Sequence Impedances and Sequence Networks
Because the transformation of Chapter 22 diagonalises every balanced element, each machine, line and transformer in the system is described by three impedances instead of one — and the three single-phase networks built from them are the working tools with which every unsymmetrical fault is solved.
- Why a balanced element has exactly three impedances, and what it means to say that \(Z_0\), \(Z_1\) and \(Z_2\) are properties of the element rather than of the disturbance.
- Where the famous \(3Z_n\) comes from — and why a neutral impedance of \(Z_n\) ohms must be entered into the zero-sequence network as three times its value.
- Why a delta winding is open to zero-sequence line current but provides a closed path for zero-sequence circulation, and how those two facts together set the topology of every transformer's zero-sequence network.
- Why the synchronous machine alone has \(Z_1 \neq Z_2 \neq Z_0\), and why \(X_0\) is the smallest of the three.
- Why a transmission line has \(Z_1 = Z_2\) exactly, while \(Z_0\) is two to five times larger because the return path is the earth.
- How to draw the positive-, negative- and zero-sequence networks of a complete system and reduce each to a Thévenin equivalent at the bus of interest.
- How the \(Z_{\text{bus}}\) algorithm of Chapter 17 is run three times to give \(Z_{kk}^{(0)}\), \(Z_{kk}^{(1)}\) and \(Z_{kk}^{(2)}\) at every bus at once.
Three Impedances for Every Element
Chapter 22 ended with a matrix identity that looked like a piece of algebra and is in fact the design brief for this chapter. For any element whose three phases are alike — equal self impedances \(Z_s\) and equal mutual impedances \(Z_m\) between every pair — the phase impedance matrix transforms into a diagonal one:
Read the three diagonal entries one at a time. The first says: if the currents in the three phases happen to form a zero-sequence set, the voltage drops they produce also form a zero-sequence set, and the ratio of one to the other is the single number \(Z_0\). The second and third say the same for the positive- and negative-sequence sets. Nothing else appears — no positive-sequence current produces a zero-sequence drop, no zero-sequence current produces a negative-sequence drop. The three sequences pass through the element without ever meeting.
That gives us a definition that does not mention matrices at all.
Each is a property of the element, measurable in a test bay, quite independent of what disturbance the system may later suffer. A generator has a zero-sequence reactance whether or not an earth fault ever occurs on it.
The second definition follows immediately. An impedance diagram of the kind drawn in Chapter 4 — every element replaced by its impedance, everything reduced to per unit on a common base — can be drawn three times over, once with each sequence impedance.
Each is an ordinary single-phase circuit, solved by ordinary single-phase methods. This is the entire dividend of Chapter 22: one intractable three-phase problem with coupling between phases has been traded for three uncoupled single-phase problems.
Three properties of these networks are settled before any element is examined in detail, and they are worth fixing now because they never change.
Only the positive-sequence network contains sources. A synchronous generator is built to produce a balanced set of internal emfs of normal phase order; that is what a rotating field winding in a symmetrical stator does. A balanced positive-sequence set has, by construction, no negative- and no zero-sequence component. So the generated emf \(E_a\) appears only in the positive-sequence network, and the negative- and zero-sequence networks are passive — they are driven only by what the fault injects into them.
Each network has its own reference bus. For the positive- and negative-sequence networks the reference is the neutral of the generators, since it is the point of zero positive- and negative-sequence potential. For the zero-sequence network the reference is ground, because zero-sequence current is precisely the component that returns through earth. This distinction is not pedantry; it is the reason the neutral impedance appears in one network and not the others.
The three networks are drawn separately and connected only at the fault. Before a fault, a balanced system carries positive-sequence quantities only; the other two networks are dead. The fault imposes constraints at one point which, when transformed, tie the three networks together at that point and nowhere else. Chapter 24 does exactly that. Everything in the present chapter is preparation for it.
The Star-Connected Load and the 3Zn Term
Take a balanced star-connected load of \(Z_Y\) ohms per phase whose neutral point \(n\) is not solidly earthed but is connected to ground through an impedance \(Z_n\). This single arrangement produces the most-quoted result in the whole of fault analysis, so it is worth deriving slowly.
