The Swing Equation
One second-order differential equation ties the rotor's inertia to the sine-shaped power the network draws from it, and every question about stability — will the machine come back, how far will it swing, how long may the breakers take — is a question about the solution of that equation.
- How the torque balance of Chapter 26 becomes the swing equation in power form, and why the change from torque to power costs nothing in a first-swing study.
- The four forms you will meet in examinations — in \(M\), in \(H\), in electrical degrees, and in per unit — and how to move between them without error.
- What the damping term \(D\,d\delta/dt\) represents physically, and why it is usually dropped for transient studies but never for small-signal ones.
- Why machines swinging coherently combine as \(H_{eq}=H_1+H_2\) while machines swinging against each other combine as \(H_{eq}=H_1H_2/(H_1+H_2)\).
- The two situations in which the swing equation has a closed-form solution — uniform acceleration and small oscillation — and why every other case needs numerical integration.
- How the point-by-point method integrates the equation by hand, including the averaging rule that handles the discontinuity when a fault is applied or cleared.
- How to read a swing curve \(\delta(t)\) and pronounce a verdict of stable or unstable from its shape.
Assembling the Equation
Chapter 26 produced three pieces and left them lying side by side. The first was Newton's second law for the rotating shaft, \(J\,d^2\theta_m/dt^2 = T_m - T_e\). The second was the change of variable \(\theta = \omega_s t + \delta\), which measures the rotor against a reference frame turning at synchronous speed and leaves the second derivative untouched. The third was the power-angle equation \(P_e = P_{max}\sin\delta\), which says how much electrical power the network takes out of the machine when the rotor sits at angle \(\delta\).
Putting the three together produces a single equation in a single unknown, and that equation is the entire subject of Part 6. The assembly is worth doing slowly, because each step discards something and it matters to know what.
Start from the torque balance and pass to electrical angle. A machine with \(P\) poles has \(\theta = (P/2)\theta_m\), so multiplying the mechanical equation through by \(2/P\) and folding the factor into the inertia gives the equation of motion in the electrical angle. Then substitute the synchronous reference.
Now multiply both sides by the rotor's angular velocity to turn torques into powers. Strictly the left side acquires \(\omega_r\) and the right side becomes \(P_m - P_e\) only if the same \(\omega_r\) multiplies both torques, which it does. The result is exact:
The one approximation of the whole derivation is made here. The coefficient contains \(\omega_r\), the actual rotor speed, which is itself a function of time — the equation as written is nonlinear in a second, unhelpful way. During a first-swing study the rotor speed departs from synchronous by well under one per cent, so replacing \(\omega_r\) by the constant \(\omega_s\) introduces an error smaller than the uncertainty in the machine data. Making that replacement, and recognising the bracket as the inertia coefficient \(M = J(2/P)^2\omega_s\times10^{-6}\) defined in Chapter 26, gives the equation in its classical form.
\(M\) is the inertia coefficient in MJ·s per electrical radian, \(\delta\) the rotor angle in electrical radians, \(P_m\) the mechanical input in MW, \(P_e\) the electrical output in MW, and \(P_a\) the accelerating power. The name is descriptive: the equation governs the swinging of the rotor about its equilibrium position, and its solution \(\delta(t)\) is called the swing curve.
What makes this innocent-looking equation hard is the right-hand side. \(P_m\) is constant on the time scale of a swing — governors act in seconds, the first swing is over in a few hundred milliseconds — but \(P_e\) is not a constant and not even an independent input. It is \(P_{max}\sin\delta\), a function of the very variable being solved for. Writing that out exposes the structure:
This is the pendulum equation with a constant driving torque. It has no solution in elementary functions. Every technique in the remainder of Part 6 — the equal-area criterion, point-by-point integration, small-signal linearisation — exists because that sentence is true.
The Working Forms of the Swing Equation
The equation is always the same; only the units of the coefficient change. Four forms recur, and confusing them is the single commonest source of a wrong answer in a stability problem — a factor of \(180/\pi\) or of \(2\) buried in an examination answer is almost always a units error here.
Start from \(M\,\ddot\delta = P_a\) with \(M\) in MJ·s per electrical radian and substitute the Chapter 26 relation \(M = GH/\pi f\), where \(G\) is the machine MVA rating, \(H\) its inertia constant in seconds and \(f\) the system frequency:
Divide throughout by \(G\). Every power becomes a per-unit power on the machine's own base, and the rating disappears from the equation entirely — which is exactly the service the per-unit system of Chapter 4 was invented to perform.
with \(\delta\) in electrical radians and \(\omega_s = 2\pi f\). This is the form to memorise. Everything else is a rearrangement of it.
