CT saturation — I'm missing something

[b_gibb]

Hi!

I want to determine if my 600:5 CT will saturate. I have all the information but I feel like my formula is not correct. I understand it to be:

Vs = ( Ifault / CTratio ) * ( Burden of CT and leads and relay ) * ( 1 + X/R )

Where,
Vs = is the voltage developed on the secondary of the CT that I can compare on the excitation graph to see if it is in the linear range
Ifault = is the primary fault current, in my case 14.738kA
CTratio = is 600/5 = 120
Burden is all the burdens added up (in my case, 0.365 ohms)
x/r is the system X/R ratio which for my system is 17.5

This all makes sense to me but the answer works out to be 830V!!!! Which isn't even close to being on the excitation graph. Even if I change it to a 1200:5 CT (double the denominator), it's not close.

What am I missing?

 

[wcaseyharman]

The factor (1+x/r) doesn’t look right to me. That ends up multiplying your fault current by 18.5, which seems way too high. I don’t recall using that factor before.

 

[b_gibb]

Agreed, but not sure how to factor in the DC offset / asymmetrical current without the x/r factor in there somewhere...

 

[wcaseyharman]

Well, absolute worst case asymmetrical offset should not exceed 2*RMS, so if you doubled it that would still be conservative.
However I don’t think the DC offset really comes through to the secondary, but it tends to saturate the CT core. I will look back at some CT models from a protection class I took and see if it sheds any light on it.

 

[wcaseyharman]

By the way, what class of CT are you using?
Because of the low ratio and pretty high fault current you might need a pretty stout CT.

 

[b_gibb]

Assuming we can figure out the saturation formula, we are considering a C200, 600:5, such as the Amran 103-601.

 

[Kiribanda]

Per IEE C57 you have to select the CT so that @ 20 times it will not saturate.
Since your fault current is 15000A your 600/5 will get saturated within no time.
Therefore, your CT should be C600 @ 800/5 at least.

 

[b_gibb]

Hi Kiribanda. My intuition is the same as yours. But I would like to show through math/formulas, which brings me back to the root question. I'll need to prove that moving to 800:5 or 1200:5 will work.

 

[wcaseyharman]

Looking at chapter 5 of Blackburn and I think you’re fine. A C200 CT is rated for a standard burden of 2 ohms at 20 times the CT ratio (12000 A), but your secondary ohms is less than .4, so 1/5 of the standard burden in IEEE C57, so you should be more than adequate.
I’ll dig up that MathCad CT model if I get a chance and play around with it.

 

[waross]

2*RMS

Not "approaches 2*Peak"?,
or 2*RMS*1.414?.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[System Protection]

Have you seen the PSRC excel saturation calculator?

https://electrical-engineering-port...s-excel-spreadsheets/ct-saturation-calculator

 

[b_gibb]

I find the PSRC Excel sheet is a great visual, but doesn't actually tell you "does the CT saturate" or give Vs=?? for a given fault current. As such, I'm looking to confirm the right formulas myself (going back to the original question in this thread).

 

[Kiribanda]

You can use PSRC excel sheet and try 800/ 5 against 1200/5 using your X/R value. You can determine the sat. time from the waveform in the excel sheet and compare with the
IEEE C57 formula given for time to saturation. If the waveform is erratic that means it is saturated.

 

[dpc]

Google Stan Zocholl, SEL, and current transformers. He wrote several papers that covered CT performance in relatively straightforward ways.

https://www.google.com/url?sa=t&rct...ownload/2436&usg=AOvVaw0v7s0F6cK2Jk-ICdoCRGKZ

 

[wcaseyharman]

My bad, yes 2*peak.