Fault Currents and Breaker Ratings

The current in per unit. With the prefault profile flat at \(V_f = 1.0\angle0^\circ\) and the fault bolted, the diagonal entry is the whole calculation:

The current lags the prefault voltage by \(90^\circ\) because resistance was neglected. Only its magnitude, 8.000 pu, matters for a rating.

The base current at 132 kV, which is the only conversion needed:

Three and a half kiloamps. Note that this is a line current, because the per-unit system was set up on three-phase MVA and line-to-line kV, so no \(\sqrt3\) appears anywhere in the conversion.

The fault level, which is the same fact expressed as apparent power:

The two routes agree exactly, and they must: the reciprocal relation between per-unit impedance and fault MVA is an identity when \(V_f = 1.0\), not an approximation.

The inverse question. A declared fault level is a complete positive-sequence model of everything upstream. Converting it to the study base is one division:

Seven ohms per phase, referred to 132 kV, replaces however many hundred kilometres of network and however many machines lie behind that point. That single number is what a utility hands to a consultant, and it is all the consultant needs.

The direction of the inequality is worth fixing in the mind. A low Thévenin impedance is a high fault level and a strong bus; a high impedance is a weak bus with a small fault current. The 2500 MVA point is three times stiffer than the 800 MVA one, so its impedance is a third as large.

Put both impedances on a 100 MVA base. The utility becomes a reactance from the reference to the 132 kV point; the transformer needs the ordinary base change of Set 3:

The transformer's percentage reactance is on its own 50 MVA rating, so it doubles when carried to a 100 MVA base. The utility's figure needed no rating at all — a fault level already carries its own.

The two are in series, because the only path from the reference to the 33 kV bus runs through both:

Fault level and current at 33 kV:

Check by the other route: \(358.97/(\sqrt3\times33) = 6.280\) kA. The two agree.

The peak making current, at the IEC 50 Hz ratio of Section 25-7:

A 36 kV, 25 kA breaker with a 62.5 kA peak making rating clears both duties with a very large margin — the 33 kV board here is a weak one, fed through a single transformer.

The infinite-bus error. Dropping \(X_{source}\) leaves the transformer alone:

The source reactance is only 11.4% of the transformer's, and it inflates the answer by exactly 11.4% — the two percentages coincide here because fault current depends on the reciprocal of a sum, so a small addition to the denominator produces a proportionally equal reduction in the result.

When the error stops being small. Repeat with a weaker utility, say 800 MVA, and \(X_{source}=0.125\) against the transformer's 0.250. Now \(X_{th}=0.375\), \(S_{sc}=266.7\) MVA, and the infinite-bus assumption overstates the fault level by 50%. The error equals the ratio of source reactance to transformer reactance, so it is negligible for a stiff grid and inadmissible for a weak one.

The network. Two machines, seven lines, all reactances in per unit on 100 MVA. Bus 2 is the hub, with four connections; bus 5 hangs off it and off bus 4.

The base current, the same for every bus because the whole network is at 132 kV:

Read the diagonal and divide. Each row of the table below is \(1/Z_{kk}\), then multiplied by \(I_{base}\), then \(100/Z_{kk}\):

The whole table is one column of arithmetic. No network reduction was performed, and none is needed: the matrix was built once, in Set 18, and answers all five questions at once.

Yes, the ranking is readable from the diagram, and checking that it is readable is the first sanity test on any completed matrix:

Buses 3 and 4 differ by 1.7% in fault level because the \(j0.03\) tie between them is an order of magnitude stiffer than anything else touching them. A matrix that ranked them differently, or ranked bus 5 above bus 3, would be evidence of an arithmetic error somewhere in the build.

The spread is only 1.7 to 1. From 833 MVA at the strongest bus to 489 MVA at the weakest — a small range, because both machines are electrically close to every bus in a network this compact. In a long radial system the ratio between the strongest and weakest bus is commonly ten to one, and the switchgear specification differs accordingly from one end to the other.