Angles in stability work are quoted in degrees far more often than in radians, and the point-by-point method of Section 27-8 is invariably worked in degrees. Converting the derivative costs one factor of \(180/\pi\):
Finally, the second-order equation is often better written as two first-order equations, because every numerical integrator — and every piece of simulation software — works on first-order systems. The natural second state variable is the speed deviation \(\Delta\omega = d\delta/dt\), in electrical radians per second.
| Form | Coefficient | Units of \(\delta\) | Units of \(P\) | Where used |
|---|---|---|---|---|
| \(M\ddot\delta = P_a\) | \(M = 2GH/\omega_s\) | elec rad | MW | Derivations; single machine in SI |
| \(\dfrac{GH}{\pi f}\ddot\delta = P_a\) | \(GH/\pi f\) | elec rad | MW | When ratings are mixed |
| \(\dfrac{2H}{\omega_s}\ddot\delta = P_a\) | \(2H/\omega_s\) | elec rad | pu | The standard form; simulation |
| \(\dfrac{H}{180 f}\ddot\delta = P_a\) | \(H/180f\) | elec deg | pu | Point-by-point solution by hand |
Damping and the Complete Equation
Chapter 26 linearised the swing equation about an operating point and obtained \(M\,\Delta\ddot\delta + P_s\,\Delta\delta = 0\), an undamped oscillator. Its prediction is that a disturbed machine oscillates forever. Real machines do not: the oscillation dies away over a few seconds. Something is missing, and it is a term proportional to the speed deviation rather than to the angle.
Three physical mechanisms produce such a term, and all three act in the same direction.
The first is the damper winding. Chapter 26 introduced it as the amortisseur that shapes the subtransient reactance. When the rotor runs at a speed different from the stator field, the damper bars cut flux, currents are induced in them, and the resulting torque opposes the relative motion — an induction motor torque, proportional to slip for small slip. Since slip is exactly \((d\delta/dt)/\omega_s\), the torque is proportional to \(d\delta/dt\).
The second is the load. Real loads are not constant-power; motor loads in particular draw power that rises with frequency. If the machine speeds up, the system it feeds absorbs more, opposing the acceleration.
The third is network resistance, neglected everywhere in this Part but never actually zero. It dissipates a fraction of the swing energy on every cycle.
\(D\) is the damping coefficient in per-unit power per per-unit speed deviation. Typical values are \(1\) to \(3\) for the damper winding alone and up to \(10\) or more when load damping is lumped in. Written with \(\Delta\omega\) in per unit the middle term is simply \(D\,\Delta\omega_{pu}\), which is where the units of \(D\) come from.
Whether to keep the term depends entirely on the question being asked, and the rule is worth stating plainly. For a transient stability study the answer is decided in the first swing, which is over in \(0.3\) to \(0.8\) s. Damping removes only a small fraction of the swing energy in that time, and neglecting it makes the calculation pessimistic — the computed critical clearing time is shorter than the true one, so the design is safe. Damping is therefore routinely dropped, and Chapter 29's equal-area criterion depends on its being dropped, because the criterion is an energy balance that no dissipative term may spoil.
For a steady-state or small-signal study, the whole question is whether an oscillation grows or decays, and that is decided by nothing but the damping term. Dropping it does not simplify the problem; it destroys it. Chapter 28 takes the linearised equation with \(D\) retained as its starting point, and Chapter 33 returns to it once more when the excitation system is shown to be capable of making \(D\) negative.
Changing Base; Machines That Swing Together
The per-unit swing equation of Section 27-2 is written on the machine's own MVA base, because that is the base on which manufacturers quote \(H\). A study of a system containing many machines needs every equation on one common base, exactly as Chapter 4 required every impedance on one common base before the network could be assembled.
The conversion rule follows from a single observation: the product \(GH\) is the stored kinetic energy in megajoules, and stored energy is a physical quantity that does not care what base an engineer chooses. Therefore \(G_{mach}H_{mach} = G_{sys}H_{sys}\), and
\(H\) scales directly with the ratio of the ratings, unlike a per-unit impedance which scales inversely. The direction is easy to remember from the physics: a large machine referred to a small base has a great deal of stored energy per unit of that base, so its \(H\) becomes large.
