Symmetrical Faults and Short-Circuit Transients

Everything in Parts 3 and 4 of this book assumed a network in its intended configuration: conductors insulated from each other and from earth, three phases balanced, currents set by load rather than by geometry. A fault is any failure of that assumption — any event that interferes with the normal flow of current. Insulation ages and punctures, a lightning stroke flashes an insulator string over, a crane arm or a falling branch bridges two conductors, ice or wind brings a span to the ground, a switching operation is carried out on the wrong circuit. On transmission lines at 115 kV and above the dominant cause by a wide margin is lightning, which ionises the air across an insulator string and leaves a low-impedance path to the earthed tower.

Two families are distinguished by what the failure does to the circuit. A shunt fault, or short circuit, creates an unintended connection — conductor to earth, or conductor to conductor. A series fault is an unintended break: one or two broken conductors, or a breaker whose three poles do not open together. The two behave in opposite ways. A shunt fault raises current and depresses voltage; a series fault reduces current in the affected phases and raises the voltage across the break. Shunt faults are far more common and far more damaging, and they are what Part 5 is about.

Only the last row leaves the system balanced. If all three phases are shorted together at the same point — with or without a connection to earth — the network retains the three-fold symmetry that Chapter 3 exploited, the three phase currents remain equal in magnitude and \(120^\circ\) apart, and the entire calculation can still be carried out on a single phase using the per-phase equivalent circuit and the per-unit impedance diagram of Chapter 4. That is why this chapter comes first: the symmetrical fault is the only one that needs no new machinery beyond what Part 4 already built.

The other three rows destroy the symmetry, and per-phase analysis collapses with it. Chapter 22 introduces the transformation that restores it, Chapter 23 builds the three networks that transformation calls for, and Chapter 24 connects them for each fault type. So the rare fault is treated first and the common ones second, which looks backwards until one notices the second reason for the ordering.

Two currents matter, and they are not the same current. The first is the current flowing in the first cycle or two after inception, which the closed breaker contacts and the busbars must survive mechanically and thermally. The second is the current still flowing three to eight cycles later, when the protective relay has decided (Chapter 36) and the breaker contacts actually part — that is the current the arc must interrupt. Between the two instants the current falls, for two independent reasons that occupy the next four sections: the circuit sheds a DC transient, and the machine raises its own reactance.

Strip the problem to its smallest honest form. An unloaded transmission line, with its shunt capacitance neglected, is a series \(R\)–\(L\) branch fed from a stiff sinusoidal source. A fault at the far end is the closing of a switch across that branch at some instant \(t=0\). Before closure the current is zero; after closure it is governed by

Here \(V\) is the rms source voltage and \(\alpha\) is the phase angle of that voltage at the instant of closure — a number the fault chooses, not the engineer. The general solution is the sum of a particular integral (the steady-state sinusoid) and a complementary function (the natural response of an \(RL\) loop, a decaying exponential):

The constant \(A\) is not free. Current in an inductance cannot change instantaneously, so \(i(0^+) = i(0^-) = 0\). Setting \(t=0\) in the general solution and solving for \(A\) gives \(A = -(\sqrt2 V/Z)\sin(\alpha-\theta)\), and with that one substitution the whole behaviour of the short-circuit current is fixed.

The DC offset is therefore not a separate physical current but the price of continuity. The steady-state sinusoid at \(t=0\) has the value \((\sqrt2 V/Z)\sin(\alpha-\theta)\), which is generally not zero; the exponential starts at exactly minus that value and cancels it, then decays away and leaves the sinusoid behind. Its initial magnitude is set entirely by \(\alpha-\theta\), the angular distance between the instant of closure and the natural zero crossing of the steady-state current.

