Set 32 — Symmetrical Faults and Short-Circuit Transients
Twenty worked problems on the one fault that leaves the network balanced. Because all three phases collapse together, the per-phase equivalent circuit and the per-unit impedance diagram of Part 1 survive intact, and no new network machinery is needed — which is why the three-phase fault is treated before the unsymmetrical ones of Sets 21 to 23 even though it accounts for barely one fault in twenty. What is new is that the current is neither constant nor sinusoidal. Two decays run at once: the circuit sheds a DC offset with time constant \(L/R\), and the machine raises its own reactance from \(X_d''\) through \(X_d'\) to \(X_d\) as its rotor circuits give up their trapped flux. This sheet works through both, then turns them into the two numbers a switchgear engineer actually buys against — the momentary duty and the interrupting duty.
The three-phase fault keeps the symmetry. Shorting all three phases at one point leaves the network balanced, so one phase still tells the whole story and the positive-sequence impedance alone determines the current. Every other fault destroys that symmetry and needs Set 21's transformation.
The DC offset is the price of continuity. Current in an inductance cannot jump, so closing onto \(\sqrt2 V\sin(\omega t+\alpha)\) forces \(i(t) = (\sqrt2 V/Z)[\sin(\omega t+\alpha-\theta) - \sin(\alpha-\theta)e^{-t/\tau}]\). The exponential exists only to cancel the sinusoid at \(t=0\); it carries no energy from the source and decays with \(\tau = L/R = X/\omega R\).
The closing instant decides everything. \(\alpha=\theta\) gives a pure sinusoid from the first instant; \(\alpha-\theta=\pm90^\circ\) gives full offset. Since \(X\gg R\) makes \(\theta\approx90^\circ\), the worst fault is one struck at a voltage zero — which the fault chooses, not the engineer.
Two multipliers, one parameter. Half a cycle after a fully offset inception, \(i_{peak}=\sqrt2 I(1+e^{-\pi R/X})\) and \(I_{rms}=I\sqrt{1+2e^{-2\pi R/X}}\). Both depend on \(X/R\) alone, and both approach 2 and \(\sqrt3\) in the lossless limit.
A machine has three reactances because it has three rotor circuits. Constant flux linkage keeps the damper and field currents alive; while they hold their flux, their leakage paths shunt \(x_{ad}\). Hence \(X_d''=x_l+(x_{ad}\|x_f\|x_D) \lt X_d'=x_l+(x_{ad}\|x_f) \lt X_d=x_l+x_{ad}\), and the ordering needs no measurement to establish.
Use \(X_d''\) for fault current, \(X_d'\) for stability, \(X_d\) for steady state. A reactance quoted without saying which of the three is meant is useless, and so is one quoted without its MVA base.
Prefault load enters through \(E''\). Set \(E''=V^{(0)}+jX_d''I_L\) for a generator and \(E''=V^{(0)}-jX_d''I_L\) for a motor. The load current then cancels in the total fault current but not in the individual branch contributions.
Fault level and Thevenin impedance are the same fact twice. \(\text{SC MVA} = \text{MVA}_{base}/|Z_{kk}|_{pu}\), so strengthening a bus and making its switchgear expensive are one action, not two. A current-limiting reactor buys fault level back at the cost of a permanent series drop.
One column of \(\mathbf{Z}_{bus}\) solves the whole network. \(I_f = V_k^{(0)}/(Z_{kk}+Z_f)\) and \(V_i = V_i^{(0)} - Z_{ik}I_f\) give every bus voltage and, through \(I_{ij}=(V_i-V_j)/z_{ij}\), every line current. Changing the fault location means reading a different column.
Classify shunt faults by their share of the total, state which of them leaves the network balanced, and justify the practice of rating equipment against the rarest of them. Then give the one common situation in which that practice understates the duty, using \(Z_1 = Z_2 = j0.15\) and \(Z_0 = j0.05\) pu at a solidly earthed generator bus.
The four shunt faults and their frequency on a transmission system at 115 kV and above, where lightning is the dominant cause:
Only the last row keeps the three-fold symmetry. A series fault — a broken conductor, or a breaker pole that fails to open — behaves oppositely: it reduces current in the affected phases and raises the voltage across the break.
Why the rarest fault sets the rating. A bolted three-phase fault collapses all three phase voltages at once, so every source in the system drives current into it through the positive-sequence impedance alone. No sequence network is in series to add impedance:
With \(Z_1=Z_2\) always, the ratio of the two is a single number.
The comparison, in one expression:
On a transmission line \(Z_0 \approx 3Z_1\) — Set 22, Problem 4 — so the ratio is \(3/5 = 0.6\) and the three-phase fault dominates comfortably. The exception is where \(Z_0\) is smaller than \(Z_1\), which happens at a machine, whose zero-sequence reactance is pure stator leakage.
The stated case. At the solidly earthed generator bus, with \(Z_1=Z_2=j0.15\) and \(Z_0=j0.05\) pu:
The earth fault exceeds the three-phase fault by 28.6%. A breaker chosen on the three-phase value alone would be 22% short. This is why generator neutrals are almost never solidly earthed — a resistance or a distribution transformer in the neutral raises \(Z_0\) by \(3Z_n\) and restores the ordering, and it is Set 22, Problem 9 that quantifies the choice.
Two currents, not one, follow from any fault study. The first is the current in the first cycle or two, which a closed breaker and the busbars must survive mechanically and thermally. The second is the current three to eight cycles later, when the relay has decided and the contacts part — that is the current the arc must interrupt:
Between the two instants the current falls for two entirely independent reasons — the circuit sheds its DC transient, and the machine raises its reactance. Problems 2 to 8 take them in turn; Problem 20 puts them back together.
A 132 kV, 50 Hz source of negligible impedance feeds a line of \(4.5+j36\ \Omega\) per phase. A bolted three-phase fault occurs at the far end. Solve the switched \(RL\) circuit from first principles, evaluate every constant, and find both components of the current for a fault struck at \(\alpha = 30^\circ\).
The governing equation. With the shunt capacitance neglected, the faulted line is a series \(R\)–\(L\) branch and the fault is the closing of a switch across it at \(t=0\):
\(V\) is the rms phase voltage and \(\alpha\) the phase angle of that voltage at the instant of closure. Nothing in the circuit chooses \(\alpha\); the insulation failure does.
The general solution is a particular integral — the steady-state sinusoid — plus the natural response of the \(RL\) loop:
The constant is not free. Current in an inductance cannot change instantaneously, so \(i(0^+)=i(0^-)=0\). Setting \(t=0\):
The DC term is not a separate physical current. It exists solely to cancel the sinusoid at \(t=0\), draws nothing from the source, and dies with the circuit's own time constant.
The constants for this line. The phase voltage is \(V = 132000/\sqrt3 = 76210.2\) V, and
The time constant, which governs how long the offset survives:
A little over one and a quarter cycles. Note that \(\tau\) in cycles is \(X/(2\pi R) = 8/6.283 = 1.273\) — the ratio \(X/R\) and the decay in cycles are the same number to within a factor of \(2\pi\), which is worth remembering as a sanity check.
The fault at \(\alpha = 30^\circ\). The angular distance from the natural zero of the steady-state current is
The two are equal and opposite, so \(i(0)=0\) as the inductance demands. The offset here is 80% of its largest possible value — and this was a perfectly ordinary closing angle, not a contrived one.
A DC source shows none of this. Replace the sinusoid by \(V_{dc}\) and the same differential equation gives
— a single rising exponential whose shape does not depend on when the switch closed, because a DC source has no phase. The whole dependence on the closing instant is a consequence of the source being alternating, and it is what makes short-circuit current a statistical quantity rather than a deterministic one.
For the same 132 kV feeder, find the closing instant that produces the largest offset, the first peak of the current, and the rms current at the end of the first half cycle. Compare with the same fault struck at the most favourable instant, and state what the difference implies for a breaker specification.
The worst instant. The offset's initial value is \(-\sqrt2 I\sin(\alpha-\theta)\), largest in magnitude when \(\alpha-\theta=\pm90^\circ\):
In time, \(7.125^\circ\) at 50 Hz is \(0.396\) ms. The fault must strike within four tenths of a millisecond of the voltage zero — and because \(X\gg R\) makes \(\theta\) nearly \(90^\circ\), "worst instant" and "voltage zero" are practically the same event in any transmission circuit.
