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

Question 6: Sequence Networks and a Double Line-to-Ground Fault (25 marks)

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 one-line diagram of Figure (4): generator $G_1$ (wye, solidly grounded) feeds the delta winding of transformer $T_1$, whose grounded-wye winding connects to bus 1 through a neutral switch S; bus 1 reaches bus 2 through line $L_1$ and bus 2 reaches bus 3 through line $L_2$; bus 3 connects to the grounded-wye winding of $T_2$, whose delta winding feeds generator $G_2$ (wye, grounded). Reactances as tabulated above; prefault voltage 1.0 pu; a double line-to-ground fault on phases B and C at bus 2.

Find. The three sequence networks including the switch, then the phase-B current in line $L_1$ with S closed and again with S open, and an interpretation of the difference.

Part (a) — constructing the three sequence networks. The positive- and negative-sequence networks have the same topology as the one-line diagram: each generator sits behind its own reactance, in series with its transformer, in series with the line reaching the fault point. The zero-sequence network is where the reading of the transformer winding symbols decides the answer.

A delta–grounded-wye transformer offers a path for zero-sequence current on its grounded-wye side only. Zero-sequence current entering that winding circulates in the delta and never appears on the machine side, so the delta bars each generator from the zero-sequence network entirely. Both transformers here carry their delta on the machine side, so both generators are cut out, and the tabulated generator zero-sequence reactance of 0.1 pu is a decoy that never enters any calculation. The only paths from the fault point to the zero-potential reference are the two transformer neutrals — and the $T_1$ neutral path exists only while switch S is closed.

Approach. Reduce each sequence network to a single reactance seen from bus 2, apply the double line-to-ground connection (positive in series with the parallel combination of negative and zero), then divide each sequence current between the two symmetric halves of the network and recombine into phase B by the symmetrical-component synthesis.

