CT Open Circuit 5

[charz]

I have read the old threads regarding the CT open circuit generating high voltages, but still I couldn't grasp it any easier. My questions are,
1. Why the acccuracy is high when the CT secondary is short circuited?
2. Why is the magnetic flux in a CT core is not constant (Varies with both open and short circuited CT secondary) as is the case with a VT?

I think mostly the answer to the question lies in the primary current of the CT not depending on the secondary of the CT? but still I'm not able to put it together.

 

[HamburgerHelper]

Isn't it more accurate to say that the secondary voltage will be proportional to the current passed through the high side of the CT? Once the CT saturates, the relationship between the primary and secondary will have heavy losses but also be very linear. You can't saturate air. If the line is heavily loaded, the open CT voltage will be higher than an unloaded line. With enough current, mathematically, you could exceed the primary voltage. The open CT voltage has nothing to do with the primary voltage and everything to do with the primary current.

v1 = L1di1/dt − M di2/dt
v2 = −L2di2/dt + M di1/dt .

i2 = 0 for open circuit

v1 = L1di1/dt
v2 = Mdi1/dt

edit: Nevermind. Scottf basically said this same thing.

 

[stevenal]

"Once the CT saturates, the relationship between the primary and secondary will have heavy losses but also be very linear"
The non-linear magnetizing branch dominates during CT saturation. The output will be distorted and misoperations can result.

"You can't saturate air"
It's the iron core of the CT that saturates.

"With enough current, mathematically, you could exceed the primary voltage."
I assume you mean the primary voltage adjusted by the CT turns ratio. The model of the CT as an ideal current controlled current source would predict the behavior you describe. Like all models, this one has limitations.

 

[marks1080]

Just want to state that this is a great thread. All protection folks should be reading this. Many thanks to all the contributes.

 

[HamburgerHelper]

Stevenal,

Once you push the CT fully into saturation, it becomes linear again. I think in text this is referenced as being supersaturated. I am just pointing out that you have two linear relationships, core linearity and air linearity. Push a CT into supersaturation and the waveform will actually look nice but won't obey the turns ratio due to heavy losses. The jaggedness of a waveform due to saturation is due to transitioning back and forth from core and air B-H slopes.

 

[edison123]

Bill

Doesn't burden come into the scene only when the secondary is closed?

Muthu

http://www.edison.co.in

 

[edison123]

Agree with marks1080. Informative thread.

Muthu

http://www.edison.co.in

 

[stevenal]

Edison:
An open secondary would generally be modeled as a very high burden.

Hamburger:

So you're saying the manufacturers' excitation curves leave off this area of supersaturation where the curve would go steep again to the right of the saturated region? And the low voltage insulation of the secondary can withstand these voltages without breaking down? References would be appreciated.

 

[HamburgerHelper]

They include it in their diagrams. If you push hard into zone three, you are in another region that is linear. The sheets that the manufacture gives are logarithmic so they go out very far.

It probably depends how much iron you have in your CT. The less iron you have the lower the saturation voltage will be and the more likely your insulation won't breakdown on open CT. I wouldn't be surprised if an opened metering CT wouldn't breakdown the insulation since it has less core iron. Protection CTs are usually sized to not saturate ,high imp. diff being an exception, for faults and probably have so much iron that the insulation rating is below the saturation voltage.

 

[stevenal]

Looks linear on a log-log scale. Is that what you mean?

Try Link and open the spreadsheet.

Change Off to 0 to zero out the DC offset for simplicity. Try an Rb of about 70 to see what a waveform from a saturated CT looks like. Leave the other values alone. If you don't think you are in what you call zone 3, increase Ip, Rb, or both.