The current in that case. Putting \(\alpha-\theta=-90^\circ\) into the general result of Problem 2:
A sinusoid riding on a decaying platform. The current never changes sign until the platform has fallen below the sinusoid's amplitude — a fact that Challenge C1 turns into a real interrupting problem.
The first peak occurs where \(\cos\omega t = -1\), half a cycle after inception. The exponent there is worth writing out, because it is where the \(X/R\) dependence comes from:
The rms value at the same instant. An AC component of rms value \(I\) and a DC component of instantaneous value \(i_{dc}\) combine in energy, not in phase:
The favourable instant, for contrast. At \(\alpha=\theta=82.875^\circ\) the closure happens exactly as the steady-state current passes through zero, no offset is required, and the current is a clean sinusoid from the very first instant:
The same fault, on the same feeder, at the same voltage, differs by 67% in first-peak current according to nothing but the microsecond at which the insulation happened to fail.
One correction, in the interest of honesty. The peak does not fall exactly at \(t=T/2\), because the exponential is still falling while the cosine is turning. Differentiating and solving numerically for this circuit puts the true maximum at \(t = 9.73\) ms, where \(i = 1.6788\sqrt2 I\) against the \(1.6752\sqrt2 I\) of the standard formula:
Two parts in a thousand, and always in the conservative direction for the engineer using the formula only if he rounds up. Every switchgear standard uses the \(T/2\) form for exactly this reason.
Tabulate the peak and rms asymmetry factors for \(X/R = 2, 5, 10, 15, 30\) and in the lossless limit. Identify the entry that the factor 1.6 in every switchgear catalogue comes from, and explain why a cable feeder is a milder duty than a generator terminal even at the same fault current.
The two multipliers, from Problem 3, each a function of the single ratio \(X/R\):
Evaluated:
The bold row is the origin of the catalogue figure. Close to a generator \(X/R\) is of order 30, and
The 1.6 used for close-and-latch duty is not a safety margin chosen by a committee; it is this square root, rounded. Problem 20 uses it as it stands.
Why the peak factor rises faster than the rms factor. The peak adds the DC value arithmetically to the AC peak; the rms adds it in quadrature:
At \(u=1\) the first gives 2 and the second \(\sqrt3=1.732\). Mechanical force on a busbar goes as the square of the instantaneous current, so it is \(k_{peak}\) that a support insulator is designed against; thermal duty follows the rms, so it is \(k_{rms}\) that sets the short-time current rating. Two different multipliers for two different failure modes.
The cable feeder against the generator terminal. Take two points with the same symmetrical fault current of 20 kA:
A 24% higher peak force and a 29% higher momentary rms, for identical symmetrical current. Two switchboards quoted at "20 kA fault level" can therefore require quite different equipment, and a specification that omits \(X/R\) is incomplete.
And the direction of the trend is worth internalising. Resistance is the friend here:
A cable's high resistance limits both the symmetrical current and the asymmetry; a generator's low resistance limits neither. It is why the most severe short circuit in a plant is almost always the one nearest the machines, on both counts at once.
A three-phase fault is struck at the instant that fully offsets phase \(a\). Show that the three DC offsets sum to zero at every instant, find the initial offset in each phase, and compute the first peak in each of the three phases for the feeder of Problem 3. Hence explain in what sense a "symmetrical" fault is not symmetrical.
The three closing angles are not independent. The three source voltages differ by \(120^\circ\), and the three phases are closed at the same physical instant, so
One instant, three different phase angles. The engineer gets no choice about any of them, and no choice about their spacing either.
The offsets sum to zero identically. The offset in phase \(k\) is \(-\sqrt2 I\sin(\alpha_k-\theta)e^{-t/\tau}\), and the three sine terms are a balanced set:
Because the three share one time constant, the sum is zero not just at \(t=0\) but permanently. This is the circuit-level reason a three-phase fault with no earth connection draws no residual current even during the transient.
The stated instant. Phase \(a\) fully offset means \(\psi = -90^\circ\), so
Sum \(= 1.0-0.5-0.5 = 0\). At most one phase can be fully offset, and when it is, the other two carry exactly half the offset, in the opposite direction.
The first peak in each phase for the feeder of Problem 3, where \(\sqrt2 I = 2970.7\) A and \(\tau = 25.465\) ms. Only phase \(a\) peaks at the half cycle; the other two peak elsewhere and must be found by searching the first cycle:
A spread of 29% between the largest and the smallest phase current, in a fault that is by definition perfectly balanced. The \(1.679\) for phase \(a\) is the true maximum discussed in Problem 3; the standard \(T/2\) formula gives \(1.675\).
In what sense the fault is symmetrical. Only in its steady-state component:
The word "symmetrical" in the phrase symmetrical fault refers to the fault's geometry — all three phases shorted at one point — and to the AC current it drives, not to the instantaneous currents in the first cycles. Nothing in the per-phase analysis of Problems 11 to 18 is invalidated by this, because that analysis computes only the AC component; the asymmetry is restored afterwards by the multipliers of Problem 4.
The practical consequence. Because the phase carrying the largest offset changes with the instant of fault inception, breaker duty is quoted for the worst phase and the worst instant, not for an average:
All three poles of a breaker are identical, so all three are designed for the one that will be unlucky. That is a 29% overdesign on two poles out of three, and it is unavoidable.
A synchronous machine has armature leakage reactance \(x_l = 0.15\), direct-axis magnetising reactance \(x_{ad} = 1.05\), field leakage \(x_f = 0.12\) and damper leakage \(x_D = 0.06\), all per unit on its own rating. Derive the three direct-axis reactances from the constant flux linkage theorem, evaluate them, and state which calculation each belongs to.
The physical argument. The flux linkage of a closed circuit of zero resistance cannot change instantaneously, and in a circuit of small resistance it changes only with that circuit's own time constant. A synchronous machine has three magnetically coupled circuits on the direct axis, and two of them are closed loops of low resistance:
At the instant of short circuit the armature current jumps and its mmf tries to demagnetise the air gap. Each rotor circuit responds with an induced current that holds its own flux linkage at the prefault value, and that forces the armature flux out into leakage paths.
The circuit consequence. Seen from the armature terminals, \(x_{ad}\) is shunted by the leakage reactance of every rotor circuit still holding its flux. Three stages follow, in the order the rotor circuits give up:
Adding a branch in parallel can only reduce a reactance, so \(X_d'' < X_d' < X_d\) is automatic and needs no measurement to establish. It is a statement about circuit topology, not about any particular machine.
The parallel combinations, evaluated:
The three reactances:
All three sit inside the usual ranges: \(X_d''\approx0.10\)–0.25, \(X_d'\approx0.15\)–0.40, \(X_d\approx1.0\)–2.0 pu. The factor of six from first to last is the whole reason a fault current decays even with the terminal voltage held constant.
Notice what dominates each stage. The damper, with the smallest leakage, dominates the subtransient combination:
Which is why a machine with no damper winding — some hydro sets, and salient-pole machines without amortisseur bars — has \(X_d''\) essentially equal to \(X_d'\), and shows no subtransient stage on its oscillogram at all. Two of the three currents then coincide.
Which calculation each belongs to, and it is not interchangeable:
Using \(X_d'\) where \(X_d''\) belongs understates the fault current by 27% here; using \(X_d\) understates it by 84%. A number quoted without saying which of the three is meant, and on what MVA base, is worthless.
A 60 MVA, 13.8 kV generator running on no load at rated terminal voltage is subjected to a bolted three-phase terminal fault. With the DC offset stripped from the trace, the symmetrical envelope gives an initial rms current of 12.60 kA, a transient value extrapolated back to \(t=0\) of 6.30 kA, and a steady value of 1.40 kA. Find the three reactances in ohms and per unit, check the per-unit arithmetic, and find the largest instantaneous current if \(X/R = 30\).
How the trace is read. Plot the symmetrical envelope on semi-logarithmic paper and strip one exponential at a time:
Each subtraction isolates one exponential, exactly as in any multi-exponential decay. The quoted "transient value extrapolated to \(t=0\)" is the intercept of the second line, not a value the current ever actually has.
The machine is on no load, so the prefault terminal voltage is the rated value and the internal emf equals it:
This is the one condition in which the naive model — a fixed emf behind a reactance — is exactly right, which is why reactance tests are done on open circuit.
