Explaining Phase Shift in Delta Wye circuit

[Do You Like Ham]

Hello… I was hoping someone could help confirm my understanding of the theory behind what is happening in a circuit… and hopefully explain it better than I can?

The question comes from this circuit

Which can be summarized as follows:

• 480/277 distribution system (277 3ph Y voltage source)
• To that, on each phase, I’ve connected what amounts to a 100A, 40A, and 25A load on phases A, B, and C, respectively. These may be verified with the green amp probes.
• Also connected to this is a set of 3 transformers. They are connected delta configuration on the primary, and wye configuration on the secondary. There is a 4:1 turns ratio, thus, producing a 120/208 Y system on the secondary.
• Like the primary, on the secondary I’ve connected what amounts to a 100A, 40A, and 25A load on phases A, B, and C, respectively. These may be verified with the yellow amp probes.

The primary phase currents resulting from the secondary loads may be determined as follows.

Table 1



The primary line currents may be determined numerically as shown below.


The cell shading color corresponds to the probe colors in the Multisim circuit. The values are roughly the same as the Irms values (rounding error due to 277/sqrt(3)/4 = 119.94 instead of nominal 120V).

Vectorially, line currents are shown below.



To get the same current on the red probes as Multisim, I found that I need to rotate the angle of the primary loads by -30 degrees. E.g., instead of A, B, and C being at 0, 240, and 120 respectively as on the secondary side in Table 2, they need to be at 330, 210, and 90 as shown in Table 3.

E.g., showing the sum numerically.



(again, blue, green, and red shading correspond to Multisim probes).

And vectorially:

I’ve seen plenty of references describing that when computing the current through delta-wye transformer there is has a phase shift. For example this shows that in balanced system, the primary currents have 30 degree difference with the secondary currents.

However, I’ve not seen any references or documentation on this phenomenon where there are additional loads on the primary side.

Can this essentially be explained by describing the primary and secondary sides having coordinate systems that are rotated from one another by 30 degrees.. thus, when adding loads on the secondary (as in Table 2) they are at 0, 240, 120, whereas on the primary (as in Table 3), they are instead 330, 120, and 90? Is there a better way to describe or explain this?

Maybe they are one-in-the same concept.. but, my initial assumption was, the phase shift is already accounted for as described in this when computing as in Table 2. However, it seems one needs to take additional care to ensure to account for the different “coordinate system” rotation, and I’m struggling a bit to wrap my head around it.

Any additional clarity you can provide would be appreciated!

 

[jim dungar]

The phase shift is real. Usually it is not important, except when tying the secondaries of different transformers together, like with dual supplies. It also a concern in some 'differential' relaying schemes where current in the primary is compared to current on the secondary.

In a typical simple 'tree' style distribution network is is almost never considered. Each transformer starts a new branch of the tree.

You only have one transformer so its phase shift is not important.

 

[Do You Like Ham]

Thanks, @jim dungar -- I wasn't really asking about the importance in this particular scenario.. I was looking to understand and validate understanding of the theory/concept.. and if there's a better way to describe it than I have.

 

[Besoeker3]

For me it was fairly simple. A delta and a star. So 11/13, 23/25. Job done.

 

[don_resqcapt19]

For me it was fairly simple. A delta and a star. So 11/13, 23/25. Job done.

I have no idea of what that means...can you expand on it?

 

[Do You Like Ham]

For me it was fairly simple. A delta and a star. So 11/13, 23/25. Job done.

Huh?

 

[Besoeker3]

I have no idea of what that means...can you expand on it?

I'll try. Usually we start with 11kV or sometimes 13.6kW for most industrial systems. This, for many cases, this is used down to 400Vac but some systems have variable speed speed systems. These can be 6, 12, or 24 systems depending the power required and the voltage. For a 12 pulse system the harmonic amplitudes are 11th and 13th. It is just a means of reducing the system harmonics, makes them a bit cleaner.

And yes, I like ham.

 

[synchro]

The L-L voltages of your wye source are at 30° from their respective L-N voltages (if the source voltages are balanced and the L-N voltages are at 120° from each other). Your transformers essentially convert the L-N loads on the secondary into equivalent L-L loads on the primary, with each L-N load resistance scaled up by a factor of 4. Therefore the currents drawn by these equivalent L-L loads will be in-phase with their corresponding L-L voltage from the wye source, and therefore will be at 30° from the L-N voltage of the source to which they are connected.

 

[wwhitney]

Can this essentially be explained by describing the primary and secondary sides having coordinate systems that are rotated from one another by 30 degrees.. thus, when adding loads on the secondary (as in Table 2) they are at 0, 240, 120, whereas on the primary (as in Table 3), they are instead 330, 120, and 90? Is there a better way to describe or explain this?

Basically, yes, but I would instead put it like this: as the primary and secondary neutral points are tied together, both voltage systems exists in a common coordinate system. It is standard to take the common neutral point as the 0V reference. Then you are free to choose that any one other voltage is at a phase angle of 0 degrees, it could be a primary conductor A, B, C or a secondary conductor A', B', C'. The only remaining choice is one of rotation: if we choose primary A is at 0 degrees, is primary B at 120 degrees or 240 degrees?

With those choices, all the conductors have a well defined voltage and phase. And what you will find is that is when the primary phase shifts are 0, 120, and 240, the secondary phase shifts are 30, 150, 270 (or possibly -30, 90, 210, depending on I assume a choice of transformer configuration).

