19.2 Symmetrical Components & Unbalanced Fault Analysis

19.2 Symmetrical Components & Unbalanced Fault Analysis

Balanced three-phase faults are the easy case and account for only about 5% of real faults. The remaining 95% are unbalanced, and the only practical way to analyse them by hand is Fortescue's method of symmetrical components, a core component of Enhanced TOS topic K, Power System Analysis (PRBEE Resolution No. 40, s. 2024 — 9.00% of the examination, 20 of the 100 Electrical Engineering items).

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1. The Symmetrical Component Transformation

Any three unbalanced phasors decompose into three balanced sets:

• Positive sequence (1) — three equal phasors 120° apart in the normal rotation $a$–$b$–$c$; this is the only set present in a healthy balanced system;
• Negative sequence (2) — three equal phasors 120° apart in the reversed rotation $a$–$c$–$b$; it produces a counter-rotating field that heats generator rotors, which is why negative-sequence relays exist;
• Zero sequence (0) — three equal phasors in phase; it can flow only where a neutral or ground return path exists.

The operator $a = 1\angle 120° = -0.5 + j0.866$ satisfies

$$a^{2} = 1\angle 240°, \qquad a^{3} = 1, \qquad 1 + a + a^{2} = 0$$

Synthesis (components to phase quantities):

$$V_a = V_{a0} + V_{a1} + V_{a2}$$ $$V_b = V_{a0} + a^{2}V_{a1} + aV_{a2}$$ $$V_c = V_{a0} + aV_{a1} + a^{2}V_{a2}$$

Analysis (phase quantities to components):

$$V_{a0} = \frac{1}{3}\left(V_a + V_b + V_c\right)$$ $$V_{a1} = \frac{1}{3}\left(V_a + aV_b + a^{2}V_c\right)$$ $$V_{a2} = \frac{1}{3}\left(V_a + a^{2}V_b + aV_c\right)$$

Two consequences are examined repeatedly:

• In a three-wire system (no neutral), $I_a + I_b + I_c = 0$, so zero-sequence current is zero by definition;
• In a four-wire system the neutral carries $I_n = 3I_{a0}$ — the neutral current is three times the zero-sequence current, not equal to it.

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2. Sequence Networks

Each sequence has its own network with its own impedance.

For static equipment — lines, cables, transformers — $Z_1 = Z_2$ always, because a static device cannot tell the direction of phase rotation. For rotating machines $Z_2 \ne Z_1$ in general, though $Z_2 \approx X_d''$ is the usual working assumption.

Zero-sequence line impedance is substantially higher than positive sequence because the return path is the earth and any overhead ground wires, enclosing a much larger flux loop.

Transformer Zero-Sequence Connection Rules

This table is the single most examinable item in the topic:

A delta winding is a zero-sequence trap: it gives circulating third-harmonic and zero-sequence currents somewhere to go, which is why a delta tertiary is added to large wye-wye autotransformers.

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3. Fault Types and Sequence Network Interconnection

Fault-current arithmetic worth memorising: with $Z_1 = Z_2$, the line-to-line fault current is

$$\frac{I_{LL}}{I_{3\phi}} = \frac{\sqrt{3}}{2} = 0.866$$

so an LL fault is always about 86.6% of the three-phase value. A single line-to-ground fault can exceed the three-phase value when $Z_0 < Z_1$, which happens on solidly grounded systems close to a grounded-wye transformer — a genuine engineering result and a favourite examination trap for candidates who assume the three-phase fault is always the worst case.

Faults ranked by frequency of occurrence: SLG 70–80%, LL about 15%, DLG about 10%, and three-phase only about 5%.

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4. System Grounding

Neutral grounding impedance $Z_n$ appears in the zero-sequence network as $3Z_n$ — three times, because the total zero-sequence current $3I_{a0}$ flows through it. Forgetting the factor of three is the standard error in grounded-neutral fault problems.

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Solved Board Exam Examples

Example 1: Resolving Unbalanced Line Currents

A four-wire feeder carries $I_a = 10\angle 0°$ A, $I_b = 10\angle 230°$ A and $I_c = 10\angle 130°$ A. Find the zero-sequence current and the neutral current.

Solution.

$$I_{a0} = \frac{1}{3}\left(I_a + I_b + I_c\right)$$

Converting to rectangular form: $10\angle 0° = 10 + j0$; $10\angle 230° = -6.428 - j7.660$; $10\angle 130° = -6.428 + j7.660$.

$$\sum = (10 - 6.428 - 6.428) + j(0 - 7.660 + 7.660) = -2.856 + j0$$ $$I_{a0} = \frac{-2.856}{3} = \boxed{-0.952 \text{ A}} = 0.952\angle 180° \text{ A}$$

The neutral carries three times this value:

$$I_n = 3I_{a0} = \boxed{2.856\angle 180° \text{ A}}$$

Had the feeder been three-wire with no neutral or ground return, $I_a + I_b + I_c$ would be forced to zero and no zero-sequence component could exist at all — a useful sanity check on any set of measured currents.

Example 2: Single Line-to-Ground Fault Current

At a bus, $Z_1 = j0.15$ pu, $Z_2 = j0.15$ pu and $Z_0 = j0.05$ pu on a 100 MVA, 13.8 kV base. Find the SLG fault current in amperes and compare it with the three-phase fault.

Solution. For an SLG fault the sequence networks are in series:

$$I_{a1} = \frac{1.0}{j(0.15 + 0.15 + 0.05)} = \frac{1.0}{j0.35} = -j2.857 \text{ pu}$$ $$I_a = 3I_{a1} = 8.571 \text{ pu}$$ $$I_{base} = \frac{100}{\sqrt{3}(13.8)} = 4.184 \text{ kA} \implies I_a = 8.571 \times 4.184 = \boxed{35.86 \text{ kA}}$$

The three-phase fault gives $1.0/0.15 = 6.667$ pu $= 27.89$ kA. Because $Z_0 < Z_1$, the ground fault is 28% more severe than the three-phase fault — precisely the case that defeats the "three-phase is always worst" assumption.

Example 3: Line-to-Line Fault Ratio

For a machine with $Z_1 = Z_2 = j0.20$ pu, find the ratio of line-to-line fault current to three-phase fault current.

Solution.

$$I_{3\phi} = \frac{1.0}{0.20} = 5.0 \text{ pu}$$

For the LL fault, $I_{a1} = \dfrac{1.0}{j(0.20+0.20)} = -j2.5$ pu, and the faulted phase current is

$$I_b = -j\sqrt{3}\,I_{a1} \implies |I_b| = \sqrt{3}(2.5) = 4.330 \text{ pu}$$ $$\frac{I_{LL}}{I_{3\phi}} = \frac{4.330}{5.0} = \boxed{0.866}$$

This √3/2 ratio holds whenever $Z_1 = Z_2$, independent of the actual impedance values.