Open 3 phase transformer

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Hey Guys!

The Forum sent me an automated email informing me of the top 30 threads from the past month. This thread struck my interest.

Don't know if any of you remember me. It's been several years since I was here. Anyway, let me put my two cents in. I hope you folks are still interested in this subject. Rattus and Iwire, I remember both of you!

Now, Dnem, I can totally understand your point and have put alot of thought in this. I also have access to a really nice multi-channel O-scope and lab equipment. I played with all this today. Here are my findings.

The entire point of the topic rests with the way the coils are connected in wye versus delta. Say each secondary coil has an X1 terminal and an X2 terminal. In the wye system, the X2 terminal of one coil is connected to the X2 terminal of the other coil. Y'all have established this above.

In the delta connection, X2 of one coil is connected to X1 of the other coil. That is the deal. The polarity of the second coil of the delta is 180 degrees out of phase to the second coil in the wye connection.

In the wye connection, we can all agree that the coil voltages are 120 degrees out of phase. Now, swap the polarity of the second coil which will now be a delta connection. Well, this also swaps the sine wave 180 degrees. And, the two voltages are now 60 degrees out of phase. Draw it on some graph paper and you will see what I am talking about.

This reversed polarity between the wye and delta coils completely changes the way the voltages add up. In the 120 degree wye, the potential difference from phase to phase will increase by a factor of 1.73. In the delta, which has sine wave voltages which are only 60 degrees out of phase, the voltages do not increase, in fact, they come out to be the same voltage as each coil.

If this is not understandable, grab some paper and pencil and make some sine wave drawings and such. It is actually quite instructive.

If I get any response, I can post some diagrams and photos of my experiments, and I can show the O-scope graphs and such.

This is way too much fun!
 
Welcome back Crossman:

Welcome back Crossman:

I must agree. This is too much fun.

Now I must disagree. The voltages in your example are defined as V at X1 relative to V and X2, right?

Swapping the leads on the transformers does not change the polarity of the voltages as you have defined them. This does change the configuration from an open wye to an open delta. Then the intersection of the two phasor arrows forms an angle of 60 degrees. But, the phase separation is still 120 degrees because the angles are measured relative to a common reference in either wye or delta.

Now, you can see that the open delta diagram comprises two sides of an equilateral triangle, ergo, the open side voltage is equal to the other two.

And, I must agree that these machinations can be confusing, but that is part of the fun, no?
 
Thanks for the "welcome back," Rattus!

Maybe I wasn't clear with my thoughts. Here are a couple of diagrams of the experiments I did today with real xfmrs and a real oscilloscope, and real voltages! :)

In both diagrams, the primary connection remained the same. The only thing that changed was the connections from the X1 and X2 terminals of XFMR 2 - I reversed these in the xfmr bank to make a delta secondary become a wye secondary. Notice that for the oscilloscope, the central point where the two secondaries were connected was used as the "reference" in both diagrams.

The fact that the phase relationship changed from 60 degrees leading in the delta to 120 degrees lagging in the wye tells me that the wye connection voltages will vectorially add together completely differently than the vectorial addition of the delta connection. We are in agreement on that.

Is there something I am missing?
 
Here is my rendition of the oscilloscope graphs I got from the above diagrams. Notice that the XFMR 2 sine wave in the delta is 180 degrees out of phase with the XFMR 2 wye sine wave. This makes sense because we didn't move the oscilloscope leads from the secondary output wires, but we did change the connection of X2 and X1 of XFMR 2, which effectively reverses the polarity of XFMR 2 secondary as related to the rest of the xfmr bank.

As for the output voltages measured from the left-hand terminal of XFMR 1 to the right-hand terminal of XFMR 2:

The delta voltage will be the potential difference between two sine waves which are 60 degrees out of phase.

The wye voltage will be the potential difference between two sine waves which are 120 degrees out of phase.

The normal delta schematic which is an equilateral triangle makes perfect vectorial sense because the sine waves are certainly 60 degrees apart. And the same thing for the wye symbol, which shows 120 degrees apart.

If this thinking isn't correct, whew, somebody please set me straight!
 
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Yes but,

Yes but,

Crossman,

Your oscillographs are correct, but you have changed he reference in your open delta. That implies that the two phases are separated by 60 degrees, and that is not the case. They are merely connected differently. The separation is still 120 degrees.

Now if you had negated the 2nd trace in the open delta, you would be seeing the 120 degree separation as before. I am saying that the delta voltages should always be measured as,

Vac, Vba, and Vcb (their sum is zero)

and,

Wye voltages should be measured as,

Van, Vbn, and Vcn (this sum is zero if the voltages are equal)

Phasor diagrams are the best way to visualzie this stuff. See my post #36 for examples.
 
RATTUS! I think I have it!

I carefully read your post, but your diagram is giving me a timeout and won't open. But, I drew my own phasors, with heads and tails in proper orientation, and I think I have it!!!

In the case of the wye, we must subtract the phasors because we have a tail-to-tail situation. In the case of the delta, we are adding the phasors.

What this means to me related to the sine wave graph, is that in the wye, we are measuring the potential difference between the two sine waves, whereas in the delta, we are actually adding the two sine waves.

Now, to a certain extent, I feel my diagrams are worthwhile, because to subtract phasors, isn't it necessary to reverse one of the phasors 180 degrees? Basically that is what I measured with the O-scope as I had it connected.... the polarity of the sine wave was reversed 180 degrees.

Hopefully I can open your diagram tomorrow. I'll give this some more thought.

I think sometimes folks like myself, while we "believe" that phasors and such are correct, we feel that they are just mathematical constructs which give results, but don't fully "explain" the phenomenon, so we "dig deeper" and sometimes make things harder than necessary. Still, an inquisitve mind is a wonderful thing. And I don't deny that the "digging deeper" sometimes results from an inadequate education though.
 
I think you have it!

I think you have it!

Crossman,

Yes, to obtain the line to line voltage in a any wye, you must subtract the two voltages, and as you say, you reverse one of the phasors and add. For example,

Vab = Van - Vbn = Van + Vnb

For a delta, you should apply Kirchoff's Voltage Law,

Vac + Vba + Vcb = 0

Then,

Vac = -Vba - Vcb = Vab + Vbc

This method will yield the correct phase angle for Vac
 
480 volt 3 ph delta with grounded B phase (weird)

480 volt 3 ph delta with grounded B phase (weird)

We popped a fuse on C phase. Originally we had 480 phase to phase on all 3 phases, with B phase grounded. Phase to ground readings were 480V, 0volt,
and 480V.

Phase to phase after the fuse blew were 480V, 380V, and 540V.

Can anyone explain this? 480volt, 3 ph 3 wire, B grounded.
 
Were you measuring the phase to phase voltage on the load side of the fuses? If so, you were likely getting feedback into C phase through any loads which were downstream. The relative impedances of the loads would affect the actual voltage readings coming back on C.
 
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