# Transformer Delta Configuration

## Main Question or Discussion Point

I have a few confusions related to the delta configuration of a transformer (single-phase transformers used to make a 3-phase transformer):

1) In a closed delta, the windings are connected such that H2 end of one winding is connected to H1 of the other winding. What will happen if one of the windings is reversed?
The flux in that winding should be opposite and thus the induced emf should also have opposite polarity..but that isn't possible! because induced emf should be equal to voltage applied across the winding. Also, what would be the effect of one reversed winding on the other windings?

How do you calculate the total power output of an open delta transformer? I don't understand the explanation given in various texts and on websites:
http://yourelectrichome.blogspot.com/2011/05/open-delta-or-v-v-connection-of-3-phase.html

The √3 ELIL formula was for closed delta because in a closed delta, we calculate the power delivered by one winding as EL*Iphase which becomes EL*(IL/√3) since phase current=IL/√3, and then multiply by three.
But why are we using this formula for open delta?
And btw, in the conductor attached to the point where the two windings are joined together, would the line current be √3Iphase?

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1) If one winding is reversed and the delta closed, a double voltage short exists. One must be careful when closing a delta and make sure the polarities are correct.

2) With open delta (V-V), the phase and line currents are equal in magnitude. Thus sqrt3*V*Iphase=sqrt3*V*Iline. But in a closed delta, Iline=sqrt3*Iphase, so that sqrt3*V*Iline would equal 3*V*Iphase. Hence the closed delta has sqrt3 times more VA capability.

The following text explains this with detailed phasor diagrams:

Electromagnetic and electromechanical machinery, by Leander Matsch, as well as

Electric Machinery, by Fitzgerald, Kingsley, and Umans. I had Dr. Umans (MIT prof. visiting Case Western Reserve Univ in Cleveland Spring 2010) for power/machinery class during my PhD course work. He really knows his stuff.

Claude

The edition of Electric Machinery on Google Books does not go into much detail:

1) winding reversed: can you please explain this in more detail?

2) But why are you using the expression sqrt(3)*V*Iline for Open Delta.. why not 2*V*Iphase (divided by 3*V*Iphase). Also, for two of the lines, it is easy to see that phase current = line current. But what about the third line that is joined to both the two secondary windings..each of the windings (individually) is not in series with this line. So I don't understand why its current would not be sqrt(3)*Iphase.

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The edition of Electric Machinery on Google Books does not go into much detail:

1) winding reversed: can you please explain this in more detail?

2) But why are you using the expression sqrt(3)*V*Iline for Open Delta.. why not 2*V*Iphase (divided by 3*V*Iphase). Also, for two of the lines, it is easy to see that phase current = line current. But what about the third line that is joined to both the two secondary windings..each of the windings (individually) is not in series with this line. So I don't understand why its current would not be sqrt(3)*Iphase.
1) Reversal of winding:
Connected correctly, the 3 windings have voltages 120 deg apart. Phases have the following voltages - angle Vab=0 deg, angle Vbc=-120 deg, angle Vca=120 deg, all magnitudes are 1.0 per unit.
Closing the delta puts these 3 voltages in series where they add together. The 3 phasors are equal in magnitude, displaced 120 deg in phase. Their sum is zero, so that the closed loops results in minimum current, that needed to support triplen harmonic magnetizing current. When a load is connected, current develops in the delta windings.

But if one phase is reversed, it gets ugly. Suppose Vbc was reversed, so that its angle is not -120 deg, but +60 deg instead. Now the summation around the delta is 3 phasors of unit magnitude with angles of 0, +60, & +120 degrees. Doing the math shows that the sum is a magnitude of 2 with a phase of +60 deg. So the delta loop is a low impedance closed path with twice the line voltage resulting in enormous circulating current.

Draw the phasor diagrams and it will be clear.

2) Best way to explain is with an example. We have 3 xfmrs each rated for 100 VA, 100 volts at 1.0 amp. The windings can only carry 1.0 amp before they overheat. When connected in delta, the power can be computed as 3*Vp*Ip, or as √3*Vl*Il. If each secondary can support 1.0 amp, then P = 3*Vp*Ip, or 3*100*1.0 = 300 W, the sum of the individual VA ratings. Likewise, with 1.0 A phase current, line current for a closed delta is √3 each phase. So we can compute as P = √3*Vl*Il = √3*100*√3 = 300W, as before.

