Question 5 of 7: Three-Phase Faults on a Meshed Network

Problem 5: Three-Phase Faults on a Meshed Network (5 + 7.5 + 7.5 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) — Main causes of short-circuit faults on Canadian systems

Lightning is the single largest cause of faults on overhead transmission and distribution in Canada. A stroke to a phase conductor, or a back-flashover from a struck shield wire or tower whose footing resistance is high, puts a steep-fronted impulse across the insulator string and causes a power follow-through arc. Most such faults are single-line-to-ground and transient, which is why single-pole and three-pole autoreclosing is standard practice and why tower-footing resistance is a design parameter on Canadian Shield rock.

Weather and mechanical loading come next and are distinctively Canadian in character. Freezing rain accretes radial ice that can exceed the mechanical design load — the January 1998 ice storm in eastern Ontario and Quebec collapsed towers outright. Asymmetric ice shedding causes conductors to jump and clash; steady crosswinds on iced conductors cause galloping, which brings phases together at mid-span; and heavy wet snow does both. Wind alone reduces phase clearances and blows debris and tree limbs into the line. In summer, conductor sag increases with load and ambient temperature and can close the clearance to vegetation.

Vegetation and wildlife account for a large share of distribution outages: tree contact under wind or growth into the right-of-way, and birds and small mammals bridging phase-to-ground clearances at poles and substations. Wildfire has become a leading cause in the British Columbia and Alberta interior, because smoke and airborne ash sharply reduce the flashover strength of the air gap even where the fire never touches the line. Contamination matters on the coasts and near industry: salt fog on the Lower Mainland and the Maritimes, and cement or fertiliser dust inland, form a conducting film that flashes over when it is wetted by light rain or dew.

The remaining causes are of the system's own making. Insulation ageing and equipment failure — moisture ingress into transformer and cable insulation, partial-discharge erosion, bushing and surge-arrester failures — produce permanent faults that no reclosing can clear. Switching and temporary overvoltages, including ferroresonance and energisation transients, can break down already weakened insulation. Third-party damage is the dominant cause on underground systems: dig-ins by excavators despite one-call locate requirements. And human error — racking in a breaker on a grounded bus, leaving safety grounds applied, incorrect switching orders — remains a real contributor, which is why the fault currents computed below matter: equipment must interrupt them and personnel must be protected from the arc flash they produce.

Parts (b) and (c) — Fault calculations

Given.

Find. The voltage at bus 1 during a bolted three-phase fault at F1 on line 1–3, and then the voltage at bus 1 during a bolted three-phase fault at F2 on line 1–4 after line 1–3 has been removed.

Approach. With both source voltages equal to $1.0\angle 0^{\circ}$ the two machines can be merged into a single reference node, after which each case reduces to a bridge that yields to one delta-to-star transformation. The fault current follows from the Thévenin reactance, and back-substitution through the reduction gives the bus voltage directly.

