Sequence Networks and a Line-to-Line Fault at the Mid-Point of a Line: 22-Elec-B7 Power Systems Engineering, May 2018, Q6

Question 6: Sequence Networks and a Line-to-Line Fault at the Mid-Point of a Line (20 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. The three-machine system of Figure (4), with each machine connected to its bus through a transformer and the two lines radiating from bus 1.

Given sequence data
Element	$X_1$ (pu)	$X_2$ (pu)	Connection in Figure (4)
Generator G1	0.12	0.12	at bus 1 through T1
Generator G2	0.24	0.24	at bus 2 through T2
Generator G3	0.12	0.12	at bus 3 through T3
Transformers T1, T2, T3	0.12	0.12	one per machine
Line 1-2	0.20	0.20	bus 1 to bus 2
Line 1-3	0.20	0.20	bus 1 to bus 3, faulted at its mid-point

Find. The positive- and negative-sequence reactance diagrams, the Thevenin equivalent of each as seen from the mid-point of line 1-3, and the phase B and phase C currents for a line-to-line fault at that point.

Approach. Assemble each sequence network from the given reactances, split line 1-3 at its mid-point to expose the fault terminal, reduce to a single Thevenin reactance by series and parallel steps, and then connect the positive- and negative-sequence networks in series, which is the interconnection that a phase-to-phase fault imposes.

Figure 6.1 — part (a): the positive-sequence reactance diagram, with line 1-3 split at its mid-point F. Each machine appears in series with its own transformer as a single radial branch to the reference bus. The negative-sequence diagram is the same network with every source short-circuited to the reference bus, because negative-sequence reactances equal the positive-sequence values for every element here.

Part (a) — build the two sequence networks. In the positive-sequence network every machine appears as its internal EMF behind its own $X_1$, in series with the reactance of its transformer, because a two-winding transformer offers the same leakage reactance to positive and negative sequence regardless of how its windings are connected — the winding connection matters only to zero sequence. Combining each machine with its transformer gives three radial branches from the reference bus, $$\begin{aligned} x_1&=0.12+0.12=0.24\ \text{pu} \\ x_2&=0.24+0.12=0.36\ \text{pu} \\ x_3&=0.12+0.12=0.24\ \text{pu} \end{aligned}$$ The two lines then join bus 1 to bus 2 and bus 1 to bus 3, each at $j0.20$ pu, and the faulted line is drawn as two halves of $j0.10$ pu with the fault terminal F between them. The negative-sequence diagram is drawn identically with every EMF replaced by a short circuit to the reference bus, since the data give $X_2=X_1$ for every element; hence $X_2=X_1$ for the whole network as well.
Part (b) — reduce the network as seen from F, first branch. Looking out from the fault terminal there are exactly two routes back to the reference bus and they meet nowhere in between, so no star-delta transformation is needed. The first route runs through the half-line to bus 1. At bus 1 the machine branch $j0.24$ stands in parallel with the whole of the bus-2 leg, which is line 1-2 in series with machine 2, $$0.20+0.36=0.56\ \text{pu}$$ so that $$x_{1,ref}=\frac{(0.24)(0.56)}{0.24+0.56}=\frac{0.1344}{0.80}=0.1680\ \text{pu}$$ and the whole first route is $$0.10+0.1680=0.2680\ \text{pu}$$
Second branch, and the Thevenin reactance. The other route runs through the far half of line 1-3 to bus 3 and then straight down machine 3's branch, $$0.10+0.24=0.3400\ \text{pu}$$ The two routes are in parallel across the fault terminal, so $$X_1=X_2=\frac{(0.2680)(0.3400)}{0.2680+0.3400}=\frac{0.09112}{0.6080}$$ $$\boxed{X_1=X_2=j0.149868\ \text{pu}}$$ A bus-impedance calculation on the untouched network returns $j0.14986842$ pu, confirming the hand reduction to eight figures. The Thevenin voltage of the positive-sequence network is the pre-fault voltage at F, taken as $1.0\,/\,0^{\circ}$ pu; the Thevenin voltage of the negative-sequence network is zero, since a balanced system generates no negative sequence.
Part (c) — recognise which networks the fault connects. A line-to-line fault between phases B and C, with no connection to earth, imposes $I_a=0$, $I_b=-I_c$ and $V_b=V_c$ at the fault point. Transforming those into symmetrical components gives $I_{a0}=0$, $I_{a2}=-I_{a1}$ and $V_{a1}=V_{a2}$, which is exactly the constraint set produced by connecting the positive- and negative-sequence networks in series across the fault terminal, with the zero-sequence network left out entirely. That is worth stating explicitly: this fault has no earth path, so no zero-sequence current can flow, and none of the transformer winding connections drawn in Figure (4) enters the answer at all. It is also why the paper supplies no zero-sequence data.
Positive-sequence current at the fault. With the two equal reactances in series across the Thevenin source, $$I_{a1}=\frac{E_{th}}{j(X_1+X_2)}=\frac{1.0}{j(0.149868+0.149868)}=\frac{1.0}{j0.299737}$$ $$\boxed{I_{a1}=3.33626\ \text{pu at }{-90^{\circ}}}\qquad I_{a2}=-I_{a1}\qquad I_{a0}=0$$
Transform back to phase quantities. With $a=1\,/\,120^{\circ}$, $$I_b=I_{a0}+a^{2}I_{a1}+aI_{a2}=(a^{2}-a)I_{a1}=-j\sqrt{3}\,I_{a1}$$ $$I_c=I_{a0}+aI_{a1}+a^{2}I_{a2}=(a-a^{2})I_{a1}=+j\sqrt{3}\,I_{a1}$$ Substituting $I_{a1}=-j3.33626$ pu, $$\boxed{I_b=5.77857\ \text{pu at }180^{\circ} \qquad I_c=5.77857\ \text{pu at }0^{\circ}}$$ and $I_a=I_{a1}+I_{a2}=0$, as the fault definition requires. The two currents are equal in magnitude and exactly in antiphase, which is the signature of a phase-to-phase fault: the current simply circulates between the two faulted conductors.
Three free checks on the answer. First, whenever $X_1=X_2$ the ratio of the line-to-line fault current to the three-phase fault current at the same point is exactly $\sqrt{3}/2$; here the three-phase value is $1.0/0.149868=6.67252$ pu and $5.77857/6.67252=0.866025$, which is $\sqrt{3}/2$ to six figures. Second, the sequence voltages at the fault must be equal, $V_{a1}=V_{a2}=1.0-(3.33626)(0.149868)=0.500$ pu, so the healthy phase holds $V_a=V_{a1}+V_{a2}=1.000$ pu exactly — the single most useful check on this class of problem, since the two transfer drops cancel identically when $I_{a2}=-I_{a1}$. Third, the two faulted phases sit at $V_b=V_c=a^{2}V_{a1}+aV_{a2}=-0.500$ pu, equal to each other as $V_b=V_c$ demands.
Distribution of the fault current between the two sources. The positive-sequence current divides between the two routes in inverse proportion to their reactances: the bus-1 route carries $3.33626\times(0.3400/0.6080)=1.86567$ pu and the bus-3 route $3.33626\times(0.2680/0.6080)=1.47059$ pu, summing back to $3.33626$ pu. Since the two machines at bus 1 and bus 2 together contribute fifty-six per cent of the fault current, breaker duty at bus 1 is the governing case on this network.

