Symmetrical Components

Every calculation in this book so far has rested on a single structural fact. In a balanced three-phase circuit the three phases carry identical quantities displaced by \(120^\circ\), so knowing one phase is knowing all three. Chapter 3 turned that into the per-phase equivalent circuit, Chapter 4 turned the per-phase circuit into a per-unit impedance diagram, Part 3 computed line performance on one phase, Part 4 solved load flow one phase at a time, and Chapter 21 analysed the three-phase fault on one phase because that fault leaves the symmetry intact.

Now consider a single conductor falling to earth. Phase \(a\) is shorted to the reference; phases \(b\) and \(c\) are untouched. The three phase currents are no longer equal in magnitude, no longer \(120^\circ\) apart, and no longer derivable from one another. Whatever is true of phase \(a\) tells us nothing about phase \(b\). The per-phase circuit has no meaning, and with it goes the per-unit diagram, the \(\mathbf{Y}_{bus}\), the \(\mathbf{Z}_{bus}\) and every result of the last eighteen chapters.

The brute-force alternative is to write the network in full three-phase detail: every line becomes a \(3\times3\) impedance matrix with mutual terms, every transformer a coupled six-winding element, and an \(n\)-bus system a \(3n\)-dimensional problem with no symmetry to exploit. It is possible, it is what a modern electromagnetic transient program actually does, and it is entirely unsuited to hand calculation or to understanding.

In 1918 Charles LeGeyt Fortescue published the alternative. Instead of abandoning balance, decompose the unbalanced problem into balanced ones. The observation is a change of variables, nothing more; its power comes from the fact that the balanced network — which is still balanced, even though the fault at its terminals is not — treats those new variables independently.

The claim is this. Given any three phasors \(V_a\), \(V_b\), \(V_c\) — arbitrary magnitudes, arbitrary angles — there exist three sets of three phasors each, whose term-by-term sum reproduces the original set, and which have the following structures.

The decomposition is unique. That is easy to believe on a counting argument: three complex phasors carry six real numbers, and the three sets are described by \(V_{a1}\), \(V_{a2}\), \(V_{a0}\) — again six real numbers, since the other two members of each set follow from the first. A transformation from six numbers to six numbers can be one-to-one, and Section 22-4 shows that this one is by exhibiting the inverse explicitly.

Writing the phase quantities as the sum of their three components,

The subscripts \(b\) and \(c\) can be eliminated at once, because each set is balanced and therefore determined entirely by its \(a\)-phase member. That is the whole reason the transformation is useful, and it is where the operator \(a\) of Chapter 3 re-enters.

Chapter 3 defined \(a\) as the unit phasor that advances an angle by \(120^\circ\), and derived the identity on which everything here depends. Collecting the properties that will be used:

The first identity is Chapter 3's statement that three equal phasors \(120^\circ\) apart sum to zero; the two that follow it are read straight off the diagram, since \(a\) and \(a^2\) are mirror images about the real axis separated vertically by \(\sqrt3\).

Now express each balanced set through its \(a\)-phase member. The positive-sequence set has \(a\)–\(b\)–\(c\) order, so \(b\) lags \(a\) by \(120^\circ\): \(V_{b1}=a^2V_{a1}\) and \(V_{c1}=aV_{a1}\). The negative-sequence set has the reverse order, so \(b\) leads: \(V_{b2}=aV_{a2}\) and \(V_{c2}=a^2V_{a2}\). The zero-sequence set has all three identical: \(V_{b0}=V_{c0}=V_{a0}\). Substituting into the superposition of Section 22-2 gives the three phase quantities in terms of three unknowns.

Synthesis is not what is needed in practice. Given a measured or computed set of unbalanced phase quantities, the sequence components must be extracted from them, which means inverting \(\mathbf{A}\). There is no need to compute a determinant and a matrix of cofactors; the identity \(1+a+a^2=0\) does the work directly.

Add the three synthesis equations. The \(V_{a1}\) terms carry the factor \(1+a^2+a=0\) and the \(V_{a2}\) terms the factor \(1+a+a^2=0\), so both vanish and only \(3V_{a0}\) survives. Now multiply the second equation by \(a\) and the third by \(a^2\) before adding. Using \(a^3=1\) and \(a^4=a\):

Multiplying instead by \(a^2\) and \(a\) isolates \(V_{a2}\) by the identical mechanism. The three results together are the analysis equations.

That last observation is worth keeping, because it says \(\mathbf{A}\mathbf{A}^*=3\mathbf{I}\). A matrix whose columns are mutually orthogonal and each of squared length three is a scaled unitary matrix, and unitary transformations preserve inner products. Section 22-6 collects the dividend: complex power comes through the transformation almost untouched.

