22-Elec-B7 Power Systems Engineering · December 2018

Question 5 of 7: Network Reduction and Three-Phase Fault Analysis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2018 — 16-Elec-B7 Power Systems Engineering, 3 hours, open book, any non-communicating calculator permitted. The paper prints seven questions, headed “Problem 1” to “Problem 7”; the rubric states that any five questions constitute a complete paper and that all questions are of equal value. Every one of the seven is solved in full here, because the set is a study resource rather than a three-hour sitting. Sub-part point values are quoted as printed; note that the printed splits total 25 points on Problems 1, 2, 4, 5 and 6, 20 points on Problem 3, and Problem 7 carries no printed split at all — under the “equal value” rubric each question is worth one fifth of the paper regardless.

Reference texts. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press / Wiley — the reference text named in the Engineers Canada syllabus for this examination code, and the source of the “cantilever” transformer model used in Problem 3. Supporting texts: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed.; J. J. Grainger and W. D. Stevenson, Power System Analysis; S. J. Chapman, Electric Machinery Fundamentals, 5th ed. Canadian practice context (protection philosophy, grounding) follows CSA and IEEE C37 relaying practice as applied by Canadian transmission utilities.

Check — two places where the printed data are not internally consistent, and what was done about each.

(1) Problem 1. The pair A = 0.95∠2°, B = 195∠87.5° forces a shunt parameter with a large real part (Re C = 3.00 × 10⁻⁴ S). No real transmission line has a shunt conductance of that size — on a physical line C lands within about a degree of +90°. The consequence is a transmission efficiency of only 80.9 %, of which about 156 MW of the 236 MW loss is charged to that fictitious conductance. The arithmetic below is exactly what the printed constants give and is what the marker will be looking for; the physical caveat is stated so the low efficiency is not read as a slip.

(2) Problem 6. Table (2) lists the line negative-sequence reactance as 0.320 pu against a positive-sequence 0.30 pu. A transmission line is a static, symmetric element, so physically X₋ must equal X₊. The printed 0.320 is used throughout (the paper governs). Taking 0.30 instead moves the answer to Problem 6(b) from 1.4169∠145.58° pu to 1.4212∠146.04° pu — a change of 0.3 %, so neither reading changes any conclusion.

Question 5: Network Reduction and Three-Phase Fault Analysis (25 marks)

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) — three protection schemes for HV transmission lines.

Distance (impedance) protection. The relay computes the apparent impedance from the measured voltage and current at its own terminal and trips if that impedance falls inside a characteristic drawn on the R–X plane. Because the impedance to a fault is proportional to the distance to it, the relay is inherently selective without needing any information from the far end. It is applied in graded zones: zone 1 covers roughly 80 % of the line and trips instantaneously, zone 2 reaches about 120 % with a short time delay to provide remote back-up, and zone 3 reaches further still. Distance protection is the workhorse of Canadian 230 kV and 500 kV networks.

Pilot / differential protection. The currents at both ends of the line are compared over a communication channel — today almost always a fibre-optic link, historically a power-line carrier or microwave path. Current differential protection sums the phasors and trips on any non-zero residual, giving absolute selectivity and instantaneous clearance over 100 % of the line. Permissive overreaching transfer trip and directional comparison blocking schemes are pilot variants that send a permission or a block signal rather than the full current phasor, and are used where channel bandwidth is limited.

Overcurrent and directional overcurrent protection. Instantaneous and time-inverse overcurrent elements, supervised by a directional element derived from the polarising voltage, form the simplest scheme. On a meshed transmission network they are too slow and too load-dependent to serve as main protection, but they remain valuable as back-up and as ground-fault protection, where a sensitive residual (zero-sequence) overcurrent element detects high-impedance earth faults that phase elements would miss.

A fourth scheme worth naming is phase-comparison protection, which compares only the phase angle of the currents at the two ends and so needs very little channel capacity. Whichever main scheme is chosen, Canadian practice follows IEEE C37 guidance in providing two fully independent protection systems on any EHV circuit, each with its own instrument transformers, d.c. supply and tripping path, so that no single failure leaves the line unprotected.

Given. The six-bus network of Figure (3-a), all reactances in per unit on a common base, with all four source voltages held at $E = 1.0$ pu. Generators at buses 1, 2, 5 and 6 each connect through $j0.02$; the tie reactances are $j0.2$ (1–2), $j0.6$ (1–3), $j0.4$ (2–3), $j0.05$ (1–4), $j0.1$ (5–4) and $j0.1$ (4–6). A three-phase bolted fault is applied at bus 3.

Find. The equivalent source reactance $X_a$ at bus 4 after buses 5 and 6 are eliminated; the equivalent source reactance $X_b$ at bus 1 after bus 4 is eliminated; the fault current at bus 3; and the voltage at bus 1 during the fault.

[Figure not reproduced: Figure 3(a) redrawn. Buses 5 and 6 each carry a generator behind j0.02 reaching bus 4 through j0.1, so the pair collapses to one equivalent source at bus 4; bus 4 then folds into bus 1 through the j0.05 tie. See the official exam paper.]

Approach. Because every source is at the same voltage and phase, sources in parallel simply combine, and the whole reduction is series–parallel arithmetic until the last step. The fault current then needs one delta–star transformation, after which back-substitution recovers every bus voltage by hand.

