Unsymmetrical Faults

The four types, with their observed frequency on transmission systems:

The commonest by a wide margin is the one that requires all three sequence networks; the rarest is the one Set 18 could already handle.

Why the LG fault dominates. The mechanisms that cause faults act on one phase at a time:

A three-phase fault requires something to bridge all three conductors at once — a conductor clash in high wind, a collapsed tower, or an earthing device left applied. Rare, and usually the result of an error rather than a natural cause.

The method, in four steps, identical for every type:

Only step 3 requires thought, and it need be done once per fault type — after which the interconnection is memorised and the calculation is a division.

What each fault needs. Three numbers from Set 22's matrices, and no more:

All three are diagonal elements at the faulted bus, computed in advance for every bus. The size of the network behind them is irrelevant to this stage.

The two conditions per fault. Six unknowns exist at the fault — three sequence currents and three sequence voltages — and the three networks supply three equations:

So exactly three further conditions are needed, and the fault supplies them. Two are independent constraints and the third is the fault impedance relation.

And a convention worth fixing now. The faulted phase is always taken as \(a\) for a single line-to-earth fault, and the faulted pair as \(b\) and \(c\) for the others. This is a choice of labelling, not a restriction — relabelling the phases rotates the answers but changes no magnitude.

In phase quantities, at the fault point:

Phases \(b\) and \(c\) are open at the fault — they carry load current, but no fault current, and the superposition network sees zero. Phase \(a\) is connected to earth through \(Z_f\).

The first two conditions, transformed. With \(I_b = I_c = 0\):

Exactly the single-phase-load result of Set 21, Problem 13 — because a single line-to-earth fault is a single-phase load, of very low impedance.

The third condition, transformed. With \(V_a = V_0+V_1+V_2\) and \(I_a = 3I_1\):

The three sequence conditions, assembled:

Two current conditions and one voltage condition — which is exactly what a series connection of three circuits looks like: one current through all of them, and their voltages adding.

The counting check. Three network equations plus three fault conditions gives six equations in six unknowns. The system is determined, and no further physical information is needed.

A note on what "at the fault" means. These are conditions on the superposition network — the change caused by the fault. The actual currents in phases \(b\) and \(c\) are their prefault load currents, which are typically 1–2% of the fault current and are neglected throughout. Where they are not negligible — a heavily loaded feeder with a low fault level — the prefault load flow must be superposed, and modern software does exactly that.

Reading the conditions as a circuit. The sequence conditions of Problem 2 are

One current common to three circuits, and their voltages adding to a fixed drop. That is a series connection of the three networks, with \(3Z_f\) completing the loop.

The circuit. Starting at the reference and going round:

A single loop containing the one source and all three impedances. The order does not matter; series elements commute.

Solving it. Substituting the network relations into the voltage condition:

And the fault current itself:

The factor 3 appearing twice, and for the same reason both times: the physical phase current is three times the sequence current, and the physical fault impedance carries three times the sequence current so appears tripled.

The sanity checks. Three limits, all correct:

The second is worth remembering: when the three sequence impedances are equal, the two fault types give identical currents. Which is worse then depends on nothing at all.

Why the three networks are in series and not parallel is worth stating plainly. The fault forces one current — the whole of \(I_a\) — to pass through all three sequence circuits in turn, because it is composed equally of the three components. The impedances therefore add, and the fault current is smaller than the three-phase one unless \(Z_0\) is small enough to compensate.

The three impedances, from Set 22:

All in per unit on 100 MVA. A solid fault, so \(Z_f = 0\).

The sum:

The sequence currents:

The fault current:

Or 519 MVA on the 100 MVA base. Lagging by 90°, the network being purely reactive.

Comparison with the three-phase fault at the same bus, computed in Set 18:

The earth fault is 10% less severe here, because \(Z_0 > Z_1\) at bus 3 — exactly as Set 22 predicted from the ratio 1.326.

The check on the formula. Using the ratio expression directly:

Matching. The ratio needs only the two impedances and never the source.

And the two zero-sequence facts that follow immediately. The current in the earth is

All of the fault current returns through earth — necessarily, since phases \(b\) and \(c\) carry none. And that 5.19 pu divides between the two transformer neutrals as Set 22's Problem 18 computed: 1.86 through bus 1's and 3.33 through bus 2's.

