Transformer Differential Protection - CT Saturation

[mdenis8218]

In the event that a transformer is subject to a catastrophic fault, one set of CTs will be exposed to very high current magnitudes which will drive them into saturation.

Is it safe to assume that since the opposing set of CTs will not enter saturation (picture a radial system), the differential relay will detect the fault and send the appropriate trip?

This seems like a no brainer to me but I can't find any documentation to confirm this...my concern is that the harmonics resulting from the saturation may cause the relay to mis-operate but it seems highly unlikely that the relay could be tricked into thinking that the currents are within tolerance for this scenario...(differential relay is an SEL 387, saturation occurs almost instantaneously)

Thanks

 

[3I0]

CT saturation calculations should be carried out, and the CT's should be sized appropriately for the application. Just because there is an internal fault does not necessarily mean the CT's will saturate, either fully or partially. There are also degrees of saturation from either transient DC offset (dependent on X/R ratio) or steady state fault current.

What a digital relay actually sees (and responds to) in the event of CT saturation is dependent on what type of filtering algorithm it utilizes to calculate the fundamental frequency current. The IEEE PSRC has a CT saturation calculation spreadsheet that helps visualize this for a DFT filter, which is common to many electronic relays. SEL also has a variety of white papers on CT saturation on their website with good information.

 

[mdenis8218]

Thanks for the reply.

I've done the calcs (this is an existing installation so I'm stuck with what I have).

The CT definitely saturates. I guess what I need to know is if I can count on the supervisory instantaneous overcurrent relay to pickup in spite of the CT saturation.
Blackburn says, "For heavy close in faults, these relays may be operating on the relative flat portion of their time curve, for which magnitude differences are not critical."
What I think this implies is that you can't count on a saturated CT providing anything useful in terms of torque controlled pickup but you can rely on getting a large magnitude on the secondary...ie. instead of an analog value that you can predict, you're getting more of a confirmation that a high magnitude fault exists...

In my case, I have a 100:5 CT that will be exposed to a 10kA fault with X/R=15...the CT saturates, I'm expecting to see somewhere around 20x rated on the secondary (ie. 100A) which would be more than enough for my 50 to pickup....

Thoughts?