Question 5 of 7: Short-circuit consequences, transformer protection, and a mid-line three-phase fault

Question 5: Short-circuit consequences, transformer protection, and a mid-line three-phase fault (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.

Part (a) — consequences of short-circuit faults. A short circuit replaces a load impedance of the order of one per unit with something close to zero, so the current rises to whatever the network's Thevenin reactance permits — commonly ten to twenty times rated, and on a strong transmission bus far more. The consequences divide into four groups.

Thermal. Conductor heating goes as $I^{2}t$, so a fault current twenty times rated deposits four hundred times the normal heat. Cable insulation, joints, current transformers and busbar supports can be destroyed in a fraction of a second, which is why fault-clearing times are specified in cycles rather than seconds and why a through-fault duration curve is part of every transformer specification.

Mechanical. Electromagnetic force between parallel conductors goes as $I^{2}$ as well. Transformer windings, switchgear busbars and cable cleats all face peak forces during the first asymmetric half-cycle, and repeated through-faults progressively loosen transformer windings until an eventual turn-to-turn failure occurs.

System-wide electrical. The faulted bus is dragged towards zero volts, and the depression propagates through the network in proportion to the electrical distance from the fault. Motors stall or drop out, contactors release, power electronic converters trip on undervoltage, and the reactive absorption of the recovering motors after clearing can cause a second, slower voltage dip. Unbalanced faults additionally inject negative-sequence current, which produces double-frequency rotor heating in generators, and zero-sequence current, which raises earth-potential rise and can interfere with communications circuits.

Stability. Because a fault suppresses the transfer power while the prime movers keep delivering mechanical power, every generator that can still see the network accelerates. If the fault is not cleared before the critical clearing time, synchronism is lost — which is precisely the calculation carried out in Problem 7 of this paper. Protection speed is therefore a stability requirement, not merely an equipment-protection one.

Part (b) — transformer protective schemes. Four functionally distinct schemes, in the order in which they would be applied to a transmission-class unit:

1. Percentage differential protection (ANSI 87T). Currents are measured at every winding, referred to a common base through the CT ratios and the vector group, and summed. In healthy operation and for any external fault the sum is nominally zero; an internal fault destroys that balance and the relay operates. Because CT saturation, tap-changer travel and magnetising inrush all produce spurious differential current, the characteristic is percentage-biased (the operating threshold rises with through current) and is restrained by second-harmonic detection for inrush and fifth-harmonic detection for overexcitation. It is the primary, unit protection: fast, and inherently selective because it looks only inside its own CT boundary.

2. Buchholz / gas-and-oil relay and sudden-pressure relay. Mounted in the pipe between tank and conservator of an oil-filled unit, the Buchholz relay collects gas liberated by incipient faults. Slow gas accumulation from partial discharge or a hot joint operates an alarm float; a surge of oil driven by a violent internal arc operates a trip vane. The sudden-pressure relay serves the same purpose by rate-of-rise of tank pressure. These are the only schemes that detect faults producing negligible terminal current, such as a single shorted turn, and they are purely mechanical, hence independent of the CT and VT circuits.

3. Restricted earth fault (ANSI 64REF) and neutral overcurrent (ANSI 51N/51G). A residual connection of the three phase CTs of a star winding is balanced against a CT in the neutral earthing connection. Any current that enters the winding zone and does not leave it must have gone to earth inside the zone, so the scheme is a high-impedance differential dedicated to earth faults and is far more sensitive than phase differential to a fault near the star point, where the driving voltage is only a small fraction of phase voltage.

4. Backup overcurrent and thermal protection (ANSI 50/51, 49). Time- and instantaneous-overcurrent elements graded with the downstream feeders give backup if the unit protection or its supply fails, and clear uncleared through-faults. Winding-temperature and top-oil devices (ANSI 49) protect against sustained overload rather than against faults, tripping or initiating cooling on a thermal replica of the winding hot spot. In Canadian practice the overall scheme, its CT requirements and its coordination are governed by the relevant IEEE C37 series, with the installation itself falling under CSA C22.1.

Given. The six-bus system of Figure (3), all reactances in per unit on a common base, every source at 1.0 pu, with a bolted three-phase fault at the mid-point of Line B.

Find. The fault current at the mid-point of Line B, and the voltage at bus 2 while the fault is on.

[Figure not reproduced: Figure 5.1 — The single-line diagram of Figure (3), redrawn. The fault F sits at the mid-point of Line B, so Line A remains a healthy path between buses 2 and 3 and the network cannot be reduced by series–parallel steps alone. See the official exam paper.]