The sequence voltages follow from the network relations with \(I_1 = 1.7300\angle-90^\circ\):

All real, since the impedances are purely reactive and the currents purely lagging. The positive-sequence voltage has been depressed to 0.70 pu, and the other two are negative.

The faulted phase:

Zero, as it must be for a solid fault. This is the arithmetic check on the whole calculation, and it costs one addition.

The healthy phases:

Both at 1.0525 pu — a rise of 5.25% above nominal, and symmetrically placed about the negative real axis.

The earthing coefficient. The ratio of the highest healthy-phase voltage to the nominal is the standard measure:

Well inside the 1.4 that defines an effectively-earthed system — consistent with \(X_0/X_1 = 1.33\) at this bus, which satisfies the criterion of Set 22.

The angle displacement is the other symptom. The healthy phases are at \(\pm124.63^\circ\) instead of \(\pm120^\circ\) — a separation of 110.7° rather than 120°. The phasor triangle has collapsed towards the faulted phase, and a relay measuring phase angles sees it.

The general result for the healthy-phase rise, worth tabulating:

And 1.33 here gives 1.053. Note that a perfect earth gives a rise below nominal, at 0.87 — the healthy phases are pulled down as well as displaced. Most of the rise occurs between ratios of 3 and infinity.

In phase quantities:

Phase \(a\) is unaffected; whatever flows into the fault in phase \(b\) returns in phase \(c\); the two are joined through \(Z_f\). No earth is involved.

The zero-sequence condition follows at once from the first two:

No zero-sequence current. The zero-sequence network takes no part in this fault, which is the single most important fact about it — a line-to-line fault is unaffected by the earthing arrangement.

The positive and negative components, using \(I_c = -I_b\):

Using \(a - a^{2} = j\sqrt3\) from Set 21. The relation \(I_1 = -I_2\) is the second of the two conditions.

The voltage condition. With \(V_0\) arbitrary (it will turn out to be zero) and using \(V_b - V_c\):

The \((a^{2}-a)\) cancels — which is why the line-to-line formula contains \(Z_f\) and not \(3Z_f\).

The three sequence conditions:

Equal and opposite currents with a common voltage difference is a parallel connection — with one of the two circuits reversed, so that the current entering one leaves the other.

And \(V_0 = 0\) follows, not as a condition but as a consequence: with \(I_0 = 0\) in a passive network, \(V_0 = -Z_0I_0 = 0\). So a line-to-line fault produces no zero-sequence voltage either, and no residual voltage relay will see it.

The connection. \(I_1 = -I_2\) with a common voltage difference means the positive- and negative-sequence networks are connected in parallel opposition — the positive network's terminal to the negative network's terminal, and their references together, so that current circulates from one into the other:

The zero-sequence network is left disconnected entirely.

Solving the loop:

using \(V_2 = -Z_2I_2 = +Z_2I_1\). Hence

The fault current itself, from the transformation:

The \(\sqrt3\) is \(|a^{2}-a|\), the chord subtending 120° on a unit circle — the same \(\sqrt3\) that relates line and phase voltages, and for the same geometric reason.

The ratio to a three-phase fault, with \(Z_1 = Z_2\) and \(Z_f = 0\):

A universal result whenever \(Z_1 = Z_2\), independent of the network. A line-to-line fault always draws 86.6% of the three-phase current at the same point — which is why it is never the switchgear rating case.

The phase \(a\) voltage during the fault is worth computing, because it is unexpected:

The healthy phase's voltage is unchanged. A line-to-line fault depresses the two faulted phases and leaves the third at its prefault value, which is quite unlike an earth fault.

And the faulted phases:

Both at half the prefault magnitude and in antiphase with \(V_a\) — because a solid line-to-line fault ties them together, and the only voltage they can share is the average of what they would otherwise have.

The sequence current:

The fault current:

And the check: \(4.9837/5.7546 = 0.8660 = \sqrt3/2\) exactly, as Problem 7 promised.

The sequence voltages:

Exactly 0.5 each, because \(Z_1 = Z_2\) makes the source voltage divide equally between the two networks.

The phase voltages:

Using \(a+a^{2} = -1\). The two faulted phases are at the same voltage — necessarily, since they are joined — and it is half the prefault magnitude, in antiphase with the healthy phase.