Figure 6.2 — the positive-sequence Thevenin equivalent at F, and the series interconnection of the positive- and negative-sequence networks that a line-to-line fault imposes.

Final results — Question 6
Quantity	Symbol	Value
Machine plus transformer, bus 1	$x_1$	$j0.24$ pu
Machine plus transformer, bus 2	$x_2$	$j0.36$ pu
Machine plus transformer, bus 3	$x_3$	$j0.24$ pu
Route from F via bus 1	—	$j0.2680$ pu
Route from F via bus 3	—	$j0.3400$ pu
Positive-sequence Thevenin reactance	$X_1$	$j0.149868$ pu
Negative-sequence Thevenin reactance	$X_2$	$j0.149868$ pu
Positive-sequence fault current	$I_{a1}$	$3.33626\,/\,{-90^{\circ}}$ pu
Phase B fault current	$I_b$	$5.77857\,/\,180^{\circ}$ pu
Phase C fault current	$I_c$	$5.77857\,/\,0^{\circ}$ pu
Healthy-phase current	$I_a$	0
Three-phase fault current at F (for comparison)	$I_{3\phi}$	6.67252 pu

Check: the data list gives “Transformers $X_{T1}=X_{T2}=0.12$” but Figure (4) draws three transformers, T1, T2 and T3, one per machine. All three are therefore taken at $j0.12$ pu, which is the only reading that closes the network; the alternative — omitting T3 altogether — would raise the fault current by twenty-four per cent, to 7.168 pu in phases B and C. The transformer winding connections shown in Figure (4) are deliberately not used: a phase-to-phase fault has no earth path, so the zero-sequence network never enters the answer, and the paper accordingly supplies no zero-sequence reactances.

Concept & Theory

Topic. Symmetrical-component analysis of an unbalanced fault, and the sequence-network interconnection that each fault type imposes.

Key relations. Line-to-line fault on phases B and C: $I_{a0}=0$, $I_{a2}=-I_{a1}$, $I_{a1}=E_{th}/j(X_1+X_2)$, $|I_b|=|I_c|=\sqrt{3}\,|I_{a1}|$; $V_{a1}=V_{a2}=E_{th}-jX_1I_{a1}$; for comparison, three-phase $I_f=E_{th}/jX_1$ and single-line-to-ground $I_a=3E_{th}/j(X_1+X_2+X_0)$; the ratio $|I_{LL}|/|I_{3\phi}|=\sqrt{3}/2$ whenever $X_1=X_2$.

Why this works. Symmetrical components work because the three sequence networks of a balanced system are mutually decoupled; the fault is the only place where they interact, and the way they interact is fixed entirely by the terminal constraints the fault imposes. A phase-to-phase fault ties $V_b$ to $V_c$ and forces $I_b=-I_c$ with $I_a=0$, and the unique circuit that reproduces those three constraints is the positive- and negative-sequence networks joined in series. No earth is involved, so the zero-sequence network stays open and every transformer neutral and winding connection becomes irrelevant. Since $X_2$ almost always equals $X_1$ for lines and transformers and is close to it for machines, the $\sqrt{3}/2$ ratio holds very generally, which is why a phase-to-phase fault is normally less severe than a three-phase fault at the same point.

Common pitfall. Building and using the zero-sequence network anyway. It is a common reflex on any question that draws transformer winding symbols, and here it wastes time and can produce a wrong answer if the zero-sequence reactance is mistakenly added to the denominator. A second pitfall is applying $\sqrt{3}$ to the wrong quantity: the factor multiplies $|I_{a1}|$, not the three-phase fault current. A third is treating the half-lines as $j0.20$ each instead of $j0.10$.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, Ch. 11 (symmetrical components and unbalanced faults); Grainger and Stevenson, Power System Analysis, Ch. 11–12; Glover, Overbye and Sarma, 6th ed., §9.1–9.5.

Question 6 of 7: Sequence Networks and a Line-to-Line Fault at the Mid-Point of a Line

Question 6: Sequence Networks and a Line-to-Line Fault at the Mid-Point of a Line (20 points)