Part (b) — combine the two remote machines into one source at bus 4. Each of the generators at buses 5 and 6 reaches bus 4 through its own $j0.02$ machine reactance in series with a $j0.1$ tie, so each presents $$0.02+0.1=0.12\ \text{pu}$$ Both sources are at $E = 1.0$ pu, so they can be paralleled directly (this is only legitimate because the source voltages are identical — unequal sources would require a Thevenin combination): $$\boxed{X_a=\frac{0.12\times 0.12}{0.12+0.12}=0.06\ \text{pu}}$$
Part (c) — fold bus 4 into bus 1. In Figure (3-b) the only connection remaining at bus 4 is the $j0.05$ tie to bus 1, so the equivalent source behind $jX_a$ now reaches bus 1 through a simple series path: $$\boxed{X_b=0.05+X_a=0.05+0.06=0.11\ \text{pu}}$$ Figure (3-c) is therefore a three-bus network: a source of 1.0 pu behind $j0.11$ at bus 1, a source of 1.0 pu behind $j0.02$ at bus 2, and ties of $j0.2$ (1–2), $j0.6$ (1–3) and $j0.4$ (2–3), with the fault at bus 3.
Part (d) — recognise that the two remaining sources share one node. Both are 1.0 pu at 0°, so their internal e.m.f. terminals are the same electrical point, call it N. The network seen from N is then a triangle N–1–2 with sides $j0.11$, $j0.02$ and $j0.2$, plus two paths onward from buses 1 and 2 to the faulted bus 3. There is no series–parallel move available on that triangle, so a delta–star transformation is unavoidable.

The delta formed by the two source arms and the 1-2 tie, and its star equivalent. The transformation preserves the voltages and currents at the three external nodes N, bus 1 and bus 2, which is what allows part (e) to be answered by back-substitution.

Transform the delta and reduce to the Thevenin reactance. With the three sides summing to $0.11+0.02+0.2 = 0.33$, the star arms are $$\begin{aligned}Z_N &= \frac{(0.11)(0.02)}{0.33}=0.0066667\\ Z_1 &= \frac{(0.11)(0.20)}{0.33}=0.0666667\\ Z_2 &= \frac{(0.02)(0.20)}{0.33}=0.0121212\end{aligned}$$ all in per unit. From the star centre S the two surviving routes to bus 3 are $Z_1 + 0.6 = 0.666667$ and $Z_2 + 0.4 = 0.412121$; in parallel they give $$\frac{(0.666667)(0.412121)}{0.666667+0.412121}=0.254682$$ and adding the $Z_N$ stem, $$\boxed{X_{th}=0.0066667+0.254682=0.261348\ \text{pu}}$$
Fault current at bus 3. For a bolted three-phase fault the prefault voltage 1.0 pu drives the Thevenin reactance: $$\boxed{I_f=\frac{1.0}{j0.261348}=3.8263\ \text{pu}\ \text{at}\ -90^\circ}$$ An independent bus-admittance solution of the untransformed three-bus network returns 3.826311 pu, agreeing to six figures.
Part (e) — back-substitute for the bus 1 voltage. The whole fault current flows through the stem $Z_N$, so the star-centre voltage is $$\begin{aligned}V_S &= 1.0-I_fZ_N = 1.0-(3.8263)(0.0066667) \\ &= 0.974491\ \text{pu}\end{aligned}$$ This voltage divides between the two arms: $$\begin{aligned}I_{via\,1} &=\frac{0.974491}{0.666667}=1.461737 \\ I_{via\,2} &=\frac{0.974491}{0.412121}=2.364574\end{aligned}$$ which sum to 3.826311 pu — a free check on the whole reduction. The bus 1 voltage is the star-centre voltage less the drop across its own star arm: $$\boxed{V_1=0.974491-(1.461737)(0.0666667)=0.8770\ \text{pu}}$$ (The same construction gives $V_2 = 0.9458$ pu; bus 2, sitting only $j0.02$ from a source, is held up much more stiffly than bus 1 behind $j0.11$.)

Part	Quantity	Result
(b)	Equivalent source reactance at bus 4, $X_a$	$j0.06$ pu
(c)	Equivalent source reactance at bus 1, $X_b$	$j0.11$ pu
(d)	Thevenin reactance at the faulted bus 3	$j0.261348$ pu
(d)	Three-phase fault current at bus 3	3.8263 pu at $-90^\circ$
(e)	Voltage at bus 1 during the fault	0.8770 pu
—	Voltage at bus 2 during the fault (by-product)	0.9458 pu

Topic. Successive network reduction for symmetrical fault studies; delta–star transformation; recovery of bus voltages by back-substitution.

Key relations. Parallel sources at equal voltage combine as ordinary parallel impedances; delta–star $Z_k = Z_{ki}Z_{kj}/\sum Z$; bolted three-phase fault current $I_f = V_{pre}/Z_{th}$; bus voltage during fault $V_k = V_{pre} - Z_{k,f}I_f$.

Why this works. The reason the reduction is legitimate is that all four sources are specified at the same magnitude and angle, so no circulating current can exist between them and their terminals are one electrical node. That single fact turns what would otherwise be a bus-impedance-matrix problem into arithmetic. Where the topology becomes meshed — the triangle of two source arms and the 1–2 tie — the delta–star transformation is the right tool precisely because it preserves the terminal behaviour at all three external nodes, so the star is not merely a device for getting $Z_{th}$ but keeps enough information to recover the individual bus voltages afterwards.

Common pitfall. Paralleling the two machine arms in part (b) while forgetting to include the $j0.1$ tie reactance in each branch gives $X_a = 0.01$ and poisons everything downstream. In part (e), the frequent error is subtracting the drop across the wrong element — the drop belongs to the star arm $Z_1$ carrying the current that flows towards bus 1, not to the whole $Z_1 + 0.6$ path or to the original $j0.11$.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 10 (faults on power systems, network reduction, three-phase fault calculations); Grainger & Stevenson, Power System Analysis, ch. 10. Protection philosophy: IEEE C37 series as applied in Canadian utility practice.

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