Synchronous Machine Dynamics and the Power-Angle Equation

Part 4 solved the network in the steady state: given the generation schedule and the loads, what voltage stands at each bus and what power flows in each line. Part 5 solved a single abnormal instant: given a short circuit, how much current flows and what must interrupt it. Both are algebraic problems. Neither says anything about time.

Yet the most consequential failures of a power system are failures in time. A line trips; the power that was flowing on it must redistribute; the generators that were supplying it suddenly find their electrical output changed while their steam or water input has not; their rotors accelerate or decelerate; and within a second or two either the machines settle into a new equilibrium or they lose synchronism and must be tripped, taking their load with them. The blackout that follows is not caused by any component exceeding its rating. It is caused by the loss of a shared rhythm.

That rhythm is what synchronism means. Every synchronous generator on an interconnected system runs at exactly the same electrical frequency, its rotor field locked to the rotating field of the network. The lock is not rigid: it is elastic, and the machine transmits power precisely in proportion to how far it is stretched. Stability is the study of whether that elastic lock survives a disturbance.

Three classes of disturbance separate the subject into three problems, and the traditional names describe the size of the disturbance rather than the mechanism.

All three rest on the same two ingredients, and this chapter builds both. The first is a mechanical equation of motion for the rotor, which needs the rotor's inertia. The second is an expression for the electrical power the machine delivers as a function of rotor position, which needs a machine model and a network. Chapter 27 combines them into the swing equation; Chapters 28 and 29 solve it in the small and in the large.

Strip away the electromagnetics and a turbine-generator set is a heavy cylinder on bearings with a torque applied at each end: the prime mover drives it, the electromagnetic reaction of the stator retards it. Newton's second law for rotation is the whole of the mechanical model.

Here \(J\) is the combined moment of inertia of turbine and generator in kg·m², \(\theta_m\) the mechanical angular position of the rotor in radians, \(T_m\) the mechanical torque of the prime mover, \(T_e\) the retarding electromagnetic torque, and \(T_a\) their difference, the accelerating torque. When \(T_a=0\) the rotor turns at constant speed and the machine is in equilibrium; any imbalance accelerates it.

Using \(J\) directly is inconvenient because it is an absolute quantity: a 500 MW machine and a 5 MW machine differ in \(J\) by three orders of magnitude, and no intuition survives that range. Power engineering therefore normalises the inertia by the machine's own rating, and the normalising quantity that turns out to be useful is the stored kinetic energy at synchronous speed.

\(G\) is the machine's three-phase MVA rating and \(\omega_{sm}\) its synchronous speed in mechanical radians per second. The units of \(H\) are MJ/MVA, which is to say seconds: \(H\) is the time for which the machine could supply its own rated output from stored kinetic energy alone, if nothing drove it. That interpretation is why \(H\) is so stable a number across machine sizes. A 5 MVA unit and a 500 MVA unit both store roughly the same energy per MVA, because both are designed to the same mechanical stress limits, and \(H\) falls in a narrow band.

To use \(H\) in the equation of motion, the mechanical angle must give way to the electrical angle. A machine with \(P\) poles turns through \(2/P\) of an electrical revolution for every mechanical revolution, so \(\theta = (P/2)\theta_m\) and \(\omega_s = (P/2)\omega_{sm}\), where \(\theta\) and \(\omega_s\) are the electrical angle and electrical angular frequency. Substituting into the kinetic energy expression and writing it in terms of the electrical speed defines the inertia coefficient \(M\):

Equating the two expressions for the stored energy, \(GH = \tfrac12 M\omega_s\), gives the relation actually used in every stability calculation. With \(\omega_s = 2\pi f\) electrical radians per second:

Two conversions recur constantly. Changing base follows the same reciprocal rule as any per-unit quantity in Chapter 4, since \(GH\) — the stored energy in MJ — is an absolute number that cannot depend on the base chosen:

The second line is the definition of coherency. Two machines on the same busbar, or on buses separated by negligible reactance, keep the same rotor angle throughout a disturbance and can be replaced by a single machine whose stored energy is the sum of theirs. Machines far apart do not, and their reduction to an equivalent — which produces \(H_1H_2/(H_1+H_2)\) rather than a sum — belongs to Chapter 27, because it depends on the swing equation itself.

