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

Problem 5: Fault Consequences, Transformer Protection and a Mid-Line Three-Phase Short (25 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.

(a) Consequences of short-circuit faults. A short circuit collapses the impedance seen by the sources, so the first consequence is a current of several thousand per cent of rating flowing until a breaker clears it. That current does mechanical and thermal damage: the $i^{2}$ electromagnetic forces on parallel conductors and transformer windings peak in the first half-cycle and can physically deform bus work and coil ends, while the $i^{2}Rt$ heating destroys insulation and, at the fault itself, produces an arc whose energy is the dominant arc-flash hazard to personnel — the reason Canadian practice requires incident-energy studies and labelling under CSA Z462. The second consequence is a voltage collapse across the faulted region: buses electrically near the fault lose most of their voltage, so motors stall or trip, contactors drop out, and sensitive process load is lost over an area far wider than the faulted element. The third is a threat to system stability: with the fault on, generators near it can export almost no power while their turbines continue to deliver mechanical torque, so their rotors accelerate, and if clearing is slow they lose synchronism — exactly the mechanism analysed in Problem 7. Fourth, unbalanced faults inject negative- and zero-sequence currents, which cause double-frequency rotor heating in generators, induce voltages in parallel communication circuits, and raise the earth-potential rise at substation ground grids to values dangerous to anyone standing nearby. Finally, every fault is followed by a switching transient and by the loss of the faulted element, which weakens the network and may cascade if the remaining paths are overloaded.

(b) Transformer protective schemes. Three functionally distinct schemes are universal on power transformers. Differential protection (ANSI device 87T) is the main unit protection: currents are measured at every winding, referred to a common base through matching ratios and phase-shift compensation, and summed. In healthy operation the sum is zero to within the ratio error, so any significant residual means the fault is inside the protected zone and the relay trips instantaneously. It is percentage-biased so that through-fault current-transformer saturation does not cause a false trip, and it is restrained against the second-harmonic content of magnetising inrush and the fifth harmonic of over-excitation, which would otherwise look like internal faults. Overcurrent and earth-fault protection (devices 51, 51N, 50) is the back-up: an inverse-time characteristic on each phase and in the neutral trips for faults the differential cannot see, notably those beyond the transformer terminals, and is graded in time with the downstream feeder relays so that the nearest device operates first. A restricted-earth-fault element gives sensitive detection of winding-to-core faults close to the neutral, where the driving voltage and hence the differential current are small. Mechanical and thermal protection is the third family: the Buchholz relay in the pipe between tank and conservator collects the gas evolved by incipient insulation breakdown and alarms, then trips on a surge of oil produced by a violent internal fault; sudden-pressure and pressure-relief devices act on the same principle; and winding-temperature devices with a thermal image of the hot spot alarm and trip on overload. To these are commonly added over-excitation (volts-per-hertz, device 24) protection, which prevents core saturation when the transformer is energised at high voltage or low frequency, and surge arresters at the bushings to limit lightning and switching overvoltages.

(c)–(d) Given.

Find. The fault current at the mid-point of line B, and the voltages that survive at buses 4 and 5 while the fault is on.

[Figure not reproduced: Figure 5 — The five-bus system of Figure (4) redrawn. Three identical generating units feed buses 2, 3 and 5; lines A and B run in parallel between buses 2 and 3, and the bolted three-phase fault sits at the mid-point of line B. See the official exam paper.]

Approach. With every internal e.m.f. at $1.0\angle0^\circ$ the three sources merge into a single node, so the problem reduces to finding the Thévenin reactance from that node to the fault point. Split line B at its mid-point, collapse the two radial source branches on bus 2, then apply one delta–star transform to the (bus 2, bus 3, F) triangle. Bus voltages during the fault follow by back-substitution.