Approach. Collapse the four generator branches onto buses 2 and 3, which leaves a delta between buses 2, 3 and the fault point sitting on two shunt arms; one delta–star transform then reduces the whole network to a single Thevenin reactance, and back-substitution through the star recovers the bus voltages.

1. Collapse the sources onto bus 2. Two independent paths reach bus 2 from the reference, and because both sources are at 1.0 pu they combine into one 1.0 pu source behind the parallel combination of their reactances: $$x_{\text{gen1}}+x_{T1}=0.2+0.1=0.3,\qquad x_{\text{gen5}}+x_{T5}+x_{25}=0.2+0.1+0.1=0.4$$ $$z_{N2}=\frac{(0.3)(0.4)}{0.3+0.4}=0.171429\ \text{pu}$$ The equal source voltages are what make this legitimate: two unequal EMFs would need a Millman combination rather than a simple parallel.
2. Collapse the sources onto bus 3. Identically, on the right-hand side, $$x_{T34}+x_{\text{gen4}}=0.3+0.2=0.5,\qquad x_{36}+x_{T6}+x_{\text{gen6}}=0.2+0.3+0.2=0.7$$ $$z_{N3}=\frac{(0.5)(0.7)}{0.5+0.7}=0.291667\ \text{pu}$$ What is left, drawn as Figure 5.2, is buses 2 and 3 tied to the reference through 0.171429 and 0.291667, joined to each other by Line A at 0.20 pu, and each joined to the fault point F by half of Line B at 0.10 pu.
3. Transform the delta to a star. The three branches 2–F, F–3 and 2–3 form a delta of 0.10, 0.10 and 0.20 pu, whose sum is 0.40. The equivalent star arms are the product of the two adjacent delta arms divided by that sum: $$\begin{aligned} Z_{s2}=Z_{s3}&=\frac{(0.10)(0.20)}{0.40}=0.0500\ \text{pu}\\ Z_{sF}&=\frac{(0.10)(0.10)}{0.40}=0.0250\ \text{pu} \end{aligned}$$ The transform is what breaks the bridge: with Line A present, the network is not series–parallel reducible, and no amount of combining will produce the Thevenin reactance.
4. Assemble the Thevenin reactance at F. Each star arm now sits in series with its own shunt path to the reference, and those two branches are in parallel: $$Z_a=Z_{s2}+z_{N2}=0.221429,\qquad Z_b=Z_{s3}+z_{N3}=0.341667$$ $$Z_a\parallel Z_b=0.134355\quad\Longrightarrow\quad X_{\text{th}}=0.134355+0.0250=0.159355\ \text{pu}$$ An independent bus-impedance-matrix inversion of the same four-node network returns $Z_{FF}=0.159355$ pu to six figures, confirming the hand reduction.
5. Compute the fault current. With a bolted three-phase fault and a pre-fault voltage of 1.0 pu at F, $$\hat I_F=\frac{1.0\angle0^\circ}{jX_{\text{th}}}=\frac{1.0}{j0.159355} =6.2753\,\angle\,-90^\circ\ \text{pu}$$ so that $$\boxed{|I_F|=6.2753\ \text{pu, lagging the pre-fault voltage by }90^\circ}$$ Because the whole network is purely reactive, the fault current is exactly in quadrature; a real system with resistance would give a slightly smaller magnitude at an angle a few degrees short of $-90^\circ$. In physical units the answer is 6.2753 times the base current of whatever common base the reactances were referred to — the paper does not state one, so per unit is the correct form for the answer.
6. Back-substitute for the branch currents. The voltage at the star centre is the fault current times the parallel combination, $$V_{\text{centre}}=|I_F|\,(Z_a\parallel Z_b)=6.2753\times0.134355=0.84312\ \text{pu}$$ which divides between the two branches as $$I_a=\frac{0.84312}{0.221429}=3.80763,\qquad I_b=\frac{0.84312}{0.341667}=2.46766\ \text{pu}$$ Their sum is 6.2753 pu, recovering the fault current, which is the free check on the division.
7. Find the bus-2 voltage during the fault. Bus 2 is fed from a 1.0 pu source behind $z_{N2}$, and the current $I_a$ flows through that reactance, so $$V_2=1.0-I_a\,z_{N2}=1.0-(3.80763)(0.171429)=1.0-0.65274$$ $$\boxed{|V_2|=0.34726\ \text{pu during the fault}}$$ For completeness the same construction gives $|V_3|=0.28027$ pu, and both are confirmed by the bus-impedance route $V_i=1-Z_{iF}/Z_{FF}$, which yields 0.34726 and 0.28027 respectively. Bus 2 retains about a third of its voltage because it is separated from the fault by half a line; the closer a bus is to F, the deeper its depression, and F itself of course sits at zero.