The Equal-Area Criterion and Critical Clearing
The swing equation cannot be solved in closed form, but it can be integrated once — and that single integration converts the whole question of first-swing stability into a comparison of two areas on the power-angle diagram, from which the critical clearing angle and time follow directly.
- How multiplying the swing equation by \(d\delta/dt\) produces a first integral that survives the nonlinearity \(P_e = P_{max}\sin\delta\), and what that integral means physically.
- Why the stability question "does the rotor stop swinging?" becomes the algebraic statement \(A_1 = A_2\) — the equal-area criterion.
- How to apply it to the three standard disturbances: a step in mechanical input, the loss of a line, and a fault cleared by switching.
- The critical clearing angle in closed form, both for the simple reclosure case and for the general three-curve case with ratios \(r_1\) and \(r_2\).
- Why \(\delta_{cr}\) does not depend on the inertia constant \(H\) while the critical clearing time \(t_{cr}\) varies as \(\sqrt{H}\).
- The exact assumptions behind the criterion, and why it stops at two machines while Chapter 27's point-by-point method does not.
A Verdict Without a Solution
Chapter 27 wrote the equation of motion of a synchronous machine and Chapter 28 stated the condition it must satisfy: after a fault is cleared, the rotor's speed deviation must fall back to zero at some angle \(\delta_{max}\) no greater than the post-fault unstable equilibrium \(\delta_u\). That is a complete criterion, and it is useless as it stands, because finding \(\delta_{max}\) appears to require solving
and this equation has no solution in elementary functions. The \(\sin\delta\) term is exactly the pendulum nonlinearity, and the pendulum's large-amplitude motion is an elliptic integral, not a sine. Chapter 27 dealt with the difficulty by abandoning closed form altogether and stepping the equation forward numerically with the point-by-point method. That works, and it is what commercial software does, but it answers only one question at a time: for this clearing time, does the machine survive? A designer setting relay times wants the boundary — the longest clearing time that can be tolerated — and finding it by repeated simulation is a search, not a formula.
The way out is to notice how little of the solution we actually need. Nothing in the stability question asks for \(\delta(t)\). It asks only whether \(d\delta/dt\) returns to zero before \(\delta\) passes \(\delta_u\), which is a statement about \(\delta\) and \(d\delta/dt\) together and says nothing about when either occurs. A relation between \(\delta\) and \(d\delta/dt\) alone — the trajectory in the phase plane, with time eliminated — is a weaker thing than the full solution, and weaker things are easier to obtain. In this case it can be had exactly, in one line of calculus.
The First Integral of the Swing Equation
Start from the per-unit swing equation of Chapter 27, with damping omitted for a reason that will become clear shortly:
The trick is to manufacture a perfect derivative on the left. Observe that
so multiplying the swing equation through by \(2\,(d\delta/dt)\) converts its left-hand side into the time derivative of the squared speed deviation:
Now multiply by \(dt\) and integrate. On the left the integrand is an exact differential, so the integral is simply the change in \((d\delta/dt)^2\). On the right, \(P_a(\delta)\,d\delta\) is a function of \(\delta\) alone — time has vanished from the problem, which is the whole point of the manoeuvre.
The machine starts from a steady operating point, so \(d\delta/dt = 0\) at \(\delta = \delta_0\) and the lower limit contributes nothing. The result holds for any \(P_e(\delta)\) whatever, including one that changes shape part-way through the swing — which is exactly what a fault does.
Read the two sides physically. The rotor's kinetic energy relative to a frame turning at synchronous speed is proportional to the square of the speed deviation, so the left-hand side is that stored energy in per-unit form. The right-hand side is the work done on the rotor by the accelerating power as the angle advances. The equation says nothing more than that work in equals kinetic energy gained — but it says it exactly, with no linearisation and no small-angle assumption anywhere.
