Remanent Flux vs CT secondary Current

[pwrtran]

May I ask someone correct me if I am wrong here.
Assume the remanent flux on a CT core is 80% of the pu value, therefore, can I say:
1) there is only 20% capacity left before it saturates
2) the secondary current will be reduced by 80%

Thanks a bunch!

 

[Skogsgurra]

It would be very rare to find remanent flux that high. Also, as soon as primary current starts flowing, it will take flux for a 'spin around the hysterisis loop' and thus bring remanent flux down.

If, however, your primary current contains a DC component, you will have problems with core saturation. And the DC component need'nt be very high to cause problems.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[veritas]

pwrtran

I guess I would not say the secondary current is reduced by 80%. It all depends on the CT burden and the primary current. Also, should you have a high enough current to drive the CT into saturation, a secondary output is still possible though it will be distorted, the degree of distortion depends on the primary current magnitude, the primary and secondary time constants and the connected burden.

80% flux is possible with any system where a high fault current has been interrupted. It is usually a problem though for (E)HV systems where high speed autoreclosure is used.

Hope this helps.

 

[pwrtran]

Hello Skogsgurra and veritas, thank you for your valued response.

IEEE C37.110 does say the remanent flux can be found as high as 80% and in Annex C it gives 27% of the CTs investigated have been found remanent flux from 61%-80%.

The point of my interesting is to calculate and determine the transformer percentage differential settings - specially the Break1 Point.

1) For a given CT the saturation voltage Vsat can be found
2) For a given relay and the known lead wiring plus the CT winding resistance the total CT secondary burden Zs is obtained.
3) Then the maximum CT secondary current before saturation caused by AC component only presented in pu will be
Ictmax=(Vsat/Zs)/5

GE has an example calculation result but no details. In the example the Ictmax was calculated at 8.34pu, which presents the Break 2 Point and beyond which the CT will saturate by the AC component only.

If the above giving CT has in the worst case having 80% remanent flux in the core, then the current that would cause saturation due the DC component and the remanent flux will be lowered to 1.668pu. 1.668 is the 20% of 8.34pu - coincident or not?

 

[Skogsgurra]

Thanks. I have never seen that high a remanent flux. But I do not deal with high-current installations like the one you seem to do. It is mostly industrial drives in the 100 - 1000 kW range.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

I need to understand one thing, is the remanent flux 80 percent of core saturation flux?

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[pwrtran]

Skogsgurra, my understanding - it is.

 

[veritas]

Hi

I am not familiar with IEEE C37.110 as I live and work in the IEC world. When it comes to CT specifications for biased differential relays, the required saturation (or kneepoint) voltage (if DC offset is ignored) is really the voltage Ek calculated by multiplying the maximum symmetrical throughfault current, say Isym, by the connected burden. Thus Ek(sym) = Isym*R (R = total connected burden).

However, an important requirement is that there should be no saturation under any throughfault condition that could give rise to spill currents that would operate the diff relay. This is overcome by multiplying Ek(sym) by what is known as a transient factor in the IEC world. Call it K. I would guess there is a similar provision in the IEEE standard. K is dependant on the X/R ratio of the system.

If the CT is never to saturate, then K = 1 +X/R should be used. This is in the GE publication as well. Quite a good one if I might add. But using K = 1 + X/R could lead to CT's being disproportionately huge and expensive. K for trfrs could easily be in excess of 15!

One then needs to decide for how long after fault inception the diff must be stable. This will then determine a reduced value of K. This is all very well explained in the GE publication and the formulas are there as well.

The first turning point (or breakpoint - BP1) is usually determined by the intersection of the minimum diff current required for operation, say Idmin which is a horizontal line, and slope 1.

BP2 is usually a bit more trickier and depends on what type of relay you are using.

Plse see following thread which I believe will be of interest to you.

http://www.eng-tips.com/viewthread.cfm?qid=245515

Kind regards.

 

[pwrtran]

It was a very informative post, great work veritas! I need to spend some time to eat and digest your post.

Have a great weekend!

 

[veritas]

You're welcome mate. Let know should you still have any more questions.

Regards.