Protection Engineering Basics and Schemes Protection Engineering | Page 213
= 1.3 ∗ I SCH = 1.3 ∗ 2799.356
= 3639.163A This current on associated 400 / 1 CT Secondary is 9.09A
Transformer Inrush current will be 8 to 12 % of full load current
= 12 ∗ I flh = 12 ∗ 349.91 = 4199A
This current on associated 400 / 1 CT Secondary is 10.4973A; we will select high set value of differential current greater than 1.3 times of fault current( we can take fault current value of HT or LT side) and 8-12 times of Full load current at HT or LT side.
We can set our I diff ≫ = 11A
Slopes of Differential Relay:
A slope of differential relay is the ratio of differential current and restraint current.