Open-Delta Operation and Vector Groups

Remove one transformer from a Δ-Δ bank and the remaining two still deliver three-phase power. The connection that results is called open delta or V-V.

• It can be used when one of the transformers in a Δ-Δ bank is disabled, and the service is to be continued until the faulty transformer is repaired or replaced.

• It can also be used for small three-phase loads where the installation of a full three-transformer bank is unnecessary.

• The total load-carrying capacity is 57.7 % of what it would be for the delta-delta connection.

The whole result turns on one change: what current each transformer carries.

In a closed delta all three transformers work at the load power factor. In an open delta they do not, because the phasor geometry displaces each transformer's voltage from its current by an extra \(\pm30^{\circ}\).

The voltage-unbalance point is worth noting. In a closed delta the three secondary voltages are held by the delta itself; in an open delta the third voltage is only the sum of the other two, so any inequality in the two internal impedance drops appears directly as unbalance.

Chapter 52 showed that a mixed connection introduces a \(30^{\circ}\) shift between primary and secondary line quantities, and Chapter 49 listed matching phase displacement as the fourth condition for parallel operation. A notation is needed to record it.

The magnitudes may match perfectly and the impedances may be identical; if the phase displacement differs, the banks must not be connected. This is why the displacement is stamped on every three-phase transformer's nameplate.

The designation has three parts, read left to right.

Hour 11 is the common one for step-down units, since a lag of \(330^{\circ}\) is a lead of \(30^{\circ}\), and it is often convenient for the LV to lead.

Two useful exceptions exist in practice. Some pairs differing by \(180^{\circ}\) — such as Yd1 and Yd7 — can be reconciled by relabelling the phases, since interchanging two terminals reverses the sense. And a Dy11 unit can be made to match a Dy1 unit by reversing the phase sequence of its connections. Both are done at the design stage, not on site.

• Open delta uses two transformers to deliver three-phase power; the third line voltage follows from \(V_{CA} = -(V_{AB} + V_{BC})\).

• In a closed delta each winding carries \(I_L/\sqrt3\); in an open delta it carries the full line current, which is what caps the capacity.

• \(S_{open} = \sqrt3\,S_{unit}\), so the bank gives \(1/\sqrt3 = 57.74\,\%\) of the closed Δ-Δ capacity — not two thirds.

• The utilisation factor \(\sqrt3/2 = 86.60\,\%\) compares the output with the two remaining units' combined rating.

• The two transformers work at \(\cos(30^{\circ}\pm\phi)\), so they heat unequally. At unity pf both run at 0.866; at 0.5 lagging one delivers no power at all.

• Open delta needs \(2/\sqrt3 = 15.47\,\%\) more transformer kVA than a closed bank, but only two units.

• Vector groups record the phase displacement: capital letter for HV, small for LV, n or N for a neutral, and a clock number giving the LV lag in units of 30°.

• Yy and Dd give even numbers; Yd and Dy give odd. Dyn11 is the standard distribution transformer.

• Two banks may be paralleled only if the clock numbers match. A difference voltage \(2V\sin(\theta/2)\) otherwise drives a large circulating current — five times rated at only 30°.

• Saying open delta gives two thirds of the capacity. It gives \(1/\sqrt3 = 57.74\,\%\), because the binding constraint is winding current, not the number of transformers.

• Confusing 57.74 % with 86.60 %. The first compares with the original three-unit bank, the second with the two units that remain.

• Assuming the two transformers share equally. They work at \(\cos(30^{\circ}\pm\phi)\) and heat unequally.

• Expecting either machine to reach full rating. Even at unity power factor neither exceeds 86.6 %.

• Thinking the open delta loses a line voltage. All three remain; only current capability is lost.

• Reading the clock number as leading. It is the lag of the LV behind the HV.

• Matching letters instead of numbers. Yy0 and Dd0 may be paralleled although their connections differ.

• Expecting an odd group to parallel with an even one. Mixed and like connections can never agree.

• Treating a 30° mismatch as minor. It drives about five times rated current.

• Skipping the voltmeter phasing check because the nameplates appear to match.

For open-delta problems decide first which quantity is capped — it is always the winding current. For vector groups, convert both designations to degrees before comparing.

1. P53.1 A Δ-Δ bank of three 200 kVA transformers loses one unit. Find the open-delta capacity and the percentage of the original.

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   \[S_{open} = \sqrt3\,(200) = 346.41~\mathrm{kVA}\]

   \[\frac{346.41}{600} = 57.74\,\%, \qquad \text{lost} = 253.59~\mathrm{kVA}\]

2. P53.2 For P53.1, find the loading of each surviving transformer.

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   \[\frac{346.41}{(2)(200)} = \frac{346.41}{400} = 86.60\,\%\]

   Neither can be fully loaded, however light the demand elsewhere.

3. P53.3 Two transformers in open delta supply a load at 0.8 lagging. Find the power factor at which each operates.

   Show answer

   \[\phi = \cos^{-1}(0.8) = 36.87^{\circ}\]

   \[\text{pf}_1 = \cos\left(30^{\circ} + 36.87^{\circ}\right) = \cos66.87^{\circ} = 0.3928\]

   \[\text{pf}_2 = \cos\left(30^{\circ} - 36.87^{\circ}\right) = \cos\left(-6.87^{\circ}\right) = 0.9928\]

   Very unequal — the second transformer does most of the real work.

