Set 22 — Sequence Impedances and Networks
Twenty worked problems on building the circuits that Set 21's transformation calls for. Because a balanced network is diagonalised by \(\mathbf{A}\), each sequence sees a network of its own, and the three can be constructed independently. The positive-sequence network is the one Part 4 already built. The negative-sequence network differs from it only in the machines. The zero-sequence network is a different object entirely: its connectivity is set by transformer windings rather than by lines, its impedances are roughly three times larger, and it exists only where an earth path does. The five-bus system of Part 4 acquires all three here.
Static elements are sequence-blind. A line, cable or reactor has \(Z_1 = Z_2\) exactly, because reversing the phase sequence of a passive symmetric element changes nothing it can detect.
Machines are not. A synchronous machine has \(Z_1 = jx'' _d\), \(Z_2 \approx jx''_d\) and \(Z_0 \approx j(0.15\text{–}0.6)x''_d\) — three different values, because the rotor sees the three fields differently.
Zero sequence returns through earth. All three phase currents flow the same way, so the return is in the earth, the earth wire or the sheath — a much larger loop, giving \(Z_0 \approx 3Z_1\) for an overhead line and 1–5 times for a cable.
Transformer connections govern the zero-sequence network. A delta blocks it from the lines and provides a circulating path; an earthed star passes it; an isolated star blocks it entirely. Six standard diagrams cover every case.
Neutral earthing enters as \(3Z_n\) in the zero-sequence network only, because that per-phase network carries \(I_0\) while the physical neutral carries \(3I_0\).
Each network is built and inverted separately. Three \(\mathbf{Y}_{\text{bus}}\) matrices, three \(\mathbf{Z}_{\text{bus}}\) matrices, and at a faulted bus \(k\) only \(Z_{0,kk}\), \(Z_{1,kk}\), \(Z_{2,kk}\) are needed.
Sources appear only in the positive-sequence network. Generators produce balanced \(abc\) voltages, which are pure positive sequence — so the negative and zero networks are passive and driven only by the fault.
Define the sequence impedances of a three-phase element, state how each would be measured, and explain why three separate networks result.
The definition. Set 21 showed that a balanced element's impedance matrix is diagonalised by \(\mathbf{A}\):
The sequence impedance is simply the ratio of the voltage to the current of that sequence, with the other two absent. No cross terms exist, so each is a well-defined single number.
How each is measured, and the tests are genuinely performed:
The zero-sequence test is the distinctive one: the three phases are joined together and energised against earth, so the measured impedance is \(V/I\) with \(I\) a third of the total current.
The zero-sequence test in detail, because the factor of three appears here too:
Three phases in parallel carry \(3I_0\) for an applied \(V_0\), so the impedance per phase is three times the ratio a single ammeter and voltmeter would suggest.
Why three networks. Each sequence sees the whole system through its own impedances, and the topologies need not match:
The third is the important one. A transformer that connects two buses in the positive-sequence network may not connect them at all in the zero-sequence network, and that is a difference of topology, not of value.
The sources appear once only. A generator produces a balanced \(abc\) set of internal voltages, which is pure positive sequence:
So the negative- and zero-sequence networks contain impedances only. They are passive, and the fault at one bus is the sole excitation they receive — which is exactly why an unsymmetrical fault is analysed by connecting the three networks at that one bus.
And the structure of the whole of Part 5 follows. Build three networks; reduce each to a Thévenin impedance at the faulted bus; connect the three according to the fault's boundary conditions; solve one small circuit. The network analysis is done three times and the fault analysis once.
Show that \(Z_1 = Z_2\) exactly for any static element, and state which elements this covers.
The argument from the matrix. Set 21 obtained, for a balanced element,
The same expression for both — so the equality is exact and not an approximation. It follows purely from the circulant structure of \(\mathbf{Z}_{abc}\).
The argument from physics, which is more memorable. Negative sequence differs from positive only in the order in which the three phases peak. A static element has no moving part and no preferred direction of rotation:
Interchanging two phases relabels the element's terminals and nothing more. Its impedance cannot change.
Which elements this covers:
The transformer case deserves care. Its zero-sequence impedance equals its positive-sequence impedance — the leakage reactance is the same whatever the sequence, because the flux path is the same. What differs is whether a zero-sequence path exists at all:
A three-limb core transformer is the exception to the first line: its zero-sequence flux has no iron return path and must go through the tank and air, giving \(Z_0 \approx 0.85Z_1\). A five-limb or shell-type core, or a bank of three single-phase units, gives \(Z_0 = Z_1\).
The practical consequence. Because \(Z_1 = Z_2\) for every line and transformer, the negative-sequence network of a real system differs from the positive-sequence one only in the machines:
And since \(x_2 \approx x''_d\) for a turbogenerator, the two networks are often taken as numerically identical — which is what this book's five-bus system will do.
The one further exception is a static var compensator or any element containing thyristors, which are switched in synchronism with the positive-sequence voltage and therefore do distinguish the two sequences. They are modelled specially, and they are the only non-rotating element that does.
Give the three sequence impedances of a synchronous machine, explain the physical origin of each, and account for the large difference between them.
Positive sequence produces a field rotating with the rotor, so the machine's response depends on how long the disturbance has lasted:
A fault study uses \(x''_d\), so \(Z_1 = jx''_d\) in every network of Part 5.
Negative sequence produces a field rotating against the rotor at \(2\omega_s\) relative to it. The damper windings and rotor body are always screening it, so there is no time dependence at all:
More precisely \(x_2 = (x''_d + x''_q)/2\), the mean of the two axes — because the backward-rotating field sweeps past both the direct and quadrature axes twice per cycle. For a round-rotor machine the two are nearly equal and \(x_2 = x''_d\).
Zero sequence is different in kind. All three phase currents are in phase, so their magnetomotive forces are displaced by 120° in space and cancel:
There is no rotating field, no fundamental air-gap flux, and therefore no reaction from the rotor. The only impedance is the leakage of the stator winding.
Hence \(x_0\) is small:
Its exact value depends on the winding pitch: a two-thirds-pitch winding produces almost no zero-sequence flux and gives the lowest \(x_0\). Machine designers choose the pitch partly for this reason.