The equation of motion is written in \(\theta\), the absolute electrical position of the rotor, and that variable is useless for computation: at \(50\) Hz it grows by \(314\) radians every second, so its interesting departures from uniform rotation are lost in the noise of an enormous linear ramp. The remedy is to observe the rotor from a frame that is itself rotating at synchronous speed.

The rotor angle \(\delta\) is constant when the machine runs at exactly synchronous speed, whatever the value of that constant, and changes only when the machine speeds up or slows down. It is therefore precisely the variable the mechanical equation wants, and the second derivative is unchanged by the substitution, so nothing is lost.

Three points about \(\delta\) cause more trouble than the rest of Part 6 combined. First, it is an electrical angle: a two-pole machine whose rotor slips one mechanical degree slips one electrical degree, but a twenty-pole machine slipping one mechanical degree slips ten. Second, \(\delta\) has no absolute meaning — only differences between machine angles, or between a machine angle and a reference bus angle, are physical, because the choice of where the synchronous reference frame's zero lies is arbitrary. Third, \(\delta\) is simultaneously a mechanical position and the phase angle of a voltage phasor, and it is this double identity that couples the two halves of the problem.

One further simplification is standard and worth stating explicitly. The equation of motion is a torque balance, but stability is discussed in terms of power. Since \(P = T\omega\), and since \(\omega\) departs from \(\omega_s\) by well under one per cent throughout a first-swing study, the per-unit power and per-unit torque are numerically equal to within that error. Dividing the torque equation by \(\omega_s\) and working in per unit therefore converts it into a power balance without further apology:

This is the swing equation, and Chapter 27 is devoted to it. Everything that remains in this chapter is the construction of the right-hand side: what \(P_e\) is, as a function of \(\delta\).

Chapter 21 established that a synchronous machine offers three different reactances to a disturbance, depending on how long ago the disturbance began: the subtransient \(X_d''\) while the damper-winding currents last, the transient \(X_d'\) while the field-winding current is still adjusting, and the synchronous \(X_d\) once everything has settled. Chapter 21 used \(X_d''\), because a breaker rating is set by the current in the first few cycles. Stability studies use \(X_d'\), and the reason is the same reason stated differently: the time window is different.

The damper windings are short-circuited copper bars with a time constant of a few tens of milliseconds; by the time the first swing is under way, a few hundred milliseconds in, their currents have gone. The field winding is a large, well-insulated coil with a time constant of several seconds; during a swing of one second its flux linkage has hardly changed at all. The constant-flux-linkage theorem then does the rest: a closed winding of negligible resistance holds its total flux linkage constant, so the field flux linkage \(\lambda_f\) is a constant of the motion over the window of interest. The voltage proportional to that flux linkage is the voltage behind transient reactance.

\(|E'|\) is not a nameplate quantity. It is computed once, from the prefault operating condition, by the same phasor arithmetic as any Thévenin source: take the terminal voltage and current delivered before the disturbance and add the drop across \(jX_d'\).

Once found, \(|E'|\) is held fixed for the duration of the study while \(\delta\) varies. The same construction with \(X_d\) in place of \(X_d'\) gives the steady-state excitation emf \(E\), which is the quantity used in the steady-state stability limit of Chapter 28.

The simplest arrangement that contains the whole of the physics is one machine connected through a purely reactive network to a bus so strong that its voltage and frequency cannot be disturbed — the single machine infinite bus or SMIB system. The infinite bus of Chapter 3 is the \(S_{sc}\to\infty\) idealisation Chapter 25 made precise; it stands in for the rest of a large interconnected system as seen by one small machine.

Reduce the whole chain — transient reactance, transformer, and the parallel lines — to a single series reactance \(X\), and let the two ends be \(V_s = V_1\angle\delta\) and \(V_r = V_2\angle0^\circ\). The current is fixed by Ohm's law, and the complex power leaving the sending end follows.