The phasor picture is worth holding. Before the fault, three phasors 120° apart. During it, \(V_a\) is unchanged and \(V_b\) and \(V_c\) have collapsed onto each other at the midpoint of where they were:

The average of the two prefault phasors, exactly. A solid short between two nodes forces them to the mean of what a source would otherwise impose.

The protection consequence. The voltage between the faulted phases is zero and the voltage to earth of every phase is 0.5 pu or more:

Which is why distance relays have separate phase-fault and earth-fault measuring elements, connected to different voltage combinations.

In phase quantities, for a solid fault:

Phase \(a\) is unaffected; both other phases are connected to earth, so both their voltages are zero. Note that this time two of the conditions are on voltages, where the line-to-earth fault had two on currents.

The current condition transforms as before:

The voltage conditions. With \(V_b = V_c = 0\):

The exact dual of the line-to-earth case, with voltages in place of currents. That duality is not a coincidence: the two faults are duals of each other, one shorting two phases to earth and the other opening two.

The three sequence conditions:

A common voltage with currents summing to zero is a parallel connection of all three networks. Compare with the line-to-earth fault's common current and adding voltages, which was a series connection.

With a fault impedance the conditions change in a specific way. If the two phases are joined to each other solidly and to earth through \(Z_g\):

So \(3Z_g\) enters in series with the zero-sequence branch only — the same factor of three, appearing for the same reason as always.

The duality, stated fully, because it makes both cases memorable:

The connection. All three networks in parallel — which in practice means the negative and zero networks in parallel with each other, and that combination in series with the positive network's source:

Because the positive-sequence network is the only one containing a source, it must supply the other two, which sit side by side across the fault point.

The positive-sequence current:

The single most complicated fault formula in Part 5, and it is still one division.

The other two follow from current division across the parallel pair. The current \(I_1\) arrives and splits in inverse proportion to the impedances, flowing out of both:

The minus signs express the condition \(I_0+I_1+I_2 = 0\): what enters through the positive network leaves through the other two. Note that the larger share goes to the smaller impedance, as always in a current divider.

The earth current is the quantity most often wanted:

And it is not the same as the current in either faulted phase, since some current circulates directly between phases \(b\) and \(c\) without reaching earth.

The two limiting cases check the formula:

The first is exactly right: with no earth path the two phases are merely shorted together. The second says a perfect earth makes the double line-to-earth fault as severe as a three-phase one in the positive sequence — and the earth current then exceeds it.

With a ground impedance \(Z_g\), the only change is to replace \(Z_0\) by \(Z_0+3Z_g\) throughout:

Consistent with every other appearance of a neutral or earth impedance in Part 5.

The parallel combination first, since everything else follows from it:

Two reactances in parallel give a reactance, as they must. It is smaller than either, which is what makes this fault severe.

The positive-sequence current:

So \(|I_{a1}| = 3.6651\) pu, appreciably larger than the \(2.8773\) pu of the line-to-line fault at the same bus, because the earth path has been added in parallel and has lowered the impedance the source sees.

Current division sends the rest into the negative- and zero-sequence networks:

Check the constraint: \(I_{a0}+I_{a1}+I_{a2} = j(1.57550+2.08957-3.66507) = j0.00000\) ✓. The negative sequence takes the larger share because it is the smaller impedance.

The earth current is three times the zero-sequence current:

In amperes, with \(I_{\text{base}} = 100\times10^6/(\sqrt{3}\times220\times10^3) = 262.4\) A, this is 1240 A returning through the earth and the two transformer neutrals.

The phase currents by \(\mathbf{I}_{abc} = \mathbf{A}\,\mathbf{I}_{012}\):

The healthy phase carries nothing, as the boundary condition demanded. The two faulted phases carry 5.5156 pu = 1447 A each, and the angle between them is \(154.63°-25.37° = 129.26°\) — not \(180°\) as in a line-to-line fault, precisely because their sum is no longer zero but flows to earth.

Arithmetic check on the earth current:

The two phase currents add vectorially to the earth current — and because they are \(129°\) apart, their sum (4.73) is much less than their arithmetic sum (11.03). This is why the earth current in a double line-to-earth fault is often smaller than the phase currents, and why relay engineers must check both.

The sequence voltages at the fault are all equal, by the boundary condition:

Three independent calculations give the same number to five figures — the strongest possible confirmation that the current division was right.

