How to compute line currents on unbalanced transformer

[Do You Like Ham]

Hello-- I stumbled across this document and have absorbed the concepts. I'd like to extend my knowledge and understand how to apply the theory to three phase transformers.

Considering the following, what are the primary currents Ia, Ib, and Ic in each of the scenarios:
1) 100A only load energized (across AB on secondary)
2) 50A only load energized (across CN on secondary)
3) both loads energized




Thanks much for any tips/guidance. I'm looking to generalize this into a spreadsheet.

 

[winnie]

The paper that you linked already gives a good introduction to using vector math to calculate currents.

Some pointers:
In a Delta:Wye transformer, remember that secondary phase A-N is electrically in phase with primary phase A-B, and so on. (As a first approximation, a single phase coil has the same phase on both primary and secondary.)
The paper already describes using vector math for calculating currents. In a spreadsheet you can represent each vector as a pair of cells, or as a complex number

-Jon

 

[Do You Like Ham]

thx jon.

on the secondary where there is 100A, am i right that there would be 25A On the primary side of the windings? (1:4 ratio). from there, it’s not clear how i’d determine each of the line currents. is it a matter of the 25A current around each of the 2 secondary loops, out of phase with each other by 120°, and in opposite directions, and doing a nodal analysis sum at each junction?

 

[synchro]

You are correct that 100A at the secondary would be 25A on the corresponding primary winding because of the 1:4 turns ratio.

For scenario 1) and 2), vector calculations are not needed because all the currents are either in-phase or 180° from each other, depending on the polarity of the transformer connections. If the three transformers are ideal, then the 1:4 current transformation applies regardless of the voltage or phase of the line inputs.
Therefore:
In scenario 1), Ia = 25A, Ic = 25A, and Ib = 50A because the polarity of top two transformers add their primary currents on line input B.
In scenario 2.), Ia = Ic = 50A/4 = 12.5A, and Ib = 0. Without any load on the top two transformers they will act like they are not even there (if ideal without magnetization currents, coupling between them etc.)

Now scenario 3) does involve vector considerations that depend on the relative phase of the 100A and 50A load currents. The bottom transformer adds its input current vectorially to the currents Ia and Ic from scenario 1). If the loads across AB and CN on the secondary are both resistive (or at least the same PF) then the current on line inputs A and C from the bottom transformer should be 90° from the current on them from scenario 1). That's because the line-line voltage across outputs A and B (and therefore load current) is shifted 30° from the L-N voltage and 120°- 30° = 90°.
Then for scenario 3) with 90° current vectors we have Ia = Ic = √(12.5² + 25²) ≈ 28A, and Ib = 50A as in scenario 1).

 

[winnie]

As synchro notes, your 100A load is connected 'line to line' on your secondary side. This means that current through coils A and B are in phase (or 180 degrees out of phase depending on how you reference things), not 120 degrees out of phase. All current in a simple loop must be in phase everywhere.

The 50A load is connected line-neutral.

Since we don't know the power factors of the loads, we don't actually know the phase angle of the currents in the loads. But if you assume resistive loads then the current in the loads will be in phase with the voltage across the loads. If you further make the approximation of balanced supply voltages and ignoring transformer impedance, then the 50A load is 90 degrees out of phase with the 100A load.

Draw out the vectors for the applied voltages, calculated the vectors for the resulting currents, reflect these currents back through the transformer coils, and see where this gets you.

-Jon

 

[Do You Like Ham]

Great.. thanks, both.. your input is much appreciated.

Starting resistive loads as a basis simplifies, and was my intent.. sorry for not being clear.

I'll need to spend some time to digest this, and start taking a stab at seeing if I can work out loads w/ non-unity power factor... but this certainly clears up the notion that I've seen on multiple spreadsheets for panel schedules over the years that simply take the power (VA) from the secondary, and apply that to the primary, then divide by the primary voltage to get primary current. I think this really only makes sense if all loads are resistive and balanced.

 

[winnie]

multiple spreadsheets for panel schedules over the years that simply take the power (VA) from the secondary, and apply that to the primary, then divide by the primary voltage to get primary current. I think this really only makes sense if all loads are resistive and balanced.