Now consider two machines on the same busbar, or close enough electrically that no appreciable angle can develop between them. Their rotors are locked together through a stiff synchronising tie and they accelerate as one. Such machines are called coherent. Write a swing equation for each, both already referred to the common system base:
The two equations simply add. The pair behaves as one machine whose mechanical input is the sum of the two inputs, whose electrical output is the sum of the two outputs, and whose inertia constant is the sum of the two inertia constants on the common base. Nothing is approximate here except the assumption of coherence itself.
Equivalently: add the stored energies in MJ and divide once by the common base. The rule extends to any number of machines and is the justification for representing an entire power station, or a whole area of a network, by a single equivalent generator in a study whose interest lies elsewhere.
Machines That Swing Against Each Other
Coherence is the friendly case. The dangerous case is the opposite one: a fault splits a system into two groups of machines, the generators near the fault accelerate while the generators far from it decelerate, and the quantity that decides whether synchronism survives is the angle between the two groups. Neither absolute angle matters; the difference does.
Reduce each group to a single equivalent machine by the coherent rule of Section 27-4, so that there are two machines with inertia constants \(H_1\) and \(H_2\) on a common base and angles \(\delta_1\) and \(\delta_2\). Write both swing equations in the form solved for the acceleration:
To make this look like a swing equation again, note that when the disturbance is internal to the pair — machine 1 gains exactly what machine 2 loses — we have \(P_{a1} = -P_{a2} = P_a\). Substituting:
The relative motion of two machines obeys an ordinary swing equation whose inertia constant is the parallel combination of the two, in exact analogy with the reduced mass of a two-body problem in mechanics. A two-machine system is therefore always reducible to one machine against an infinite bus, which is why the single-machine analysis of Chapters 26, 28 and 29 covers far more ground than it appears to.
The two rules pull in opposite directions and it is worth seeing why. Coherent machines add their inertias, so a group is harder to accelerate than any of its members — good for stability. Machines swinging in opposition combine as a parallel sum, which is always smaller than the smaller of the two, so the relative angle responds more violently than either machine alone would suggest. If one of the two is an infinite bus, \(H_2\to\infty\) and \(H_{eq}\to H_1\): the machine swings against a wall, and its own inertia is all that resists.
The Multi-Machine Swing Equations
A real system has hundreds of machines and no infinite bus. The formulation generalises without difficulty; what grows is the amount of network algebra needed to supply \(P_{ei}\) at each step.
Give machine \(i\) its classical representation from Chapter 26 — a constant voltage \(E_i'\angle\delta_i\) behind \(jX_{di}'\) — and absorb the machine reactances and the loads into the network. Loads are converted to constant admittances \(Y = P/|V|^2 - jQ/|V|^2\) taken from the pre-fault load flow of Part 4, so that the whole network becomes passive and can be reduced by the methods of Chapter 16 to a bus admittance matrix seen only from the \(n\) internal machine nodes. The result is the reduced \(Y_{bus}\), of order \(n\times n\), whose elements \(Y_{ij}=|Y_{ij}|\angle\theta_{ij}\) already contain everything about the transmission system.
The power injected by machine \(i\) then follows from the same phasor algebra that produced the two-machine power-angle equation:
The first term is the power lost in the machine's own driving-point conductance, which now includes the converted loads. The sum is the transfer of power to every other machine, and each term reduces to \(P_{max}\sin\delta_{ij}\) when the network is treated as lossless, \(G_{ij}=0\). One swing equation is then written for each machine:
with every \(H_i\) and every power on one common MVA base. The set is \(n\) coupled nonlinear second-order equations — \(2n\) first-order equations in state form — and it must be integrated numerically. The coupling is entirely through the angle differences \(\delta_i - \delta_j\), which is why a common time reference is arbitrary and only differences are ever plotted.
Three practical points govern how such a study is actually run. First, the network changes twice — at fault inception and at clearing — so three reduced \(Y_{bus}\) matrices are built in advance, and the integrator simply switches from one to the next at the appointed instants. Second, the reference is usually taken as the centre of inertia, \(\delta_{COI} = \sum H_i\delta_i / \sum H_i\), whose motion represents the drift of the whole system's frequency and carries no information about synchronism; plotting \(\delta_i - \delta_{COI}\) separates the machines that are genuinely pulling away from those that are merely riding the common drift. Third, the verdict is read off the angle spread: if \(\max_{i,j}|\delta_i-\delta_j|\) stops growing and turns, the system is stable for that disturbance.
Two Cases That Solve in Closed Form
The nonlinear swing equation resists analytic solution, but two special cases yield, and between them they cover a surprising amount of examination work and a good deal of physical understanding.
