Chapter 15 — The synchronous machine

Every element so far has had one thing in common: its parameters sit still. A resistor is a resistor, a line's surge impedance does not care what time it is, and even the saturating transformer of Chapter 14 keeps a constant stamped conductance with the nonlinearity pushed into an injection. A synchronous machine breaks that. Its stator windings are magnetically coupled to a rotor that is physically turning, so the inductance seen between two stator terminals is a function of rotor position, and rotor position changes every step.

This chapter is about the transformation that rescues the situation, the model it produces, the equations that turn the model into something the network solver can accept, and the awkward business of starting a machine in balance. It is the longest chapter in the course, because the synchronous machine is the most elaborate component in the library.

Learning objectives

By the end of this chapter you should be able to:

• Say why a rotating machine cannot be stamped as a fixed inductance matrix, and what Park's transformation does about it.
• Write the $dq$ stator voltage equations and identify which terms a stability program drops and an EMT program keeps.
• Name the rotor circuits behind $X_d^{\prime }$, $X_d^{″}$ and the open-circuit time constants, and explain the two parameter-entry routes.
• Apply the swing equation, and say when a single lumped inertia is not enough.
• Describe the voltage-behind-subtransient-reactance interface and the one-step lag it carries.
• Recover a machine's reactances and time constants from a short-circuit oscillogram.

15.1 Why a machine is not just another branch

Take a salient-pole machine and measure the self-inductance of stator phase $a$ as you turn the rotor by hand. When the rotor's direct axis lines up with the phase-$a$ winding the flux path is nearly all iron and the inductance is high; a quarter of a pole-pitch later the flux crosses the interpolar gap and the inductance is low. The variation is approximately

$$L_{aa}(\theta )=L_{s0}+L_{s2}\cos 2\theta ,$$

with the mutual terms varying the same way around their own reference angles,

$$L_{ab}(\theta )=-M_{s0}-M_{s2}\cos (2\theta +\frac{\pi }{3}),$$

and the stator-to-rotor mutuals varying at $\cos \theta$ rather than $\cos 2\theta$. The doubling is geometric: a rotor pole looks the same whichever end faces you, so the reluctance seen by a stator phase repeats twice per electrical revolution.

Now recall what Chapter 2 does with an inductance. Trapezoidal integration of $v=Ldi/dt$ gives a companion conductance $G=\mathrm{\Delta }t/2L$, and Chapter 3 puts that conductance into $\mathbf{G}$ and factorizes once for the whole run. If $L$ is a function of $\theta$ and $\theta$ advances $\omega \mathrm{\Delta }t$ radians every step, then $\mathbf{G}$ changes every step and the factorization is worthless. This is the same objection Chapter 14 raised against restamping a saturating inductance, except that there the nonlinearity was occasional and here the variation is relentless — the machine never stops turning.

There is a second problem. The rotor windings are short-circuited coils buried in a moving frame; their currents are not network unknowns and their voltages are not measurable at any terminal. Even if you were willing to refactorize, you would still have to carry them somewhere.

15.2 Park's transformation

The trick, published by R. H. Park in 1929, is to stop describing the stator in terms of three physical windings and start describing it in terms of two fictitious windings that rotate with the rotor, plus a third that carries whatever is left over. Define

$$\left[\begin{array}{c}{f}_d\\ f_q\\ f_0\end{array}\right]=\frac{2}{3}\left[\begin{array}{ccc}\cos \theta & \cos (\theta -\frac{2\pi }{3})& \cos (\theta +\frac{2\pi }{3})\\ -\sin \theta & -\sin (\theta -\frac{2\pi }{3})& -\sin (\theta +\frac{2\pi }{3})\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}{f}_a\\ f_b\\ f_c\end{array}\right],$$

