Calculation of Unbalanced Transformer Loads 1

[7JLAman4]

I'm attempting to calculate the Line currents (with no sucess) of a 120V Delta transformer secondary with 3,4 & 5 kVA on phase windings A, B & C respectively with a phase sequence ABC. I originally started out with a 15kVA 3Ph load (5kVA per phase) and knowing how to calculate for a balanced system, tried to achieve the currents using nodal analysis which I was then going to apply to the problem above. I can't seem to calculate the line currents from the phase currents using vectors. I had started out with:
Ia=3kVA/120V=25A@0deg, Ib=4kVA/120V=33.3A@-120deg, Ic=5kVA/120V=41.7A@120deg. IA=Ia-Ic, IB=Ib-Ia, IC=Ic-Ib. IA=(25cos(0)+i25sin(0)) - (41.7cos(120)+i41.7sin(120))
I'm not sure if this is the correct method. However using a sample that I know what the outcome should be, Ican't seem to get near close enough to the answer.

 

[waross]

A single phase load current will divide equally between the two current paths in a delta transformer bank.
respectfully

 

[jghrist]

stevenal,
Well, I've slept on it. Forget what I said about solving the equations. You meant to force the sum of the winding currents to zero, not the line currents. Problem with the original PO terminology. Actually, I tried this and get:

Iaw = (2·IA + IC)/3
Ibw = -(IA + 2·IC)/3
Icw = (IC - IA)/3

where IA, IB, IC are the line currents.

Using this with the line currents calculated in my resistor example, I get:

Iaw = 29.27 < -4.72°
Ibw = 33.68 < -111.79°
Icw = 37.58 < 116.33°

As far as sequential component analysis goes, I think it would apply as you say theoretically. The tools available are inadequate for an arbitrary unbalanced phase-to-phase connected load, however. I don't think Aspen could model this. There are some unbalanced load sequence diagrams in the Westinghouse T&D Manual, but they do not allow all three phases to be different.

 

[stevenal]

jghrist,
No special tools are needed like the sequence connection diagrams. You just proceed this way: Do a delta wye conversion to find the wye equivalent of your delta load to find the line currents. Find the positive and negative sequence components of the line currents. To move the sequence components into the delta windings, divide each component by sqrt(3) and shift it by 30 degrees. The positive sequence will shift in the normal direction while the negative sequence shifts in the opposite direction. Then reassemble the winding currents from the sequence components. I was required to do this so many times in college, that I can recite the procedure years later on a Saturday morning before coffee. For some reason I lack the desire to jump in and actually do the math. The same assumption applies, no zero sequence current circulating in the delta windings.

 

[stevenal]

I need to correct my previous postings. I wanted to use a delta wye conversion to make finding the line currents easier. This fails to work as intended, since the unbalance causes the wye point to shift away from neutral. Once I gave up on this plan and used node equations at the load end to find the line currents, shifting the sequence components as stated above yielded jghrist's results from his last post.

 

[stevenal]

Update,

Jghrist and I took it offline for a bit. We agree that the OP's starting point of winding kVA is suspect. A more likely starting point would be the line to line load connected kVA. If this was actually the starting point, my trig method of June 8 and 9 won't work.

Once the sequence component method to work jghrist's example is in a worksheet, it's easy enough to alter the problem to to solve for a single line to line connected load. Just raise two of the input resistances to a very high value, and the solution falls out: the in phase current splits evenly between the two paths. If currents are forced to be 120 degrees out in each winding, one transformer will carry the load as if the others were not present.

 

[waross]

Hi stevenal
I don't quite understand this statment.
"If currents are forced to be 120 degrees out in each winding, one transformer will carry the load as if the others were not present."
It is my understanding that the current in an open delta transformer connection is not forced 120 degrees out in each winding.
The voltages are 120 degrees out, but the current is in phase with the resultant voltage and that voltage is the vector sum of the 120 degree displaced winding voltages.
The proof is to solve the phase angle and voltage drops and resulting terminal voltage of a loaded open delta transformer bank and compare the result with a single phase transformer in place of the missing phase.
Rather than combining impedances, it is more understandable to compare voltage drops.
Just as the vector sum of the voltages in an open delta are equal to the original voltage but at a different phase angle, the voltage drops due to resistance, reactance and impedance, if added vectorily will result in equal voltage drops at different phase angles. The resultant phase angles will be found to be the same as the corresponding phase angles in the single phase transformer.
The open delta transformer will be found to be equivalent to the single transformer and if connected in parallel, (A closed delta bank) the current will divide equally.
The joker is that although the open delta is equivalent to the single transformer for the purposes of paralleling and/or current sharing the losses are double. This is one of the factors that limits this connection to 2/3 of the three phase KVA rating when a three phase generator is used on single phase.
respectfully

 

[stevenal]

Waross,

You need to read back to the beginning of the thread. A false assumption was made that the delta winding currents feeding an unbalanced load were unequal yet 120 degrees apart in phase. I was accepting that assumption briefly in order to disprove it. There is no magical force that keeps them at 120 degrees. I agree with you that current divides equally with the single line to line connected load.

 

[stevenal]

For you sequence component fans, here's how it's done:

http://www.webfilehost.com/?mode=viewupload&id=3905865

The zip file contains an mcd and also an rtf for those without Mathcad. Jghrist's delta connected resistances are used, but any value of impedance may be substituted. I've deviated somewhat from the convention of the thread, and used both nodes to describe phase to phase quantities. I avoided the use matrices.

 

[stevenal]

Guess I should have tested the link. This one seems to work:

http://www.savefile.com/files.php?fid=4564358

 

[waross]

Thanks stevenal
respectfully

 

[BlownUp]

I found your post by looking for information on or a tool to do something very similar (mostly because I hate vector math). I was unable to find anything and wound up putting together a spreadsheet to do the calculations for what I needed. Here is that spreadsheet:

http://www.savefile.com/files/6776522

Maybe it will do what you need. If not, you should be able to modify it.

 

[7JLAman4]

I have also created a "panel transformer" worksheet

http://www.savefile.com/files/4426260

that will calculate the line currents for each of the four types of transformer configurations (D-D, D-Y, Y-Y, Y-D) based on individual loads (Hp and/or KW single and three phase). I have received so much information from this post and emails directly that I had to create this. The only limitation this worksheet has at the moment is not incorporating any centertaps from any windings.