The phase voltages:

The two faulted phases sit at earth potential, and the healthy phase rises to 1.089 pu — a 9% rise, much milder than the 5% seen in the single line-to-earth case at the same bus, because with two phases clamped to earth the neutral cannot shift nearly as far.

The boundary conditions are the simplest of the four types — all three phases are connected together and to earth:

Note that no condition is imposed on the currents at all. That is the reverse of every other fault, where the current conditions did the work.

Transform the voltage conditions:

All three sequence voltages vanish. A zero vector transforms to a zero vector — the only case where this happens, and the reason the three-phase fault is the trivial one.

Apply the three network equations with these voltages:

The second and third networks have no source, so setting their terminal voltage to zero forces their current to zero as well. Only the first, which contains \(E\), produces anything.

Hence the phase currents are a balanced set:

Which is exactly what "symmetrical fault" means, and confirms that the entire pre-symmetrical-components method of Set 21 was a special case of this framework all along. The three-phase fault needs no sequence networks — but it is reassuring that the general machinery reproduces it.

Evaluated at every bus, using the positive-sequence driving-point impedances:

Fault MVA is simply \(100\times|I_f|\) here, since \(|E| = 1.0\). Bus 2 is the strongest point in the network and bus 5 the weakest, exactly as the Z-bus diagonal of Set 18 predicted.

Why the three-phase fault is still the design case despite being the rarest:

The second point deserves emphasis: in every fault type, \(I_1 = E/(Z_1+Z_{\text{extra}})\) with \(Z_{\text{extra}} \ge 0\), so the three-phase value is a strict upper bound on the positive-sequence current. It is not an upper bound on the phase current or the earth current, as the next problem shows.

The three formulas with \(Z_f\) included:

The fractional effect of \(Z_f\) depends on what it is being added to, and those three denominators are very different in size.

Numerically at bus 3 (\(Z_1=Z_2=j0.17377\), \(Z_0=j0.23047\)):

All currents in pu on 100 MVA. Every column falls, but at very different rates.

Reduction relative to the solid fault:

The line-to-line fault is by far the least sensitive, and the three-phase fault the most.

The explanation is the size of the denominator each \(Z_f\) is added to:

The line-to-line case has the best of both: a base twice the three-phase one, and no factor of three on \(Z_f\). That is why \(Z_f = j0.20\) still leaves it at 63% of its solid value while the three-phase fault has fallen below half.

A crossover is the practical consequence. At \(Z_f = 0\) the three-phase fault is the largest at this bus; by \(Z_f = j0.10\) the line-to-line fault has overtaken it:

The ranking of fault severity is not a fixed property of the bus. It depends on the arc resistance, which is unknown and variable. This is one reason protection settings are checked against the solid-fault case for breaker duty and against a high-impedance case for relay sensitivity.

Where \(Z_f\) comes from physically, and typical magnitudes:

On a 220 kV base of \(220^2/100 = 484\ \Omega\), \(Z_f = j0.05\) pu is 24 Ω — a plausible tower-footing figure. Note that real fault impedance is overwhelmingly resistive, so taking it as reactive here is a simplification made only to keep the arithmetic in a single quantity; the qualitative conclusions are unaffected.

Form the ratio of the two currents directly:

The source EMF cancels — the ratio is a pure property of the network, independent of loading and of the prefault voltage.

Use \(Z_2 = Z_1\), which holds to good accuracy in any system without salient-pole machines:

Everything now depends on a single dimensionless number: the ratio of zero- to positive-sequence impedance at the faulted bus. For a purely reactive network this is \(X_0/X_1\).

The condition follows by inspection:

A remarkably clean result. The earth fault is the worse of the two exactly when the zero-sequence network is stiffer than the positive-sequence one at that point.

Applied to the five-bus system:

The system divides cleanly into two regions, and the boundary is exactly where the earthed transformer neutrals are.

Why buses 1 and 2 behave differently from the rest:

A solidly earthed star point is a very low-impedance zero-sequence source — lower than the generator's \(x''_d\) in this case. Move away from it and the tripled line reactances take over rapidly, pushing \(X_0/X_1\) well above unity.

