Current Transformer

saw · Jun 29, 2012

Hi there, Hope you, smart engineers, do not get tired of my quetions. I am planning to pass the exam this time and I do not want to leave any doubt behind me. I am hoping that my quetions not only will help me to understand the material better, It also help the others by discussing the questions here.

Here is my new dilemma,

A three phase, delta, wye connected, 30 MVA, 33/ 11 KV transformer is protected by a differential relay. Calculate the relay current setting for faults drawing up to 200 percent of the rated current. The CT current ratio on the primary side is 500:5, and that on the secondary side is 2000:5.

The primary line current is: Ip = (30 MVA)/ (1.73 X 33 KV) = 525 A

The secondary line current is: Is = 3 Ip = 3 X 525 = 1574.6 A

The CT current on the primary is: I1 = 525 / (500/5) = 5.25 A

CT on the secondary side is: I2 = 1574.5 X (5 / 2000) X (1.73)

why do we multiply by 1.73? It can't be base on the type of the connection since the right side of transformer is Y connection which means the I phase and I line are equal and we consider the primary side as 3 phase not single phase.

Thank you

gte636i · Jun 29, 2012

In order to account for the 30 degree phase shift in a delta-wye transformer the ct's on the delta side are connected in a wye configuration and the ct's on the wye side are connected in a delta configuration.

saw · Jul 2, 2012

Thanks for the respond. I thought about the 30 degree change too. But why don't we consider the 30 degree difference in Primary side? Why it is only considered in secondary side. I am just trying to understand the concept.

slowhokie · Aug 5, 2012

If I understand right, the sqrt(3) for the secondary isn't necessarily for the 30 deg shift, but it's because the secondary CT's are in a delta configuration. The fact that the CT's are wye-delta accounts for the 30 deg shift.

This is an interesting problem, is it out of a particular book?