Two special cases bracket everything. If \(\alpha = \theta\), the closure happens exactly when the steady-state current passes through zero: no offset is needed, and the current is a pure sinusoid from the first instant. If \(\alpha - \theta = \pm 90^\circ\), the closure happens when the steady-state current would be at its peak: the offset is a full \(\sqrt2 V/Z\), and the current is as asymmetric as it can be. In a power network \(X \gg R\), so \(\theta \approx 90^\circ\), and the worst case corresponds to \(\alpha \approx 0\) — a fault struck at the instant the driving voltage passes through zero.

Take the worst case, \(\alpha-\theta=-90^\circ\), and write \(I=V/Z\) for the rms symmetrical current. The expression collapses to

The current never changes sign — it is a sinusoid riding on a positive exponential platform — and it reaches its largest value when \(\cos\omega t = -1\), that is half a cycle after inception, at \(t = T/2 = \pi/\omega\). Evaluating the exponential there and noting that \(t/\tau = \pi\omega R/(\omega X) = \pi R/X\) gives the peak directly.

The limiting values are instructive. If the circuit had no resistance at all the exponential would never decay, the peak would be \(2\sqrt2 I\) — the familiar doubling effect — and the rms would be \(\sqrt3\,I = 1.732\,I\). Real circuits fall short of this, and by how much is governed by \(X/R\) alone:

The bold entry is the origin of a rule that every switchgear catalogue uses. Close to a generator, where \(X/R\) is of the order of \(30\), the rms current including the DC offset at the end of the first half cycle is about \(1.6\) times the symmetrical value. The factor \(1.6\) is not an arbitrary safety margin; it is \(\sqrt{1+2e^{-2\pi/30}}\), and Section 21-9 uses it exactly as it stands.

One warning about three-phase systems. The three source voltages differ in phase by \(120^\circ\), so the three phases are closed at three different values of \(\alpha\). At most one phase can be fully offset; the other two carry smaller offsets, and the three offsets always sum to zero. A "symmetrical" three-phase fault is therefore symmetrical only in its AC component. The DC offsets are unequal, which is why the phase currents differ during the first few cycles even when the fault is perfectly balanced, and why breaker duty is quoted for the worst phase.

The RL model of the last two sections assumed a source of fixed emf behind a fixed impedance. Applied to a real generator it is wrong, and wrong in a way that matters. Oscillograms of a machine short-circuited at its terminals show the AC component itself decaying — the envelope falls steeply for the first few cycles, then more gently, then settles. Since the terminal voltage was constant beforehand, the only way the current can fall is for the effective internal reactance to rise with time.

The mechanism is the constant flux linkage theorem: the flux linkage of any closed circuit of zero resistance cannot change instantaneously, and in a circuit of small resistance it changes only slowly, with that circuit's own time constant. A synchronous machine has three magnetically coupled circuits on the direct axis — the armature winding, the field winding, and the damper (amortisseur) bars — and the last two are closed loops of low resistance.

At the instant of short circuit the armature current jumps and its mmf tries to demagnetise the air gap. The damper circuit and the field circuit each respond by developing an induced current that holds their own flux linkage at its prefault value, which forces the armature flux to divert into leakage paths instead. Seen from the armature terminals, the magnetising reactance \(x_{ad}\) is therefore shunted by the leakage reactances of whichever rotor circuits are still holding their flux. Three stages follow, in the order in which the rotor circuits give up.

where \(x_l\) is the armature leakage reactance, \(x_{ad}\) the direct-axis magnetising reactance, \(x_f\) the field leakage reactance and \(x_D\) the damper leakage reactance. Adding a branch in parallel can only reduce a reactance, so the ordering is automatic and needs no measurement to establish:

Each of the three stages is entered and left exponentially, with a time constant set by the resistance of the rotor circuit that is decaying. Superposing them gives the armature current after a three-phase short circuit at the terminals of a machine running at no load with terminal voltage \(V_t\):

Read the bracket as an envelope. \(T_d''\), the subtransient short-circuit time constant, is of the order of \(0.03\) s — a couple of cycles — and is fixed by the damper winding resistance. \(T_d'\), the transient short-circuit time constant, is of the order of \(0.5\)–\(2\) s and is fixed by the field winding resistance. The armature resistance has been neglected, so the current lags the voltage by exactly \(90^\circ\); that is the \(-\pi/2\) in the sine.

