wye wye transformer neutral impedance

[winnie]

I am trying to understand how 'stiff' the line-neutral voltage will be in a wye-wye transformer built on a common 3 leg core, with only 3 wires supplied to the primary.

I know that in the case of a wye-wye transformer built out of 3 separate single phase transformers, the 'neutral impedance' would be very high. The reasoning is to consider the arrangement with a single phase load connected line-neutral on the secondary. This secondary current requires primary current on the connected leg. However there is no secondary current on the other two legs of the wye, and thus the primary current on the other two legs of the wye is just magnetizing current.

But in the case of a common 3 leg core, each secondary coil interacts with all three primary coils, so current on only one of the secondary coils can cause current on all of the primary coils.

I was hoping for a pointer to how to calculate this case.

Thanks
Jon

 

[gar]

181211-1419 EST

winnie:

I don't know. I suspect there may be some difference whether it is the center coil that is loaded, vs one of the outer coils.

My approach would be to get a small transformer and run some experiments, and from this try to analyze a likely model.

.

 

[kwired]

181211-1419 EST

winnie:

I don't know. I suspect there may be some difference whether it is the center coil that is loaded, vs one of the outer coils.

My approach would be to get a small transformer and run some experiments, and from this try to analyze a likely model.

.

You have me confused with "center coil" I am seeing three coils 120 dergee apart within a continuous circle. to have a center you must designate "ends".

 

[winnie]

You have me confused with "center coil" I am seeing three coils 120 dergee apart within a continuous circle. to have a center you must designate "ends".

Not the schematic where the coils are drawn with their phase angle, but instead the actual position of the coils on the steel core.

For common small transformers you have 3 coils on 3 'legs' with the legs joined top and bottom with cross pieces. One leg is in the center, and presumably couples differently to the outer legs than the outer legs with each other.

I've simply decided to test this using a single phase variac and measuring output voltage (and short circuit currents) on the various terminals.

-Jon

 

[gar]

191212-0909 EST

minnie:

In your post numbered 4 you have described the structure of a three phase transformer like I am describing..

The core has three legs. On each leg is wound a primary and secondary for one phase.

The magnetic circuit for an outer coil pair goes thru its leg to the center leg, and then the other outer leg which is in parallel with the other outer leg. This is an unbalanced magnetic path.

The magnetic path from the center leg is a balanced pair of outer legs in parallel with the center leg.

So we have similar magnetic circuits for the three legs, but not identical.

.

 

[Sahib]

I am trying to understand how 'stiff' the line-neutral voltage will be in a wye-wye transformer built on a common 3 leg core, with only 3 wires supplied to the primary.

I know that in the case of a wye-wye transformer built out of 3 separate single phase transformers, the 'neutral impedance' would be very high. The reasoning is to consider the arrangement with a single phase load connected line-neutral on the secondary. This secondary current requires primary current on the connected leg. However there is no secondary current on the other two legs of the wye, and thus the primary current on the other two legs of the wye is just magnetizing current.

But in the case of a common 3 leg core, each secondary coil interacts with all three primary coils, so current on only one of the secondary coils can cause current on all of the primary coils.

I was hoping for a pointer to how to calculate this case.

Thanks
Jon

It is not separate cores or common core that makes a difference. It is the neutral shift that causes no heavy current on primary or secondary side of a star/star connected transformer with a floating neutral when there is a single phase short circuit on its secondary side.

 

[winnie]

It is not separate cores or common core that makes a difference. It is the neutral shift that causes no heavy current on primary or secondary side of a star/star connected transformer with a floating neutral when there is a single phase short circuit on its secondary side.

Yes, however I am trying to quantify the magnitude of the neutral shift, and understand how the separate or common core will change the magnitude of this neutral shift.

One can theorize two idea conditions; first a wye-wye transformer with separate cores, infinite core permeability and perfect primary to secondary coupling. In this case you essentially have the equivalent of a 3 wire feeder with no neutral.

The second is a wye-wye transformer where you have perfect 1:1 coupling with coils on the same leg, and perfect 1:0.5 coupling with coils on the other legs. In this case, I believe that the 'derived neutral' would have fairly low impedance.

A real world transformer with separate cores will look quite a bit like ideal 1. With a common core it will look more like ideal 2. I am trying to quantify just how much more like 2 a real world common core transformer will be.

-Jon

 

[gar]

181212-1701 EST

From some Internet searching:

This was the first that I looked at. Some good broad basics. English is probably not the author's native language. But you can still read thru that problem and get some useful information.
https://www.electronicshub.org/three-phase-transformer/

The second one I looked at has some useful information. I only took a quick look.
http://journal.telfor.rs/Published/Vol6No1/Vol6No1_A8.pdf
A repeat of the same article.
https://www.researchgate.net/public...Characteristic_with_Experimental_Verification

I have not found a reference with a direct discussion of winnie's question. But with a open input neutral on a wye primary with unbalanced loading on the secondary I think things will get wild.

.

 

[gar]

181212-1941 EST

winnie:

Look at the 3 phase transformer as 3 single phase transformers. What is the input impedance to a single phase transformer primary as a function of the load impedance on the secondary? Input Z = N^2*Load Z. Thus, the input impedance is infinite for no load on the secondary.

For a short on one secondary and no load on the other two secondaries, and a wye to wye with the input center point (neutral) floating, then we have a short in series with one single other primary. The over voltaged primary will see 1.7 times its rated voltage, and this will certainly drive it into saturation, and burn up that over voltaged primary.

