Question 6 of 7: Unsymmetrical Faults Seen at Breaker B 2

Problem 6: Unsymmetrical Faults Seen at Breaker B₂ (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.

Find. The fault current for a three-phase, a single-line-to-ground, a double-line-to-ground and a line-to-line fault at bus 3, and the identification of the least severe of the four.

Approach. Build the three sequence networks looking back from bus 3, then apply the standard sequence-network interconnections: series-parallel for each fault type, transform back to phase currents. Because the system is radial — source, transformer, bus 1, line, bus 2, line, bus 3 — every ampere reaching a fault at bus 3 must pass through breaker B₂, so the currents asked for are the total fault currents.

[Figure not reproduced: Figure 6.1 — the three sequence networks seen from bus 3. The delta winding of the transformer blocks zero-sequence current from the source, so the grounded wye is the only zero-sequence source and the transformer contributes only its own j5 Ω. See the official exam paper.]

1. Assemble the sequence impedances at the fault point. Positive and negative sequence see the transformer and both line sections in series: $$Z_1=Z_2=j5+j10+j10=j25\ \Omega .$$ Zero sequence is different. The transformer is delta on the source side and grounded-wye on the line side, so zero-sequence current can circulate in the delta but cannot pass through to the generator: looking back from the line, the zero network sees the transformer's own $j5\ \Omega$ to the reference bus, and nothing beyond. Adding the two line sections at their much larger zero-sequence value, $$Z_0=j5+j30+j30=j65\ \Omega .$$ The prefault voltage at bus 3 is the phase value $V_f = 34.5/\sqrt{3} = 19919$ V.
2. Part (a) — symmetrical three-phase fault. A balanced fault excites only the positive-sequence network: $$I_f^{(3\phi)}=\frac{V_f}{Z_1} =\frac{19919}{j25} \;\Longrightarrow\;\boxed{I_f^{(3\phi)}=796.7\ \text{A} \ \text{(lagging }90^\circ)}$$
3. Part (b) — single line to ground. A phase-a-to-ground fault puts the three sequence networks in series, so $I_{a1}=I_{a2}=I_{a0}=V_f/(Z_1+Z_2+Z_0)$ and the faulted-phase current is three times that: $$I_a=\frac{3V_f}{Z_1+Z_2+Z_0} =\frac{3\times19919}{j(25+25+65)} =\frac{59756}{j115}$$ $$\boxed{I_f^{(\text{SLG})}=519.6\ \text{A}}$$ It is smaller than the three-phase value precisely because $Z_0 = j65\ \Omega$ is much larger than $Z_1$ — the long zero-sequence path of the two lines dominates the series sum.
4. Part (c) — double line to ground, sequence components. A b–c–ground fault puts the negative and zero networks in parallel and that combination in series with the positive network: $$I_{a1}=\frac{V_f}{Z_1+\dfrac{Z_2Z_0}{Z_2+Z_0}} =\frac{19919}{j\left(25+\dfrac{25\times65}{90}\right)} =\frac{19919}{j43.056} =462.6\ \text{A}\ \angle-90^\circ$$ The other two sequence currents follow from the current-division rule at that parallel pair: $$I_{a2}=-I_{a1}\frac{Z_0}{Z_0+Z_2}=334.1\ \text{A} \ \angle+90^\circ,\qquad I_{a0}=-I_{a1}\frac{Z_2}{Z_0+Z_2}=128.5\ \text{A} \ \angle+90^\circ .$$
5. Double line to ground, phase currents. Recombining with $a = 1\angle120^\circ$, $$I_b=I_{a0}+a^{2}I_{a1}+aI_{a2},\qquad I_c=I_{a0}+aI_{a1}+a^{2}I_{a2}$$ $$\boxed{|I_b|=|I_c|=716.4\ \text{A}}$$ and the current returning through earth is $$I_g=3I_{a0}=385.5\ \text{A}.$$ The healthy phase carries $I_a = I_{a0}+I_{a1}+I_{a2} = 0$, as it must. Both numbers matter in practice: the phase elements of the relay at B₂ see 716 A while the ground element sees 386 A.
6. Part (d) — line to line. A b–c fault clear of ground excites no zero sequence; the positive and negative networks are connected in opposition, so $I_{a1}=-I_{a2}=V_f/(Z_1+Z_2)$ and $$I_b=-I_c=-j\sqrt{3}\,\frac{V_f}{Z_1+Z_2} =-j\sqrt{3}\times\frac{19919}{j50}$$ $$\boxed{|I_f^{(\text{LL})}|=690.0\ \text{A}}$$ which is $\sqrt{3}/2 = 0.866$ of the three-phase value — the classic result for a fault fed through equal positive and negative sequence impedances.
7. Part (e) — which fault is the least severe? Collecting the four magnitudes of the current that flows in the faulted phase conductors, and therefore through breaker B₂:

$$\boxed{\text{smallest fault current: single line-to-ground, } 519.6\ \text{A}}$$

The reason is worth stating rather than merely quoting the number. In most solidly grounded systems close to a generator, $Z_0$ is smaller than $Z_1$ and the single-line-to-ground fault is the most severe of the four. Here the opposite holds: the zero-sequence impedance of an overhead line is typically two to three times its positive-sequence value (the return path is the earth and the shield wires, not the other two phases), and with two line sections in the path $Z_0 = j65\ \Omega$ against $Z_1 = j25\ \Omega$. The series sum $Z_1+Z_2+Z_0 = j115\ \Omega$ is then so much larger than $Z_1$ alone that the ground fault is the mildest case. Note also that the double-line-to-ground ground current, 386 A, is lower still — but that is the earth-return current, not the current the breaker's phase contacts must interrupt.