The general picture, which every fault study reproduces:

Substation buses are the dangerous case, because they are precisely where transformers are earthed and where breakers must be rated. The habit of rating switchgear on three-phase fault MVA alone is therefore not conservative near an earthed neutral.

The extreme case deserves a number. As \(X_0/X_1 \to 0\):

So an earth fault can be up to 50% worse than a three-phase fault, but never more. That bound is a useful sanity check on any fault study: a computed ratio above 1.5 means an error in the zero-sequence data, almost always a forgotten \(3Z_n\) or a delta winding modelled as a star.

The four formulas, collected for reference, with \(Z_2 = Z_1\) throughout:

Only the L-L case has a fixed ratio to the three-phase fault; the two earth faults depend on \(Z_0\).

The complete table, all values in pu on 100 MVA:

"LLG phase" is the current in each of the two faulted conductors; "LLG earth" is \(3I_0\). Bold marks the largest of the three phase-conductor currents at each bus — which is what a breaker must interrupt.

Three separate rankings emerge, and confusing them is the commonest error in a fault study:

At bus 2, for example, the breaker sees 9.97 pu but the earth grid must carry 12.42 pu. Those are different components sized from different columns of the same table.

The pattern across the network:

And note that at buses 3–5 the LLG phase current (5.5156, 5.4093, 4.5956) sits just below the three-phase value (5.7546, 5.6560, 4.8884), so the three-phase fault governs — but only by 4–6%, close enough that neglecting the LLG case would be careless if any parameter were uncertain.

Ratings in engineering units. With \(I_{\text{base}} = 262.4\) A and 100 MVA base:

The "standard rating" column picks the next standard short-circuit rating above the computed duty at 220 kV. Note that these currents are modest for a transmission bus — the 5-bus test system is deliberately small — but the selection procedure is exactly that used on a real network.

A margin check. Fault levels grow as generation is added, so a study is normally repeated for the forecast horizon:

Applying 25% to bus 2's 997 MVA gives 1246 MVA, which still fits the 1200 MVA class only marginally — a flag that any new generation at bus 2 should trigger a re-check. This is exactly the calculation performed in Problem 20 of Set 18, where adding a machine at bus 5 raised bus 5's duty from 489 to 1156 MVA.

The four voltage signatures at the fault point, computed from the currents of Problems 4, 8, 11 and 12:

Each row is a direct restatement of that fault's network connection: series connection forces the voltages to sum to zero; parallel connection forces them equal. The table is the voltage dual of the current table.

The diagnostic content. A relay measuring all three sequence voltages can identify the fault type without any current measurement at all:

And \(V_0\) alone separates the earth faults from the phase faults — which is exactly what the broken-delta VT connection, producing \(3V_0\) directly, was invented to measure.

The retained positive-sequence voltage is the quantity that matters for stability and for motor contactor dropout:

A three-phase fault collapses the useful voltage completely; a single line-to-earth fault leaves 70% of it. This is why the three-phase fault dominates transient-stability studies even where it is not the largest current — the machines lose all their electrical output.

The system-wide profile for the L-G fault at bus 3, computed with the off-diagonal Z-bus terms \(V_{s,i} = -Z_{s,i3}I_s\):

All sequence voltages are real here because the network is purely reactive and \(E\) is real. The faulted-phase voltage collapses to zero only at bus 3 itself; bus 4, one short line away, still holds 0.10 pu, and the generating buses hold more than half.

Three observations from the profile that recur in every fault study:

The first is because the positive-sequence sources are at buses 1 and 2; the second because the negative- and zero-sequence networks have their only source at the fault and are passive elsewhere. Negative-sequence quantities are therefore inherently directional, pointing at the fault — the basis of negative-sequence directional relaying.

Why \(V_0\) at bus 5 is so small (0.1715) compared with bus 4 (0.3368):

Zero-sequence voltage decays much faster than positive-sequence voltage because the tripled line reactances make the zero-sequence network electrically much longer. An earth-fault relay at bus 5 therefore sees a weak signal from a fault at bus 3 — a real limitation of residual-voltage polarisation on long lines.

The zero-sequence voltages found in Problem 16 drive the whole distribution. Restating them as magnitudes (all are negative real):

The fault bus has the largest magnitude, as it must: it is the injection point of the zero-sequence network.