Each reactance from its own current, by \(X = V_t/I\):
In per unit on the machine rating:
The check. A per-unit fault current on a machine at rated voltage must be the reciprocal of the per-unit reactance:
The two agree, so no base has been dropped. Do this check on every machine before its reactance is entered into a study — it costs one division and catches the single commonest data error in the subject.
The largest instantaneous current. The AC peak at inception is \(\sqrt2 I''\), and the worst phase adds a nearly full offset. With \(X/R=30\) at the machine terminals:
The lossless bound would be \(2\sqrt2(12.60)=35.64\) kA, so the machine's own armature resistance removes 5% of the worst case and no more. Rounding the 1.6192 to the catalogue 1.6 gives \(20.16\) kA, a 1.2% difference — within the tolerance of every published reactance.
The machine of Problem 7 has \(T_d'' = 35\) ms and \(T_d' = 1.1\) s. Write the envelope of the symmetrical armature current, evaluate it at \(t = 50\) ms, 0.5 s and 2 s, and identify which of the three currents governs a three-cycle breaker, an eight-cycle breaker and the thermal rating of the busbars.
The envelope. Each stage is entered and left exponentially, with a time constant set by the rotor circuit that is decaying, and the three superpose:
Read the bracket as a reciprocal reactance that grows with time. At \(t=0\) it telescopes to \(1/X_d''\); once \(t\gg T_d''\) but \(t\ll T_d'\) it is \(1/X_d'\); long after \(T_d'\) only \(1/X_d\) survives.
The coefficients, in per unit with \(V_t = 1.0\):
The check at \(t=0\): \(2.5097+1.9520+0.5577 = 5.0194 = 1/X_d''\), as it must.
At \(t = 50\) ms — two and a half cycles, the earliest a fast breaker's contacts can part:
Already down to 60% of \(I''\). The subtransient term has lost three quarters of its value while the transient term has barely moved — the separation of time constants is what makes the two stages distinguishable at all.
At \(t = 0.5\) s and \(t = 2\) s:
A factor of nine from the first instant to the steady state, with the terminal voltage held constant throughout. Nothing external changed; the machine simply stopped screening its own air gap.
Which current governs which duty:
Standards do not ask the engineer to evaluate this envelope; they ask for \(I''\) and supply a multiplier that falls as the breaker gets slower. The envelope is what the multiplier encodes, and knowing that is what makes a table of multipliers intelligible rather than arbitrary.
The thermal check, since it is the one calculation the multiplier does not cover. A one-second withstand rating \(I_{th}\) must satisfy
and with a fault cleared in 160 ms the integral here evaluates to \(1.44\) pu²·s — an equivalent constant current of \(\sqrt{1.44/0.16}=3.00\) pu — against a typical 1 s rating of 25 kA, which on the 2.51 kA base is \(9.96^2 = 99\) pu²·s. Nearly two orders of margin — thermal duty is almost never the binding constraint on a transmission busbar, and almost always is on a cable screen.
A 25 MVA, 11 kV generator with \(X_d''=0.15\) pu is delivering 20 MVA at 0.85 power factor lagging with its terminals at 1.0 pu when a bolted three-phase fault occurs at those terminals. Compute \(E''\), then the subtransient fault current both with and without the prefault load current, and state the error incurred by neglecting it.
The model. During the subtransient period the machine is a constant emf \(E''\) behind a constant reactance \(jX_d''\), with \(E''\) chosen so that the model reproduces the prefault terminal conditions exactly:
The sign difference is nothing more than the sign convention for \(I_L\), taken positive out of a generator and into a motor. Once \(E''\) is fixed, the machine is an ordinary source and the network is linear.
The prefault current, on the machine's own 25 MVA base with the terminal voltage as reference:
The internal emf:
Note that the lagging load has raised the internal emf 6.8% above the terminal voltage. That excess is exactly what the extra fault current comes from.
The fault current with the load included. The terminal bus goes to zero potential, so the emf drives its own reactance alone:
The real part is exactly \(I_L\)'s real part — the load current is still flowing, superposed on the fault current, and the imaginary part is the pure-fault component.
Without the prefault current, the flat-voltage assumption:
In amperes, which is what the breaker sees:
A difference of 595 A, or six per cent. Small in the context of switchgear steps, which come in ratings of 16, 20, 25, 31.5 kA — but not small enough to ignore when a machine is loaded to its limit at a poor power factor, which is exactly when a fault is most likely.
Why the neglect is usually defensible, and where it stops being so:
Adding a unit vector to a vector six times longer and at right angles to it changes the magnitude by only \(\sqrt{1+0.12^2}-1 = 0.7\%\). The 6.4% found above is larger than that because the load is not quite in quadrature — the reactive part of \(I_L\) adds directly to the fault current. A leading load would have reduced the fault current instead, which is the sign check worth carrying: lagging load raises \(E''\), leading load lowers it.
An 11 kV industrial bus is supplied from a 33 kV system of 500 MVA fault level through two 10 MVA, 33/11 kV transformers of 8% reactance in parallel. Connected to the bus are induction motors totalling 8 MVA with a locked-rotor reactance of 0.20 pu on their own rating. Find the subtransient fault current at the bus with and without the motors, and state how long the motor contribution lasts.
Base and conversions. Take 10 MVA and 11 kV at the industrial bus. The supply system's reactance follows from its fault level, which is the reciprocal statement of Problem 13:
Why the motors appear at all. During the fault the bus voltage collapses, the motor's rotor flux does not — it decays with the rotor's own time constant — so the machine's internal emf momentarily exceeds its terminal voltage and it feeds the fault instead of drawing from it:
An induction motor is represented as an emf behind its locked-rotor reactance for exactly that reason — locked rotor being the condition in which it draws its starting current, which is what it now delivers.
Without the motors, a single series path from the 33 kV system:
With the motors, which sit directly on the faulted bus and therefore add in parallel with everything upstream:
In amperes and MVA:
The motors add 24% to the first-cycle current. Neglecting them would not merely be inaccurate — with 11 kV switchgear commonly rated at 250 MVA it could be the difference between an adequate board and one that is not.
How long it lasts, and what that means for the two duties:
Standards handle this by applying a multiplier to the motor reactance rather than by re-solving the network: typically \(1.0\times X''\) for the momentary calculation and \(1.5\times X''\) or infinity for the interrupting calculation, according to motor size. Here the interrupting duty falls back to the 16.667 pu of the supply alone.
A sanity check on the motor figure. A locked-rotor reactance of 0.20 pu means a starting current of five times full load:
Identical to the 2.10 kA above. The rule of thumb "motors contribute their starting current" is not a rule of thumb at all — it is the same calculation stated without per unit.
A 40 MVA, 13.8 kV generator with \(X_d''=0.18\) pu supplies, through a 40 MVA 13.8/132 kV transformer of 0.09 pu, a 132 kV line of \(25\ \Omega\) and a 40 MVA 132/13.2 kV transformer of 0.09 pu, a 20 MVA 13.2 kV synchronous motor with \(X_d''=0.18\) pu on its own rating. The motor is drawing its rated 20 MVA at 0.8 lagging with its terminals at 1.0 pu when a bolted three-phase fault strikes those terminals. Find the fault current and each machine's contribution, by the direct route and by superposition, and reconcile the two.
The base and the conversions. Take 40 MVA throughout, with 13.8 kV in the generator zone, 132 kV on the line and 13.2 kV in the motor zone. The transformer ratios define the zones, so only the motor and the line need converting:
The prefault current, with the motor terminal voltage as reference and the current taken into the motor:
The two internal emfs. The motor absorbs \(I_L\), the generator delivers it — hence the opposite signs:
The motor's internal emf is below 1.0 pu and the generator's above it — the difference between the two drives the load current through the intervening 0.777 pu of reactance.
The direct route. With the motor bus at zero potential, each machine drives its own emf through its own reactance into the fault:
The real parts are \(\pm0.4\) and cancel exactly — that is the load current, still circulating between the two machines and contributing nothing to the fault.
The superposition route. Kill both sources, insert \(-V_k^{(0)}\) at the fault point, and the passive network is two reactances in parallel:
The reconciliation. The totals are identical, as they must be; the individual contributions are not, and restoring the load current branch by branch fixes them:
Restoring means \(I_G''=I_{G0}+I_L\) and \(I_M''=I_{M0}-I_L\), with \(I_L\) in its own direction. The fault current is exact either way; the branch currents are wrong by 11–12% if the restoration is skipped — which matters, because it is branch currents that relays measure.
