Question 6 of 7: Per-Unit Sequence Networks and a Line-to-Line Fault Mid-Line

Problem 6: Per-Unit Sequence Networks and a Line-to-Line Fault Mid-Line (25 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.

Find. All three sequence networks in per unit on the stated base, then the current in the faulted phases for a line-to-line fault between phases B and C at the mid-point of the transmission line.

Approach. Establish the voltage base in each of the three zones from the transformer turns ratios, convert every reactance with $X_{\text{new}}=X_{\text{old}}\left(\frac{S_{\text{new}}}{S_{\text{old}}}\right) \left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^{2}$, build the networks, then apply the line-to-line fault connection $I_{a1}=-I_{a2}=E/(X_1+X_2)$.

1. Part (a) — establish the voltage bases in all three zones. The base is 25 MVA throughout and 11 kV on the generator side. Crossing T2, whose ratio is 10.8 : 121, the line-side base becomes $$V_{b,\text{line}}=11\times\frac{121}{10.8}=123.24\ \text{kV}$$ and crossing T1 back down, whose ratio is the same, the motor-side base returns to $123.24\times\frac{10.8}{121}=11.00$ kV. The corresponding impedance bases are $$Z_{b,\text{gen}}=\frac{11^{2}}{25}=4.84\ \Omega,\qquad Z_{b,\text{line}}=\frac{123.24^{2}}{25}=607.53\ \Omega$$
2. Convert the machines and transformers. The generator is rated at exactly the chosen base, so its reactance is unchanged at 0.20 pu. Each transformer must be moved from 30 MVA and 10.8 kV to 25 MVA and 11 kV: $$X_T=0.10\left(\frac{25}{30}\right)\left(\frac{10.8}{11}\right)^{2}=0.08033\ \text{pu}$$ The motors are rated at 10 kV, not the 11 kV base of their zone: $$\begin{aligned} X_{m1} &= 0.25\left(\frac{25}{15}\right)\left(\frac{10}{11}\right)^{2}=0.34435\ \text{pu}\\ X_{m2} &= 0.25\left(\frac{25}{7.5}\right)\left(\frac{10}{11}\right)^{2}=0.68871\ \text{pu} \end{aligned}$$ and the line reactance converts by dividing by its own impedance base: $$X_{\text{line}}=\frac{100}{607.53}=0.16460\ \text{pu}$$ Every one of these agrees with the figure to within its printed precision (0.0805, 0.164, 0.345 and 0.69 respectively), which is what part (a) asks to be verified.
Figure 6.1 — the positive-sequence network on the 25 MVA, 11 kV base, with the line split at the fault point F. The negative-sequence network is identical with the internal voltage sources removed.
3. Zero-sequence network, and why the printed machine data is a decoy. The zero-sequence line reactance converts the same way, $300/607.53=0.49380$ pu, and each half-line is $0.24690$ pu. The connections now decide everything. Figure (3-a) shows T2 with its delta on the generator side and T1 with its delta on the motor side, both line-side windings being solidly earthed wye. A delta winding provides a circulating path for zero-sequence current but passes none of it through, so it is an open circuit looking from the machine and a short to the reference looking from the line. Consequently both machines are cut out of the zero-sequence network altogether: the 0.06 pu machine zero-sequence reactances and both 2.5 ohm neutral reactors (which would have contributed $3Z_n/Z_b = 7.5/4.84 = 1.550$ pu) never appear in any answer. The only paths to the reference are the two earthed transformer neutrals: $$X_0\ \text{at }F=\frac{0.24690+0.08033}{2}=0.16362\ \text{pu}$$ the two branches being equal because F is exactly mid-line.
Figure 6.2 — the zero-sequence network. Both delta windings isolate their machines, leaving only the two earthed transformer neutrals as paths to the reference bus.
4. Part (b) — reduce the positive-sequence network to the fault point. Looking left from F: the generator, transformer T2 and half the line in series, $$X_{\text{left}}=0.20000+0.08033+0.08230=0.36263\ \text{pu}$$ Looking right: the two motors in parallel behind transformer T1 and the other half-line, $$X_{m1}\parallel X_{m2}=\frac{(0.34435)(0.68871)}{0.34435+0.68871}=0.22957\ \text{pu}$$ $$X_{\text{right}}=0.22957+0.08033+0.08230=0.39220\ \text{pu}$$ In parallel these give the Thevenin reactance at F: $$\boxed{X_1=X_2=\frac{(0.36263)(0.39220)}{0.36263+0.39220}=0.18842\ \text{pu}}$$ $X_2$ equals $X_1$ because the negative-sequence network has the same branch values with the sources shorted, and for these machines the negative-sequence reactance is taken equal to the subtransient reactance.
5. Apply the line-to-line fault connection. A fault between phases B and C involves no earth, so no zero-sequence current can flow and $I_{a0}=0$; the positive- and negative-sequence networks are connected in parallel opposition at the fault point, giving $$I_{a1}=-I_{a2}=\frac{E}{j(X_1+X_2)}=\frac{1.0}{j(0.18842+0.18842)} =\frac{1.0}{j0.37684}=2.6537\angle -90^{\circ}\ \text{pu}$$ The phase currents follow from the symmetrical-component synthesis with $a=1\angle 120^{\circ}$: $$I_b=a^{2}I_{a1}+aI_{a2}+I_{a0}=(a^{2}-a)I_{a1}=-j\sqrt3\,I_{a1}$$ so the magnitude of the current in each faulted phase is $$|I_b|=|I_c|=\sqrt3\,(2.6537)=4.5963\ \text{pu}$$ and $I_a = 0$, as it must be for a phase-to-phase fault.
6. Convert to amperes at the fault point. The fault is on the transmission line, so the relevant base current is the line-side one: $$I_{b,\text{line}}=\frac{25\times10^{6}}{\sqrt3\,(123\,240)}=117.12\ \text{A}$$ $$\boxed{|I_b|=|I_c|=4.5963\times117.12=538.3\ \text{A}}$$ The result carries its own check: the three-phase fault at the same point would be $1/X_1=5.3074$ pu, and the ratio of the two is $4.5963/5.3074=0.866=\sqrt3/2$ exactly, which is the signature of a line-to-line fault on a network where $X_1=X_2$.

Final Results.