The three together, for a typical 200 MVA turbogenerator:
Induction machines follow the same pattern with different numbers. A running induction motor has
The ratio \(Z_1/Z_2\) is about 6, which is the factor behind the motor derating figures of Set 21.
Derive the zero-sequence impedance of a transmission line from its self and mutual impedances, and justify the rule \(Z_0 \approx 3Z_1\).
From Set 21's diagonalisation:
A single ratio, determined entirely by how large the mutual is relative to the self.
The values for an overhead line with earth return. Working in inductance per kilometre, the self and mutual terms both refer to the earth-return conductor at Carson's equivalent depth \(D_e\):
With \(D_e = 931\) m for average soil at 50 Hz, from Problem 5.
A numerical case. GMR \(= 0.0114\) m and GMD \(= 7.56\) m, the line of Set 15:
The sequence inductances and reactances:
The \(D_e\) cancels — which is why the positive-sequence reactance of Set 15 could be computed without any reference to the earth at all.
The rule of thumb, derived. Here \(D_e\) does not cancel, which is the whole difference between the two sequences.
And the resistances. The earth-return resistance \(r_e = 0.0493\ \Omega\)/km appears once in each self term and once in each mutual, so:
So the zero-sequence resistance is nearly doubled as well — which matters for the damping of earth-fault transients and for the arc-resistance component of an earth-fault loop.
The physical reason, stated without algebra. Positive-sequence current returns through the other two phases, a few metres away; zero-sequence current returns through the earth, hundreds of metres below:
A much larger loop, hence a much larger inductance. The earth's resistivity sets the return depth, so \(x_0\) depends on the soil — which is why it is quoted as a range rather than a number.
The observed range:
An earth wire lowers \(x_0\) by providing a nearer return path in parallel with the earth. A double circuit raises it, for the reason of Problem 7.
Explain Carson's equivalent-depth model of the earth return, evaluate the depth for three soil resistivities, and state its effect on the zero-sequence impedance.
The problem Carson solved. Earth-return current does not flow in a wire; it spreads through a semi-infinite conducting half-space, with a distribution that depends on frequency and resistivity. Carson's 1926 solution replaces it by a fictitious conductor at an equivalent depth:
With \(\rho\) in Ω·m and \(f\) in hertz. That single substitution turns an electromagnetic field problem into a circuit one.
The values at 50 Hz:
The 931 m used in Problem 4 corresponds to average soil. Note the range: a hundredfold change in resistivity gives a tenfold change in depth, because of the square root.
The effect on \(x_0\), which is much smaller than that range suggests:
A hundredfold change in soil resistivity changes \(x_0\) by 23%. The logarithm again — and it is why a single value of \(x_0\) can be used for a line crossing varied terrain.
The earth resistance term is the surprise, and it does not depend on the soil at all:
Depending only on frequency. As the resistivity rises the current spreads deeper and wider, and the two effects cancel exactly in Carson's approximation. A remarkable result, and it means the resistive part of \(z_0\) is known even when the soil is not.
Where the approximation fails. Carson's series is truncated after the first terms, which requires
— satisfied comfortably at power frequency, where \(D \sim 10\) m and \(D_e \sim 900\) m. At the frequencies of a lightning surge, \(D_e\) falls to metres and the approximation collapses, which is why travelling-wave studies use a full frequency-dependent earth model.
And the practical consequence for protection. The earth-fault loop impedance depends on soil, so a distance relay's earth-fault reach is less certain than its phase-fault reach:
Which is why earth-fault zones are set with a larger margin, and why \(Z_0\) is measured on commissioning rather than calculated where the setting is critical.
Explain why a cable's zero-sequence impedance behaves differently from an overhead line's, and give the range of \(Z_0/Z_1\) for the common constructions.
The difference is the metallic sheath. A cable's conductors are surrounded by an earthed screen only millimetres away, which offers a return path far closer than the earth:
A factor of about twenty thousand in the return distance — which the logarithm compresses, but not to nothing.
The zero-sequence current divides between the sheath and the earth in inverse proportion to the two paths' impedances:
And the sheath, being close, takes most of it — typically 70–95%. That is why sheath bonding arrangements matter so much for zero-sequence behaviour.
The bonding arrangements, and each gives a different \(Z_0\):
Cross-bonding is standard for long high-voltage circuits because it eliminates sheath circulating current under normal balanced load — which is a positive-sequence consideration — while still providing an earth-fault return.
The ratios in practice:
A bonded cable's \(Z_0\) can be lower than three times \(Z_1\) and occasionally close to it — the opposite of the overhead-line rule, and a trap for anyone applying the factor of three by habit.
The resistive part is the striking one. The sheath is aluminium or lead of substantial cross-section but much higher resistivity than the conductor, so
Far higher than an overhead line's 1.9. An earth fault on a cable is therefore a much more resistive fault than one on a line, which changes both the fault current's angle and the arc energy.
The consequence for a mixed circuit. A line that is partly overhead and partly cable has a zero-sequence impedance that is not a simple weighted average, because the earth-return mechanisms differ:
And the distance-relay earth-fault reach, which depends on \(Z_0/Z_1\) through the residual compensation factor, must be set for a composite that is different in each section. This is one of the harder settings problems in practice.
Two circuits share a tower. Show that they are mutually coupled in the zero-sequence network only, quantify the coupling, and state what it does to earth-fault calculations.
Why the coupling is zero-sequence only. Consider the flux linking circuit B from a balanced current in circuit A:
A balanced set of currents produces a field that falls off rapidly with distance — the three contributions largely cancel a few metres away. A zero-sequence set produces a field equivalent to \(3I_0\) in a single conductor with an earth return, which falls off only logarithmically.
The magnitudes:
Two orders of magnitude apart. The first is universally neglected; the second cannot be.
The zero-sequence mutual, computed. For two circuits whose conductors have a geometric mean separation \(D_{AB}\):
The factor 0.6 rather than 0.2 because all three phases of each circuit participate. With \(D_e = 931\) m and \(D_{AB} = 10\) m:
Against a self \(x_0\) of 1.32 Ω/km — a coupling of 65%.