The current returning through the neutral connection is the sum of the three line currents, and Chapter 22 identified that sum:
Each balanced set sums to zero — that is what balanced means. Only the zero-sequence set, whose three members are identical, survives, and it sums to \(3I_{a0}\). Neither positive- nor negative-sequence current ever leaves the star point.
The neutral is therefore not at ground potential. It is pushed above ground by
Now write the line-to-ground voltage of each phase as the line-to-neutral voltage plus this displacement, and collect the three into a matrix statement:
Pre-multiply by \(\mathbf{A}^{-1}\). The first term is easy: \(Z_Y\) is a scalar times the identity, so it commutes with everything and simply reappears. The second term needs one small evaluation, and it is the pivot of the whole argument:
A voltage that is common to all three phases is, by definition, pure zero sequence — the identity \(1+a+a^2=0\) removes it entirely from the other two rows. Hence
The neutral impedance appears in the zero-sequence network alone, and it appears multiplied by three. Solid earthing (\(Z_n=0\)) gives \(Z_0=Z_Y\); an isolated neutral (\(Z_n\to\infty\)) gives \(Z_0\to\infty\), which is the circuit statement of the physical fact that with no return path no zero-sequence current can flow at all.
Students meet the factor of three, memorise it, and forget where it came from. It is not a fudge. The zero-sequence network is a single-phase circuit that carries \(I_{a0}\), one third of the current that actually flows in the physical neutral conductor. For the model to produce the correct voltage drop, the impedance it contains must be scaled up in the same proportion:
Delta Connections and the Trapped Zero Sequence
A delta-connected load has no neutral and no earth connection, so one expects something special of its zero-sequence behaviour. Two separate facts are involved, and confusing them is the commonest error in this topic.
Fact one: no zero-sequence current can enter or leave a delta through the lines. Each line current is the difference of two phase currents,
The cancellation is complete and unconditional: it does not depend on the load being balanced, only on the delta having no fourth terminal. From the line terminals, a delta is an open circuit to zero sequence.
Fact two: zero-sequence current can nevertheless circulate inside the delta. A zero-sequence set of phase currents \(I_{ab}=I_{bc}=I_{ca}=I_{\Delta0}\) chases itself round the closed loop without disturbing a single line current, because the three subtract out. Whether such a circulation exists depends on whether anything drives it. For a passive load nothing does. For a transformer delta winding something does — the flux produced by zero-sequence current in a magnetically coupled winding — and that is why a delta winding is the one thing that makes zero-sequence current possible on the other side of a transformer.
The positive- and negative-sequence behaviour of a delta load holds no surprises. Kirchhoff's voltage law round the loop gives \(V_{ab}+V_{bc}+V_{ca}=0\) always, so \(Z_\Delta(I_{ab}+I_{bc}+I_{ca})=0\) for the passive load, and for the two balanced sequences the delta is converted to its star equivalent in the usual way.
In the zero-sequence network the three line terminals of a delta load are simply left disconnected. The zero-sequence "impedance" is infinite in the sense of an open circuit — not because the copper is resistive, but because the topology offers no return.
Sequence Impedances of the Synchronous Machine
The generator is the only element in the system for which all three sequence impedances are genuinely different, and the reasons are worth understanding, because they explain the numbers a manufacturer supplies.
Begin with the static picture. Let each stator phase have resistance \(R\) and self inductance \(L_s\), with mutual inductance \(M_s\) between any two phases; let the machine's internal emfs be the balanced positive-sequence set \(E_{an},E_{bn},E_{cn}\). For phase \(a\),
Written for all three phases at once this is \(\mathbf{V}_{abc}=\mathbf{E}_{abc}-\mathbf{Z}_{abc}\mathbf{I}_{abc}\) with the familiar symmetric structure, self term \(Z_s=R+j\omega L_s\) and mutual term \(Z_m=-j\omega M_s\). Pre-multiplying by \(\mathbf{A}^{-1}\) and using the diagonalisation of Section 23-1 — together with the fact that a balanced positive-sequence emf set transforms into \([\,0,\;E_{an},\;0\,]^{\mathsf T}\) — gives the three machine equations directly.
Three conclusions can be drawn straight from these lines. The emf appears in the positive-sequence equation and nowhere else, as promised in Section 23-1. The zero-sequence impedance is smaller than the other two, because the mutual term enters with the opposite sign — physically, three identical in-phase currents in three windings spaced 120° apart in space produce fundamental magnetomotive forces that cancel, so zero-sequence current sets up no rotating field at all and meets only leakage flux. And, if the machine were a static three-coil object, \(Z_1\) would equal \(Z_2\).