where $f$ stands for voltage, current or flux linkage alike and $\theta$ is the rotor's electrical angle. The $\frac{2}{3}$ scaling is the amplitude-invariant convention: a balanced set of phase currents with peak $\hat{I}$ maps to $d$ and $q$ components whose magnitude is also $\hat{I}$. NumaSim uses this convention throughout, which is why its per-unit bases are peak quantities — $V_{\text{base}}=V_{\text{rated}}\sqrt{2/3}$ peak line-to-neutral and $I_{\text{base}}=\frac{2}{3}{S}_{\text{base}}/V_{\text{base}}$ peak phase. Other texts use $\sqrt{2/3}$ instead to make the transformation power-invariant; the physics is identical and only the bookkeeping differs, but mixing the two is a reliable source of factor-of-1.5 errors.

Apply the transformation to the position-dependent inductance matrix and the $\theta$ dependence disappears. What comes out is

$${\psi }_d=-L_d{i}_d+L_{ad}(i_{fd}+i_{1d}),{\psi }_q=-L_q{i}_q+L_{aq}(i_{1q}+i_{2q}),{\psi }_0=-L_0{i}_0,$$

with every inductance a constant. The physical explanation is worth a sentence: the $d$ and $q$ windings turn with the rotor, so from the rotor's point of view they are stationary, and the reluctance between a stationary winding and a stationary pole does not change. The negative signs are the generator convention — positive stator current flows out of the machine.

Zero sequence separates cleanly. ${\psi }_0$ couples to no rotor circuit at all, so it carries its own impedance and its own story, and NumaSim gives it its own parameters (X0, R0) rather than deriving it from the others. Real machines have $X_0$ smaller than the positive-sequence subtransient reactance, which is impossible to reproduce with a passive neutral element and forces the coupled stamp described in §15.6.

15.3 The $dq$ voltage equations

Transform the stator voltage equations and two terms appear where you might have expected one:

$$v_d=-R_a{i}_d+\underset{\text{transformer emf}}{\underbrace{\frac{d{\psi }_d}{dt}}}-\underset{\text{speed emf}}{\underbrace{{\omega }_r{\psi }_q}},v_q=-R_a{i}_q+\frac{d{\psi }_q}{dt}+{\omega }_r{\psi }_d.$$

These are in per unit with time also in per unit, so the derivatives carry no ${\omega }_0$; write them against seconds, as the code does, and each $d\psi /dt$ picks up a $1/{\omega }_0$.

The ${\omega }_r\psi$ terms are the speed emf — the voltage generated by moving a conductor through a field, which is what a generator is for. The $p\psi$ terms are the transformer emf — the voltage generated by a changing flux linkage in a winding that is not moving relative to it. In the steady state the $dq$ fluxes are constant and the transformer terms vanish, leaving the familiar $v_q=e_q-R_a{i}_q-X_d{i}_d$ picture that phasor analysis assumes.

This is the sharpest instance in the whole course of the distinction Chapter 1 drew. Stability programs discard $p{\psi }_d$ and $p{\psi }_q$. They do it deliberately and for good reason: dropping the stator transients removes the 60 Hz DC-offset dynamics from the model, which lets a stability program take 10 ms steps and still track a rotor swing that plays out over seconds. The cost is that the model can no longer represent the offset current after a fault, the second-harmonic torque that rides on it, or anything faster than a few hertz.

An EMT program keeps them. That is the entire difference in this context, and it is why the terminal short circuit in §15.9 looks like this:

Stator currents in the first 110 ms of a bolted three-phase fault on an unloaded 100 MVA machine; the dashed lines mark the rated peak of $\pm 5.92$ kA. Each phase carries the same AC component on its own DC offset, and phase $a$ reaches 9.2 times rated. Everything asymmetric here comes from the transformer-emf terms; a stability formulation would draw three symmetric sinusoids.

If you have ever wondered what "EMT" buys you over a transient-stability run, the answer for machines is exactly these two terms.