This is also why the damping term \(D\,d\delta/dt\) of Chapter 27 had to be dropped. A damping torque dissipates energy, and once energy leaks out of the system the bookkeeping between the two sides is no longer a clean equality; the right-hand integral would depend on how fast the angle was traversed, restoring the very time-dependence we have just eliminated. Neglecting damping is therefore not laziness. It is a necessary condition for the method to exist, and it is conservative: real damping removes energy from the swing, so a machine judged stable without damping is stable with it.
The criterion follows immediately. The rotor stops swinging outward at the angle \(\delta_{max}\) where \(d\delta/dt\) returns to zero, and by the first integral that happens where the integral itself vanishes:
The machine is stable on the first swing if such a \(\delta_{max}\) exists with \(\delta_{max}\le\delta_u\), where \(\delta_u\) is the post-disturbance unstable equilibrium. Note that \(H\) has cancelled out of the condition entirely — inertia decides when the rotor reaches a given angle, never whether it reaches it.
The Criterion on the Power-Angle Diagram
An integral of \((P_m - P_e)\,d\delta\) is an area on the power-angle diagram of Chapter 26 — the region between the horizontal line \(P = P_m\) and the curve \(P_e(\delta)\), measured along the \(\delta\) axis. Where \(P_m > P_e\) the integrand is positive and the rotor gains energy; where \(P_e > P_m\) it is negative and the rotor gives energy back. Splitting the integral at the angle where \(P_a\) changes sign turns the condition into a statement about two ordinary, positive areas.
\(A_1\) is the accelerating area, swept out while the rotor gains speed; \(A_2\) is the decelerating area, swept out while it gives that speed back. \(\delta_1\) is the angle at which \(P_e\) crosses \(P_m\) — the post-disturbance stable equilibrium. The machine survives the first swing if an \(A_2\) equal to \(A_1\) can be found before the curve falls back below \(P_m\) at \(\delta_u\).
Three outcomes are possible, and the diagram makes them visible without any arithmetic. If the area available between \(\delta_1\) and \(\delta_u\) exceeds \(A_1\), the rotor runs out of excess energy at some \(\delta_{max}\) short of \(\delta_u\) and the machine is stable, with the leftover area serving as a margin. If the two are exactly equal the machine is critically stable — it reaches \(\delta_u\) with zero speed deviation and balances there, a state no real system would be operated in. If the available area falls short, \(\delta\) sweeps past \(\delta_u\) with speed to spare, \(P_e\) drops below \(P_m\) again, the accelerating power turns positive and never turns back, and the machine loses synchronism.
It is worth being precise about what happens after a stable first swing. At \(\delta_{max}\) the rotor is momentarily at synchronous speed but sits at an angle where \(P_e > P_m\), so it is still decelerating; it therefore falls back through \(\delta_1\), overshoots to the left, and oscillates. In the undamped model the oscillation is permanent and the areas balance on every half-cycle. Any real machine has damping, so the successive swings shrink and the rotor settles at \(\delta_1\). The first swing is the largest, and that is why the first-swing test of Chapter 28 is the one that matters.
A Sudden Increase in Mechanical Input
The simplest disturbance is a step in the governor setting: a machine delivering \(P_{m0}\) at angle \(\delta_0\) has its steam valve opened so that the input jumps to \(P_{m1}\). Nothing electrical has changed, so a single power-angle curve \(P_e = P_{max}\sin\delta\) serves throughout, and the whole disturbance is the horizontal line moving up.
The rotor cannot respond instantaneously, so immediately after the step the machine still sits at \(\delta_0\) with \(P_e = P_{m0} < P_{m1}\). It accelerates, \(\delta\) rises along the curve, and the accelerating power falls to zero at the new equilibrium \(\delta_1 = \sin^{-1}(P_{m1}/P_{max})\). By then the rotor is turning faster than synchronous speed, so it carries on past \(\delta_1\) into the decelerating region. Applying the criterion with both areas measured against the same sine curve:
Setting \(A_1 = A_2\), the terms in \(\cos\delta_1\) and \(P_{m1}\delta_1\) appear on both sides and cancel, leaving a relation between the two angles that bracket the swing and no reference at all to the intermediate equilibrium:
A transcendental equation in \(\delta_2\), solved by trial or by Newton's method in two or three iterations. The cancellation is no accident: it is the statement \(\int_{\delta_0}^{\delta_2}P_a\,d\delta=0\) written out, and splitting that integral at \(\delta_1\) was only ever a convenience.