The circuit consequence. The zero-sequence voltage drop along circuit A depends on circuit B's current:
So the two circuits' zero-sequence networks are not independent, and the singular transformation of Set 16 — with off-diagonal terms in \([\mathbf{y}]\) — is the only formation method that handles it.
What it does to an earth-fault calculation. Three effects, all of them significant:
The middle case is the dangerous one for personnel: a circuit taken out of service and earthed at both ends carries substantial induced current from a fault on its neighbour, and the earthing arrangements must be rated for it.
The relay-reach problem is the commonest practical consequence. A distance relay computes the earth-fault loop impedance using a residual compensation factor:
Which assumes the only zero-sequence current is the circuit's own. With mutual coupling, part of \(3I_0\) belongs to the parallel circuit, the compensation is wrong, and the relay sees the fault at the wrong distance. The direction of the error depends on whether the parallel circuit is in service, out of service, or out and earthed — so a single setting cannot be right for all three.
Set out the zero-sequence equivalent circuit for each of the six standard transformer connections, and state the rule that generates them.
The two rules, applied to each winding independently:
And a zero-sequence current can flow through the transformer only if both windings offer a path. Everything else follows.
The six connections:
Only the first passes zero-sequence current through; only the third provides a source of it at a bus. The other four are open circuits.
The \(Y_g\)–\(\Delta\) case is the important one, and it deserves the physical explanation. Zero-sequence current flowing in the earthed star winding must be balanced by ampere-turns in the delta:
The delta provides that path, so current can flow in the star winding — but it never leaves the delta side. In the sequence diagram this appears as an impedance from the star-side bus to the reference: a shunt, not a series element.
Which makes an earthed-star/delta transformer a zero-sequence source. It supplies earth-fault current to the star-side network without any connection to what lies beyond the delta. That is why a delta-star unit transformer at a generating station provides the earth-fault infeed even though the generator itself is delta-connected and cannot.
The three-winding case adds one further element. A \(Y_g\)–\(Y_g\)–\(\Delta\) autotransformer has all three:
Modelled as a star equivalent of three impedances \(Z_H\), \(Z_M\), \(Z_L\) with the tertiary's branch going to the reference. The delta tertiary of a large autotransformer exists partly for this purpose — without it the zero-sequence path would be poor and the earth-fault current low.
The commonest error is to draw the zero-sequence network with the same topology as the positive-sequence one. In a network of the sort in Problem 12, most transformers are open circuits to zero sequence, and the zero-sequence network is a set of disconnected islands where the positive-sequence network is fully meshed.
Compare the four neutral earthing arrangements, give the zero-sequence impedance each produces, and state the trade being made.
The four arrangements, with the impedance each contributes to the zero-sequence network:
And a fifth, resonant (Petersen coil) earthing, in which \(3X_n\) is tuned to resonate with the system's total capacitance to earth — which cancels the fault current rather than limiting it.
The earth-fault current each gives, for a system with \(Z_1 = Z_2 = j0.15\) and \(Z_0 = j0.05\) pu at the machine:
Three orders of magnitude between the extremes.
The trade being made is between fault damage and overvoltage:
An isolated system continues to run with one phase earthed — an advantage in a continuous process — but the other two phases sit at full line voltage to earth indefinitely, so the insulation must be rated for it, and a second earth fault on another phase becomes a phase-to-phase fault through two lengths of cable.
The effectively-earthed criterion formalises the boundary:
A system meeting this is called effectively earthed and may use surge arresters rated at 80% of line-to-line voltage rather than 100% — a real saving on insulation and arrester cost at transmission voltages, and the reason transmission systems are solidly earthed.
The practice by voltage level:
The generator case is distinctive. A generator's stator earth fault burns iron, which is expensive to repair, so the earth-fault current is limited to a few amperes — typically by a distribution transformer with a resistor on its secondary, giving an effective \(R_n\) of hundreds of ohms. The fault is then detected by voltage rather than current.
State the positive-sequence network of the five-bus system, and say what has and has not changed from Set 18.
It is the network of Set 18, unchanged. The fault network built there — generators behind their subtransient reactances, lines as pure reactances — is the positive-sequence network:
Its impedance matrix is the one built in nine steps in Set 18:
What has changed is only the label. Set 18 called it "the fault network" and computed three-phase fault currents from its diagonal. It is now called the positive-sequence network, and its diagonal is \(Z_{1,kk}\) — one of the three quantities every unsymmetrical fault formula needs.
What has not changed is the modelling. The same four assumptions of Set 18 apply, and their direction of error is unchanged:
And one further point specific to sequence work. The generators' internal voltages are pure positive sequence, so this is the only network with a source in it. Reducing it for a fault at bus \(k\) gives a Thévenin source \(E = 1.0\) pu behind \(Z_{1,kk}\); the other two networks reduce to a bare impedance.
The Thévenin impedances at the five buses, which are all that Set 23 will use:
Construct the negative-sequence network of the five-bus system and state precisely how it differs from the positive-sequence one.
Two changes only, and one of them is often nil:
The lines are unchanged, because \(Z_1 = Z_2\) exactly for a static element (Problem 2). The topology is identical.
The machine values. For the turbogenerators of this system, taking \(x_2 = x''_d\) as is usual for a round-rotor machine:
Identical to the positive-sequence values. So for this system the two networks are numerically the same network.
Hence:
Which spares a whole matrix inversion, and is the usual situation in practice.
When the two do differ. Three cases, in increasing order of importance:
The second is the one that bites: a system with substantial motor load has a negative-sequence network appreciably stiffer than its positive-sequence one, because every motor is a low-impedance shunt to negative sequence.
The critical structural point. The negative-sequence network is passive. No source appears anywhere in it:
Because a generator's internal EMFs are a balanced \(abc\) set and therefore pure positive sequence. Any negative-sequence current that flows is driven entirely by the fault, entering the network at the faulted bus.
Which means the reduction is simpler. Reducing the negative-sequence network to the faulted bus gives a single impedance \(Z_{2,kk}\) to the reference, with no source in series. The same is true of the zero-sequence network. Only the positive-sequence network reduces to a source behind an impedance, and that source drives the whole fault.