It does not, and the reason is the rotor. A positive-sequence stator current produces a field rotating forwards at synchronous speed; the rotor runs with it, the relative slip is zero, and the reactance the stator sees is governed by the direct-axis behaviour of the field and damper circuits — which is exactly the \(X_d''\), \(X_d'\), \(X_d\) sequence developed in Chapter 21. A negative-sequence stator current produces a field rotating backwards at synchronous speed; relative to the rotor it sweeps past at twice synchronous speed, alternately aligning with the direct and quadrature axes. The stator therefore sees an average of the two subtransient reactances.
Add \(3Z_n\) to \(X_0\) if the machine neutral is earthed through an impedance. For a turbo-alternator with a solid cylindrical rotor \(X_d''\approx X_q''\) and \(X_2\approx X_d''\); for a salient-pole machine the two differ and \(X_2\) is a genuine average. \(X_0\) is always the smallest of the three.
The choice of \(X_1\) deserves a sentence, because Chapter 21 offered three candidates. Which one is used depends on the instant being studied: \(X_d''\) for the first cycle or two, which is what circuit-breaker interrupting duty and relay settings need; \(X_d'\) for a few tenths of a second; \(X_d\) for the sustained condition. Fault studies almost always want the first, so \(X_1=X_d''\) unless the problem says otherwise. \(X_2\) and \(X_0\) have no such time dependence — they do not decay, because the fluxes that produce them do not link the field winding in a way that lets the field current respond.
Measuring these numbers is straightforward in principle. \(X_0\) is obtained by connecting the three phases in parallel and applying a single-phase voltage \(V\) between that common point and the neutral; every phase then carries the same current in the same direction, which is zero sequence by definition. If the total current drawn is \(I\), then each phase carries \(I/3\) and
\(X_2\) is obtained by driving the machine at rated speed with the field short-circuited and applying a reduced balanced voltage of reversed phase sequence, then taking the ratio of voltage to current. Both tests are, in effect, the definition of Section 23-1 carried out with instruments.
Sequence Impedances of the Transmission Line
A transmission line is static: copper, aluminium and air, with nothing that rotates. It cannot distinguish a forward-rotating set of currents from a backward-rotating one, and so
This is not an approximation. Reversing the phase order of a set of currents in a passive symmetric network reverses the phase order of the voltage drops and leaves every impedance unchanged.
The value of \(Z_1\) is the ordinary series impedance computed in Chapters 6 and 7: \(R+j\omega L\) per phase, with \(L\) from the GMD/GMR formula for a transposed line. What remains is \(Z_0\), and the whole difficulty is that zero-sequence current does not return through the other two phase conductors. It returns through the earth and through any overhead ground wires.
Set up the calculation with the neutral (earth) path explicit. Let \(Z_{aa}\) be the self impedance of a phase conductor, \(Z_{ab}\) the mutual between two phase conductors, \(Z_{nn}\) the self impedance of the earth-return path, and \(Z_{an}\) the mutual between a phase conductor and that path. The voltage drop along phase \(a\), measured between the two ends of the line and referred to the return conductor, is
The earth-return terms have been absorbed into effective self and mutual impedances, and the resulting \(3\times3\) matrix has the balanced form Section 23-1 requires. Applying the diagonalisation and substituting back,
The earth-return terms cancel out of \(Z_1\) and \(Z_2\) — which is why a positive-sequence calculation never needs to mention the earth — and survive, tripled, in \(Z_0\).
Two consequences follow. First, \(Z_0\) is always substantially larger than \(Z_1\): the earth-return loop encloses far more area than the loop formed by two phase conductors a few metres apart, so its inductance is much greater. Typical ratios are \(Z_0/Z_1\) between 2 and 3.5 for a single-circuit line without earth wires, falling towards 2 when continuous earth wires are strung, and rising to 3 – 5.5 for a double-circuit line on shared towers. Second, \(Z_0\) has a much larger resistive component, because the return current spreads through soil of finite resistivity; the classical treatment is Carson's, which replaces the diffuse earth return by an equivalent conductor at a depth \(D_e = 658.4\sqrt{\rho/f}\) metres for soil resistivity \(\rho\) in ohm-metres.
