IEEE CT Class

[burntcoil]

I use to do CT sizing based on the available short circuit and see if the voltage developed on the secondary is less then my knee point voltage.

I have come accross instances where the above criteria meets but my CT does not meet the IEEE definition of "C" class CT. Please see below a practicle example.

CT Ratio 3000/5, Vk = 120V, IEEE Class C50

Fault Current = 35kA, Voltage developed on secondary for this fault current is 83.85V, which is much lower then the Knee Point voltage of the CT, so the CT is fine as per my understanding.

As per deifination C50 means for 20times rated current and 5A secondary my CT is suitable if my burden is 0.5ohm. i.e. 20 x 5 x 0.5 = 50

But in actual case my burden is 1.44Ohm, 35000 (11.66 of rated 3000A) and a secondary of 5A. So using above deifination of Class "C" CTs, I can determine the required IEEE Class i.e. 1.44 x 11.66 x 5 =84, so now I shall choose a CT of Class C100.

My question is that if my CT meets the Knee Point criteria, even then shall I look at the relaying class of the CT or that is not important.

OR

I have to meet both the Knee point criteria as well as Relaying Class as well.

Thanks in advance for your responses.

 

[DTR2011]

Stan Zocholl from SEL has many good papers on this topic..

http://www.selinc.com/search/SearchPage.aspx?searchtext=stanley

There are several papers available which discuss these topics.

One significant part of one of the papers(CURRENT TRANSFORMER CONCEPTS, pg 9), states

"IEEE/ANSI Standard C57.13 suggests that cts for relaying be applied on the basis that the maximum symmetrical fault current not exceed 20 times the ct current rating and that its burden
voltage not exceed the accuracy class voltage of the ct.

Applying cts for relaying is an art rather than a science because the engineer is left to choose the specific operating point on the excitation curve. However, there is a rationale for choosing a ct to produce the knee-point excitation at the maximum symmetrical fault current since the magnetizing reactance is at a maximum.

Observe that the knee-point of a typical excitation curve is about 46% of excitation voltage corresponding to 10 amperes excitation current. A popular rule-of-thumb suggests that the C-rating be twice the excitation voltage developed by the maximum fault current. By good planning or dumb luck, the rule-of-thumb guarantees operation near the knee-point of the excitation curve for the maximum symmetrical fault."

 

[LionelHutz]

It's possible the CT uses a standard core and is a C50 just because it can't quite reach the C100 level.

 

[randyman]

Relay class is just a definition - not an application. It is not uncommon to see what you describe in small window-type CTs with very high secondary winding impedance. In this case it could be well over an ohm and therefore require 150V or more to meet class (by definition).

It is not uncommon to do what you are doing. The only question I would have here is did you include the CT winding impedance in with your burden calculation??