Case 1 — Uniform acceleration
If \(P_e\) is constant during an interval, the right-hand side is constant and the equation integrates twice by inspection. The important instance is a three-phase fault at the machine terminals or at the sending bus, where Chapter 26 showed the transfer reactance becomes infinite and \(P_e\) collapses to zero. The machine then accelerates under its full mechanical input.
The rotor angle grows as \(t^2\) — a parabola, not an exponential — and the speed deviation grows linearly. Both start from zero because the rotor cannot change position or speed instantaneously. Two consequences are used constantly. The angle reached at clearing time \(t_c\) is \(\delta_c = \delta_0 + (\pi f P_m/2H)t_c^{2}\), which Chapter 29 inverts to obtain the critical clearing time. And the time to reach any given angle scales as \(\sqrt{H}\): doubling the inertia buys a factor \(\sqrt2\) in permissible clearing time, not a factor of two.
Case 2 — Small oscillation
If the disturbance is small enough that \(\delta\) never strays far from \(\delta_0\), expand \(P_e\) to first order as in Chapter 26 and keep the damping term of Section 27-3. Writing \(\delta = \delta_0 + \Delta\delta\) and using \(P_m = P_e(\delta_0)\):
This is the standard second-order system of control theory, and everything known about it transfers directly. The natural frequency \(\omega_n\) rises with the synchronising stiffness and falls with inertia; putting typical numbers in gives \(0.5\) to \(2\) Hz, the electromechanical oscillation band. The damping ratio is small — a few per cent at best — so a disturbed machine rings for several seconds. The response to a small step \(\Delta P_m\) in mechanical input settles at \(\Delta\delta_{ss} = \Delta P_m/P_s\) after an overshoot of nearly \(100\%\), because \(\zeta\) is so small. Chapter 28 works with this equation throughout.
Neither covers the general case, in which \(P_e\) is a nonzero sine of an angle that moves far from its initial value. That case is the business of the next section and of Chapter 29.
Step-by-Step Solution: The Point-by-Point Method
The general swing equation is integrated numerically. Before digital computers this was done by hand with a method deliberately arranged so that each step needs only one sine and two additions, and that method — the point-by-point or step-by-step method — is still the one taught, because working three or four steps of it by hand teaches more about the dynamics than any amount of watching a simulation run.
Divide time into equal intervals \(\Delta t\), typically \(0.05\) s at \(50\) Hz, and label the instants \(t_0, t_1, \dots\) The scheme rests on two assumptions, each an ordinary numerical-integration choice.
The accelerating power \(P_a\) is assumed constant over the interval preceding each instant, held at the value computed at the beginning of that interval. Therefore the speed deviation, which is the integral of the acceleration, changes linearly; and the angle, which is the integral of the speed, changes quadratically. Working the two integrations over one interval and eliminating the intermediate speed gives the recurrence.
The speed is evaluated at the midpoints of the intervals and the angle at the ends, an arrangement that makes the scheme second-order accurate at no extra cost — it is the leapfrog method under another name. The intermediate speed never has to be written down, and the working reduces to a two-column table.
With \(\delta\) in electrical degrees and powers in per unit, \(M = H/180f\) and the constant \(k = (\Delta t)^2/M = 180 f(\Delta t)^2/H\) is computed once at the start. Every subsequent line of the table is one evaluation of \(P_a = P_m - P_{max}\sin\delta\) followed by two additions.
One detail decides whether the answer is right, and it is the detail students most often miss. The recurrence assumes \(P_a\) is a well-defined number at the instant \(t_{n-1}\). At the moment a fault is applied or cleared, \(P_a\) is discontinuous: \(\delta\) cannot change instantaneously, but \(P_{max}\) does, so \(P_e\) jumps and \(P_a\) has one value just before the switching instant and another just after. Using either alone biases the whole subsequent solution. The correct treatment is the average of the two, which is exactly what a proper integration of the step discontinuity over the interval straddling it gives.