Two further points matter in practice. A double-circuit line on common towers has a zero-sequence mutual coupling \(Z_{0m}\) between the two circuits that cannot be ignored: both circuits carry zero-sequence current in the same direction, so the coupling adds. If both circuits are identical, each with zero-sequence impedance \(Z_0\), and both are in service between the same two buses, each carries half the total and
which is larger than the \(Z_0/2\) that two uncoupled parallel circuits would give. There is no corresponding effect in the positive-sequence network, because the positive-sequence currents in the two circuits produce fields that very nearly cancel at the distance of the neighbouring circuit. Finally, the shunt capacitance of the line also splits into sequences: \(C_1=C_2\) as always, while \(C_0\) is markedly smaller than \(C_1\) because zero-sequence charging current has no phase-to-phase component and depends only on capacitance to earth. In fault studies both are usually neglected, but \(C_0\) is the whole story in an isolated-neutral system, where earth-fault current is capacitive and is what a Petersen coil is tuned to cancel.
Transformers: the Connection Decides the Zero-Sequence Network
The positive- and negative-sequence networks of a transformer are dull, and that is welcome news. With magnetising current neglected — as Chapter 4 established it may be — the transformer is a series leakage impedance \(Z\), the same for both sequences because the transformer is static. The one refinement is the phase shift of a star–delta bank: on the ANSI convention the high-voltage positive-sequence quantities lead the low-voltage ones by \(30^\circ\), and the negative-sequence quantities lag by \(30^\circ\). In a per-unit study of a single fault the shift often cancels out of the magnitudes and is ignored, but it is not optional when currents on one side of the bank must be related to currents on the other — as Example 6 of Chapter 24 shows.
The zero-sequence network is where the whole interest lies, and it is decided not by the impedance but by the connection. Three physical requirements govern it, and everything else is bookkeeping.
Requirement one — a path into the winding. Zero-sequence current is identical in all three phases and must return somewhere. A star winding admits it only if its neutral is earthed; an unearthed star or a delta presents no fourth terminal to the line, so no zero-sequence current can enter from outside.
Requirement two — ampere-turn balance. Current in one winding must be matched by current in the other, or the core saturates and the transformer refuses to behave as a transformer. So zero-sequence current in an earthed-star winding is possible only if the other winding can carry the balancing zero-sequence current — either out through its own line terminals (a second earthed star) or round and round inside itself (a delta).
Requirement three — the delta circulates but does not conduct. As Section 23-3 established, a delta gives zero-sequence current a closed path internally while presenting an open circuit to the lines. In the network model this is drawn as a connection from the transformer's series impedance to the reference bus on the delta side, with the delta-side bus itself left disconnected.
Those three rules are captured by one universal circuit with four switches. Draw the leakage impedance \(Z_0\) in the middle, with a lead to each side; on each side, one switch connects the lead to that side's bus and another connects it to the reference bus. Close a bus-side switch if that winding is an earthed star; close a reference-side switch if that winding is a delta; leave both open if the winding is an unearthed star.
Working the switches through the five connections that occur in practice produces the table every fault study needs.
| Connection | Switches closed | Resulting zero-sequence network | Comment |
|---|---|---|---|
| Yg – Yg | S₁, S₂ | \(Z_0\) in series between bus P and bus Q | Zero sequence passes straight through; earth-fault current on one side is fed from the other |
| Yg – Δ | S₁, S₄ | \(Z_0\) from bus P to the reference bus; bus Q open | The commonest arrangement. The delta is an earth-fault current source for the star side and a barrier for everything beyond it |
| Y – Δ | S₄ only | Open circuit at both buses | No path into the unearthed star, so nothing flows anywhere |
| Yg – Y | S₁ only | Open circuit | Current could enter the earthed star but nothing can balance its ampere-turns; one earthed neutral is not enough |
| Δ – Δ | S₃, S₄ | Open circuit at both buses | Zero sequence circulates in each delta independently and never reaches a line |
The leakage impedance itself is augmented by any neutral earthing impedance, exactly as in Section 23-2 and for the same reason. Working in per unit on a common base, where the turns ratio has already been absorbed (Chapter 4),
Every row of the table above is this sentence applied to a particular pair of connections. Memorising the table is unnecessary once the sentence is understood.