How mutual fluxes will change this I don't know. It is a complex problem.

.

 

[Sahib]

181212-1941 EST

For a short on one secondary and no load on the other two secondaries, and a wye to wye with the input center point (neutral) floating, then we have a short in series with one single other primary. The over voltaged primary will see 1.7 times its rated voltage, and this will certainly drive it into saturation, and burn up that over voltage

.

The over voltages on primary side are with respect to floating neutral but primary side line voltages remain normal. As such, no saturation and burning up of transformer would occur.

 

[gar]

181212-2209 EST

Sahib:

My understanding of what winnie said was ---
1. A true 3 phase transformer is under consideration.
2. The primary is a wye configuration.
3. The center point, neutral point, of the primary is floating.
4. A load is placed on only one secondary winding. Or at least an unbalanced load exists on the secondaries.

If a short is placed on one secondary, then its primary looks like a short. Thus, we have one wye primary essentially placed line to line, and therefore supplied with a large over voltage.

.

 

[Sahib]

181212-2209 EST



If a short is placed on one secondary, then its primary looks like a short. Thus, we have one wye primary essentially placed line to line, and therefore supplied with a large over voltage.

.

Actually what happens is the corresponding primary wye is almost shorted to floating neutral when a phase to neutral is shorted on secondary side. So it is line to neutral over voltage for other two primary side phases and no line to line over voltage on primary side.

 

[gar]

181212-2344 EST

Sahib:

A wye load on a wye source with the wye source having a neutral, but the mid point of the wye load is not connected to the neutral of the source, but the midpoint is floating, is the circuit I believe winnie is talking about.

If I replace one spoke of the wye load with a short circuit, then each of the other spokes are subjected to line to line voltage instead of line to neutral.

.

 

[Sahib]

181212-2344 EST


A wye load on a wye source with the wye source having a neutral, but the mid point of the wye load is not connected to the neutral of the source, but the midpoint is floating, is the circuit I believe winnie is talking about.

Yes

If I replace one spoke of the wye load with a short circuit, then each of the other spokes are subjected to line to line voltage instead of line to neutral.

.

The other spokes (primary or secondary other phases) are subjected to line to line voltage with reference to floating neutral, as the phase to neutral short on secondary side causes similar short circuiting effect on the corresponding primary wye with the result that other primary side phases are subjected to line to line voltage with reference to floating neutral. But the voltages between primary lines remain normal at line to line voltage.

 

[winnie]

181212-1941 EST

winnie:

Look at the 3 phase transformer as 3 single phase transformers. What is the input impedance to a single phase transformer primary as a function of the load impedance on the secondary? Input Z = N^2*Load Z. Thus, the input impedance is infinite for no load on the secondary.

For a short on one secondary and no load on the other two secondaries, and a wye to wye with the input center point (neutral) floating, then we have a short in series with one single other primary. The over voltaged primary will see 1.7 times its rated voltage, and this will certainly drive it into saturation, and burn up that over voltaged primary.

How mutual fluxes will change this I don't know. It is a complex problem.

.

Exactly.

Modeled as 3 separate cores is pretty easy to understand. The shorted secondary reflects as a _single phase_ shorted primary.

Modeled as 3 cores with all flux in the core and perfectly divided on the two paths is also easy. On the same leg your turns ratio is N, on different legs your turns ratio is N/2.

-Jon

 

[winnie]

The over voltages on primary side are with respect to floating neutral but primary side line voltages remain normal. As such, no saturation and burning up of transformer would occur.

I am asking about a wye:wye transformer, so if the line-neutral short on the secondary side causes the _primary_ neutral to shift, then the primary coils do see over-voltage even though the line-line voltage remains constant.

-Jon

 

[gar]

181213-0803 EST

winnie:

Consider this:

Assume ideal transformers, thus no leakage flux. Then secondary impedance is exactly reflected to the primary via N^2. Thus, whatever impedances exist on the secondaries can be viewed as a wye impedance network on the primary side.

Next assume that we can use single phase transformer theory for an equivalent circuit of each transformer of the true three phase transformer. Now we have an equivalent circuit load as seen at the primary side where each spoke is a series circuit of the transformer internal impedance added to its load impedance.

Measurement of each internal impedance of a real transformer, and then an experiment with the transformer to see how this modeling would correlate with the calculated results might indicate whether this would be an adequate model.

Here we have to deal with loads that do not cause core saturation, meaning assume a linear system.

.

 

[winnie]

So I am working from the point of view of a linear system and ignoring saturation. I am also simplifying by ignoring the magnetizing branch of the equivalent circuit.

Building the model as 3 separate single phase units is a mistake, because I am specifically interested in the interaction created by the shared core.

Instead, I am taking Z (pri) = N^2*Z(sec) (thanks for the reminder; I knew that one but wasn't thinking down that path), and adding that when coils are on two separate legs only half the flux will couple from primary to secondary, essentially halving the turns ratio.

I know that this is a horrible oversimplification, but I think it is a good enough back of the envelope for me to understand the tests that will be run.

-Jon

 

[Sahib]

I am asking about a wye:wye transformer, so if the line-neutral short on the secondary side causes the _primary_ neutral to shift, then the primary coils do see over-voltage even though the line-line voltage remains constant.

-Jon

The line to line voltage remains constant on the primary side; may not be so on secondary side.

 

[winnie]

The line to line voltage remains constant on the primary side; may not be so on secondary side.

Agreed. But what I am trying to figure out is the what happens to the neutral on the secondary side. For simplifying I am assuming the primary line-line voltage is constant.