Unsymmetrical Faults

Chapter 21 solved the symmetrical three-phase fault, and the solution was short because the fault preserved the symmetry of the system: one network, one Thévenin equivalent, one division. Every other fault destroys that symmetry. The three phase currents are unequal, the three phase voltages are unequal, and no per-phase circuit exists in which to work.

Unsymmetrical faults are also the ones that actually happen. Field statistics from transmission systems put single line-to-ground faults at roughly 70 to 80 per cent of all faults, line-to-line at 10 to 15 per cent, double line-to-ground at 5 to 10 per cent, and the balanced three-phase fault — the one that is easiest to calculate — at only 2 to 5 per cent. The chapter that is hardest to write is the one that covers almost everything that occurs.

Chapter 22 supplied the transformation and Chapter 23 supplied the networks. What remains is the connection between them, and it rests on a single observation: a fault is a set of algebraic constraints imposed at one point of an otherwise healthy system. Two conductors intact means two currents are zero. A short to earth means one voltage is fixed by the fault path. Those constraints are statements about phase quantities; transform them, and they become statements about sequence quantities; and statements about sequence quantities are instructions for wiring three circuits together.

Before any algebra, the assumptions must be stated, because every formula in this chapter inherits them. They are the assumptions of Chapter 21, extended to three networks.

Two notational conventions run through the chapter. The fault is always taken to be on phase \(a\), or on phases \(b\) and \(c\), because the transformation of Chapter 22 is written with phase \(a\) as reference; any real fault is relabelled to fit. And the fault currents flowing out of the system into the fault are written \(I_{fa}, I_{fb}, I_{fc}\), with their sequence components \(I_{a0},I_{a1},I_{a2}\). Older texts, Stevenson among them, write these as \(I_{fa}^{(0)},I_{fa}^{(1)},I_{fa}^{(2)}\); the meaning is identical.

Section 23-8 reduced each sequence network of the system to a Thévenin equivalent at the bus \(k\) where the fault is to occur. Those three equations are the fixed half of every problem in this chapter, and they are worth writing once more, because everything else is added to them.

Here \(Z_1=Z_{kk}^{(1)}\), \(Z_2=Z_{kk}^{(2)}\) and \(Z_0=Z_{kk}^{(0)}\) are the driving-point impedances of the three sequence networks at the fault bus, obtained either by series–parallel reduction or as diagonal entries of the three bus impedance matrices of Chapter 17. The signs record that positive current is defined as flowing out of the network into the fault, so it produces a drop in each network.

Three equations, six unknowns. The fault supplies the missing three. Whatever the fault type, the procedure is the same, and it is worth stating as a recipe because the four cases differ only in step 2.

A conductor touches a tower, a lightning stroke flashes an insulator string over, a tree grows into a phase — the result is a connection from one phase to earth through whatever impedance \(Z_f\) the arc or the contact provides. Phases \(b\) and \(c\) remain intact, so no current leaves them, and phase \(a\) is tied to earth through \(Z_f\).

Transform the two current conditions. With \(I_{fb}=I_{fc}=0\), the analysis equations of Chapter 22 read

Every entry of the first column of \(\mathbf{A}^{-1}\) is \(\tfrac13\), so a current in phase \(a\) alone splits into three equal thirds. Three networks carrying the same current can only be in series. That is the whole of the topology; what remains is to find what else is in the loop.

Add the three Thévenin equations, using \(I_{a0}=I_{a1}=I_{a2}\):

and equate it to the voltage the fault path imposes. Here the factor of three appears for the second time in two chapters, and for exactly the same reason as in Section 23-2: the physical fault impedance carries the whole fault current \(I_{fa}=3I_{a0}\), while each sequence network carries only \(I_{a0}\).

The voltages follow at once. Substituting the common current back into the three network equations gives \(V_{a1}=V_f-Z_1I_{a0}\), \(V_{a2}=-Z_2I_{a0}\) and \(V_{a0}=-Z_0I_{a0}\); the synthesis matrix then produces \(V_{ka}\) (which must equal \(3Z_fI_{a0}\), and is zero for a bolted fault) together with \(V_{kb}\) and \(V_{kc}\). The healthy-phase voltages are the interesting ones, and Section 24-6 examines what they do.

One structural point deserves emphasis. Because the three networks are in series, the fault current is limited by the sum of the three impedances, including \(Z_0\). If the zero-sequence network is open at the fault bus — an isolated neutral, or a fault beyond a delta winding — then \(Z_0=\infty\) and the earth-fault current is zero. That is not an artefact of the algebra: with no return path, no earth-fault current can flow.