The fault current first, from the diagonal:

The superposition statement. Before the fault every bus stands at \(V_f = 1.0\). The fault is an injection of \(-I_f\) at bus 5, and a single injection at bus \(k\) changes bus \(i\) by \(-Z_{ik}I_f\):

The fault current cancels out of the ratio, so the voltage profile depends only on the shape of the column, not on its scale. That is why the same profile is obtained whether the prefault voltage is taken as 1.00 or 1.05 pu.

One division per bus, using column 5 \((0.10160,\ 0.11872,\ 0.13064,\ 0.13626,\ 0.20457)\) over \(Z_{55}=0.20457\):

The relay question. With a pickup at 0.45 pu, only bus 1 at 0.503 pu stays above the setting:

So an undervoltage element at bus 1 does not pick up while three other substations are in a deep sag. That is the mechanism behind a familiar complaint: a fault at one end of a system trips motor contactors at the far end and not at the near one, because electrical distance is not geographical distance.

The ordering is not the ordering of the diagram. Bus 4 sags more deeply than bus 2, although bus 2 is joined to bus 5 by the stiffest of the two paths (\(j0.12\) against \(j0.24\)). The reason is that bus 2 has a machine on it holding it up, and bus 4 has none:

The ratio is a dimensionless distribution factor. It is 1 at the fault itself and would be 0 at a bus completely disconnected from the fault — for example, at any bus on the far side of a delta winding in the zero-sequence network of Set 22.

A check that costs nothing. Every voltage must lie between 0 and \(V_f\), and the faulted bus must come out at exactly zero. Setting \(i=k=5\) gives \(1 - Z_{55}/Z_{55} = 0\) identically, which confirms that the right column was used.

Ohm's law on each branch, using the voltages of Problem 4:

The bus voltages are the only thing the matrix was needed for. Once they exist, the branch currents are elementary — and this is the reason the whole study is arranged around \(\mathbf{Z}_{\text{bus}}\) rather than around any direct current calculation.

Line 1–2, worked in full as the pattern for the rest:

Positive in the direction 1 to 2, so current flows from bus 1 towards bus 2 — from the machine that is holding its voltage up, towards the fault.

All seven branches, with the kA column at \(I_{base}=0.43739\) kA:

The verification at the fault point. Only two lines enter bus 5, and everything that arrives must go into the fault:

Agreement to five figures. This check is worth doing every time, because it tests the voltage profile, the line data and the diagonal entry simultaneously; a transposed matrix element or a mis-keyed reactance will fail it immediately.

A second check at an unfaulted bus. Kirchhoff's current law must also hold at bus 3, which has no machine and no fault, so its three branch currents must sum to zero:

Which breaker carries the most. The largest branch current is 3.497 pu, or 1.530 kA, in line 2–5:

No breaker in the network interrupts 4.888 pu. The fault current is shared, and each breaker's duty is its own branch current — which for the bus-5 fault is at most 72% of the total. Rating every breaker at a bus for the full bus fault level is conservative and is what is actually done for busbar switchgear, but for a line breaker it can be a substantial over-specification.

Confirm the machine branches first. On the 100 MVA base:

Which are the two shunt branches used in Set 18's build — \(j0.25\) at bus 1 and \(j0.20\) at bus 2. The data is consistent, and that consistency is worth checking before any current is computed.

Method one: current out of each machine branch. Each generator is an emf of 1.0 pu behind its own reactance, and the bus it stands on has fallen to the value found in Problem 4:

Method two: the branch currents leaving each generator's bus. Everything \(G_1\) produces must leave bus 1 along its two lines, and everything \(G_2\) produces must be the net outflow at bus 2:

Note the sign on \(I_{12}\): line 1–2 brings current into bus 2, so it is subtracted from what leaves. Two entirely different routes through the arithmetic give the same two numbers, which is as strong a check as a hand calculation admits.

The shares:

The smaller-reactance machine at the better-connected bus supplies more, as it must. \(G_2\) reaches bus 5 through \(j0.12\) directly; \(G_1\) reaches it only through bus 2 or the long way round through buses 3 and 4.

In terms of each machine's own rating. Convert to the machine base by the ratio of ratings:

Four to five times full load, which is what \(1/x''_d\) of 8 and 8.33 would give for a fault at the machine's own terminals, reduced by the network between. Both figures are within the short-circuit withstand of a normally specified machine, and both are far above anything the stator could carry for more than a few seconds.