1. Part (b) — reduce the three sequence networks with S closed. Each side of the positive-sequence network presents generator plus transformer plus line, and the two sides are symmetric about the fault: $$X_1=\tfrac{1}{2}\left(0.20+0.25+0.30\right)=\tfrac{1}{2}(0.75)=0.375\ \text{pu}$$ The negative-sequence network is identical in shape with the negative-sequence data: $$X_2=\tfrac{1}{2}\left(0.15+0.25+0.320\right)=\tfrac{1}{2}(0.72)=0.360\ \text{pu}$$ In the zero-sequence network the generators are absent, so each side is only the transformer neutral in series with its line: $$X_0=\tfrac{1}{2}\left(0.25+0.30\right)=\tfrac{1}{2}(0.55)=0.275\ \text{pu}$$
2. Apply the double line-to-ground connection. For a DLG fault the positive-sequence network drives the parallel combination of the negative- and zero-sequence networks. The parallel term is $$X_2\!\parallel\!X_0=\frac{(0.360)(0.275)}{0.360+0.275}=0.155906\ \text{pu}$$ so the positive-sequence current at the fault is $$\boxed{I_{a1}=\frac{1.0}{j(0.375+0.155906)}=1.88357\angle -90^\circ\ \text{pu}}$$
3. Recover the sequence voltage and the remaining two sequence currents. All three sequence voltages are equal at a DLG fault: $$\begin{aligned}V_{a1} &= V_{a2} = V_{a0} = 1.0-I_{a1}(j0.375) \\ &= 1.0-0.706340 = 0.293660\ \text{pu}\end{aligned}$$ Each of the other two networks is then driven by that voltage: $$I_{a2}=\frac{-V_{a2}}{jX_2}=\frac{-0.293660}{j0.360}=0.815721\angle +90^\circ\ \text{pu}$$ $$I_{a0}=\frac{-V_{a0}}{jX_0}=\frac{-0.293660}{j0.275}=1.067853\angle +90^\circ\ \text{pu}$$ The check is that the three must sum to zero, because the healthy phase A carries no current in a B–C–ground fault: $-1.88357+0.81572+1.06785 = 0$ to nine decimal places. The total current into earth is $3I_{a0}=3.2036\angle 90^\circ$ pu.
4. Divide each sequence current into line $L_1$. Line $L_1$ carries only the contribution flowing in from the $G_1$ side. Because the two halves of every sequence network are identical in this system, each sequence current splits exactly in half: $$\begin{aligned}I_1^{L1} &=0.941787\angle -90^\circ \\ I_2^{L1} &=0.407861\angle +90^\circ \\ I_0^{L1} &=0.533927\angle +90^\circ\ \text{pu}\end{aligned}$$ The zero-sequence share is present only because S is closed; that is the whole point of part (c).
5. Synthesise the phase-B current in $L_1$. With $a = 1\angle 120^\circ$, the phase-B current is $$I_B=I_0+a^{2}I_1+aI_2$$ Term by term, $$\begin{aligned}a^{2}I_1 &= 0.941787\angle 150^\circ = -0.815611+j0.470894\\ aI_2 &= 0.407861\angle 210^\circ = -0.353218-j0.203931\\ I_0 &= 0.533927\angle 90^\circ = 0+j0.533927\end{aligned}$$ Summing, $$\boxed{I_B^{L1}=-1.168829+j0.800890=1.4169\angle 145.58^\circ\ \text{pu}}$$ Two free checks confirm it: phase C in $L_1$ comes out at the mirror value $1.4169\angle 34.42^\circ$ pu, and the two full-network phase currents sum to $I_B+I_C = 3I_{a0} = j3.2036$ pu exactly, as they must.
6. Part (c) — reopen the switch and rebuild only the zero-sequence network. Opening S breaks the $T_1$ neutral connection, so the entire left-hand branch of the zero-sequence network disappears. The positive- and negative-sequence networks are untouched. The only surviving path from bus 2 to the reference is through $L_2$ and the $T_2$ neutral: $$X_0'=0.30+0.25=0.55\ \text{pu}$$ The zero-sequence reactance has therefore doubled. Repeating the connection, $$X_2\!\parallel\!X_0'=\frac{(0.360)(0.550)}{0.910}=0.217582\ \text{pu}$$ $$I_{a1}'=\frac{1.0}{j(0.375+0.217582)}=1.687529\angle -90^\circ\ \text{pu}$$ $$V_{a}'=1.0-(1.687529)(0.375)=0.367177\ \text{pu}$$ $$\begin{aligned}I_{a2}' &=1.019935\angle 90^\circ \\ I_{a0}' &=0.667594\angle 90^\circ\ \text{pu}\end{aligned}$$
7. Recompute the phase-B current in $L_1$ — noting that its zero-sequence share is now zero. This is the step where the answer stops being a simple rescaling of part (b). With $T_1$'s neutral open there is no zero-sequence path through $L_1$ at all, so $I_0^{L1} = 0$; the whole of $I_{a0}'$ flows through $L_2$. The positive- and negative-sequence currents still split evenly: $$\begin{aligned}I_1^{L1} &=0.843764\angle -90^\circ \\ I_2^{L1} &=0.509968\angle +90^\circ\ \text{pu}\end{aligned}$$ $$\begin{aligned}a^{2}I_1 &= 0.843764\angle 150^\circ = -0.730721+j0.421882\\ aI_2 &= 0.509968\angle 210^\circ = -0.441645-j0.254984\end{aligned}$$ $$\boxed{I_B^{L1}=-1.172366+j0.166898=1.1842\angle 171.90^\circ\ \text{pu}}$$

8. Interpret the effect of grounding the transformer. Opening the $T_1$ neutral reduces the phase-B current in line $L_1$ from 1.4169 pu to 1.1842 pu, a fall of 16.4 %, and reduces the total current into earth at the fault from $3I_{a0} = 3.2036$ pu to 2.0028 pu, a fall of 37.5 %. Grounding the transformer therefore increases ground-fault duty. It does so for a specific reason worth stating: the neutral connection adds a second zero-sequence path in parallel with the first, which lowers $X_0$, and because $X_0$ then falls below $X_1$ the earth-fault current is driven up rather than down.

   The engineering trade is the classical one. Solid grounding buys a large, unambiguous earth-fault current that protective relays detect quickly and selectively, and it clamps the healthy-phase voltage rise during an earth fault to about 1.4 times normal, which lets insulation and surge arresters be rated lower. The price is greater fault energy — more arc damage at the fault, more mechanical stress on the windings, and a higher step-and-touch-potential hazard around the station ground grid. Leaving the neutral open reduces the damage but leaves the earth-fault current small and hard to detect, and permits the healthy phases to rise to full line-to-line voltage. Canadian transmission practice at 230 kV and above is effectively grounded for exactly the reasons above; resistance grounding is reserved for industrial and distribution systems where limiting the damage matters more than the speed of detection.