Now push the step until the machine only just survives. The largest \(\delta_2\) that is any use is \(\delta_u = \pi-\delta_1\), the unstable equilibrium on the far side of the peak; beyond it the curve has fallen back below \(P_{m1}\) and no further decelerating area exists. Setting \(\delta_2 = \pi-\delta_1\) and writing \(P_{m1}=P_{max}\sin\delta_1\):
One equation in the single unknown \(\delta_1\), after which the greatest sustainable input is \(P_{m1}=P_{max}\sin\delta_1\). For the extreme case of a machine floating on the bus with no load at all, \(P_{m0}=0\) and \(\delta_0=0\), the equation collapses to \(\sin\delta_1(\pi-\delta_1)=1+\cos\delta_1\), whose root is \(\delta_1 = 46.44^\circ\).
Chapter 28's steady-state stability limit assumes the load is raised in infinitesimal steps, each allowed to settle. Applied all at once, barely three-quarters of it can be reached. Every limit in stability work is therefore tied to a particular manner of loading, and quoting one without the other is meaningless.
Switching Without a Fault
The second standard disturbance keeps \(P_m\) fixed and changes the network instead. A generator feeds an infinite bus over two parallel lines; one is taken out of service — for maintenance, or because a protection scheme tripped it on a false signal — and the transfer reactance rises. By the Chapter 26 relation \(P_{max}=E'V/X\), the power-angle curve shrinks from \(P_{max,1}\) to \(P_{max,3}\), the numbering chosen to leave room for a during-fault curve in the next section.
The switching is instantaneous compared with rotor motion, so the operating point drops vertically from \((\delta_0, P_m)\) on the tall curve to \((\delta_0, P_{max,3}\sin\delta_0)\) on the short one. Since \(P_{max,3} < P_{max,1}\), the electrical output has fallen below the unchanged mechanical input and the rotor accelerates — the same situation as a step in \(P_m\), reached from the other direction. The algebra is identical to Section 29-4 with \(P_{max}\) replaced by \(P_{max,3}\):
The machine is stable if the root \(\delta_2\) exists and satisfies \(\delta_2 \le \delta_u\); equivalently, if the accelerating area \(A_1\) taken from \(\delta_0\) to \(\delta_1\) is no larger than the entire area available between \(\delta_1\) and \(\delta_u\). Testing that inequality directly avoids solving for \(\delta_2\) when only a verdict is wanted, and it is the quicker route in an examination.
Two limiting conditions deserve separate mention. If \(P_{max,3} < P_m\), the short curve never reaches the mechanical input line at all, no equilibrium exists after switching, and the machine loses synchronism regardless of inertia or damping — this is a steady-state failure exposed by a switching event, not a transient one. If \(P_{max,3}\) exceeds \(P_m\) but only barely, an equilibrium exists yet the decelerating area is a thin sliver, and the transient limit is reached long before the steady-state one.
A Fault Cleared by Reclosure
Now the case the criterion was invented for. A three-phase fault appears at the sending-end bus of a single line. With the fault directly on the machine terminals the transfer reactance to the infinite bus is effectively infinite, so \(P_e = 0\) throughout the fault: the machine still turns, still carries full mechanical torque, and delivers nothing. The rotor accelerates under the whole of \(P_m\).
The breaker opens at angle \(\delta_c\), the fault is extinguished, and the breaker recloses onto a healthy line, restoring the original curve. This is the classic idealisation — the fault was transient, as the great majority of overhead-line faults are, and high-speed autoreclosure returns the system to exactly its pre-fault configuration. Both areas are then measured against the same sine curve, one below \(P_m\) and one above it.