An observation the ratings make possible. \(G_2\) is 20% larger than \(G_1\) but delivers 46% more fault current. Fault contribution scales as \(\mathrm{MVA}/x''_d\) and not as MVA, so a machine with a low subtransient reactance is a disproportionately large contributor. That is a purchasing decision: specifying \(x''_d\) a few per cent higher costs a little efficiency and buys fault-level headroom for the life of the plant.

The fault impedance is in series with the Thévenin impedance, because the whole network reduces to one branch at the fault point:

The bolted case for comparison, from Problem 3:

A fault impedance of 0.05 pu — five per cent of one — cuts the current by 22%. The arithmetic is not subtle: 0.05 is not 5% of anything relevant, it is 28% of \(Z_{44}\), and it is the ratio to \(Z_{44}\) that matters.

The general statement. Write \(\lambda = Z_f/Z_{kk}\):

A reduction of 22.05%, matching the direct calculation. The lesson generalises: at a strong bus, where \(Z_{kk}\) is small, a given arc impedance matters more, not less. A 0.05 pu arc at bus 2 (\(Z_{22}=0.12009\)) would remove 29% of the current.

The faulted bus is no longer at zero volts. The fault impedance carries the whole fault current:

Twenty-two per cent of nominal remains at the fault point — nearly 17 kV of arc voltage on a 132 kV system. A disturbance recorder that shows a residual voltage at the fault point is showing an arcing fault, and the recorded value is the direct measurement of \(Z_f I_f\).

The rest of the profile, from \(V_i = 1 - Z_{i4}/(Z_{44}+Z_f)\) with column 4:

Every voltage is lifted, and by more at the buses nearest the fault. Bus 3, which sits behind the stiff \(j0.03\) tie to bus 4, rises from 0.084 to 0.286 pu — the sag it suffers is very nearly the sag at the fault, arc or no arc.

Why rating calculations still assume a bolted fault. Every effect of \(Z_f\) is to reduce the current, so the bolted assumption is the conservative one and no argument about the arc's resistance can invalidate a rating based on it. The arc matters in the other direction — for protection, where an underestimate of the fault current can leave a relay short of pickup, and for arc-flash energy, where the arc voltage is precisely the term that makes the incident energy nonzero.

The tap point is a new bus. A fault at a point \(p\) that is not a bus is handled by making it one: line 2–4 is deleted and replaced by two sections in series through \(p\). Three modifications are needed, in this order:

Removing an element is adding one of negative impedance — the same Type 4 formula with the sign reversed. That is the whole trick, and it is why the building algorithm is worth having in a planning study where lines come and go.

Step A, removing line 2–4. The fictitious diagonal is

Negative, as it must be when an element is deleted; the network's own impedance between buses 2 and 4 is \(j0.06493\), smaller than the line being removed, so the removal will raise every diagonal.

The matrix with line 2–4 out:

Bus 4 is worst hit: \(Z_{44}\) rises 18%, from \(j0.17680\) to \(j0.20895\), because it has lost one of its three connections. Bus 1 barely moves.

Step B, the tap bus. A Type 2 addition from bus 2 to the new bus \(p\): the new row and column copy row and column 2, and

At this stage \(p\) is a dead end fed only from bus 2, so its Thévenin impedance is bus 2's plus the section — exactly what a first-year calculation would give.

Step C, closing the loop to bus 4:

The correction term uses \(Z_{pp}-Z_{p4}=j0.07843\), the difference between the two ends of the closing branch. Squared and divided by \(Z_{ll}\), it removes 11% of the tap bus's impedance.

The fault at the 40% point:

An independent check: rebuilding a six-bus \(\mathbf{Y}_{\text{bus}}\) with the split line and inverting it gives \(Z_{pp}=j0.170393\), agreeing to six figures.