Construct the zero-sequence network of the five-bus system. Each generator connects through its own unit transformer, delta on the machine side and earthed star on the network side, with zero-sequence reactances of \(j0.10\) at bus 1 and \(j0.08\) at bus 2. Line zero-sequence reactances are three times their positive-sequence values.
The generators disappear. Their unit transformers are delta on the machine side, and a delta blocks zero sequence:
However low the machine's own zero-sequence reactance, no zero-sequence current can reach it. The machine is simply absent from this network.
The transformers become shunts. Each is a \(Y_g\)–\(\Delta\), which by Problem 8 gives an impedance from the star-side bus to the reference:
These are the network's only earth connections, and therefore the only sources of earth-fault current in the whole system.
The lines are tripled:
The network, described: the same seven lines with three times the reactance, two shunt paths to the reference at buses 1 and 2, and nothing at all at buses 3, 4 and 5. Compare with the positive-sequence network, which has shunt sources at buses 1 and 2 of \(j0.25\) and \(j0.20\):
The shunts are stiffer and the lines are weaker — the reverse of the positive-sequence situation, and it changes the character of the network completely.
The consequence, before any arithmetic. Buses 1 and 2 have a very stiff local earth connection and will have a low \(Z_0\); buses 3, 4 and 5 have none and must reach the earth through tripled line reactances, so their \(Z_0\) will be high. Expect
Which is exactly what Problem 13 finds, and it means earth faults will be more severe than three-phase faults at the generator buses and less severe elsewhere.
If the unit transformers had been star–star with both neutrals earthed, the machines' own \(x_0\) would appear in series with the transformer's, giving shunts of perhaps \(j0.16\) and \(j0.14\) instead. If they had been star–star with the machine neutrals isolated — the usual arrangement — the network would have no earth connection at all and earth-fault current would be zero everywhere.
Build the zero-sequence impedance matrix of the five-bus system and present it.
The method is that of Set 18, applied to a different network. Nine elements again: two shunts and seven lines, so one Type 1, one Type 3, and the rest as before.
The matrix:
A check on bus 1's diagonal. The transformer's \(j0.10\) is in parallel with the path through the network to bus 2's \(j0.08\). The dominant alternative path is line 1–2 at \(j0.18\) in series with \(j0.08\):
Against the computed 0.06985. The small difference is the further parallel paths through buses 3, 4 and 5 — which contribute little, since they lead nowhere except back to the same two earths.
The diagonal, in order:
The prediction of Problem 12 confirmed: the ratio is about 0.5 at the transformer buses and rises to 1.55 at bus 5, the furthest from any earth.
The off-diagonals are revealing too. \(Z_{0,34} = j0.19466\) is 84% of \(Z_{0,33}\) — buses 3 and 4 are strongly coupled in the zero-sequence network as well, by the \(j0.09\) tie. But \(Z_{0,12} = j0.02412\) is only 35% of \(Z_{0,11}\), against 80% in the positive-sequence matrix:
Buses 1 and 2 are much less coupled in zero sequence, because each has its own strong local earth and does not need the other.
The three matrices are now complete. \(\mathbf{Z}_1\) from Set 18, \(\mathbf{Z}_2 = \mathbf{Z}_1\) from Problem 11, and \(\mathbf{Z}_0\) here. Every unsymmetrical fault calculation of Set 23 needs three numbers from them — one diagonal element from each — and nothing more.
Compare the three networks bus by bus and predict, without computing any fault current, at which buses an earth fault will exceed a three-phase fault.
The criterion. The two fault currents are
With \(Z_1 = Z_2\), which holds here:
A single comparison. The earth fault is more severe wherever the zero-sequence impedance is lower than the positive-sequence one.
Applying it:
At the two transformer buses the earth fault is 17–20% more severe; elsewhere the three-phase fault is worse by 10–15%.
The physical reason. At buses 1 and 2 the earth connection is the transformer — \(j0.10\) and \(j0.08\) — which is stiffer than the machine's \(j0.25\) and \(j0.20\) in the positive-sequence network. The zero-sequence network is locally the stronger of the two.
The bounds of the ratio are worth stating, because they set the range of what is possible:
The upper bound of 1.5 is reached only with a perfect earth at the fault point — never in practice, but a solidly-earthed transformer terminal approaches it. Values of 1.2 to 1.4 at generator and transformer buses are common.
The practical consequence, and it is a real one. Switchgear is normally rated on the three-phase fault level, which is the standard assumption:
So a breaker selected on the three-phase level alone would be under-rated at those two buses. IEC and IEEE standards require the earth-fault level to be checked separately at solidly-earthed transformer terminals for exactly this reason, and it is the commonest place where the two differ enough to matter.
Reduce all three networks to their Thévenin equivalents at bus 3, and set out what the resulting three-terminal picture represents.
The three reductions, each a single diagonal element:
Three two-terminal circuits, each with one terminal at bus 3 and one at the reference.
What each represents. The whole five-bus system, seen from one bus, in one sequence:
Two generators, seven lines and two transformers, reduced to three numbers — and no information relevant to a fault at bus 3 has been lost.
Why only the positive network has a source. Thévenin's theorem gives the open-circuit voltage at the terminal:
A healthy balanced system has no negative- or zero-sequence voltage anywhere, by definition. So those two Thévenin sources are zero and only their impedances remain.
The three-terminal picture. Draw three boxes, each with a terminal pair — one marked \(F_1\)–\(N_1\), one \(F_2\)–\(N_2\), one \(F_0\)–\(N_0\):
The fault is a connection between these six terminals, and which connection depends entirely on the fault type. That is the whole content of Set 23.
The three sequence voltages and currents at the fault satisfy the network relations:
Three equations, six unknowns. The fault type supplies the other three — for example \(I_1 = I_2 = I_0\) and \(V_1+V_2+V_0 = 0\) for a solid single line-to-earth fault, which is two independent conditions plus the definition.
And the reduction must be redone for each faulted bus. The three numbers above apply to bus 3 alone; a fault at bus 5 needs \(Z_{1,55}\), \(Z_{2,55}\) and \(Z_{0,55}\). That is why the three matrices are built rather than three single equivalents — each diagonal element is one bus's Thévenin impedance, computed in advance.
