Three-Phase Transformer Connections

A three-phase transformer can be constructed in two different ways:

• Using one pre-assembled and balanced three-phase transformer, consisting of three pairs of single-phase windings mounted onto one single laminated core.

• Connecting together three single-phase transformers, known as a transformer bank.

The construction is best understood as an evolution in three steps.

1. Three single-phase cores are positioned at \(120^{\circ}\) to one another, sharing a common central return limb.
2. For balanced three-phase sinusoidal voltages applied, the fluxes \(\Phi_a\), \(\Phi_b\) and \(\Phi_c\) are also sinusoidal and balanced, so the total combined flux in the merged limb becomes zero. The central limb can therefore be removed, since it carries no flux.
3. That structure is still not convenient to build. The practical form is a core of three limbs in the same plane, built from stacked laminations, each limb carrying the low-voltage and high-voltage windings of one phase.

• The shell type is constructed by stacking three single-phase shell transformers one above another.

• The winding direction of the central unit b is made opposite to that of units a and c.

• If the system is balanced with phase sequence a-b-c, the flux will also be balanced.

• The magnitude of the combined flux in the sections joining adjacent units is then equal to the magnitude of each of its components.

The trade is therefore between first cost, which favours the single unit, and continuity of supply, which favours the bank. Example 52.5 puts numbers on both sides.

Each side of a three-phase transformer may be connected in star or in delta, giving four combinations. Everything follows from two elementary relations.

Note that the kVA is the same in every case. Three transformers of 10 kVA form a 30 kVA bank however they are connected; the connection redistributes voltage and current but cannot create capacity.

• Generally used for small, high-voltage transformers. Because of the star connection the number of turns per phase is reduced, since \(V_{ph} = V_L/\sqrt3\), and the amount of insulation required is reduced with it.

• This connection can be used only if the connected load is balanced.

This is what vector groups exist to record. The notation Dyn11, Yd1 and so on states the connection of each winding and the phase displacement in units of 30°, so that compatibility can be checked at a glance. Chapter 53 takes up the notation in full.

• Generally used for large, low-voltage transformers. The number of turns per phase is relatively greater than for the Y-Y connection, but the conductor is correspondingly thinner.

• This connection can be used even for unbalanced loading.

• Even if one transformer is disabled, the system can continue to operate in open-delta connection, though with reduced available capacity.

• The closed delta gives the third-harmonic magnetising current a path to circulate, so the flux stays sinusoidal and no harmonic voltages are induced.

• Three-phase transformation is obtained either from one three-phase unit on a single core or from a bank of three single-phase transformers.

• For balanced excitation \(\Phi_a + \Phi_b + \Phi_c = 0\), so the central limb carries no flux and is removed. The practical core has three limbs in one plane, which is magnetically asymmetrical — the outer phases draw more magnetising current.

• In the shell type the middle winding is reversed, replacing \(\left|\Phi_a - \Phi_b\right| = \sqrt3\Phi_m\) by \(\left|\Phi_a + \Phi_b\right| = \Phi_m\) — 42.3 % less iron in the joining sections.

• A single unit is about 15 % cheaper, smaller, lighter and simpler to connect; a bank is easier to transport and repair, and keeps running on two phases.

• Star: \(V_L = \sqrt3V_{ph}\), \(I_L = I_{ph}\). Delta: \(V_L = V_{ph}\), \(I_L = \sqrt3I_{ph}\).

• Line ratios: Y-Y and Δ-Δ give \(K\); Y-Δ gives \(\sqrt3K\); Δ-Y gives \(K/\sqrt3\). The kVA is \(3\times\) the unit rating in every case.

• Mixed connections introduce a 30° phase shift, so they cannot be paralleled with Y-Y or Δ-Δ banks.

• Y-Y is troubled by unbalanced loading and by third harmonics, which are in phase in all three phases and cannot sum to zero at the star point. Remedy: solid earthing of the neutral, or a tertiary delta.

• Δ-Δ tolerates unbalance and harmonics because the closed loop gives circulating current a path, and survives the loss of one phase as an open delta.

• Standard practice: star on the HV side to save insulation, delta on the LV side to save copper and to carry the harmonic.

• Thinking a connection changes the kVA. Three 10 kVA units give 30 kVA in every connection.

• Applying \(\sqrt3\) to voltage and current on the same side. Star takes it on voltage, delta on current — never both.