Assembling the Three Networks of a System
With every element characterised, drawing the three networks of a complete system is a matter of following a fixed procedure. The single-line diagram of Chapter 3 is the starting point; per-unit conversion on a common base, from Chapter 4, is the essential preliminary, because the three networks must eventually be connected together and cannot be if they are in different units.
| Step | Positive-sequence network | Negative-sequence network | Zero-sequence network |
|---|---|---|---|
| Sources | All generator emfs, in series with \(jX_d''\) | None — passive throughout | None — passive throughout |
| Generators | \(jX_d''\) (or \(jX_d'\), \(jX_d\)) | \(jX_2\) | \(jX_0+3Z_n\), to the reference bus |
| Transformers | Leakage \(Z\), \(+30^\circ\) shift if Δ–Y | Leakage \(Z\), \(-30^\circ\) shift if Δ–Y | Per the switch table of Section 23-6 |
| Lines | \(Z_1\) | \(Z_2=Z_1\) | \(Z_0\approx 2\text{ to }3.5\,Z_1\) |
| Star loads / earthed neutrals | \(Z_Y\) | \(Z_Y\) | \(Z_Y+3Z_n\) |
| Delta loads and windings | \(Z_\Delta/3\) | \(Z_\Delta/3\) | Open at the line terminals |
| Reference bus | Generator neutral | Generator neutral | Ground |
A worked skeleton makes the pattern concrete. Consider a generator feeding a transmission line through a delta–star transformer, with the fault to be studied at the far end of the line. Take a 25 MVA base throughout, 11 kV on the generator side and 66 kV on the line side.
The zero-sequence network is the interesting one. Bus 2 sits on the earthed-star side of T1, so zero-sequence current enters the transformer there, passes through its leakage impedance, and finds its balancing ampere-turns in the delta winding, which returns it to the reference bus. The generator's own zero-sequence branch — \(j0.06\) in series with \(3\times j0.05=j0.15\), a total of \(j0.21\) — sits on the far side of that delta and takes no part whatever in an earth fault at bus 2. A student who includes it has misunderstood the delta; a student who omits the generator from the positive-sequence network has misunderstood the source.
Thévenin Equivalents, the Sequence Z-Bus and Typical Values
Chapter 21 solved a three-phase fault by reducing the positive-sequence network to a Thévenin equivalent at the faulted bus: an emf \(V_f\) equal to the pre-fault voltage there, in series with the driving-point impedance \(Z_{kk}\). Nothing about that argument used the symmetry of the fault; it used only that the network is linear. So it applies to each of the three sequence networks separately.
\(V_f\) is the pre-fault line-to-neutral voltage of phase \(a\) at bus \(k\), normally taken as \(1.0\angle0^\circ\) per unit. The negative- and zero-sequence equivalents have no source term because those networks are passive: their voltages exist only as a response to the currents the fault injects, and the minus signs record that the current flows into the network at the fault point.
These three one-line equations are the complete interface between this chapter and the next. Chapter 24 supplies the constraints that the fault type imposes, solves them together with these three, and every unsymmetrical fault falls out.
Obtaining the three driving-point impedances is exactly the problem Chapter 17 solved. Build the bus impedance matrix of each sequence network by the \(Z_{\text{bus}}\) building algorithm — three separate runs, one per sequence, using the same bus numbering — and the diagonal entries \(Z_{kk}^{(0)}\), \(Z_{kk}^{(1)}\), \(Z_{kk}^{(2)}\) are the Thévenin impedances at every bus at once. The off-diagonal entries are needed too, because Chapter 24 uses them to find the voltage at every other bus during the fault. For a system of a few elements, series–parallel reduction by hand is quicker and is what the examples below do; for a real system the three \(Z_{\text{bus}}\) matrices are what a fault program stores.
A note of caution about the zero-sequence \(Z_{\text{bus}}\): because delta windings break the zero-sequence network into islands, that network is often not connected. A bus with no zero-sequence path to the reference has infinite \(Z_{kk}^{(0)}\), and a program must handle this rather than invert a singular matrix. Physically it simply means no earth-fault current can flow at that bus, which is precisely what an isolated-neutral section of system does.
