Question 5 of 7: Consequences of Short Circuits and a Bolted Fault on a Meshed Network

Problem 5: Consequences of Short Circuits and a Bolted Fault on a Meshed Network (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) — what a short circuit does to a power system

A short circuit collapses the impedance between a phase conductor and either another phase or earth, so the current is limited only by the source and network reactances and rises to many times the rated value — on the network in this question, more than seven times. The consequences fall into four families, and a competent answer treats them separately because the protection strategy for each is different.

Thermal and mechanical damage. Fault current heats conductors adiabatically, since there is no time for the heat to escape; the limit is expressed as an $I^{2}t$ withstand and is what sizes cable screens, busbars and current transformers. Simultaneously the electromagnetic force between parallel conductors scales with the square of the current, so a seven-per-unit fault imposes roughly fifty times the normal force on busbar supports and on transformer windings; winding deformation from through-faults is one of the commonest causes of subsequent transformer failure. At the fault point itself the arc energy damages insulation and, in switchgear, is the arc-flash hazard that governs personal protective equipment selection under CSA Z462.

Voltage depression. As parts (b) and (c) show quantitatively, a bolted fault holds the faulted bus at zero volts and drags every other bus down with it — here buses 1 and 2 fall to 0.42 and 0.47 pu, less than half of nominal, across the whole network rather than just near the fault. Industrial processes trip on undervoltage, motor contactors drop out, and variable-speed drives shut down; the economic loss from the ride-through failure of a large process plant routinely exceeds the cost of the damaged equipment. Induction motors also lose torque as the square of the voltage, so they decelerate and then draw a large re-acceleration current when the fault clears.

Loss of synchronism and system instability. A fault close to a generator reduces the electrical power it can export to nearly zero while the turbine keeps delivering mechanical power, so the rotor accelerates. If the fault is not cleared before the critical clearing angle — the subject of Problem 7 — the machine pulls out of step, and the resulting out-of-step condition must itself be detected and separated. Frequency and voltage excursions can then cascade through the interconnection.

Personnel safety and secondary effects. Earth-fault current returning through the station earth grid raises the ground potential and creates step and touch voltages, which is the design basis of IEEE Std 80 and of the grounding requirements in CSA C22.3. Induced voltages appear on parallel communication and pipeline circuits. The countermeasures are correspondingly layered: limiting the available fault current (neutral earthing reactors, as in Problem 6, and split bus operation), clearing it quickly with graded protection and adequately rated interrupting devices, and designing every component to survive the maximum through-fault it will ever see.

Parts (b) and (c) — the fault calculation

Given.

Find. The bolted three-phase fault current at bus 4, then the voltages held at buses 1 and 2 while that fault persists.

Approach. With both sources at $1.0\angle 0^{\circ}$ and no prefault load, every bus sits at 1.0 pu before the fault, so the fault current is simply $1/Z_{th}$ at bus 4. Reduce the passive network to that Thevenin reactance by series–parallel combination plus one delta–star transform, then back-substitute to recover the bus voltages.

1. Part (b) — collapse the path that bypasses bus 4. Buses 1 and 2 are joined by two routes that do not touch bus 4: the direct diagonal $j0.20$, and the path through bus 3, $j0.15+j0.10=j0.25$. These are in parallel: $$x_{12}^{\text{eq}}=\frac{(0.25)(0.20)}{0.25+0.20}=\frac{0.05}{0.45}=0.11111\ \text{pu}$$ What is left is a four-node network: the reference $N$, buses 1 and 2, and the faulted bus 4.
2. Transform the reference–1–2 delta into a star. The three sides are $x_{N1}=0.15$, $x_{N2}=0.15$ and $x_{12}^{\text{eq}}=0.11111$, whose perimeter is $0.41111$. The star arms are the products of the two adjacent sides divided by the perimeter: $$\begin{aligned} Z_N &= \frac{(0.15)(0.15)}{0.41111}=0.054730\ \text{pu}\\ Z_1 &= Z_2 = \frac{(0.15)(0.11111)}{0.41111}=0.040541\ \text{pu} \end{aligned}$$ $Z_1$ and $Z_2$ are equal because the two source branches are equal; the network is symmetric about the diagonal even though bus 4 is not.
3. Combine the two remaining paths into bus 4. Bus 4 reaches the star centre by two routes, one through bus 1 and one through bus 2: $$\begin{aligned} \text{via bus 1: } & 0.10+Z_1=0.140541\ \text{pu}\\ \text{via bus 2: } & 0.15+Z_2=0.190541\ \text{pu} \end{aligned}$$ In parallel these give $0.080883$ pu, and adding the star arm to the reference, $$Z_{th}=0.080883+0.054730$$ $$\boxed{Z_{th}=j0.13561\ \text{pu}}$$ An independent bus-impedance-matrix inversion of the full five-node network returns the same 0.135612 pu, confirming the reduction.
4. Fault current at bus 4. With every bus at 1.0 pu before the fault, $$I_f=\frac{V_{\text{prefault}}}{Z_{th}}=\frac{1.0\angle 0^{\circ}}{j0.13561}$$ $$\boxed{I_f=7.374\ \text{pu at }-90^{\circ}}$$ The current lags by exactly ninety degrees because the network is purely reactive; on a 100 MVA, 230 kV base this would be about 1851 A, and the corresponding fault level at bus 4 is 737 MVA.
5. Part (c) — work back to the star centre. The whole fault current flows through the arm $Z_N$, so the star-centre node sits at $$V_{\text{centre}}=1.0-I_fZ_N=1.0-(7.374)(0.054730)=1.0-0.40357=0.59643\ \text{pu}$$ The current then divides between the two routes in inverse proportion to their reactances: $$\begin{aligned} I_{\text{via 1}} &= 7.374\times\frac{0.190541}{0.331082}=4.2439\ \text{pu}\\ I_{\text{via 2}} &= 7.374\times\frac{0.140541}{0.331082}=3.1300\ \text{pu} \end{aligned}$$ These sum back to 7.374 pu, which is the free check on the division.
6. Bus voltages during the fault. Each bus voltage is the star-centre voltage less the drop in its own star arm: $$\begin{aligned} V_1 &= 0.59643-(4.2439)(0.040541)=0.59643-0.17205\\ V_2 &= 0.59643-(3.1300)(0.040541)=0.59643-0.12689 \end{aligned}$$ $$\boxed{V_1=0.4244\ \text{pu},\qquad V_2=0.4695\ \text{pu}}$$ Bus 3, not asked for, follows the same way at 0.4515 pu, and bus 4 is of course zero. Two sanity checks close the problem. Bus 1 is electrically nearer the fault than bus 2 (0.10 pu against 0.15 pu of line), so it must be the more depressed of the two, and it is. And the two source contributions, $(1-V_1)/0.15 = 3.837$ pu and $(1-V_2)/0.15 = 3.536$ pu, add to 7.374 pu — exactly the fault current, as they must.

Final Results.