Question 5 of 7: Transmission-Line Protection Schemes, and a Mid-Line Three-Phase Fault

Question 5: Transmission-Line Protection Schemes, and a Mid-Line Three-Phase Fault (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.

Part (a) — three protection schemes for high-voltage transmission lines.

Distance (impedance) protection. The relay measures the ratio of the local voltage to the local current and compares the resulting apparent impedance with the known impedance of the line, so it responds to the electrical distance to the fault rather than to the fault current alone. Its characteristic is plotted on the R-X diagram — mho, offset-mho or quadrilateral — and it is applied in stepped zones: zone 1 is set to reach about eighty to eighty-five per cent of the line and trips instantaneously, zone 2 reaches roughly one hundred and twenty per cent with a short time delay to cover the remaining line end while coordinating with the downstream relay, and zone 3 provides remote backup with a longer delay. Its great advantage is that the setting is independent of the source impedance behind the relay, which changes with system configuration; its limitation is that zone 1 can never be set to cover the whole line, and that fault-arc resistance and infeed from an intermediate source can move the measured impedance out of the characteristic.

Pilot (unit) protection using a communication channel. Because no single-ended measurement can distinguish a fault just inside the far bus from one just outside it, the two line ends are made to exchange information over a fibre, power-line carrier or microwave channel. Line current differential protection compares the phasor currents at the two ends directly and trips on the vector difference, which is a true unit scheme with sharply defined boundaries. Directional comparison schemes — permissive over-reaching or under-reaching transfer trip, and directional comparison blocking — instead send a simple permit or block signal derived from the local directional or distance elements. Both give instantaneous tripping for one hundred per cent of the protected circuit, which is what makes them mandatory on the extra-high-voltage backbone where slow clearing would threaten transient stability.

Overcurrent and directional earth-fault protection, with automatic reclosing. Time-graded inverse-definite-minimum-time overcurrent relays, made directional where the circuit can be fed from both ends, and a sensitive directional earth-fault element for high-resistance ground faults, form the backup layer beneath the two schemes above. Since the great majority of overhead-line faults are transient flashovers, single-shot high-speed auto-reclosing is normally applied with these schemes, with synchronism check on interconnecting circuits. In Canadian practice the relaying, coordination and reclosing requirements for these circuits follow the IEEE C37 relay standard series together with the utility standards written around CSA C22.3 for overhead systems.

Given. The four-bus system of Figure (3), with four identical generating units and four transmission circuits, all reactances in per unit on a common base.

Find. The voltage at bus 3 while the bolted three-phase fault at F1 persists.

Approach. All four sources are at $1.0\,/\,0^{\circ}$ pu, so they may be merged into one node; the network then reduces to the fault point by a series-parallel step, one star-to-delta transformation at bus 3 and one delta-to-star transformation on the remaining triangle. Back-substituting the fault current gives the bus voltages, and a single nodal equation at bus 3 completes the answer.

