Protection CT - Choosing a CT based on your Max Fault Current

[CorenJ]

Good Afternoon,

I need help in determining whether the CT we have meets the fault requirements.
The CT is a 1000:5 5P20 30VA Protection CT - I have a saturation curve with a knee point at 140V. Rct (Secondary Winding res.) = 0.245 @75C
The max fault I need to design for is 32 kA.
I understand that the correct CT will need to operate at 32kA without saturating i.e. before the knee point.

How can I determine this ?
I appreciate all the help.
Thanks Again
Coren

 

[scottf]

To figure this out, you also need to know the burden of the relay and round-trip leads (Rb).

Vsec = 32 x 5A x (Rct + Rb) < 140V

Works out to being ok if Rb < 0.63 ohms.

 

[CorenJ]

Thank you very much for the info Scott - just a quick question for you:

If the knee point is measured @140V with I=0.124A giving the measured CT burden at 17.36VA.
As the CT is rated at a 30VA burden - is it safe to estimate that the CT can handle a current 34.56 times the Primary (20 * 30/17.36 = 34.56) =~34kA prior to saturation ?

Unfortunately at this time I dont yet know Rb - would the above derive a good estimate/indication ?

Thanks Again
Coren

 

[scottf]

It all boils down to what you call "saturation".

The rating of 5P20 - 30 VA means that the error won't exceed 5% at 20 time over-current when loaded at 30 VA.

20 x 5A = 100A
5% of 100A is 5A, which means excitation current can't exceed 5A up to a secondary voltage of (30/(5^2)) * 100A = 120 V

The CT was measured with an Iexc=0.124A at 140V, so it meets the class by a good margin.

With all that said, you still need to know your Rb to confirm what over-current the CT can handle without saturation. There really isn't any other way to do it correctly. You can certainly approximate it for a good estimate. You need the lead wire size, the round-trip lead length, and the relay input impedance. If it's a modern electronic relay, then it can really be ignored, since the leads will dominate the burden calculation.