Question 6 of 7: Single line-to-ground fault on an industrial bus

Problem 6 — Single line-to-ground fault on an industrial bus (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. A radial industrial system — generator, transformer T2, a transmission circuit, transformer T1 and two synchronous motors — with all three sequence networks already drawn. All internal emfs are 1.0 pu and pre-fault load current is neglected. The winding connections that matter are read from Figure (4-a): T2 has its delta on the generator side and its grounded wye on the bus-2 side; T1 has its grounded wye on the bus-3 side and its delta on the motor side; motor 1 is an ungrounded wye and motor 2 is grounded through a reactor.

Find. The current at bus 3 — that is, the phase currents flowing in the bus-2/bus-3 circuit at the bus-3 end — when a single line-to-ground fault occurs on phase a at bus 2, together with the total fault current itself.

[Figure not reproduced: Figure 8 — the zero-sequence network of Figure (4-d), redrawn to show what actually carries current. Two whole branches (greyed) are decoys: the generator is barred by the T2 delta and motor 1 by its ungrounded wye, and the entire bus-4/5/6 island is cut off from the fault by the T1 delta. See the official exam paper.]

Approach. Reduce each sequence network to its Thevenin reactance seen from bus 2, connect the three in series as a single line-to-ground fault requires, then divide each sequence current between the two paths that meet at bus 2 and recombine the shares that flow in the bus-3 circuit into phase quantities.

1. Reduce the positive-sequence network at bus 2. Two paths reach the reference from bus 2. To the left, the generator and T2 in series: $0.2+0.1=0.3000$ pu. To the right, the line, T1 and the two motors in parallel: $$0.15+0.1+\frac{(0.35)(0.7)}{0.35+0.7}=0.15+0.1+0.233333=0.483333\ \text{pu}$$ and these two in parallel give $X_1=\dfrac{(0.3000)(0.483333)}{0.783333}=0.185106$ pu.
2. The negative-sequence network is identical. Figure (4-c) prints the same reactances as Figure (4-b) with the sources removed, so $X_2=X_1=0.185106$ pu. That equality is worth noticing: it is what makes the final phase-current combination collapse so neatly in step 7.
3. Reduce the zero-sequence network — and identify the decoys. Transformer T2 presents its delta to the generator, so no zero-sequence current can reach the machine and the printed $j1.55+j0.05$ generator branch never enters any answer. Transformer T1 presents its delta to the motors, so the whole bus-4/5/6 island, including motor 2’s grounded neutral, is likewise barred. What remains at bus 2 is the T2 grounded-wye winding, $j0.1$ pu to the reference, in parallel with the line and the T1 grounded-wye winding in series, $j0.5+j0.1=j0.6$ pu: $$X_0=\frac{(0.1)(0.6)}{0.1+0.6}=\boxed{0.085714\ \text{pu}}$$ Two of the four reactances printed in Figure (4-d) are therefore never used, which is the point of the question.
4. Connect the three networks in series for the ground fault. A single line-to-ground fault on phase a imposes $I_b=I_c=0$ and $V_a=0$, which forces $I_{a0}=I_{a1}=I_{a2}$ and puts the three Thevenin reactances in series across the pre-fault voltage: $$I_{a1}=\frac{E}{j(X_1+X_2+X_0)}=\frac{1.0}{j(0.185106+0.185106+0.085714)}=2.193333\angle -90^{\circ}\ \text{pu}$$
5. Get the total fault current. The faulted phase carries the sum of the three equal sequence currents: $$I_f=3I_{a1}=\boxed{6.5800\angle -90^{\circ}\ \text{pu}}$$ This exceeds the three-phase fault current at the same bus, $1/X_1=5.4023$ pu, and it must: the single line-to-ground fault is the more severe of the two whenever $X_0 \lt X_1$, which is the case here because the two grounded-wye transformer neutrals form a much stiffer path to earth than the generator and motors do to the positive-sequence network.
6. Divide each sequence current between the two paths at bus 2. The positive- and negative-sequence currents split in inverse proportion to the two arms: the share flowing in from the bus-3 side is $\dfrac{0.3000}{0.783333}=0.382979$, so $I_1=I_2=0.8400$ pu in that circuit. The zero-sequence current splits between the $0.1$ pu T2 neutral path and the $0.6$ pu path through the line, so only $\dfrac{0.1}{0.7}=0.142857$ of it, that is $I_0=0.3133$ pu, flows in the bus-3 circuit.
7. Recombine into phase currents at bus 3. With $I_1=I_2$ the operator combination collapses, since $a^{2}+a=-1$: $$I_a=I_0+2I_1,\qquad I_b=I_c=I_0-I_1$$ Substituting the two shares, $$\boxed{|I_a|=1.9933\ \text{pu},\qquad |I_b|=|I_c|=0.5267\ \text{pu}}$$ Note that the healthy phases are not current-free: the zero-sequence share returning through the T1 neutral flows equally in all three conductors, and it does not cancel against the positive- and negative-sequence pair in phases b and c.
8. Close the answer with a balance check. The remaining share arrives from the bus-1 side, where the phase-a current is $(1-0.1429)I_{a1}+2(1-0.3830)I_{a1}=4.5867$ pu. Adding the two contributions to the faulted phase gives 6.5800 pu, which is $I_f$ — the sum rule is satisfied and the division is therefore right. For completeness the bus-2 voltages during the fault are $V_a=0$ by definition and $|V_b|=|V_c|=0.9108$ pu, so the healthy phases are barely depressed.