Protection Engineering Basics and Schemes Protection Engineering | Page 141

5.3.6 Case study # 3: We need to make protection setting of relay from below data.

% Z = 20.33 % MVA ∗ 1000

I Primary Full load = I PFL = 1.732 ∗ V HT

1000 = 65 ∗

1.732 ∗ 110 pg. 141