| Instant | What to use for \(P_a\) | Reason |
|---|---|---|
| \(t=0\), fault applied | \(\tfrac12\big[0 + (P_m - P_{max,2}\sin\delta_0)\big]\) | Before the fault \(P_a=0\); after it, the during-fault value |
| During the fault | \(P_m - P_{max,2}\sin\delta\) | Single-valued, no averaging |
| \(t=t_c\), fault cleared | \(\tfrac12\big[(P_m-P_{max,2}\sin\delta_c)+(P_m-P_{max,3}\sin\delta_c)\big]\) | \(\delta\) continuous, \(P_{max}\) discontinuous |
| After clearing | \(P_m - P_{max,3}\sin\delta\) | Single-valued again |
| Clearing between grid points | Step to the next grid instant, or shorten \(\Delta t\) | The scheme needs switching to land on a grid point |
Two refinements are worth knowing. The modified Euler method works on the state form of Section 27-2: predict \(\delta\) and \(\Delta\omega\) with a plain Euler step, evaluate the derivatives again at the predicted point, and advance with the average of the two derivative estimates. It has the same second-order accuracy and handles the state variables symmetrically, which matters when a damping term is present. Fourth-order Runge–Kutta takes four derivative evaluations per step and is what production software uses; it tolerates a step of \(0.01\) to \(0.02\) s comfortably.
On step size, the guidance is simple: \(\Delta t\) must be small compared with the period of the oscillation being followed. With \(\omega_n\) around \(1\) Hz the period is \(1\) s, and \(\Delta t = 0.05\) s gives twenty points per cycle — adequate for hand work and for the first swing, which is all a transient study needs. Halving \(\Delta t\) and confirming that the answer barely moves is the standard check.
Reading a Swing Curve
The output of any of these methods is a table of \(\delta\) against \(t\), and plotting it gives the swing curve. The verdict is read from its shape, and only two shapes occur.
If the curve rises, reaches a maximum, and turns back, the rotor has been decelerated enough to arrest its outward swing before it passed the point of no return. The machine remains in synchronism and the system is stable for that disturbance and that clearing time. With damping neglected the curve then oscillates about the post-fault equilibrium indefinitely; with damping included it spirals into it. Either way the question has been answered at the first maximum.
If the curve rises without turning — if \(d\delta/dt\) is still positive when \(\delta\) passes the post-fault unstable equilibrium \(180^\circ - \delta_{eq}\) — then beyond that point \(P_e<P_m\) again, the accelerating power becomes positive once more, and \(\delta\) grows without bound. The machine has pulled out of step. In reality it does not run away to infinity; it slips poles, draws enormous currents, and its protection trips it.
In practice a computed swing curve is run for one second or so. A curve that has turned once has answered the question; a curve still climbing at \(\delta = 180^\circ\) has answered it the other way. Everything Chapter 29 does with the equal-area criterion is a way of predicting which of the two will happen without integrating at all.
Worked Examples
Problem. A \(100\) MVA, \(50\) Hz turbo-alternator with \(H = 5\) s is delivering \(90\) MW to an infinite bus at a rotor angle of \(30^\circ\). A three-phase fault occurs at its terminals, so that its electrical output falls to zero. Write the swing equation in per unit and in electrical degrees, find the initial acceleration, the rotor angle \(0.1\) s after the fault, and the time the rotor would take to reach \(120^\circ\) if the fault persisted.
Solution. Work on the machine's own \(100\) MVA base, so \(P_m = 90/100 = 0.90\) pu and \(P_e = 0\), giving \(P_a = 0.90\) pu throughout the fault.
The swing equation is therefore \(0.03183\,\ddot\delta = 0.90 - P_e\) with \(\delta\) in radians, or \(5.556\times10^{-4}\,\ddot\delta = 0.90 - P_e\) with \(\delta\) in degrees. With the fault on:
Because \(P_a\) is constant, Case 1 of Section 27-7 applies exactly: \(\delta(t) = 30^\circ + \tfrac12(1620)t^{2} = 30 + 810\,t^{2}\) degrees.
The speed deviation at \(0.1\) s is \(1620\times0.1 = 162\) elec deg/s, which is \(0.45\) electrical revolutions per second, or a slip of \(0.9\%\) — comfortably within the range over which replacing \(\omega_r\) by \(\omega_s\) in Section 27-1 was justified. Note also that a fault lasting a third of a second would carry this machine to \(120^\circ\), and Chapter 29 will show that it cannot recover from anything like that.
Problem. Generator \(A\) is rated \(500\) MVA with \(H_A = 4.0\) s and generator \(B\) is rated \(250\) MVA with \(H_B = 6.0\) s, each on its own base. Find the equivalent inertia constant on a \(100\) MVA system base (a) when the two machines swing coherently, and (b) when they swing against each other. In case (b), find the initial relative acceleration if a disturbance leaves machine \(A\) with \(+0.40\) pu accelerating power and machine \(B\) with \(-0.40\) pu, both on the \(100\) MVA base, at \(50\) Hz.
