Open 3 phase transformer

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Try this:

Try this:

dnem, I think I see your problem.

After removing one transformer the phasor diagram is a triangle, but in comparing it with a delta, you will note that the sides are not equal, and the arrows are not arranged head to tail as they are in the delta.

Draw out this triangle with two equal sides including a 120 degree angle. The third side is 1.732 time either of the other sides. That is your 480V side.
 
dnem said:
?If one transformer is removed from the wye, the "neutral" still carries unbalanced current?
I don?t believe that. . You can?t have balance without all phases present. . I see the ?former neutral? as having full current, same level as the 2 remaining ungrounded phases. . Full current on a grounded conductor makes it a grounded phase.
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Where have you ever seen a definition that says if a conductor carries full current it is no longer a neutral? Are you saying that if there is 100% unbalance on a 120/240V MWBC the neutral is no longer a neutral even though it is carrying full current?

And what about a 2-wire 208V load, doesn't it have balanced current even without the third phase?
 
Lost Single Phase Transformer

Lost Single Phase Transformer

The question does not give enough information, so answering assuming Delta to Wye (answer is different if Wye to Wye)

In this configuration the Delta is created by combining a start with a finish on the primary side.

The wye is created by combining all the starts together.

So what happens - the primary can not support the open delta and becomes two single phase inputs, leaving two single phase outputs - so you only have voltage Line to neutral.

In a normal Open Delta the Starts are Tied together creating the phase relationship, but when you tie a start to a finish you do not obtian this relationship.
 
fordaputz said:
The question does not give enough information, so answering assuming Delta to Wye (answer is different if Wye to Wye)

In this configuration the Delta is created by combining a start with a finish on the primary side.

The wye is created by combining all the starts together.

So what happens - the primary can not support the open delta and becomes two single phase inputs, leaving two single phase outputs - so you only have voltage Line to neutral.

In a normal Open Delta the Starts are Tied together creating the phase relationship, but when you tie a start to a finish you do not obtian this relationship.
Click to expand...

Don't worry about the primaries. Just consider three separate transformers fed by three phases from the poco. The phase separation is determined by the poco. For the moment, just consider the secondaries.

In ANY delta, the start is ALWAYS connected to the finish. The phasor diagram comprises three arrows each pointing 120 degrees away from its neighbor. Connect the arrows head to tail and you have an equilateral triangle which gives the illusion of 60 degrees separation, but it is a mistake to believe that.

In any wye, the starts are connected to the neutral. The phasor diagram then is in the shape of a wye. If we draw in the line to line phasors we have three isosceles triangles, with the phase voltage arrows subtending 120 degree angles. A little trig tells us that the line to line voltage is 1.732 times the phase voltage.

That is all there is to it. Removing one transformer has absolutely no effect on the remaining phase voltages or line to line voltage! Ideally that is.
 
jim dungar said:
Page 12 talks about a Wye-Delta with one unit missing, but there is no discussion of an open Y-Y.
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I would imagine an open Y primary would simply be mirrored in the secondary. Line-to-line loads would still place some voltage on the open-primary's secondary, though.
 
dnem,

Consider the open wye secondary. If you swap the terminals on one of the secondaries, it becomes an open delta, and you will see 277V on the open side!

That is the difference--head to tail or tail to tail.
 
Thanks for your picture.
That was what I was thinking about

coulter said:
dnen- -
I'm sure we are describing the same thing. I moved your coils around, but left them pointing the same direction, and connected them the same direction. Note the 60 degree relationship. We're on the same page - same book:cool:
Click to expand...

The only 60? angle that I'm aware of is on the visual triangle way of displaying a delta which is visual only, it's not electrically correct. . That's why I asked you to draw it the electrically correct way.

coulter said:
For the Y connected coils, the angular relationship between the two coils still alive is 120 degrees.
Click to expand...

The 277v open delta has two coils still alive at 120? apart and has 277v phase to phase. . The 277v open wye has two coils still alive at 120? apart so why would the phase to phase be 480v ?

David
 
rattus said:
Let's start with your idea that all three phases must be present to obtain 480V. To do this, just remove completely one transformer from an open wye and assume no load to make it simple. In my example, I assume,

Van = 277 @ 0; this voltage is determined by transformer A and no other.
Vbn = 277 @ -120; this voltage is determined by transformer B and no other

Vcn will float because it is a floating node, not connected to anything..

The line to line voltage, by definition, is the potential difference between nodes A and B. That is, you subtract one voltage from the other. The missing transformer has absolutely nothing to do with it. You can do this graphically, or you can use phasor math. The result will be the same.

Or in other words, treat the three transformers as three totally independent sources.

If we can get over this hurdle, the rest should fall in place.
Click to expand...

"The missing transformer has absolutely nothing to do with it."

So if the missing transformer means nothing then you're left with 2 transformers 120? apart if the original was a 3 transformer delta formation OR a 3 transformer wye formation.

