Fault Current Computation and Breaker Ratings

Chapters 21 through 24 built the theory of faults one layer at a time. Chapter 21 replaced every machine by an emf behind \(jX_d''\) and solved the balanced three-phase fault. Chapter 22 supplied the transformation that turns three unbalanced phasors into three balanced sets. Chapter 23 gave each element three impedances and assembled them into three networks. Chapter 24 wrote down, for each kind of unbalanced fault, the interconnection of those networks and the resulting fault current. What remains is to run the machinery on a real system and to say what the answer is for.

A fault study is not an academic exercise, and its output is not a single number. The switchgear engineer needs the current through each breaker, not the current into the fault, because a breaker interrupts only what flows through it. The relay engineer needs the current seen by each current transformer and the voltage seen by each potential transformer, at every relaying point, for a fault at every location — that is how a distance relay's reach is set and how a differential scheme's stability is checked. The civil and mechanical side needs the peak current, because busbars are braced against the \(i^2\) force between conductors. The earthing designer needs the ground return current \(3I_{a0}\), because that is what raises the earth-mat potential. And the arc-flash calculation needs the current together with the clearing time.

All of these come from the same computation, run once per fault location and fault type. The inputs are fixed by the preceding chapters.

The one assumption worth defending is the flat prefault profile. A load-flow solution (Chapters 18 to 20) would give the true prefault voltages, and a rigorous study uses them. But fault currents are dominated by the machine and transformer reactances, which are small; a prefault voltage that is 2 % low changes the fault current by 2 %, whereas the choice between \(X_d''\) and \(X_d'\) changes it by 40 %. For rating purposes the flat profile at \(1.0\) per unit — or at the highest expected operating voltage, typically \(1.05\) — is both simpler and safer.

Each sequence network is a passive network with a well-defined reference node, so each has a bus admittance matrix built exactly as in Chapter 16, and a bus impedance matrix obtained either by inverting it or by the building algorithm of Chapter 17.

What distinguishes them is not the algebra but the network each is built from. The positive-sequence network contains every machine as \(jX_d''\) to the reference, every transformer and line as its series reactance, and has the same topology as the single-line diagram. The negative-sequence network has identical topology; only the machine branches change, since \(Z_1=Z_2\) for every static element and \(X_2\approx X_d''\) for a machine with damper windings. The zero-sequence network is a different network altogether: transformer winding connections open some branches and earth others, a delta winding blocks zero-sequence current from crossing it, and a neutral impedance \(Z_n\) appears as \(3Z_n\) in series with the machine or winding it grounds.

The consequence for the zero-sequence Z-bus is dramatic and easy to spot. Every delta winding severs the zero-sequence network into disconnected islands, so \(\mathbf{Z}_{bus}^{(0)}\) comes out block diagonal: a bus on one side of a delta has zero transfer impedance to every bus on the other side. Physically the statement is simply that zero-sequence current cannot cross a delta, and mathematically it is what makes zero-sequence fault currents so local. A generator whose step-up transformer is delta on the machine side contributes nothing at all to the ground current of a fault on the transmission network — its zero-sequence emf has nowhere to go.

Fix the fault at bus \(k\). Chapter 24 gave the sequence currents \(I_{a0}, I_{a1}, I_{a2}\) flowing out of the network into the fault; with the Thévenin impedances now read off the diagonals, the four standard cases collect into one table. Write \(Z_1=Z_{kk}^{(1)}\), \(Z_2=Z_{kk}^{(2)}\), \(Z_0=Z_{kk}^{(0)}\).

These give the fault point. Everything else follows from superposition, exactly as in Chapter 21, but now applied sequence by sequence. Before the fault only the positive-sequence network is alive, and it carries \(V_f\) at every bus. The fault is modelled as an injection of \(-I_{a1}\) into bus \(k\) of the positive network, \(-I_{a2}\) into bus \(k\) of the negative network, and \(-I_{a0}\) into bus \(k\) of the zero network. Since \(\mathbf{V}=\mathbf{Z}_{bus}\mathbf{I}\), a single injection at bus \(k\) produces at bus \(i\) a change \(-Z_{ik}I\).

Branch currents follow by Ohm's law in each sequence network, using the element's own sequence impedance:

Two practical points close the loop. First, when a machine or a group of machines feeds the fault through a single path, its contribution is obtained more quickly by a current divider on the sequence network than by differencing bus voltages; both give the same answer and the divider is easier to check. Second, when the path from a machine to the fault crosses a Y–Δ transformer, the \(30^\circ\) phase shift of Chapter 24 must be applied — positive sequence advanced by \(30^\circ\) toward the high-voltage side, negative sequence retarded by \(30^\circ\), zero sequence absent. Ignoring the shift leaves the magnitudes correct and the phase angles wrong by \(30^\circ\), which does not matter for a breaker rating but matters a great deal for a phase-comparison relay.

