Transformer Voltage Drop

[steve66]

Does anyone have a formula to figure the voltage drop across a transformer secondary?? I assume it would only depend on the %Z of the transformer, and the secondary rated current, and actual secondary current.

For example, if a 750KVA transformer has a 5% Z, and the open circuit secondary voltage is 480V, what would be the secondary voltage at 180 amps of load?

Thanks:
Steve

[bob]

http://www.skm.com/support/kb/120001.pdf

[Cold Fusion]

steve -
I attached a couple of pages from the ieee redbook that shows the phasor relationships described in bob's post.

cf

[Besoeker]

Originally Posted by steve66

Does anyone have a formula to figure the voltage drop across a transformer secondary?? I assume it would only depend on the %Z of the transformer, and the secondary rated current, and actual secondary current.

For example, if a 750KVA transformer has a 5% Z, and the open circuit secondary voltage is 480V, what would be the secondary voltage at 180 amps of load?

Thanks:
Steve

As the others here have indicated, there is no single answer to that.
I think the phasor diagram posted by Cold Fusion is probably as succinct an answer as you can get.
The load PF as well as the current it draws determines the voltage drop. Or increase for that matter if the load is capacitive.

[templdl]

I found this to be an good explaination of transformer regulation:
http://www.allaboutcircuits.com/vol_2/chpt_9/6.html

[jbelectric777]

i always use: 2k X length X Current divided by CM, k= 12.9 for copper (resistivity per 1000 feet) current= load/amps CM = circular mil area of conductor. aluminum is higher its in conductor properties in the codebook if you need it. in three phase remember to use 1.732 factor with K (instead of a two wire circuit 2K its KX1.732) example below:

3PH 480V 180A with 250kcmil 400 feet

kx1.732(12.9x1.732)X400X180/250000 if it were 1PH just use 2k or 12.9

hope that helped.....

[steve66]

Originally Posted by Besoeker

As the others here have indicated, there is no single answer to that.

Yes, it does look like it is more complex than I assumed.

Originally Posted by templdl

I found this to be an good explaination of transformer regulation:
http://www.allaboutcircuits.com/vol_2/chpt_9/6.html

I was assuming I could use something similar to the equation in templdl's link: Vreg = (E noload - E fullload)/E noload x 100%.

I figured there was a simple relationship between %Z and E full load.

I did find one reference that said the voltage drop was equal to the %Z. For example, a transformer with a 5% Z would have a 5% voltage drop at full load. But SKM gives a much smaller voltage drop. So that reference was apparently incorrect.

Steve

[ron]

Steve,
Using the %Z is pretty darn close to the real calculated value.

[Besoeker]

Originally Posted by steve66

I did find one reference that said the voltage drop was equal to the %Z. For example, a transformer with a 5% Z would have a 5% voltage drop at full load. But SKM gives a much smaller voltage drop. So that reference was apparently incorrect.
Steve

Yes. Voltage regulation and impedance are two different things.
I have a spreadsheet I use for estimating voltage drop*.

I took the model and loaded the transformer (Z at 5%) to its full current rating for different PF loads.
At 0.8 lag, the voltage drop is 3.9%. At unity pf it is 1.34%.

*I'm an indolent old fellow. If I do the mathematics once and put it in a form where I can make use of it again it saves me time and effort.

[winnie]

At full load current, the voltage across the transformers internal impedance is essentially what you expect from the % impedance _However_, the voltage across the internal impedance will likely be at a different phase angle than the load voltage, so you need to use vector math to determine the voltage drop actually seen by the load.

The transformer impedance has a fairly low inductive power factor. The most loads have a slightly inductive high power factor. The voltage drop seen by such a load will be a fraction of the voltage across the transformer impedance.

If the load has a poor inductive power factor, then it will see greater voltage drop; if the load has a significant capacitive power factor, it can even see its voltage _rise_.

-Jon