Evaluating the envelope at three instants recovers three currents that the rest of the chapter uses constantly. At \(t=0\) all three exponentials are alive and the bracket telescopes to \(1/X_d''\). Once \(t\) is a few multiples of \(T_d''\) but still small compared with \(T_d'\), the first term has vanished and the second is still essentially unity, so the bracket is \(1/X_d'\). Long after \(T_d'\) only the constant survives.

This is also how the reactances are measured. Short-circuit a machine running at no load with a known terminal voltage, record the current, and remove the DC offset from each phase. Plot the resulting symmetrical envelope on semi-logarithmic paper. The last, flat part gives \(I\) and hence \(X_d\). Subtract that constant and the remainder is a straight line on the log plot whose intercept at \(t=0\) is \(I'-I\), giving \(X_d'\), and whose slope gives \(T_d'\). Subtract that in turn and the residue gives \(I''-I'\) and \(T_d''\). Each stripping operation isolates one exponential, exactly as in any multi-exponential decay.

Sections 21-4 and 21-5 described a machine on no load. In practice the fault interrupts a system that was carrying load, and the model has to account for that. The device used is standard and deceptively simple: represent the machine during the subtransient period by a constant emf \(E''\) behind the constant reactance \(jX_d''\), and choose \(E''\) so that the model reproduces the prefault terminal conditions exactly.

\(V^{(0)}\) is the prefault terminal voltage and \(I_L\) the prefault load current, taken positive out of a generator and into a motor — which is the whole reason for the sign difference. Once \(E''\) is computed the machine is a fixed source and the network is linear, so all the tools of Part 4 apply unchanged.

Synchronous motors and large induction motors must be included, and this is easy to forget. During a fault the system voltage collapses, the motor's internal emf momentarily exceeds its terminal voltage, and the machine feeds current into the fault instead of drawing it. A synchronous motor does this indefinitely; an induction motor does it for two or three cycles, until its rotor flux dies away, and is represented by an emf behind its locked-rotor reactance for exactly that reason. Neglecting motor contribution can understate the first-cycle fault current at an industrial bus by a third or more.

With every source reduced to an emf behind a reactance, the faulted network is a linear circuit with several sources, and there are two ways to solve it. The direct way computes each \(E''\) from the prefault load flow, closes the switch representing the fault, and solves the resulting circuit; each machine's contribution then comes out complete, load current included. The second way is superposition, and it is the one that scales.

Split the faulted network into two networks whose responses add to the true one. The first is the prefault network, with all its sources and its known bus voltages \(V^{(0)}\) and load currents — nothing has happened in it yet. The second is the pure fault network: the same network with every source killed, driven by a single voltage source of value \(-V_k^{(0)}\) inserted at the fault point. Adding the two reproduces the faulted condition, because at the fault point the two sources \(V_k^{(0)}\) and \(-V_k^{(0)}\) sum to zero — which is precisely the statement that the bus has been shorted to the reference.

\(Z_{th}\) is the Thevenin impedance of the passive network looking into the fault point, and \(Z_f\) is the fault impedance, zero for a bolted fault. The great convenience is that the pure fault network has no load, no generation and exactly one source, so it can be solved by ordinary series–parallel reduction or by current division. The great trap is that its currents are not the answer: they must be added back to the prefault currents branch by branch.

In hand calculation the addition is often skipped, and the justification is worth stating rather than assuming. Prefault load current at a bus is of the order of \(1\) per unit; the pure-fault current is of the order of \(1/Z_{th}\), which for a strong bus is five to twenty per unit. Adding a unit vector to a vector of length ten changes its magnitude by at most ten per cent, and usually much less because the two are nearly in quadrature — load current is close to real, fault current is almost purely inductive. Example 3 carries out both calculations so the size of the error can be seen.

