Transformer Voltage Drop

[steve66]

Originally Posted by 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

That makes everything a lot clearer. Thanks.
Steve

[steve66]

Originally Posted by Besoeker

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.

With SKM software, I also get about 3.9% at 0.8 lag.

But for unity, I get about 1.1%. If I give the xformer a very high X/R ratio, that drops to almost 0%.

For very low PF loads, the voltage drop is close to the %Z. For a PF =0.5, the voltage drop is very close to 5%. For a PF = 0, the voltage drop is just a little higher at about 5.3%.

So it seems like it would be pretty safe to assume the voltage drop is normally going to be something ess than the %Z.

Steve

[mivey]

Z^2 = R^2 + X^2 for percent impedance, resistance & reactance

R = load loss / 10 / kVA

% voltage regulation = sqrt[R^2 + X^2 + 200 x (Xsin@ + Rcos@) + 10000] - 100
where @ is the power factor angle (+ for inductive) and impedance components are in %.

ex. 37.5 kVA 7200/12470Y-120/240 with no-load loss of 155, rated loss = 568, 1.86% impedance:

R = (568-155) / 10 / 37.5 = 1.1%
X = sqrt(1.86^2 - 1.1^2) = 1.5%

For 50% p.f.:
Reg = sqrt[1.1^2 + 1.5^2 + 200 x (1.5 x 0.866 + 1.1 x 0.50) + 10000] - 100 = 1.849%

For 80% p.f.:
Reg = sqrt[1.1^2 + 1.5^2 + 200 x (1.5 x 0.6 + 1.1 x 0.80) + 10000] - 100 = 1.781%

For unity p.f.:
Reg = sqrt[1.1^2 + 1.5^2 + 200 x (1.5 x 0 + 1.1 x 1.0) + 10000] - 100 = 1.111%

[bob]

% voltage regulation = sqrt[R^2 + X^2 + 200 x (Xsin@ + Rcos@) + 10000] - 100
where @ is the power factor angle (+ for inductive) and impedance components are in %.

Miley
I am not following this formula. What are the 200, 10000 and 100?

[mivey]

Originally Posted by bob

% voltage regulation = sqrt[R^2 + X^2 + 200 x (Xsin@ + Rcos@) + 10000] - 100
where @ is the power factor angle (+ for inductive) and impedance components are in %.

Miley
I am not following this formula. What are the 200, 10000 and 100?

I did not derive the formula as it came out of a transformer handbook.

[Cold Fusion]

Originally Posted by mivey

...R = load loss / 10 / kVA ...

And I need help figuring where this one comes from. Is this an assumption that the X/R ratio is 10.

cf

[mivey]

Originally Posted by Cold Fusion

And I need help figuring where this one comes from. Is this an assumption that the X/R ratio is 10.

cf

No. It is the conversion between watts, kVA and percent. The loss was in watts and the transformer size in kVA and R in %.

[Cold Fusion]

Originally Posted by mivey

No. It is the conversion between watts, kVA and percent. The loss was in watts and the transformer size in kVA and R in %.

Ahhh - thank you

cf

[Besoeker]

Originally Posted by steve66

With SKM software, I also get about 3.9% at 0.8 lag.

But for unity, I get about 1.1%.

I used an X/R ratio that is typical for many of the power transformers we use.
It's also that which use for calculating system losses.
I don't think we can fall out over a difference of about 0.2%.