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

Question 5 (Problem 5): Consequences of Short Circuits, and a Mid-Line Three-Phase Fault (5 + 10 + 10 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 the sources and the fault point, so the immediate consequence is a current of five to twenty times rated flowing from every machine that can reach the fault. That current does mechanical and thermal damage first: the electromagnetic forces between adjacent conductors and between the turns of a transformer winding scale with the square of the current and can distort or tear windings and busbars in the first few cycles, while the $I^{2}t$ energy melts contacts, vaporises conductors at the arc root and ignites insulation. An arcing fault releases enough energy to constitute an arc-flash hazard to anyone working nearby, which is why Canadian practice under CSA Z462 requires the incident energy to be calculated from exactly this fault current.

The system-wide consequences matter as much. The bus voltages collapse over a wide area — as this problem shows, the healthy buses of a meshed network fall to a third or a half of nominal — so motors stall or drop out, sensitive process loads trip, and the reactive demand of recovering motor load makes the depression worse. Generators lose their ability to export electrical power while the turbines keep delivering mechanical power, so the machines accelerate; if the fault is not cleared before the critical clearing time they pull out of step, which is the subject of Problem 7. Unbalanced faults inject negative-sequence current that heats generator rotors, and zero-sequence current that flows in the earth and couples into communication circuits. Finally, every fault stresses the protection: a mis-graded or slow scheme turns a local fault into a cascading outage.

Given. A three-bus network in which generator 1 (0.2 p.u.) reaches bus 1 through a 0.125 p.u. transformer and generator 2 (0.15 p.u.) reaches bus 2 through a 0.035 p.u. transformer; line 1–2 is 0.225 p.u. and lines 1–3 and 2–3 are 0.24 p.u. each. Both source voltages are 1.0 p.u.

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

Approach. Reduce the network to a single Thevenin reactance seen from the fault point using one delta-to-star transform, then back-substitute the current division to recover the bus voltages.

1. Part (b) — collapse the two radial branches. Each machine and its transformer form a single series path from the common source node N to its bus: $$\begin{gathered}x_{N1}=0.2+0.125=0.3250, \\ x_{N2}=0.15+0.035=0.1850\end{gathered}$$ Splitting line 1–3 at its mid-point creates the fault node F, so the fault sees 0.120 p.u. towards bus 1 and, through the remaining half plus line 2–3, $$x_{F\text{-}3\text{-}2}=0.120+0.24=0.360$$ towards bus 2. Bus 3 is a passive junction and disappears into that sum.
2. Transform the (1, 2, N) triangle to a star. The three nodes 1, 2 and N are joined in a delta by line 1–2 and the two radial branches, and the fault current has to pass through it. With $\Sigma=x_{12}+x_{N1}+x_{N2}=0.7350$, $$\begin{gathered}Z_{1}=\dfrac{x_{12}x_{N1}}{\Sigma}=0.0994898, \\ Z_{2}=\dfrac{x_{12}x_{N2}}{\Sigma}=0.0566327, \\ Z_{N}=\dfrac{x_{N1}x_{N2}}{\Sigma}=0.0818027\end{gathered}$$
3. Combine the two arms and add the common branch. The fault now sees two parallel routes to the star point, and the star point reaches the source node through $Z_{N}$: $$\begin{gathered}\text{arm A}=0.120+Z_{1}=0.2194898, \\ \text{arm B}=0.360+Z_{2}=0.4166327\end{gathered}$$ $$\begin{aligned}&X_{th} \\ &=\dfrac{(\text{arm A})(\text{arm B})}{\text{arm A}+\text{arm B}}+Z_{N} \\ &=0.1437563+0.0818027=\boxed{j0.225559\ \text{p.u.}}\end{aligned}$$
4. Fault current. With a bolted fault the terminal voltage at F is zero and the prefault voltage is 1.0 p.u., so $$\boxed{I_{F}=\dfrac{1.0}{jX_{th}}=\dfrac{1.0}{j0.225559}=-j4.433429\ \text{p.u.} \;\;(\,|I_{F}|=4.433429\ \text{p.u.})}$$ An independent bus-impedance-matrix inversion of the same five-branch network returns $Z_{FF}=j0.225559$ to six figures, which confirms the hand reduction.
5. Part (c) — back-substitute for the star-point voltage. The fault current flows out of the source node through $Z_{N}$ first, so $$\begin{aligned}V_{N'}&=1.0-I_{F}Z_{N} \\ &=1.0-(4.433429)(0.0818027) \\ &=0.637333\ \text{p.u.}\end{aligned}$$ It then divides between the two arms in inverse proportion to their reactances: $$\begin{gathered}I_{A}=\dfrac{V_{N'}}{\text{arm A}}=2.903704, \\ I_{B}=\dfrac{V_{N'}}{\text{arm B}}=1.529725\end{gathered}$$ and the two add back to 4.433429 p.u., which is the fault current again — a free check on the division.
6. Bus voltages under the fault. Each bus sits one star arm away from the star point, carrying its own share of the current: $$\begin{aligned}V_{1}&=V_{N'}-I_{A}Z_{1} \\ &=0.637333-(2.903704)(0.0994898) \\ &=\boxed{0.348444\ \text{p.u.}}\end{aligned}$$ $$\begin{aligned}V_{2}&=V_{N'}-I_{B}Z_{2} \\ &=0.637333-(1.529725)(0.0566327) \\ &=\boxed{0.550701\ \text{p.u.}}\end{aligned}$$ Both are real because every element is a pure reactance and both sources were taken at the same angle. The arithmetic closes on itself: continuing from bus 1 through the 0.120 p.u. half-line gives $0.348444-(2.903704)(0.120)=0$, which is the fault point, and the same is true along the other route. For completeness the same substitution gives $V_{3}=0.183567$ p.u.