The procedure is best fixed by running it once, in full, on a system small enough to check by hand yet large enough to contain every feature that matters: two rotating machines, two delta–star transformers, a transmission line with a distinct zero-sequence reactance, and a neutral reactor. The following four-bus system is used for the rest of this chapter.

Choose \(50\) MVA as the system base, with \(20\) kV in the machine circuits and therefore \(110\) kV on the line. The generator and both transformers are already on this base and need no conversion. The motor is rated \(40\) MVA, so Chapter 4's rule \(X_{new}=X_{old}\times(\text{MVA}_{new}/\text{MVA}_{old})\) applies; the line reactances are in ohms and are divided by the base impedance of the \(110\) kV circuit.

Now assemble the networks. The positive-sequence network is a chain: the reference through the generator's \(j0.20\) to bus 1, then \(j0.10\) to bus 2, \(j0.10\) along the line to bus 3, \(j0.10\) to bus 4, and finally the motor's \(j0.25\) back to the reference. Because both machines are sources, the reference node appears at both ends, and the network is a single loop of total reactance \(0.20+0.10+0.10+0.10+0.25=j0.75\).

The zero-sequence network is unrecognisable by comparison. \(T_1\) is delta on the machine side, so bus 1 is severed from bus 2: the generator's \(j0.075\) stands alone as an island, while on the bus-2 side the transformer reactance \(j0.10\) goes straight to the reference through the delta. The same happens at \(T_2\), leaving bus 4 with only the motor's branch \(j(X_{M0}+3X_n)=j(0.125+0.1875)=j0.3125\) — the factor of three on \(X_n\) being the result of Chapter 23, since all three zero-sequence phase currents return through the one neutral conductor.

Writing the bus admittance matrices from these two pictures is mechanical. In the positive network the branch admittances are \(1/0.20=5\) from bus 1 to reference, \(10\) for each of the three series branches, and \(1/0.25=4\) from bus 4 to reference, giving

Every entry can be checked without inverting anything, because the network is a single loop. Number the nodes by their distance around the loop from the reference through the generator: \(\alpha_1=0.20\), \(\alpha_2=0.30\), \(\alpha_3=0.40\), \(\alpha_4=0.50\), with total loop reactance \(T=0.75\) and \(\beta_i=T-\alpha_i\) the distance the other way. Injecting one ampere at bus \(i\) sends it around both arms in inverse proportion, so

The zero-sequence matrix is easier still, because its three islands never talk to one another:

The pattern is worth reading rather than merely computing. Bus 1 has \(Z_{11}^{(0)}=j0.075\), smaller than its positive-sequence \(j0.1467\), so an earth fault at the generator terminals is more severe than a three-phase fault there — the price of a solidly earthed machine. Bus 4 has \(Z_{44}^{(0)}=j0.3125\), nearly twice its positive-sequence value, because the neutral reactor was installed precisely to make it so. And the zero rows for buses 1 and 4 contain nothing but their own diagonal: an earth fault anywhere on the \(110\) kV network produces no zero-sequence current at all in either machine.

Section 21-9 introduced the short-circuit level of a bus as the apparent power delivered into a bolted three-phase fault there. With a flat prefault profile of \(1.0\) per unit it reduces to a single reciprocal:

with \(V_L\) the line voltage in kV and \(I_{f,L}\) the line current in kA. The two forms are the same fact seen from two sides: a bus of low Thévenin impedance is a bus of high fault level. Because the reciprocal relation is exact, the fault level is also the most convenient way to import a piece of network that is not modelled in detail. A utility declares "the supply point has a fault level of 800 MVA"; that statement is a complete positive-sequence model of everything upstream, and converting it to an impedance on the study base takes one line:

For the four-bus system the four fault levels follow immediately from the diagonal of \(\mathbf{Z}_{bus}^{(1)}\). The base current is \(50/(\sqrt3\times20)=1.443\) kA in the machine circuits and \(50/(\sqrt3\times110)=0.2624\) kA on the line.

Two features of the table are general. First, the strongest bus is the one closest to the largest source, and the fault level falls monotonically as one moves away from generation and out along impedance. Second, the currents in kA are an order of magnitude apart between the \(20\) kV and \(110\) kV buses even though the fault levels differ by only a quarter — which is the reason transmission is done at high voltage and the reason the physically largest switchgear is not always at the highest fault level.

Where all internal emfs are equal to \(V_f\) and each source reaches the fault by one path, the contribution of a source is simply the reciprocal of its own path impedance, with no divider algebra needed. At bus 4 the motor's own branch is \(j0.25\), so it supplies \(1/0.25 = 4.0\) per unit; the generator reaches bus 4 through \(0.20+0.10+0.10+0.10=0.50\) and supplies \(2.0\) per unit; and \(4.0+2.0=6.0\) is precisely \(1/Z_{44}\). The motor, rated at four-fifths of the generator and standing at the fault, delivers twice the generator's contribution.

