Harmonics in Three-Phase Transformers

Chapter 42 established the origin, and it is worth restating exactly.

1. The applied voltage is sinusoidal, so from \(v \approx -N\,\mathrm{d}\Phi/\mathrm{d}t\) the flux must be sinusoidal.
2. The relation between flux and current is the B-H curve, which is non-linear and hysteretic.
3. To force a sinusoidal flux through a non-linear characteristic, the current must be non-sinusoidal — and it comes out peaked, because near the flux peak the core is entering saturation and disproportionately more current is needed.

Contrast this with Chapter 50's inrush current, which is asymmetrical and therefore rich in even harmonics, principally the second. The steady-state magnetising current and the switching transient have opposite harmonic signatures, and protection exploits the difference.

In a balanced three-phase system the fundamental currents are displaced by \(120^{\circ}\). The third harmonic of each is at three times that displacement.

Everything in this chapter follows from that single line. A quantity that is identical in all three phases is called a zero-sequence quantity, and zero-sequence current needs a fourth path — a neutral conductor, an earth connection, or a closed delta — or it cannot flow at all.

Repeating the calculation for each harmonic order gives a simple pattern.

The rule explains why the fifth and seventh, though present, cause none of these difficulties. Being negative- and positive-sequence they sum to zero at a star point exactly as the fundamental does, so they need no special path and simply flow as ordinary balanced currents.

The core needs a peaked magnetising current to produce a sinusoidal flux. If the circuit refuses to supply the third-harmonic component of that current, something has to give.

This is the worst case, and it is why the Yy connection is rarely used alone.

• There is no path for zero-sequence current — no fourth wire, no delta.

• The magnetising current is forced sinusoidal, so the flux becomes flat-topped.

• Third-harmonic EMFs of up to 45 % appear in every phase winding, stressing the insulation.

• The star point oscillates at three times supply frequency relative to earth.

• Line voltages remain sinusoidal, so the trouble is invisible from the terminals.

Earthing the star point, or running a fourth wire, provides the missing path. The third-harmonic current can now flow, the flux stays sinusoidal, and the harmonic EMFs vanish.

The cure has its own cost. That neutral current returns through the earth, and a current at 150 Hz flowing through the ground induces voltages in any parallel conductor — which is the origin of the telephone-interference problem of Section 54-9.

A delta solves the problem more neatly than an earthed neutral, because the circulating current never leaves the transformer.

This is the harmonic half of the case for Dyn11 made in Chapter 53. The HV delta contains the circulating harmonic entirely within the transformer, while the LV star supplies the four-wire neutral — and neither function interferes with the other.

Where both main windings must be star — usually because neutrals are wanted on both sides for earthing at two voltage levels — a third winding is added in delta purely to provide the harmonic path.

A tertiary is conventionally rated at about one third of the main winding. That rating is not set by the harmonic duty at all — the circulating harmonic is only about 1 % of rated current, which is thermally trivial. It is set by the fault and auxiliary duties, and by the mechanical forces a through-fault imposes.

• A sinusoidal voltage forces sinusoidal flux; the non-linear B-H curve then forces a peaked, non-sinusoidal magnetising current containing odd harmonics, the third being 10 to 40 %.

• The distortion barely affects the RMS value — 36.7 % THD raises it by only 6.5 % — which justifies the equivalent sinusoidal current.

• Third harmonics are at \(3\times120^{\circ} = 360^{\circ} \equiv 0^{\circ}\), so they are in phase in all three lines — zero sequence.

• Sequence rule: \(h = 3k\) zero, \(3k+1\) positive, \(3k+2\) negative. Only triplens cause difficulty.

• Zero-sequence current needs a fourth path: a neutral, an earth, or a closed delta. Without one it cannot flow.

• The dilemma: allow the harmonic current and the flux is sinusoidal; block it and the flux flat-tops.

• Since \(E_n \propto n\Phi_n\), a 15 % flux harmonic gives a 45 % EMF harmonic — peak phase voltage 1.45 pu though RMS rises under 10 %.