The unit transformer at bus 2 is replaced by a star–star unit with both neutrals isolated. Rebuild the zero-sequence matrix and assess the consequences.
The change. An isolated-star winding offers no zero-sequence path, so bus 2 loses its earth connection entirely:
The network now has exactly one connection to earth, at bus 1.
The new diagonal:
Bus 1's value becomes exactly \(j0.10\) — the transformer alone, since no other earth path exists and the lines lead nowhere. Bus 2 rises fourfold.
The earth-fault currents fall correspondingly. At bus 5:
A 19% reduction from changing one transformer's winding connection — with no change whatever to any line, any machine, or any impedance value.
The consequences, and they cut both ways:
The second is decisive at transmission voltages: \(X_0/X_1 > 3\) puts the system outside the effectively-earthed definition and forces 100%-rated surge arresters instead of 80%.
The fourth is the operational objection. With one earth point, taking bus 1's transformer out of service leaves the network with no earth reference:
Earth-fault current becomes negligible, earth-fault protection stops working, and the healthy phases rise to full line voltage on any earth fault. Systems are therefore designed with at least two earth points and an operating rule that one must always remain.
The general lesson about switching. The zero-sequence network's topology changes when plant is switched, in a way the positive-sequence network's does not:
Which is why earth-fault studies are repeated for every credible switching arrangement, and why the "minimum earthing" case is a standard study alongside the maximum-plant one.
An earthing transformer of zero-sequence reactance \(j0.12\) pu is installed at bus 5. Recompute the zero-sequence matrix and assess whether it is worth installing.
The modification is a Type 3 addition to \(\mathbf{Z}_0\) — a branch from bus 5 to the reference:
The result:
A dramatic local effect and a modest remote one — which is the signature of a shunt element in any network.
The earth-fault current at bus 5:
A 46% increase — and it now exceeds the three-phase fault current at bus 5, which is 4.888 pu.
Is it worth installing? The arguments for:
And against:
The first is the binding one. If bus 5's switchgear was rated on the 4.888 pu three-phase level, the new 6.046 pu earth-fault level exceeds it and the plant must be replaced.
The verdict depends on why it was proposed. If earth-fault detection at bus 5 is inadequate — and at \(Z_0/Z_1 = 1.55\) it may well be — the earthing transformer solves that. But a smaller one would too:
A \(j0.30\) unit gives most of the detection benefit and keeps the earth-fault level below the three-phase one. Sizing an earthing transformer is a matter of choosing where on this curve to sit, not of making the earth as strong as possible.
For an earth fault at bus 3 drawing \(3I_0 = 5.190\) pu, find the zero-sequence voltage at every bus and the zero-sequence current in every line, and verify the balance.
The zero-sequence current per phase at the fault is
Drawn out of bus 3 in the zero-sequence network. Everything else follows by superposition, exactly as the positive-sequence fault profile did in Set 18.
The zero-sequence bus voltages, from column 3 of \(\mathbf{Z}_0\):
Largest at the fault and falling with electrical distance from it. Note that a zero-sequence voltage of 0.40 pu exists at bus 3 — this is the neutral displacement, and it is what a residual voltage relay measures.
The line currents, each \((V_{0,a}-V_{0,b})/z_{0,ab}\):
All three of bus 3's branches carry current away from it — the fault is the source in this network and the two transformer neutrals are the sinks. Bus 5, with no earth of its own, is a pure through-route: it receives 0.2295 from bus 4 and passes the same on to bus 2.
The earth-path currents, through the two transformers:
Every ampere of zero-sequence current entering the network at the fault leaves through one of the two transformer neutrals. The balance closes exactly.
Kirchhoff at bus 3. All three branches carry current away from the fault:
Exactly the injected zero-sequence current. The 0.6884 on the stiff 3–4 tie is the largest single flow in the network, and it goes the long way round to bus 2's transformer via buses 4 and 5.
The physical picture, stated plainly. An earth fault at bus 3 injects zero-sequence current into the zero-sequence network at that bus. It spreads through the tripled line reactances towards buses 1 and 2, and leaves the network through the two transformer neutrals in the ratio 36:64 — bus 2's transformer taking the larger share, being both stiffer and better connected.
Explain why only the positive-sequence network contains sources, and examine what would follow if a generator produced an unbalanced internal EMF.
The argument. A synchronous machine's internal EMFs are generated by a rotating field cutting three identical windings displaced by 120°:
A balanced \(abc\) set, so pure positive sequence — by Problem 6 of Set 21. The negative- and zero-sequence networks contain no source of any kind.
The structural consequence. The three networks reduce differently at a faulted bus:
Which is why the fault current always has \(E\) in its numerator and a combination of all three impedances in its denominator. Every fault formula in Set 23 has that structure.
And the prefault condition. Before the fault the system is balanced, so
The negative- and zero-sequence networks are entirely dead until the fault occurs. That is what makes superposition so clean: the fault's contribution is the whole of the negative- and zero-sequence quantities.
If a machine's EMF were unbalanced, which happens in practice for a small residue:
A source in the negative-sequence network would drive negative-sequence current continuously, even with no fault — which is exactly what a machine with a winding fault does, and why negative-sequence current is used to detect one.
The magnitude of the effect. A \(0.3\%\) negative-sequence EMF driving the network's negative-sequence impedance of, say, \(j0.12\) pu:
Which is a third of a turbogenerator's continuous negative-sequence capability, produced by nothing but manufacturing tolerance. It is one reason machines are tested for negative-sequence emission before commissioning.
The same argument covers loads. A balanced load is a passive impedance and contributes to all three networks; an unbalanced load — a traction feeder, a large single-phase furnace — behaves as a negative-sequence source and injects \(I_2\) into the network continuously. Modelling it requires abandoning the assumption that only the positive-sequence network is driven, and it is done by injecting a specified \(I_2\) at the load's bus.
Present the three sequence networks of the five-bus system as a finished data set, state every assumption, and identify which of them most affects the answer.
