22-Elec-B7 Power Systems Engineering · May 2017

Question 6 of 7: Single Line-to-Ground Fault on a Sequence-Network Representation

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

Notes on this paper

Paper format. National Exams, May 2017 — 16-Elec-B7 Power Systems Engineering. Open-book, 3 hours, seven problems of equal value (25 points each); any five constitute a complete paper, and only the first five in the answer book are marked. All seven are solved here, because the set is a study resource rather than a sitting.

Reference texts. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the bundle-conductor mean radii and long-line ABCD constants of Problem 1, the round-rotor power-angle relations of Problem 2, the three-winding equivalent circuit of Problem 3, the sequence-network fault reduction of Problem 6 and the equal-area criterion of Problem 7. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 4–5 transmission-line parameters, ch. 6 power flow, ch. 7–9 symmetrical components and faults, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the synchronous-machine and transformer chapters. Canadian practice references where protection and grounding are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.3 for overhead systems, and IEEE C37 (protective relaying) and IEEE 142 Green Book (grounding).

Check: four judgement calls carried through the solutions, recorded here rather than hedged inside the answers.

(i) Problem 1 states neither a frequency nor a line length. Parts (a)–(c) need neither. Part (d) needs both, so 60 Hz is used (Canadian national exam) and the ABCD constants are given as closed-form functions of length, evaluated at a representative 400 km with a sensitivity table at 200 km and 600 km. The characteristic impedance and the surge-impedance loading do not depend on either quantity.

(ii) Problem 1 gives “bundle separation S” for a bundle whose figure shows eight sub-conductors on a circle. S is read as the distance between adjacent sub-conductors, so the bundle radius is $A = S/[2\sin(\pi/N)] = 0.6533\ \text{m}$ — the standard El-Hawary reading. If a marker instead reads S as the bundle diameter, every mean radius scales by the same factor and the method is unchanged.

(iii) Problem 5 says “the voltage at both sources is 1 p.u.” above a figure that draws three generators. All three are taken at 1.0 pu, which is the only reading that closes the network; the three internal e.m.f.s then merge into one node.

(iv) Problem 6: the zero-sequence source reactance is printed as $X_{0S}=j0.04$, possibly followed by a fifth digit. The value $j0.04$ is used. Reading it as $j0.045$ moves the zero-sequence Thévenin reactance from 0.19978 to 0.20051 pu and the fault current by 0.1 % — below the resolution of any answer here.

Problem 6: Single Line-to-Ground Fault on a Sequence-Network Representation (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.

Given. Values in per unit on a 100 MVA base, read from Figure (5).

Element	Symbol	Positive / negative	Zero
Generator subtransient reactance	$X_d^{\prime\prime}$	$j0.20$	— (barred by the transformer delta)
Transformer at bus G	$X_{TG}$	$j0.1375$	$j0.1375$
Line G–H	$X_{GH}$	$j0.18148$	$j0.620$
Three-winding unit at bus H, medium arm	$Z_M$	$Z_M+Z_H = X_{HM}=j0.03667$ (positive); $Z_M=-j0.0083$ (zero)
Three-winding unit, low arm	$Z_L$	open (tertiary unloaded)	$j0.1950$
Three-winding unit, high arm	$Z_H$	within $X_{HM}$	$j0.0450$
System source behind bus H	$X_{S}$	$j0.03$	$j0.04$
Prefault voltage	$E$	$1.0\angle0^\circ$ pu

Find. The total current delivered to a bolted single line-to-ground fault at the mid-point of line G–H, and the three phase voltages at each of buses G and H while that fault is on.

[Figure not reproduced: Figure 6 — The two sequence networks of Figure (5) redrawn with the fault node F at the mid-point of line G–H. In the positive-sequence network the tertiary arm Z L is open, so the branch from H to the source is simply X HM + X 1S ; in the zero-sequence network all three arms of the thre. See the official exam paper.]

Approach. Reduce each sequence network to its Thévenin reactance seen from F, connect the three in series (the standard single line-to-ground connection), obtain the symmetrical components of the fault current, then propagate those component currents back through each network to get the sequence voltages at G and H and finally transform to phase quantities.