The rotor equations have no such subtlety — the rotor windings really are stationary in the $dq$ frame, so they are plain $R$-$L$ circuits:

$$v_{fd}=R_{fd}{i}_{fd}+\frac{d{\psi }_{fd}}{dt},0=R_{1d}{i}_{1d}+\frac{d{\psi }_{1d}}{dt},0=R_{1q}{i}_{1q}+\frac{d{\psi }_{1q}}{dt},$$

with a fourth for $2q$ on a round rotor. The dampers are shorted, hence the zeros. NumaSim integrates these flux linkages — four on a round rotor, three on a salient-pole one — with the trapezoidal rule, the same integrator as every other dynamic element in the program and for the same reason: second-order accurate and A-stable, so a seeded steady state stays put.

15.4 Rotor circuits and the standard parameters

How many rotor circuits should the model carry? The honest answer is that a solid steel rotor is a continuum and any lumped circuit is a fit, but the industry has settled on a small standard set:

Rotor type	$d$-axis circuits	$q$-axis circuits
Round rotor (turbo-generator)	field $fd$, damper $1d$	dampers $1q$, $2q$
Salient pole (hydro)	field $fd$, damper $1d$	damper $1q$

That is precisely the component's rotor_type switch. The extra $q$-axis circuit on a round rotor stands in for eddy currents in the solid rotor body, which have no counterpart in a laminated salient-pole machine with a physical damper cage.

Nobody publishes $L_{ad}$, $L_{fd}$ and $R_{1q}$. What a manufacturer puts on a datasheet is the standard parameter set, defined by how the machine responds rather than by what is inside it:

• $X_d$, $X_q$ — synchronous reactances, seen once all rotor transients have died.
• $X_d^{\prime }$, $X_q^{\prime }$ — transient reactances, seen while the field still holds its flux but the dampers have given up.
• $X_d^{″}$, $X_q^{″}$ — subtransient reactances, seen in the first cycle while every rotor circuit holds its flux.
• $X_l$, $R_a$ — stator leakage and armature resistance.
• $T_{d0}^{\prime }$, $T_{d0}^{″}$, $T_{q0}^{\prime }$, $T_{q0}^{″}$ — the corresponding open-circuit time constants.

The nesting $X_d^{″}<X_d^{\prime }<X_d$ is not a convention, it is physics: flux trapped in a closed rotor circuit cannot cross the air gap, so it looks to the stator like a short-circuited turn that lowers the effective reactance. The more circuits still holding flux, the lower the reactance. Typical values for a large turbo-generator are $X_d=1.8$, $X_d^{\prime }=0.3$, $X_d^{″}=0.2$ pu — the subtransient reactance is nine times smaller than the synchronous one, which is why a generator's fault current starts high and ends up unimpressive.

Between the datasheet set and the circuit set is a closed-form inversion. In the $d$ axis, with $L_{ad}=X_d-X_l$,

$$L_{fd}=\frac{L_{ad}(X_d^{\prime }-X_l)}{L_{ad}-(X_d^{\prime }-X_l)},R_{fd}=\frac{L_{ad}+L_{fd}}{{\omega }_0{T}_{d0}^{\prime }},$$

and a similar pair for $1d$ built from $X_d^{″}$ and $T_{d0}^{″}$. The component does this for you and exposes both ends: param_input_mode selects Standard (datasheet) or Fundamental. Use the first unless you are reproducing a published model that quotes the circuit parameters directly.

The short-circuit time constants

Datasheets quote open-circuit time constants because those are what the decay test measures. What appears in a fault waveform is the short-circuit constants, related by the reactance ratio:

$$T_d^{\prime }=T_{d0}^{\prime }\frac{X_d^{\prime }}{X_d},T_d^{″}=T_{d0}^{″}\frac{X_d^{″}}{X_d^{\prime }}.$$

For the numbers above, $T_{d0}^{\prime }=8$ s becomes $T_d^{\prime }=1.33$ s and $T_{d0}^{″}=30$ ms becomes $T_d^{″}=20$ ms. Confusing the two is the most common error in reading a short-circuit oscillogram, and §15.9 measures both.