Finally, orders of magnitude. Fault calculations are checked far more often by whether the numbers look plausible than by repeating the algebra, so the following ranges are worth carrying in the head.
| Element | \(X_1\) (pu on own rating) | \(X_2\) | \(X_0\) |
|---|---|---|---|
| Turbo-alternator (2-pole) | \(X_d''\approx 0.09-0.15\) | \(\approx X_d''\) | \(0.02-0.08\), plus \(3Z_n\) |
| Salient-pole hydro generator | \(X_d''\approx 0.15-0.25\) | \(\approx 1.0-1.3\,X_d''\) | \(0.03-0.10\), plus \(3Z_n\) |
| Synchronous motor / condenser | \(0.15-0.25\) | \(\approx X_1\) | \(0.04-0.10\) |
| Induction motor (as a source for a few cycles) | \(\approx 0.17-0.25\) | \(\approx X_1\) | usually no path (delta or unearthed star) |
| Two-winding transformer | leakage \(0.05-0.15\) | \(=X_1\) | \(\approx X_1\), topology per Section 23-6 |
| Overhead line, single circuit | \(X_1\) | \(=X_1\) | \(2.0-3.5\,X_1\) |
| Overhead line, double circuit | \(X_1\) | \(=X_1\) | \(3.0-5.5\,X_1\) including mutual |
| Underground cable | \(X_1\) | \(=X_1\) | \(1.0-3.0\,X_1\), sheath-dependent |
Worked Examples
Problem. A balanced star-connected load of \(Z_Y = 6+j8\ \Omega\) per phase has its neutral earthed through \(Z_n=j2\ \Omega\). Find its three sequence impedances, and the neutral displacement voltage when the zero-sequence line current is \(I_{a0}=1.5\angle 0^\circ\) A. The same three impedances are then reconnected in delta; find the sequence impedances seen from the lines.
Solution. The star case follows Section 23-2 directly. The neutral impedance touches only the zero-sequence path:
The zero-sequence impedance is half as large again as the other two, purely because of a small reactor in the neutral. The neutral itself carries \(I_n = 3I_{a0}=4.5\angle0^\circ\) A, so
Reconnected in delta, the load presents \(Z_\Delta = 6+j8\ \Omega\) between each pair of lines. Its star equivalent is one third of that:
Two things changed. The positive- and negative-sequence impedances fell by a factor of three, so the delta draws three times the balanced current at the same line voltage — the familiar result of Chapter 3. And the zero-sequence path vanished entirely: with no neutral there is nothing for zero-sequence current to return through, however the impedances are valued.
Problem. A fully transposed 100 km line has, with the earth return already folded in, an effective self impedance \(Z_s = 0.15+j0.72\ \Omega/\text{km}\) and an effective mutual impedance \(Z_m = 0.09+j0.35\ \Omega/\text{km}\) between phases. Find \(Z_1\), \(Z_2\) and \(Z_0\) for the whole line and the ratio \(|Z_0|/|Z_1|\).
Solution. Apply the two formulas of Section 23-5 per kilometre and then multiply by the length.
The ratio sits at the upper end of the range quoted in Section 23-8, which is what one expects when the earth return is doing all the work and no continuous earth wire is present to share it. Notice also what happened to the resistance: \(R_0=0.33\) against \(R_1=0.06\ \Omega/\text{km}\), a factor of five and a half, all of it earth-path loss. Earth-fault currents are therefore much more strongly damped than phase-fault currents, which matters when Chapter 25 estimates the d.c. offset.
Problem. A 25 MVA, 11 kV generator has \(X_d''=0.20\), \(X_2=0.25\) and \(X_0=0.06\) per unit on its own rating, and its neutral is earthed through a \(0.5\ \Omega\) reactor. Express all three sequence impedances on a 100 MVA, 11 kV system base, and state the value of the branch that represents the machine in the zero-sequence network.
Solution. Chapter 4 gives the base-change rule: at unchanged voltage base, per-unit impedance scales with the MVA base.
The neutral reactor is given in ohms, so it must be converted with the base impedance at 100 MVA, 11 kV:
A single half-ohm reactor has turned a zero-sequence reactance of \(j0.24\) into one of \(j1.48\) — six times larger, and now much the biggest of the three. That is exactly what a neutral earthing reactor is fitted to do, and Chapter 24 will put a number on the earth-fault current it suppresses.
