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 (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 collapses the impedance between an energised conductor and either another phase or earth, and the immediate consequence is a current limited only by the source and network reactance — commonly ten to fifty times the rated current of the affected circuit. That current does damage on three different timescales. Within the first cycle the electromagnetic force between parallel conductors, which grows as the square of current, can bend busbars, tear transformer windings off their supports and destroy switchgear compartments; the asymmetrical first peak, which includes the decaying d.c. offset, is the quantity switchgear must be rated to close onto. Over the following tenths of a second the $I^{2}t$ heating of conductors, cable insulation and current transformers accumulates, and the arc at the fault itself — tens of megajoules in a large system — vaporises copper, burns transformer oil and creates the arc-flash hazard that dominates modern workplace electrical safety practice.

The system-wide consequences are just as important as the local ones. Voltage collapses across a wide area around the fault point, roughly in proportion to how electrically close each bus is to it; motors decelerate or stall, contactors drop out, and sensitive process loads trip on undervoltage even when they are many buses away. Generators near the fault suddenly find their electrical output reduced to almost nothing while their mechanical input is unchanged, so their rotors accelerate: this is the transient-stability problem of Problem 7, and if the fault is not cleared inside the critical clearing time the machine loses synchronism and must be tripped. Unbalanced faults inject negative-sequence current, which induces double-frequency rotor currents and rapid rotor-surface heating in synchronous and induction machines, and zero-sequence current, which raises earth-potential rise at the fault and can drive dangerous step and touch voltages in substation grounding systems and induce interference in parallel telecommunication circuits. Finally there is the service consequence: the protection must disconnect the faulted element, so some load is lost, and if the protection is slow, mis-graded or fails to operate, back-up protection removes a much larger part of the network. All of this is why fault levels are calculated at every bus and why equipment is specified with explicit making, breaking and short-time withstand ratings.

Part (b) — three transformer protective schemes, by function

Differential (unit) protection — ANSI 87T. Current transformers on every winding are connected so that the relay measures the difference between the currents entering and leaving the protected zone, after correction for the turns ratio, the winding connection phase shift and the CT ratios. Under any external condition the corrected currents balance and the differential quantity is nominally zero; an internal fault destroys the balance and the relay operates instantaneously with no grading delay, which makes it the primary protection of any transformer above a few MVA. The scheme is percentage-biased so that CT saturation and tap-changer travel do not cause false operation, and it is restrained by second-harmonic content to ride through magnetising inrush and by fifth-harmonic content to ride through overexcitation.

Gas and oil-surge (Buchholz) protection — ANSI 63. A mechanical device mounted in the pipe between the tank and the conservator responds to the by-products of an internal fault rather than to its current. Slow gas evolution from incipient faults — partial discharge, a shorted turn, overheating of a joint — collects in the upper float chamber and raises an alarm long before the fault is large enough to unbalance a differential relay; a violent internal arc produces an oil surge towards the conservator which strikes a lower vane and trips the transformer at once. Because it senses the fault medium directly, Buchholz protection detects interturn faults that draw almost no terminal current, which is exactly the blind spot of the differential scheme. Pressure-relief devices and winding- or oil-temperature devices belong to the same family of non-electrical protection.

Overcurrent and earth-fault back-up protection — ANSI 50/51 and 51N, with restricted earth fault 64REF. Time-graded inverse or definite-time overcurrent relays on the high-voltage and low-voltage sides provide back-up for faults inside the transformer and primary protection for uncleared faults on the downstream system; they are graded to be slower than the downstream feeder protection so that they only act if that protection fails. Restricted earth-fault protection is the important refinement on a star winding: it compares the residual current of the three phase CTs with the current in the neutral CT, is therefore sensitive to earth faults very close to the star point where the driving voltage and hence the fault current are small, and is stable for through faults. In Canadian practice the whole scheme is coordinated under the requirements of CSA C22.1 and the relevant IEEE C37 standards, with the transformer’s through-fault withstand curve as the grading limit.

