Open 3 phase transformer

[dnem]

The attached sketches here show the math.

Let me try another angle here because you keep posting math without supplying the reasoning behind it. . I don?t dispute the math. . I question the choice of which math.

Scenerio:
You?re looking at 2 transformer coils connected together at one end but not at the other. . So you see 3 leads leaving the transformer. . You don?t see a third coil because it burned up sometime in the past and was removed. . So you know that you originally had 3 phase therefore the coils have always been 120? apart.

The customer wants to power the remaining 2 coils. . You need to know what voltage you will have between any 2 chosen leads.

Now here are the first 2 lines of each of your posted equations:

V? = 277? + 277? ? 2 . . . (277)(277) cos (120)

V1 = -480 + f0

Which equation do you choose and why ? . What do you base your decision on ? . And don?t say whether it was formerly a 3 coil delta or a 3 coil wye because you don?t have that information. . All you have is what you?re looking at now, 2 coils 120? apart.

What you base your decision of which equation to use ?

 

[rattus]

I agree that Kirchoff's current law still holds which is why the neutral is no longer a neutral when you lose one of the 3 phases. . There’s nothing magical about it. . When you lose one of the 3 phases, your neutral becomes a grounded phase.



“there will be variations among the phase voltages which means the neutral is no longer the midpoint!”
Variations among the phase voltages isn’t how you define neutral. . The vector sum is how you define neutral [2008NEC neutral point definition FPN].

“The schematic may look like an open, corner grounded delta, but it is still an open wye. The phasing is wrong for a delta.”
The phasing is wrong for a delta. . That’s what I need details about.

You have 2 coils 120?. . How is the phasing wrong ?

David

David, you are being overly legalistic about this. What was designed to be the neutral is still the neutral. It is pointless to redefine it.

I challenge you to produce a reference that says the neutral becomes a grounded phase. If you can't do this, your claim is baseless.

And, if the voltages are not perfectly balanced, the phasor sum is no longer zero, and "poof" your neutral disappears?

Perhaps I should say that the transformer connections, and hence the phasor connections are wrong for a delta. Look at the open wye phasor diagram. The phasors are connected tail to tail which forms an isosceles triangle, not an equilateral triangle. That is the point that you are missing.

If you don't understand trig and phasors, you can still scale the diagrams to obtain the ratio, 1.732.

I challenge you also to provide a reference which says that 480V is only possible with all three phases.

Now I will answer for Carl.

With the two remaining phases of the wye, the remaining line to line voltage is simply,

Vl-l = 2 x 277V x cos(30) = 277v x 1.732 = 480V

That is about as simple as I can make it.

 

[coulter]

Let me try another angle here because you keep posting math without supplying the reasoning behind it. . I don?t dispute the math. . I question the choice of which math.

... So you see 3 leads leaving the transformer. . You don?t see a third coil because it burned up sometime in the past and was removed. . So you know that you originally had 3 phase therefore the coils have always been 120? apart.

The customer wants to power the remaining 2 coils. . You need to know what voltage you will have between any 2 chosen leads.

Now here are the first 2 lines of each of your posted equations:

V? = 277? + 277? ? 2 . . . (277)(277) cos (120)

V1 = -480 + f0

Which equation do you choose and why ? . What do you base your decision on ? ...

I think I highlighted the pieces in your thought string.

Here's my translation: There are three leads coming out of the xfmr. We don't know if the original was Y or D connected. How can we tell what the voltage will be when energized?

My answer would be, "I need one additional piece of information. "I can't tell the sense (direction) of the coil vectors by looking at the leads. As has already been said, (rattus, jim) the choice in the model is made depending if the coils are connected "tail to tail" or "nose to tail"

Without turing it on, the only way I can think of is to do resistance measurements. That would tell me if the coils were connected Y or D. Then I could pick the model.

carl

 

[rattus]

Summary:

Summary:

Forget all this smoke and dust.

The open side voltage of the open delta is given by,

V = 2 x Vphase x cos(60) = Vphase

The voltage between ANY two phases of a wye is given by,

V = 2 x Vphase x cos(30) = 1.732 x Vphase

This much should be obvious from the phasor diagrams. Any argument to the contrary is baseless. Any argument that the neutral somehow becomes a grounded phase is baseless.

