Question 5 of 6: Zero-sequence network and shunt-fault currents

Question 5: Zero-sequence network and shunt-fault currents (20 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.

Find. (A) the zero-sequence network of the system, with every determinable impedance labelled on the 100 MVA base; and (B) the three-phase and single-line-to-ground fault magnitudes at a separate bus, plus a short explanation of how zero-sequence current is excluded from a machine and why zero- and negative-sequence currents are harmful to one.

Approach. Convert the one datum that is in ohms, then walk the one-line diagram element by element and ask a single question of each: does this element offer a path for a set of three equal, in-phase currents to reach earth? Delta windings and ungrounded wyes do not; grounded-wye windings and grounded machine neutrals do, with any neutral impedance counted three times.

1. Put the line onto the per-unit base. The base impedance on the 138 kV side is $$Z_{base}=\frac{V_{base}^{2}}{S_{base}}=\frac{(138\times 10^{3})^{2}}{100\times 10^{6}} =190.44\ \Omega$$ so the positive-sequence line reactance is $17.4/190.44=0.09137$ pu. Every machine and transformer datum is already expressed on 100 MVA and its own rated voltage, so no other conversion is needed. For the zero-sequence network the line reactance is taken at the customary $X_{0,line}\approx 3X_{1,line}$, giving $0.2741$ pu (see the callout).
2. Handle the generator neutrals. A machine grounded through a reactance $X_g$ appears in the zero-sequence network as its own $X_0$ in series with three times the neutral impedance, because the full residual current $3I_{a0}$ flows in the neutral connection: $$X_{0,branch}=X_0+3X_g=0.3+3(0.1)=0.60\ \text{pu per machine}$$ G1 and G2 sit on the same 20 kV bus, so they parallel to $0.60/2=0.30$ pu. G3 has the same $0.60$ pu branch on its own bus.
3. Read the transformer connections, because they decide the whole topology. The one-line diagram draws T1 as grounded-wye on both sides, so zero-sequence current passes straight through it as a series reactance of $j0.15$ pu joining the 20 kV bus to the 138 kV bus. It draws T2 as grounded-wye on the 138 kV side and delta on the 20 kV side, so zero-sequence current entering from the line circulates in the delta and returns to earth through the grounded neutral: T2 appears as a $j0.15$ pu branch from the 138 kV bus to the reference, and everything beyond it — G3 and load B — is completely severed from the rest of the zero-sequence network. The transformers' own $X_g=0$ means their neutrals are solidly grounded and add nothing.
4. Deal with the two loads, which are decoys. Load A is delta connected, and a delta has no connection to earth, so it presents an open circuit to zero-sequence current. Load B is stated as wye connected with no neutral-to-earth connection shown on the diagram, so it is an ungrounded wye and is likewise an open circuit. Neither load appears in the zero-sequence network at all. Their megawatt ratings are therefore never used — a deliberate trap. $$\boxed{\text{Zero-sequence network: } X_{0}^{G1\|G2}=0.30,\;X_{T1}=0.15,\; X_{0,line}=0.2741,\;X_{T2}=0.15\ \text{pu; G3 island } 0.60\ \text{pu}}$$

Part B(i) — solid three-phase fault. A balanced three-phase fault excites only the positive-sequence network, which is a single source of 1.0 pu behind $X_1$: $$|I_f^{3\phi}|=\frac{|V_f|}{|X_1|}=\frac{1.0}{0.23}=\boxed{4.348\ \text{pu}}$$

Part B(ii) — single line-to-ground fault. An SLG fault connects the three sequence networks in series, so $$I_{a1}=I_{a2}=I_{a0}=\frac{V_f}{X_1+X_2+X_0}=\frac{1.0}{0.23+0.23+0.45}=1.0989\ \text{pu}$$ and the faulted-phase current is the sum of the three components: $$|I_f^{SLG}|=|3I_{a0}|=\frac{3\times 1.0}{0.91}=\boxed{3.297\ \text{pu}}$$ Here the ground-fault current is smaller than the three-phase current, and the reason is visible in the data: $X_0=0.45>X_1=0.23$. That inequality is the standard sanity check on this class of problem — whenever $X_0 < X_1$ the single-line-to-ground current must exceed the three-phase current, and whenever $X_0 > X_1$ it must fall below it. On the 138 kV base the two currents are $4.348\times 418.4=1\,819$ A and $3.297\times 418.4=1\,379$ A respectively.

Part B(iii) — blocking zero-sequence current in a generator. Zero-sequence current is a set of three equal, in-phase currents, so it can only flow if there is a return path through the neutral to earth. Break that path and the current cannot exist. In practice there are three ways to do it, in order of how common they are.

The usual arrangement on a unit-connected machine is to interpose a delta-connected transformer winding between the generator and the system: the generator step-up transformer is built grounded-wye on the high-voltage side and delta on the machine side, so system zero-sequence current circulates within the delta and never reaches the stator. Second, the machine neutral can simply be left ungrounded (isolated), which removes the return path at the machine itself; this is rare on large units because it makes the machine's own ground-fault voltage difficult to detect and produces high transient overvoltages during arcing faults. Third, and most common in practice for the machine's own protection, the neutral is grounded through a high impedance — a distribution transformer with a secondary loading resistor, or a neutral grounding reactor as the $X_g=0.1$ pu of part A — which does not strictly block zero-sequence current but limits it to a few amperes, small enough to avoid burning the stator iron while still large enough to be detected by a neutral overvoltage relay. A delta-connected stator winding would also block it, but stators are wound in wye so that the neutral is available for exactly this protection.

Part B(iv) — why zero- and negative-sequence currents must be minimised. Neither component does useful work, and both damage the machine, but by different mechanisms.

Negative-sequence current produces a stator m.m.f. wave that rotates backwards at synchronous speed. Relative to a rotor turning forwards at synchronous speed that wave sweeps past at twice synchronous speed, so it induces double-frequency (120 Hz) currents in the rotor body, the retaining rings, the wedges and the amortisseur winding. Those currents flow in the rotor surface, where the skin depth is small and the resistance therefore high, and the resulting heating is intense and very local: it can anneal the retaining rings or melt the wedges in minutes. The interaction also produces a double-frequency pulsating torque that shakes the shaft and excites torsional modes. Machine standards therefore limit continuous negative-sequence current to about 5–10 % of rated and limit short-time exposure by an $I_2^{2}t\le K$ criterion, which is what a negative-sequence (46) relay implements.

Zero-sequence current produces no net rotating air-gap flux at all — the three components are in phase, so their m.m.f. contributions cancel around the periphery — and therefore produces no torque. What it does produce is $I^{2}R$ heating in the stator, third-harmonic flux that circulates in the end regions and induces eddy currents in the core clamping structure, and above all a ground-fault current path. A stator ground fault that is allowed to carry appreciable current burns the laminated core at the fault point, and the damage escalates from a repairable winding fault to a core restack — a distinction worth hundreds of thousands of dollars and months of outage. That, rather than the heating, is the real reason large machines are high-impedance grounded.