Solution. Convert both to the common base first, using \(H_{sys}=H_{mach}\,G_{mach}/G_{sys}\):
The stored-energy check: \(500\times4.0 + 250\times6.0 = 2000+1500 = 3500\) MJ, and \(3500/100 = 35.0\) s, which is the coherent equivalent.
For the relative acceleration, use the two-machine swing equation \(\dfrac{H_{eq}}{\pi f}\ddot\delta_{AB} = P_a\) with \(P_a = 0.40\) pu:
Check it the long way: \(\ddot\delta_A = \pi f(0.40)/20.0 = 3.142\) rad/s² and \(\ddot\delta_B = \pi f(-0.40)/15.0 = -4.189\) rad/s², whose difference is \(7.33\) rad/s². The coherent equivalent is about four times the opposed one, so the same power imbalance opens the angle between two opposed groups four times as fast as it would move a coherent group as a whole. This is the quantitative statement of why splitting a system is dangerous.
Problem. A \(50\) Hz machine with \(H = 5\) s operates with a synchronising power coefficient \(P_s = 1.5\) pu per electrical radian and a damping coefficient \(D = 2.0\) pu. Find the undamped natural frequency of oscillation in hertz, the period, the damping ratio, the damped frequency, and the time constant of the decay envelope.
Solution. The linearised equation of Section 27-7, Case 2, with \(M = 2H/\omega_s = 10/314.16 = 0.03183\):
The machine oscillates at \(1.09\) Hz — squarely in the electromechanical band — and the envelope of the oscillation decays with a ten-second time constant, so roughly eleven cycles of swing are visible before the disturbance has substantially died away. The damped and undamped frequencies differ in the fourth decimal place, which is why the distinction is almost always ignored in stability work: \(\zeta\) is small enough that \(\omega_d\simeq\omega_n\), but not small enough to make the decay irrelevant.
Problem. A \(50\) Hz generator with \(H = 4.0\) MJ/MVA delivers \(P_m = 1.0\) pu to an infinite bus. The power-angle curves are \(P_{max,1}=2.0\) pu before the fault, \(P_{max,2}=0.5\) pu during it, and \(P_{max,3}=1.5\) pu after the faulted circuit is cleared. A three-phase fault occurs at \(t=0\) and is cleared at \(t=0.15\) s. Compute the swing curve to \(t=0.30\) s with \(\Delta t = 0.05\) s and state whether the machine is stable.
Solution. The initial angle comes from the pre-fault curve, and the step constant from the degree form of the swing equation:
At \(t=0\) the fault is applied, so the accelerating power is discontinuous and the averaging rule applies:
Each subsequent line applies \(\Delta\delta_n = \Delta\delta_{n-1} + 5.625\,P_{a(n-1)}\). At \(t=0.15\) s the fault is cleared, so the average of the during-fault and post-fault values is used at that instant.
| \(t\) (s) | \(\delta\) (deg) | \(\sin\delta\) | \(P_e\) (pu) | \(P_a\) (pu) | \(k P_a\) | \(\Delta\delta\) (deg) |
|---|---|---|---|---|---|---|
| \(0^-\) | 30.000 | 0.5000 | 1.000 | 0 | — | — |
| \(0^+\) | 30.000 | 0.5000 | 0.250 | 0.750 | — | — |
| \(0_{avg}\) | 30.000 | — | — | 0.375 | 2.109 | 2.109 |
| 0.05 | 32.109 | 0.5316 | 0.266 | 0.734 | 4.130 | 6.239 |
| 0.10 | 38.348 | 0.6204 | 0.310 | 0.690 | 3.880 | 10.119 |
| \(0.15^-\) | 48.467 | 0.7484 | 0.374 | 0.626 | — | — |
| \(0.15^+\) | 48.467 | 0.7484 | 1.123 | −0.123 | — | — |
| \(0.15_{avg}\) | 48.467 | — | — | 0.251 | 1.415 | 11.534 |
| 0.20 | 60.001 | 0.8660 | 1.299 | −0.299 | −1.682 | 9.852 |
| 0.25 | 69.853 | 0.9388 | 1.408 | −0.408 | −2.296 | 7.556 |
| 0.30 | 77.409 | 0.9761 | 1.464 | −0.464 | −2.611 | 4.945 |
The increment \(\Delta\delta\) has been falling steadily since clearing — \(11.53^\circ\), \(9.85^\circ\), \(7.56^\circ\), \(4.95^\circ\) — because the accelerating power has been negative throughout the post-fault period. Continuing the table gives \(\delta = 82.35^\circ\) at \(0.35\) s and \(84.56^\circ\) at \(0.40\) s, after which \(\Delta\delta\) turns negative and the angle falls back. The maximum swing is about \(84.6^\circ\), comfortably short of the post-fault unstable equilibrium at \(180^\circ - \sin^{-1}(1.0/1.5) = 138.2^\circ\). The machine is stable.