Separating real and imaginary parts, and noting that a purely reactive network consumes no real power so that \(P_s = P_r = P_e\):

Three consequences deserve to be read off immediately. Power flows from the leading machine to the lagging one, so a generator delivering power must run with its rotor ahead of the network and \(\delta>0\); a synchronous motor absorbs power with \(\delta<0\). The transmissible power is inversely proportional to the reactance, which is why Chapter 14 cared about series compensation, why the reactors of Chapter 25 cost stability, and why losing one of two parallel lines nearly halves \(P_{max}\). And no matter what \(\delta\) does, \(P_e\) can never exceed \(P_{max}\) — a hard ceiling on how much a given machine can push through a given network at a given excitation.

When the connecting network is not purely reactive the sine law generalises but does not change character. Reduce everything between the internal node and the infinite bus to a two-port, with driving-point impedance \(Z_{11}=|Z_{11}|\angle\theta_{11}\) and transfer impedance \(Z_{12}=|Z_{12}|\angle\theta_{12}\). Then

Resistance does two things: it adds a constant term \(P_c\), the power dissipated in the network by the machine's own emf, and it shifts the curve sideways by \(\gamma\) so that \(P_e\ne0\) at \(\delta=0\). Both effects are small on transmission networks, where \(X/R\) is 10 or more, and the lossless form is used throughout Part 6 with the understanding that neglecting resistance is slightly pessimistic — resistance is a damping mechanism, and dropping it removes damping the real system has.

The derivation above treated the machine as a single reactance, which is exact for a round-rotor turbo-alternator whose air gap is uniform. A salient-pole machine — every hydro set, and every large synchronous motor — has a rotor with pronounced poles, so the air gap under a pole is much smaller than the air gap between poles. The armature flux therefore meets a different magnetic path depending on where the rotor happens to be, and the machine has two reactances: \(X_d\) along the direct (pole) axis and \(X_q\) along the quadrature (interpolar) axis, with \(X_q\) between one half and two thirds of \(X_d\).

A two-reaction analysis resolves the armature current into direct- and quadrature-axis components and adds their contributions. The result is one extra term:

The first term is the familiar excitation power, proportional to the field. The second is the reluctance power, and it is independent of excitation entirely: it exists because the rotor, being magnetically unsymmetrical, is pulled toward the position of minimum reluctance by the stator field alone. It vanishes when \(X_d=X_q\), recovering the round-rotor result, and it repeats twice per revolution — hence \(\sin2\delta\) — because a two-pole rotor has two positions of minimum reluctance per turn.

The picture explains the two effects saliency has. Because the reluctance term is positive for \(0<\delta<90^\circ\) and negative beyond, it adds to the excitation term on the rising side of the curve and subtracts on the falling side. The maximum therefore rises — typically by five to ten per cent — and moves to an angle below \(90^\circ\), usually near \(70^\circ\). And because the reluctance term survives with no field current at all, a salient-pole machine can hold synchronism unexcited, which is exactly what a reluctance motor is.

For transient stability studies the distinction largely disappears. During the first swing the machine is represented by \(E'\) behind \(X_d'\), and \(X_q'\approx X_d'\) for practical salient-pole machines, so the classical model of Section 26-4 is used for salient and round-rotor machines alike. Saliency matters for the steady-state limit of Chapter 28, for excitation planning, and for the small-signal analysis of Chapter 33 — which is why the reluctance term is derived here and then set aside.

Draw the power-angle curve and lay across it the horizontal line \(P_e = P_m\), the constant mechanical input the prime mover is delivering. The two curves intersect twice in \(0\le\delta\le180^\circ\), at \(\delta_0=\sin^{-1}(P_m/P_{max})\) and at \(\delta_{max}=180^\circ-\delta_0\). At either point the accelerating power is zero and the rotor turns at exactly synchronous speed, so both are equilibria of the swing equation. They could hardly be more different.

Perturb the rotor slightly at \(\delta_0\), on the rising side of the curve. A small advance \(+\Delta\delta\) increases \(P_e\) above \(P_m\), so the accelerating power \(P_m-P_e\) becomes negative and the rotor is pulled back. A small retardation does the reverse. The equilibrium restores itself, and the machine oscillates about it. Perturb the rotor at \(\delta_{max}\), on the falling side. A small advance now decreases \(P_e\), the accelerating power becomes positive, the rotor advances further, \(P_e\) falls further still, and the machine runs away. Synchronism is lost.