Positive sequence. Generators \(j0.25\) at bus 1 and \(j0.20\) at bus 2 to the reference; seven lines at \(j0.06\), \(j0.24\), \(j0.18\), \(j0.18\), \(j0.12\), \(j0.03\), \(j0.24\); source \(E = 1.0\angle0^\circ\).
Negative sequence. Identical, sources removed.
Zero sequence. Transformer earths \(j0.10\) at bus 1 and \(j0.08\) at bus 2; the same seven lines at three times their reactance; no generators.
The assumptions, in the order they were made:
Which matters most. Not the numerical ones. Assumptions 1 to 7 together move the answers by perhaps 30%, and each is defensible against a standard. Assumption 8 changes the answer by a factor of four (Problem 16) or removes it entirely:
A modelling decision with a hundredfold range of consequence, made by reading a nameplate rather than by calculation.
Assumption 6 is the second most important, and the least often examined. Taking \(x_0 = 3x_1\) when the true value is 2.0 or 3.5:
A 21% spread in earth-fault current from a ratio that is rarely measured and usually assumed. On a system where earth-fault relay grading is tight, that is enough to matter.
The data set is now complete. Set 23 needs three numbers per faulted bus and nothing else:
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.
P1. Why is \(Z_1 = Z_2\) exactly for a transmission line?
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A static element has no way to distinguish \(abc\) from \(acb\) — reversing the sequence just relabels its terminals. Problem 2.P2. A machine has \(x''_d = 0.18\). Estimate \(x_2\) and \(x_0\).
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\(x_2 \approx \mathbf{0.18}\); \(x_0 \approx 0.15\text{--}0.6\) of that, so \(\mathbf{0.03\text{--}0.11}\) pu. Problem 3.P3. A line has \(Z_s = 0.30+j1.10\) and \(Z_m = 0.05+j0.40\) Ω/km. Find \(Z_0\) and \(Z_1\).
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\(Z_0 = Z_s+2Z_m = \mathbf{0.40+j1.90}\); \(Z_1 = Z_s-Z_m = \mathbf{0.25+j0.70}\). Ratio \(x_0/x_1 = 2.71\).P4. Find Carson's equivalent depth for \(\rho = 200\ \Omega\)m at 50 Hz.
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\(658.4\sqrt{200/50} = \mathbf{1317}\) m.P5. Which transformer connections pass zero-sequence current straight through?
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Only \(\mathbf{Y_g\text{--}Y_g}\). \(Y_g\)–\(\Delta\) gives a shunt path but no through current; the other four are open. Problem 8.P6. A generator's neutral is earthed through \(j0.05\) pu and its \(x_0 = 0.06\). What appears in the zero-sequence network?
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\(x_0 + 3x_n = 0.06+0.15 = \mathbf{j0.21}\) pu.P7. A bus has \(Z_1 = Z_2 = j0.15\) and \(Z_0 = j0.10\). Which is worse, an earth fault or a three-phase fault?
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\(Z_0 < Z_1\), so the earth fault: ratio \(3(0.15)/(0.30+0.10) = 1.125\). Problem 14.P8. What is the theoretical maximum of \(I_{LG}/I_{3\phi}\)?
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\(\mathbf{1.5}\), reached as \(Z_0 \to 0\) with \(Z_1 = Z_2\).P9. Why is zero-sequence mutual coupling between parallel circuits so much larger than positive-sequence coupling?
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A balanced set's field cancels within metres; a zero-sequence set behaves as \(3I_0\) in one conductor with an earth return, whose field falls off only logarithmically. 60% against 1–5%. Problem 7.P10. A cable's sheath is bonded at both ends. Is \(Z_0/Z_1\) larger or smaller than for an overhead line?
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Smaller — 1–2 rather than 3–3.5, because the sheath is a much closer return than the earth. Problem 6.P11. Which sequence networks contain sources, and why?
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Only the positive. A machine's EMFs are a balanced \(abc\) set, hence pure positive sequence. Problem 19.P12. A system's only earthed transformer is switched out. What happens to earth-fault current?
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\(Z_0 \to \infty\), so the current becomes negligible — and the healthy phases rise to full line voltage on any earth fault. Problem 16.
Challenge Problems
Three problems on the zero-sequence network's awkward corners — the ones that produce wrong answers in real studies.
C1 — The earth wire. An overhead line carries an earth wire bonded to every tower. Show how it enters the zero-sequence impedance, quantify the reduction, and explain why it does not affect the positive sequence.
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Why the positive sequence is unaffected. The three phase currents sum to zero, so their net field at the earth wire's position is small and the induced current is negligible. The earth wire is electrically invisible to positive- and negative-sequence current.
The zero-sequence case. All three phases carry \(I_0\) in the same direction, inducing a substantial voltage along the earth wire. Being bonded at every tower, it carries a circulating current \(I_w\) that opposes the inducing flux — a shorted turn.
The circuit. Writing the two loop equations for the phase group and the earth wire:
\[ V_0 = Z_{0}'I_0 + Z_{0w}I_w \qquad 0 = Z_{0w}I_0 + Z_{ww}I_w \]Eliminating \(I_w\):
\[ Z_0 = Z_0' - \frac{Z_{0w}^{2}}{Z_{ww}} \]A reduction, always — the subtracted term is positive, being a squared quantity over an impedance of similar angle.
The magnitude. With a steel earth wire of \(Z_{ww} \approx 3+j2\) Ω/km, \(Z_{0w} \approx 0.05+j0.8\) and \(Z_0' = 0.31+j1.32\):
\[ \frac{Z_{0w}^{2}}{Z_{ww}} = \frac{(0.80\angle86^\circ)^{2}}{3.61\angle34^\circ} = 0.177\angle138^\circ = -0.132+j0.118 \]\[ Z_0 = 0.31+j1.32 - (-0.132+j0.118) = 0.442 + j1.202 \]The reactance falls by 9% and the resistance rises by 43% — the second being the loss in the earth wire itself, which the phase conductors now see as added resistance. With an aluminium-clad or OPGW earth wire of much lower \(Z_{ww}\) the reactance reduction reaches 25–30%.
