set 23 — read from the positive-sequence terminals, each fault type simply presents whatever the other two networks look like from there.

The negative-sequence network at the fault point. From \(F\), one path goes through half the faulted circuit to bus \(A\) and thence to the machine or the healthy circuit; the other goes directly to the infinite bus:

The 0.30 is the machine's \(x_2 = 0.20\) plus its transformer 0.10; the 0.40 is the healthy circuit; the infinite bus has \(x_2 = 0\).

The zero-sequence network, with every line reactance tripled:

The machine itself does not appear: its delta winding blocks zero sequence entirely, so only the transformer's 0.10 connects bus \(A\) to the reference. This is the Set 22 result reused without change.

The four shunt reactances:

The order is fixed and general: \(0 < X_2\parallel X_0 < X_2 < X_2+X_0\) whatever the network. That single inequality determines the entire stability ranking.

Each shunt is placed at node \(F\) and the network reduced again. Repeating the elimination with \(X_f\) in place of the solid earth:

A larger shunt reactance means less current diverted to earth and therefore more power still reaching the infinite bus. The line-to-earth fault barely disturbs the transfer at all.

The stability verdict at \(P_m = 1.000\):

Only the three-phase fault threatens stability at this loading. For the other three, the during-fault curve still peaks above \(P_m\), so an equilibrium exists throughout the fault, the rotor merely oscillates about it, and the machine would stay in step even if the fault were never cleared.

Raise the loading to \(P_m = 1.200\) to expose the full ranking, with \(\delta_0 = 36.87°\) and \(\delta_{\max} = 126.87°\):

The three-phase fault gives 0.186 s; the double line-to-earth fault gives 0.520 s, nearly three times as long. The other two are not stability events at all at this loading.

The ranking is the exact reverse of the current ranking. Compare with Set 23, where at buses 1 and 2 the earth fault drew 20% more current than the three-phase fault:

The reason is that the two questions measure opposite things. Fault current is set by how low the impedance to earth is; stability is set by how much power still reaches the far end. An earth fault on a system with a stiff earth draws enormous current precisely because the zero-sequence path is short — but that path is in parallel with, not in series with, the transfer path, so the transfer barely notices.

The practical consequence. Since 70–85% of faults are single line-to-earth and those are the mildest for stability, two design conclusions follow:

Single-pole tripping opens only the faulted phase for a line-to-earth fault, leaving the other two carrying power throughout the dead time. On a critical EHV interconnector it can be the difference between surviving and not, and it is standard practice on long single-circuit ties.

Three stages now, not two. The machine sees a sequence of power-angle curves:

The dead time \(t_d = t_r - t_c\) must be long enough for the arc to de-ionise — typically 0.3–0.5 s at transmission voltages, and longer at higher voltage because the trapped charge takes longer to decay.

The equal-area statement now has three regions, and the third has the largest decelerating power:

And since reclosure restores the prefault curve, the final limit angle is \(\delta_{\max} = \pi-\arcsin(1.000/2.000) = 150°\) rather than 138.19° — a wider window as well as a taller curve.

But \(\delta_r\) is fixed by time, not by angle, which is what makes this problem awkward. The reclosure happens at \(t_r = t_c + t_d\), and where the rotor is at that instant depends on the whole history. The calculation must be done in the time domain.

Integrating the swing equation through all three stages and searching for the clearing time at which the machine just fails:

The result:

The benefit is real but modest, and it decays quickly as the dead time lengthens.

Why the gain is so small. The critical first swing is over in about half a second:

A reclosure at \(t_d = 0.30\) s arrives at \(t = 0.63\) s — after the rotor has already passed its peak. It helps the second swing, not the first, and the first swing is where instability occurs. Only a very fast reclosure, arriving while the rotor is still swinging outward, adds meaningful decelerating area.

What auto-reclosing is actually for is therefore not first-swing stability:

The first two rows are the reason every transmission line has it. The stability contribution is a bonus, not the justification.