Two conductors touch — a bird bridges them, conductors clash in a gale, a jumper swings across — and phases \(b\) and \(c\) are joined through \(Z_f\) with no connection to earth. Phase \(a\) is intact.

The current conditions transform in two lines:

No zero-sequence current at all, and the positive- and negative-sequence currents are equal and opposite. That is the signature of two networks connected in parallel opposition: current leaves the positive-sequence network at the fault point and returns into the negative-sequence network.

The zero-sequence network deserves a sentence of its own. It carries no current, and being passive it contains no source, so every voltage in it is zero. The zero-sequence network is dead throughout an LL fault and plays no part in the calculation whatever the earthing arrangement. This is physically obvious — the fault does not involve earth — but it is worth stating because it means a line-to-line fault current is the same on a solidly earthed system and on an isolated-neutral one.

Now the voltage condition, which must be shown to be satisfied by the parallel connection rather than merely assumed. Expand each side separately. Since \(V_{a0}\) is zero,

The common factor \((a^2-a)\) cancels — it is non-zero, being \(-j\sqrt3\) — and what survives is a statement purely about sequence quantities:

which is precisely the loop equation of the positive- and negative-sequence Thévenin equivalents joined in opposition through \(Z_f\). Substituting \(V_{a1}=V_f-Z_1I_{a1}\) and \(V_{a2}=-Z_2I_{a2}=+Z_2I_{a1}\) gives the answer.

Two conductors touch and the pair also reaches earth, through a common impedance \(Z_f\). This is the fault a fallen tower or a flashover of two insulator strings to the structure produces, and it is the one that combines both previous mechanisms.

Because phase \(a\) carries no current, the sum of the other two is the residual current:

Sequence currents that sum to zero are three branches meeting at a node: the networks are all three in parallel. The voltage conditions fix what is in each branch. Transform \(V_{ka},V_{kb},V_{kb}\) — the second and third phase voltages being equal — and take the second and third rows:

using \(a+a^2=-1\) in both lines. The first row supplies the link to the zero-sequence branch:

Read the last line as a circuit. The positive- and negative-sequence terminals are at the same potential; the zero-sequence terminal is at that potential plus the drop across \(3Z_f\). So the negative-sequence network and the series combination of the zero-sequence network with \(3Z_f\) hang in parallel from the positive-sequence network's terminals, and the three currents leaving that node sum to zero, as required.

Two limits check the result. Setting \(Z_f=0\) gives the bolted double line-to-ground fault. Letting \(Z_f\to\infty\) opens the zero-sequence branch, so \(I_{a0}\to0\) and \(I_{a1}\to V_f/(Z_1+Z_2)\) — the line-to-line fault of Section 24-4, which is what a double line-to-ground fault becomes when the path to earth is removed. The formulas of the three fault types are not independent results; they are one family.

A switchgear engineer sizing a circuit breaker, and a protection engineer setting a relay, both need to know which fault at a given bus produces the largest current. The formulas of the last three sections answer the question, and the answer turns out to depend on a single ratio.

Take the common approximation \(Z_2=Z_1\), write \(k=Z_0/Z_1\), and set \(Z_f=0\). Then the three bolted fault currents, all measured as the largest phase current, are

This is why solidly earthed generators are rare. Earthing the machine neutral through even a modest impedance adds \(3Z_n\) to \(Z_0\), pushes \(k\) above unity, and brings the earth-fault current below the three-phase value, which is what the machine's bracing and the breaker's rating were designed for. Example 5 sizes such a reactor.

The same ratio governs the other half of the earthing question — what happens to the two healthy phases. For a bolted LG fault on phase \(a\) with \(Z_2=Z_1\), substituting the sequence voltages into the synthesis equation and simplifying gives a compact result:

The quantity on the left, divided by \(V_f\), is the earth-fault factor, and it is what insulation and surge-arrester ratings are chosen against. Evaluating it:

The two columns move in opposite directions, and that is the whole of the earthing dilemma. Raising \(Z_0\) reduces earth-fault current — good for equipment damage, for step-and-touch potentials, and for breaker duty — but raises the voltage the healthy phases must withstand, so the insulation and the arresters must be rated higher. A system is called effectively earthed when \(X_0/X_1\le3\) and \(R_0/X_1\le1\), which keeps the earth-fault factor below about 1.4, or 80 per cent of the \(\sqrt3\) that an isolated neutral would give. That threshold is why transmission systems at 132 kV and above, where insulation is the dominant cost, are almost always solidly or effectively earthed, while distribution and industrial systems, where continuity of supply matters more, are often resistance-earthed. Chapter 37 works through the choice in full.