Set \(A_1=A_2\) and the terms in \(\delta_c\) collect on the left:
The critical case is the one that uses every scrap of decelerating area, so put \(\delta_{max}=\delta_u=\pi-\delta_0\) — since the post-fault curve is the pre-fault curve, the unstable equilibrium is the mirror image of \(\delta_0\) about \(90^\circ\). Using \(P_m/P_{max}=\sin\delta_0\) and \(\cos(\pi-\delta_0)=-\cos\delta_0\):
The whole answer depends on one number, the pre-fault angle \(\delta_0\), and through it on the loading \(P_m/P_{max}\) alone. Neither \(H\) nor \(f\) nor the absolute size of \(P_{max}\) appears. The more heavily the machine is loaded, the larger \(\delta_0\), the smaller \(\delta_{cr}\), and the less time the protection has.
Two sanity checks confirm the formula. A machine on no load has \(\delta_0=0\), giving \(\cos\delta_{cr}=-1\) and \(\delta_{cr}=180^\circ\): it can never be pushed out of step by a fault, because there is no mechanical power to accelerate it. A machine loaded to the steady-state limit has \(\delta_0=90^\circ\), giving \(\cos\delta_{cr}=\sin 90^\circ(\pi-\pi)-0=0\) and \(\delta_{cr}=90^\circ=\delta_0\): the permissible clearing angle has collapsed onto the operating angle and no fault of any duration can be tolerated. Between these extremes the formula interpolates smoothly.
The General Three-Curve Case
Real clearing is rarely a reclosure onto the same network. A fault on one of two parallel lines is cleared by opening that line at both ends, which leaves the system permanently weaker than before; and a fault out along a line, rather than at the bus, still permits some power to flow while it persists. Both effects are captured by keeping three separate power-angle curves, one for each stage of the event:
| Stage | Transfer reactance | Peak power | Ratio |
|---|---|---|---|
| Pre-fault | \(X_1\), both lines in service | \(P_{max,1}=E'V/X_1\) | \(1\) |
| During fault | \(X_2\), from a star–delta reduction of the faulted network | \(P_{max,2}=E'V/X_2\) | \(r_1=P_{max,2}/P_{max,1}\) |
| Post-fault | \(X_3\), faulted line removed | \(P_{max,3}=E'V/X_3\) | \(r_2=P_{max,3}/P_{max,1}\) |
Always \(X_2 > X_3 > X_1\), hence \(r_1 < r_2 < 1\). Obtaining \(X_2\) is the only laborious part, and it is pure network reduction of the kind Chapter 16 systematised: the shunt fault grounds an interior node, and a star–delta conversion of the three branches meeting at the sending bus produces the direct generator-to-infinite-bus branch whose reactance is \(X_2\). Example 5 carries the conversion out in full.
With the three curves in hand the areas are written down exactly as before, the accelerating area now sitting above the during-fault curve rather than above zero, and the decelerating area under the post-fault curve rather than under the original:
Equating them, gathering the \(\cos\delta_c\) terms on one side and everything else on the other, gives the general result. Note that \(\delta_{max}\) is now set by the post-fault curve, which is lower than the pre-fault one, so the unstable equilibrium has moved inward and the available area has been squeezed from both sides.
with \(\delta_0=\sin^{-1}\!\big(P_m/P_{max}\big)\) and \(\delta_{max}=\pi-\sin^{-1}\!\big(P_m/r_2P_{max}\big)\), all angles in radians. Putting \(r_1=0\) and \(r_2=1\) recovers Section 29-6 exactly, since then \(\delta_{max}=\pi-\delta_0\) and the expression reduces to \(\sin\delta_0(\pi-2\delta_0)-\cos\delta_0\).
