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How can I select CT(Current Transformer) capacity?

How can I select CT(Current Transformer) capacity?

How can I select CT(Current Transformer) capacity?

(OP)
Design Conditions
-Full Load Current(Motor 1790kw, 6kv, 50Hz) : 200A

How can I select CT capacity(primary side)? ..CT Ratio?

Is there a related Code or Standard?

Please! Please! Tell me any information...

Thank you!

RE: How can I select CT(Current Transformer) capacity?

The CT ratio should be slightly higher to the prim current and if possible to the next standard ratio. So in your case it should be say 250. I assume that the secondary current is known to you as being either 1A or 5A.
Note that a 225A prim current would also suffice.

RE: How can I select CT(Current Transformer) capacity?

For the same voltage rating 125 percent of the rated current i.e 300Amp  will be available.

RE: How can I select CT(Current Transformer) capacity?

Dear Imh4534,
    The CT you select should meet following conditions:-
1) Assuming that only load on the CT is motor and no other load then your CT should be able to carry 1.2 times the full load current.
2) Be sure to see what is the short time current rating of the CT and for how many seconds 1 sec or 3 sec?
3) On the secondary side the CT current can be selected as 5A or 1A based on your requirement, but this choice of yours (i.e 1A or 5A) will also have cost implication.
4) The class of core you are selecting is also to be scrutinized.

    Finally there is an IS which supports my first point stated above.
Bye.
Nihar

RE: How can I select CT(Current Transformer) capacity?


Might find some guidance in IEEE C37.96, C37.110 and C57.13.
  

RE: How can I select CT(Current Transformer) capacity?

There's another key rating that hasn't been discussed yet here:  burden.  This is about the same thing as the secondary load on a distribution transformer.

The motor is not the load or burden on the CT.  The protective relaying being driven by the CT is the burden.  Every relay you add burns a bit of the power available from the CT.  You need to make sure that the CT is rated to handle the burden imposed by all relaying.

Unless you are going crazy with relaying, though (such as using LOTS of mechanical relays), any CT designed for use in relaying should have sufficient capacity.

By the way, I'm gonna add this little piece of advice since it sounds like you're not too familiar with CT's:  consider getting a "shorting terminal block" for connection of the CT wiring.  This facilitates maintenance and can help prevent shocks or equipment damage from open-circuited CT's.  Not required, just a nice option.  Some might consider this "old school," so I'd be interested to see what anyone thinks.

RE: How can I select CT(Current Transformer) capacity?

I agree with you peebee, shorting terminal blocks should be provided for all CT installations.

Regards
 

RE: How can I select CT(Current Transformer) capacity?

Peebee: I'll give you a red star for that piece of advice.

Electronic relays (Static, Electronic and microprossesors) have normally a very low burden, electromechanical relays is the buggers with a high burden. When you are working with a electromechanical relay, make extra sure that your CT's will drive the relay under fault conditions.

RCC

RE: How can I select CT(Current Transformer) capacity?

Dont forget the wire that leads from the CT to the relays. This is often where you run into problems with burdens with modern equipment. So if have an abnormal length of wire(say >15'), don't forget to add it to the overall burden.

RE: How can I select CT(Current Transformer) capacity?

(OP)
Thanks everyone!
so many good views & infomations..I'm so happy!

I want to see IEEE C37.96, C37.110 and C57.13. stated above. Could anyone help me?

M.H.Lee
mhlee@seahsp.co.kr

RE: How can I select CT(Current Transformer) capacity?

http://shop.ieee.org/store/product.asp?mscssid=HQVNVHSBVJTA9PJ11XGUHEAGXBM0DSQB&CID_LDAP=&PROMO_SRC_CODE=&BUYER_TYPE=MEMBER&account_type=&member_grade=&hftype=all&AUTH=&pfid=6687&dDN=General%2FOther+%28Power+and+Energy%29&dTN=Power+and+Energy&src=search&loc=http%3A%2F%2Fshop%2Eieee%2Eorg%2Fstore%2Fresults%2Easp%3Fmscssid%3DHQVNVHSBVJTA9PJ11XGUHEAGXBM0DSQB%26CID%5FLDAP%3D%26PROMO%5FSRC%5FCODE%3D%26BUYER%5FTYPE%3DMEMBER%26account%5Ftype%3D%26member%5Fgrade%3D%26hftype%3Dall%26AUTH%3D%26ct%3Dproduct%26so%3Dmeta%255FIEEE%255FSTANDARD%255FNO%252CDocTitle%252CRank%255Bd%255D%26qu%3D%2528%2524all%2Bc37%2E96%252D2000%2529%26mh%3D10%26sh%3D0%26ae%3D1&pic=130900

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