• Using \(K\) as the line ratio for mixed connections. Y-Δ gives \(\sqrt3K\) and Δ-Y gives \(K/\sqrt3\).

• Getting Y-Δ and Δ-Y the wrong way round. The first letter is the HV winding.

• Ignoring the 30° shift when paralleling. It makes a mixed bank incompatible with a like one whatever the ratios.

• Saying third harmonics cancel at the star point. They are in phase, so they add rather than cancel.

• Believing a three-limb core is magnetically symmetrical. The outer limbs have longer paths.

• Forgetting the shell type's reversed middle winding. Without it the joining flux is \(\sqrt3\) times larger.

• Assuming a single unit is always cheaper. Once a spare is held, the bank usually wins.

• Expecting a Y-Y bank to run on unbalanced load without a neutral. The phase voltages shift badly.

Work always from the winding voltage and current of one transformer, then apply the star or delta relation to each side separately.

1. P52.1 Three 25 kVA, 1100/220 V transformers are connected Δ-Y. Find the line voltages, line currents and total kVA.

   Show answer

   \[I_{HV} = \frac{25\,000}{1100} = 22.727~\mathrm{A}, \qquad I_{LV} = \frac{25\,000}{220} = 113.64~\mathrm{A}\]

   HV in delta, LV in star:

   \[V_{L,HV} = 1100~\mathrm{V}, \qquad I_{L,HV} = \sqrt3\,(22.727) = 39.36~\mathrm{A}\]

   \[V_{L,LV} = \sqrt3\,(220) = 381.05~\mathrm{V}, \qquad I_{L,LV} = 113.64~\mathrm{A}\]

   \[S = \frac{\sqrt3\,(381.05)(113.64)}{1000} = 75.00~\mathrm{kVA}\]

2. P52.2 For P52.1, find the line-voltage ratio and check it against \(K/\sqrt3\).

   Show answer

   \[\frac{1100}{381.05} = 2.8868, \qquad K = \frac{1100}{220} = 5, \qquad \frac{K}{\sqrt3} = \frac{5}{1.7321} = 2.8868 \quad\checkmark\]

3. P52.3 The same three transformers are connected Y-Δ. Find the line voltages and the line ratio.

   Show answer

   \[V_{L,HV} = \sqrt3\,(1100) = 1905.3~\mathrm{V}, \qquad V_{L,LV} = 220~\mathrm{V}\]

   \[\text{ratio} = \frac{1905.3}{220} = 8.6603 = \sqrt3\,K = \sqrt3\,(5) \quad\checkmark\]

4. P52.4 A three-limb core has a mean flux path of 1.10 m for the centre limb and 1.35 m for each outer limb, with \(\mu_r = 5000\) and a cross-section of 0.030 m². Find the two reluctances and their ratio.

   Show answer

   \[\mu = (5000)(4\pi\times10^{-7}) = 6.2832\times10^{-3}~\mathrm{H/m}\]

   \[S_{centre} = \frac{1.10}{(6.2832\times10^{-3})(0.030)} = 5835.7~\mathrm{A/Wb}\]

   \[S_{outer} = \frac{1.35}{(6.2832\times10^{-3})(0.030)} = 7162.0~\mathrm{A/Wb}\]

   \[\frac{S_{outer}}{S_{centre}} = 1.2273 = \frac{1.35}{1.10}\]

   Only the path length differs, so the reluctance ratio is simply the length ratio.

5. P52.5 For P52.4, with 500 turns per phase and a peak flux of 0.020 Wb, find the RMS magnetising current in each limb and the percentage unbalance.

   Show answer

   \[\hat{F}_{centre} = (0.020)(5835.7) = 116.71~\mathrm{At} \quad\Longrightarrow\quad \hat{I} = \frac{116.71}{500} = 0.23343~\mathrm{A}\]

   \[I_{centre} = \frac{0.23343}{\sqrt2} = 0.16506~\mathrm{A}\]

   \[\hat{F}_{outer} = (0.020)(7162.0) = 143.24~\mathrm{At} \quad\Longrightarrow\quad I_{outer} = \frac{143.24/500}{\sqrt2} = 0.20257~\mathrm{A}\]

   \[\frac{0.20257}{0.16506} - 1 = 22.73\,\%\ \text{higher in the outer limbs}\]

6. P52.6 A Δ-Δ bank of three 300 kVA units loses one transformer. Find the open-delta capacity and the loading of each survivor.

