Calculation of Unbalanced Transformer Loads

[JLann]

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.

 

[jtester]

It is not tirvial to go from doing calculations on a balanced delta system to an unbalanced high leg delta. For example, the two power transformers each are loaded with 1/3 of the 120 volt power while the lighting transformer will be loaded with 2/3 of the power. So arbitrarily assigning wattages to a transformer make the problem very difficult. The currents just don't distribute arbitrarily. You can take this approach and calculate some connected load, but it won't be the 15 kva you had on a balanced system.

It is much easier for me to visualize that you have 5 kva of three phase load, and 10 kva of single phase load, and then see how it divides up in the transformers, than to say I have so much much load on each of these transformers, and it ought to be 15 kva of connected load.

Does that make sense?

Jim T

 

[charlie b]

. . . 3, 4 & 5 kVA on phase windings A, B & C respectively. . . . I originally started out with a 15kVA 3Ph load (5kVA per phase). . . .

Why 15? Why not 12?

I don't know if you can use vector addition here. For figuring out unbalanced loading conditions, I would use the analysis tool called "symmetrical components." It is the same tool used for analyzing fault conditions, and an imbalance is just a special case (i.e., far less severe) of a fault condition.

That is, I would if I could remember how to do it. It's been a while.

 

[RUWired]

If your (AB=3k) (BC=4k) (CA=5k), Then [A=8k/207=38.6amp] [B=7k/207=33.8amp] [C=9k/207=43.47amp] Using E=120 volt I=120x1.73.
Rick

 

[charlie b]

When you are looking at the relationship between current and voltage in a single phase of a three-phase transformer, the 1.732 factor does not come into play. It is only when looking at the combined effect of the three currents in the transformer windings, as they join to create the line currents seen at the secondary, that the 1.732 factor applies.

 

[jtester]

Charlie is correct. While nodal analysis won't fail you, symmetrical components require that you use 3 balanced systems connected in various ways to model unbalanced 3 phase systems. There you can do either loop or node equations.

Loop and node equations are very misleading if you try to do them one time to model a completely unbalanced system.

Jim T

 

[rattus]

Yes, one can use vector (phasor) addition here. The factor 1.732 does not apply because the currents are not equal.

One must convert the polar quantities into rectangular form, add them, and then convert back to polar form. It is a bit tricky because one must conisder the "direction" of the currents. One must subract one current from another to obtain the correct line currents. I will do this after the Dog (notice the caps) takes me for a walk.

OK, the Dog wasn't any help, so I will just wing it.

I calculate:

Ia = 51A /35
Ib = 65A /-86
Ic = 54.1A /151

Someone double check me on this. The Dog sometimes has trouble with the small keys on the calculator.

 

[JLann]

Thanks for the replies.

 

[steve66]

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.

I think that is exactly the method you should use. I get 58 amps for IA, but I'm not 100% sure that is right. (YOU DID SAY 120V - THAT'S 120V LINE-LINE, OR VAB = 120V. A normal delta would be 240V line-line).

You could also do a delta to wye conversion on this. For example, Za=(Zab*Zac)/ (Zab+Zbc+Zca) and Zb = (Zab*Zbc)/(Zab+Zbc+Zca). Then use 120V/sqrt(3) for the line to neutral voltage.

Steve

 

[steve66]

Rattus:

I assume your 51/35 means 51 amps at 35 degrees? I can't seem to verify the source power equals the 12KW disipated on the load. I've tried to do this with both your numbers and my numbers and neither seem correct

For example, the line to neutral voltage is 120V/sqrt(3) = 69.28V. For your numbers, I think 69.28*(51+65+54) should equal 12000. Right? But I get 11787.

With the currents I got, 69.28*(57.5+45.5+77.7) = 12532.

Your's seem too low, and mine seem too high?

Steve

 

[charlie b]

For your numbers, I think 69.28*(51+65+54) should equal 12000. Right?