1. Part (b) — Collapse the radial and series paths first. Bus 2 carries nothing but the source and the transformer, so those add in series: $0.125+0.05=0.175$ from the reference to bus 3. Bus 5 lies on a simple path from bus 1 to bus 4, so it also collapses: $$x_{1\text{-}5\text{-}4}=0.175+0.150=0.3250$$ which is in parallel with the direct line 1–4: $$x_{14}^{eq}=\frac{(0.3250)(0.250)}{0.3250+0.250}=0.141304$$ Adding the remaining line 3–4 gives a single equivalent branch from bus 1 to bus 3 that bypasses the faulted line: $$x_{1\text{-}3}^{alt}=0.141304+0.150=0.291304$$
2. Transform the remaining delta into a star. What is left is a bridge: the reference node, bus 1 and bus 3 form a triangle of $0.175$, $0.175$ and $0.291304$, with the fault node hanging between buses 1 and 3. With $\Sigma=0.175+0.175+0.291304=0.641304$, the star arms are $$\begin{aligned} Z_N&=\frac{(0.175)(0.175)}{\Sigma}=0.047754\\ Z_1&=\frac{(0.175)(0.291304)}{\Sigma}=0.079492\\ Z_3&=\frac{(0.175)(0.291304)}{\Sigma}=0.079492 \end{aligned}$$
3. Assemble the Thévenin reactance at the fault. From the fault point F there are now exactly two paths to the star centre O, and from O a single arm to the reference: $$\begin{aligned} Z_{F\text{-}1\text{-}O}&=0.225+0.079492=0.304492\\ Z_{F\text{-}3\text{-}O}&=0.100+0.079492=0.179492 \end{aligned}$$ $$X_{th}=Z_N+\frac{(0.304492)(0.179492)}{0.304492+0.179492}=0.047754+0.112925=\boxed{j0.160679\ \text{pu}}$$ $$I_F=\frac{1.0}{j0.160679}=6.2236\angle-90^{\circ}\ \text{pu}$$
4. Back-substitute to the voltage at bus 1. The whole fault current flows through $Z_N$, so the star-centre voltage is $$V_O=1.0-I_F Z_N=1.0-(6.2236)(0.047754)=0.702797\ \text{pu}$$ That voltage then drives the two arms in parallel, and the share taken by the arm containing bus 1 is $$I_{F\text{-}1\text{-}O}=\frac{0.702797}{0.304492}=2.308101\ \text{pu}$$ Since bus 1 sits between that arm and the bolted fault (which is at zero volts), the bus voltage is simply the drop across the 0.225 section: $$\boxed{V_1=(2.308101)(0.225)=0.5193\ \text{pu}}$$ The two arm currents sum to $2.308101+3.915490=6.2236$, which reproduces 6.2236 — a free check that the star transformation and the division were both done right.
5. Part (c) — Rebuild the network with line 1–3 out. With line 1–3 removed, bus 3 can reach bus 1 only through bus 4, and bus 5 again collapses into a single path: $$\begin{aligned}x_{N\text{-}4}&=0.175+0.150=0.3250 \\ x_{1\text{-}4}&=0.175+0.150=0.3250\end{aligned}$$ The triangle is now the reference, bus 1 and bus 4, with the fault F2 hanging between buses 1 and 4. With $\Sigma'=0.175+0.3250+0.3250=0.8250$, $$\begin{aligned}Z_N'&=0.068939 \\ Z_1'&=0.068939 \\ Z_4'&=0.128030\end{aligned}$$
6. Compute the second fault current and the bus voltage. The two arms from the fault to the star centre are $0.150+0.068939=0.218939$ and $0.100+0.128030=0.228030$, so $$X_{th}'=0.068939+\frac{(0.218939)(0.228030)}{0.218939+0.228030}=\boxed{j0.180636\ \text{pu}}$$ $$\begin{aligned}I_F'&=\frac{1.0}{0.180636}=5.5360\ \text{pu} \\ V_O'&=1.0-(5.5360)(0.068939)=0.618351\ \text{pu}\end{aligned}$$ $$I'_{F\text{-}1\text{-}O}=\frac{0.618351}{0.218939}=2.824302\ \text{pu} \;\Rightarrow\;\boxed{V_1'=(2.824302)(0.150)=0.4236\ \text{pu}}$$
7. Interpret and cross-check both results. The second fault is electrically closer to bus 1 — 0.150 pu of line instead of 0.225 pu — and the loss of line 1–3 has weakened the network, so although the total fault current is smaller (5.5360 against 6.2236 pu, because the Thévenin reactance rose from 0.160679 to 0.180636), the voltage at bus 1 is lower: 0.4236 against 0.5193 pu. That is the point of the question: fault-current magnitude and depth of voltage depression are governed by different things, the first by the Thévenin impedance at the fault and the second by the electrical distance between the fault and the bus of interest. Both results were confirmed independently by inverting the full five-node bus admittance matrix and forming $V_k = 1 - Z_{kF}/Z_{FF}$; the two methods agree to twelve significant figures.

Final results.