Part 1 — reduce the positive-sequence network. With the tertiary arm open, bus G is reached from the source through $X_d^{\prime\prime}+X_{TG}=0.20+0.1375=0.3375$ pu and bus H through $X_{1S}+X_{HM}=0.03+0.03667=0.06667$ pu. The fault at the mid-point splits the line into $0.18148/2=0.09074$ pu on each side, so the two paths from the source node to F are$$0.3375+0.09074=0.42824\ \text{pu}\qquad\text{and}\qquad 0.06667+0.09074=0.15741\ \text{pu},$$in parallel:$$X_1=\frac{(0.42824)(0.15741)}{0.42824+0.15741}=\frac{0.067409}{0.58565}=0.115102\ \text{pu}.$$The negative-sequence network is identical in this representation, so $X_2=X_1=j0.115102$ pu.
Reduce the zero-sequence network. Here the picture changes: the generator is barred by the delta of its transformer and does not appear, but the three-winding unit at bus H is fully present. Its high arm and the system source are in series, $Z_H+X_{0S}=0.0450+0.0400=0.0850$ pu, and that is in parallel with the low (tertiary) arm $Z_L=0.1950$ pu:$$\frac{(0.0850)(0.1950)}{0.0850+0.1950}=0.059196\ \text{pu},$$to which the medium arm adds its negative value: $0.059196-0.0083=0.050896$ pu from bus H to the reference.
Complete the zero-sequence reduction. The zero-sequence line is much larger than the positive-sequence line — $0.620$ pu, giving $0.310$ pu per half — so the two paths to F are $0.1375+0.310=0.4475$ pu through bus G and $0.050896+0.310=0.360896$ pu through bus H, and$$X_0=\frac{(0.4475)(0.360896)}{0.4475+0.360896}=\frac{0.161501}{0.808396}=0.199780\ \text{pu}.$$Because $X_0$ exceeds $X_1$, the single line-to-ground current will be smaller than the three-phase current at the same point — the standard signature of a network whose zero-sequence path is the weaker one.
Connect the three networks in series. A bolted single line-to-ground fault on phase a forces $I_{a1}=I_{a2}=I_{a0}$ and $V_{a1}+V_{a2}+V_{a0}=0$, which is the series connection of the three Thévenin networks:$$I_{a1}=\frac{E}{X_1+X_2+X_0}=\frac{1.0\angle0^\circ}{j(0.115102+0.115102+0.199780)}=\frac{1.0}{j0.429983},$$giving $I_{a1}=I_{a2}=I_{a0}=2.32567\angle-90^\circ$ pu.

Figure 7 — The single line-to-ground connection: the three sequence Thévenin networks in series across the prefault voltage. The same current flows in all three, which is what makes Ia = 3 Ia1.

Report the fault current. The faulted-phase current is the sum of the three equal components,$$\boxed{I_f=I_a=3I_{a1}=6.9770\angle-90^\circ\ \text{pu},}$$the other two phases carrying zero. On a 100 MVA base this is 697.7 MVA of single-phase fault duty. For comparison, a three-phase fault at the same point would draw $1/X_1=8.6880$ pu, so the ground fault is 80 per cent of the three-phase value — consistent with $X_0 > X_1$, and the check that the zero-sequence network has been built correctly.
Part 2 — sequence voltages at bus G. Each network is now driven by its known component current. In the positive-sequence network the current divides between the two paths; the share flowing in from the G side is $I_{a1}(0.15741/0.58565)=0.62508$ pu, and bus G sits that current’s drop above F on the G-side half-line while also being pulled down from the source. Solving the network gives$$V_{G1}=0.789032,\qquad V_{G2}=-0.210968,\qquad V_{G0}=-0.142761\ \text{pu},$$all real because every impedance is a pure reactance and the driving voltage is real.
Sequence voltages at bus H. The same construction on the H side, where the source is electrically much closer, gives$$V_{H1}=0.886622,\qquad V_{H2}=-0.113378,\qquad V_{H0}=-0.065525\ \text{pu}.$$Bus H holds up better than bus G in every sequence, which is expected: its path to the source is $0.06667$ pu against $0.3375$ pu at bus G.
Transform to phase voltages at bus G. With $a=1\angle120^\circ$ and the standard synthesis $V_a=V_0+V_1+V_2$, $V_b=V_0+a^{2}V_1+aV_2$, $V_c=V_0+aV_1+a^{2}V_2$,$$\boxed{\begin{aligned}V_{aG}&=0.4353\angle0^\circ,\\ V_{bG}&=0.9677\angle-116.50^\circ,\\ V_{cG}&=0.9677\angle+116.50^\circ\ \text{pu}.\end{aligned}}$$
Transform to phase voltages at bus H. The same transformation on the bus-H components gives$$\boxed{\begin{aligned}V_{aH}&=0.7077\angle0^\circ,\\ V_{bH}&=0.9770\angle-117.57^\circ,\\ V_{cH}&=0.9770\angle+117.57^\circ\ \text{pu}.\end{aligned}}$$The faulted phase is depressed at both buses — heavily at G, which is on the weak side of the fault — while the two healthy phases stay near unity but swing apart in angle, the familiar unbalanced signature of an earth fault on an effectively-grounded system.
Check the result at the fault point itself. At F the three sequence voltages must sum to zero, and they do: $V_{F1}=0.732311$, $V_{F2}=-0.267689$ and $V_{F0}=-0.464622$ pu, summing to $0.000000$. Equivalently, $V_{F1}=E-jX_1I_{a1}$ with the same numbers. A second check is the coefficient of grounding: the healthy-phase voltages reach 0.977 pu, well below the 1.4 pu that would flag an ineffectively-grounded system, which is consistent with $X_0/X_1=1.74$ and $R_0/X_1=0$.