The formula returns a value of \(\cos\delta_{cr}\), which must then be checked for physical sense. If it comes out less than \(-1\) there is no critical angle: even clearing at \(\delta_u\) leaves the decelerating area larger than the accelerating one, and the machine survives any clearing time whatever — a situation that arises with a light load and a strong post-fault network. If it comes out greater than \(\cos\delta_0\), the required clearing angle is smaller than the pre-fault angle, which is impossible, and the machine cannot be stabilised for this fault at this loading by any breaker. Between those bounds the arccosine gives the answer, and it must satisfy \(\delta_0 < \delta_{cr} < \delta_{max}\).
From Critical Angle to Critical Clearing Time
A protection engineer cannot set a relay to an angle. What is needed is the critical clearing time \(t_{cr}\) — the elapsed time from fault inception to breaker contact separation at which the rotor reaches \(\delta_{cr}\). Converting one to the other means re-introducing time, which is to say solving the swing equation during the fault. In general that is a numerical problem again, but in the important special case \(P_e=0\) it is a single integration.
With the machine's electrical output suppressed, the accelerating power is the constant \(P_m\) and the swing equation is that of a body under constant force:
using \(d\delta/dt=0\) at \(t=0\). The angle grows as the square of time — the parabolic swing curve of Chapter 27. Inverting at \(\delta=\delta_{cr}\) gives the clearing time directly.
Both forms take \(P_m\) in per unit on the machine base and \(H\) in seconds, with \(\omega_s=2\pi f\). The two are the same equation; the second absorbs the \(180/\pi\) conversion of Chapter 27 so that angles can be left in degrees.
The structure of the result repays attention. The critical angle was shown in Section 29-6 to be independent of \(H\); the critical time is proportional to \(\sqrt H\). Doubling the inertia of a machine buys a factor \(\sqrt2\approx1.41\) in permissible clearing time and not one degree of extra angle. That is the precise sense in which inertia contributes to stability, and it explains why the displacement of large synchronous units by converter-interfaced wind and solar plant — which presents no rotating mass to the system at all — is treated in Chapter 39 as a stability problem rather than merely an energy one.
When \(P_e\ne0\) during the fault there is no closed form, and the conversion is done by integrating the during-fault swing equation forward with the point-by-point method of Chapter 27 until \(\delta\) reaches the \(\delta_{cr}\) already found from the areas. The two methods are complementary rather than competing: the criterion supplies the target angle in one line of algebra, the numerical method supplies the time to reach it.
What the Criterion Assumes
Every result of this chapter descends from one integration performed on one equation, so the assumptions buried in that equation are the assumptions of the whole method. They are worth listing explicitly, because a criterion applied outside its range gives an answer that looks perfectly respectable and is wrong.
The classical machine model. \(P_e=P_{max}\sin\delta\) with \(P_{max}=E'V/X\) requires a constant voltage \(E'\) behind transient reactance — the Chapter 26 model. Flux decay in the field winding over a second or two, saliency, and the action of the exciter are all excluded. Chapter 28's remark that a fast automatic voltage regulator raises \(E'\) during the swing means the real \(P_{max}\) is larger than the model's, so the criterion errs on the safe side here.
Constant mechanical input. The turbine governor is slower than the swing by an order of magnitude — seconds against a few hundred milliseconds — so holding \(P_m\) fixed through the first swing is sound. It is not sound for the second and later swings, which is one more reason the method is confined to the first.
No damping. Required, as Section 29-2 showed, for the energy bookkeeping to close. Conservative.
One angle. The decisive limitation. With three or more machines there is no single \(\delta\), no single power-angle curve, and no area to compute. The energy idea survives — it generalises into the transient energy function methods that compute a system-wide potential energy and compare it against the energy injected by the fault — but the elegant picture does not.