But now we remove one xfmr using an open delta. Each winding carries but 1.0 A, as before. Got a meeting to attend, I'll finish later.

Claude

• 1 person
Thank you, Claude, for the reply.

1) I drew the phasor diagrams and spent quite some time trying to figure out how the delta works. I've understood your explanation for the first question. Thank you very much.

2) Waiting anxiously for the remaining part.

2) Sorry for the delay. We use 2 xfmrs connected in V-V or open delta. The load is balanced 3 phase supported by just 2 xfmrs. Consider the load a closed delta of resistors. Here is the open delta limitation we have to deal with.

In the closed delta load resistor bank, the line current is √3 times the phase current. But the phase current for the open delta xfmr secondary must equal the line current since the delta is not closed and the 2 secondaries directly power the load bank. Let's say phases A-B and B-C are the 2 open delta windings. The outer terminals are A & C, and the junction of the 2 windings is B. Since the line currents from A & C cannot exceed 1.0 amp. that is the limitation we have.

The closed delta load resistor bank cannot draw more than 1/√3 amp of phase current, approx. 0.577 amp. This assures that the line current is 1.0 amp, which we must observe as the limit. Inside the load resistor closed delta, the phase currents are Iab=1/√3 with angle 0 deg, Ibc has same magnitude with angle -120 deg, Ica has same magnitude with +120 deg angle.

A good text will illustrate that line currents entering the closed delta load will lag the phase currents by 30 deg. Thus Ia=1.0 w/ angle -30 deg, Ib=1.0 w/ angle -150 deg, Ic=1.0 w/ angle +90 deg. For the open delta secondary the windings carry the full line current, so that 1.0 amp is the magnitude of line current for all 3 phases. Since phase A current, Ia is 1.0 amp at -30 deg, and Ic is 1.0 amp at +90 deg, phase B current Ib must equal the negative of the sum of Ia & Ib, since currents at a junction must sum to 0. Ia and Ic sum to 1.0 amp w/ angle of +30 deg. The negative of this sum is 1.0 amp with angle of -150 deg.

Thus for a closed delta, we have 2 xfmrs instead of 3. The limitation occurs due to the fact that w/o a closed path, the xfmr secondary windings must support full line current. With a closed delta, line current = phase current X √3. Without the closed path Iline = Iphase. Thus we lose the benefit of the times √3 multiplication. With a closed delta we can load the system with a 100 ohm delta resistor bank. The phase currents inside the load delta and xfmr secondary delta are 1.0 amp, but the line currents are √3 amp, hence 300 W dissipation.

But in the open delta, we must limit load current in closed delta load to 1/√3 amp instead of 1.0 amp. The resistor bank load must be 173.2 ohms in closed delta, not 100 ohms. The open delta must carry the full line current, and the √3 factor is no longer enjoyed. So the power here is √3*100V*1.0A = 100√3 W= 173.2 W dissipation.

Though we have two 100 watt xfmrs, we do not get 200 W capability, but only 173.2 W. Hence we must observe this limitation when using 2 xfmrs connected for open delta operation. As long as we limit our loading demands all should be fine.

Claude

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• 1 person
Thank you once again for spending time and effort to help me understand things. You had given the example of delta-configured load while the text I have had given the example of wye-configured load. I was wondering why we were restricting ourselves to only these two configurations. And then it suddenly occurred to me:

Can all balanced loads be represented as delta or wye (as delta and wye themselves can be converted into each other)?

Yes all balanced loads can be represented as either a delta or a wye. Of course they can be converted interchangeably. The impedance in each leg of a balanced delta is 3 times that of its equivalent wye. So we can use either representation for balanced loads. With unbalanced loads, there is a more complicated set of equations for converting between the two. A search should turn it up, if not I can post them.

Claude

• 1 person
My problem seems to be solved. Feels great.