Electronic – Sizing a transformer circuit breaker UK

What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -

Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -

\$V_P\cdot I_P = V_S\cdot I_S\$

The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.

This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.

The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.

I've never seen a seven-winding transformer but I have dealt with three-winding transformers.

The nameplate information for one particular three-winding transformer is as per below. (Graphic redrawn from photo to avoid showing the full extent of the very large nameplate.)

The impedances given are all on 60 MVA base. In this case tap no. 5 is the principal tapping, so consider the impedances as HV-LV 19.32%, HV-TV 33.08%, LV-TV 9.93%.

The impedance quoted for each pair of windings is with a voltage applied to the first winding, the second winding shorted, and all other windings open circuit. Refer IEC 60076.1:2005 Power Transformers - Part 1: General:

Therefore, assuming --

1. your transformer nameplate gives impedance for each individual winding
2. the testing was done as described above
3. your fault condition will be three phase fault on one winding, other windings effectively open circuit

-- your fault current depends only on the impedance from the HV winding to the particular winding under fault.

In this example, a three-phase fault on the 33 kV winding would produce a maximum three-phase fault current of 60 MVA ÷ 19.32% ÷ 33000 V ÷ √3 = 5.43 kA. A three phase fault on the 11kV winding would produce 60 MVA ÷ 33.08% ÷ 11000 V ÷ √3 = 9.51 kA.