   Show answer

   \[S_{open} = \sqrt3\,(300) = 519.62~\mathrm{kVA}\]

   \[\frac{519.62}{900} = 57.74\,\%\ \text{of the original bank}\]

   \[\frac{519.62}{2(300)} = 86.60\,\%\ \text{of each survivor's rating}\]

7. P52.7 Prove that the central limb of a three-phase core carries no flux, and state the condition required.

   Show answer

   With balanced excitation the three fluxes are equal in magnitude and 120° apart:

   \[\Phi_a = \Phi_m\angle0^{\circ}, \quad \Phi_b = \Phi_m\angle-120^{\circ}, \quad \Phi_c = \Phi_m\angle+120^{\circ}\]

   Resolving,

   \[\text{real}: \Phi_m\left(1 - 0.5 - 0.5\right) = 0, \qquad \text{imaginary}: \Phi_m\left(0 - 0.86603 + 0.86603\right) = 0\]

   so \(\Phi_a + \Phi_b + \Phi_c = 0\). Equivalently, three equal phasors at 120° laid tip to tail form a closed equilateral triangle.

   The merged limb carries the sum of the three fluxes, which is zero, so it can be removed — saving its iron, weight and cost. This is the magnetic counterpart of balanced three-phase currents needing no neutral conductor.

   The condition is balance. If the excitation is unbalanced the sum is no longer zero, and in a three-limb core the residual flux must return through the oil, the tank and the structural steel, causing stray loss and local heating.

8. P52.8 Explain why the middle winding of a shell-type three-phase transformer is reversed.

   Show answer

   The flux in the core section joining two adjacent units is the combination of their fluxes, and the winding sense decides whether that combination is a difference or a sum.

   \[\text{same sense}: \left|\Phi_a - \Phi_b\right| = \sqrt3\,\Phi_m = 1.732\,\Phi_m\]

   \[\text{middle reversed}: \left|\Phi_a + \Phi_b\right| = \left|-\Phi_c\right| = \Phi_m\]

   The second line follows immediately from \(\Phi_a + \Phi_b + \Phi_c = 0\): the sum of two of the three balanced fluxes is the negative of the third, whose magnitude is \(\Phi_m\) by definition.

   The benefit is that the joining sections need only \(1/\sqrt3 = 57.7\,\%\) of the cross-section they would otherwise require — a saving of 42.3 % of the iron there, obtained simply by connecting one winding the other way round.

9. P52.9 Explain the third-harmonic problem of the star-star connection and its remedies.

   Show answer

   The magnetising current of any iron-cored transformer is non-sinusoidal, with a strong third harmonic (Chapter 42).

   The difficulty. Fundamental components in the three phases are 120° apart and sum to zero at the star point. Third harmonics are at three times that displacement, 360° apart — so they are all in phase with one another and their sum at the star point is not zero.

   With no neutral and no delta, that current has nowhere to flow. It is therefore suppressed, forcing the magnetising current to be sinusoidal — which, through the non-linear B-H curve, forces the flux to be non-sinusoidal instead. The distorted flux induces harmonic voltages in every winding, and the phase voltages become distorted and unbalanced.

   Remedies: solidly ground the star neutral, giving the harmonic a path; or provide a tertiary delta winding, in which the harmonic can circulate without leaving the transformer. The tertiary is standard on large Y-Y units and often does useful work supplying station auxiliaries.

10. P52.10 Why is star normally used on the HV side and delta on the LV side?

    Show answer

    Because the dominant cost is different at the two voltage levels.

    Star on HV. The phase voltage is \(V_L/\sqrt3\), so the winding stands at 57.7 % of the line voltage. That means 42.3 % fewer turns and 42.3 % less insulation. At high voltage, insulation dominates the cost of a winding, so star is the economical choice. A neutral is also available for earthing and for surge protection.

    Delta on LV. The phase current is \(I_L/\sqrt3\), so the conductor need only be 57.7 % of the cross-section — a 42.3 % saving in copper, and at low voltage copper dominates the cost.

    Two further reasons for the delta: it provides the closed path the third-harmonic magnetising current requires, keeping the flux sinusoidal; and it tolerates unbalanced loading, since zero-sequence current can circulate within it.

    Hence Y-Δ for step-down at a substation and Δ-Y for step-up at a generating station. Where a four-wire distribution neutral is needed the LV side must be star instead, and the delta moves to the HV side.