Everything so far has produced a steady phasor. The actual current in a faulted circuit is not steady: Chapter 21 showed that it starts with a DC offset determined by the instant of fault inception, and that the AC part itself decays as the machine flux redistributes from the subtransient to the transient to the steady-state condition. A breaker meets that current twice, at two different times, and the two encounters are physically different problems.

The first encounter is at the very beginning. If the fault already exists when the breaker closes — a maintenance earth left on, or a reclosure onto a persistent fault — the breaker must close onto the full asymmetrical current and then latch, holding its contacts together against the electromagnetic repulsion that tries to blow them apart. This is a mechanical and thermal problem, it peaks within the first half cycle, and nothing has yet decayed. The second encounter comes several cycles later, when the relay has operated and the contacts part: now an arc must be extinguished, and what matters is the current still flowing at that instant, by which time part of the DC offset has died and the machines have moved from \(X_d''\) toward \(X_d'\).

The ANSI/IEEE procedure handles the decay of the AC component without solving any differential equation. Instead of one network it builds two, identical in topology but with the rotating machine reactances multiplied by different factors. The first-cycle network models what is present at \(t=0\); the interrupting network models what survives to contact parting. Generators are unchanged between the two because their flux decays slowly, but motors — whose stored flux is small and whose excitation is the stator field they have just lost — are heavily de-rated.

With each network reduced to its Thévenin reactance at the bus of interest, the duties follow from Chapter 21's asymmetry factors. The symmetrical rms current is \(E/X\) with \(E=V_f\); the momentary duty multiplies it by the first-half-cycle asymmetry factor, and the interrupting duty by a multiplying factor that depends on how fast the breaker is and how close the fault is to generation.

The slower the breaker, the smaller the interrupting duty — a fact that reads backwards until one remembers what is being asked. A breaker with a five-cycle rating parts its contacts about three cycles after inception; an eight-cycle breaker parts at four cycles, by which time more of the offset and more of the subtransient contribution have gone. Speed costs money at the interrupter and saves it in every downstream consequence of a long fault, above all in stability (Chapter 29), which is why modern transmission breakers are two-cycle machines even though the current they must clear is thereby larger.

A breaker nameplate is a specification of survivable duty, and each entry answers a question the fault study has just posed.

Two of these deserve their arithmetic spelled out. The making current is not an independent number: it is the breaking current re-expressed as the crest of a fully offset wave. IEC 62271-100 fixes the ratio at \(2.5\) for \(50\) Hz systems and \(2.6\) for \(60\) Hz, the difference being the extra decay of the DC offset during the longer \(50\) Hz half cycle.

The second is the ANSI voltage range factor \(K\) introduced in Section 21-9. A breaker on the older total-current rating basis carries a rated maximum voltage and a rated short-circuit current at that voltage; used below it, its interrupting capability rises inversely with voltage, but never beyond \(K\) times the rated value, because at some point the mechanism rather than the dielectric becomes the limit:

Modern IEC practice avoids the factor entirely by defining a single symmetrical breaking current valid over the whole rated voltage range, with a separately declared DC component percentage at contact separation. The engineering content is unchanged; only the bookkeeping differs, and a specification written against one standard must be translated carefully before it is compared with a study run against the other.

Selecting from such a list is a matter of clearing every row of the nameplate table at once, and of remembering that the study which produced the duty is itself approximate. A margin of fifteen to twenty per cent between the computed interrupting duty and the rated breaking current is normal practice; it absorbs future generation added to the system, the difference between assumed and actual machine reactances, and the fact that \(1.05\) per unit prefault voltage is not impossible.

Fault level grows with every megawatt of generation added and every kilometre of parallel circuit built. Switchgear, once installed, does not. A station commissioned with 350 MVA breakers eventually finds itself in a system whose fault level has drifted past that figure, and the choice is then between replacing every breaker in the station and raising the Thévenin impedance seen from the busbar.

The cheapest instrument for the second option is a current-limiting reactor: an air-cored coil, deliberately without iron so that it cannot saturate under fault current and lose the very reactance it was installed to provide. Where it is placed decides what it protects.

Every one of these buys fault-level headroom with the same currency: series reactance in the normal operating path. The bill arrives in three places. Regulation worsens, because the voltage drop \(I X\) is present at full load and not only during faults — and the percentage reactance of a reactor on the rating of the circuit it sits in is, to a good approximation, exactly the percentage voltage drop it imposes at that circuit's full load. Losses rise, since the reactor is not perfectly lossless. And transient stability degrades, because Part 6 will show that the maximum power a machine can deliver is \(|E||V|/X\); anything that raises \(X\) lowers that ceiling.

Unless stated otherwise take \(V_f = 1.0\angle0^\circ\) per unit, neglect load current, resistance and line charging, and represent machines by an emf behind their subtransient reactance.