What distinguishes the two is nothing but the sign of the slope, and the slope is important enough to have its own name.

The word "stiffness" is exact, not metaphorical. Substitute \(\delta=\delta_0+\Delta\delta\) into the swing equation of Section 26-3, expand \(P_e\) to first order, and use \(P_m = P_e(\delta_0)\):

A mass on a spring, with \(M\) the mass and \(P_s\) the spring constant. If \(P_s>0\) the solution is a sustained oscillation of natural frequency \(\omega_n\); if \(P_s<0\) the roots are real and one is positive, giving exponential divergence. Real machines oscillate at \(0.5\) to \(2\) Hz by this formula, and those are the electromechanical oscillations seen on interconnected systems. The undamped result is an artefact of neglecting resistance and damper-winding torque; Chapter 27 restores a damping term, and Chapter 33 shows how an excitation controller can add damping deliberately — or, if badly tuned, subtract it.

Notice finally what \(P_s\) depends on. Increasing excitation raises \(|E'|\), which raises \(P_{max}\), which lowers \(\delta_0\) for the same \(P_m\) and raises \(\cos\delta_0\) — so \(P_s\) rises twice over. Adding series reactance does the opposite. Loading the machine more heavily pushes \(\delta_0\) toward \(90^\circ\) and drives \(P_s\) toward zero: a heavily loaded machine is a weak spring, oscillates slowly, and has little margin left. All three statements are quantitative versions of operating experience, and all three come from one derivative.

A short circuit does not change the machine, its inertia, or its mechanical input. It changes the network, and therefore the transfer reactance, and therefore \(P_{max}\). Since the fault is applied and later cleared, the system passes through three distinct networks in a few hundred milliseconds, and each has its own power-angle curve.

The during-fault reactance is the one that requires work, because the faulted point is a third node grounded in the middle of the network, and the transfer reactance from the machine's internal node to the infinite bus must be found with that node shorted. The standard tool is the star–delta transformation. Take the three branches meeting at the sending bus — toward the machine, toward the healthy line, and toward the fault — and convert that star to a delta; the delta side joining the machine node to the infinite bus is the transfer reactance sought, and the two branches that terminate on the shorted node carry current that never reaches the receiving end and so do not appear in \(P_e\).

Two limiting cases are worth carrying in the head. A three-phase fault at the sending bus itself gives \(X_F=0\), hence \(X_{AR}=\infty\) and \(P_{max,2}=0\): the machine delivers nothing at all while the fault is on, and accelerates under the full mechanical input. This is the worst case and is the one used to define critical clearing time in Chapter 29. At the other extreme an unsymmetrical fault does not short the positive-sequence network to earth directly; the sequence networks of Chapter 23 appear at the fault point as a shunt impedance \(Z_2\), \(Z_2+Z_0\), or \(Z_2\parallel Z_0\) depending on fault type, so \(X_F\) is finite and rather large, and some power continues to flow. That is the quantitative form of a fact stated in Chapter 24 and used ever since: a single line-to-ground fault is the mildest disturbance to stability and a three-phase fault the harshest, in exactly the reverse order of their frequency of occurrence.

The figure shows the whole of the transient stability problem in one picture. Before the fault the machine sits where the mechanical input line crosses the tall curve. The instant the fault appears the operating point drops vertically to the low curve — the rotor cannot move instantaneously, so \(\delta\) is unchanged while \(P_e\) collapses — and the machine now has more mechanical input than electrical output. It accelerates, and \(\delta\) grows. When the breakers clear the faulted circuit the operating point jumps up to the post-fault curve; if it lands where \(P_e>P_m\), the rotor begins to decelerate, and whether it decelerates enough before reaching the unstable equilibrium of the post-fault curve is precisely the question the equal-area criterion of Chapter 29 answers.

Take \(f=50\) Hz unless stated otherwise, work in per unit on the stated base, and use the classical model with \(|E'|\) held constant. Neglect resistance and machine damping.