 

[dnem]

I challenge you to produce a reference that says the neutral becomes a grounded phase. If you can't do this, your claim is baseless.

I gave you the reference already and you discounted it. . You don't like the answer so you ask again but the answer is still the same no matter how many times you ask. . The FPN in the neutral definitions in NEC2008.

And, if the voltages are not perfectly balanced, the phasor sum is no longer zero, and "poof" your neutral disappears?

2008NEC Neutral Point FPN, "..... the vector sum of the nominal voltages .....".
2008NEC Voltage, Nominal, "A nominal voltage assigned ..... for the purpose of conveniently ..... The actual voltage at which a circuit operates can vary from the nominal within a range ....."

"perfectly" balanced or not is irrelevant.

You're not going to win this neutral definition battle so why don't you stick to the subject of how an open wye is different from an open delta.

Perhaps I should say that the transformer connections, and hence the phasor connections are wrong for a delta. Look at the open wye phasor diagram. The phasors are connected tail to tail which forms an isosceles triangle, not an equilateral triangle. That is the point that you are missing.

If you don't understand trig and phasors, you can still scale the diagrams to obtain the ratio, 1.732.

"phasor connections are wrong"
I can't follow the logic of your phaser connections because the delta ones show 60 degee angles ! . I don't know what else to say to you.

We're not on the same wavelength and I can't understand what you're talking about.

My conclusion up to this point:
Because several of you are talking phasers and giving an explanation of the difference between open delta and open wye, I can safely assume that there is, in fact, a difference between the 2. . But until I understand why you keep referring to 60 degree angles and saying stuff like, "the 60 degree angle is the same as the 120" until I understand what you mean, I doubt that I'm going to be able to see your point.

"The phasors are connected tail to tail which forms an isosceles triangle, not an equilateral triangle. That is the point that you are missing."
Yep, I am missing it !
Why are we talking triangles when the power phases are 120 degrees apart ? . Triangle angles must add up to 180 degrees. . Three phases 120 degrees apart add up to 360 degrees. . How can you use a triangle for 360 degrees ?

"If you don't understand trig and phasors ....."
I don't understand why you use a triangle or use 60 degrees when you're calculating delta but don't use the triangle when you're doing wye. . Why do you use 120 degree angles to calculate wye but use 60 degree angles to calculate delta ?

David

 

[dnem]

Forget all this smoke and dust.

The open side voltage of the open delta is given by,

V = 2 x Vphase x cos(60) = Vphase

The voltage between ANY two phases of a wye is given by,

V = 2 x Vphase x cos(30) = 1.732 x Vphase

This much should be obvious from the phasor diagrams. Any argument to the contrary is baseless. Any argument that the neutral somehow becomes a grounded phase is baseless.

We need a no spin zone. . You could be a politician.

You said 2 things that have nothing to do with each other.
#1

Forget all this smoke and dust.

The open side voltage of the open delta is given by,

V = 2 x Vphase x cos(60) = Vphase

The voltage between ANY two phases of a wye is given by,

V = 2 x Vphase x cos(30) = 1.732 x Vphase

This much should be obvious from the phasor diagrams. Any argument to the contrary is baseless.

#2

Any argument that the neutral somehow becomes a grounded phase is baseless.

#2 has already been proven to be wrong by the NEC, so let's just stick to #1.

"This much should be obvious from the phasor diagrams"
What's "obvious" is that you use 120 degree angles to define wye phase relationships and 60 degree angles to define delta phase relationships and I haven't seen your explanation yet.

David

 

[dnem]

My answer would be, "I need one additional piece of information. "I can't tell the sense (direction) of the coil vectors by looking at the leads. As has already been said, (rattus, jim) the choice in the model is made depending if the coils are connected "tail to tail" or "nose to tail"

"(direction) of the coil vectors"
When/If you know the direction of the vector coils what will you do with that information ? . Does this have any bearing on your choice of 120 degree vectors or 60 degree vectors and why ?

"tail to tail" or "nose to tail"
Can you explain what you would do with two, 120 degrees apart, 277volt coils that are connected "tail to tail" ?
Can you explain what you would do with two, 120 degrees apart, 277volt coils that are connected "nose to tail" ?

Do you use your 60 degree math on one of them ? . Which one ? . And why ?

