Sequence Networks and a Single Line-to-Ground Fault: 22-Elec-B7 Power Systems Engineering, December 2014, Q6

Question 6 (Problem 6): Sequence Networks and a Single Line-to-Ground Fault (15 + 10 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. Two generators with X₁ = X₂ = 0.20 and X₀ = 0.05 p.u. feed a two-bus system through transformers of 0.05 p.u.; two parallel lines join the buses, with 0.10 p.u. positive and negative sequence and 0.30 p.u. zero sequence. From Figure (3), T1 is grounded wye on both sides and T2 has its delta on the generator side. Prefault voltage 1.0 p.u.

Given data
Quantity	Symbol	Value
Generators, positive and negative sequence	X₁ = X₂	0.20 p.u.
Generators, zero sequence	X₀	0.05 p.u.
Transformers	X_T1 = X_T2	0.05 p.u.
Lines, positive and negative sequence	X₁₂	0.10 p.u.
Lines, zero sequence	X₁₂	0.30 p.u.
Transformer connections	T1 / T2	grounded Y – grounded Y / grounded Y – delta
Fault	—	single line to ground, mid-point of the lower line 1–2

Find. The Thevenin equivalent of each of the three sequence networks seen from a fault at the mid-point of the lower line, and the resulting single line-to-ground fault current in per unit.

Figure 7 — single-line diagram for the unsymmetrical fault study. Only the transformer neutrals and the grounded-wye generator can supply zero-sequence current.

Figure 8 — the two sequence networks seen from the fault point. The delta winding of the second transformer isolates the far generator from the zero-sequence network.

Figure 9 — the three Thevenin equivalents in series, which is the connection a single line-to-ground fault imposes on the sequence networks.

Approach. Build each sequence network according to what the transformer connections admit, reduce each to its Thevenin reactance at the fault point, then connect the three in series as a ground fault requires.

Part (a) — read the connections before drawing anything. The sequence networks differ only in what the transformer connections allow. Transformer T1 is grounded wye on both sides, so zero-sequence current passes through it and generator G1 — itself a grounded wye machine — is part of the zero-sequence network. Transformer T2 has its delta on the machine side, which provides a circulating path for zero-sequence current but blocks it from reaching G2. The generator zero-sequence reactance 0.05 therefore enters the calculation once, on the G1 side only; on the G2 side the only path to the reference is the grounded neutral of T2 itself.
Positive- and negative-sequence Thevenin reactance. Each machine reaches its bus through $X_{N1}=X_{N2}=0.20+0.05=0.25$, the upper line contributes 0.10 between the buses, and the faulted lower line is split into two halves of 0.050 each. Because the two sides are identical, the two buses are at the same potential when current is injected at F, so no current flows in the upper line and it drops out: $$\begin{aligned}X_{1}&=X_{2}=\dfrac{(0.050+0.25)}{2} \\ &=\boxed{j0.1500\ \text{p.u.}}\end{aligned}$$ A bus-impedance inversion that keeps the upper line in the network returns the same value, confirming that the symmetry argument is exact and not an approximation.
Zero-sequence network — the two shunt arms are now unequal. On the bus-1 side the path to the reference is $0.05+0.05=0.10$; on the bus-2 side it is the transformer alone, $0.05=0.05$. The lines carry their zero-sequence value 0.30, so the halves are 0.150 each. The symmetry argument no longer applies, and the (1, 2, reference) triangle must be transformed. With $\Sigma_{0}=0.30+0.10+0.05=0.45$, $$\begin{gathered}Z_{1}^{(0)}=0.06666667, \\ Z_{2}^{(0)}=0.03333333, \\ Z_{N}^{(0)}=0.01111111\end{gathered}$$
Combine the zero-sequence arms. $$\begin{gathered}\text{arm A}=0.150+0.06666667=0.21666667, \\ \text{arm B}=0.150+0.03333333=0.18333333\end{gathered}$$ $$\begin{aligned}&X_{0} \\ &=\dfrac{(\text{arm A})(\text{arm B})}{\text{arm A}+\text{arm B}}+Z_{N}^{(0)} \\ &=0.09930556+0.01111111 \\ &=\boxed{j0.110417\ \text{p.u.}}\end{aligned}$$ The zero-sequence reactance is smaller than the positive-sequence value, which is the usual situation on a system with solidly grounded transformer neutrals close to the fault, and it is the reason the ground fault current below exceeds the three-phase value.
Part (b) — connect the three networks for a single line-to-ground fault. A bolted single line-to-ground fault on phase a forces $I_{b}=I_{c}=0$ and $V_{a}=0$, which in sequence quantities means the three networks are connected in series and carry equal currents: $$\begin{aligned}I_{a1}&=I_{a2}=I_{a0}=\dfrac{E}{j\left(X_{1}+X_{2}+X_{0}\right)} \\ &=\dfrac{1.0}{j(0.1500+0.1500+0.110417)} \\ &=-j2.436548\ \text{p.u.}\end{aligned}$$
Faulted-phase current. The phase current is the sum of the three equal sequence components: $$\boxed{I_{f}=I_{a}=3I_{a1}=-j7.309645\ \text{p.u.},\qquad |I_{f}|=7.309645\ \text{p.u.}}$$ For comparison, a three-phase fault at the same point would draw $1/X_{1}=6.6667$ p.u. The ground fault is the more severe of the two here, purely because $X_{0}\lt X_{1}$; on a system whose transformers were both delta-connected on the bus side the zero-sequence network would be open and the ground fault current would be zero instead.

Problem 6 — results
Quantity	Symbol	Result
(a) positive-sequence Thevenin reactance	X₁	j0.150000 p.u.
(a) negative-sequence Thevenin reactance	X₂	j0.150000 p.u.
(a) zero-sequence Thevenin reactance	X₀	j0.110417 p.u.
(b) sequence current	I_a1 = I_a2 = I_a0	2.4365 p.u. (lagging 90°)
(b) fault current	\|I_f\| = 3\|I_a1\|	7.3096 p.u.
Three-phase fault at the same point (comparison)	\|I_F,3φ\|	6.6667 p.u.

Concept & Theory

Topic. Sequence networks and the single line-to-ground fault, including which machines a transformer connection admits to the zero-sequence network.

Key relations. $I_{a1}=I_{a2}=I_{a0}=E/(Z_1+Z_2+Z_0)$ for a bolted SLG fault; $I_f=3I_{a1}$; grounded-Y–grounded-Y passes zero sequence; grounded-Y–delta connects the wye side to the reference and opens the delta side.

Why this works. Symmetrical components turn one unbalanced three-phase problem into three balanced single-phase problems that interact only at the fault point, where the fault type dictates how the three networks are interconnected. A single line-to-ground fault imposes zero current in two phases, and that constraint translates into equal sequence currents, hence a series connection. The zero-sequence network is the one that differs structurally from the other two, because zero-sequence current is a set of three in-phase currents that must return through earth or through a delta winding; a transformer connection therefore either passes it, blocks it, or provides a source of it.

Common pitfall. Using the tabulated generator zero-sequence reactance on both sides. Behind a delta winding it is a decoy that never enters the answer, and including it makes the zero-sequence reactance too large and the ground fault current too small — a non-conservative error when breaker duty is being checked.

Source. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press/Wiley, Ch. 11 (symmetrical components and unbalanced faults); J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed., Cengage, §8.1–9.4; CSA C22.1 Section 10 for the Canadian grounding requirements that decide these connections in practice.

Question 6 of 7: Sequence Networks and a Single Line-to-Ground Fault

Question 6 (Problem 6): Sequence Networks and a Single Line-to-Ground Fault (15 + 10 points)