4. P53.4 A 90 kVA three-phase load is to be supplied in open delta. Find the minimum rating of each transformer.

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   \[S_{unit} = \frac{90}{\sqrt3} = 51.96~\mathrm{kVA} \quad\Longrightarrow\quad \text{choose 60 kVA units}\]

   \[\text{capacity} = \sqrt3\,(60) = 103.92~\mathrm{kVA}, \qquad \text{loading} = \frac{90}{103.92} = 86.60\,\%\]

5. P53.5 Two banks are Yd11 and Dy11. May they be paralleled?

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   Yes. Both have clock number 11, so both displace the LV by 330°. The connections differ, but what must match is the phase displacement, not the winding arrangement — provided the ratios, polarities and percentage impedances also agree.

6. P53.6 A Dyn11 bank is inadvertently paralleled with a Yd1 bank of the same rating, the combined impedance being 0.08 pu. Estimate the circulating current.

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   \[\left|11 - 1\right|\times30^{\circ} = 300^{\circ} \equiv 60^{\circ}\]

   \[\Delta V = 2\sin30^{\circ} = 1.0000~\mathrm{pu}\]

   \[I_{circ} = \frac{1.0000}{0.08} = 12.5~\mathrm{pu} = 1250\,\%\ \text{of rated}\]

   Both transformers would be destroyed unless protection operated within a cycle or two.

7. P53.7 Derive the capacity of the open-delta connection and explain why it is not two thirds of the closed bank.

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   Closed delta. Each winding carries the phase current \(I_{ph} = I_L/\sqrt3\), so one unit is rated

   \[S_{unit} = V_{ph}I_{ph} = \frac{V_LI_L}{\sqrt3}\]

   and the bank delivers \(\sqrt3V_LI_L = 3S_{unit}\).

   Open delta. With the third leg removed there is no parallel path, so each winding carries the full line current, \(I_{ph} = I_L\). To stay within rating,

   \[I_L \le \frac{S_{unit}}{V_L} \quad\Longrightarrow\quad S_{open} = \sqrt3V_LI_L = \sqrt3\,S_{unit}\]

   \[\frac{S_{open}}{S_{\Delta\Delta}} = \frac{\sqrt3S_{unit}}{3S_{unit}} = \frac{1}{\sqrt3} = 0.5774\]

   Why not two thirds. That figure counts transformers, but the binding constraint is winding current. Each survivor must carry \(\sqrt3\) times the current it carried in the closed bank, so the line current must be reduced by \(1/\sqrt3\) to keep within rating. Two thirds of the machines remain, but each is worked \(\sqrt3\) times harder per ampere of line current.

8. P53.8 Show that the two transformers of an open delta operate at different power factors, and find the load power factor at which one delivers no power.

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   The phasor geometry displaces each transformer's voltage from the line current by an extra \(\pm30^{\circ}\), so

   \[P_1 = V_LI_L\cos\left(30^{\circ} + \phi\right), \qquad P_2 = V_LI_L\cos\left(30^{\circ} - \phi\right)\]

   Their sum recovers the correct three-phase power:

   \[P_1 + P_2 = 2V_LI_L\cos30^{\circ}\cos\phi = \sqrt3\,V_LI_L\cos\phi\]

   Zero output occurs when \(30^{\circ} + \phi = 90^{\circ}\), that is \(\phi = 60^{\circ}\), a load power factor of 0.5 lagging. At that point \(P_1 = 0\) while \(P_2 = V_LI_L\cos30^{\circ}\) carries the whole load.

   The first transformer still carries current and voltage, and still dissipates copper loss — its voltage and current are simply in quadrature. At unity power factor both run at 0.866, which is the utilisation factor, so neither machine ever reaches its full rating.

9. P53.9 Explain the clock notation and why odd and even groups cannot be paralleled.

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   The designation gives the HV connection as a capital letter (D, Y, Z), the LV as a small letter, an n or N where a neutral is brought out, and a clock number giving the phase displacement:

   \[\theta = (\text{clock number})\times30^{\circ}\ \text{lagging}\]

   The HV phasor is imagined at twelve o'clock and the LV phasor at the stated hour. Thus Dyn11 is HV delta, LV star with neutral, LV lagging by 330° — equivalently leading by 30°.

   Odd versus even. Like connections (Yy, Dd) introduce no inherent shift, so their displacement is a multiple of 60° and the clock number is even. Mixed connections (Yd, Dy) necessarily bring in the 30° of the star-delta transformation, so their number is odd.

   An odd and an even group therefore differ by an odd multiple of 30°, which can never be zero. Paralleling them leaves a difference voltage \(2V\sin(\theta/2)\) round the loop — 0.5176 pu at 30° — driving several times rated current through the two transformers' impedances alone.

10. P53.10 Why is Dyn11 the standard distribution transformer connection?

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    Because one connection answers three separate requirements at once.

    • The HV delta gives the third-harmonic magnetising current a closed path in which to circulate, so the flux stays sinusoidal and no harmonic voltages are induced — the problem that spoils the Yy connection.

    • The LV star provides a four-wire supply, so that single-phase loads at the phase voltage and three-phase loads at the line voltage can be served from the same transformer.

    • The brought-out neutral gives a solid earth reference for protection and for consumer safety.

    The delta is on the HV side where the current is small, so the conductor saving matters less than the harmonic path it provides; and the star is on the LV side where the neutral is needed.

    The 11 rather than 1 is convention: a 330° lag is a 30° lead, and standardising on it means that distribution transformers across a network share one group and can be paralleled freely.