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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// to calculate fault currentin feeder lines,primary and secondary lines of receiving end transformers
+
+clc;
+
+r=60; //kva rating of 3-ph common base
+s=200; //kva rating of 3ph transformer
+//sending end
+X_Tse=.06*r/s; //.06= reactance of transformer based on its own rating
+//in 2 kv feeder
+V_B=2000/sqrt(3); //line to neutral
+I_B=r*1000/(sqrt(3)*2000);
+Z_B=V_B/I_B;
+X_feeder=0.7/Z_B; //feeder reactance=0.7
+//receiving end
+X_Tre=0.0051;
+X_tot=X_Tse+X_feeder+X_Tre;
+V_se=20/20;
+I_fc=V_se/X_tot; //feeder current
+
+I_f=I_fc*I_B; disp(I_f,'current in 2kv feeder(A)');
+I_t1=I_f/sqrt(3); disp(I_t1,'current in 2kv winding of transformer(A)');
+I_t2=I_t1*10; disp(I_t2,'current in 200kv winding of transformer(A)');
+I_l=I_t2*sqrt(3); disp(I_l,'current at load terminals(A)'); \ No newline at end of file