Wye-Delta, Delta-Wye Transfomers

[engineeringthoughts]

Source to transformer stepped-up from 480/277V to 13.2 kV wye - delta (corner grounded). Then stepped down via delta (corner grounded) to wye (grounded) from 13.2 kV back to 480 V. The load is about 630 kVA and .9 pf. I only have line currents on the load (wye) side of A = 200, B = 16, and C=190 A, I need the neutral current on both wye sides and the line currents traced back to the source. This is a problem I have been working on (purely for myself).

 

[wwhitney]

The load is about 630 kVA and .9 pf.

What exactly do you mean by the a 0.9 power factor? In the context of single phase, or balanced 3 phase, the meaning is clear. But with unbalanced 3 phase, each line current could have a different phase shift.

I only have line currents on the load (wye) side of A = 200, B = 16, and C=190 A, I need the neutral current on both wye sides and the line currents traced back to the source.

I don't see how those currents can square with your voltages of 480Y/277V and 13,200V delta and the load figure of 630 kVA. For a balanced loading (just for comparison and order of magnitude), that 630 kVA would be 758A at 480Y/277V or 28A at 13,200V delta.

Anyway, given the magnitudes of the 3 line currents on a wye, you can't calculate the neutral current unless you also know the relative phase shifts of those 3 line currents.

For example, say you know the phase shifts are all pairwise 120 degrees. Call the A current 0 degrees and the B current 120 degrees. Then in vector form the A current is (200,0); the B current is (16 cos 120, 16 sin 120); and the C current is (190 cos 240, 190 sin 240). The sum of the 3 vectors is (97, -151). That's the negative of the neutral current, so its magnitude is sqrt(97²+151<sup>2[/sup) = 179A.

Alternatively, suppose the 3 line current were all in phase (perhaps very unlikely, as that would mean one of the current represents power flowing away from the load). Then the neutral current would be the simple sum of the magnitudes of the line currents, or 406A.

Lastly, suppose the B and C line currents are 180 degrees out of phase from the A line current. Then the neutral current would have magnitude |200-16-190| = 6A.

These last two cases are the two extremes--all you can say with mathematical certainty without relative phase shift information on the line currents is that the neutral current will be somewhere between 6A and 406A.

Cheers, Wayne</sup>

 

[electrofelon]

A few comments about the setup not relating to your specific question. The first transformer you should not land the 480 neutral on the transformer. I've never heard of anybody corner grounding medium voltage, I would probably just leave it ungrounded. A better choice for the first transformer would have been a Delta -> wye

 

[Elect117]

I think it is more of a thought provoking idea.

If we label the first transformer T1 and the second Transformer T2, would a imbalance or neutral current on the low side of T2 appear on the 480/277 side of T1.

I do not believe so. With Delta MV sides, as you step up and back down the neutral current would appear in the delta as phase current, and respectively line current between the two transformers. I do not believe you would have a T1 277/480V neutral current from imbalance on T2. I also do not think the neutral is tied into a winding other than to be the bonded point. The return neutral current will want to complete the circuit at the low side of the transformer. I also imagine it would depend on the type of transformer core.

Now to get actual values, see Wayne's answers above. It might also help by drawing it out with the transformer windings to make it look like a KCL/KVL system.

https://www.3phaseee.com/static/docs/wye-delta-transformer-phasing-explained.pdf

https://www.belfuse.com/resources/whitepaper/signaltransformer/img-blog-3-phase_Figure1.jpg

I guess a good follow up question, if I am on the right track, is to bond the T1 277/480V neutral to the T2 low side neutral and see what happens. I imagine you would get a coupling effect that could create a odd effect on a fault or open open neutral.

P.S. I don't see the corner ground doing anything electrically during this scenario. It would only serve as a method to limit voltage rise and provide a reference for safety.

 

[wwhitney]

If we label the first transformer T1 and the second Transformer T2, would a imbalance or neutral current on the low side of T2 appear on the 480/277 side of T1.

I do not believe so.

Depends on how transformer T1's primary wye side is wired. If it is supplied by a 3P4W wye supply, with the neutral connected to X0, then the load currents on the T1 primary will be identical (for idealized transformers) to the load currents on the T2 secondary. Of course, with this arrangement, if the 3P4W supply's neutral conductor voltage is not at the exact average voltage (as phasors) of the other 3 voltages, in addition to the load current the T1 primary will see a steady and possibly quite large current attributable to that voltage mismatch.

If on T1 X0 is left unconnected, as is more common to avoid the possibility of such a current due to voltage mismatch, then the question is how does such a wye-delta transformer with X0 unconnected handle unbalanced current on the delta? And the answer is that the sum of the delta currents must be zero, so that the sum of the primary coil currents into X0 is also zero. So there will be an additional delta circulating coil current of -1/3 of the sum (as phasors) of the delta load currents. Once you add that delta circulating coil current, the wye line currents are just equal to the delta coil currents time the turns ratio.

So with T2 delta-wye and T2 secondary = 480Y/277V, T1 primary supplied with 480V 3P3W with X0 unconnected, if you load T2 secondary with say 3A A-N, on the T1 primary you will see 2A on line A, -1A on lines B and C, with all current waveforms in phase with each other (rather than in phase with their respective L-N voltages).

Cheers, Wayne