Now sweep the fault along the line. Repeating the three steps with \(\alpha\) in place of 0.4:

Why the minimum is inside the line. Bus 4 is not a dead end: it is tied to bus 3 by \(j0.03\) and to bus 5 by \(j0.24\), so the network reaches the fault from both ends of line 2–4. Moving the fault away from bus 2 raises the impedance of one path and lowers the other, and the sum is maximised where the two are comparable:

A parabola-like curve with an interior maximum. It degenerates to a monotonic rise only when the far end is a dead end, \(Z_B\to\infty\) — which is the radial case every student pictures.

What it costs to get this wrong. Assuming the far bus gives the smallest current would predict 2.474 kA as the minimum; the true minimum is 2.375 kA, 4.0% lower. An earth-fault or overcurrent relay graded on the larger figure is 4% short of where it thinks it is — small here, but on a long line fed strongly from both ends the discrepancy reaches 15 to 20% and decides whether the far end of a line is covered at all.

The current after inception, for a fault applied at the instant that produces full offset:

Two terms: a decaying unidirectional component whose initial value equals the crest of the AC term, and the AC term itself. The DC term exists because the current in an inductive circuit cannot jump, so a component must be added to make \(i(0)=0\).

The decay in cycles. One cycle is \(t = 2\pi/\omega\), so

The decay depends on \(X/R\) alone, not on the impedance magnitude — which is why \(X/R\) is the one extra number a fault study must carry beyond the reactances.

The two factors. Half a cycle after inception the AC term is again at its crest and the DC term has decayed by \(e^{-\pi R/X}\). The crest of the total is the sum, and the rms of an AC component plus a DC component is the root of the sum of squares:

The 2 inside the root is \((\sqrt2)^2\): the DC component's contribution to the rms is its own instantaneous value squared, and that value is \(\sqrt2\) times the rms of the AC part.

Evaluated:

Why 1.6 and 2.7. The ANSI figures are the values at \(X/R \approx 30\), rounded down slightly:

Thirty is the ratio near large generation, which is where the momentary duty is most severe. Applying the same 1.6 at an industrial board with \(X/R=15\) overstates the duty by 5%, which is accepted as the price of having one number instead of a table.

The lossless limit is worth knowing, because it bounds everything: as \(R\to0\) nothing decays, \(k_{crest}\to2\sqrt2 = 2.828\) and \(k_{rms}\to\sqrt3 = 1.732\). No breaker anywhere ever meets more than 2.83 times the symmetrical rms current as a crest, and the IEC making-current ratio of 2.5 sits comfortably below that because a real circuit has resistance and a real closing takes finite time.

Every branch on the 100 MVA base. The utility from its declared fault level, the machines by the ratio of ratings:

All five stand between the bus and the reference, so the network is five branches in parallel and nothing else. The topology is identical in both cases; only the values change.

The de-rating factors, from the ANSI table of Section 25-6:

The generator is untouched between the two networks because its field circuit keeps it excited; a motor loses its excitation the moment the bus voltage collapses, and the smaller its rotor time constant the faster its contribution dies.

The first-cycle network, five parallel admittances:

Only the small induction motors are altered here, from 4.25 to \(1.2\times4.25=5.10\). Working in admittances rather than impedances is not a convenience but a necessity — parallel branches add only in that form, and the sum is also the fault current in per unit.

The interrupting network, with three of the five branches enlarged:

What has actually changed. The two non-motor branches contribute 3.78125 in both networks; the three motor branches fall from 0.85784 to 0.51961, a loss of 39.4% of the motor contribution:

Under 8% change in the network impedance, because the motors were never the dominant branch here. In a plant fed by a weak utility with a large motor load the same calculation can move \(X\) by 30%.

A structural point about the method. Nothing has been integrated, no time constant has been used, and no differential equation has been solved. The standards replace a continuously decaying source by two frozen snapshots taken at the two instants the hardware cares about, and the whole of the machine physics has been reduced to a column of multipliers.

The base current at 11 kV:

Large, because 11 kV is a low voltage to move 100 MVA through. That is the reason industrial switchgear carries the biggest fault currents in a power system while the transmission network carries the biggest fault MVA.