Problem. Repeat Example 4 with the fault cleared at \(t = 0.30\) s instead of \(0.15\) s, and show that the machine now loses synchronism.
Solution. Everything up to \(t = 0.15\) s is unchanged, since the two systems are identical until then: \(\delta = 30.000^\circ,\ 32.109^\circ,\ 38.348^\circ,\ 48.467^\circ\) at \(t=0,\,0.05,\,0.10,\,0.15\) s, with the last increment \(\Delta\delta_3 = 10.119^\circ\). The fault now continues, so no averaging is done at \(0.15\) s and the during-fault accelerating power is used:
| \(t\) (s) | \(\delta\) (deg) | \(P_e\) (pu) | \(P_a\) (pu) | \(k P_a\) | \(\Delta\delta\) (deg) |
|---|---|---|---|---|---|
| 0.15 | 48.467 | 0.374 | 0.626 | 3.520 | 13.639 |
| 0.20 | 62.106 | 0.442 | 0.558 | 3.139 | 16.778 |
| 0.25 | 78.884 | 0.491 | 0.509 | 2.865 | 19.643 |
| \(0.30^-\) | 98.527 | 0.494 | 0.506 | — | — |
| \(0.30^+\) | 98.527 | 1.483 | −0.483 | — | — |
| \(0.30_{avg}\) | 98.527 | — | 0.011 | 0.062 | 19.705 |
| 0.35 | 118.232 | 1.322 | −0.322 | −1.810 | 17.896 |
| 0.40 | 136.128 | 1.039 | −0.039 | −0.221 | 17.675 |
| 0.45 | 153.803 | 0.662 | +0.338 | 1.901 | 19.576 |
| 0.50 | 173.379 | 0.173 | +0.827 | 4.653 | 24.229 |
Follow the sign of \(P_a\) down the table. After clearing at \(98.5^\circ\) the accelerating power is negative and the increment does begin to shrink — \(19.71^\circ\), \(17.90^\circ\), \(17.68^\circ\) — but far too slowly. At \(\delta = 138.2^\circ\), the post-fault unstable equilibrium, the deceleration runs out; beyond it \(P_e\) falls below \(P_m\) once more, \(P_a\) turns positive at \(153.8^\circ\), and the increments start growing again. The machine has pulled out of step.
The two examples together bracket the answer: cleared in \(0.15\) s the machine survives, cleared in \(0.30\) s it does not, so the critical clearing time lies between. Locating it by repeated integration is tedious, and Chapter 29 obtains it in one line from an area balance.
Problem. Three generators run in parallel on a \(50\) Hz system: \(G_1\), \(500\) MVA, \(H_1=5.0\) s; \(G_2\), \(300\) MVA, \(H_2=4.0\) s; \(G_3\), \(200\) MVA, \(H_3=3.0\) s, all on their own bases. A fault near \(G_2\) leaves the machines, at the first instant, with accelerating powers of \(0.30\), \(0.85\) and \(0.10\) pu respectively on a \(100\) MVA system base. Find each machine's initial angular acceleration, the acceleration of the centre of inertia, and the initial accelerations relative to it.
Solution. Refer every inertia constant to the \(100\) MVA base:
Each machine obeys \(\ddot\delta_i = \omega_s P_{ai}/2H_i'\):
| Machine | \(H'\) (s, 100 MVA) | \(P_a\) (pu) | \(\ddot\delta\) (elec rad/s²) | \(\ddot\delta\) (elec deg/s²) |
|---|---|---|---|---|
| \(G_1\) | 25.0 | 0.30 | 1.885 | 108.0 |
| \(G_2\) | 12.0 | 0.85 | 11.126 | 637.5 |
| \(G_3\) | 6.0 | 0.10 | 2.618 | 150.0 |
Subtracting the common motion gives the accelerations that actually threaten synchronism:
Every machine accelerates in absolute terms, because the system as a whole has lost load and is speeding up; only \(G_2\) pulls away from the group. It has both the largest accelerating power, being nearest the fault, and a modest inertia, and the product of those two facts identifies it immediately as the machine at risk. The other two, seen against the centre of inertia, fall back — which is the correct physical picture, since they are being asked to supply what \(G_2\) is not.