Parts (c) and (d) — the mid-line three-phase fault

Given. The single-line diagram of Figure (3), all reactances in per unit on a common base: three generators of 0.2 pu subtransient reactance; three transformers of 0.1 pu; lines A and B of 0.2 pu each joining bus 2 to bus 3; and a 0.1 pu circuit from bus 2 down to bus 5. Every source emf is 1.0 pu and pre-fault load current is neglected.

Find. (c) the current in a bolted three-phase fault at the mid-point of line B; (d) the voltages at buses 4 and 5 while that fault persists.

Approach. Merge the two sources that reach bus 2, reduce the resulting three-terminal mesh with one delta-to-star transform, obtain the Thevenin reactance at the fault point, then back-substitute to recover every bus voltage and every machine contribution.

1. Part (c) — reduce each source to a single reactance to its bus. Because every internal emf is $1.0\angle 0^{\circ}$, the three sources can be treated as one node. The generator at bus 1 reaches bus 2 through $0.2+0.1=0.3$ pu; the machine below bus 5 reaches bus 2 through $0.2+0.1+0.1=0.4$ pu; the generator at bus 4 reaches bus 3 through $0.2+0.1=0.3$ pu.
2. Parallel the two arms that land on bus 2. $$X_L=\frac{(0.3)(0.4)}{0.3+0.4}=0.171429\ \text{pu}$$ so the reduced network is a triangle: reference to bus 2 through $0.171429$, reference to bus 3 through $0.3$, and bus 2 to bus 3 through line A, $0.2$ pu — with line B hanging between buses 2 and 3 as two 0.1 pu halves meeting at the fault point F.
3. Transform that triangle to a star. With $\Sigma = 0.171429+0.3+0.2=0.671429$, the star arms are $$\begin{aligned} Z_N &= \frac{(0.171429)(0.3)}{\Sigma}=0.076596 \\ Z_2 &= \frac{(0.171429)(0.2)}{\Sigma}=0.051064 \\ Z_3 &= \frac{(0.3)(0.2)}{\Sigma}=0.089362 \end{aligned}$$ all in per unit, with $Z_N$ joining the star centre to the reference and $Z_2$, $Z_3$ joining it to buses 2 and 3.
4. Collapse the two paths from the star centre to the fault. Through bus 2 the path is $Z_2+0.1=0.151064$ pu and through bus 3 it is $Z_3+0.1=0.189362$ pu; in parallel these give $0.084029$ pu, so the Thevenin reactance at F is $$X_{th}=Z_N+0.084029=\boxed{j0.160625\ \text{pu}}$$
5. Compute the fault current. With a pre-fault voltage of 1.0 pu at the fault point, $$I_f=\frac{1.0\angle 0^{\circ}}{j0.160625}=\boxed{6.2257\angle -90^{\circ}\ \text{pu}}$$ On a 100 MVA base that is 622.6 MVA of fault level at the mid-point of line B.
6. Part (d) — work back to the bus voltages. The current divides between the two arms in inverse proportion to them, so the bus-2 arm carries $3.4630$ pu and the bus-3 arm carries $2.7626$ pu (they sum to $I_f$, which is the check). Since the fault holds F at zero volts, each bus voltage is that arm current times the 0.1 pu half-line: $$V_2=0.346304,\qquad V_3=0.276265\ \text{pu}$$
7. Bus 4 — behind the bus-3 transformer. The generator at bus 4 contributes $I_B=(1.0-V_3)/j0.3=2.4125$ pu, and bus 4 sits one generator reactance below the source: $$V_4=1.0-|I_B|(0.2)=1.0-0.482490=\boxed{0.5175\ \text{pu}}$$
8. Bus 5 — on the spur from bus 2. The third machine contributes $I_C=(1.0-V_2)/j0.4=1.6342$ pu through the 0.4 pu path, and bus 5 lies 0.3 pu (generator plus transformer) below its source: $$V_5=1.0-|I_C|(0.3)=1.0-0.490272=\boxed{0.5097\ \text{pu}}$$ As a closure check the three machine contributions are 2.1790, 2.4125 and 1.6342 pu, summing to 6.2257 pu — the fault current, as required.