Protection Engineering Basics and Schemes Protection Engineering | Page 139
5.3.5 Case study # 2:
We have 132 / 13.8KV power system, 67 MVA DYn11 Transformer, Z % is 29.48.
1000
I PFL = 67 ∗ 1.732 ∗ 132 1000
= 67 ∗
1.732 ∗ 132 I PFL = 293A
1000 I PSC = 67 ∗
1.732 ∗ 29.48 % ∗ 132 IPSC = 994A
1000 I SFL = 67 ∗
1.732 ∗ 13.8 I SFL = 2803.16A I SSC = 67 ∗ 1000 /( 1.732 ∗ 29.48 % ∗ 13.8) I SSC = 9508. 683A
CT ratio for primary is 400 / 1 and secondary is 4000 / 1, for working on over current protection scheme of 13.8KV feeder. We have below mention calcultion