In amperes, on the 13.2 kV motor zone:
Ten times the prefault load current, which is the ratio that makes the flat-voltage approximation tolerable for fault-current purposes and intolerable for branch-current purposes at the same time.
Two generators — 30 MVA with \(X''=0.15\) pu and 45 MVA with \(X''=0.20\) pu, both 11 kV — operate in parallel on a common bus. Find the short-circuit MVA at that bus on a 100 MVA base, the fault current in kA, and the contribution of each machine. Then show that the fault MVA could have been written down without choosing a base at all.
Convert both to the common base. A per-unit reactance moves between bases as the ratio of the MVA bases, the voltage base being unchanged here:
The smaller machine ends up with the larger per-unit reactance on the system base, which is the usual outcome and the usual source of confusion.
The Thevenin impedance at the bus is the two in parallel:
The fault level and the current:
Cross-check from the line quantities directly: \(\sqrt3\times11\times22.31 = 425\) MVA. The two routes agree, which confirms the base current.
Each machine's contribution:
The 45 MVA machine contributes more current despite its higher per-unit reactance, because per unit is measured against its own larger rating. Per-unit values are never comparable between machines until they are on a common base.
Without choosing a base at all. The fault MVA a machine can deliver into a bolted fault at its own terminals is its rating divided by its per-unit reactance on that rating:
Fault levels in parallel add; impedances in parallel do not. That is the whole reason fault level is a more convenient currency than impedance for anything connected in parallel, and impedance the more convenient one for anything in series.
Where the shortcut fails, and it fails often enough to state:
Insert even a small reactor between the two machines and the addition is wrong, because the two contributions no longer see the same fault point. Problem 15 is precisely that case, and the fault level there is not the sum.
A 33 kV bus has a three-phase fault level of 900 MVA. Find its Thevenin impedance in per unit on a 100 MVA base and in ohms, the bolted fault current in kA, and the voltage drop the bus suffers when a 40 MVA load at 0.9 lagging is switched onto it. Explain why a strong bus and cheap switchgear are incompatible requirements.
Fault level and Thevenin impedance are reciprocal descriptions of one fact. With flat prefault voltage,
Neither number contains information the other lacks. Which one is quoted is a matter of who is asking: a planner asks for MVA, a relay engineer for ohms.
In ohms:
The fault current, both ways:
The load-switching drop, which is the same impedance doing its everyday work. A 40 MVA load is 0.4 pu, at \(\cos\phi=0.9\) so \(\sin\phi=0.4359\). For a purely reactive source impedance the drop in magnitude is dominated by the reactive component of the current:
A two-per-cent step every time that load is switched. On a 300 MVA bus the same load would cause 5.8%, which is outside most supply-quality limits — this is the calculation that decides whether a large motor can be started direct-on-line.
The incompatibility, stated numerically. Both quantities move together and in the same direction:
A bus that holds its voltage well against load switching and motor starting is, by the same arithmetic, one that demands expensive switchgear. There is no arrangement of the network that improves both, because they are the same impedance read in two directions.
Which is why the compromise is bought, not designed. The engineer chooses the fault level he wants and then either accepts the switchgear it implies, or inserts a current-limiting reactor and accepts the regulation it costs:
Problem 14 costs the second option out, and Problems 15 and 16 compare two ways of arranging the reactors so that the penalty is paid only during the fault.
The fault level at the 33 kV bus of Problem 13 is to be reduced from 900 MVA to 600 MVA by a series current-limiting reactor. Find its reactance in per unit and in ohms, its continuous current rating for the 40 MVA load, the reactive power it absorbs, and the additional voltage regulation it imposes. State the one thing a reactor cannot do.
The target impedance follows from the target fault level:
The reactor is the difference, because it sits in series with the existing source impedance:
Note that a 50% reduction in fault level required only a 50% increase in impedance, not a doubling of it — reactor sizing looks cheap in impedance terms and expensive in every other term.
The check:
The continuous rating. The reactor carries the whole load current all its life, not just during faults:
The reactor must be insulated for the system voltage but is rated for this drop — which is why a current-limiting reactor is physically small compared with a transformer of the same current rating.
What it costs, in the two currencies that matter:
Just under one per cent of extra regulation, permanently, and 0.89 MVAr of reactive absorption that has to come from somewhere. The regulation penalty is the reason the tie-bar arrangement of Problem 16 exists: it places the reactors where load current does not normally flow.
What a reactor cannot do. It limits the current a fault draws; it does not limit the energy the arc dissipates before clearing, nor does it help at all with a fault on its source side:
And it raises \(X/R\): the added reactance comes with almost no resistance, so the asymmetry factors of Problem 4 go up even as the symmetrical current comes down. Here \(X/R\) rises from 15 to 22.5, and the peak multiplier from 1.811 to 1.870 — a 3% partial recovery of what the reactor just removed, and it must be carried through to the momentary duty in Problem 20.
Three 15 MVA, 11 kV generators with \(X''=0.15\) pu each feed their own busbar section, and the three sections are joined in a ring by three identical reactors of 0.10 pu on 15 MVA. Find the fault MVA and fault current for a three-phase fault on section 1, identify every contribution, and state what the reactors bought. Then find what happens when a fourth identical set is added to the ring.
The arrangement. Each generator is tied to its own section; the sections are connected section-to-section by reactors, forming a closed ring. A fault on one section is fed by its own machine directly and by the other two through reactors.
The symmetry that makes it tractable. All three emfs are equal, so their internal nodes may be joined into one reference node. Sections 2 and 3 then stand in identical relation to section 1, so they are at the same potential and
A reactor that carries no current may be deleted from the circuit. This is the step that turns a mesh into two series paths, and it works only because the machines are identical — a point Challenge C2 returns to.
The three paths from the reference node to section 1:
The fault level and current:
Cross-check: \(\sqrt3\times11\times11.547 = 220\) MVA.
Every contribution accounted for:
The faulted section's own machine supplies 45% of the current and is not limited at all — no reactor stands between it and the fault. That is the ring system's structural weakness and it cannot be designed away.
What the reactors bought. With the three sections solidly bussed, the three machines are simply in parallel:
Three reactors of 0.10 pu bought a 27% reduction. Not much, and the reason is the unlimited local contribution — no amount of reactance in the ring can take the fault level below the 100 MVA that \(G_1\) alone delivers.
Adding a fourth set. With four sections in a square ring the symmetry argument no longer deletes anything, and the mesh must be solved. Nodal analysis with the four generator branches as shunts to the reference gives
Adding 15 MVA of generation — a third more plant — raised the fault level by only 10.7%, because the new machine reaches the fault through two reactors, one at each end of its route round the ring. The ring's fault level grows sub-linearly with the number of sections, and that, rather than the modest reduction above, is the arrangement's real merit.
The same three generators are now connected in the tie-bar arrangement: each section reaches a common tie-bar through its own 0.10 pu reactor, and no section connects directly to another. Find the fault level on section 1, compare it with the ring, work out what happens when a fourth set is added, and state the operational drawback that the numbers do not show.
The arrangement. The tie-bar is a bus that carries no generation and no load of its own; it exists only as a meeting point. Every section reaches every other section through two reactors instead of one.
The reduction. Join the three equal emfs into one reference node. Sections 2 and 3 reach the tie-bar through \(0.15+0.10 = 0.25\) pu each:
The extra 0.10 is section 1's own reactor, which the remote machines must also pass through. That second reactor is the whole difference from the ring.
In parallel with the local machine:
Against the ring, for identical machines and identical reactors:
The same three reactors, differently arranged, give 220 MVA or 166.7 MVA. The tie-bar is 24% better for no extra equipment — the remote contributions each pass through two reactors instead of one, while the local machine's unlimited path is unchanged in both.
Adding a fourth set. Three remote sections now reach the tie-bar in parallel:
Against the ring's \(+10.7\%\) for the same addition. Both arrangements grow sub-linearly, the tie-bar slightly more slowly, and both approach the same asymptote — the fault level can never fall below the local machine's own 100 MVA however many reactors are installed.
The economy that does not appear in the impedance table. A tie-bar reactor normally carries only the interchange between one section and the rest, which on a station with equally loaded machines is close to zero:
Less current means less \(I^2X\) loss, less reactive absorption and less voltage drop in normal running — which is why the tie-bar arrangement dominates modern station design and the generator reactor has almost vanished. Challenge C2 puts numbers on the comparison.