The symmetrical currents. With \(E = V_f = 1.0\), the per-unit current is simply the admittance sum already formed:

The momentary duties, at the ANSI factors of Problem 9:

The crest figure is the one checked against a breaker's rated short-circuit making current, and it is also the number the busbar bracing is designed to, since the force between conductors goes as the square of the instantaneous current.

The interrupting duty:

\(\mathrm{MF}=1.0\) is justified here because two thirds of the current arrives from the utility through transformers and cable, so the composite \(X/R\) at the bus is well below 17 and the DC offset has essentially gone by the time the contacts part.

With \(\mathrm{MF}=1.1\):

Two situations force this. The first is a fault fed predominantly from local generation, where \(X/R\) reaches 30 and the offset survives; the second is a fast breaker, whose contacts part in two cycles rather than four and therefore meet a current that has had less time to decay. Both are cases where the DC component has not finished dying, and MF is the factor that accounts for it.

Collecting the three numbers a purchase specification needs:

Note that the largest of the three is a peak and the other two are rms, so they cannot be compared with one another. Problem 14 checks each against its own nameplate row.

Each branch is its own current source. With all internal emfs equal to \(V_f\) and every branch running straight from the bus to the reference, no divider algebra is needed — each contribution is the reciprocal of its own reactance:

The two tables, at \(I_{base}=5.2486\) kA:

Each column sums to the totals of Problem 11, which is the check: the sum of the reciprocals of the branch reactances must equal the reciprocal of the parallel combination.

What the motors are worth:

Two fifths of the motor contribution has gone in three cycles, and the loss is very unevenly distributed: the small induction motors lose 60% of theirs, the large ones and the synchronous machine a third.

The study that treated motors as load. Delete all three motor branches:

The momentary crest would come out at \(2.7\times19.85 = 53.6\) kA instead of 65.7 kA — and Problem 14 will show that the difference between those two figures is exactly the difference between a breaker that is adequate and one that is not.

Why the error is worse in the first cycle than at interruption. Because the motors decay and the utility does not, the motor share falls from 18.5% to 12.1%. Neglecting motors therefore does the most damage to the making and bracing checks, which are the ones least often revisited. A plant that grows by adding motors moves its momentary duty faster than its interrupting duty, and can outgrow its switchgear's close-and-latch rating while still comfortably inside its breaking rating.

A rule of thumb worth carrying. An induction motor contributes roughly \(1/X''_d \approx 5\) to 6 times its own rated current for the first few cycles. A plant with 16 MVA of motors on a bus fed by 300 MVA of utility therefore adds about \(16\times5.5 = 88\) MVA of first-cycle contribution — 29% on top of the utility — which is the order of the 22.7% found above.

Making current is not an independent rating. It is the breaking current re-expressed as the crest of a fully offset wave:

Where 2.5 and 2.6 come from. IEC 62271-100 fixes the DC time constant of the test circuit at \(T = 45\) ms for both frequencies, so the difference is entirely in the length of the half cycle:

The 60 Hz crest arrives sooner, so less of the offset has died. A 45 ms time constant corresponds to \(X/R = 2\pi f T = 14.1\) at 50 Hz — a deliberately moderate figure, which is why installations near large generation are given a separate, higher-offset test duty.

Breaking capacity in MVA, the older way of quoting the same rating:

Note the rated voltage, 12 kV, not the nominal system voltage of 11 kV. Using 11 kV would understate the capacity by 8.3% and is the commonest slip in this calculation.

The reverse question. A study demanding a 65.74 kA peak — the momentary duty of Problem 11 — implies a breaking rating of at least

Which is more than 25 kA. The 25 kA breaker fails the making check although its breaking rating exceeds the 22.57 kA interrupting duty by 11%. The two checks use different currents — the first-cycle current and the interrupting current — so passing one does not imply passing the other, and the making check is the one more often binding in plants with heavy motor load.

A short comparison across the standard classes, to fix the orders of magnitude:

The current rises by a factor of only 2.5 from the smallest to the largest entry while the capacity rises by a factor of 88 — because capacity carries the voltage. That is why a 12 kV interrupter is a small device handling a large current and a 420 kV one is a large device handling a moderate one.