Close enough. The number "51" has two significant digits. Therefore, the results of your calculation must be expressed to no more than two significant digits. To achieve this, you must take the number 11787 and round up to 12000. Similarly, your numbers are expressed to no more than three significant digits, so your 12532 must be rounded down to 12500.

Please note that "his numbers" give a result that is less than 2% away from 12000, and that "your numbers" give a result that is about 4% away.

The difference is probably in the degree of precision used in the intermediate steps of your respective calculations, coupled with "round off error."

 

[JLann]

Steve -
The output of the transformer VAB, VBC, VCA is 120V. The transformer "bank" is comprised of (3) single phase 450/120V transformers connected via phase sequence ABC in a delta configuration. There are normally no center-tapped windings in this configuration.

 

[steve66]

OK, with the help of Excel (my trig must be really rusty) I basically came up with this:

IA = 50.69
IB = 65.09
IC = 58.33

So I think Rattus only had one of them off by 4 amps. That gives 12,062 watts, which is pretty close. But I still not sure where the extra 62 watts came from. (I used Excel this time Charlie, so there shouldn't be any round off errors??)

Still not sure why the delta to wye conversion didn't work right, or why I couldn't get the same numbers with my calculator. I wish I had more time to look at this to see what I did wrong.

STeve

 

[rattus]

Steve,

Yes, the "/" indicates the angle. Now to calculate IA.

IA = Ia -Ib = 25A/0 - 33.3A/-120

= 25A/0 +33.3A/60

= 25A +33.3A*cos(60) + j0 + 33.3A*jsin(60)

=25A + 16.65A + j28.85A

= 41.65A +j28.85A

= 50.7A/34.7

This is an example of the use of phasors. In this case both the polar and rectangular forms are used.

 

[JLann]

Just curious as to how -120deg turned to 60deg and how is IA = Ia-Ib if phase sequence is ABC.

 

[bob]

Steve,
Yes, the "/" indicates the angle. Now to calculate IA.
IA = Ia -Ib = 25A/0 - 33.3A/-120
= 25A/0 +33.3A/60
= 25A +33.3A*cos(60) + j0 + 33.3A*jsin(60)
=25A + 16.65A + j28.85A
= 41.65A +j28.85A
= 50.7A/34.7QUOTE]
Rattus
Should not jsin(-120)"you used 60" be -j28.85

 

[jtester]

If a node equation says the sum of all currents at a node =0, then
IA+Ia+Ib=0 or IA=-Ia-Ib
Rattus how did you get IA=Ia-Ib?

Jim T

 

[rattus]

Yep, IC = 58A/-141. Since I received someone else's corneas, I need some new reading glasses!

Power is simply the summation ov Vphase*Iphase.

Ptotal = (25 + 33.3 + 41.7)Ax120V = 12,000 watts

Or one could do it with line currents.

Ptotal = (51 + 65 + 58)A*120V/1.73 = 12,000 watts

Here, we pretend the load is connected in a wye.

 

[charlie b]

If a node equation says the sum of all currents at a node =0, then . . . .

Not true. It is not the sum of all currents AT a node that equals zero, but rather the sum of all currents ENTERING (or if you prefer, LEAVING) a node that equals zero. Before you get started, you have to define the positive direction of all currents that pass through that node. I suspect that in Rattus' model, the currents "Ia" and "Ib" are defined as positive entering the node, and "IA" is defined as positive leaving the node. So the node equation would be,
-IA +Ia +Ib = 0.

 

[steve66]

If a node equation says the sum of all currents at a node =0, then
IA+Ia+Ib=0 or IA=-Ia-Ib
Rattus how did you get IA=Ia-Ib?

Jim T

You draw all the reference currents in the delta as flowing clockwise. That gives opposite signs for Ia and Ib.

Actually, I have Ia and Ic connected to node A. You almost have to draw this out with everything labeled to make sure everyone is on the same page.

-120 degrees turned into 60 degrees because Rattus changed the sign of the magnitude. A@0 degrees is -A@180 degrees. So A@-120 degrees is -A@60 degrees.

Steve