Everything so far has been a shunt fault: a connection made between conductors, or between a conductor and earth, that did not exist before. The other family of unbalance is the series fault, in which a connection that should exist is broken. A conductor snaps; a fuse in one phase clears while the other two hold; a circuit breaker's three poles fail to close together and one contact stays open. The system is then unbalanced not at a point on the line but across a gap in it, and the sequence method handles it with the same machinery, applied to the two ends of the gap instead of to a bus and earth.

Let \(V_{aa'},V_{bb'},V_{cc'}\) be the voltages across the break in each phase, and \(I_a,I_b,I_c\) the currents through it. For one conductor open — phase \(a\) broken, phases \(b\) and \(c\) continuous — the conditions are that no current flows in \(a\) and no voltage appears across the unbroken conductors:

Equal voltages across three branches whose currents sum to zero: the three sequence networks are connected in parallel across the break. For two conductors open — phases \(b\) and \(c\) broken, phase \(a\) continuous — the conditions are the mirror image:

Equal currents in three branches whose voltages sum to zero: the three networks are connected in series across the break. Laid beside the shunt faults, a duality appears that is worth tabulating, because it halves the amount to be remembered.

The impedances used are different, though. For a series fault the relevant quantities are the Thévenin impedances seen looking into the two ends of the gap in each sequence network, with the broken branch removed — not the driving-point impedances to the reference bus. The driving quantity is supplied by the compensation theorem: if the branch carried a pre-fault current \(I^{\text{pre}}\), then opening it leaves an open-circuit voltage \(V_{th}=I^{\text{pre}}Z_1^{\,pp'}\) across the gap in the positive-sequence network, and none in the other two.

A single numerical case shows the scale of the effect. Suppose a line carries a pre-fault current of \(1.0\angle{-20^\circ}\) per unit and that, with the phase-\(a\) conductor removed, the impedances across the gap are \(Z_1^{\,pp'}=Z_2^{\,pp'}=j0.50\) and \(Z_0^{\,pp'}=j1.20\) per unit. With the three networks in parallel,

The load current in the two surviving phases has fallen only from 1.000 to 0.904 per unit, which no overcurrent relay would notice. But half a per unit of residual current now circulates, and every earth-fault relay in the vicinity sees it. This is why a broken conductor lying on dry ground — drawing almost no fault current, yet lethal to anyone who touches it — is detected by unbalance rather than by magnitude, and why negative-sequence and broken-conductor relays exist at all. Chapter 36 sets them.

The sequence currents at the fault bus are only the first answer. A protection study needs the current in every line and the voltage at every bus, because those are the quantities the relays and the instrument transformers actually measure. Chapter 17 supplies the machinery: the off-diagonal entries of the three bus impedance matrices.

Line currents follow from the sequence voltage differences. In a branch between buses \(p\) and \(q\) with sequence impedances \(z_1^{pq}\), \(z_2^{pq}\), \(z_0^{pq}\),

and the three phase currents in that branch are recovered by \(\mathbf{A}\) in the usual way. One warning applies to the zero-sequence part: because delta windings break the zero-sequence network into islands, \(V_{j0}\) is identically zero at every bus outside the island containing the fault, and no residual current flows there at all.

Crossing a delta–star transformer requires care of a different kind. Chapter 23 recorded that the positive-sequence quantities on the high-voltage side lead those on the low-voltage side by \(30^\circ\) while the negative-sequence quantities lag by \(30^\circ\). The two shifts are equal and opposite, so they do not cancel when the sequence components are recombined — they rotate the positive- and negative-sequence phasors in opposite directions, and the phase pattern that emerges on the far side is quite different from the one that went in.

Finally, a practical note on measurement. The three line current transformers of a feeder are almost always connected in residual, so that the relay in the common return receives \(I_a+I_b+I_c=3I_{a0}\). That single connection is the reason zero-sequence current is the workhorse quantity of earth-fault protection: it is available directly, without any computation, and it is zero in a healthy balanced circuit no matter how heavily loaded. Everything this chapter computes about \(I_{a0}\) is a prediction about what that one relay will see.

Take \(V_f=1.0\angle0^\circ\) per unit and neglect pre-fault load current unless a problem says otherwise. In every case draw the network interconnection before computing anything, and check at the end that the sequence currents recombine to the phase currents the fault type demands.