Transformer Voltage Drop

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steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
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
 
Besoeker

Besoeker

Senior Member
Location
UK
steve66 said:
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
Click to expand...
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.
 
J

jbelectric777

Senior Member
Location
NJ/PA
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

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Besoeker said:
As the others here have indicated, there is no single answer to that.
Click to expand...

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

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

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
 
Besoeker

Besoeker

Senior Member
Location
UK
steve66 said:
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
Click to expand...
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.
 
W

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
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
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
winnie said:
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
Click to expand...

That makes everything a lot clearer. Thanks.
Steve
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Besoeker said:
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.
Click to expand...

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

mivey

Senior Member
Formulas

Formulas

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

bob

Senior Member
Location
Alabama
% 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

mivey

Senior Member
bob said:
% 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?
Click to expand...
I did not derive the formula as it came out of a transformer handbook.
 
Besoeker

Besoeker

Senior Member
Location
UK
steve66 said:
With SKM software, I also get about 3.9% at 0.8 lag.

But for unity, I get about 1.1%.
Click to expand...
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%.
:)
 
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