The operational drawback. The tie-bar is a single point through which every inter-section transfer must pass:
A ring degrades to a chain; a tie-bar degrades to an island. The usual answer is to duplicate the tie-bar or to section it with a bus coupler, which restores the redundancy at the cost of the extra switchgear the arrangement was chosen to avoid.
Station A has two 60 MVA, 11 kV generators of \(X''=0.16\) pu on a common bus feeding a 120 MVA, 11/132 kV transformer of 0.10 pu. Station B has one 100 MVA, 11 kV generator of \(X''=0.20\) pu feeding a 100 MVA, 11/132 kV transformer of 0.10 pu. The two 132 kV buses are joined by a line of \(30\ \Omega\). Find the fault level and fault current for a bolted three-phase fault at each 132 kV bus, with the contribution of each station, on a 100 MVA base.
Every reactance onto the 100 MVA base. The transformer ratios define the voltage zones, so only the MVA bases and the line's ohms need attention:
Note that station A's transformer, being rated above the generators it serves, converts to a smaller per-unit value than its nameplate 10% — a place where sign-of-the-correction errors are common.
The two station branches, each seen from its own 132 kV bus:
Two sources behind two impedances, joined by one line — the smallest network in which the fault location genuinely matters.
Fault at bus A. Station A feeds it directly; station B feeds it through the line:
In amperes, and split between the stations:
Check: \(4.615+2.118 = 6.733\), and \(\sqrt3\times132\times2.945 = 673.3\) MVA. Station A supplies 69% of the fault, having only 55% of the plant — the line's reactance is what makes the difference.
Fault at bus B, the mirror calculation:
Bus B is the weaker of the two by 12%, because its local plant is 100 MVA behind 0.30 pu against A's 120 MVA behind 0.2167 pu.
What the line contributes, which is the question a planner actually asks. Remove it and each bus is fed by its own station alone:
A single 30 Ω interconnector raises the weaker bus's fault level by more than three quarters. Every fault study must therefore state the switching state it assumes, and switchgear must be rated for the strongest credible configuration — usually all lines in, all plant running.
And the reverse question, worth asking once. If bus B's switchgear were rated at 500 MVA, the interconnector as it stands is inadmissible. The remedy is a series reactor in the line:
That is \(0.211\times174.24 = 36.8\ \Omega\), more than the line itself — which tells the planner immediately that reactoring is the wrong answer here, and that the switchgear should be uprated instead.
A three-bus 132 kV system on a 100 MVA base has generators at buses 1 and 2 behind \(j0.20\) and \(j0.30\) pu, and lines \(z_{12}=j0.25\), \(z_{13}=j0.15\), \(z_{23}=j0.20\) pu. Build \(\mathbf{Z}_{bus}\), verify two of its entries independently, and solve a bolted fault at bus 3 completely — fault current, every bus voltage, every line current, and both consistency checks. Then read off the fault level at all three buses.
Form \(\mathbf{Y}_{bus}\) with the generator branches as shunts to the reference, and invert. Doing so gives
Symmetric, as any passive reciprocal network's must be, and every diagonal entry larger than the off-diagonals in its row — both are cheap structural checks worth making before any arithmetic depends on the matrix.
Verify \(Z_{33}\) independently, because a wrong \(\mathbf{Z}_{bus}\) invalidates everything after it. Convert the delta of lines 1–2–3 (values 0.25, 0.15, 0.20; sum 0.60) into a star:
And \(Z_{11}\). Bus 3 carries no source, so \(Z_3'\) dangles and carries no current when bus 1 is driven; delete it:
Two entries confirmed by hand from two different routes. The remaining seven follow from symmetry and from the same star network, and the reader who wants a third check will find \(Z_{13}\) from the current division \(0.383333/0.645833 = 0.59355\) times the 0.20 generator branch, giving 0.118710.
The fault at bus 3, bolted, with flat prefault voltages:
The off-diagonal entries measure how far the disturbance travels: bus 1 is held up slightly better than bus 2 despite being electrically nearer the fault, because it has the stiffer generator behind it.
The line currents, from the bus voltages and the primitive line impedances — not from the matrix:
All currents are \(-j\) times these magnitudes, the network being purely reactive; the magnitudes are quoted for readability.
Both consistency checks. First, everything arriving at the faulted bus must equal \(I_f\):
Second, Kirchhoff's current law at each healthy bus:
Both pass. A fault study that does not close on these two checks has an error in the matrix, in the primitive impedances, or in the sign of a line current — and it is far quicker to find it here than in the results.
Every bus at once, which is the whole point of the method. Read a different diagonal entry:
With \(I_{base}=437.39\) A at 132 kV. Bus 1, with the stronger generator directly on it, is 49% more severe than bus 3, which has no generation at all. Changing the fault location meant reading a different column, not repeating a calculation — and for a five-hundred-bus system that is the difference between a study and an impossibility.
Repeat the bus-3 fault of Problem 18 three more ways: through an arcing fault impedance \(Z_f = j0.04\) pu; with a prefault voltage of 1.05 pu instead of 1.0; and with both. Quantify each effect separately, verify the faulted-bus voltage two ways, and say which assumption is the dangerous one to get wrong.
Both assumptions enter the same formula, one in the numerator and one in the denominator:
So prefault voltage scales the answer and fault impedance shifts its denominator. They are independent, and their effects are of opposite sign — which is why they are so often assumed to cancel, and why they do not.
Fault impedance alone, with \(Z_{33} = 0.205806\) pu:
An arc impedance of only 0.04 pu — about \(7.0\ \Omega\) at 132 kV — removes a sixth of the fault current, because it is a fifth of the Thevenin impedance it adds to.
The faulted-bus voltage, computed two ways as the check that the fault impedance has been entered consistently:
The faulted bus sits at 16% of nominal rather than at zero, and every other bus is held higher too. This is precisely why an impedance fault is harder for a distance relay to see than a bolted one — the measured impedance includes the arc, and the voltage the relay sees has not collapsed.
Prefault voltage alone. Being a pure scaling of the numerator:
Exactly 5%, and it must be — the network is linear and the fault current is proportional to the driving voltage. Fault studies for breaker selection commonly use 1.05 or 1.1 pu for this reason: the largest credible prefault voltage gives the largest credible duty.
All four cases together:
The two effects do not cancel: 5% up and 16% down leaves 12% down, and the combined case is nowhere near the bolted flat-voltage one. Fault MVA here is \(|V^{(0)}||I_f|\times\text{MVA}_{base}\), which is why the 1.05 pu bolted case shows 535.7 rather than 510.2.
Which assumption is dangerous. They fail in opposite directions, and only one of them fails unsafely:
For rating equipment, a bolted fault at the highest operating voltage is the right pessimism. For setting protection, the pessimism reverses: the relay must still see the smallest credible fault current, which means an arcing fault at the lowest operating voltage with the weakest plausible generation. The same network gives two studies, and using one study's numbers for the other's purpose is the error this problem exists to prevent.
A note on scale. Arc impedance is not negligible on distribution systems and is nearly so at EHV, because it is a fixed number of ohms against a base that grows as the square of the voltage:
Sixteen times more significant at 33 kV than at 132 kV. This is the arithmetic behind the rule that arc resistance may be neglected in transmission fault studies and must not be in distribution earth-fault settings.
A breaker is to be chosen for the 33 kV bus of Problems 13 and 14, where \(I''=15.75\) kA and \(X/R = 15\). The candidate is rated: nominal 33 kV, rated maximum voltage 36 kV, \(K=1.0\), rated short-circuit current 16 kA, three-cycle interrupting time, peak withstand 2.5 times the rated short-circuit current. Assess it before and after the reactor of Problem 14 is installed, and state the decision.
The two duties the breaker must meet, from Problem 1's two instants and Problem 4's two multipliers:
The momentary duty is a mechanical and thermal stress on a closed breaker; the interrupting duty is what the arc must clear. Different failure modes, different numbers, and a breaker can pass one and fail the other.
Before the reactor, with \(X/R = 15\):
The breaker's capability at the 33 kV operating voltage is the smaller of the inverse-voltage value and the \(K\)-factor ceiling:
A modern breaker with \(K=1\) gains nothing from operating below its rated maximum voltage. Older designs with \(K=1.2\) or more did, and the ceiling exists because the interrupter's dielectric recovery, not its arc energy, eventually becomes the limit.