David

 

[coulter]

David -
First: I'm not upset that you don't agree nor understand why I'm in favor of particular math models for different system connections. Another poster might be, but I'm not.

Second: I'm pretty sure I'm dead right about the models I have chosen and the application to the different systems.

Third: I am equally sure my teaching talents are close to poor. Sometimes I don't give very good explanations of my thinking.

I am trying - I promise.

...When/If you know the direction of the vector coils what will you do with that information ? . Does this have any bearing on your choice of 120 degree vectors or 60 degree vectors and why ?...

Use that imformation to determine which math model fits - thereby knowing how to make the calculations that will match up with what actually happens when the system is energized.

Yes it would have a bearing on my choice. The "why" is because the model that fits delta systems is not a good fit for Wye systems.

...Can you explain what you would do with two, 120 degrees apart, 277volt coils that are connected "tail to tail" ?

There are two ways to come up with the voltage between the ends of two coils.
Vectors: With two vectors connected tail to tail, the voltage measured between them can be calculated by subtracting the two vectors. but that gets messy. As you recall, I showed that in one of my previous sketches.

The other method is to use trig to calculate the distance (in volts) between the two vector endpoints. The triangle formed by the two 277 legs, 120 deg apart, connected at the tails. I showed that math in a previous sketch.

Can you explain what you would do with two, 120 degrees apart, 277volt coils that are connected "nose to tail" ...

Redraw them so they are shown the way they are connected. As you recall, I did this in one of my earlier sketches. Even so, I have attached another sketch showing the redraw. I didn't change the electrical connections, I didn't change the sense. I just arranged the drawing to better see how the math fits. This one fits the 60 degree math model.

...Do you use your 60 degree math on one of them ? . Which one ? . And why ? ...

I'm not sure what you are after here. I use the model that is the best fit to the real world.

If one leg of a 480 Y opens up, the other two are still two vectors, 120 degrees apart, connected at the tail. The voltage across the open ends was 480V, and it's still 480V. The system does not inexplicapably collapse to a 60 deg equalateral triangle. The angles are 120, 30, 30.

Why are the Delta angles 60 degrees? Cause that's the way they are connected - three vectors, connected in a triangle, nose to tail, equal angles

Why are Wyes 120 degrees? Cause that is the way they are connected. Three vectors, connected at the tails, equal angles around a full circle.

I hoped this helped. I starting to repeat myself. I've always figured that if I didn't explain well enough the first time, saying it louder and again wouldn't make it any easier to understand.

edited to clean up some formating

carl

 

[rattus]

You are not listening!

You are not listening!

dnem,

The 60 degree angles in a delta phasor diagram DO NOT indicate the angle of separation. Phasors are always drawn in the direction dictated by their phase angles. You need to find a text and read up on phasors.

Head to tail phasors which intersect at 120 degrees do not occur in 3-phase systems! Tail to tail phasors intersect at 120 degrees in wye diagrams.

Now, if you draw them head to tail, the intersection of the two lines does form a 60 degree angle, but that is NOT the angle of separation. that is merely the complement of 120 degrees. If you draw them tail to tail, the included angle is 120 degrees. This is high school geometry!

Refer to the diagrams posted by me and by Carl. That should say it all.

As far as the grounded phase is concerned, you should provide a solid reference to back up your argument. You have only provided a vague reference to the NEC. You reverse interpretation of the neutral definition hardly suffices as a solid reference. Furthermore, whether the line is grounded or not is immaterial, and it doesn't change anything to rename the neutral, the phasor diagrams are unchanged. Even if you were right, it doesn't make any difference!

You seem to be trying to equate the delta phasor diagram to the wye phasor diagram. Can't be done!

 

[steve066]

I have a bunch of questions.

3phases from 3phase conductors can come from 2 single phase transformers that are 120 degrees apart. . This situation can be purposeful or it can come from a single transformer failure in a 3phase delta configuration.

That much I understand.

But what I can't wrap my brain around no matter how much I contemplate it is:
What happens to a 3 single transformer wye configuration that losses one transformer ?

You have 2 transformers 120 degrees apart from each other. . Is there anything different about this former wye that would cause it to act differently than the former delta ?

Obviously one of the 3 wires attached to the 2phases that are still working was once a neutral. . Is it still a neutral or is it now a phase conductor ? . Do the voltages change ? . Is the phase to former neutral now equal to the phase to phase voltage ? . One single voltage or does it still have 2 voltages somehow ?