• Third-harmonic EMFs cancel in the line voltages and appear between phase and neutral, giving the oscillating neutral.

• An earthed star neutral carries \(3I_3\) at 150 Hz even at balanced no load, causing telephone interference.

• A delta confines the circulating harmonic within the transformer and is the clean solution; a tertiary delta supplies it where both mains are star.

• Saying third harmonics are 120° apart. Three times 120° is 360°, so they are in phase.

• Assuming a balanced system needs no neutral current. True for the fundamental; false for triplens.

• Expecting the fifth and seventh to cause the same trouble. They are negative and positive sequence and sum to zero normally.

• Adding the third harmonic to get a peaked wave. It must be subtracted; adding it flat-tops the wave.

• Taking \(E_3/E_1 = \Phi_3/\Phi_1\). Differentiation multiplies by the order, so it is three times as large.

• Judging the overvoltage by RMS. RMS rises under 10 % while the peak rises 45 %, and insulation responds to the peak.

• Expecting to see the distortion in the line voltages. It cancels there; look between phase and neutral.

• Thinking the delta current reaches the external system. It circulates inside the closed loop and never leaves.

• Sizing the tertiary from the harmonic duty. That needs only a few percent; the one-third rating comes from fault and auxiliary duties.

• Omitting the tertiary from a star-star transformer. It also provides the zero-sequence path that earth-fault protection depends on.

For any harmonic question, first classify the order by the sequence rule — that decides at once whether a path is needed.

1. P54.1 A magnetising current has \(I_1 = 1.00\), \(I_3 = 0.30\), \(I_5 = 0.12\). Find the THD and the RMS value.

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   \[\text{THD} = \frac{\sqrt{0.0900 + 0.0144}}{1.00} = \sqrt{0.1044} = 0.3231 = 32.31\,\%\]

   \[I_{rms} = \sqrt{1 + 0.1044} = \sqrt{1.1044} = 1.05090\]

   Only 5.09 % above the fundamental despite 32 % distortion.

2. P54.2 Classify the 15th and 17th harmonics by sequence, and state which needs a special path.

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   \[15 = 3(5) \Rightarrow \text{zero sequence (a triplen)}\]

   \[17 = 3(5) + 2 \Rightarrow \text{negative sequence}\]

   The 15th needs a neutral, earth or delta; the 17th flows as an ordinary balanced set.

3. P54.3 A flat-topped flux contains a third harmonic of 12 % of the fundamental. Find the third-harmonic EMF and the peak phase voltage.

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   \[\frac{E_3}{E_1} = (3)(0.12) = 0.36\]

   \[\hat{V}_{ph} = 1 + 0.36 = 1.36~\mathrm{pu} \quad (36\,\%\ \text{overvoltage})\]

   \[V_{rms} = \sqrt{1 + (0.36)^{2}} = 1.06283~\mathrm{pu} \quad (6.28\,\%\ \text{up})\]

4. P54.4 A 50 MVA, 66 kV earthed star winding has a magnetising current of 2.5 % of rated, of which 30 % is third harmonic. Find the neutral current.

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   \[I_{rated} = \frac{50\times10^{6}}{\sqrt3\,(66\times10^{3})} = 437.39~\mathrm{A}\]

   \[I_3 = (0.30)(0.025)(437.39) = 3.280~\mathrm{A}\]

   \[I_N = 3I_3 = 9.841~\mathrm{A}\ \text{at }150~\mathrm{Hz} = 2.25\,\%\ \text{of rated}\]

5. P54.5 Why does a 15 % third-harmonic flux not raise the RMS phase voltage by 45 %?

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   Because harmonics add in quadrature for RMS purposes but arithmetically at the instant their peaks coincide.

   \[V_{rms} = \sqrt{1 + (0.45)^{2}} = 1.0966 \quad\text{vs}\quad \hat{V} = 1 + 0.45 = 1.45\]

   A 45 % component contributes only \((0.45)^{2} = 0.2025\) to the sum of squares, hence under 10 % on the RMS. Insulation responds to the peak, so the 45 % figure is the one that matters.