Problem. A 50 MVA, 132/33 kV transformer has a leakage reactance of \(0.10\) per unit on its rating. Describe its three sequence networks (a) when connected \(\Delta\) on 132 kV and \(Y\) solidly earthed on 33 kV, (b) when both windings are star with both neutrals solidly earthed, and (c) as in (a) but with the 33 kV neutral earthed through a \(1\ \Omega\) resistor.
Solution. In all three cases the positive- and negative-sequence networks are identical: a series branch of \(j0.10\) per unit between the 132 kV and 33 kV buses. In case (a) and (c) the positive-sequence quantities on the 132 kV side additionally lead those on the 33 kV side by \(30^\circ\), and the negative-sequence quantities lag by \(30^\circ\). Only the zero-sequence network differs.
(a) Δ – Yg. Switches S₁ (earthed star, on the 33 kV side) and S₄ (delta, on the 132 kV side) close. The zero-sequence network is a branch of \(j0.10\) from the 33 kV bus down to the reference bus; the 132 kV bus is left open. An earth fault on the 33 kV system draws zero-sequence current through the transformer, but nothing of it reaches the 132 kV system.
(b) Yg – Yg. Switches S₁ and S₂ close, S₃ and S₄ stay open. The zero-sequence network is a series branch of \(j0.10\) joining the two buses, exactly like the positive-sequence branch. Zero-sequence current now passes from one system to the other, and an earth fault on the 33 kV side is fed partly from the 132 kV side.
(c) Δ – Yg through 1 Ω. The topology is that of (a), but \(3Z_n\) is added in series. Converting the resistor with the base impedance on the 33 kV side,
The one-ohm resistor is worth \(0.138\) per unit in the zero-sequence branch — more than the transformer's own reactance — and it makes \(Z_0\) predominantly resistive. Resistance earthing does two jobs at once: it limits the earth-fault current, and by damping the fault it prevents the transient overvoltages that arcing earth faults can produce on a lightly earthed system.
Problem. For the sample system of Section 23-7 — generator \(X_1=j0.20\), \(X_2=j0.25\), \(X_0=j0.06\) with \(X_n=j0.05\); transformer T1 \(\Delta\)–\(Y_g\), \(j0.08\); line \(X_1=X_2=j20\ \Omega\), \(X_0=j60\ \Omega\); 25 MVA base, 66 kV on the line — find \(Z_{22}^{(1)}\), \(Z_{22}^{(2)}\) and \(Z_{22}^{(0)}\) at bus 2, the far end of the line.
Solution. Convert the line first, since it alone is given in ohms:
The positive- and negative-sequence networks are simple series chains from the reference bus, through the machine, transformer and line, to bus 2:
The zero-sequence network stops at the delta. Looking into bus 2, the current passes along the line and through the transformer, and the delta winding returns it to the reference bus. The generator branch \(j(0.06+3\times0.05)=j0.21\) is on the far side of that delta and is not in the circuit at all:
Three numbers — \(j0.3948\), \(j0.4448\), \(j0.4244\) — now describe everything the system can do at bus 2, whatever kind of fault occurs there. Chapter 24 uses precisely these three in its Example 4. Notice that here \(Z_0 > Z_1\), which by the sanity test of Section 23-8 means the three-phase fault at bus 2 will be more severe than the earth fault.
Problem. Two identical circuits run between the same pair of buses on shared towers. Each has \(Z_1=j0.12\) and \(Z_0=j0.30\) per unit, and the zero-sequence mutual coupling between them is \(Z_{0m}=j0.15\) per unit. Find the effective positive- and zero-sequence impedances of the pair with both circuits in service, and again with one circuit switched out. Comment on what the coupling does to an earth-fault calculation.
Solution. The positive-sequence case is the ordinary one: the mutual coupling between circuits is negligible for positive sequence, because the three phase currents of one circuit produce fields that very nearly cancel at the other circuit's distance. Two equal impedances in parallel give
For zero sequence the currents in the two circuits are in phase and equal, so the coupling reinforces rather than cancels. Write the drop along circuit A:
Without the coupling the answer would have been \(j0.15\); the mutual term has raised the effective zero-sequence impedance by 50 %. With one circuit out of service and earthed at both ends the surviving circuit sees its own \(Z_0=j0.30\), modified by whatever current the coupling drives round the earthed loop of the dead circuit; with the dead circuit isolated it is simply \(j0.30\).
