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Could 0.5FS5 IPCT be replaced by 0.5FS2 ?

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shalhoob109

Electrical
Aug 21, 2009
16

hi all ,

I found during my commissioning work in substation that
Interposong CT used with an Ameter has the following class
0.5FS2( Burden 5 VA) but the required one as per drawing is 0.5FS5 with burden 2 VA. they match in other ch's
Ratio 1/1 , Rct = 1.3 ohm etc.contractor engineer told me
that will not affect! but I would like to ask experts
before accepting that especially , our design and planning
division replay will spend long time.

there is somthing I would like to mention that IPCT is
connected to Main CT used for( protection+ metering )
Ratio 1200/1 Class x which exist in GIS.

question now , is there any problem using that ipct ?
 
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If I understood you correctly and the only difference is the burden, 5 VA vs 2 VA required, it should be very much acceptable.

Just curious are your model nos. mixed up? FS2 (5VA) and FS5(2VA)?

Rafiq Bulsara
 
Fs5 - 2 VA is functionally equivalent to Fs2 - 5VA, in that the security factor performance at the same connected burden is the same. This assumes the connected burden is <2 VA.
 

rbulsara

Not only burden is the difference. 0.5fs5 has security factor 5 where 0.5fs2 has different security fator which is
clearly 2

they are different in
1) VA Burden
2) security factor
 

scottf

why you conclude that they are functionally equivalent?
what equation You used to relate FS by Burden ??


shalhoob
 
shalhoob109-

Security factor is basically the saturation point, which is a function of flux density.

Burden and security factory are inversely proportional. The security factory (saturation point) is rated based on rated burden. Therefore, if a CT has a security factory of 2 and a burden rating of 5 VA, at half of rated burden (2.5VA), the security factory performance will be 4...half the burden will allow the CT to go twice as high before saturating.

So, a unit rated 2 VA Fs5 could be substituted for a unit rated 5VA Fs2, as long as the connected burden is less than 2 VA (so the unit will still meet accuracy class).

2 VA * Fs 5 = 10

5 VA * Fs 2 = 10
 
scottf , thank you for replay


you mentioned ,
(so the unit will still meet accuracy class).
2 VA * Fs 5 = 10
5 VA * Fs 2 = 10

so , why the manafacturer make to type of IPCT security
factor and burden if they will end up by the same Class as
you stated!.and why do you think designer choosen that
specified value ?

thanks alot

shalhoob
 
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