Each line current follows from Ohm's law across it, using \(x_0 = 3x_1\):

There is no shunt in the zero-sequence line model here — line charging is omitted for fault studies — so the series current is the whole current.

Line 3–4 worked in full as the pattern:

Magnitude 0.6884 pu, flowing from bus 3 towards bus 4. The very low \(x_0 = j0.09\) of this short, heavily built line makes it the largest single zero-sequence path in the system.

All seven lines:

Note the direction convention: current flows from the fault bus and returns through the earthed neutrals, so most arrows point towards buses 1 and 2.

Kirchhoff's law at bus 3. All three of bus 3's branches carry current away from it:

Which is exactly \(I_0 = I_f/3 = 5.1901/3 = 1.7300\) pu, the zero-sequence current injected at the fault ✓. Every ampere injected at bus 3 leaves through a line — there is no earthed neutral at bus 3 itself.

The transformer neutrals are where the current re-enters the reference:

These are the currents that a neutral CT would measure, and they are what an earthing-transformer or neutral-earthing resistor must be rated to carry. Check: \(0.6184+1.1116 = 1.7300\) ✓ — the two neutrals between them return the whole zero-sequence current.

Bus 2 carries more than bus 1 despite bus 1 being the slack. The reason is entirely zero-sequence:

In positive sequence, bus 1 is the stronger source. In zero sequence, the ranking reverses. The earth-fault current distribution has nothing to do with generation and everything to do with where the star points are earthed — one of the most useful practical facts in this whole subject.

Consequence for earthing practice. Suppose transformer 2's neutral were opened:

Earth-fault current would drop substantially, which sounds attractive until you note that the healthy-phase voltage rise increases with \(X_0/X_1\) (Problem 5) and the fault becomes harder to detect. Multiple earthing points is a deliberate compromise between limiting current and keeping the system detectably and safely earthed.

Series faults need different boundary conditions. Let \(p\) and \(p'\) be the two sides of the break, and let \(V_a, V_b, V_c\) be the voltages across it. With phase \(a\) open and \(b, c\) intact:

The intact conductors carry current with no voltage drop across the break; the broken one has voltage across it but no current. Exactly the reverse of a shunt fault, where the fault point had voltage conditions on the faulted phases.

Transform both conditions:

Equal voltages and currents summing to zero — the signature of a parallel connection. The one-open-conductor case connects the three sequence networks in parallel across the break, exactly as the double line-to-earth fault connected them in parallel at the fault bus.

The dual case, two conductors open (\(b\) and \(c\) broken, \(a\) intact):

Equal currents, voltages summing to zero — a series connection, the mirror of the single line-to-earth fault. The whole of Part 5 reduces to two connection patterns used four ways.

The Thévenin equivalent across the break. With the break fully open, the network seen from \(pp'\) in each sequence is:

Both computed on the network with line 4–5 broken. The zero-sequence value is roughly three times the positive-sequence one, which is the tripled-reactance rule showing up again.

The driving voltage is the open-circuit voltage that would appear across \(pp'\) if all three conductors were opened. By superposition it is the prefault current times the positive-sequence Thévenin impedance:

Sanity check: with all three conductors intact, the impedance across \(pp'\) is zero, and \(I_1 = E_{th}/Z_{th,1} = I_{\text{pre}}\) ✓. The equivalent reproduces the prefault state, as it must.

One conductor open — the parallel connection gives:

The positive-sequence current has fallen from 0.06196 to 0.03542 pu — the line still carries 57% of its prefault power on two conductors, because the remaining phases have a return path through the rest of the network.

The phase currents:

All in pu on 100 MVA. Note that with one conductor open the two survivors each carry less than the prefault current (0.0553 vs 0.0620) — the load simply redistributes to other paths rather than overloading the remaining conductors.

Why open conductors are so hard to detect. The earth current here is:

Seven amperes, against an earth-fault relay that may be set at 200 A and a phase relay set well above load. No overcurrent element will ever see this. Yet a broken conductor lying on the ground is one of the most dangerous conditions on a distribution system.

The quantity that does detect it is the negative-sequence ratio:

A ratio of 75% — enormous, since a healthy system runs below 2%. And it is a ratio, so it is independent of loading: the same 0.750 whether the line carries 6 MW or 60. Broken-conductor protection is built on exactly this measurement, typically alarming above \(I_2/I_1 = 0.2\). For two conductors open the ratio is 1.000, higher still.

