Consequences of short circuits and a bolted three-phase fault: 22-Elec-B7 Power Systems Engineering, May 2015, Q5

Question 5: Consequences of short circuits and a bolted three-phase fault

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 a source and the fault point, so the immediate consequence is a current many times rated — commonly ten to fifty times, and determined almost entirely by the machine subtransient reactances and the network between them. Four families of consequence follow.

Thermal and mechanical damage. The $I^{2}t$ let-through melts conductors, vaporises contacts and carbonises insulation at the fault itself, while the electromagnetic forces between parallel conductors, which scale with the square of current, distort busbars, deform transformer windings and can tear switchgear apart. An arcing fault in a switchroom releases enough energy to be lethal, which is the basis of the arc-flash hazard assessment required under provincial occupational health and safety regulation and CSA Z462.

Voltage collapse in the surrounding network. The fault holds its own bus at or near zero volts and depresses voltage across a wide area, dropping out contactors, stalling induction motors, tripping sensitive process loads and causing converters to commutate incorrectly. Voltage sags propagate far beyond the faulted feeder and are the single largest cause of industrial process interruption.

Loss of synchronism and system instability. While the fault persists the electrical power a generator can export falls sharply — to zero in the extreme case of Question 7 — while mechanical input is unchanged, so rotors accelerate. If clearing is slower than the critical clearing time the machine cannot decelerate within its post-fault characteristic and pulls out of step, which can cascade into islanding and load shedding.

Equipment stress beyond the fault point. Transformers experience through- fault forces, generators see negative-sequence rotor heating during unbalanced faults, circuit breakers must interrupt the asymmetrical current within their rated duty, and repeated faults age insulation cumulatively. Ground faults also drive earth-potential rise at substations and can induce dangerous voltages in nearby communication circuits.

These consequences set the whole design chain: short-circuit studies fix breaker interrupting ratings and bus bracing, protection settings and coordination fix clearing times, and the resulting clearing time is what the transient-stability study of Question 7 has to live with.

Parts (b) and (c) — the three-phase fault at bus 4

Given. The four-bus network of Figure (2), with ideal 1.0 pu sources at buses 1 and 2 and all reactances on a common base.

Branch reactances (per unit, common base)
Branch	Reactance	Branch	Reactance
Bus 1 – bus 3	j0.15	Bus 1 – bus 4	j0.10
Bus 3 – bus 2	j0.10	Bus 4 – bus 2	j0.15
Bus 1 – bus 2 (diagonal)	j0.20	Source voltages	1.0 pu at buses 1 and 2

Find. The current in a bolted three-phase fault at bus 4, and the voltages at buses 1 and 2 while that fault is on the system.

Figure 5 — The network of Figure (2). Both machines are represented as ideal 1.0 pu sources directly on their buses.

Approach. Because both sources are specified as ideal 1.0 pu voltages with no internal reactance shown, buses 1 and 2 are held at the same potential. No current can therefore flow in any branch that connects them to each other, and the Thevenin impedance at bus 4 reduces to the two branches that reach it from those buses.