Why does the two 277v transformers, 120? apart, and originally part of a delta configuration have 277v phase to phase but
two 277v transformers, 120? apart, and originally part of a wye configuration have 480v phase to phase ?

You have two 277v transformers, 120? apart how are you going to get two different phase to phase voltages ?

David
 
rattus said:
dnem,

Consider the open wye secondary. If you swap the terminals on one of the secondaries, it becomes an open delta, and you will see 277V on the open side!

That is the difference--head to tail or tail to tail.
Click to expand...

"If you swap the terminals on one of the secondaries, ....."
So I'm flip flopping the 2 terminals on one of the remaining 2 coils.
"..... it becomes an open delta, and you will see 277V on the open side!"
277v phase to phase ?
( because I think everybody is in agreement that no matter what delta wye open closed formation you use, you will continue to have 277v across any one coil )

And therefore if I DON'T flip flopping the 2 terminals on one of the remaining 2 coils I will have 480v phase to phase ?

Is the switching of the terminations on one of the two remaining coils the key to it being open delta or open wye, 277v phase to phase or 480v phase to phase ?

David
 
jim dungar said:
Where have you ever seen a definition that says if a conductor carries full current it is no longer a neutral?
Click to expand...

You're right, it's not current level, it's whether or not it's unbalanced current.
Full current doesn't matter. . Unbalanced matters. . Vector sums matter.

NEC2008 page 28, neutral conductor and neutral point
Has to be one of 3:
1] Midpoint of a 3? wye
2] Midpoint of a single phase
3] Midpoint of a single phase portion of a 3? system [center tapped delta]

plus the FPN says that the total vector sum from all other phases must be zero or the same as the neutral point. . When you lose a phase, your vector sum is no longer the same as the former neutral point, therefore the former neutral point is no longer the present neutral point but rather is now classified as a grounded phase.

jim dungar said:
Are you saying that if there is 100% unbalance on a 120/240V MWBC the neutral is no longer a neutral even though it is carrying full current?
Click to expand...

The neutral would still be carrying the unbalanced current only even during a scenerio where the full current is the same value as the unbalanced current.

One phase/leg of a 120/240v circuit has full current and the other phase/leg has no current, the neutral will carry the unbalanced current which in that example only would be identical to the full current but in most real world examples the unbalanced is less than full current.

Phase/Leg A = 20 amps full circuit current
Phase/Leg B = 20 amps full circuit current
Neutral = 0 amps unbalanced current only

Phase/Leg A = 20 amps full circuit current
Phase/Leg B = 15 amps full circuit current
Neutral = 5 amps unbalanced current only

Phase/Leg A = 20 amps full circuit current
Phase/Leg B = 0 amps full circuit current
Neutral = 20 amps unbalanced current only

jim dungar said:
And what about a 2-wire 208V load, doesn't it have balanced current even without the third phase?
Click to expand...

The definition of neutral depends on the source not the circuit.

In a 2 wire circuit that has one black wire and one white wire supplied from a wye system, the white wire is still a neutral even while carrying full current and even tho only one phase is present in that circuit. . The neutral is a neutral when connected to the neutral point of the system.

David
 
From dnem:

From dnem:

"plus the FPN says that the total vector sum from all other phases must be zero or the same as the neutral point. . When you lose a phase, your vector sum is no longer the same as the former neutral point, therefore the former neutral point is no longer the present neutral point but rather is now classified as a grounded phase."

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.

Furthermore, we can still use the term "neutral" even if it no longer fits the definition. In practice, there will be variations among the phase voltages which means the neutral is no longer the midpoint! Do we now call it the "former neutral". It is too confusing to call it anything else.

The schematic may look like an open, corner grounded delta, but it is still an open wye. The phasing is wrong for a delta.
 
dnem said:
NEC2008 page 28,

FPN says that the total vector sum from all other phases must be zero or the same as the neutral point. . When you lose a phase, your vector sum is no longer the same as the former neutral point, therefore the former neutral point is no longer the present neutral point but rather is now classified as a grounded phase.

The definition of neutral depends on the source not the circuit.

The neutral is a neutral when connected to the neutral point of the system.
Click to expand...

David,

You are correct in saying it is the source that defines if a neutral point exists in the system. In a grounded WYE- grounded WYE transformer connection the neutral point on the source side does not change even if one of the wye transfromers is removed therefore the secondary neutral point does not change.
 
dnem said:
Thanks for your picture.
That was what I was thinking about...
Click to expand...
You're welcome. I was pretty sure it matched what you are saying. At least I spend plenty of time trying to make sure I understood.

dnem said:
...The only 60? angle that I'm aware of is on the visual triangle way of displaying a delta which is visual only, it's not electrically correct. . That's why I asked you to draw it the electrically correct way. ...
Click to expand...
As you asked, I drew it the electrical correct way.

