It has already been discussed that 3-phase short-circuit faults result in symmetrical fault currents i.e.
fault currents in the three phases are equal in magnitude but displaced 120o electrical from one an-other. Therefore, problems involving such faults can be solved by considering one phase only as the same conditions prevail in the other two phases. The procedure for the solution of such faults in-volves the following steps :
(i) Draw a single line diagram of the complete network indicating the rating, voltage and per-
centage reactance of each element of the network.
(ii) Choose a numerically convenient value of base kVA and convert all percentage reactances to this base value.
(iii) Corresponding to the single line diagram of the network, draw the reactance diagram show- ing one phase of the system and the neutral. Indicate the % reactances on the base kVA in the reactance diagram. The transformer in the system should be represented by a reactance in series.
(iv) Find the total % reactance of the network upto the point of fault. Let it be X%.
(v) Find the full-load current corresponding to the selected base kVA and the normal system
voltage at the fault point. Let it be I.
(vi) Then various short-circuit calculations are :
Example 17.1. Fig. 17.8 (i) shows the single line diagram of a 3-phase system. The percentage
reactance of each alternator is based on its own capacity. Find the short-circuit current that will flow into a complete 3-phase short-circuit at F.
reactance of each alternator is based on its own capacity. Find the short-circuit current that will flow into a complete 3-phase short-circuit at F.