1. Part (b) — merge the sources and eliminate the dead-end bus. Every generator is represented by $E=1.0\,/\,0^{\circ}$ pu behind $j0.175$, so the four internal nodes are at identical potential and can be joined into one reference source node N. Bus 2 then carries no load or fault: it is reached from N directly through $j0.175$, and round through line 1-2 in series with bus 1's own source, $j(0.175+0.300)=j0.475$. Those two paths are in parallel between N and bus 1, $$x_{N1}=\frac{(0.175)(0.475)}{0.175+0.475}=\frac{0.083125}{0.650}=\boxed{0.127885\ \text{pu}}$$ Bus 2 has now been removed from the reduction; its voltage is recovered at the end.
2. Convert the star at bus 3 into a delta. Bus 3 is a junction of three branches and carries nothing else: $j0.175$ to N, $j0.300$ to bus 1 and $j0.150$ to bus 4. Labelling those arms $a=0.175$, $b=0.300$ and $c=0.150$, the sum of the pairwise products is $$ab+bc+ca=0.0525+0.0450+0.02625=0.12375$$ and each delta branch is that sum divided by the opposite arm, $$\begin{aligned} x_{N1}'&=\frac{0.12375}{0.150}=0.8250 \\ x_{14}'&=\frac{0.12375}{0.175}=0.707143 \\ x_{N4}'&=\frac{0.12375}{0.300}=0.4125 \end{aligned}$$ The transformation is exact for every quantity external to bus 3, which is why bus 3 must be restored by a separate step later.
3. Combine the parallel branches that the delta created. The new N-to-1 branch parallels the one found in step 1, and the new N-to-4 branch parallels bus 4's own source reactance: $$\begin{aligned} x_{N1}''&=0.127885\,\|\,0.825=\frac{(0.127885)(0.825)}{0.952885}=0.110721 \\ x_{N4}''&=0.175\,\|\,0.4125=\frac{0.0721875}{0.5875}=0.122872 \end{aligned}$$ leaving a four-node network: N, bus 1, bus 4 and the fault point F, with $x_{14}=0.707143$ bridging the two buses and $j0.25$ from F to each of them.
4. Convert the remaining triangle into a star. The triangle N-1-4 has branches $0.110721$, $0.122872$ and $0.707143$, summing to $0.940736$. Each star arm is the product of the two delta branches meeting at that terminal divided by the sum, $$\begin{aligned} x_1&=\frac{(0.110721)(0.707143)}{0.940736}=0.083228 \\ x_4&=\frac{(0.122872)(0.707143)}{0.940736}=0.092362 \\ x_N&=\frac{(0.110721)(0.122872)}{0.940736}=0.014462 \end{aligned}$$ Call the new centre node S. This is the step that unlocks the problem: the bridge is now a simple two-branch parallel path from F back to S, plus one series arm from S to the source.
5. Thevenin reactance at the fault point. Working back from F, the two arms are $$0.25+x_1=0.333228 \qquad 0.25+x_4=0.342362$$ in parallel, $\frac{(0.333228)(0.342362)}{0.675590}=0.168867$, and in series with $x_N$: $$\boxed{X_{th}=0.014462+0.168867=0.183328\ \text{pu}}$$
6. Fault current. With a bolted three-phase fault at F and the pre-fault voltage taken as $1.0\,/\,0^{\circ}$ pu everywhere, $$I_F=\frac{1.0}{jX_{th}}=\frac{1.0}{j0.183328}=\boxed{I_F=5.4547\ \text{pu at } {-90^{\circ}}}$$ On a 100 MVA base this is a fault level of 545 MVA at the mid-point of line 1-4.
7. Back-substitute to the voltages at buses 1 and 4. The voltage at the star centre is $$V_S=1.0-I_FX_N=1.0-(5.454694)(0.014462)=0.921116\ \text{pu}$$ and the fault current divides between the two arms in inverse proportion to them, $$\begin{aligned} I_1&=I_F\frac{0.342362}{0.675590}=2.764219\ \text{pu} \\ I_4&=I_F\frac{0.333228}{0.675590}=2.690474\ \text{pu} \end{aligned}$$ Each bus voltage is the star-centre voltage less the drop in its own arm, $$V_1=0.921116-(2.764219)(0.083228)=0.691055\ \text{pu}$$ $$V_4=0.921116-(2.690474)(0.092362)=0.672619\ \text{pu}$$ The two arm currents sum back to $5.454693$ pu, which is a free check on the division.
8. Restore bus 3 with one nodal equation. Bus 3 was eliminated by the star-delta step, so it is recovered from Kirchhoff's current law at that node. It connects to the source through $j0.175$, to bus 1 through $j0.300$ and to bus 4 through $j0.150$, and carries no injection of its own, so $$\frac{V_3-1.0}{0.175}+\frac{V_3-V_1}{0.300}+\frac{V_3-V_4}{0.150}=0$$ $$V_3=\frac{1/0.175+V_1/0.300+V_4/0.150}{1/0.175+1/0.300+1/0.150}=\frac{5.714286+2.303516+4.484124}{15.714286}$$ $$\boxed{V_3=0.795577\ \text{pu}}$$ The common factor $j$ cancels from every term because all branches are purely reactive and every voltage in this network is therefore real.
9. Recover bus 2 and check the whole solution. The same nodal treatment at the dead-end bus gives $$V_2=\frac{1/0.175+V_1/0.300}{1/0.175+1/0.300}=\frac{5.714286+2.303516}{9.047619}=0.886178\ \text{pu}$$ Solving the untouched four-node network directly by nodal analysis, with no reduction at all, returns $V_1=0.691055$, $V_2=0.886178$, $V_3=0.795577$ and $V_4=0.672619$ pu and a fault current of $5.454694$ pu — agreement to every figure quoted. The pattern is physically sensible: the two buses adjacent to the faulted circuit are depressed hardest, and bus 2, furthest from the fault and backed by its own machine, holds up best.