The verdict before the reactor:
It fails on the peak, and the interrupting margin is inside the uncertainty of the machine reactances themselves. Notice which duty failed: the symmetrical calculation said 15.75 against 16 and looked adequate, and only the asymmetry factor exposed the problem. A study that stops at the symmetrical current has not finished.
After the reactor of Problem 14, which cut the fault level to 600 MVA. The reactor adds reactance and almost no resistance, so \(X/R\) rises:
The verdict after:
Both duties pass with room to spare, which is what a switchgear engineer wants — the fault level will only rise as plant is added, and a margin bought once is cheaper than a replacement later.
The decision, and its alternative. Two ways to reach a compliant installation:
A 25 kA board would cover 15.75 kA interrupting and \(2.5\times25 = 62.5\) kA peak against 40.33 kA required, with margin for future growth and no permanent loss. On a bus with many bays the reactor usually wins; on one with two or three, the uprated switchgear does — and the calculation above is what the comparison rests on.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.
P1. An 11 kV, 50 Hz source of negligible impedance feeds a line of \(0.5+j5\ \Omega\) per phase. Find the symmetrical rms current for a bolted three-phase fault at the far end.
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\(V=6350.9\) V, \(Z=5.0249\ \Omega\), \(I=\mathbf{1264}\) A. Problem 2.P2. For the same line, find \(\tau\), the worst-case first peak and the first half-cycle rms.
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\(\tau = \mathbf{31.83}\) ms, \(e^{-\pi/10}=0.7304\); \(i_{peak}=\mathbf{3093}\) A, \(I_{rms}=\mathbf{1817}\) A. Problem 3.P3. At what closing angle does that same fault produce no DC offset at all?
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\(\alpha=\theta=\tan^{-1}10 = \mathbf{84.29^\circ}\) — the instant at which the steady-state current would pass through zero. Problem 3.P4. A machine on open circuit at rated voltage has \(X_d''=0.12\), \(X_d'=0.20\), \(X_d=1.50\) pu. Give the three symmetrical currents in per unit.
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\(I''=\mathbf{8.333}\), \(I'=\mathbf{5.000}\), \(I=\mathbf{0.667}\) pu — a factor of 12.5 from first to last. Problems 7 and 8.P5. A 100 MVA, 15 kV generator has \(X''=0.16\) pu. Find the fault MVA and current for a bolted terminal fault.
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\(100/0.16=\mathbf{625}\) MVA; \(625\times10^6/(\sqrt3\times15000)=\mathbf{24.06}\) kA. Problem 12.P6. Two generators, 50 MVA with \(X''=0.20\) and 75 MVA with \(X''=0.25\), share a bus. Fault MVA?
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\(50/0.20+75/0.25 = 250+300 = \mathbf{550}\) MVA. Fault levels on a common bus add. Problem 12.P7. A 132 kV bus has a fault level of 1500 MVA. Find its Thevenin impedance in ohms and the bolted fault current.
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\(132^2/1500 = \mathbf{11.62}\ \Omega\); \(I_f = \mathbf{6.56}\) kA. Problem 13.P8. A reactor is to reduce that bus from 1500 MVA to 1000 MVA. Find its reactance in ohms.
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\(132^2/1000 - 132^2/1500 = 17.424-11.616 = \mathbf{5.81}\ \Omega\). Problem 14.P9. Find the peak and rms asymmetry factors at \(X/R = 12\).
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\(k_{peak}=1+e^{-\pi/12}=\mathbf{1.770}\), \(k_{rms}=\mathbf{1.478}\). Problem 4.P10. A bus whose supply contributes 20 kA also carries 6 MVA of 11 kV induction motors that start at five times full load. Find the first-cycle total.
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Motor full load 314.9 A, contribution \(5\times314.9=1.57\) kA; total \(\mathbf{21.57}\) kA. Problem 10.P11. A generator with \(X''=0.20\) pu delivers 1.0 pu at 0.8 lagging with terminals at 1.0 pu. Find \(E''\).
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\(E''=1.0+j0.2(0.8-j0.6)=1.12+j0.16 = \mathbf{1.1314\angle8.13^\circ}\) pu. Problem 9.P12. A bus has \(Z_{kk}=j0.16\) pu on a 100 MVA base. Find the fault current through \(Z_f=j0.04\) pu, and the bolted fault MVA.
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\(I_f = 1/(0.16+0.04)=\mathbf{5.00}\) pu; bolted fault level \(100/0.16=\mathbf{625}\) MVA. Problems 18 and 19.
Challenge Problems
Three problems that need an idea rather than a formula — the corners of symmetrical-fault work where the standard recipe gives an answer that is wrong for a reason worth understanding.
C1 — The current that has no zero. A circuit breaker interrupts only at a current zero. For the machine of Problems 7 and 8 (\(I''=5.0195\), \(T_d''=35\) ms, \(T_d'=1.1\) s) faulted at its terminals with an armature time constant \(T_A = 0.25\) s, determine whether a current zero exists in the worst phase during the first ten cycles. Explain the mechanism, and state what is done about it.
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The condition for a zero crossing. The instantaneous current in the worst phase is the AC component plus the DC offset:
\[ i(t) = \sqrt2\,I_{ac}(t)\sin(\omega t - \tfrac{\pi}{2}) + \sqrt2\,I''e^{-t/T_A} \]A zero exists in a given cycle only if the sinusoid's amplitude reaches down past the platform:
\[ I_{ac}(t) \;\ge\; I''e^{-t/T_A} \]The trap is that both sides decay. In a passive circuit the AC component is constant and only the DC decays, so a zero always appears within a cycle or two. At a machine's terminals the AC component decays too — and faster, because \(T_d''=35\) ms is far shorter than \(T_A=250\) ms:
\[ \begin{array}{lccc} t & I_{ac}(t)\ (\text{pu}) & I''e^{-t/T_A}\ (\text{pu}) & \text{zero?} \\ \hline 0 & 5.019 & 5.019 & \text{marginal} \\ 20\ \text{ms} & 3.892 & 4.634 & \text{no} \\ 60\ \text{ms} & 2.858 & 3.949 & \text{no} \\ 100\ \text{ms} & 2.484 & 3.365 & \text{no} \\ 150\ \text{ms} & 2.296 & 2.755 & \text{no} \\ 200\ \text{ms} & 2.194 & 2.255 & \text{no} \\ 250\ \text{ms} & 2.115 & 1.847 & \textbf{yes} \end{array} \]Solving \(I_{ac}(t) = I''e^{-t/T_A}\) numerically puts the first zero at \(t = 209\) ms — 10.4 cycles. For the whole of the first ten cycles the current is one-signed and the breaker, however fast, cannot interrupt.
The physical picture. The armature's DC offset is a trapped flux linkage in the stator circuit, and it decays with the stator resistance, which is very low. The AC component decays with the rotor time constants, which are shorter. So the machine sheds its AC faster than the circuit sheds its DC, and for a while the total never crosses zero. This is the classic generator-circuit-breaker problem and it has nothing to do with the breaker.
What is done about it. Four measures, in increasing order of cost:
\[ \begin{array}{ll} 1 & \text{Delay tripping deliberately, so the contacts part after the first zero} \\ 2 & \text{Use a generator circuit breaker rated for delayed current zeros (IEEE C37.013)} \\ 3 & \text{Add resistance — an interrupter whose arc voltage forces a zero} \\ 4 & \text{Site the breaker on the HV side of the unit transformer, whose resistance shortens } T_A \end{array} \]The last is the usual answer in practice and explains a piece of station layout that otherwise looks arbitrary: many unit-connected generators have no breaker at their terminals at all. The transformer's resistance is in the fault path for any fault beyond it, \(T_A\) collapses, and the problem disappears.
The general lesson. Every number in this sheet is an envelope, and the envelope is enough for rating a breaker's capability. Whether the breaker can operate at all is a question about the waveform, and the waveform can behave in ways the envelope never reveals.
C2 — Where to spend the reactance. A 33 kV station has four 25 MVA generators, each \(X''=0.20\) pu, on a common bus. The switchgear is rated 350 MVA. Compare a generator-reactor scheme with a tie-bar scheme: find the reactor size each needs, the total reactance installed, and the operating penalty each imposes. Then explain why feeder reactors are not an answer at all.
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The present fault level, on a 25 MVA base:
\[ Z_{th} = \frac{0.20}{4} = 0.05\ \text{pu} \quad\Rightarrow\quad \text{SC MVA} = \frac{25}{0.05} = 500\ \text{MVA} \]Required: 350 MVA, so \(Z_{th} \ge 25/350 = 0.071429\) pu.