Instantaneous switching. The fault appears, and later disappears, at a point in time; the curves jump. Breaker arcing time, unequal pole clearing and the finite duration of the fault arc are all compressed into the single number \(t_c\).
| Method | Machines | Gives | Cost | Chapter |
|---|---|---|---|---|
| Linearised swing equation | 1 (small signal) | Oscillation frequency, damping, steady-state limit | Trivial | 28 |
| Equal-area criterion | 1 or 2 | \(\delta_{max}\), \(\delta_{cr}\), margin — no time | One integral, one transcendental equation | 29 |
| Point-by-point / modified Euler | Any | Full swing curve \(\delta(t)\) | Repeated arithmetic per time step | 27 |
| Transient energy function | Any | Stability margin without simulation | Network reduction plus a controlling-mode search | Beyond this book |
Worked Examples
Problem. A generator connected to an infinite bus has \(P_{max}=1.5\) pu and delivers \(0.6\) pu. The mechanical input is suddenly raised to \(1.0\) pu. Find the maximum rotor angle reached and confirm that the machine stays in step.
Solution. The two equilibrium angles come straight from the sine curve:
Apply the result of Section 29-4, \(P_{m1}(\delta_2-\delta_0)=P_{max}(\cos\delta_0-\cos\delta_2)\), with \(\delta_0=0.4115\) rad and \(\cos\delta_0=0.9165\):
Trial: at \(\delta_2=60^\circ=1.0472\) rad the left side is \(1.0472+0.750-1.7863=+0.0109\); at \(65^\circ=1.1345\) rad it is \(1.1345+0.6339-1.7863=-0.0179\). The root lies between; interpolating and refining once gives \(\delta_2=62.0^\circ\). Since \(62.0^\circ < 138.19^\circ\) the machine is stable, with a wide margin.
Check by computing the areas separately, with all angles in radians:
The areas agree to four decimal places, as they must.
Problem. The same machine \((P_{max}=1.5\) pu\()\) operates at \(\delta_0=20^\circ\). What is the greatest sudden increase in mechanical input it can withstand?
Solution. The initial loading is \(P_{m0}=1.5\sin20^\circ = 0.513\) pu. The limiting condition of Section 29-4 is \(\sin\delta_1(\pi-\delta_1-\delta_0)=\cos\delta_0+\cos\delta_1\) with \(\delta_0=0.3491\) rad and \(\cos\delta_0=0.9397\):
Trial at \(\delta_1=50^\circ=0.8727\) rad: left \(=0.766\times1.9198=1.4706\); right \(=0.9397+0.6428=1.5825\). Left is short. At \(60^\circ=1.0472\) rad: left \(=0.8660\times1.7453=1.5115\); right \(=0.9397+0.500=1.4397\). Left now exceeds right, so the root lies between. At \(55.7^\circ=0.9721\) rad: left \(=0.8261\times1.8204=1.5039\); right \(=0.9397+0.5635=1.5032\). Hence \(\delta_1=55.67^\circ\).
The rotor then swings to \(\delta_2=180^\circ-55.67^\circ=124.33^\circ\) and just holds. Compare with the steady-state limit of \(1.5\) pu: applied gradually the machine would carry \(21\%\) more than it can be given at a stroke.
Problem. A machine delivers \(1.0\) pu to an infinite bus over two identical parallel lines, for which \(P_{max,1}=2.0\) pu. One line is switched out with no fault, leaving \(P_{max,3}=1.4\) pu. Find the maximum angle reached, and the largest load for which switching out a line is survivable.
Solution. Before switching, \(\delta_0=\sin^{-1}(1.0/2.0)=30^\circ=0.5236\) rad. After switching the operating point falls onto the lower curve, where the output at \(30^\circ\) is only \(1.4\sin30^\circ=0.70\) pu — a deficit of \(0.30\) pu, and the rotor accelerates. The new equilibrium and its mirror are
The swing condition is \(P_m(\delta_2-\delta_0)=P_{max,3}(\cos\delta_0-\cos\delta_2)\):
At \(\delta_2=60^\circ\): \(1.0472+0.700-1.7360=+0.0112\). At \(65^\circ\): \(1.1345+0.5916-1.7360=-0.0099\). Interpolating and refining, \(\delta_2=62.8^\circ\), comfortably below \(134.42^\circ\).
