Primary current calc question for unbalanced Delta / Y

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Hi,
If I have a 480V delta to 120/208V Y transformer, and the loads on the secondary are unbalanced, will the currents on the primary side also be unbalanced?

Please excuse me if I'm not following the proper forum protocol. I'm new to the online forum scene.

Thanks in advance.
 

jim dungar

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Hi,
If I have a 480V delta to 120/208V Y transformer, and the loads on the secondary are unbalanced, will the currents on the primary side also be unbalanced?

Please excuse me if I'm not following the proper forum protocol. I'm new to the online forum scene.

Welcome. and you are doing fine.

Yes, the individual primary side line currents will also be unbalanced.
Do you want to see the math, or are you just curious?:cool:
 

jghrist

Senior Member
Yes, the individual primary side line currents will also be unbalanced.
Do you want to see the math, or are you just curious?:cool:
Agreed, but the unbalance will be less in general.

Example-
Transformer: 480-208Y/120V
Secondary amps: Ia = 100A, Ib = 100A, Ic = 50A (all same pf)
Primary amps: IA = 43.3A, IB = 33.1A, IC = 33.1A
 

mull982

Senior Member
Agreed, but the unbalance will be less in general.

Example-
Transformer: 480-208Y/120V
Secondary amps: Ia = 100A, Ib = 100A, Ic = 50A (all same pf)
Primary amps: IA = 43.3A, IB = 33.1A, IC = 33.1A

Can you give a quick explanation of these results?

Does it have to do with the fact that current in wye configuration is the same as the line current where as current in a delta configuration is 1.73 times less the line current?
 
More math

More math

Thanks,
I guess I am interested in the math.

Looking at a delta / Wye transformer connection schematic, the results above (by JGHRIST) seem sensible, but I haven't seen equations in print for calculating individual primary side line currents based on unbalanced loads.

Where the loads have been balanced, I've always just added up the KVA on the load side, then I used that KVA at the primary voltage to calculate the primary current.

Also, check out this article:
http://en.wikipedia.org/wiki/Three-phase_electric_power

Under the section "Single-Phase Loads", in the 4th paragraph.
It seems to indicate that primary side somehow becomes automatically balanced by the transformer. I'm not sure what they're saying in the article applies for 3-phase Y secondary transformers, and I'm not sure what they are saying is even correct.

Dave
 

wirenut1980

Senior Member
Location
Plainfield, IN
I'm curious why all three phases of your transformer have different turns ratios...?

Not different turns ratios, vector addition. What helps me solve problems like this is to draw out the phasor diagram of the delta primary and wye secondary and draw in the amps on the secondary. Using the above example, through turns ratio, you get primary winding currents of 25 Amps angle, 25Amps, and 12.5 Amps. Then those are added vectoraly to obtain the primary line current.

By the way, I get 21.7 Amps instead of 33.1.

-12.5 @ 60 degrees + 25 @ 120 degrees = 21.7 Amps
-12.5 @ 60 degrees + 25 @ 0 degrees = 21.7 Amps
25 @ 0 degrees - 25 @ 120 degrees = 43.3 Amps.

Depending on how you show current flow direction, angles might be different. I had currents flow out of the center of the wye and kept the same direction in the delta windings.

Now I might be confused on my vector addition...:-?
 

drbond24

Senior Member
What the heck are you guys doing? :-?

480/208 = 2.308:1 turns ratio

100, 100, 50 on primary
43.3, 43.3, 21.7 on secondary

A three phase transformer is just three single phase transformers wired together. The turns ratio determines what the magnitude of the currents will be in relation to each other. If you don't get the same answer with vectors, there is a problem because there is only one right answer.

This is getting a little drawn out. For simplicity's sake, I will also answer the OP's question so he doesn't get confused with our nonsense. See below for the answer:

YES
 

jghrist

Senior Member
Not different turns ratios, vector addition. What helps me solve problems like this is to draw out the phasor diagram of the delta primary and wye secondary and draw in the amps on the secondary. Using the above example, through turns ratio, you get primary winding currents of 25 Amps angle, 25Amps, and 12.5 Amps. Then those are added vectoraly to obtain the primary line current.

By the way, I get 21.7 Amps instead of 33.1.

-12.5 @ 60 degrees + 25 @ 120 degrees = 21.7 Amps
-12.5 @ 60 degrees + 25 @ 0 degrees = 21.7 Amps
25 @ 0 degrees - 25 @ 120 degrees = 43.3 Amps.

Depending on how you show current flow direction, angles might be different. I had currents flow out of the center of the wye and kept the same direction in the delta windings.

Now I might be confused on my vector addition...:-?

Currents in 120V windings
Ia = 100 @ 0?, Ib = 100 @ -120?, Ic = 50 @ 120?

Turns ratio = 4:1 (voltage ratio = 4/1.732)

Currents in 480V windings:

IAC = Ia/4 = 25 @ 0? = 25 + j0
IBA = Ib/4 = 25 @ -120? = -12.5 - j21.65
ICB = Ic/4 = 12.5 @ 120? = -6.25 + j10.825

Current in 480V lines:

IA = IBA-IAC = -37.5 - j21.65 = 43.3 @ -150?
IB = IBA - ICB = -6.25 - j32.475 = 33.07 @ -100?
IC = ICB - IAC = -31.25 + j10.825 = 33.07 @ 160.9?
 

mull982

Senior Member
Ahhh. That clears it up for me too... Do your results show the expected 30deg phase shift that occurs on this type of transformer?
 

jghrist

Senior Member
Ahhh. That clears it up for me too... Do your results show the expected 30deg phase shift that occurs on this type of transformer?
It would if the currents were balanced. Unbalanced currents throw off the primary phase angles a lot.
 

drbond24

Senior Member
Currents in 120V windings
Ia = 100 @ 0?, Ib = 100 @ -120?, Ic = 50 @ 120?

Turns ratio = 4:1 (voltage ratio = 4/1.732)

Currents in 480V windings:

IAC = Ia/4 = 25 @ 0? = 25 + j0
IBA = Ib/4 = 25 @ -120? = -12.5 - j21.65
ICB = Ic/4 = 12.5 @ 120? = -6.25 + j10.825

Current in 480V lines:

IA = IBA-IAC = -37.5 - j21.65 = 43.3 @ -150?
IB = IBA - ICB = -6.25 - j32.475 = 33.07 @ -100?
IC = ICB - IAC = -31.25 + j10.825 = 33.07 @ 160.9?

Ok, I see what you're doing now. I was doing it for a Y-Y effectively because I was saying turns ratio = voltage ratio.
 
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