The fault current at bus 1, from the driving-point impedances of Set 22:

9.3709 pu = 2459 A, and each sequence current is \(I_0 = I_1 = I_2 = 3.12364\) pu. Larger than the three-phase value of 7.9909 pu, because \(X_0/X_1 = 0.558 < 1\) here (Problem 14).

Sequence voltages at bus 1:

Sum zero ✓. The phase voltages are \(V_a = 0\) and \(V_b = V_c = 0.9258\angle\mp110.70°\) — an earth-fault factor of only 0.926, below nominal, which is the hallmark of a strongly earthed bus.

The transformer's own contribution is found from the branch equations. In positive and negative sequence the machine and its transformer are lumped into \(x = j0.25\); in zero sequence only the transformer appears, \(x_0 = j0.10\):

Note the asymmetry: this branch supplies half the positive- and negative-sequence current (1.564 of 3.124) but 70% of the zero-sequence current (2.182 of 3.124). It is the nearer of the only two earthing points.

The star-side line currents by recombination:

The healthy phases are not current-free on this branch — only the total at the fault has \(I_b = I_c = 0\), and the two contributions from transformer 1 and from the network cancel there. Each individually carries 0.618 pu. The neutral current is \(3I_0 = 6.546\) pu = 1718 A.

Crossing to the delta side. Two things happen, and both matter:

The \(\pm30°\) is the standard Yd1/YNd1 phase displacement. Positive sequence advances, negative sequence retards — by the same angle, in opposite directions. In a balanced study the shift can be ignored; here it cannot, because the two sequences no longer rotate together.

The generator-side line currents:

A completely different pattern from the star side. One phase carries nothing, and the other two carry equal and opposite currents.

The classic result, worth committing to memory:

A single line-to-earth fault on the star side looks like a line-to-line fault from the delta side, in the ratio \(1:0:1\). The generator's own protection therefore sees an unbalanced two-phase condition, not an earth fault — and no earth-fault relay on the generator side will operate at all, because \(3I_0 = 0\) there.

Two checks:

The first confirms that no zero sequence crossed the delta; the second is the general rule that for this pattern the delta-side line current is \(\sqrt3\) times the positive-sequence current, because \(I_1'\) and \(I_2'\) are 60° apart.

Step 1 — assemble the data. Everything needed is three numbers from Set 22:

And the ratio that governs everything: \(X_0/X_1 = 0.06071/0.12009 = 0.506\). Being well below unity, we know before computing anything that the earth fault will be the worst.

Step 2 — the three-phase fault:

2185 A at 220 kV. All sequence voltages zero at the fault; \(I_0 = I_2 = 0\).

Step 3 — the single line-to-earth fault:

9.9705 pu = 2617 A, the largest phase-conductor current at this bus. The earth-fault factor is 0.917 — the healthy phases actually fall, the signature of a solidly and stiffly earthed bus.

Step 4 — the line-to-line fault:

1892 A, and exactly \(0.866\times8.3270\) ✓ — the universal \(\sqrt3/2\) ratio holds regardless of the network, because \(Z_2 = Z_1\).

Step 5 — the double line-to-earth fault:

The earth current of 12.42 pu = 3260 A is the largest current anywhere in this study — 49% above the three-phase fault. With \(X_0\) only half of \(X_1\), the parallel branch is very low impedance and the earth path takes a large share.

Step 6 — the summary table, which is the actual deliverable:

Breaker duty: 9.9705 pu = 2617 A. Earth-grid duty: 3260 A. Two different numbers, from two different rows, for two different pieces of equipment.

Step 7 — the consistency checks. Five identities that any correct study must satisfy:

Each of these is computed by a route independent of the number it checks, so agreement to five figures is strong evidence, not a tautology.

Step 8 — the sanity bounds that catch data errors rather than arithmetic ones:

A violation of any of these means the input data is wrong — most often a delta winding entered as an earthed star, a forgotten \(3Z_n\), or a zero-sequence line reactance left equal to the positive-sequence value.

Step 9 — what the study is for. The numbers feed four distinct downstream decisions:

Note the third and fourth rows use the minimum and maximum earth current respectively. Sensitivity is set by the smallest credible fault; thermal rating by the largest. Using one number for both is the commonest error in a first fault study.