Rated voltage first, because it is a yes-or-no test and disposes of nothing else. The system's highest voltage is 12 kV against a nominal 11 kV, so the 12 kV class is correct; a 7.2 kV unit is inadmissible whatever its current rating.

Rated normal current. The 1250 A load is a continuous thermal duty unrelated to the fault:

Standard values run 630, 1250, 1600, 2000, 2500, 3150 A. Choosing 1250 A exactly leaves no room for emergency transfers, which are the reason a bus-section breaker in a two-incomer board must carry the whole board.

The 25 kA candidate, checked row by row:

Rejected on the making current. The breaking rating is adequate; the close-and-latch rating is not. That is the case Problem 13 anticipated, and it happens because the first-cycle current is 8% larger than the interrupting current while the two nameplate figures are tied together by the fixed ratio 2.5.

The 31.5 kA candidate:

The thermal row in figures, since "ample" is not an engineering statement:

Why not go straight to 40 kA. Its margins are 77% and 52%, and margins that large are not free: a 40 kA cubicle is physically larger, its busbar bracing is heavier, and the whole switchboard grows. Fifteen to twenty per cent is the accepted band, and 19.9% sits at the top of it — which is right for a plant likely to add motor load.

What the choice would have been on a careless study. Problem 12 showed that neglecting the motors gives 19.85 kA first cycle and a crest duty of \(2.7\times19.85 = 53.6\) kA. Against a 62.5 kA making rating that is a 17% margin, and the 25 kA breaker would have been bought. The motors are the difference between the right answer and a switchboard that cannot survive a reclosure onto a standing fault.

What the factor means. Used below its rated maximum voltage, such a breaker can interrupt more, because the recovery voltage across the gap is lower — but not without limit, since beyond some point the mechanism rather than the dielectric becomes the constraint:

The two candidates at 11 kV:

The \(K\) ceiling governs. Taking the inverse-voltage figure alone would credit the breaker with 1.15 kA it does not have.

The transition voltage, where the two expressions cross:

Above 11.54 kV the capability rises inversely with voltage; below it, the capability is flat at 23.40 kA however far the voltage falls. An 11 kV system sits just inside the flat region — which is why the 15 kV class was built with \(K=1.3\) in the first place.

Against the duty:

It passes, and only just. Three and a half per cent of margin is inside the uncertainty of the study itself — the prefault voltage was taken at 1.00 rather than 1.05 pu, and at 1.05 pu the duty is \(1.05\times22.57 = 23.70\) kA, which the breaker would fail. The unit is not acceptable for this bus.

Modern practice has removed the factor. IEC declares one symmetrical breaking current valid across the whole rated voltage range, with the permitted DC component at contact separation stated separately. The engineering is unchanged; only the bookkeeping differs — and a specification written against one standard must be translated before it is compared with a study run against the other.

The rating is an energy, not a current. Adiabatic heating over a short fault makes the temperature rise proportional to \(\int i^2\,dt\), so the nameplate pair \(I_k / t_k\) is shorthand for

Both clearing times pass on this basis: 0.35 s and 1.2 s are well inside 3.671 s. The backup time is the one that matters, because it is the duration the breaker must survive while a different breaker fails to clear.

Part 2 — the DC component adds heat. With a fully offset wave the current is \(i = \sqrt2 I\big[e^{-t/T} - \cos\omega t\big]\). Integrating the square over \([0,t]\) and averaging out the cross term, the AC part contributes \(I^2t\) and the DC part

The DC time constant at \(X/R = 15\):

Over 0.35 s the exponential has long since vanished, so \(1-e^{-2t/T}\to1\) and \(m\) reduces to \(T/t\).

The equivalent currents:

Both far below 1875. Note that the DC contributes proportionally more for a fast clearance, because the offset's energy is a fixed quantity delivered in the first 50 ms whatever happens afterwards — so a 6.6% uplift at 0.35 s becomes only 2.0% at 1.2 s. For clearances of a few cycles, which is where a differential scheme operates, the uplift reaches 30 to 50% and cannot be ignored.

