Power Transformers

""The percentage impedance of a transformer is the voltage drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage."

"It is also the percentage of the normal terminal voltage at one side required to circulate full-load current under short circuit conditions on other side." The impedance of a transformer has a major effect on system fault levels. It determines the maximum value of current that will flow under fault conditions.It is easy to calculate the maximum current that a transformer can deliver under symmetrical fault conditions.

By way of example, consider a 2 MVA transformer with an impedance of 5%. The maximum fault level available on the secondary side is: 2 MVA x 100/5 = 40 MVA and from this figure, the equivalent primary and secondary fault currents can be calculated.

A transformer with a lower impedance will lead to a higher fault level (and vice versa).

The figure calculated above is a maximum. In practice, the actual fault level will be reduced by the source impedance, the impedance of cables and overhead lines between the transformer and the fault, and the fault impedance itself."

https://www.quora.com/Why-is-the-transformer-impedance-given-in-percentage

The operating parameters of a transformer must be known to determine the ability to handle the load and where to set safety device settings...

There are many power system studies that go into choosing the appropriate impedance for a transmission transformer. There are regulatory, operational, steady-state, dynamic, and transient considerations that are necessary to ensure the safe and reliable performance of the grid; the transformer's parameters must meet all of them.

In no particular order of importance: load-flow, transient/dynamic stability, voltage drop/profile, BIL, short-circuit, fault contingency, overload capacity/capability, system security, protective relaying, regulatory/grid codes, etc., etc. There are many others, it is definitely not a random choice.

In my opinion, what you defined as p.u. it has to be short-circuit rated voltage[uk% IEC standard] or Zk% [IEEE standard] .This series of values could be for medium voltage [6 to 35 kV] and high voltage 110-220 KV transformer primary voltage.

P.u. it is referred to a base apparent power usually [Sbase]10 KVA up to 100 MVA and a voltage base[VLLbase]-usually

the System [or transformer] rated voltage.

Zbase=VLLbase^2/Sbase the base impedance in ohm

Let's say you have a 110 KV 40 MVA transformer 10% zk%. Actually, the short-circuit impedance of the transformer will be:

Zxfr=110^2/40*10/100=30.25 ohm

Let's say Sbase=100 MVA and VLLbase=110 KV Zbase=110^2/100=121 ohm.

Zp.u.=30.25/121=0.25

Z (p.u.) - doesn't that stand for power utilization impedance at maximum load? I don't know, but I am here to learn.

Not quite. Per Unit (pu) is a scaling system that makes it easier for power engineers to perform calculations despite differences in the relative ratings of the equipment, especially transformers because of the transformation ratio.

A complete introduction to the per unit system is more cut and paste than the CR4 editor system can handle, but there are plenty of tutorials on the web.

Thanks for foreword to the introduction.