The "open" transformer bank consisting of 2 transformers 120 degrees apart that supply 3 phases of power is always called an "open delta". . But is it a delta or a wye or neither, just an open 3 phase transformer ?

What part of the picture am I missing ?

David

\

Wow, lots of discussion, it would take me a month to read and catch up. I'm hoping maybe I can shed some light if you are still confused.

From your post, I'm not sure if you understand that the coils of wire on a transformer are not 120 degrees apart. Only the applied line to line voltages (L-L) are 120 degrees apart. Remember, these are just coils of wire, and they don't spin like motors, so as far as the coils go, they are all the same.

If a wye looses one transformer, (here I am assuming the primary of a stepdown transformer is a wye, and that is what has an open winding) the first two phases work exactly like they did before. Remember, the power company sends out a constant L-L voltage, and this would be the same voltage across two transformers. That voltage doesn't change at all. (say 4160 on a 4160/2400 volt system).

The neutral voltage also doesn't change. It is still connected to the center of the wye transformer at the power company. And it is probably grounded also. So the voltage at this point is essentially still zero. And you still get 2400 volts from the line side of each transformer to this point.

The primary current also stays the same in the first two coils (if we ignore single phasing motors and such). The voltage is the same, so the current is the same. However, the current on the neutral going back to the power company will increase. If all three phases are running normal, the currents from each phase combine in the neutral, and they all cancel each other out (three eaual voltages each 120 degrees out of phase add to zero). With one lost, the canceling doesn't occur the same. It is easy to see that 2 equal voltages 120 degrees out of phase add up to some number greater than zero.

In the 3rd winding, no current flows since it is open. Therefore, there is no voltage induced in the 3rd phase of the secondary of the transformer. (This would be different if all the windings are wound on the same core as in a single three phase transformer.)

No voltage on the 3rd phase of the secondary means no current flows in that winding.

 

[steve066]

As far as your open delta question goes, picture this:

Picture 3 single phase transformers, arranged in a delta (triange), but all leads are unconnected.

Now place a single phase voltage across each primary. And place a single phase load across each secondary. (We still have three completely independent transformers, no two transfomers are connected together.) If we make our 3 voltages 4160V, and make them 120 degrees out of phase, we have a typical 3 phase source, and a typical 3 phase load.

However, note that we need 6 wires from the power company to our load. And each wire carries the full line current. Not very efficient.

We can go ahead and connect our transofmer secondaries in either a delta or wye. Since the voltage around the delta adds to zero, and the voltage in the center of the wye is zero, we are basically connecting wires with zero voltage differnce. Nothing changes except we need fewer wires going to our load. Oh, and each wire actually has less current on it because of the phase currents partially canceling each other out.

But to the load and the source at the power company, nothing has changed.

Going one step farther, we can connect our transformer primarys in delta. Once again, since the voltages around the delta add to zero, everything looks the same to the source and the load. But now we only need 3 wires from the power company to our transformer. And again, the current on each wire is less since they partially cancel. Much more efficient.

If you have followed this far, you have a typical poco generator, 3 phase distribution, a delta primary transformer, either a delta or wye seconday, and a three phase load.

What happens voltage wise if we take out one single transformer. Just open the windings on the primary and secondary. Maybe we even have a DP-DT switch that takes out the transformer with one flip of the switch. What do the voltages do when we flip the switch?

The voltages don't do anything. They all stay the same. You can convince yourself this is true. The POCO sits the L-L primary voltage. That won't change. And two transformer windings on the secondary are all that are needed to ensure the third voltage would be the same.

Picture the 3rd transformer drawn in in dashed lines. It could be there, it could not be there. The voltages all stay the same either way. Flip it on, flip it off. Again, the voltages are the same. That's why you should still think of it as a delta.

ON the other hand, taking the third transformer out does change the currents. By opening the delta, you open one path for current. And basically the other two windings have to generate enough power to make up for the missing winding. They do this by having their currents increased.

 

[dnem]

From your post, I'm not sure if you understand that the coils of wire on a transformer are not 120 degrees apart.