Part 3 — why the rating cannot be traded upward. The proposal is

Thermally the arithmetic is sound: 1875 kA²s either way. Electrically it is nonsense, because a 43.3 kA symmetrical fault brings with it a peak of

The contacts would be blown apart by the electromagnetic force long before the heating became the issue. The short-time withstand may be traded from \(t_k\) to longer durations at lower current, but never to shorter durations at higher current, because the peak withstand caps it. IEC states the limit explicitly: the rating is valid for durations up to \(t_k\), and the current is never to exceed the rated short-circuit breaking current.

The same rule applies to everything else in the circuit. A current transformer, a cable and a busbar all carry \(I^2t\) ratings, and all three are checked against the backup clearing time, not the main one:

With \(k=94\) for XLPE-insulated copper, a 22.6 kA fault held for 1.2 s needs \(22600\sqrt{1.2}/94 = 263\ \text{mm}^2\) of copper. The breaker was never the binding constraint; the cable is.

The new branch on the study base:

A shunt from an existing bus to the reference — a Type 3 addition, which closes a loop and therefore changes every element of the matrix.

The Type 3 modification, a single rank-one correction using column 5:

One denominator for the whole matrix, and the numerator is the outer product of column 5 with itself. That is the entire cost of adding a machine — no inversion, and the arithmetic is the same for a five-bus network as for a five-thousand-bus one, per column.

The five diagonals, worked:

The independent check at bus 5, where the modification is a simple parallel combination:

Exact agreement. The check works only at the bus the element was added to, and it is worth doing every time, because it catches a wrong column or a wrong denominator immediately.

The consequences, bus by bus:

Which switchgear is at risk. Against a 1000 MVA busbar rating:

The bus that fails is not the bus the machine was connected to. Bus 5 rose the most in percentage terms and remains the weakest bus in the system; bus 2 rose by less and broke the rating, because it started closest to it. A fault-level assessment must be run at every bus, not at the point of connection.

Why the effect reaches so far. Compare with the earthing-transformer result of Set 22, where adding a zero-sequence earth point changed one bus by 73% and the others by 2 to 13%. Here every bus moves by at least 19.5%:

In the positive-sequence network every bus is coupled to every other, because there is no delta winding to sever anything. A change in the positive-sequence network is a system-wide event; a change in the zero-sequence network is usually a local one.

The planning consequence. Sixty megavolt-amperes of new generation — a machine whose own contribution is 333 MVA against the 833 MVA already present at bus 2 — raised that bus's fault level by 30%. Fault level does not grow in proportion to installed capacity; it grows in proportion to installed capacity divided by reactance, and a modern machine has a lower \(x''_d\) than the ones it joins. Every connection study must therefore include the fault-level check, and it is the reason a distribution utility can refuse a generator connection on switchgear grounds alone.

Everything on a 100 MVA base at 11 kV:

The arrangement as it stands. Three identical shunt branches in parallel:

Over by 7.1%. Not a large exceedance, but a breaker is not a component one operates 7% beyond its rating — the failure mode is a restrike and a busbar fire, not a gradual degradation.

Sizing the reactor. With \(X_r\) in each lead the three branches remain in parallel, each of \(0.560+X_r\), so the required Thévenin impedance follows directly from the rating:

Write the new Thévenin impedance symbolically in terms of \(X_r\), set the resulting fault level equal to the switchgear rating, and solve. Everything else in the problem is a consequence of the 0.0400 obtained here.

In ohms and as a percentage:

One per cent. The percentage reactance of a reactor on the rating of the circuit it occupies is, to a good approximation, exactly the full-load voltage drop it imposes — so this reactor costs 1% of regulation, which is unobtrusive. A 10% reactor would not be.

The drop, computed directly to confirm the shortcut:

The fault between a machine and its own reactor. That machine's own contribution does not pass through its lead breaker; what the breaker interrupts is everything arriving from the busbar, which is the other two branches in parallel plus the faulted machine's own reactor in series:

Comfortably inside the 500 MVA rating. The reactor that was installed to protect the busbar also relieves the machine-lead breakers, which is a free benefit of the generator-reactor arrangement and one that a feeder-reactor arrangement does not provide.