Now, what I am asking is: look at the second half of the same page. I moved the coils on the paper. I didn't change the connections, directions, or sense. No swapping of ends required. I contend that what you asked for and what I drew are electrically identical.

So why the change from 120 deg to 60 deg. Well, it didn't. It's a trig issue. It has to do with 120 deg and 60 deg being complimentary (supplimentary? - aarg can't remember) angles. The way they are shown in your drawing have the tails physically in the same place on the paper, but the tails are not electrically connected. As for if the arrows are electrically 120 out, or 60 out, it is as the rattus says (paraphrased), "It's in how you connect the scope." And that is also true of what a motor sees. The rotating field a motor sees is entirely dependent on how the coils are connected.

dnem said:
...The 277v open delta has two coils still alive at 120? apart and has 277v phase to phase. . The 277v open wye has two coils still alive at 120? apart so why would the phase to phase be 480v ?...
Click to expand...
You need to perform the exercise I recommended. Draw the two lines out on paper and measure them.

This is an exercise in mathematical models. The test is if the model calculations match the measured results.

carl

I didn't have time to preview - please excuse the typos
 
dnem said:
...The 277v open delta has two coils still alive at 120? apart and has 277v phase to phase. . The 277v open wye has two coils still alive at 120? apart so why would the phase to phase be 480v ?...
Click to expand...

The attached sketches show the math models as understand them. Before you land on me about the "Delta coils are 120 degrees apart", I would ask you review a previous sketch showing the translation between what you had me draw and what I put on the same page. What I put down is the way the coils are connected - same as your representation.

The attached sketches here show the math. (Just so you know - I'm pretty sure you already knew the math:smile: )

carl
 
Look at it this way:

Look at it this way:

Consider the attached phasor diagrams of an open delta and an open wye. The arrow heads correspond to the polarity markings on the transformer diagram.

Clearly, Vac and Vba combine to provide the open side voltage, Vcb in the open delta.

Now consider the open wye. Clearlly, Vab is the phasor DIFFERENCE, Van - Vbn. A little trig shows that |Vab| is 1.732 times the magnitude of either phase voltage, or you can do it graphically.

To summarize, the phasor diagrams of the open delta and the open wye are vastly different! There is no basis for claiming otherwise.
 
rattus said:
"plus the FPN says that the total vector sum from all other phases must be zero or the same as the neutral point. . When you lose a phase, your vector sum is no longer the same as the former neutral point, therefore the former neutral point is no longer the present neutral point but rather is now classified as a grounded phase."

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.
Click to expand...

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.

rattus said:
Furthermore, we can still use the term "neutral" even if it no longer fits the definition. In practice, there will be variations among the phase voltages which means the neutral is no longer the midpoint! Do we now call it the "former neutral". It is too confusing to call it anything else.

The schematic may look like an open, corner grounded delta, but it is still an open wye. The phasing is wrong for a delta.
Click to expand...

?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
 
jim dungar said:
In a grounded WYE- grounded WYE transformer connection the neutral point on the source side does not change even if one of the wye transfromers is removed therefore the secondary neutral point does not change.
Click to expand...

I disagree. . There is no information about the primary that has any bearing on determining the neutral point of the secondary. . The secondary neutral point [if any] is based exclusively on vector sum of secondary phase coils.

A primary neutral can be connected to a secondary neutral.
A primary neutral can be connected to a secondary grounded phase.

The primary is not a determining factor in identifying the secondary neutral.

David
 
coulter said:
So why the change from 120 deg to 60 deg. Well, it didn't. It's a trig issue. It has to do with 120 deg and 60 deg being complimentary (supplimentary? - aarg can't remember) angles. The way they are shown in your drawing have the tails physically in the same place on the paper, but the tails are not electrically connected. As for if the arrows are electrically 120 out, or 60 out, it is as the rattus says (paraphrased), "It's in how you connect the scope." And that is also true of what a motor sees. The rotating field a motor sees is entirely dependent on how the coils are connected.
Click to expand...

I'm not objecting to the 60? just to be difficult. . I don't see how you multiple 60 x 3 and get 360?.

If one phase hits maximum voltage and then 60? later in the cycle a second phase hits maximum voltage and then 60? later in the cycle a third phase hits maximum voltage, what happens 60? later ? . You're only half way thru the cycle.

60? separate requires 6 phases to complete the cycle whether you?re on 60 hertz or any other chosen cycle speed.

Someone could dream up a 6 phase system for their private commune or private island and install their private power plant generator coils 60? apart and everything would be fine. . But we don?t have 6 phases so I don?t understand how delta can be referred to as phases 60? apart.

If you have a 6 cylinder car and you remove 3 spark plugs, how is it going to run ?

David
 
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agree....
diagrams and calculation from "coulter" said it all.....

why need two transformers (Y)....when one could do....
 
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