1. Part (c) — merge the sources and collapse the radials. Every machine sits behind $X_g+X_T=0.20+0.10=0.30$ pu of its own bus, and all three e.m.f.s are equal, so they become one reference node $N$ at $1.0\angle0^\circ$. From $N$ there are two independent paths to bus 2 — the unit at bus 1 directly at 0.30 pu, and the unit at bus 5 through its own 0.30 pu plus the 0.10 pu tie, i.e. 0.40 pu — so$$\begin{aligned}X_{N2}&=\frac{(0.30)(0.40)}{0.30+0.40}=\frac{0.12}{0.70}=0.171429\ \text{pu},\\ X_{N3}&=0.30\ \text{pu}.\end{aligned}$$
2. Split line B at the fault. A mid-line fault divides line B into two halves of $0.20/2=0.10$ pu each, creating a new node F. What remains is a four-node network: $N$–2 at 0.171429, $N$–3 at 0.300, 2–F at 0.100, F–3 at 0.100, and line A bridging 2–3 at 0.200.
3. Delta–star the (2, 3, F) triangle. With arms $z_{2F}=0.10$, $z_{3F}=0.10$, $z_{23}=0.20$ and $\Sigma=0.40$, the equivalent star arms are$$\begin{aligned}z_2&=\frac{z_{2F}z_{23}}{\Sigma}=0.050\ \text{pu},\\ z_3&=\frac{z_{3F}z_{23}}{\Sigma}=0.050\ \text{pu},\\ z_F&=\frac{z_{2F}z_{3F}}{\Sigma}=0.025\ \text{pu}.\end{aligned}$$The star point is the natural centre of the mesh; call it $M$.
4. Reduce to the Thévenin reactance. The two paths from $N$ to $M$ are now in parallel, and $z_F$ is in series with the result:$$\begin{aligned}X_{th}&=\frac{(0.171429+0.050)(0.300+0.050)}{(0.171429+0.050)+(0.300+0.050)}+0.025\\&=\frac{(0.221429)(0.350)}{0.571429}+0.025=0.135625+0.025,\end{aligned}$$so that$$\boxed{X_{th}=j0.160625\ \text{pu}\quad\Longrightarrow\quad I_F=\frac{1.0\angle0^\circ}{j0.160625}=6.2257\angle-90^\circ\ \text{pu}.}$$On a 100 MVA base that is 622.6 MVA of fault level at the mid-point of line B.
5. Part (d) — work back to the bus voltages. Holding $V_F=0$ and $E=1.0$, the star-point voltage is$$V_M=I_Fz_F=(6.2257)(0.025)=0.155642\ \text{pu (in phase with }E).$$The fault current divides between the two branches reaching $M$ in inverse proportion to their reactances:$$\begin{aligned}I_{G2}&=I_F\frac{0.350}{0.571429}=3.8133\ \text{pu},\\ I_{G3}&=I_F\frac{0.221429}{0.571429}=2.4124\ \text{pu},\end{aligned}$$and the two sum back to 6.2257 pu, which is the free check on the division.
6. Recover buses 2 and 3. Each bus sits one star arm above $M$:$$V_2=V_M+I_{(via\ 2)}z_2=0.155642+(3.8133)(0.050)=0.34630\ \text{pu},$$$$V_3=V_M+I_{(via\ 3)}z_3=0.155642+(2.4124)(0.050)=0.27626\ \text{pu}.$$Bus 3 is lower than bus 2 because it is fed by only one machine while bus 2 is fed by two.
7. Bus 4 — the generator terminal on the right. Bus 4 lies between its machine and its transformer. The current that machine contributes is$$I_{g2}=\frac{E-V_3}{0.30}=\frac{1-0.27626}{0.30}=2.4125\ \text{pu},$$and bus 4 is one generator reactance below the e.m.f.:$$\boxed{V_4=E-I_{g2}X_g=1-(2.4125)(0.20)=0.51751\ \text{pu}.}$$
8. Bus 5 — the tie-fed generator bus. The unit at bus 5 sees bus 2 through the 0.10 pu tie, so its contribution is$$I_{g3}=\frac{E-V_2}{0.30+0.10}=\frac{1-0.34630}{0.40}=1.63424\ \text{pu},$$and bus 5 sits one machine-plus-transformer reactance below the e.m.f.:$$\boxed{V_5=E-I_{g3}(X_g+X_T)=1-(1.63424)(0.30)=0.50973\ \text{pu}.}$$An independent nodal solution of the whole five-bus network with $V_F$ forced to zero returns $V_1=0.56420$, $V_2=0.34630$, $V_3=0.27626$, $V_4=0.51751$ and $V_5=0.50973$ pu, confirming every figure above.
9. Interpret the result for protection purposes. The generator terminals hold just over half voltage while the fault is on, so undervoltage-based back-up would be marginal there, but the transmission buses fall to a third of nominal — ample discrimination for distance or overcurrent relaying on lines A and B. Note also that opening line B at both ends clears this fault completely, since the fault lies on the line itself; the remaining network is then intact, which is exactly the situation Problem 7 examines from the stability side.

Final Results