Question 5 of 7: Short-Circuit Consequences, Transformer Protection, and a Mid-Line Three-Phase Fault

Problem 5: Short-Circuit Consequences, Transformer Protection, and a Mid-Line Three-Phase Fault (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — consequences of short-circuit faults. A short circuit replaces the load impedance with something close to zero, so the current rises to a value limited only by the source and network reactances — typically ten to fifty times rated. The damage that follows falls into four groups.

Thermal and mechanical damage to plant. Fault current heats conductors adiabatically ($I^{2}t$), so cable insulation, busbar joints and transformer windings can be destroyed in a fraction of a second. The simultaneous electromagnetic forces vary as the square of the current and act to throw parallel conductors apart; transformer windings distort or telescope, and switchgear bracing fails, if the fault is not cleared inside the equipment's rated withstand time. At the arc itself the energy release is an arc-flash hazard governed in Canada by CSA Z462.

Voltage collapse over a wide area. During the fault the voltage at and near the fault point falls to nearly zero and depresses voltage across a large region — part (d) below shows buses two and three network elements away held at only 0.36 to 0.49 pu. Motors stall or trip on undervoltage, electronic and drive loads drop out, and induction generation disconnects; the economic loss to industrial customers from a voltage sag is often far larger than the utility's repair cost.

Loss of synchronism and system instability. With the terminal voltage depressed, the electrical power a generator can export collapses while the mechanical input is unchanged, so the rotor accelerates. If the fault is not cleared before the critical clearing angle is reached the machine pulls out of step — which is exactly the calculation of Problem 7. Unbalanced faults additionally inject negative-sequence current, whose double-frequency rotor currents heat the rotor body and damper windings very quickly, and zero-sequence current, which raises earth-potential rise and induces voltage into parallel telecommunication and pipeline plant.

Protection and equipment-rating consequences. The fault level dictates the interrupting rating of every breaker and the withstand rating of every bus in the station, and it changes as generation is added, so a short-circuit study is repeated for every system change. Faults also cause protection to operate, and an incorrect or slow operation cascades: an overreaching zone-2 element or a stuck breaker turns a single line fault into a multiple-element outage.

Part (b) — three transformer protective schemes.

• Percentage differential protection (ANSI 87T). Currents are measured at every winding, referred to a common base by ratio and phase-shift compensation, and summed. In the healthy transformer the compensated sum is essentially zero; an internal fault destroys that balance and the relay trips. The percentage restraint characteristic makes the pickup proportional to the through current so that CT saturation and tap-changer travel do not cause false trips, and second-harmonic (inrush) and fifth-harmonic (overexcitation) restraint prevent operation on energisation and on overfluxing. It is the primary unit protection of any transformer above a few MVA.
• Overcurrent and restricted-earth-fault backup (ANSI 50/51, 51G, 64REF). Time-graded phase and ground overcurrent relays on each winding provide backup for the differential and cover faults beyond the transformer. Restricted earth fault compares the neutral current with the residual of the three phase currents on one grounded-wye winding, which gives it far better sensitivity to a winding-to-core fault near the neutral end than the phase differential can achieve.
• Mechanical (gas and pressure) protection — Buchholz and sudden-pressure relays (ANSI 63). An incipient internal fault decomposes oil into gas. The Buchholz relay, mounted in the pipe between tank and conservator, collects that gas: slow accumulation raises an alarm, a violent oil surge trips. The sudden-pressure relay responds to the rate of rise of tank pressure. Both detect faults that produce very little terminal current — inter-turn faults, core bolt failures, partial discharge — and so are complementary rather than redundant to the differential.

Also standard are winding- and oil-temperature protection (49) with alarm and trip stages, and volts-per-hertz overexcitation protection (24) on generator step-up units.

Given. The Figure (3) network, with every reactance in per unit on a common base and both sources at 1.0 pu.

Find. The bolted three-phase fault current at the mid-point of line 1–3, and the resulting voltages at buses 1 and 2.

Approach. Split the faulted line at its mid-point to create a new node F, kill both (identical) sources to find the Thévenin reactance seen from F, divide 1.0 pu by it for the fault current, then back-substitute through the reduction to recover the bus voltages.