For the limiting load, set \(\delta_2=\delta_u=\pi-\delta_1\) and solve \(P_m(\pi-\delta_1-\delta_0)=P_{max,3}(\cos\delta_0+\cos\delta_1)\) with \(\delta_0=\sin^{-1}(P_m/2.0)\) and \(\delta_1=\sin^{-1}(P_m/1.4)\), both functions of \(P_m\). Trial gives \(P_m=1.268\) pu; at that loading \(\delta_0=39.35^\circ\), \(\delta_1=64.94^\circ\) and \(\delta_u=115.06^\circ\). Above \(1.268\) pu the line must not be switched out on load, even though the post-switching steady-state limit is \(1.4\) pu.
Problem. A \(50\) Hz generator with \(H=5\) MJ/MVA delivers \(0.9\) pu to an infinite bus, with \(P_{max}=1.8\) pu. A three-phase fault at the sending-end bus reduces the output to zero, and the breaker recloses onto the healthy line. Find the critical clearing angle and the critical clearing time.
Solution. The pre-fault angle is \(\delta_0=\sin^{-1}(0.9/1.8)=\sin^{-1}0.5=30^\circ=0.5236\) rad, so \(\delta_u=150^\circ\). Section 29-6 gives
The clearing angle is available in closed form because the areas are elementary; the time requires the during-fault motion, which is uniformly accelerated since \(P_e=0\):
That is \(12.4\) cycles at \(50\) Hz — generous by modern standards, where a transmission breaker with its relay clears in three to five cycles. The margin exists because the machine is loaded to only half of \(P_{max}\); Example 6 shows how quickly it disappears.
Problem. A generator with \(x_d'=0.20\) pu feeds an infinite bus at \(1.0\angle0^\circ\) through a transformer of \(0.10\) pu and two parallel lines of \(0.40\) pu each. The internal voltage is \(E'=1.2\) pu and the machine delivers \(1.0\) pu. A three-phase fault occurs at the midpoint of one line and is cleared by opening that line at both ends. Find the critical clearing angle.
Solution. Take the three reactances in turn. Before the fault the two lines are in parallel, \(0.40\parallel0.40=0.20\), so
During the fault the network is a star centred on the high-voltage bus, with branches of \(0.30\) pu to the internal node, \(0.40\) pu to the infinite bus along the healthy line, and \(0.20\) pu to the faulted midpoint, which is held at zero volts. Converting that star to a delta, the branch joining the internal node to the infinite bus is the one opposite the \(0.20\) pu arm:
The other two delta branches connect to the fault point and carry no power to the infinite bus, so they are discarded. The ratios and angles follow:
Substituting into the general formula of Section 29-7, with \(\cos\delta_0=0.9091\) and \(\cos\delta_{max}=-0.8123\):
The result lies properly between \(\delta_0=24.6^\circ\) and \(\delta_{max}=144.3^\circ\). It is larger than the \(79.6^\circ\) of Example 4 for two reasons that work together: the machine is less heavily loaded relative to its pre-fault curve, and the fault out on the line still permits \(0.923\sin\delta\) pu to flow, which reduces the accelerating power throughout. A bus fault is always the more severe case, which is why it is the one used for design.
Problem. The machine of Example 4 has its fault cleared in \(0.20\) s. Find the clearing angle, the maximum angle reached, and the ratio of available to required decelerating area. What happens if the breaker takes \(0.25\) s?
Solution. With \(P_e=0\) during the fault the angle grows as \(t^2\). In degrees, \(\delta = \delta_0 + 90 f P_m t^2/H\):
Since \(\delta_{cr}=79.56^\circ\), clearing in \(0.20\) s is safe and clearing in \(0.25\) s is not — the machine loses synchronism, in the manner of the uppermost swing curve in Section 29-8. For the safe case, the maximum angle follows from \(P_m(\delta_m-\delta_0)=P_{max}(\cos\delta_c-\cos\delta_m)\):
At \(\delta_m=95^\circ=1.6581\) rad: \(0.8290-0.0872-0.7251=+0.0167\). At \(100^\circ=1.7453\) rad: \(0.8727-0.1736-0.7251=-0.0260\). Interpolating gives \(\delta_m=96.9^\circ\), well short of \(\delta_u=150^\circ\).
