short calculation with solved example

Check here. This is a straight forward example.

http://ecmweb.com/mag/electric_shortcircuit_calculation_methods/?smte=wr

Assuming that the transformer secondary is 433V and the percentage impedance of the transformer is 4%, the short circuit current for a three-phase fault would be:

(1000 x 1000) / (Root 3 x 433 x 0.04) = 33,335A or 33.33kA.

Note: It is assumed that the 72m is the length of the bus dust whose impedance is neglected in the above calculation. If the bus bar impedance has to be included or if the 72 m shown is that of a cable from the transformer to the panel, then the cable impedance has to be considered and the fault current will reduce considerably. But, the above value can be taken as a safe design.

For short circuit rating you have to take into account %age impidence of the transformer, rated voltage is to be known as well.

Length of cable or bus duct provides additional limiting impidence however motors contribue and if the motors are of large size there will be considerable effect.

Without going into detailed calculations 30 KA short circuit rating at 400 volts works well with 1000 KVA and is widely practiced.

Symmetrical short circuit current formula with only the transformer under consideration is as follows:

Symmetrical Short Circuit KA = In x 100/%Zt

See this pdf and it will help you step by step.

http://www.bussmann.com/library/docs/EPR_Booklet.pdf