Scheme A — a reactor in series with each generator. The four branches stay in parallel, each of \(0.20+x\):
\[ \frac{0.20+x}{4} = 0.071429 \quad\Rightarrow\quad x = 4(0.071429)-0.20 = 0.085714\ \text{pu} \]\[ x = 0.085714\times\frac{33^2}{25} = 3.734\ \Omega\ \text{each}, \qquad \text{total installed} = 4x = 0.3429\ \text{pu} \]Scheme B — four bus sections on a tie-bar. With a fault on section 1, the other three reach the tie-bar through \(0.20+y\) each and then pass through section 1's own reactor:
\[ Z_{th} = 0.20 \,\Big\|\, \left[\frac{0.20+y}{3}+y\right] = 0.071429 \]Solving: the bracket must equal \(0.111111\), so \(0.066667+1.33333y = 0.111111\) and
\[ y = 0.033333\ \text{pu} = 1.452\ \Omega\ \text{each}, \qquad \text{total installed} = 4y = 0.1333\ \text{pu} \]The comparison:
\[ \begin{array}{lccc} & \text{each reactor} & \text{total reactance} & \text{normal current carried} \\ \hline \text{Generator reactors} & 0.0857\ \text{pu} = 3.73\ \Omega & 0.343\ \text{pu} & \text{full machine output, always} \\ \text{Tie-bar reactors} & 0.0333\ \text{pu} = 1.45\ \Omega & 0.133\ \text{pu} & \text{only that section's net export} \end{array} \]The operating penalty. A generator reactor at rated current absorbs
\[ Q = I^2x = 1^2(0.085714)(25) = 2.14\ \text{MVAr each} = 8.57\ \text{MVAr for the station} \]and drops \(0.0857\) pu \(=1633\) V per phase permanently, every hour the machine runs. A tie-bar reactor on a station with equally loaded machines carries close to nothing and absorbs close to nothing. The tie-bar scheme needs 61% less reactance and imposes almost no running cost — which is why generator reactors have essentially disappeared from new stations.
Why feeder reactors are not an answer. A reactor in each outgoing feeder limits the current that feeder can deliver into a fault on itself, and does nothing whatever for a fault on the busbar:
\[ \begin{array}{ll} \text{Fault on a feeder} & \text{limited} \\ \text{Fault on the bus} & \text{still } 500\ \text{MVA — the bus is upstream of every feeder reactor} \end{array} \]Since the busbar fault is the one that sizes the switchgear, feeder reactors leave the specification untouched. They have their uses — protecting an old cable, or limiting the duty on a specific consumer's board — but they are not a fault-level measure for the station.
The rule that generalises. A reactor reduces the fault level at a point only if it lies between the sources and that point. Placing it anywhere else is expensive decoration, and the question "which side of the fault is it on?" answers more reactor problems than any formula.
C3 — The study that came out low. A study of an 11 kV industrial bus predicts 18 kA symmetrical in the first cycle. A recorded fault gives 23 kA symmetrical and a first peak of 55 kA. Identify the causes in order of probability, quantify each, and state how each would be confirmed.
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Two independent discrepancies, and they must be diagnosed separately:
\[ \frac{23}{18} = 1.278 \quad(\text{28\% in the symmetrical current}) \]\[ \frac{55}{\sqrt2(23)} = 1.691 \quad\Rightarrow\quad 1+e^{-\pi R/X} = 1.691 \;\Rightarrow\; \frac{X}{R} = 8.5 \]So the measured asymmetry implies \(X/R \approx 8.5\). If the study assumed 15, it predicted a peak of \(\sqrt2(18)(1.811)=46.1\) kA; the true peak is 55 kA, and both errors contributed.
Cause 1: motor contribution omitted. The commonest error at an industrial bus and the only one that produces exactly 20–30% on the symmetrical current. An 11 kV bus with 8 MVA of motors on a 10 MVA base contributes 24% (Problem 10), and nothing else in a routine study is that size. Confirm by listing every motor above 100 kW on the board and summing their locked-rotor currents; if the sum is 4–6 kA the diagnosis is closed.
Cause 2: transformer impedance at tolerance. IEC allows \(\pm7.5\%\) on a two-winding transformer's nameplate impedance, and a unit at \(-7.5\%\) on a bus whose impedance is transformer-dominated raises the fault current by 8%. Compounds with cause 1 but cannot explain 28% alone. Confirm by the works test certificate rather than the nameplate.
Cause 3: prefault voltage. An industrial bus is commonly run at 1.03–1.05 pu to give motor-starting headroom. That is 3–5% directly on the fault current, in the unconservative direction if the study assumed 1.0 (Problem 19). Confirm by the SCADA record of bus voltage in the minutes before the fault.
Cause 4: \(X/R\) assumed too high. This one affects only the peak, not the symmetrical value. A study that carried the supply system's \(X/R\) of 15 down to a bus fed through a resistive cable run will overstate \(X/R\) — but here the measurement shows 8.5 against an assumed 15, which makes the peak prediction too high, not too low. Confirm by recomputing \(X/R\) at the bus rather than inheriting it from the source.
Reconciling the two discrepancies. This is the step that closes the diagnosis. Combining causes 1 to 3:
\[ 18 \times 1.24\ (\text{motors}) \times 1.04\ (\text{voltage}) = 23.2\ \text{kA} \;\approx\; 23\ \text{kA} \;\;\checkmark \]and the peak then follows from the measured \(X/R\) of 8.5 without any further assumption. The transformer tolerance is not needed and should not be invoked — three causes account for the observation, and a fourth added to an already-closed reconciliation is a sign the diagnosis is being fitted rather than found.
The general lesson, and it is the opposite of the one for unsymmetrical faults. Set 22's Challenge C3 concluded that topology errors give factors and impedance errors give tens of per cent, so topology should be suspected first. Here there was no topology error at all: the network was right and the inventory was wrong. A symmetrical fault study fails most often by omitting a source — a motor, a small embedded generator, a second incomer that was closed — because the topology of a balanced fault is too simple to get wrong.
Multiple-Choice Questions
MCQ 1. The DC offset in a short-circuit current exists because:
(a) the source has a DC component (b) current in an inductance cannot change instantaneously (c) the fault is unbalanced (d) of transformer saturationShow answer
(b). The exponential exists only to cancel the steady-state sinusoid at \(t=0\); it draws nothing from the source, which rules out (a). Problem 2.MCQ 2. Maximum DC offset occurs when the fault strikes at:
(a) a voltage maximum (b) a voltage zero (c) \(\alpha=\theta\) (d) any instant — it does not depend on timingShow answer
(b), because \(X\gg R\) makes \(\theta\approx90^\circ\) and the worst condition \(\alpha-\theta=-90^\circ\) then puts \(\alpha\approx0\). Answer (c) is the condition for no offset at all. Problem 3.MCQ 3. The peak asymmetry factor \(1+e^{-\pi R/X}\) approaches 2 when:
(a) \(R\to\infty\) (b) \(X/R\to0\) (c) \(X/R\to\infty\) (d) the fault is remoteShow answer
(c) — a lossless circuit, in which the offset never decays. Options (a), (b) and (d) all describe high resistance, which drives the factor toward 1. Problem 4.MCQ 4. The factor 1.6 used for close-and-latch duty is:
(a) an arbitrary safety margin (b) \(\sqrt{1+2e^{-2\pi/30}}\) (c) \(\sqrt e\) (d) the ratio \(X_d'/X_d''\)Show answer
(b) \(=1.619\), the rms asymmetry factor at a generator terminal where \(X/R\approx30\). Not a margin at all — a computed quantity, rounded. Problem 4.MCQ 5. In a three-phase fault the three DC offsets:
(a) are all equal (b) are all zero (c) sum to zero at every instant (d) sum to \(3\sqrt2 I\)Show answer
(c). The three closing angles form a balanced set and the three offsets share one time constant, so the sum is zero permanently, not merely at \(t=0\). At most one phase can be fully offset. Problem 5.MCQ 6. \(X_d'' < X_d' < X_d\) because:
(a) saturation increases with time (b) rotor circuits holding their flux shunt the magnetising reactance (c) the armature heats up (d) the speed fallsShow answer
(b). Adding a parallel branch can only reduce a reactance, so the ordering follows from circuit topology alone and needs no measurement. Problem 6.MCQ 7. Which reactance is used to compute breaker duty?