That's the global picture. Whenever you are working solely on one side of the transformer, you are free to make a different coordinate choice for convenience. E.g. if working on the secondary side only, you can think of the secondary voltages as having phases 0, 120, and 240 degrees. But that choice is incompatible with the choice that the primary side voltages have phases 0, 120, and 240 degrees, and in any computation that involves both sides of the transformer, you must choose one coordinate system or the other.

Cheers, Wayne

 

[david luchini]

Can this essentially be explained by describing the primary and secondary sides having coordinate systems that are rotated from one another by 30 degrees..

It's not the different coordinate systems from the primary to the secondary, per se.

It's that on the primary side, the secondary loads are connected line-to-line, and the primary loads are connected load-to-neutral.

The line-to-line voltage is at a different phase angle from the line-to-neutral voltage. Therefore the load currents are at different phase angles.

 

[Do You Like Ham]

I'm not really sure I follow the explanations offered by @david luchini or @synchro .. namely, that there is some rationale related to the line to line vs. line to neutral... I don't see how that has anything to do with it... and here's why.. if I modify the system to have basically the same scenario, but instead of 100, 40, and 25 A loads at 120 and 277v, have them at 208 and 480v... the same phenomena exists.

See this circuit.


Each set of values is computed in the table below... (the dark yellow/brown values notated by step 6 below are not represented in image above).

Notice that the yellow values (1) and green values (5) are both L-L... yet the green values on the primary are rotated 30deg relative to those on the secondary. Thus, the shift, as I think @wwhitney is verifying, is in fact due to the primary vs. secondary... rather than some notion related to L-L or L-N... thus, the loads connected directly to primary need to be shifted relative to the secondary by 30 degrees.. e.g., in this case, the AB, BC, and CA phases are 0, 240, 120 on the secondary, and 330, 210, 90 on the primary. This needs to be applied in addition to the shift that happens during the computation of the current from the primary onto the secondary (steps 1->2->3->4)... I don't yet understand the physics of it.. but the mathematics is pretty clear.

If someone could elaborate on the physics here.. that would be amazing!





Vector analysis of each step is below...

 

[wwhitney]

I'm not really sure I follow the explanations offered by @david luchini or @synchro .. namely, that there is some rationale related to the line to line vs. line to neutral...

Those explanations are correct. They have to do with the transformer itself having its primary coils connected L-L to the supply, but its secondary coils providing L-N. If you have a wye-wye transformer (where the primary and secondary coils are both connected L-N) or a delta-delta transformer (where the primary and secondary coils are both connected L-L), there is no phase shift.

I don't see how that has anything to do with it... and here's why.. if I modify the system to have basically the same scenario, but instead of 100, 40, and 25 A loads at 120 and 277v, have them at 208 and 480v... the same phenomena exists.

Again, it's not how the loads are connected to the two voltage systems, it's how the coils of the transformer itself are connected to the two voltage systems.

If someone could elaborate on the physics here.. that would be amazing!

The physics here is simple to visualize: draw a delta of coils (an equilateral triangle, where each side is at an angle that represents the phase shift of that coil). Those are the L-L primary voltages.

Now draw a wye of coils next to it, making sure that the each side of the wye is parallel to a side of the delta, and all the wye legs are the same length (each pair of one primary and one secondary coil on a transformer leg will have the no phase shift across it, i.e. the voltages on each of the pair will have the same phase). Those are the L-N secondary voltages.

So comparing L-L primary to L-N secondary there is no phase shift. But you want to compare L-L primary to L-L secondary. So draw in the L-L voltages on the secondary wye, i.e. add the 3 sides that make an equilateral triangle circumscribing the wye. Now that triangle is tilted 30 degrees from the original delta triangle.

BTW, my comments have all been about voltages, but since you are using resistive loads, the load current will be in phase with the supply voltage. So you'll have the same phase relationship between the currents.

Cheers, Wayne

 

[GoldDigger]

I think david lucini said it best, but I will try to say it using a few more words to see if that helps.
First, we need to assume for a start that all of the loads are resistive, so that their power factor does not confuse the situation. We can add in the effects of power factor after we have the basics down.
Second, we will ignore the magnetizing current which flows through the primary into the transformer inductance even when there is no load on the secondary.

Given that, on each secondary winding the current will be directly proportional to the voltage. And within the transformer, the applied voltage on the primary winding will be exactly in phase with the output voltage on the secondary. As long as you measure the voltages from terminal to terminal on the two windings, this will be true. And those two voltages will be exactly in phase.
When we measure the voltage on the secondary side and calculate the phase angles, we are measuring the Line to Neutral voltages
As simply as possible, the line to line voltages on the primary will be in phase with the line to neutral voltages of the secondary. This means that the line to line voltages on the secondary, being out of phase with the line to neutral voltages on the secondary, cannot be in phase with the line to line voltages on the primary.
This tells you that the voltages cannot be in phase. It takes the math to tell us exactly what that phase shift is. That voltage phase shift will not vary as we vary the secondary loads. That same phase shift must then apply between the primary line to line currents and the secondary line to line currents.

 

[Do You Like Ham]

As simply as possible, the line to line voltages on the primary will be in phase with the line to neutral voltages of the secondary. This means that the line to line voltages on the secondary, being out of phase with the line to neutral voltages on the secondary, cannot be in phase with the line to line voltages on the primary.

AH! Yes.. This helps immensely!

 

[GoldDigger]

AH! Yes.. This helps immensely!

Because....Physics.
Merry Christmas.

 

[Do You Like Ham]

Thanks, @wwhitney , @GoldDigger, @synchro, & @david luchini !