The practical points. The earth wire's material matters as much as its presence: a steel wire gives a small reduction and a large resistance increase, an ACSR or OPGW wire a large reduction and a small increase. And an earth wire that is not bonded at every tower — or whose bonds have corroded — behaves quite differently, which is why \(Z_0\) is measured after major line refurbishments.
C2 — The autotransformer. A 400/220 kV autotransformer with a delta tertiary connects two networks. Derive its zero-sequence equivalent, explain what the tertiary does, and show what happens if it is omitted.
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The three-winding star equivalent. Measured on three short-circuit tests, the impedances \(Z_{HM}\), \(Z_{HL}\), \(Z_{ML}\) give
\[ Z_H = \tfrac{1}{2}(Z_{HM}+Z_{HL}-Z_{ML}) \quad Z_M = \tfrac{1}{2}(Z_{HM}+Z_{ML}-Z_{HL}) \quad Z_L = \tfrac{1}{2}(Z_{HL}+Z_{ML}-Z_{HM}) \]A star of three branches meeting at a fictitious point. In the zero-sequence network:
\[ \begin{array}{ll} Z_H & \text{from the 400 kV bus to the star point} \\ Z_M & \text{from the 220 kV bus to the star point} \\ Z_L & \text{from the star point to the } \textbf{reference}\ \text{(the delta)} \end{array} \]What the tertiary does. Three things, and only the first is usually mentioned:
\[ \begin{array}{ll} 1 & \text{Provides a zero-sequence path, so the autotransformer earths both systems} \\ 2 & \text{Traps triple harmonics, which are a zero-sequence set} \\ 3 & \text{Stabilises the neutral against unbalanced loading} \end{array} \]Without the tertiary, an autotransformer is a \(Y_g\)–\(Y_g\) device with the two windings physically joined, so:
\[ \begin{array}{ll} \text{Zero sequence passes through} & \text{but with no path to the reference} \\ \text{The two systems' zero-sequence networks are joined} & \text{whether or not that is wanted} \end{array} \]That coupling is the real problem. An earth fault on the 220 kV system draws zero-sequence current from the 400 kV system through the autotransformer, and the earth-fault levels of the two systems become interdependent. Adding a delta tertiary provides a local return and largely decouples them.
And a subtlety in \(Z_L\). The star-equivalent impedances are not physical: \(Z_L\) is frequently negative for an autotransformer, because the three measured impedances do not satisfy the triangle relations a physical star would require. It is nonetheless the correct circuit model, and a negative branch impedance in a transformer equivalent is normal rather than an error.
C3 — The study that gave the wrong answer. An earth-fault study of a 132 kV network predicts 8 kA at a substation; the measured value on a real fault is 5.2 kA. Identify the four most likely causes in order of probability, and state how each would be confirmed.
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Cause 1: a transformer connection wrong in the data. The commonest, and it produces errors of exactly this size. A \(Y_g\)–\(\Delta\) entered where the plant is \(Y_g\)–\(Y\), or an earthing switch modelled as closed when it is open, removes an earth path and raises \(Z_0\). Confirm by physically checking the nameplates and the neutral earthing arrangement of every transformer within two buses.
Cause 2: \(x_0/x_1\) assumed as 3 when it is higher. A ratio of 3.5 rather than 3.0 raises \(Z_0\) by 15% and reduces the earth-fault current by around 8%. Not enough alone, but it compounds. Confirm by a zero-sequence impedance measurement on the line, or by the line's design data.
Cause 3: mutual coupling with a parallel circuit neglected. If the faulted circuit shares a tower with another, and that other was out of service and earthed, the coupling changes the effective \(Z_0\) substantially. Confirm by checking the switching state at the time of the fault against the study's assumption.
Cause 4: fault impedance. The measurement is of a real fault with an arc; the study assumes a solid fault. An arc of 1–2 Ω on a 132 kV system is 0.1–0.2 pu, which is comparable to the whole \(Z_0\). Confirm by the recorded voltage at the fault point — a solid fault gives zero, an arc gives a few kilovolts.
The diagnosis strategy. Compute the \(Z_0\) the measurement implies:
\[ \frac{8.0}{5.2} = 1.54 \quad\Rightarrow\quad 2Z_1 + Z_0^{\text{true}} = 1.54\left(2Z_1+Z_0^{\text{assumed}}\right) \]and see which of the four causes could account for that much. A 54% increase in the total sequence impedance is too large for causes 2 or 4 alone and is characteristic of a missing earth path — so cause 1 is the working hypothesis, and it is the one to check first because it costs nothing to check.
The general lesson: when a fault study disagrees with a measurement, suspect the topology before the impedances. Impedance errors give tens of per cent; topology errors give factors.
Multiple-Choice Questions
MCQ 1. For a transmission line:
(a) \(Z_1 = Z_2 = Z_0\) (b) \(Z_1 = Z_2 \ne Z_0\) (c) \(Z_1 \ne Z_2 = Z_0\) (d) all three differShow answer
(b). Static elements cannot distinguish the two rotating sequences, but zero sequence returns through earth. Problems 2 and 4.MCQ 2. A synchronous machine's zero-sequence reactance is small because:
(a) the winding resistance is low (b) the zero-sequence MMF cancels in space (c) the rotor screens it (d) of saturationShow answer
(b) — no rotating field is produced, so only leakage remains. Problem 3.MCQ 3. For an overhead line, \(x_0/x_1\) is typically:
(a) 1 (b) 2 (c) 3 (d) 10Show answer
(c), because the earth-return loop is far larger than the phase-to-phase one. Problem 4.MCQ 4. Carson's equivalent depth \(D_e\) depends on:
(a) conductor size (b) soil resistivity and frequency (c) phase spacing (d) line lengthShow answer
(b) — \(658.4\sqrt{\rho/f}\). But the earth resistance term depends on frequency alone. Problem 5.MCQ 5. A delta winding in the zero-sequence network is drawn as:
(a) an open circuit to the line (b) a short to the line (c) a connection to the reference bus (d) both (a) and (c)Show answer
(d). It blocks the line and provides a circulating path — drawn as a connection from the transformer's internal node to the reference. Problem 8.MCQ 6. Which connection passes zero-sequence current from one side to the other?