The practical consequence is that an earth-fault current computed by treating the two circuits as uncoupled parallel branches would be too large by roughly a third. Worse, the coupling induces a zero-sequence voltage in the healthy circuit while the faulted one carries earth-fault current, which is why distance relays on double-circuit lines need a mutual-compensation input — a matter Chapter 36 takes up.
Chapter Summary
\(Z_0=Z_s+2Z_m\), \(Z_1=Z_2=Z_s-Z_m\) for any balanced element; each is a measurable property.
Only the positive-sequence network contains emfs. The other two are passive and driven by the fault.
\(Z_0=Z_Y+3Z_n\), because the real neutral carries \(3I_{a0}\) while the model carries \(I_{a0}\).
Open to zero-sequence line current, but a closed path for circulation — the trap that bounds every zero-sequence network.
\(X_1=X_d''\), \(X_2\approx(X_d''+X_q'')/2\), \(X_0\) smallest of all because the zero-sequence mmf cancels.
\(Z_1=Z_2\) exactly. Only something that rotates can tell forward from backward phase order.
Two to five times \(Z_1\), with a much larger resistance, because the return path is the earth.
\(V_{a1}=V_f-Z_{kk}^{(1)}I_{a1}\), \(V_{a2}=-Z_{kk}^{(2)}I_{a2}\), \(V_{a0}=-Z_{kk}^{(0)}I_{a0}\) — the input to Chapter 24.
Practice Problems
Work in per unit on a stated common base wherever more than one voltage level appears. For every zero-sequence question, draw the network before computing anything — the topology, not the arithmetic, is where the marks and the mistakes are.
- A balanced star-connected load of \(10+j6\ \Omega\) per phase has its neutral earthed through \(2+j1\ \Omega\). Find \(Z_0\), \(Z_1\) and \(Z_2\), and the neutral displacement voltage when \(I_{a0}=2\angle{-30^\circ}\) A.
- The same three impedances of Problem 1 are reconnected in delta and supplied from the same three lines. State the sequence impedances now seen from the lines and explain, without algebra, why \(Z_0\) changed as it did.
- A transposed 150 km line has effective self impedance \(0.12+j0.80\ \Omega/\text{km}\) and mutual impedance \(0.06+j0.30\ \Omega/\text{km}\). Find \(Z_1\), \(Z_2\) and \(Z_0\) for the whole line and the ratio \(|Z_0|/|Z_1|\). On a 100 MVA, 220 kV base, express all three in per unit.
- A 50 MVA, 13.8 kV generator has \(X_d''=0.18\), \(X_2=0.20\) and \(X_0=0.07\) per unit on its own rating, with its neutral earthed through a \(0.4\ \Omega\) reactor. Express the three sequence impedances on a 100 MVA, 13.8 kV base and give the value of the machine's branch in the zero-sequence network.
- Sketch the zero-sequence network of a three-winding transformer connected \(Y_g\) on the primary, \(\Delta\) on the tertiary and \(Y\) (unearthed) on the secondary, given the three star-equivalent leakage impedances \(Z_P\), \(Z_S\) and \(Z_T\). Which buses are open, and what impedance does an earth fault on the primary bus see?
- A generator (\(X_1=j0.15\), \(X_2=j0.18\), \(X_0=j0.05\), solidly earthed) feeds bus 1. Transformer T1 (\(Y_g\)–\(Y_g\), \(j0.09\)) connects bus 1 to bus 2, and a line (\(X_1=X_2=j0.10\), \(X_0=j0.28\)) connects bus 2 to bus 3. Draw all three sequence networks and find the three Thévenin impedances at bus 3. Then repeat with T1 reconnected \(\Delta\)–\(Y_g\), the delta on the generator side, and state what changed and why.
- Two identical circuits with \(Z_0=j0.45\) each and zero-sequence mutual coupling \(Z_{0m}=j0.20\) run between the same two buses. Find the effective zero-sequence impedance with both in service, and the percentage error made by ignoring the coupling.
- Explain why a \(Y_g\)–\(Y\) transformer, in which one neutral is solidly earthed and the other is not, presents an open circuit in the zero-sequence network even though an earthed neutral is present. What single change to the second winding would turn it into a zero-sequence source for the first?