Determining Transformer Resistance & Reactance

[GreyGoose]

Given the following test data:
Transformer base rating (in VA)
Transformer no-load (or iron) losses (in W)
Transformer load (or copper) losses (in W)
Transformer percent impedance (%Z)

How do you determine %X & %R values?

some confusion on my part...is it:

a) load losses/rating x 100 = %R

or

b) (no-load losses + load losses)/rating x 100 = %R

or some other way entirely?

The rest is then easy given %Z = SQRT(%X^2 + %R^2)

 

[electricuwe]

Version a) of your formulas is closest to the the correct one:

SQRT(load losses/rating)=%R

 

[JBD]

It is not possible to correctly identify %R and %X from the data you have provided.

However using Conductor (also called copper or load) Losses only will approximate the R of larger (>500KVA) transformers.

One formula I've seen is W=[(Conductor losses in KW)F^2]/(Rated KVA) x F. Where F = Load Factor (Fraction of Rated KVA used). Then use Ohm's Law to solve for R.

 

[stevenal]

a

 

[electricpete]

It seems like a) can be shown to be correct.

assume the losses occur at rated (base) current.
R=[I^2*R]/Ib^2=[Ib^2*R]/(Sb^2/Vb^2)

Zb=Vb^2/Sb

Rpu=R/Zb =[I^2*R]/[(Sb^2/Vb^2)*(Vb^2/Sb)
Rpu = [I^2*R]/Sb

 

[230842]

In my oppinion, confusion arise from "load loses" definition. If it sould correspond to short circuit loses (at rated current) also called "copper losses", then a) is right. Otherwise if it sould correspond to total load losses at rated voltage and current then b) should be:
%R = 100x(Total load loses - no-load losses)/rated power
I think a) is the correct option. When internal impedances are given in per unit or per cent, bases values must correspond to rated values. Then:
Zb = (Ub^2) / Sb = Sb / (Ib^2) Or Sb = Zb*Ib^2 = rated power

puR = R / Zb = (R*Ib^2) / (Zb*Ib^2) = (copper losses) / (rated power)

That all is only applicable to one phase transformers. For three phase trasformers that shall be applicable to their "equivalent one phase transformer". The easier way to do that is:

1.- Take as Base Voltage the rated voltage (line to line voltage) divided by square root of 3
2.- Take as Base Power the rated power divided by 3
3.- No modification for Base Current or Impedance.
4.- Consider as copper loses its third only

Hope to help you Julian