15.5 The swing equation and the shaft

The electrical side produces a torque; the mechanical side decides what the rotor does about it. In per-unit,

$$2H\frac{d{\omega }_r}{dt}=T_m-T_e-D({\omega }_r-1),\frac{d\delta }{dt}={\omega }_0({\omega }_r-1),$$

with the air-gap torque falling straight out of the $dq$ fluxes and currents:

$$T_e={\psi }_d{i}_q-{\psi }_q{i}_d.$$

The inertia constant $H$ is stored kinetic energy at synchronous speed divided by the machine's rating, in seconds. Its physical meaning is sharp: a machine with $H=3.5$ s that suddenly loses all its electrical load and keeps full mechanical torque accelerates at $1/(2H)=0.143$ pu/s, so it would reach 110 % speed in about 0.7 s. Large steam units land between 2.5 and 6 s; hydro sets run lower, and the difference is why hydro governors are tuned differently.

$D$ is a lumped damping coefficient, and it is worth being honest about it: in an EMT model the physical damping is already there, produced by the damper windings through $T_e$. The $D$ term exists for turbine windage, friction and load-frequency effects, and setting it to zero (as every sample in this course does) is defensible.

When one mass is not enough

A 600 MW steam turbine-generator is not a flywheel. It is a high-pressure turbine, two or three low-pressure turbines, the generator rotor and often an exciter, strung along a shaft tens of metres long and coupled by a spring stiff enough to look rigid at 60 Hz but not at 20. Replace the scalar swing equation with a chain:

$$2H_i\frac{d{\omega }_i}{dt}=T_{m,i}-[i=g]T_e-D_i({\omega }_i-1)-\sum _j[K_{ij}({\delta }_i-{\delta }_j)+D_{ij}({\omega }_i-{\omega }_j)],$$

each mass with its own inertia and driving torque, adjacent masses coupled by a torsional stiffness $K_{ij}$ and a mutual damping $D_{ij}$. Only the generator mass feels $T_e$; only the generator mass's angle drives the $dq$ frame. NumaSim's shaft_model switches this on, giving up to four turbine masses plus the generator and an optional exciter, each with H_*, D_* and a K_* spring between neighbours.

The reason to bother is torsional resonance. A chain of five masses has four torsional modes, typically between 10 and 45 Hz. A series-compensated transmission line has an electrical resonance at $f_0\sqrt{X_C/X_L}$, often 20 to 40 Hz, which appears to the rotor as a torque at the complementary frequency $f_0-f_{\text{er}}$. When that lands on a torsional mode the two resonances feed each other, and subsynchronous resonance can tear a shaft apart — as it did at Mohave in 1970 and 1971. A single-mass model cannot see any of it, because a single mass has no torsional modes at all.

15.6 Interfacing the machine to the network

The $dq$ model is complete, but it lives in a rotating frame and the network solver lives in phase coordinates. Something has to join them.

The standard device is voltage behind subtransient reactance. Choose an average $X^{″}=(X_d^{″}+X_q^{″})/2$ and treat the machine, as far as the network is concerned, as a fixed series $R_a+j{X}^{″}$ branch from an internal emf node to each terminal. That impedance is constant, so it discretizes into a constant companion conductance by the Chapter 2 recipe and stamps into $\mathbf{G}$ once. Everything that the constant impedance fails to capture — saliency, saturation, and the whole rotor dynamics — is carried by the internal emf, which becomes a per-step Norton current injection. The pattern is exactly Chapter 14's: constant matrix, moving injection.

The internal emf comes from the subtransient fluxes,

$$e_d^{″}=\frac{1}{{\omega }_0}\frac{d{\psi }_d^{″}}{dt}-{\omega }_r{\psi }_q^{″},e_q^{″}=\frac{1}{{\omega }_0}\frac{d{\psi }_q^{″}}{dt}+{\omega }_r{\psi }_d^{″},$$

— both emf terms of §15.3, kept — and is projected back to phase quantities by the inverse Park transformation. The saliency the average $X^{″}$ threw away is added back into the injection as a correction term proportional to $\mathrm{\Delta }x=(L_d^{″}-L_q^{″})/2$ times the stator current, which is why a salient machine with $X_d^{″}\ne X_q^{″}$ still behaves correctly even though its stamped conductance is a single average.