The risk of unsuccessful reclosure. If the fault is permanent, reclosing re-applies it:

The machine is subjected to a second acceleration when it may already be swinging outward. Reclosing onto a permanent fault is materially worse for stability than not reclosing at all, and this is why heavily loaded critical circuits are often given single-shot reclosing only, or none. The check must be made explicitly.

Single-pole auto-reclosing is the refinement that does help stability substantially, and Problem 15 explains why:

With only the faulted phase opened, the other two continue to transmit throughout the dead time. Since 70–85% of faults are single line-to-earth, this covers most events. It is standard on long single-circuit EHV ties, where losing the whole circuit even briefly would be unacceptable.

Inertia does not affect the critical clearing angle at all. This deserves emphasis, because it is the key to the whole problem:

\(H\) appears nowhere. The equal-area criterion is a statement about energy per unit inertia, and inertia cancels from both sides. So \(\delta_{cc} = 80.74°\) for every machine on this network, whatever its size.

Inertia affects only how fast the rotor gets there. The during-fault swing equation:

Doubling \(M\) halves every acceleration, and the rotor takes \(\sqrt2\) times as long to cover the same angle — because displacement under constant acceleration goes as \(t^2\).

The scaling law, derived exactly. Substitute \(\tau = t/\sqrt{M}\):

\(M\) has vanished. So the trajectory \(\delta(\tau)\) is identical for every inertia, and real time is simply rescaled:

Hence:

Exactly, not approximately, and independent of fault type or location.

Confirmed numerically by integrating each case from 30° to 80.74°:

The last two columns agree to four decimal places — the square-root law is exact. The clearing angle column is constant, as predicted.

The engineering reading. Halving inertia costs 29% of the available clearing time:

Still comfortable against a 4-cycle scheme. But the square root cuts both ways: to double the clearing time available you must quadruple the inertia, which is not something that can be bought. Inertia is a weak lever.

The low-inertia problem, which is the defining stability challenge of modern grids:

Converter-interfaced generation is decoupled from system frequency and contributes no natural inertia. Rate of change of frequency \(= f\Delta P/(2H)\) rises in proportion, and many distributed generators trip on RoCoF — a positive feedback that has caused real system separations.

Synthetic inertia is the mitigation, and its limitation is worth stating precisely:

Real inertia is instantaneous and passive — a physical consequence of stored angular momentum. Synthetic inertia is a measured-and-responded control action, and 100 ms of delay is a substantial fraction of a 325 ms critical clearing time. It helps frequency response well; it helps first-swing transient stability much less. That distinction is at the centre of current grid-code development.

Three things change together as the loading rises, and all three work against stability:

The window \(\delta_{\max}-\delta_0\) closes from both ends while the rate of travel across it increases. This is why loading is by far the strongest lever on transient stability.

The sweep, for the mid-line three-phase fault with \(H = 5.0\):

The collapse is dramatic. Between 1.00 and 1.35 pu — a 35% increase in loading — the available clearing time falls by 79%, from 16.2 cycles to 3.4.

The 0.60 pu case is qualitatively different. There \(P_2 = 0.750 > P_m = 0.600\), so an equilibrium exists during the fault:

The rotor oscillates about 53.13° for as long as the fault lasts and never reaches the critical angle of 147.85°. A machine loaded to 60% here would survive a permanent uncleared mid-line fault indefinitely — which is exactly the situation Problem 15 found for the unsymmetrical fault types.

The transient stability limit for a 5-cycle scheme (\(t_{\text{clear}} = 0.10\) s). Solving \(t_{cc}(P_m) = 0.10\):

So with modern protection this machine may be loaded to 132 MW. Any more and a mid-line three-phase fault would pull it out of step.

The absolute ceiling, for a hypothetical instantaneous breaker. Set \(\delta_c = \delta_0\) so \(A_1 = 0\):

This is the post-fault transient stability limit — the largest loading from which the machine can survive the switching alone, computed in Problem 9 by the same equation. No amount of protection speed can exceed it, because it corresponds to zero fault duration.