The zero-sequence line deserves emphasis on its own. It says that \(V_{a0}\) is one third of the sum of the three phase quantities. If the three sum to zero — which for a balanced set they do, by the very identity used above — there is no zero-sequence component at all. Zero sequence is precisely the measure of how far a three-phase set departs from summing to zero, which is why it is intimately tied to earth and neutral paths.

Apply the analysis equations to currents in a four-wire system. Kirchhoff's current law at the star point gives the neutral current as the sum of the three line currents, and the zero-sequence equation gives one third of that same sum. The two statements combine into the single most-used result in unbalanced analysis.

Three corollaries follow, and each of them decides the shape of a sequence network in Chapter 23.

If the star point is isolated, there is no fourth path, so \(I_n=0\) and therefore \(I_{a0}=0\). The line currents of an ungrounded star connection contain no zero-sequence component whatever the unbalance. If the star point is earthed through an impedance \(Z_n\), the current \(3I_{a0}\) flows through it and produces a voltage drop \(3I_{a0}Z_n\) between the star point and earth; only the zero-sequence network feels it, and it feels it as an impedance \(3Z_n\) — a factor of three that Section 22-7 derives rather than asserts.

If the load or winding is delta-connected, the line currents are differences of phase currents, \(I_a = I_{ab}-I_{ca}\) and so on, and those differences necessarily sum to zero. So again \(I_{a0}=0\) in the lines. Inside the delta, however, a zero-sequence current may perfectly well circulate: if \(I_{ab}=I_{bc}=I_{ca}=I_0^{\Delta}\), each line current is unaffected but a real current runs round the closed loop. A delta winding is therefore a trap for zero-sequence current — it absorbs it and hides it from the line — which is exactly why delta tertiary windings are fitted to large transformers.

Two further facts are used constantly and are worth stating plainly. A perfectly balanced system in \(a\)–\(b\)–\(c\) sequence contains positive-sequence quantities only; \(I_{a1}\) equals the phase current and \(I_{a2}=I_{a0}=0\). Consequently the appearance of any negative- or zero-sequence current is by itself evidence of an abnormal condition, which is the entire basis of negative-sequence and earth-fault relaying in Chapter 36. And the labelling is a convention with content: if the resultant set rotated in the reverse order, one would name the larger balanced set "positive", so in normal usage \(|I_{a2}|\) does not exceed \(|I_{a1}|\).

A change of variables that mangles power would be of limited use, since every engineering answer eventually has to be expressed in megawatts. Fortunately \(\mathbf{A}\) is a scaled unitary matrix, and the calculation is three lines long. Start from the three-phase complex power written as a matrix product:

Everything now hangs on the matrix \(\mathbf{A}^T\mathbf{A}^*\). Since \(\mathbf{A}\) is symmetric, \(\mathbf{A}^T=\mathbf{A}\), and since \(a^*=a^2\), \(\mathbf{A}^*\) is \(\mathbf{A}\) with \(a\) and \(a^2\) interchanged. The diagonal entries of the product are sums like \(1+a^2a+aa^2 = 1+1+1 = 3\); the off-diagonal entries are sums like \(1+a+a^2=0\).

The absence of cross terms is a genuine physical statement, not bookkeeping. A negative-sequence current in a machine cannot draw power from the positive-sequence voltage; it draws power from the negative-sequence voltage alone, and in a generator that power is a loss, delivered into the rotor surface as double-frequency heating. Section 22-8 returns to the point.

A word on the alternative convention. Some texts define the transformation with a factor \(1/\sqrt3\) built into \(\mathbf{A}\), which makes it exactly unitary and removes the \(3\). That convention makes power literally invariant but destroys the pleasant property that a balanced set has \(V_{a1}\) equal to the actual phase voltage. Power systems practice keeps the convention used here, and pays for it with the factor of three.

Everything so far has been a change of variables applied to phasors. Nothing yet explains why the change is useful, and the explanation is the subject of this section: applied to a balanced network, the transformation diagonalises the impedance matrix.