Zero sequence needs care. Because a real machine's $X_0$ is usually below $X^{″}$, you cannot get the right zero-sequence impedance by adding anything to the neutral — a passive neutral element can only raise$Z_0$. NumaSim therefore stamps a coupled $3\times 3$ admittance whose symmetric-circulant structure gives one impedance to the positive/negative sequence and an independent one to zero sequence. In that form the matrix–vector product collapses to a per-phase scalar plus a common-mode correction, so the coupling costs almost nothing.

The one-step lag

There is a loose end. The internal emf for this step depends on the rotor fluxes, which depend on the stator current $(i_d,i_q)$ — and the stator current is what the network solve is about to produce. The injection must be built before the solve, so it uses the previous step's current. That is a one-step lag, and it is a genuine approximation, not a bookkeeping detail.

At sensible time steps it is harmless. At coarse steps, or when the machine sits on a very stiff bus, the lag can push the interface into a growing oscillation at the Nyquist frequency — the model appears to "blow up" for no visible physical reason. The classic remedies are to reduce $\mathrm{\Delta }t$, to add a little damping to the interface, or to iterate the machine and the network to convergence within the step. If you ever see a machine current growing by a constant ratio every step, suspect the interface before you suspect the machine.

15.7 Starting in balance

Give a machine model a set of parameters and press run, and it will not start on an operating point unless you tell it one. This causes more confusion than any other aspect of machine modelling, so it is worth being explicit about what goes wrong.

The rotor flux linkages are state variables. Left at zero they represent an unexcited machine, which then has to magnetize itself through the field circuit — a first-order process with time constant $T_{d0}^{\prime }$, around 8 seconds. Meanwhile the rotor angle has to walk from wherever it started to the angle that carries the applied torque. The existing synchronous machine load sample shows this plainly: with a 1.05 pu field and 0.4 pu of shaft torque, the rotor angle passes 6.4° at 0.2 s, 28.6° at 2 s, 37.1° at 8 s, and only reaches its final 39.8° after 20 seconds of simulated time. Nothing is wrong. The machine is magnetizing.

NumaSim offers three tools, and they solve different problems:

1. Flux seeding. The efd0, id0, iq0 and init_angle parameters place the rotor fluxes directly on their steady-state values: ${\psi }_{fd}=(L_{ad}+L_{fd})i_{fd0}-L_{ad}{i}_{d0}$ and its companions. Left at their defaults this reduces exactly to the open-circuit point, which is what the short-circuit lab wants.
2. Soft start (startup_tau). Seeding the rotor does not seed the stator companion history, so the armature still takes a "switch closes at $t=0$" inrush that rings down over the armature time constant. A positive startup_tau ramps the internal emf in with a first-order law so the network energizes smoothly.
3. Rotor lock. Freezes the mechanical dynamics — speed held, angle held — while the electrical model keeps running. Use it to let electrical transients settle before allowing the rotor to swing, or, as in §15.9, to enforce the constant-speed assumption a textbook analysis depends on.

What none of these do is compute the operating point for you. That is the job of a load-flow solution, which fixes each machine's terminal voltage and power and back-calculates the $\delta$, $E_{fd}$, $i_{d0}$ and $i_{q0}$ that produce them. Chapter 18 covers the matching question for controllers — an AVR seeded inconsistently with its machine will fight the seed and produce a transient that looks physical and is not.

15.8 Lab A: open-loop on a stiff bus

Open the synchronous machine load sample in simulator →

A 100 MVA round-rotor machine shares a 13.8 kV bus with a near-ideal source and a 40 MW + $j$10 MVAr load. Field voltage is pinned at efd0 = 1.05 pu and shaft torque at tm0 = 0.4 pu, with no regulator on either. Run it for the full 20 s.