The four limits in order:

Each row is more restrictive than the one above, and the binding constraint is the last. Note that it is only 66% of the prefault steady-state limit — which is the usual situation, and the reason transmission systems are almost never operated near their steady-state limit.

Why loading is the strongest lever. Comparing the three sensitivities computed in this set:

And unlike inertia or fault location, loading is a quantity the operator controls minute by minute. This is why real-time stability limits — computed continuously and enforced as transfer limits on interconnectors — are the standard operational tool, rather than fixed thermal ratings.

Every reactance appears three times, which is why the effect is not obvious in advance:

Reducing a reactance raises all three curves, but not by the same proportion — and the critical clearing angle depends on their ratios, not their magnitudes.

Varying the machine's transient reactance, with \(x_T = 0.10\), \(x_L = 0.40\) and the fault at mid-line:

A one-third reduction in \(x'_d\) buys 20° of clearing angle — a 24% improvement. The mechanism is straightforward: a lower \(x'_d\) raises every curve, so the machine starts at a smaller angle and has a taller post-fault curve to decelerate against.

Varying the line reactance:

A comparable effect, and for the same reason. Note that the two 1.333 rows — \(x'_d = 0.40\) and \(x_L = 0.50\) — give almost identical critical angles (66.86° and 66.93°). What matters is the total post-fault reactance, not where it sits.

That observation generalises. Plotting \(\delta_{cc}\) against \(P_3\) for both sweeps gives one curve:

The post-fault peak is the dominant parameter. This is a useful screening rule: to a first approximation, whatever raises the post-fault power-angle curve improves transient stability, regardless of which element you change.

Ranking the design measures available for improving transient stability:

The top row wins decisively on cost-effectiveness, which is why the history of transmission stability is largely the history of faster protection and faster breakers — from 20 cycles in the 1930s to 2 cycles today.

Series compensation quantified, since it is the classic transmission-level remedy. A 40% series capacitor on each circuit:

A 21° improvement — comparable to halving the fault clearing time, and achievable on an existing line. The penalty is the subsynchronous resonance risk discussed in Set 14, which is why compensation levels above 50% are approached cautiously.

Step 1 — define the contingency set. A stability study is only as good as the list of events it examines:

The convention is that all N−1 events must be survived with margin, and stuck-breaker or failed-protection cases must be survived at all. Simultaneous independent faults are excluded as non-credible.

Step 2 — evaluate each at the nominal loading of 1.00 pu:

The binding case is unambiguous: a three-phase fault at the sending bus, giving 9.0 cycles.

Step 3 — specify the protection with an appropriate margin. A 30% margin on clearing time is conventional:

A factor of two on the binding case. Comfortable, and standard for a 220 kV transmission line.

Step 4 — the backup case. If the main protection fails, backup clears in 0.35 s:

The machine would lose synchronism. Two responses are available: accept it (backup failure is rare, and out-of-step protection will trip the machine cleanly), or fit breaker-failure protection with a 0.15 s timer, which restores stability by clearing at 0.24 s total. On a 220 kV system supplying a single machine the first is usually accepted; on a critical interconnector the second is mandatory.

Step 5 — set the operating limit. From Problem 18, with the 4.5-cycle scheme applied to the binding bus fault:

So the machine may be despatched to 117 MW, against a rating of 100 MVA. Stability is not the binding constraint at rated output — the machine's own thermal rating is — which is the desired outcome of a good design. Note that the mid-line fault of Problem 18 permitted 1.317 pu; it is the sending-bus fault that sets the real limit, as it sets everything else.

Step 6 — the consistency checks, five identities that any correct study must satisfy:

Each is computed by a route independent of the quantity it checks.

Step 7 — the limitations, which must be stated with any classical study:

The first four err safe; the last two do not. A real study replaces the infinite bus with a network equivalent, models exciters and governors, and integrates for several seconds — but it uses the classical result above as its sanity check, and the answers rarely differ by more than 10–20% on the first swing.