The margin is the ratio of the total decelerating area available up to \(\delta_u\) to the accelerating area actually incurred:
Twice the required area is available, and yet the permissible clearing time is only \(24\%\) longer than the one used. The reason is the quadratic growth of angle with time: area margin is a weak lever on time, because time enters as a square root. A stability study that reports comfortable margins can still be sensitive to a slow breaker.
Chapter Summary
\(\dfrac{H}{\omega_s}\big(\dot\delta\big)^2=\displaystyle\int_{\delta_0}^{\delta}P_a\,d\delta\) — kinetic energy equals work done, exactly.
The swing stops where \(A_1=A_2\); stability needs that balance to occur at or before \(\delta_u\).
\(P_m(\delta_2-\delta_0)=P_{max}(\cos\delta_0-\cos\delta_2)\); an unloaded machine takes at most \(0.724P_{max}\).
Same algebra with \(P_{max,3}\); if \(P_{max,3}<P_m\) no equilibrium exists and nothing can help.
\(\cos\delta_{cr}=\sin\delta_0(\pi-2\delta_0)-\cos\delta_0\) — a function of loading alone.
\(\cos\delta_{cr}=\big[\tfrac{P_m}{P_{max}}(\delta_{max}-\delta_0)+r_2\cos\delta_{max}-r_1\cos\delta_0\big]/(r_2-r_1)\).
\(t_{cr}=\sqrt{4H(\delta_{cr}-\delta_0)/\omega_sP_m}\) when \(P_e=0\); \(\delta_{cr}\) is free of \(H\), \(t_{cr}\propto\sqrt H\).
Classical model, no damping, constant \(P_m\), one or two machines, first swing only.
Practice Problems
Work all angles in radians inside the formulae and convert to degrees only when quoting an answer. Take \(f=50\) Hz and per-unit powers on the machine base unless stated otherwise.
- A machine with \(P_{max}=1.6\) pu delivers \(0.7\) pu to an infinite bus. Find the maximum rotor angle reached if the mechanical input is suddenly raised to \(1.1\) pu, and state the stability margin in terms of \(\delta_u\).
- For the machine of Problem 1, find the greatest sudden increase in mechanical input it can absorb, and compare the result with the steady-state limit.
- A generator supplies \(0.8\) pu over two parallel lines with \(P_{max,1}=1.9\) pu. One line is opened without a fault, leaving \(P_{max,3}=1.25\) pu. Determine whether the machine stays in step, and find \(\delta_{max}\) if it does.
- A \(50\) Hz machine with \(H=4\) MJ/MVA delivers \(1.0\) pu with \(P_{max}=1.6\) pu. A three-phase fault at its terminals is cleared by reclosure. Find \(\delta_{cr}\) and \(t_{cr}\).
- Repeat Problem 4 for \(H=8\) MJ/MVA and confirm that \(\delta_{cr}\) is unchanged while \(t_{cr}\) increases by a factor \(\sqrt2\). Explain both results physically.
- A machine with \(x_d'=0.25\) pu and \(E'=1.15\) pu feeds an infinite bus at \(1.0\) pu through a transformer of \(0.12\) pu and two parallel lines of \(0.45\) pu each, delivering \(1.0\) pu. A three-phase fault occurs at the sending end of one line and is cleared by opening that line. Find \(r_1\), \(r_2\) and \(\delta_{cr}\).
- For the system of Problem 6, the same fault is instead placed at the midpoint of the line. Recompute \(r_1\) and \(\delta_{cr}\), and account for the difference between the two answers.
- A machine delivering \(P_m\) pu with \(P_{max}=2P_m\) suffers a bus fault cleared at \(\delta_c=70^\circ\). Show by computing both areas that it is stable, then find the largest \(P_m/P_{max}\) for which clearing at \(70^\circ\) would still be acceptable.