The approach is also correct if all loads have the same power factor, and is probably 'good enough'. However you are correct that it is not exactly correct.

-Jon

 

[Do You Like Ham]

OK.. let's see if I understand this right.

Since the PF for both loads is 1.0, on the secondary, the currents are in sync w/ the voltages. In the general case, the currents for 100% resistive loads would be as represented by the vector diagram on the right.

These currents are reflected through the coils, thus:

• Ibc on the primary is at the same angle as Iab on the secondary
• Iac on the primary is at the same angle as Icn on the secondary

The resulting current on primary line C is 27.95 ∠ 26.57°
That logic seems to fall apart a bit when I consider the current on A or B.. so I must be missing something.

 

[winnie]

I've not checked in detail, but you are using the correct approach.

Remember that the current through the Ian has the inverse sense of Ibn in the circuit as shown; if current is flowing out of secondary terminal A it must be flowing into secondary terminal B, reverse half a cycle later.

-Jon

 

[Do You Like Ham]

@winnie - i don’t follow... there is no current Ian or Ibn, is there? Are you referring to the notion that the current on secondary from A to B would mean that current on primary is A to B, and C to B?

i’m lost. i don’t see how i determine the vector angles I’d sum to determine primary current a, b, c... e.g., current on AC + AB and AC + BC.

I think I have some of this messed up.. e.g., if current is from C->N on secondary, then it seems the current on primary should be C->A on primary, but I drew the vector as Iac.

 

[winnie]

I think you have the logic and math correct, but feel lost because the answer doesn't fit your intuition.

On the secondary side there is no current external to the transformer flowing from a to n or b to n. But there certainly is current on the n to a and n to b coils. You've drawn this correctly as 100A for each.

This 100A current is in phase with Vab. Important to recognize: even though there is a resistive load, this current is _not_ in phase with Vna or Vnb. Even though there is a resistive load connected, the transformer coils see a power factor; one coil sees a 30 degree leading power factor, the other a 30 degree lagging power factor.

The 25A flowing from A to B on the primary side is thus perfectly in phase with the 25A flowing from C to B, and they sum to 50A just as you show.

Try drawing out the voltage and current vectors for each connected coil to see if this helps your intuition.

-Jon

 

[Do You Like Ham]

Thanks, again, Jon (@winnie)

" feel lost because the answer doesn't fit your intuition. " - bingo!

Let's see if I have this...

1. I've updated the transformer diagram with arrows showing the various currents, using the diagram on the top/right, again, from this document on page 39.
2. From that, on the far left, show the current vector analysis at each line.
3. The "Secondary Voltages" diagram reflect the diagram on page 39, color coded.
4. From that, determined the current vectors in "Secondary Currents" diagram... since the loads are resistive (1.0 PF), they are in sync w/ the voltages.
5. Reflected those to create the "Primary Currents" vector diagram.
6. Using that, along w/ the vector analysis (from #2 above), determined resulting values for each line current.
7. Graphed "Primary Currents and Voltages"

 

[winnie]

I've not checked the signs, but I believe you have it correct.

-Jon

 

[Julius Right]

After reading the last winnie's post I also checked the phase diagrams.
Sorry. The numbers are correct but the angles are not.
The voltage Va is 30 degrees after the voltage Vab- is not leading- and from here the angle of the current Ia and Ic are changed between they.

 

[Do You Like Ham]

Thanks for taking another look, @Julius Right !

I interpret what you're saying as the magnitudes of the currents are correct, but the Ia and Ic angles are swapped.

I think the error was introduced from my interpretation of Fig 30... and to be honest, I am a bit uncertain there... and not sure I completely understand. Note that in Fig 30 (top/right corner) the line-to-neutral voltages (and currents) are rotated 180 degrees from other sources (such as this)... but they're also labeled, e.g., Vda (instead of Vad... where d=neutral).

After adjusting in the 'Secondary Voltages' diagram (I've highlighted the change in the left hand column, and adjusted in the right hand column), I've highlighted, adjusted through the other steps... resulting in the Ia and Ic swapping as you suggest.