Recognise what the ideal sources do to the network. Buses 1 and 2 are both clamped at $1.0\angle0^\circ$ pu. Two nodes at identical potential can carry no current between them, so the direct branch 1–2 ($j0.20$) and the series pair 1–3–2 ($j0.15 + j0.10$) are inactive for this fault. Bus 3 lies between two points of equal potential and therefore also sits at 1.0 pu with no current in either of its branches.
Form the Thevenin reactance at the fault point. Only the branches 1–4 and 4–2 remain, and they appear in parallel between the 1.0 pu source potential and bus 4: $$X_{th}=\frac{(j0.10)(j0.15)}{j0.10+j0.15}=j0.060\ \text{pu}$$
Compute the fault current. A bolted three-phase fault places zero impedance from bus 4 to reference, so $$I_f=\frac{V_{th}}{X_{th}}=\frac{1.0\angle 0^\circ}{j0.060}$$ $$\boxed{I_f = 16.667\ \text{pu}\ \angle-90^\circ}$$ The current lags by exactly 90 degrees because the network is purely reactive; a real network with resistance would give a slightly smaller magnitude at an angle a little less than 90 degrees, and the d.c. offset would add asymmetry in the first cycles.
Split the fault current between the two sources. Each branch carries the full 1.0 pu driving voltage across its own reactance: $$\begin{aligned} I_{1\to4} &= \frac{1.0}{j0.10}=10.000\angle-90^\circ\ \text{pu} \\ I_{2\to4} &= \frac{1.0}{j0.15}=6.667\angle-90^\circ\ \text{pu} \end{aligned}$$ and $10.000+6.667 = 16.667$ pu confirms the parallel reduction. This split is what a protection engineer actually needs, because it fixes the duty on each of the two breakers at bus 4.
Part (c), evaluate the bus voltages. An ideal voltage source holds its terminal voltage whatever current it delivers, so $$\boxed{V_1 = V_2 = 1.0\angle 0^\circ\ \text{pu}}$$ This is a direct consequence of how Figure (2) models the machines: no subtransient reactance is drawn between either generator and its bus, so neither bus can be depressed by the fault.
Complete the voltage profile and check it by nodal analysis. The fault itself forces $V_4 = 0$, and bus 3 — carrying no current, as Step 1 established — remains at $V_3 = 1.0\angle0^\circ$ pu. Solving the four-node admittance network directly with $V_1 = V_2 = 1$ and $V_4 = 0$ imposed returns $V_3 = 1.000\angle0^\circ$ pu and a fault current of $16.667\angle-90^\circ$ pu, reproducing both results from an independent route.
State the engineering caveat. Real generators have subtransient reactances of 0.1 to 0.3 pu, and if, say, $X_d'' = 0.20$ pu were placed behind each source the same fault would give $I_f = 6.225$ pu with $V_1 = 0.357$ pu and $V_2 = 0.398$ pu — a 63 per cent reduction in fault current and a severe area-wide voltage depression. The answer above is therefore the correct answer to the question as drawn, and it is also a useful upper bound on fault duty, but it should not be read as a realistic voltage profile.

Question 5 — final results
Quantity	Result
Thevenin reactance at bus 4	j0.060 pu
Fault current at bus 4	16.667 pu ∡−90°
Contribution through branch 1–4	10.000 pu ∡−90°
Contribution through branch 4–2	6.667 pu ∡−90°
Voltage at bus 1	1.000 ∡0° pu
Voltage at bus 2	1.000 ∡0° pu
Voltage at bus 3	1.000 ∡0° pu
Voltage at bus 4	0 (bolted fault)

Topic. Balanced three-phase fault analysis by Thevenin reduction, and the role of source representation.

Key relations. $I_f = V_{th}/Z_{th}$ for a bolted fault; $V_i^{\text{fault}}=V_i^{\text{pre}}-Z_{ik}I_f$ in the bus-impedance formulation, with $Z_{th}=Z_{kk}$ at the faulted bus $k$; branch currents by current division.

Why this works. For a balanced three-phase fault only the positive-sequence network is excited, so a single-phase equivalent suffices. Superposition lets the faulted network be treated as the pre-fault network plus a current source of $-I_f$ injected at the fault bus, and it is that step which makes the Thevenin impedance at the bus, or equivalently the diagonal entry of the bus impedance matrix, the whole answer. Recognising equipotential nodes before building any matrix — as in Step 1 here — often removes most of the network by inspection.

Common pitfall. Adding the 1–2 and 1–3–2 branches into the reduction. They connect two equipotential nodes and carry nothing; including them in a series-parallel combination gives a smaller Thevenin impedance and an overstated fault current. The other pitfall is expecting the source buses to be depressed — they cannot be, when the sources are drawn without internal reactance.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, Ch. 7 (Faults on Power Systems, balanced three-phase faults and Z-bus methods); Glover, Overbye and Sarma, Power System Analysis and Design, 6th ed., Ch. 7 (Symmetrical Faults). Arc-flash consequences: CSA Z462, Workplace Electrical Safety.

Question 5 of 7: Consequences of short circuits and a bolted three-phase fault

Question 5: Consequences of short circuits and a bolted three-phase fault

Part (a) — consequences of short-circuit faults

Parts (b) and (c) — the three-phase fault at bus 4