1. Confirm the power-angle relation. Steady power across the synchronous reactance goes as $P\approx (E_{fd}/X_d)\sin \delta$, which for these values is $0.58\sin \delta$ pu. Read delta and Pgen at 20 s: the angle settles at 39.8°, so the relation predicts $0.58\times 0.640=0.371$ pu against the 0.397 pu the machine is actually delivering. Seven per cent is about the accuracy the expression deserves — it ignores $R_a$, saliency, and the fact that the terminal is not quite an infinite bus.
2. Watch the field transient, not the swing. The angle's slow walk from 6.4° to 39.8° is the $d$-axis flux decaying on $T_{d0}^{\prime }$, not a mechanical oscillation. Confirm by noting the time scale — an electromechanical swing for $H=3.5$ s would have a period around a second, and this takes twenty.
3. Move the field. Raise efd0 to 1.3 pu. Reactive output climbs from −12.7 MVAr through zero to +4.6 MVAr while the angle falls to 31.1°. A stronger field both exports vars and buys back rotor angle, because a larger $E_{fd}$ needs less $\sin \delta$ for the same $P$. This is the lever an AVR pulls automatically.
4. Approach pull-out. Raise tm0 to 0.62 pu. The machine carries 61.7 MW but needs more than 70° to do it, and takes far longer to get there — the $P$–$\delta$ curve flattens as $\delta$ approaches 90°, so each extra megawatt costs more angle and more settling time.

15.9 Lab B: a three-phase terminal short circuit

Open the machine short-circuit test in simulator →

This is the classic parameter-identification test, and it justifies everything in §15.4. The same 100 MVA machine runs open-circuited at rated terminal voltage; at $t=0.5$ s a bolted three-phase fault closes across the terminals and stays closed. The rotor is locked at synchronous speed for the whole run, so the constant-$\omega$ assumption the analysis depends on is enforced rather than hoped for.

1. Check the starting point. Before the fault, terminal voltage is 0.9999 pu and stator current is zero. efd0 = 1.0 really does put an unloaded machine at rated volts.
2. Read the first peak. Phase $a$ reaches 54,546 A, which against a rated peak of 5,917 A is 9.22 pu. The ceiling is 10 pu: the AC component starts at $E/X_d^{″}=5$ pu and a full DC offset can double it. Check that the three offsets sum to zero — no fault can give every phase a full offset.
3. Recover the reactances. Take the AC envelope (half the peak-to-peak swing, cycle by cycle), subtract its final value, and plot on a log axis. Two straight lines appear — see below.
4. Read the armature time constant. The DC offset decays with $\tau =0.210$ s. The textbook prediction $X_2/(\omega R_a)$ gives 0.265 s — until you notice that the 1 mΩ fault path adds 0.000525 pu to the loop resistance, which brings the prediction to 0.210 s exactly. The measurement reads the whole loop, not just the machine.
5. Look at the torque. Air-gap torque peaks at 4.85 pu and reverses to −3.72 pu half a cycle later. That oscillation is the DC offset beating against the rotor field, it exists only because §15.3's $p\psi$ terms were kept, and it is why §15.5's multi-mass shaft exists.
6. Break the assumption on purpose. Set rotor_lock to Off (free from t=0). With tm0 = 0 the rotor now coasts down during the fault, sliding the waveform off 60 Hz and invalidating the envelope analysis. Seeing it fail once is the best way to remember that the analysis assumed constant speed.

Step 3 is the heart of the exercise, so here is what the plot looks like when you draw it:

Left: the AC envelope with the steady-state component removed, on a log axis, where the transient term is a straight line. Right: what is left of the first 100 ms once that fitted line is subtracted — a second straight line, one point per half cycle. Two slopes give $T_d^{\prime }$ and $T_d^{″}$; the two intercepts at $t=0$ give $X_d^{\prime }$ and $X_d^{″}$.