If all three phases are running normal, the currents from each phase combine in the neutral, and they all cancel each other out (three eaual voltages each 120 degrees out of phase add to zero). With one lost, the canceling doesn't occur the same. It is easy to see that 2 equal voltages 120 degrees out of phase add up to some number greater than zero.

You're right, I don't understand.
First you seem to be saying that the transformer coils are not 120 degrees apart [which I know isn't true].
Then you seem to be saying that there are three equal voltages each 120 degrees out of phase [which I know is true].

No matter what speed you wish to cycle an AC system, 50 hertz for Europe, or 60 for the USA, or any other speed, you will always have 360 degrees in a given cycle. . If you want 3 phase power and you don't want a "dead spot" in your cycle [like a car with one bad spark plug], then you must have 120 degrees between phases.

"I'm not sure if you understand that the coils of wire on a transformer are not 120 degrees apart."
You're right, I don't understand that nor believe that.

David

 

[dnem]

dnem,

The 60 degree angles in a delta phasor diagram DO NOT indicate the angle of separation. Phasors are always drawn in the direction dictated by their phase angles. You need to find a text and read up on phasors.

Head to tail phasors which intersect at 120 degrees do not occur in 3-phase systems! Tail to tail phasors intersect at 120 degrees in wye diagrams.

Now, if you draw them head to tail, the intersection of the two lines does form a 60 degree angle, but that is NOT the angle of separation. that is merely the complement of 120 degrees. If you draw them tail to tail, the included angle is 120 degrees. This is high school geometry!

Refer to the diagrams posted by me and by Carl. That should say it all.

And I say again, the geometry was never in question. . The reason for choosing one equation for delta and a different one for wye was the question. . I recognize high school geometry when I see it. . But dry math equations are meaningless without a way of connecting them to the application. . There may be some people that are content with looking for the equation that is labeled "delta" and looking for another equation that's labeled "wye" but I wanted to know why the equations applied.

As far as the grounded phase is concerned, you should provide a solid reference to back up your argument. You have only provided a vague reference to the NEC. You reverse interpretation of the neutral definition hardly suffices as a solid reference. Furthermore, whether the line is grounded or not is immaterial, and it doesn't change anything to rename the neutral, the phasor diagrams are unchanged. Even if you were right, it doesn't make any difference!

If you consider the NEC to be a "vague" reference, that's your choice. . Without a way to define words, they become meaningless. . You can name your cat "Neutral" for all I care but it won't help you in a meaningful conversation about electrical theory.

It's amazing that you can't disconnect your false conclusions about neutrals from the delta/wye conversation.

"You reverse interpretation of the neutral definition hardly suffices as a solid reference."
I didn't provide any interpretation. . I quoted definitions. . The fact that you don't like the definitions isn't my fault.

"Furthermore, whether the line is grounded or not is immaterial, and it doesn't change anything to rename the neutral, the phasor diagrams are unchanged. Even if you were right, it doesn't make any difference!"
And that's why I can't figure out why you keep bringing it up. . It does make any difference, so stop bringing it up.

You seem to be trying to equate the delta phasor diagram to the wye phasor diagram. Can't be done!

I'm trying to understand the difference.
At least I'm trying to understand what I don't currently understand while you stubornly insist that the NEC definitions are wrong. . Why do you resist the NEC definitions so strongly ? . Can't you ever learn and grow or do you need to come into the conversation with all of the answers ?

David

 

[dnem]

I WANT TO THANK EVERYONE THAT HAS RESPONDED !

Even tho it didn't result in a "meeting of the minds", it was a worthwhile attempt. . I'm going to turn to a book explanation and see if I can find the missing piece or pieces of the puzzle that I was unable to see in these posts.

David

 

[rattus]

dnem,

A little high school trig is all you need. If you understand sines and cosines, the derivation of the formulas will be obvious from the phasor diagrams. Oh, and Kirchoff's voltage law too.

 

[dnem]

post #2
Just look at the phasor diagrams for the two systems. With the delta, the remaining phases add to provide the missing phase.

post #55
dnem,

A little high school trig is all you need. If you understand sines and cosines, the derivation of the formulas will be obvious from the phasor diagrams. Oh, and Kirchoff's voltage law too.

You've been talking about the phaser diagrams since your first post, but that never has been the issue. . You can give me any phaser diagram you wish and I'll do the trig. . I was doing that trig 25 years ago.