1. Part (c) — introduce the fault node. A fault at the mid-point of line 1–3 splits that line into two equal halves, so $$x_{1F}=x_{F3}=\tfrac12(0.225)=0.1125\ \text{pu}.$$ Because the two internal emfs are identical at $1.0\angle0^\circ$, no circulating current flows between them before the fault, the pre-fault voltage is 1.0 pu everywhere, and in the passive (source-killed) network the two internal nodes become a single reference node.
2. Collapse the radial branches that carry no choice. Bus 2 has only one connection to the rest of the network, through the transformer to bus 3, so the source behind it and that transformer are in series: $$x_{\text{ref}\to3}=x_{g2}+x_{32}=0.150+0.035=0.185\ \text{pu}.$$ Similarly bus 4 lies between buses 1 and 3 with nothing else attached, so it is a plain series path: $$x_{1\to3\ \text{(via 4)}}=x_{14}+x_{34}=0.215+0.125=0.340\ \text{pu}.$$
3. Convert the surviving delta to a star. What remains between the reference node, bus 1 and bus 3 is a delta of $0.200$ (reference–1), $0.185$ (reference–3) and $0.340$ (1–3). Its sum is $0.725$, so the equivalent star arms are $$Z_1=\frac{(0.340)(0.200)}{0.725}=0.093793,\ Z_3=\frac{(0.340)(0.185)}{0.725}=0.086759,\ Z_{\text{ref}}=\frac{(0.200)(0.185)}{0.725}=0.051034.$$ Call the star centre N.
4. Combine the two paths from F to the star centre. Node F reaches N through $0.1125+Z_1$ one way and $0.1125+Z_3$ the other, and those two are in parallel: $$\frac{(0.206293)(0.199259)}{0.206293+0.199259}=\frac{0.041105}{0.405552}=0.101357\ \text{pu}.$$
5. Add the last series arm to get the Thévenin reactance. The reference node reaches N through $Z_{\text{ref}}$, so $$\boxed{X_{\text{Th}}=0.101357+0.051034=j0.152392\ \text{pu}}$$
6. Divide to get the fault current. With pre-fault voltage 1.0 pu at F and a bolted (zero-impedance) three-phase fault, $$\boxed{I_F=\frac{1.0\angle0^\circ}{j0.152392}=6.5620\angle-90^\circ\ \text{pu}}$$ The current lags 90° because every element in the model is a pure reactance; in a real network with resistance the angle would be a few degrees less. On a 100 MVA base this is 656 MVA of fault duty at the mid-line point.
7. Part (d) — back-substitute for the voltage at the star centre. The whole fault current flows through $Z_{\text{ref}}$, so $$V_N=1.0-I_F Z_{\text{ref}}=1.0-6.56203(0.051034)=0.665106\ \text{pu}.$$ (The $j$ of the reactance and the $-j$ of the current cancel, leaving a real result in phase with the sources.)
8. Divide the fault current between the two paths and recover $V_1$. The share flowing through the bus-1 side, and the drop it causes in the star arm $Z_1$, give the bus-1 voltage directly: $$I_{F,1}=\frac{V_N}{0.206293}=3.22412\ \text{pu},\ I_{F,3}=\frac{V_N}{0.199259}=3.33792\ \text{pu},$$ $$\boxed{V_1=V_N-I_{F,1}Z_1=0.665106-3.22412(0.093793)=0.3627\ \text{pu}}$$ The two shares add to 6.56204 pu, which is the fault current again — a free check that the division is right.
9. Recover $V_3$, then walk out to bus 2. The same construction on the other arm gives $$V_3=V_N-I_{F,3}Z_3=0.665106-3.33792(0.086759)=0.375514\ \text{pu}.$$ Bus 2 is fed radially from its own source through $0.150+0.035$, so the current in that path and the drop across the machine reactance are $$I_{g2}=\frac{1.0-V_3}{0.185}=\frac{0.624486}{0.185}=3.37560\ \text{pu},$$ $$\boxed{V_2=1.0-I_{g2}(0.150)=1.0-0.506340=0.4937\ \text{pu}}$$
10. Sanity-check the voltage profile. The depressed voltages should fall away with electrical distance from F, and they do: $V_1=0.363$ and $V_3=0.376$ at the two ends of the faulted line, $V_4=0.371$ on the parallel path, and $V_2=0.494$ — the highest of the four because it sits behind the whole $0.150$ pu of its own machine reactance. Nowhere in the network does the voltage exceed the pre-fault 1.0 pu, as it must not for a bolted shunt fault.