Estimate the short-circuit at point A for a system with the following specifications for the main transformer and utility:
P_T=1000 \ kVA, Z_T=4\%, P_{SC,U}
First, the combined utility-transformer per-unit impedance is calculated:
Z_{UT}=Z_U+Z_Twith
Z_u=\frac{P_T}{P_{SC,u}}=\frac{1000 \ kVA}{100,000 \ kVA}=0.01Thus,
Z_{UT}(pu)=4\%+1\%=0.05The application of Ohm’s law using the per-unit value provides
I_{SC}(pu)=\frac{E(pu)}{Z_{UT}(pu)}=\frac{1}{0.05}I_{SC}(pu)=20
Without any motor contribution, the short-circuit at point A can be estimated:
I_{SC}(A)=I_{SC}(pu)*I_{L,S}(A)with
I_{L,S}(A)=\frac{1000 \ kVA}{0.48 \ V\sqrt{3} }=1202 \ AThus, the short-circuit current at point A without any motor contribution is
I_{SC}=24,056 \ A