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

Problem 6: Sequence Networks and Single Line-to-Ground Faults (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 Figure (4) system — G1 – T1 – bus 1 – L1 – bus 2 – L2 – bus 3 – T2 – G2 — with T1 delta on the machine side and grounded wye (through switch S) on the bus-1 side, and T2 grounded wye on the bus-3 side and delta on the machine side.

Find. The three sequence networks, then the phase-A fault current for a single line-to-ground fault at bus 2 with S closed and with S open, and the current for a three-phase fault at bus 2.

Approach. Build each sequence network from the one-line diagram — the transformer winding connections alone decide the zero-sequence topology — reduce each to a single reactance at bus 2, then connect the three networks in series for the single line-to-ground fault and use the positive network alone for the three-phase fault.

1. Part (a) — build the positive-sequence network. Every element appears in series along its path, and each generator drives its own internal emf of 1.0 pu behind its positive-sequence reactance. From bus 2, the two sides are identical chains: $$X_{+,\text{left}}=x_{G}+x_{T1}+x_{L1}=0.200+0.275+0.225=0.700\ \text{pu},$$ and the right-hand chain through $L_2$, $T_2$ and $G_2$ has the same three values. The two are in parallel at bus 2, so $$\boxed{X_1=\tfrac12(0.700)=0.350\ \text{pu}}$$
2. Build the negative-sequence network the same way. The topology is identical; only the machine reactance changes, and there is no internal emf because a balanced generator produces only positive-sequence voltage: $$X_{-,\text{each side}}=0.150+0.275+0.225=0.650\ \text{pu},$$ $$\boxed{X_2=\tfrac12(0.650)=0.325\ \text{pu}}$$
3. Part (b) — build the zero-sequence network from the winding connections. Zero-sequence current is in phase in all three conductors, so it needs a return path through earth. A delta winding provides a closed path for zero-sequence current to circulate but none for it to leave the winding, so it acts as an open circuit to the rest of the network; a grounded-wye winding connects to the reference bus through its neutral. Here both $T_1$ and $T_2$ have their delta on the machine side, so both generators are isolated from the zero-sequence network entirely, and the only connections to the reference are the two transformer neutrals — T1's through the switch S, T2's solidly.
4. Reduce the zero-sequence network with S closed. From bus 2 the left path is line $L_1$ then $T_1$ to reference, and the right path is line $L_2$ then $T_2$ to reference: $$X_{0,\text{left}}=x_{L1,0}+x_{T1,0}=0.225+0.275=0.500\ \text{pu} = X_{0,\text{right}},$$ $$\boxed{X_0=\tfrac12(0.500)=0.250\ \text{pu (S closed)}}$$
5. Part (c) — connect the three networks in series for the single line-to-ground fault. A bolted single line-to-ground fault on phase A imposes $I_b=I_c=0$ and $V_a=0$, which forces the three sequence currents to be equal and the three networks into series: $$I_{a1}=I_{a2}=I_{a0}=\frac{E}{j(X_1+X_2+X_0)}=\frac{1.0}{j(0.350+0.325+0.250)}=\frac{1.0}{j0.925}=1.0811\angle-90^\circ.$$
6. Sum the sequence components to get the phase current. Since $I_a=I_{a1}+I_{a2}+I_{a0}=3I_{a1}$, $$\boxed{I_{a}=\frac{3(1.0)}{j0.925}=3.2432\angle-90^\circ\ \text{pu (S closed)}}$$
7. Part (d) — open the switch and rebuild the zero-sequence network. With S open, T1's neutral no longer reaches the reference, so the entire left branch of the zero-sequence network is dead and the only path left is through $L_2$ and $T_2$: $$X_0'=0.225+0.275=0.500\ \text{pu},$$ $$\boxed{I_{a}'=\frac{3(1.0)}{j(0.350+0.325+0.500)}=\frac{3}{j1.175}=2.5532\angle-90^\circ\ \text{pu}}$$
8. State the effect of the transformer grounding. Opening S raises the zero-sequence reactance from 0.250 to 0.500 pu and drops the ground-fault current from 3.243 to 2.553 pu, a reduction of 21.3 %. That is the general result and the general design trade-off: more grounded-wye neutrals means a stiffer zero-sequence source, hence higher ground-fault duty on the breakers but also a well-defined neutral, ground-fault current large enough for ground relays to see reliably, and neutral-to-earth voltage on the healthy phases held near normal. Fewer grounded neutrals reduces the fault duty but pushes the healthy-phase voltages towards $\sqrt3$ times normal during a ground fault and can leave the ground fault too small to detect. Canadian transmission practice is effectively grounded systems ($X_0/X_1\le3$, $R_0/X_1\le1$), obtained by grounding enough transformer neutrals; distribution feeders are multi-grounded on the same reasoning. Note also that a single line-to-ground fault here draws more current than a three-phase fault (step 9), which is the signature of a strongly grounded system with $X_0\lt X_1$.
9. Part (e) — use the positive network alone for the three-phase fault. A balanced three-phase fault produces no negative- or zero-sequence current at all, so only $X_1$ appears: $$\boxed{I_F=\frac{1.0}{jX_1}=\frac{1.0}{j0.350}=2.8571\angle-90^\circ\ \text{pu}}$$ The switch position is irrelevant to this answer, because no zero-sequence current flows in a balanced fault.