I've tried to be clear about the fact that I'm questioning the diagram supplied, not the trig itself. . Delta is displayed by a triangle with 60 degree angles, why ? . Until that question is answered, the trig is worthless to me because the whole question doesn't revolve around how to perform trig functions. . The question is why is the formula chosen, not how do you plug numbers into the formula and do the calculation.

I can't believe after 55 posts, that fact still isn't clear.

?Oh, and Kirchoff's voltage law too.?
You just can?t give up your battle with the NEC definitions, can you !?
You started that in post #32

post #32
This is an illogical inference. Kirchoff's current law still holds if you lose a phase. Even so it is illogical to infer that the neutral magically becomes a grounded phase with the loss of one transformer.

Claiming that the problem is my inability to do high school trig won?t answer the question that I?ve asked. . You just want to continue to try to make yourself feel better by taking shots at other people.
Claiming that the problem is my inability to understand Kirchoff?s law won?t make the NEC go away. . You just want to continue to try to make yourself feel better by taking shots at other people.

I don?t believe there is anything that can be accomplished by continuing this conversation. . It was worth a try to post this thread but I can see that I need to go to plan B.

David

 

[rattus]

Case Closed:

Case Closed:

dnem,

I enjoy helping those who want to be helped, but you seem to be more interested in attacking me than in understanding what I am saying. I have answered your 60 degree question, but you don't seem to be listening. Rather than asking for clarification, you criticize. Your attitude is not conducive to learning, and it is certainly not appropriate for this forum.

You remind of "The Sniper" who criticized me harshly a couple of years ago. He could not answer my challenges, and he even referred me to a diagram I had helped develop. He now asks my opinion. Case closed!

 

[dnem]

dnem,

I enjoy helping those who want to be helped, .....

And of course you don't need any help because you have everything already figured out. . Everybody that disagrees with you is wrong and should be belittled, including the guys that wrote the NEC definitions.

..... but you seem to be more interested in attacking me than in understanding what I am saying.

That claim might have been effective if we were verbally conversing and the conversation wasn't being taped.

But here, I don't have to respond to your accusation because the whole thread is posted here. . Everybody can see who has been on the attack. . After you attempt to belittle someone, you apparently subscribe to the concept: "The best defense is a good offense". . Just accuse the other person of doing what you were doing. . But it's all recorded here for anybody to read.

David

 

[coulter]

David -
I did think of additional reasoning as to why I use a different model on the two systems.

First, your are right, Y or D, the coils are 120 degrees apart. (Note I said "coils", not phasors or vectors - this may matter later - I hope)

Second, a couple of things I know you already know:
1. Drawing the vectors of a delta in a nose to tail doesn't show the phase angle relationships for the rotating fields. Your drawing (I did for you on one of the previous sketches) does show the correct phase angle relationships

2. Drawing the vectors of a delta N-T-T is convenient for showing how the coils are connected. The three coils have to sum to 0 volts, cause that's the way they are connected. For example, a change in one coil of a few percent sets up a huge circulating current in the delta. Again, It does not show the phase relationships - it is just a convenient way to show the coil connections and that they add to 0V. The practice of using arrows instead of coils is just a convenience.

3. The ends of the Y connected coils are sort of hanging in free air. A few percent change in one coil doesn't really do anything (except the motors don't like it a lot). The three coils don't have to sum to 0V.

Summary:
1. As you stated, drawing the vector triangle for a delta is absolutely incorrect for showing the phase angle relationships for the rotating fields.

2. Drawing the vector triangle (and using 60 deg math) is absolutely useful for calculating the voltages and currents in a delta ckt. This is the "why".

Two things that didn't change, but I didn't say them right:
1. Losing a phase on a 480 Y: You still have four leads. One is still the N wire, connected to ground. One is now just dead - not connected to the others. The two phases that are still alive are still 120 deg apart, still connected tail to tail, still have 480V between them.

2. Losing a phase on 480 D: You still have three leads. All three are still live. And voltage measurements around the delta still have to sum to 0V.

again, hopefully this helps.

edited to fix some grammar

carl

 

[dnem]

Carl,

Thanks for setting aside the trig and just talking concepts. . Your points are helpful. . I need to think about them.

I'm going to be meeting with a fellow teacher sometime this week. . I'm going to run this by him and see if he has anything to add.

David