The busbar duty and the making check. At the limiting 500 MVA:

And the reactors themselves must carry 26.24 kA for the full short-time duration without their air-cored windings deforming — a mechanical specification quite separate from their reactance.

The alternative: split the busbar. Operating the board as two sections, two machines on one and one on the other:

Far below 500 MVA, achieved with no hardware and no voltage drop at all. The price is redundancy: each section must carry its own load unaided, a fault on one section loses its machines outright, and the open tie must be closed for maintenance — at which moment the fault level returns to 536 MVA. Bus splitting is free in capital and expensive in operational flexibility; a reactor is the reverse.

The base current at 33 kV:

The three-phase fault is the same at both buses, because it involves the positive-sequence network alone:

The earth fault puts the three networks in series (Set 23), and the phase current is three times the sequence current:

The ratio rule confirms both without any of that arithmetic. With \(k = Z_0/Z_1\):

The rule crosses unity at \(k=1\): the earth fault is worse wherever \(Z_0 < Z_1\) and milder wherever it is not. That single comparison, made before any calculation, tells you which answer to expect.

Which governs:

A breaker at bus A rated on the three-phase fault would be 14% short. The mechanism is not exotic: a solidly earthed transformer is a stiffer zero-sequence source than the network is a positive-sequence one, so the earth path is the lower impedance — and that is precisely the situation at every transformer terminal, which is where busbars are.

The theoretical bound. As \(Z_0\to0\) the ratio approaches \(3/2 = 1.5\), so an earth fault can never exceed a three-phase fault by more than 50%. Bus A's 14.3% is unremarkable; ratios of 20 to 30% are common at large solidly earthed generator-transformers, and reaching 1.4 requires a zero-sequence impedance close to zero.

What the neutral reactor bought and what it cost. At bus B the earth fault is 18% below the three-phase value, so the switchgear is sized by the balanced fault and the earth-fault duty on the transformer neutral is modest. The cost is that the healthy-phase voltage rise is larger and the earth-fault current may fall so low that sensitive earth-fault protection is needed to detect it at all — the design trade that Set 22's effective-earthing criterion \(X_0/X_1 \le 3\) is written to bound.

One caution about motors. The positive- and negative-sequence figures include every rotating machine on the bus; the zero-sequence figure usually does not, because a motor is fed through a delta winding or has an unearthed star. So motor contribution inflates \(I_{3\phi}\) and leaves \(I_{LG}\) nearly untouched, and a bus with heavy motor load is more likely to be governed by the three-phase fault than these impedances alone suggest.

Part 1 — the impedances. The grid becomes a reactance from its declared fault level; the transformers convert from their own rating:

The motors, in both networks. Above 1000 hp, so \(1.0X''_d\) in the first cycle and \(1.5X''_d\) at interruption:

The three duties:

\(\mathrm{MF}=1.0\) is justified: everything except the motors arrives through two transformers from a remote grid, so the composite \(X/R\) at the board is well under 17.

Part 2 — the selection as things stand. Against the 25 kA unit:

Every row passes, the smallest margin being 20.5% on the making current. On today's system the 25 kA breaker is the correct and economical choice.

Part 3 — with the third transformer. Three in parallel:

The 25 kA breaker now fails:

And 4.7% of breaking margin is not a margin at all: raising the prefault voltage from 1.00 to 1.05 pu, which is a legitimate operating condition, takes the interrupting duty to 25.07 kA and the breaker is over its rating on that row too.

The 31.5 kA unit against the future case:

All rows pass with the smallest margin at 15.6%, which is the accepted lower bound of normal practice.

The recommendation, and its arithmetic. Buy 36 kV, 31.5 kA now. The premium over a 25 kA board is a few per cent of the switchgear cost; replacing an entire indoor board five years after commissioning costs the equipment again plus an outage of the whole substation. The making current is the row that decides it, and it is the row that would have been passed over by anyone checking breaking capacity alone.

A closing check on the motors. Delete them and the future first-cycle current falls from 25.25 to 21.13 kA and the crest from 68.16 to 57.05 kA — inside the 25 kA breaker's ratings. The entire decision turns on 40 MVA of motor load that no supply-side calculation would ever see.