Series–parallel reduction works for a network of five branches and fails for one of five hundred, and it has to be repeated from scratch for every candidate fault location. Chapter 17 removed both objections. The bus impedance matrix was built there precisely so that \(Z_{kk}\) is the Thevenin impedance seen at bus \(k\) — one matrix, computed once, containing the Thevenin equivalent of every bus in the system simultaneously.

Recall the argument, since Part 5 leans on it repeatedly. In the pure fault network all sources are dead, and the only injection is \(-I_f\) at bus \(k\). By definition of \(\mathbf{Z}_{bus}\), the bus voltage changes are

so that \(V_i = V_i^{(0)} - Z_{ik}I_f\) for every bus. One equation is still missing, and the faulted bus supplies it: there the voltage during the fault is the drop across the fault impedance, \(V_k = Z_f I_f\). Equating this to \(V_k^{(0)}-Z_{kk}I_f\) and solving closes the problem.

Three consequences follow at once. First, the faulted bus voltage is \(V_k = Z_f V_k^{(0)}/(Z_{kk}+Z_f)\), which is zero for a bolted fault as it must be. Second, the voltage depression at a remote bus \(i\) is in the ratio \(Z_{ik}/Z_{kk}\) — the off-diagonal entries measure how far the disturbance propagates, which is exactly the electrical distance between buses. Third, the arithmetic is self-checking: the currents arriving at bus \(k\) along every connected line, plus any generator connected there, must add up to \(I_f\), and Kirchhoff's current law must hold at every unfaulted bus. Example 4 uses both checks.

The flat prefault assumption simplifies this further. Setting every \(V_i^{(0)}=1.0\angle 0^\circ\) makes \(I_f = 1/(Z_{kk}+Z_f)\) and \(V_i = 1 - Z_{ik}/(Z_{kk}+Z_f)\), so a whole fault study reduces to reading one column and dividing. For an \(n\)-bus system this gives the fault level at all \(n\) buses from a single matrix, which is why the \(\mathbf{Z}_{bus}\) formulation, rather than the \(\mathbf{Y}_{bus}\) formulation of Chapter 16, is the natural one for fault work — and why load flow, which needs sparsity, uses the other.

The number that summarises a bus in one figure is its short-circuit level or fault level: the apparent power that would be delivered into a bolted three-phase fault there, at the prefault voltage. With flat prefault voltage and per-unit quantities on a chosen base,

with \(V_L\) in kV and \(I_{f,L}\) in kA. The last form is the useful one: fault level and Thevenin impedance are reciprocal descriptions of the same fact. A strong bus has small \(Z_{kk}\), holds its voltage well against load changes and motor starting, and demands expensive switchgear; a weak bus is the reverse. Every decision about adding generation, adding a line, or splitting a busbar moves both quantities together, and the tension between them runs through system planning.

Turning the fault level into a breaker specification requires the two currents identified in Section 21-1, and the two multipliers of Section 21-3.

Breakers are catalogued by nominal voltage, rated maximum voltage, continuous current, rated short-circuit current at rated maximum voltage, operating time, and a voltage range factor \(K\). The last is the ratio of rated maximum voltage to the lowest voltage at which the breaker retains full interrupting capability. Within that range the breaker's interrupting capability rises as the voltage falls, in inverse proportion, but never beyond \(K\) times the rated short-circuit current:

The corresponding statement in apparent power, which is how fault duty is usually quoted in a specification, is \(\text{SC interrupting MVA} = \sqrt3\,|V_{pf,L}|\,|I_{f,L}|\) with the prefault line voltage in kV and the interrupting current in kA. Chapter 25 works through the complete computation for a multi-machine system, Chapter 35 describes what happens inside the interrupter during those few milliseconds, and Chapter 36 sets the relay that gives the trip command in the first place.

Work in per unit throughout, on a stated base, and assume flat prefault voltages of \(1.0\angle0^\circ\) unless the problem says otherwise. Difficulty rises down the list.