Fault Current

Divide 630kVA by the voltages to obtain amperes. Anything above each of those two numbers would indicate an overload fault.

On the 11kV side, available fault current depends on the system upstream, either the utility's source or the larger transformer that feeds this one, plus the effects of lengths of cable and bus and any contribution from generation or motors on-site. It is not trivial to analyze.

On the 415V, it can be approximated by calculating the Full-Load Amps of the transformer kVA / (kV * sqrt 3) = 630 / (0.415 * 1.732) = 876 amps. Divide this by the transformer's percent impedance expressed as a decimal (e.g., 5.5%Z = 0.055) to get a rough estimate of max fault current at the transformer secondary. Assuming this unit is about 5.5%Z, available secondary fault current is about 16kA. To this would also be added contribution due to motors on the circuit, which feed back into the fault while they are still spinning as generators.

Again, not a trivial analysis. If you want to calculate it all out by hand, it is possible - the procedure is documented in any textbook on power system analysis, or the IEEE Brown Book (IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis). Easier way is to get software such as the SKM Power*Tools Analysis, but that entails cost and a training/learning curve.