Take the most general three-phase element that is symmetric in its phases — a transposed line, a symmetrical machine, a static load with equal branches. Its three phases each have the same self impedance \(Z_s\), and each pair has the same mutual impedance \(Z_m\). In phase variables,

The three phases are thoroughly coupled: a current in phase \(a\) produces a voltage in phases \(b\) and \(c\). Substituting the transformation on both sides and premultiplying by \(\mathbf{A}^{-1}\) gives \(\mathbf{V}_{012}=\big(\mathbf{A}^{-1}\mathbf{Z}_{abc}\mathbf{A}\big)\mathbf{I}_{012}\), so the sequence impedance matrix is a similarity transform. To evaluate it, split \(\mathbf{Z}_{abc}\) into a diagonal part and an all-ones part \(\mathbf{J}\):

The middle step is where the identity does its work once more: the columns of \(\mathbf{A}\) sum to \(3\), \(1+a^2+a=0\) and \(1+a+a^2=0\) respectively, so \(\mathbf{J}\mathbf{A}\) has only its first column non-zero. Assembling the two pieces gives a diagonal matrix, and the diagonal entries are the three sequence impedances.

Add a neutral earthing impedance \(Z_n\) and the same machinery delivers the factor of three promised in Section 22-5. With the star point returned to earth through \(Z_n\), the phase equation gains the neutral drop \(Z_nI_n = Z_n(I_a+I_b+I_c)\) in every row, which is precisely \(Z_n\mathbf{J}\mathbf{I}_{abc}\). Transforming, that adds \(Z_n\,\text{diag}(3,0,0)\) to \(\mathbf{Z}_{012}\):

The components are not merely algebraic conveniences; each corresponds to something a machine or an instrument can feel.

Positive sequence is the normal system. Applied to the stator of a machine it produces a magnetic field rotating forward at synchronous speed, in step with the rotor. It carries essentially all the useful power, and it is the only sequence present when nothing is wrong. Its impedance is the ordinary impedance used throughout Parts 3 and 4 — a generator's positive-sequence reactance is exactly the \(X_d''\), \(X_d'\) or \(X_d\) of Chapter 21, according to the instant of interest.

Negative sequence produces a field rotating backwards at synchronous speed. Relative to a rotor turning forwards at synchronous speed, that field sweeps past at twice synchronous speed, inducing double-frequency currents in the rotor body, the damper bars and the wedges. The consequences are a braking torque of no use to anyone and, far more seriously, concentrated surface heating that can damage a large turbo-alternator within seconds. This is why negative-sequence overcurrent relays exist and why continuous negative-sequence current is limited to a few per cent of rating.

Zero sequence produces no rotating field at all: three identical currents in three windings spaced \(120^\circ\) apart give three mmf contributions that cancel in the air gap. It produces only leakage flux, so a machine's zero-sequence reactance is small — typically below its subtransient reactance. What zero sequence does instead is flow in earth and neutral paths, and its magnitude depends entirely on how the system is earthed. That dependence, not the winding, dominates \(Z_0\) in practice.

The transformation earns its place only if the fault conditions themselves become simple in the new variables, and they do. Each fault type imposes two or three constraints on the phase quantities at the fault point; transforming those constraints yields relations among the sequence quantities that dictate how the three networks are to be interconnected. This section states the constraints and their transforms; Chapter 24 turns them into circuit connections.

Take a single line-to-ground fault on phase \(a\) through an impedance \(Z_f\). Two currents are zero because those conductors are intact, and the third is fixed by the fault path: \(I_b=I_c=0\) and \(V_a=Z_fI_a\). Substituting the first pair into the analysis equations,

Three equal sequence currents flowing in three networks: that is the signature of elements connected in series, and it is why the three sequence networks are joined in series for an LG fault. Now a line-to-line fault between phases \(b\) and \(c\), where \(I_a=0\) and \(I_b=-I_c\):

so \(I_{a1}=-I_{a2}\) with no zero sequence at all — which is the signature of two elements connected in parallel (or, more precisely, in opposition), and the zero-sequence network takes no part because the fault does not involve earth. The remaining cases follow the same pattern, and the table below is the summary Chapter 24 works from.

One row of that table is a useful sanity check on everything in this chapter. A bolted three-phase fault sets all three phase voltages to zero; three zeros sum to zero and have no unbalance, so \(V_{a0}=V_{a1}=V_{a2}=0\) and the currents contain positive sequence only. Symmetrical components applied to a symmetrical fault return exactly the single-network calculation of Chapter 21, as any correct generalisation must.

A second consequence is worth noting for Chapter 30. A single-phase load connected between one line and neutral of a three-phase supply satisfies \(I_b=I_c=0\) — mathematically identical to the LG fault condition — and therefore draws equal positive-, negative- and zero-sequence currents. Every single-phase traction load or domestic feeder is, in this precise sense, a permanent low-level unsymmetrical fault on the system that feeds it.

Refer every set of components to phase \(a\), and check each answer by synthesising at least one phase quantity back from the components. Difficulty rises down the list.