6. P54.6 A star-star transformer has its neutral earthed on the primary only. Does this cure the harmonic problem?

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   Yes. The magnetising current is drawn on the primary side, so a zero-sequence path on that side is sufficient — the third harmonic flows out through the earthed neutral and the flux stays sinusoidal.

   The secondary neutral is irrelevant to this particular problem, since the harmonic is a magnetising current and never needed to appear in the secondary at all.

   But the cure brings the earth-return current of Section 54-9, with its interference risk, which is why a tertiary delta is usually preferred.

7. P54.7 Explain why third-harmonic currents in the three phases are in phase, and what follows.

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   The fundamentals are displaced by 120°. The third harmonic of each is at three times that displacement:

   \[3\times0^{\circ} = 0^{\circ}, \quad 3\times\left(-120^{\circ}\right) = -360^{\circ} \equiv 0^{\circ}, \quad 3\times\left(-240^{\circ}\right) = -720^{\circ} \equiv 0^{\circ}\]

   All three therefore coincide — the third harmonic is a zero-sequence quantity.

   What follows. At a star point the fundamentals cancel but the third harmonics add, giving \(3I_3\). Since a three-wire star offers nowhere for that current to go, it cannot flow at all, and the magnetising current is forced sinusoidal.

   The core then cannot produce a sinusoidal flux, so the flux flat-tops and induces third-harmonic EMFs instead. The trouble is transferred from the current to the voltage.

   A neutral, an earth connection or a closed delta gives the current a path and restores the sinusoidal flux.

8. P54.8 Show that third-harmonic EMFs do not appear in the line voltages, and say where they do appear.

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   Any line voltage is the difference of two phase EMFs. Since the third harmonics are equal and in phase,

   \[V_{AB,3} = E_{A3} - E_{B3} = 0\]

   and the same for the other two lines. Line-to-line voltages therefore remain sinusoidal.

   Where they appear. Being common to all three phases, they act between the star point and earth, displacing the neutral by a voltage at three times supply frequency — the oscillating neutral.

   Two practical consequences follow. Phase-to-earth insulation is stressed by the sum of the fundamental and this oscillation; and any single-phase load between line and neutral receives a distorted supply. Since terminal line-voltage measurements look perfectly clean, the condition is easily missed.

9. P54.9 Explain how a delta winding solves the harmonic problem, and why it is preferable to earthing the neutral.

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   The three third-harmonic EMFs are equal and in phase, so round a closed delta they add rather than cancel, giving a driving voltage of \(3E_3\) round the loop. This drives a circulating third-harmonic current inside the delta, which supplies exactly the magnetising harmonic the core requires.

   The flux therefore stays sinusoidal, no harmonic EMF appears at any terminal, and the phase voltages are clean.

   Why it is preferable. An earthed neutral also provides a path, but the current then leaves the transformer, flows through the earth at 150 Hz and returns through the ground — inducing voltages in parallel conductors and causing telephone interference, as well as biasing sensitive earth-fault relays.

   The delta current never leaves the transformer. It circulates in a closed loop within the tank, so the harmonic is dealt with entirely internally — which is why almost every large transformer has a delta winding somewhere, buried if not brought out.

10. P54.10 Why is a tertiary rated at a third of the main winding when the harmonic current is only about 1 %?

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    Because the harmonic duty is the least demanding of its four functions.

    • Harmonic path — needs only a few percent of rated current, thermally trivial.

    • Earth-fault current — must provide a zero-sequence path carrying enough current to operate protection, and must withstand it.

    • Auxiliary supply — often brought out at 11 kV to feed station services, a real continuous load.

    • Reactive compensation — reactors or capacitors may be connected to it.

    The rating must also cover the mechanical forces of a through-fault, since the tertiary carries circulating current whenever any zero-sequence condition exists anywhere on the system.

    The one-third convention is set by the worst of these duties, not the mildest. A tertiary sized for the harmonic alone would fail the first time an earth fault occurred.