Fitting the tail from 120 ms onwards gives $T_d^{\prime }=1.34$ s and, extrapolated back to $t=0$, an amplitude of 2.81 pu above a steady 0.556 pu. Subtracting that fitted tail from the first 50 ms leaves a second exponential decaying in $T_d^{″}=20.1$ ms from 1.69 pu. Invert the running sums:

$$X_d=\frac{1}{0.556}=1.80,X_d^{\prime }=\frac{1}{3.37}=0.297,X_d^{″}=\frac{1}{5.06}=0.198,$$

against nameplate values of 1.8, 0.3 and 0.2. The time constants are the short-circuit pair of §15.4: $T_{d0}^{\prime }{X}_d^{\prime }/X_d=1.333$ s and $T_{d0}^{″}{X}_d^{″}/X_d^{\prime }=20.0$ ms. Everything comes back within about a per cent, which is the model read backwards agreeing with the model read forwards.

15.10 Summary

• Stator inductances vary with rotor position, so a machine cannot be stamped as a constant matrix in phase coordinates.
• Park's transformation moves the stator onto rotor-fixed $d$, $q$ and $0$ axes, where every inductance is constant. NumaSim uses the amplitude-invariant $\frac{2}{3}$ scaling, so per-unit bases are peak quantities.
• The $dq$ voltage equations split into speed emf ($\omega \psi$) and transformer emf ($p\psi$). Stability programs drop the second; EMT keeps it, and that is precisely what puts the DC offset in a fault current.
• Rotor circuits are a field winding plus one or two dampers per axis. Datasheets give the derived standard parameters $X_d$, $X_d^{\prime }$, $X_d^{″}$, $T_{d0}^{\prime }$…; the model inverts them to circuit values. Open-circuit and short-circuit time constants differ by the reactance ratio.
• The swing equation converts $T_m-T_e$ into acceleration through $2H$. Steam turbine-generators need a multi-mass shaft to represent torsional modes and subsynchronous resonance.
• The machine reaches the network as a constant $R_a+j{X}^{″}$ branch behind a moving internal emf: constant matrix, per-step injection, with a one-step lag as the price.
• Left unseeded, a machine takes tens of seconds to magnetize. Flux seeding, soft start and rotor lock each solve a different part of that problem; only a load flow computes the operating point itself.

15.11 Problems

Problem 15.1. A machine has $X_d=1.8$, $X_d^{\prime }=0.3$, $X_d^{″}=0.2$ pu, $T_{d0}^{\prime }=8$ s and $T_{d0}^{″}=0.03$ s. Sketch the symmetrical fault-current envelope from an open-circuit start at rated voltage, giving the value at $t=0$, at $t=0.1$ s, and as $t\to \mathrm{\infty }$.

Solution 15.1

Use the three-term decomposition with $E=1$ pu:

$$I_{\mathrm{ac}}(t)=(\frac{1}{0.2}-\frac{1}{0.3})e^{-t/T_d^{″}}+(\frac{1}{0.3}-\frac{1}{1.8})e^{-t/T_d^{\prime }}+\frac{1}{1.8}$$

with $T_d^{″}=0.03\times 0.2/0.3=20$ ms and $T_d^{\prime }=8\times 0.3/1.8=1.33$ s, i.e. $1.67e^{-t/0.02}+2.78e^{-t/1.33}+0.556$ pu.

At $t=0$ this is $5.0$ pu $=1/X_d^{″}$. At $t=0.1$ s the subtransient term has fallen to $e^{-5}=0.7\mathrm{\%}$ of itself, leaving $2.78\times 0.928+0.556=3.13$ pu. As $t\to \mathrm{\infty }$ it settles at $0.556$ pu $=1/X_d$, or 56 % of rated current.

Problem 15.2. Why must the three phase DC offsets in a three-phase fault sum to zero, and what follows about the worst possible first peak?