(a) \(Y_g\)–\(\Delta\) (b) \(Y_g\)–\(Y_g\) (c) \(\Delta\)–\(\Delta\) (d) \(Y\)–\(Y\)Show answer
(b), and only that one. Problem 8.MCQ 7. An earth fault exceeds a three-phase fault when:
(a) \(Z_0 > Z_1\) (b) \(Z_0 < Z_1\) (c) \(Z_0 = Z_1\) (d) neverShow answer
(b). The ratio is \(3Z_1/(2Z_1+Z_0)\), which exceeds 1 exactly when \(Z_0 < Z_1\). Problem 14.MCQ 8. Zero-sequence mutual coupling between two circuits on one tower is about:
(a) 1% of the self (b) 10% (c) 60% (d) 200%Show answer
(c) — against 1–5% in the positive sequence. It moves distance-relay reach by up to 30%. Problem 7.MCQ 9. Which network contains sources?
(a) all three (b) positive only (c) positive and negative (d) zero onlyShow answer
(b). Machine EMFs are a balanced \(abc\) set. Problem 19.MCQ 10. A system is "effectively earthed" if:
(a) \(X_0/X_1 \le 3\) and \(R_0/X_1 \le 1\) (b) \(Z_n = 0\) (c) every neutral is earthed (d) \(Z_0 = Z_1\)Show answer
(a), which bounds the healthy-phase rise at 1.4 and permits 80%-rated arresters. Problem 9.MCQ 11. An isolated-neutral system on an earth fault has healthy phases at:
(a) nominal (b) \(1.4\times\) nominal (c) \(\sqrt3\times\) nominal (d) zeroShow answer
(c), sustained — which is why the insulation must be rated line-to-line. Problem 9.MCQ 12. Adding an earthing transformer at one bus:
(a) affects that bus most and others little (b) affects all buses equally (c) affects only remote buses (d) has no effectShow answer
(a) — 73% at the bus itself, 2–13% elsewhere. That locality is what makes it a usable design lever. Problem 17.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Sequence impedances | \(Z_0 = Z_s+2Z_m\), \(Z_1 = Z_2 = Z_s-Z_m\) | Balanced element |
| Static element | \(Z_1 = Z_2\) exactly | Lines, cables, transformers |
| Machine | \(Z_1 = jx''_d\), \(Z_2 = j(x''_d+x''_q)/2\) | \(Z_0 \approx 0.15\)–0.6 of \(x''_d\) |
| Line inductances | \(L_s = 0.2\ln(D_e/\text{GMR})\), \(L_m = 0.2\ln(D_e/\text{GMD})\) | mH/km |
| Carson depth | \(D_e = 658.4\sqrt{\rho/f}\) m | 294–2944 m over normal soils |
| Earth resistance | \(r_e = 9.869\times10^{-4}f\) Ω/km | Independent of soil |
| Zero-sequence resistance | \(r_0 = r_c + 3r_e\) | About twice \(r_1\) |
| Zero-sequence mutual | \(L_{0M} = 0.6\ln(D_e/D_{AB})\) | 50–70% of the self |
| Neutral earthing | appears as \(3Z_n\) | Zero-sequence network only |
| Effectively earthed | \(X_0/X_1 \le 3\), \(R_0/X_1 \le 1\) | Healthy-phase rise \(\le 1.4\) |
| Fault ratio | \(I_{LG}/I_{3\phi} = 3Z_1/(2Z_1+Z_0)\) | Maximum 1.5 |
| Earth wire effect | \(Z_0 = Z_0' - Z_{0w}^{2}/Z_{ww}\) | Lowers \(x_0\), raises \(r_0\) |
| Three-winding star | \(Z_H = \tfrac{1}{2}(Z_{HM}+Z_{HL}-Z_{ML})\) | \(Z_L\) is often negative |
| Residual compensation | \(k_0 = (Z_0-Z_1)/3Z_1\) | Distance relay setting |
Common Mistakes
Drawing the zero-sequence network with the positive-sequence topology. The connectivity comes from winding connections, not from the single-line diagram — Problems 8 and 12.
Including a machine in the zero-sequence network behind a delta. The delta blocks it entirely; the machine is absent — Problem 12.
Writing \(Z_n\) rather than \(3Z_n\). Still the commonest error in Part 5 — Problem 9.
Applying \(Z_0 = 3Z_1\) at a bus. The rule is for lines; at a network bus the ratio ranged from 0.51 to 1.55 — Problem 13.
Applying the overhead-line rule to a cable. A bonded cable's ratio is 1–2, not 3 — Problem 6.
Neglecting zero-sequence mutual coupling on a double-circuit line. It is 60%, not the 1–5% of the positive sequence — Problem 7.
Assuming a three-phase fault is always the worst. At a solidly-earthed transformer terminal the earth fault can exceed it by 20% — Problem 14.
Putting sources in the negative- or zero-sequence networks. Machine EMFs are pure positive sequence — Problem 19.
Forgetting that a transformer's \(Z_0\) equals its \(Z_1\). The question is whether a path exists, not what the impedance is — Problem 2.
Using the same zero-sequence network for every switching state. Losing one earthed transformer can make \(Z_0\) infinite — Problem 16.
Sizing an earthing transformer as small as possible. A very low \(Z_0\) can push the earth-fault level above the switchgear rating — Problem 17.
Suspecting the impedances when a study disagrees with a measurement. Topology errors give factors; impedance errors give tens of per cent — Challenge C3.
Three networks now exist for the five-bus system, and they are strikingly unalike. The positive-sequence one is Part 4's, unchanged. The negative-sequence one is the same network with the sources removed. The zero-sequence one has three times the line reactance, two earth connections where the positive-sequence network has two generators, and no machines at all — and its diagonal ranges from half the positive-sequence value at the transformer buses to one and a half times it at bus 5.
Set 23 connects them. Each fault type imposes two boundary conditions on the six terminals of Problem 15, and each pair of conditions corresponds to one interconnection: series for a line-to-earth fault, parallel opposition for a line-to-line fault, and a three-way parallel for a double line-to-earth fault. The arithmetic that follows is one series or parallel combination, and the four standard faults are then computed at every bus of the system.