The impedance, in per unit of the transformer, unless stated otherwise, is determined on it's own base. So if, Z = 8%, the per unit impedance is 0.08 pu. It is acceptable to display it as a percentage or as per unit (p.u.)
If the transformer base rating is say 1250 KVA and 480V, then the impedance of the transformer in ohms would be:
Zbase = (Vbase)^2/KVA
Zbase = 480^2/1250KVA
Zbase = 0.184 ohms
Zxmfr = Zpu x Zbase
Zxfmr = 0.08 x 0.184 ohms
Zxfmr = 0.01475 ohms
The maximum available short circuit current of the transformer i.e. through the transformer is:
KVAsc = KVAxfmr/Zpu
KVAsc = 1250KVA/0.08
KVAsc = 15,625 KVA
then;
Isc = KVAsc/(Vll*sqrt 3)
Isc = 15,625KVA/(480*sqrt 3)
Isc = 18.8kA
You can also just take the Irated and divide by the Zpu, it will give you the same result.
Unfortunately, without either the components of Z, i.e. X and R, or the X/R ratio, you cannot get the actual values. However, you can use a couple approximations. The first is by looking in IEEE Std 141 (Red Book) Chapter 6 that has a graph of typical X/R ratios of transformers based on IEEE C37.010.
From the graph, a typical 1250KVA transformer has an X/R of approx. 7. This is a unitless number.
With this you can determine the X and R components of Z as follows:
|Z| = (|X|^2 + |R|^2) ^0.5
substituting;
|Z| = (|7R|^2 + |R|^2) ^0.5
|Z| = (49R^2 + R^2) ^ 0.5
Solving for R in pu:
|R| = 0.011pu or approx. 1.1%
Therefore Xpu = 0.0792 pu
From observation you can see that the r component of the impedance is very small in comparison to X and in lieu of having the X/R chart, the second way of getting the X and R components is you can very closely approximate the X and R components of Z by using the fact that R is typically around 1%.
This will hold true for most preliminary calcs. Once actual equipment is bought and testing and nameplate data is available, it is always a good idea to go back and check your calcs for validation.