Solution 15.2

The DC offsets are the constants needed to make each phase current continuous at the fault instant. In a three-wire connection the phase currents sum to zero at every instant, both before the fault and after, so their DC components must sum to zero too.

It follows that at most one phase can carry a full offset while the other two share the negative of it. The worst first peak is therefore close to $2\times E/X_d^{″}$ in one phase — 10 pu for the machine above — and never in more than one phase at a time. The short-circuit lab measures 9.22 pu, the shortfall being the offset's own decay during the first half cycle.

Problem 15.3. A transient-stability program and an EMT program are given the same machine and the same fault. Name two features that appear in the EMT stator current and not in the stability one, and identify the term responsible.

Solution 15.3

The DC offset in the phase currents, and the resulting second-harmonic component in air-gap torque. Both come from the transformer-emf terms $d{\psi }_d/dt$ and $d{\psi }_q/dt$ in the stator voltage equations, which a stability formulation sets to zero (Kundur §5.1.1). A stability run shows the envelope of the fault current and the rotor swing; it cannot show what the current does inside a cycle.

Problem 15.4. A machine with $H=3.5$ s carrying 0.9 pu of mechanical torque suffers a fault that drops its electrical torque to zero. How long before the rotor is 10 electrical degrees ahead, ignoring damping?

Solution 15.4

The acceleration is $d\omega /dt=(T_m-T_e)/2H=0.9/7=0.1286$ pu/s. Angle advances as $d\delta /dt={\omega }_0(\omega -1)$, so with a constant acceleration, $\delta (t)=\frac{1}{2}{\omega }_0(0.1286)t^2$ radians. Ten degrees is 0.1745 rad, so

$$t=\sqrt{\frac{2\times 0.1745}{377\times 0.1286}}=0.085\text{s},$$

about five cycles. This is why fault clearing times are quoted in cycles and why "critical clearing time" is a meaningful quantity.

Problem 15.5. Explain why the machine interface uses an average $X^{″}=(X_d^{″}+X_q^{″})/2$ for the stamped conductance rather than the two axis values separately, and what is done about the difference.

Solution 15.5

The stamped conductance lives in phase coordinates, where the $d$ and $q$ axes are not separable — a phase winding sees a mixture of the two that changes as the rotor turns. Using the individual values would make the stamp rotor-position-dependent, which is exactly the refactorization problem Park's transformation was introduced to avoid.

The average is used for the constant part, and the difference $\mathrm{\Delta }x=(L_d^{″}-L_q^{″})/2$ is folded into the per-step current injection as a correction proportional to the stator current. The network sees the right impedance on average and the injection makes up the saliency, so the matrix stays constant and the model stays exact.

Problem 15.6. A user models a 600 MW steam turbine-generator on a series-compensated line with shaft_model set to single-mass. The run is stable. Should they be reassured?

Solution 15.6

No. A single lumped inertia has no torsional modes, so a subsynchronous-resonance interaction between the line's electrical resonance and a shaft mode cannot appear in the results at all — the model is not capable of producing the failure being looked for. A stable single-mass run is evidence about the electromechanical swing and nothing else.

To assess SSR the shaft must be represented as a chain of masses with realistic stiffnesses, and the resulting torsional mode frequencies compared against $f_0-f_{\text{er}}$ for the compensated line.

15.12 References

• P. Kundur, Power System Stability and Control, McGraw-Hill — Chapter 3 for Park's transformation, per-unit conventions, saturation and the swing equation; Chapter 4 for the standard parameters and their inversion to fundamental circuit values; Section 5.1 for what stability formulations discard; Section 15.1 for subsynchronous resonance.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — Chapter 7 for the machine-network interface, the multimass shaft equations and the one-step-lag instability.
• R. H. Park, "Two-reaction theory of synchronous machines," AIEE Transactions, vol. 48, 1929 — the original.
• IEEE Std 1110, Guide for Synchronous Generator Modeling Practices — how many rotor circuits to use, and how to justify the choice.

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