# Tranformer protection and CT



## eng787 (Oct 2, 2010)

Problem :

A 3 phase Delta-Y connected 15 MVA, 33/11 KV transformer is protected by CTs. determine CT ratio for differential protection such that circulating current ( through transformer delta ) does not exceed 5 A.

Solution Provided : I_delta=262.44 A

I_Y= 787.30A

If CT on high voltage side are connected in Y then CT ratio on high voltage side is 787.30/5

CT ratio on low voltage side is 262.44x5/sqrt(3).

*If transformer is Delta-Y , do we always connect CTs in Y-Delta ?????*

How they come up with CT ratio.

Can somebody explain ???????????


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## cableguy (Oct 2, 2010)

First off, yes, if your transformer line connection is wye, your CT connection is delta, and the reverse is also true - delta line, wye CT's. So for a delta-wye transformer, you do wye-delta CT's.

That said, this is a bad problem. I just worked through it this morning also. I could not arrive at their solution via any rational method. And their "circulating current" addition in the problem bugs me. I get delta current of 151.5 amps, so would 5 amps of circulating current make it 156.5 amps? but where does that really play in with their solution? The problem is not explicit enough, IMO.

I am following the method outlined in a book called Protective Relaying: Principles and Applications, chapter 9 on transformer differential protection. A little Google time for you (looking for pdf) will be well spent. 

You can also look at Chapter 11 here:

http://www.gedigitalenergy.com/multilin/no...rtsci/index.htm

and read about transformer differential protection.

Also note that their answer for #21 afternoon is wrong, I got 3.15 - and so did they, until they wrote 3.31.


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## knight1fox3 (Oct 4, 2010)

cableguy said:


> I am following the method outlined in a book called Protective Relaying: Principles and Applications, chapter 9 on transformer differential protection. A little Google time for you (looking for pdf) will be well spent.
> You can also look at Chapter 11 here:
> 
> http://www.gedigitalenergy.com/multilin/no...rtsci/index.htm


Would this be the reference you mentioned?


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## cableguy (Oct 4, 2010)

Yes.


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## knight1fox3 (Oct 4, 2010)

cableguy said:


> Yes.


Great! Thanks for the tip. I'll probably add Ch. 9 and maybe 11 to my repertoire.


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## Chickman (Oct 6, 2010)

I haven't seen the problem, but agree that the solution seems a little confusing. I know they recommend connecting the wye side CTs in a delta configuration to remedy zero sequence circulating currents, which could cause the relay to pick up on say a ground fault (nuasance trip).

I think the relays now days have capabilities of selecting any magnitude and phase angle shifts through the transformers, which allow the CTs to be placed outside of the transformer windings in one configuration (of course it isn't practicle to place the ct's inside the windings).

Just thought I would mention that ...

15MVA/(sqrt(3)*33kV) = 262.432 amps

15MVA/(sqrt(3)*11kV) = 787.296 amps

Cts on high side are wye connected (implying delta high side connection) then they give the ratio for the rating of the low voltage current size down. (i.e. low voltage current is 787.296 so 787.296:5A ... low good luck finding that CT ratio)

Cts on the low side are delta connected (implying wye high side connection) are the ratio of the delta side winding or phase currents

I'm guessing you mean 262.432/sqrt(3):5amps which at least somewhat makes sense

Maybe someone can fill in the gaps by recommending why they recommend doing this???


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## Chickman (Oct 6, 2010)

Was hoping to hear a reply back, but I figured I might also mention that schweitzer solid state relays allow for differential transformer protection connections on any CT wiring configuration (wye-wye, delta-wye ... etc). Those relays are sofisticated enough to differentiate the zero sequence currents (by turning off RTS setting in the relay).

I don't expect any questions on the exam on this topic, though there was one conceptual on the practice exam (maybe two). I think it said that you could get nuisance trips by not being concerned with the the resultant current phase shifts or something. I think a practicle angular/phasor three phase review might be essential for the test since they also stated that a wye source and delta source shouldn't be paralleled.

I once had a professor recommend showing the phasor relationships every time a transformer was drawn, a good idea in my mind.


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## z06dustin (Oct 7, 2010)

Chickman said:


> Was hoping to hear a reply back, but I figured I might also mention that schweitzer solid state relays allow for differential transformer protection connections on any CT wiring configuration (wye-wye, delta-wye ... etc). Those relays are sofisticated enough to differentiate the zero sequence currents (by turning off RTS setting in the relay).


Excactly. I had an SEL application engineer, who I trusted a great deal, recommend that I never design another Y/delta CT scheme. He said that SEL recommends Y/Y connections of the CTs, regardless of how the primary is configured. This was for a xfmr differential 487.


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## Nik (Oct 23, 2010)

Where did you get this problem from ?

Thanks,



Baljit Gill said:


> Problem :
> A 3 phase Delta-Y connected 15 MVA, 33/11 KV transformer is protected by CTs. determine CT ratio for differential protection such that circulating current ( through transformer delta ) does not exceed 5 A.
> 
> Solution Provided : I_delta=262.44 A
> ...


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## cableguy (Oct 23, 2010)

Kaplan EE Power Sample Exam. Afternoon Problem #20.


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## cabbagekid2 (Oct 27, 2010)

Reviving this post...muahahahaha (evil laugh-going crazy)

So the problem was:

A 3 phase Delta-Y connected 15 MVA, 33/11 KV transformer is protected by CTs. Determine CT ratio on the high voltage side for differential protection such that circulating current ( through transformer delta ) does not exceed 5 A.

Answer: 158

I_Y (secondary)=15MVA/(sqrt(3)*11kV) = 787.296 amps

CT ratio on high voltage side: 787.3/5=158.

I know this problem is worded really bad but could a reason for their answer of using the low side current instead of the high side current for the CT ratio be . . .

*The circulating current (through the transformer delta) should not exceed 5A. *Could they mean the current in the CTs connected in delta (which is on the secondary side of the transformer) should not exceed 5A. And not the current in the transformer delta? So you have to find the CT ratio that would make the current in the delta connected CTs less than 5A?

I'm just trying to make some sense out of this. Thanks.


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## cableguy (Oct 28, 2010)

It's too late to make sense of Kaplan problems. Close that book before you make yourself crazy. Seriously. It's so full of mistakes and questionable stuff that, one day from the exam, you should not be using it.

I can't justify the Kaplan answer on this one.


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## agabee (Oct 28, 2010)

that answer is definitely wrong. cross it out! the question asks for CT ratio of the HV side. CT ratio has nothing to do with the connection of the transformer, it has to do with only the load current. when a transformer is delta-Y, the CT is Y-delta, but that has nothing to do with the CT ratio selection. this comes into pla only when you are calculating the currents going into the relay differential elements. in the solution, the answer does not even ralate to the HV side, where the current is about 262A. A CT ratio on the HV side would be 300/5A or 60.


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