Chapter Summary
\(M\ddot\delta = P_m - P_e\), with the one approximation \(\omega_r\simeq\omega_s\) in the coefficient.
\(\dfrac{2H}{\omega_s}\ddot\delta = P_a\) in radians; \(\dfrac{H}{180f}\ddot\delta = P_a\) in degrees.
\(P_e = P_{max}\sin\delta\) makes it a driven pendulum — no elementary solution.
\(D\dot\delta/\omega_s\) decides small-signal decay; safely dropped in first-swing work.
\(H\) scales directly with rating: \(H_{sys}=H_{mach}G_{mach}/G_{sys}\).
Machines swinging together add: \(H_{eq}=H_1+H_2\) on a common base.
Machines swinging against each other combine in parallel: \(H_{eq}=H_1H_2/(H_1+H_2)\).
\(P_e=0\) gives \(\delta = \delta_0 + (\pi f P_m/2H)t^2\); small signals give \(\omega_n=\sqrt{\omega_sP_s/2H}\).
\(\Delta\delta_n = \Delta\delta_{n-1}+kP_{a(n-1)}\) with \(k = 180f(\Delta t)^2/H\).
At fault inception or clearing, use the mean of the two one-sided values of \(P_a\).
One equation per machine, coupled through \(\delta_i-\delta_j\) in the reduced \(Y_{bus}\).
A swing curve that turns is stable; one that climbs past \(180^\circ-\delta_{eq}\) is not.
Practice Problems
Take \(f = 50\) Hz unless stated otherwise, neglect damping unless it is given, and work in per unit on the base indicated. Where a swing curve is asked for, use \(\Delta t = 0.05\) s and remember the averaging rule at every switching instant.
- A \(150\) MVA, \(50\) Hz, four-pole generator has \(H = 5.2\) s. Write its swing equation on its own base with \(\delta\) in electrical degrees, and find the initial acceleration when the accelerating power is \(0.6\) pu. Repeat for a \(60\) Hz machine of the same \(H\) and comment on the difference.
- A generator delivering \(0.75\) pu suffers a three-phase fault at its terminals. Its inertia constant is \(H = 3.5\) s and the fault is cleared in \(0.12\) s. Using the constant-acceleration solution, find the rotor angle at clearing if \(\delta_0 = 25^\circ\), and the speed deviation in rev/min at that instant if the machine has two poles.
- Two units in one station are rated \(400\) MVA, \(H = 3.5\) s and \(600\) MVA, \(H = 4.5\) s. Find the equivalent inertia constant on a \(200\) MVA base for a disturbance external to the station, and on the same base for a fault on the busbar between them.
- A machine has \(H = 6\) s and operates at \(\delta_0 = 35^\circ\) on a power-angle curve with \(P_{max}=1.8\) pu. Find the synchronising power coefficient, the natural frequency of oscillation in hertz, and the period. If the damping coefficient is \(D = 3.0\) pu, find the damping ratio and the number of visible cycles before the envelope falls to \(1/e\).
- For the system of Example 4 (\(H=4\) s, \(P_m=1.0\), \(P_{max}\) of \(2.0/0.5/1.5\)), compute the swing curve for a clearing time of \(0.225\) s. Choose \(\Delta t\) so that clearing lands on a grid point, and state whether the machine is stable.
- A generator with \(H = 4.5\) s supplies \(0.9\) pu through a double-circuit line. A permanent three-phase fault at the middle of one circuit gives \(P_{max,1}=1.9\), \(P_{max,2}=0.55\) and \(P_{max,3}=1.35\) pu. Compute the swing curve to \(t=0.25\) s for clearing at \(0.10\) s, and find the angle and accelerating power at each step.
- Show that if the pre-fault and post-fault curves are identical (a fault cleared by reclosing onto a healthy line), the maximum swing angle depends only on \(P_{max,2}\), the clearing time, and \(H\). Verify numerically with \(H = 5\) s, \(P_m = 0.8\), \(P_{max}=1.6\) pu throughout, \(P_{max,2}=0.4\) pu, and a clearing time of \(0.15\) s.
- Three machines have \(H\) of \(20\), \(12\) and \(8\) s on a \(100\) MVA base. Immediately after a disturbance their accelerating powers are \(-0.20\), \(+0.55\) and \(+0.25\) pu. Find each machine's acceleration, the acceleration of the centre of inertia, and which machine is most at risk of losing synchronism. Explain why the machine with the largest \(P_a\) is not automatically the answer.