How to calculate 480V line currents for a single 208V load on a delta-wye

[dinos]

If a 208V load is connected to phase b and c of the transformer and it draws 2A per phase, what are the line currents produced on the 480V side of the transformer?

My "gut " feeling is 0.866A on Phase B and C (208V*2A=480V*0.866A=416VA) but if that is correct, how is that calculated via vector analysis?

 

[LarryFine]

I'm too lazy to do the math, but I think converting to kva will at least allow you to check your math, if not do the actual calculation.

Oh, you did. Sorry I can't be of more help; I'm more electrician than engineer.

 

[winnie]

Think of it as a set of 3 single phase transformers, each 480:120.

The H1-H2:X0-X2 coil is supplying 2A on the secondary side. That means 0.5A on the primary side.
Same for H2-H3:X0-X3

But now the kicker: what are the phase angles? If there are no other connected loads, and we ignore transformer magnetizing effects, then the secondary currents must be in phase and thus the primary currents must be in phase. That means 1A on the primary H2 terminal.

-Jon

 

[Beaches EE]

While Engineering is a precise science, we do at times use engineering approximations to get us in range then decide if higher precision is needed. Jon's approximation in his context makes sense.

 

[kwired]

Think of it as a set of 3 single phase transformers, each 480:120.

The H1-H2:X0-X2 coil is supplying 2A on the secondary side. That means 0.5A on the primary side.
Same for H2-H3:X0-X3

But now the kicker: what are the phase angles? If there are no other connected loads, and we ignore transformer magnetizing effects, then the secondary currents must be in phase and thus the primary currents must be in phase. That means 1A on the primary H2 terminal.

-Jon

1 amp on H2 and probably also on H3?

H2 has to have a complete circuit to something else or there won't be any current flowing at all.

 

[winnie]

I get 0.5A on H1, 0.5A on H3 and 1A on H2, with no current flowing on the H1 to H3 coil (to repeat, ignoring magnetizing current and assuming the load has unity power factor).

 

[synchro]

Think of it as a set of 3 single phase transformers, each 480:120.

The H1-H2:X0-X2 coil is supplying 2A on the secondary side. That means 0.5A on the primary side.
Same for H2-H3:X0-X3

But now the kicker: what are the phase angles? If there are no other connected loads, and we ignore transformer magnetizing effects, then the secondary currents must be in phase and thus the primary currents must be in phase. That means 1A on the primary H2 terminal.

-Jon

Very insightful and concise analysis.
Just to restate this argument with a little bit more detail to clarify it, at least in my mind:
The currents going through the X0-X2 and X0-X3 are identical because they are in series (and no current through X0-X1) and so as Jon said these currents are obviously in phase alignment with each other. Therefore the currents through the corresponding windings H1-H2 and H2-H3 are also in phase alignment with each other.

Now the polarities on delta windings are connected so that their vector voltages effectively subtract and the delta can be closed on itself. This also makes the winding currents add at their respective node connections. Normally, these currents are at some phase angle relative to each other. So, for example, there's a 0.866 scaling factor if the currents are at 60 degrees to each other. But in the OP's problem the H1-H2 and H2-H3 currents are in exact phase alignment as Jon said, and so you can simply add the two RMS currents directly and 0.5A + 0.5A = 1.0A flows through the line connected to H2.
There's 0.5A at both H1 and H3 as Jon noted because of the transformers' 480:120 ratio.

 

[dinos]

Very insightful and concise analysis.
Just to restate this argument with a little bit more detail to clarify it, at least in my mind:
The currents going through the X0-X2 and X0-X3 are identical because they are in series (and no current through X0-X1) and so as Jon said these currents are obviously in phase alignment with each other. Therefore the currents through the corresponding windings H1-H2 and H2-H3 are also in phase alignment with each other.

Now the polarities on delta windings are connected so that their vector voltages effectively subtract and the delta can be closed on itself. This also makes the winding currents add at their respective node connections. Normally, these currents are at some phase angle relative to each other. So, for example, there's a 0.866 scaling factor if the currents are at 60 degrees to each other. But in the OP's problem the H1-H2 and H2-H3 currents are in exact phase alignment as Jon said, and so you can simply add the two RMS currents directly and 0.5A + 0.5A = 1.0A flows through the line connected to H2.
There's 0.5A at both H1 and H3 as Jon noted because of the transformers' 480:120 ratio.

I've revised the sketch based on the answer but this raises a 2nd question in my mind.
If the 208V load is 416VA (208V*2A) wouldn't 416VA also have to be supplied by the 480V windings?
Are the 480V windings supplying 0.5A*480V*2=480VA?

 

[david luchini]

I've revised the sketch based on the answer but this raises a 2nd question in my mind.
If the 208V load is 416VA (208V*2A) wouldn't 416VA also have to be supplied by the 480V windings?
Are the 480V windings supplying 0.5A*480V*2=480VA?
View attachment 2552228

The voltage and the current in the 480V winding don't have the same phase angle. They are 30deg apart.

So the 480V windings are supplying 480V*0.5A*cos(30) * 2.

 

[winnie]

Also, the phase angle in the 208V windings is not the same as in the load.

The load is a 208V load, and thus 416 VA. The transformer windings are 120V each, so each coil (and the transformer heating) is loaded with 240 VA.

-Jon

 

[david luchini]

The load is a 208V load, and thus 416 VA. The transformer windings are 120V each, so each coil (and the transformer heating) is loaded with 240 VA.

Do you mean 208 VA?

 

[jaggedben]

Also, the phase angle in the 208V windings is not the same as in the load.

The load is a 208V load, and thus 416 VA. The transformer windings are 120V each, so each coil (and the transformer heating) is loaded with 240 VA.

-Jon

Are you saying that 64VA is lost to transformer heating? Just trying to follow.

 

[dinos]

Do you mean 208 VA?

I calculate 208V*2A=416VA as supplied to the load

 

[david luchini]

I calculate 208V*2A=416VA as supplied to the load

So 208VA from each connected transformer secondary winding

 

[dinos]

So 208VA from each connected transformer secondary winding

That's my take on it. I haven't thought about how power factor would complicate all of this

 

[winnie]

Each coil has 120V across it and and has 2A going through it. So each coil is _delivering_ 240VA.

Because of the phase angle between the two series connected coils, the net voltage delivered to the load is 208V rather than 240V.

So 416VA is delivered to the load, but the transformer heating is as if 480VA were flowing through the transformer. There isn't 64VA being lost as heat; the actual heating of the transformer depends on its efficiency. But the transformer is heating as though it were supplying 480VA.

Take this to an even greater extreme. Imagine you have a 480V open delta secondary, with a 1A load across the open jaw and no other loads. Each coil is delivering 480V and 1A, so the transformer heating is as if 960VA is being supplied. But only 480VA is being delivered to the load.

-Jon