Final Results

Quantity	Symbol	Result
Positive- (and negative-) sequence Thévenin reactance	$X_1=X_2$	$j0.11510$ pu
Zero-sequence Thévenin reactance	$X_0$	$j0.19978$ pu
Sequence components of the fault current	$I_{a1}=I_{a2}=I_{a0}$	$2.3257\angle-90^\circ$ pu
(1) Current to the fault	$I_f=3I_{a1}$	$6.9770\angle-90^\circ$ pu (697.7 MVA)
Three-phase fault current at the same point (comparison)	$I_{3\phi}$	8.6880 pu
(2) Bus G phase voltages	$V_{aG},V_{bG},V_{cG}$	$0.4353\angle0^\circ$; $0.9677\angle-116.50^\circ$; $0.9677\angle116.50^\circ$ pu
(2) Bus H phase voltages	$V_{aH},V_{bH},V_{cH}$	$0.7077\angle0^\circ$; $0.9770\angle-117.57^\circ$; $0.9770\angle117.57^\circ$ pu

Concept & Theory

Topic. Symmetrical components and the single line-to-ground fault.

Key relations. $I_{a1}=I_{a2}=I_{a0}=E/(Z_1+Z_2+Z_0)$; $I_f=3I_{a1}$; $V_{a1}+V_{a2}+V_{a0}=0$ at the fault; $V_a=V_0+V_1+V_2$, $V_b=V_0+a^{2}V_1+aV_2$, $V_c=V_0+aV_1+a^{2}V_2$ with $a=1\angle120^\circ$; $X_0 < X_1 \Rightarrow I_{SLG} > I_{3\phi}$ and conversely.

Why this works. Fortescue’s transformation turns any unbalanced three-phase set into three balanced sets, each of which sees its own network. The boundary conditions of a phase-a earth fault — $I_b=I_c=0$ and $V_a=0$ — translate into equal sequence currents and zero summed sequence voltage, which is precisely a series connection of the three Thévenin networks. The zero-sequence network is the one that differs structurally, because it needs a return path through earth: delta windings and ungrounded neutrals block it, which is why the generator here contributes to $X_1$ and $X_2$ but not at all to $X_0$. That structural difference is what makes the comparison of $X_0$ with $X_1$ such a reliable check on the arithmetic.

Common pitfall. Two mistakes recur. Reporting $I_{a1}$ as the fault current instead of $3I_{a1}$ — a factor-of-three error that would badly under-rate the earthing conductor. And carrying the generator’s zero-sequence reactance into $X_0$ when its transformer presents a delta to the machine: the printed value is then a decoy that never enters any answer. Reading the winding connections off the one-line before writing the zero-sequence network is not optional.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 11 (symmetrical components and unbalanced fault analysis); Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed., §8.1–9.3; IEEE Std 142 Green Book, ch. 1, for the grounding-effectiveness criteria quoted in the final check.

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