(a) \(X_d\) (b) \(X_d'\) (c) \(X_d''\) (d) \(X_2\)Show answer
(c). The first few cycles are what the equipment must survive. Using \(X_d'\) instead would have understated the current by 27% in Problem 6, and \(X_d\) by 84%.MCQ 8. An induction motor on a faulted bus contributes for about:
(a) no time at all (b) two to four cycles (c) half a second (d) indefinitelyShow answer
(b) — until its rotor flux decays. A synchronous motor, having its own excitation, is (d), which is why the two are treated differently in the interrupting calculation. Problem 10.MCQ 9. Neglecting prefault load current in a fault study:
(a) makes the fault current wrong (b) makes the branch currents wrong (c) makes both wrong (d) makes neither wrongShow answer
(b). The load current circulates between machines and cancels in the total, but not in the individual contributions — 11% and 12% errors in Problem 11. This is the comfortable kind of error: the number you check is right.MCQ 10. Short-circuit MVA at a bus equals:
(a) \(\text{MVA}_{base}\times Z_{kk}\) (b) \(\text{MVA}_{base}/Z_{kk}\) (c) \(\sqrt3 V Z_{kk}\) (d) \(V^2/Z_{kk}\) in per unitShow answer
(b), with \(Z_{kk}\) in per unit on that same base. Fault level and Thevenin impedance are reciprocal descriptions of one fact. Problem 13.MCQ 11. For identical machines and identical reactors, the tie-bar arrangement compared with the ring gives:
(a) a higher fault level (b) a lower fault level (c) the same fault level (d) it depends on the number of sectionsShow answer
(b) — 166.7 MVA against 220 MVA in Problems 15 and 16, because every remote contribution passes through two reactors instead of one. Same equipment, different topology, 24% better.MCQ 12. A fault study for setting protection should assume:
(a) bolted fault, 1.05 pu, all plant running (b) arcing fault, 0.95 pu, minimum plant (c) bolted fault, 1.0 pu (d) whatever the rating study assumedShow answer
(b). The relay must still see the smallest credible fault current, so the pessimism reverses. Option (a) is the correct assumption for rating, and using one study's numbers for the other's purpose is a real and common failure. Problem 19.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| Switched RL current | \(i=\frac{\sqrt2 V}{Z}[\sin(\omega t+\alpha-\theta)-\sin(\alpha-\theta)e^{-t/\tau}]\) | From \(i(0)=0\) |
| Circuit time constant | \(\tau = L/R = X/\omega R\) | \(X/2\pi R\) cycles |
| No offset | \(\alpha=\theta\) | Pure sinusoid from \(t=0\) |
| Full offset | \(\alpha-\theta=\pm90^\circ\) | Fault at a voltage zero |
| Peak asymmetry factor | \(i_{peak}=\sqrt2 I(1+e^{-\pi R/X})\) | Max 2, at \(t=T/2\) |
| RMS asymmetry factor | \(I_{rms}=I\sqrt{1+2e^{-2\pi R/X}}\) | Max \(\sqrt3\); 1.619 at \(X/R=30\) |
| Three offsets | \(\sum_k \sin(\alpha_k-\theta)=0\) | Sum zero at every \(t\) |
| Machine reactance ladder | \(X_d''=x_l+(x_{ad}\|x_f\|x_D)\), \(X_d'=x_l+(x_{ad}\|x_f)\) | \(X_d=x_l+x_{ad}\) |
| The three currents | \(I''=V_t/X_d''\), \(I'=V_t/X_d'\), \(I=V_t/X_d\) | rms symmetrical, no load |
| Decrement envelope | \(I(t)=(\tfrac{1}{X_d''}-\tfrac{1}{X_d'})e^{-t/T_d''}+(\tfrac{1}{X_d'}-\tfrac{1}{X_d})e^{-t/T_d'}+\tfrac{1}{X_d}\) | Times \(V_t\); telescopes at \(t=0\) |
| Voltage behind \(X_d''\) | \(E''=V^{(0)}\pm jX_d''I_L\) | + generator, − motor |
| Thevenin fault current | \(I_f = V_k^{(0)}/(Z_{th}+Z_f)\) | Add prefault branch currents back |
| Z-bus solution | \(I_f=\frac{V_k^{(0)}}{Z_{kk}+Z_f}\), \(V_i=V_i^{(0)}-Z_{ik}I_f\) | Line current \((V_i-V_j)/z_{ij}\) |
| Short-circuit MVA | \(\text{SC MVA}=\text{MVA}_{base}/|Z_{kk}|_{pu}\) | \(=\sqrt3 V_LI_{f,L}\) |
| Machines on one bus | \(\text{SC MVA}=\sum_k S_k/X_k''\) | Only if on the same bus |
| Current-limiting reactor | \(X_R = \text{MVA}_{b}\big(\tfrac{1}{S_{new}}-\tfrac{1}{S_{old}}\big)\) | Raises \(X/R\) as well |
| Ring vs tie-bar | \(Z_{ring}=X_g\|\tfrac{X_g+X_r}{2}\), \(Z_{tie}=X_g\|(\tfrac{X_g+X_r}{2}+X_r)\) | Three identical sets |
| Breaker duties | \(I_{mom}\approx1.6I''\), \(I_{int}=m\,I''\) | \(m\) falls as the breaker slows |
| Interrupting capability | \(I_{cap}=\min(I_{rated}V_{max}/V,\;KI_{rated})\) | Must exceed \(I_{int}\) |
| Delayed current zero | no zero while \(I_{ac}(t) < I''e^{-t/T_A}\) | Generator terminals only |
Common Mistakes
Quoting a machine reactance without its base, or without saying which of the three it is. The two halves of the same error. In Problem 12 the 30 MVA machine's 0.15 pu became 0.50 pu on the system base, leaving the smaller machine with the larger per-unit value; and in Problem 6 the three reactances differed by a factor of six, so \(X_d'\) used where \(X_d''\) belongs understates the current by 27% — Problems 6, 7, 11, 12 and 17.
Stopping at the symmetrical current, and expecting the reactor to lower \(X/R\). The breaker of Problem 20 passed on 15.75 kA against 16 kA and failed on the 40.33 kA peak, which only the asymmetry factor revealed. And the reactor raises \(X/R\) — it adds reactance and almost no resistance — so here 15 became 22.5 and 3% of what the reactor removed came straight back — Problems 14 and 20.
Forgetting the motors. Twenty-four per cent at the industrial bus of Problem 10, and the most likely single cause when a study comes out low — Challenge C3. Static load contributes nothing; every rotating machine contributes something.
Taking superposition's branch currents as the answer. The total is exact and the branches are 11–12% wrong until the prefault current is added back branch by branch — Problem 11.
Adding fault MVAs for sources that are not on the same bus. The rule \(\sum S_k/X_k''\) holds only for machines sharing one bus. Insert any impedance between them and it fails — Problems 12 and 15.
Using the rating study's assumptions to set protection. Rating wants a bolted fault at the highest voltage with all plant in; protection wants an arcing fault at the lowest voltage with minimum plant. The same network, two studies, opposite pessimism — Problem 19.
Believing the envelope tells you the breaker can operate. It tells you what it must withstand. Whether a current zero exists at all is a question about the waveform, and at generator terminals the answer can be no for ten cycles — Challenge C1.
Sizing reactors before deciding where they go. The same reactance gave 220 MVA in a ring and 166.7 MVA on a tie-bar. Placement is worth more than sizing, and a feeder reactor is worth nothing at all against a busbar fault — Problems 15, 16 and Challenge C2.
Everything on this sheet was possible because the fault left the network balanced. One phase carried the whole story, the per-unit impedance diagram of Part 1 was reused without modification, and the only new physics was the two decays — the circuit's and the machine's. Nineteen faults in twenty are not like that.
Set 21 introduces the transformation that restores symmetry to an unbalanced system, Set 22 builds the three networks that transformation calls for, and Set 23 connects them according to each fault's boundary conditions. The machine reactances of Problems 6 to 8 reappear there as \(Z_1\), with \(Z_2\approx X_d''\) and \(Z_0\) a fraction of it — and the observation of Problem 1, that an earth fault at a solidly earthed machine terminal can exceed the three-phase fault, is proved there rather than asserted. Set 33 then takes the fault currents computed here and turns them into complete breaker and switchgear specifications for a multi-machine system.