a question on CT's performance 2

[charlierod]

Hello everybody.

Current transformers normally are specified for a maximum transformation ratio error under certain conditions.

1. Does this ratio error always tend to give a lower value of secondary current than expected? (due to excitation current)

For example if i have a C800 10P20 6000/5 current i would expect that the transformation ratio due to errors could be as high as 1320 but will never be lower than 1200 for symetrical fault currents not exceding 20 times rated values and burden not exceding 8 ohms.

2. If i don't have information about core characteristics, can i make an estimation of the errors introduced based on knoweledge of rms values of asymmetrical fault currents?

Thanks for your comments

 

[scottf]

1. In general, the ratio error of CTs is always negative (except when parallel turns correction is used for some metering CTs).

As the current rises above rated current, the accuracy of the CT continues to improve (i.e. less negative error) until saturation starts to occur.

You are correct that for a 10P20 rated cores, the maximum composite error at 20 times rated current is 10% and this will be a negative error.

2. Are you saying that you would like to calculate the exact error at a particular fault current level? This is tough to do unless you have all the details on the core design. Also, keep in mind that the error is a function of the connected burden. The lower the burden, the less the error. Rountine accuracy tests are made with the rated burden. In most modern cases with electronic relays, protection CTs are rarely loaded to their maximum burden.

 

[charlierod]

Scott:

thanks for your contributions, when you talk about negative errors are you saying that the ratio is higher than rated as i say in the original post?

Of course i don't expect to get a high degree of accurary in results without knowing core characteristics. I'm aware that a time domain simulation should be carried out in this case. But there are some approximations like the one suggested in the following paper:

http://www.selinc.com/techpprs/6025.pdf

They make an estimation of CT performance under asymmetrical fault currents based on external system X/R ratio. I just wanted to know the validity of these kind of approaches.

 

[busbar]

Pardon me for invoking social status or “pecking order” in power-systems engineering, but the lead author of the cited paper seems to have a lot of resources, experience and practical knowledge under his belt.

 

[scottf]

charlierod-

Yes...the ratio is higher than rated, meaning that the CT has a negative error or the secondary current is lower than it should be (based on the rated ratio).

I'm familiar with the paper you linked and I think it's fairly accurate as far as the CTs are concerned. What part in particular are you concerned with.

Also, to address Busbar's comments...I don't think it's bad to question things you see in papers. One should work to understand the information and be able to feel confident in its assertions. For example, there are plenty of papers floating around out there that are WAY off base concerning capacitor voltage transformers.

If it makes Busbar feel any better, I work for an instrument transformer manufacturer and sometimes get calls from SEL asking questions on IT's.

 

[charlierod]

Busbar:

When i said "i wanted to know the validity" i really meant to say "i'd like to know how accurate is the approach". I think SEL is number one in relaying industry and they have published excellent papers like the one cited above.

I'd like to apologize in advance if some "Mr. Electricity" got offended.

Scott:

I'm evaluating CTs performance in a percent differential scheme prior to transformer commissioning. Transient performance of CTs was not analyzed, their ratings were selected so that secondary symmetrical fault currents are about a half of maximum allowable:

For a 10P20 C800 CT secondary current is about 50 A for a three phase fault condition. I'm on the way to check CTs burden.

Any comments about this?. This is a public web site so all comments are welcome even if they don't contribute to the original post.

 

[veritas]

charlierod - could you explain the C800 to me please? I'm not familiar with this term.

thanks.

 

[charlierod]

C800 for a CT means maximum secondary voltage value is
800 V (according to ANSI). If a higher voltage is impressed significant errors could be experienced as a result of core saturation.

Example:

for C800 CT if relay maximum secondary current is 50 A, relay burden should not exceed 16 ohms.

All this analysis is based on symmetrical currents. Due to asymmetrical currents that flow during faults a detailed analysis should include transformer core characteristics as stated above in this forum.

 

[light1]

C800 means "C" CT accuracy Class classification, 800 stands for 800V associated with 8 Ohm Relaying Burden, 5A secondary winding current and 20 times rated secondary current design withstand secondary winding, i.e. 8 Ohms x 5Amps x 20times = 800 (secondary Volts). This forms C800. See ANSI Std. C57.13.

 

[scottf]

for what it's worth...the 'C' in C800 stands for calculated. There is also a 'T' class which stands for tested. The only difference has to do with the core/coil design. Some designs cannot be calculated due to the winding design.

 

[alpha75]

I think there is some mixed up about the C800 10P20. The C800 relay class is based on ANSI/IEEE while the 10P20 protection class (with the burden defined, e.g., 10va 10P20, 15va 10P20, etc) is based on IEC standard.

Metering class in ANSI is defined as, e.g., 0.6B0.9, 0.3B1.8,etc. While the IEC measurement class can be defined as, example: 5va CL0.2, 15va CL0.5, etc.

The ANSI 0.3B1.8 metering class (this is a revenue grade) means that the CT has a 0.3% ratio error up to a burden of 1.8 ohms (or 45va).

The IEC 15va CL0.2 is defined as having a 0.2% error up to a burden of 15va.

Note that the phase angle errors are also considered in the above definitions.

;-)

 

[charlierod]

As far as i'm concerned 10P20 means a maximum 10% ratio error for currents up to 20 times secondary rated current. If secondary rated current is 5 A and CT is C800 class, then for 100 A maximum burden would be 8 ohms. Any mistake in this way of reasoning?

 

[charlierod]

alpha 75:

I've seen CTs rated with a "C" class and a XPY class at the same time, for example a C800 10P20 6000/5 CT.

 

[scottf]

As listed above-

Technically a rating of 10P20 is an IEC rating and the burden in VA must also be given.

'C' class is an IEEE rating. The accuracy limiting voltage (i.e. 800) can be converted into a burden (i.e. 8 ohms).

There are some minor differences in how IEC and IEEE specify testing and some other minor class details. In practice a 5A rated secondary CT with 10P20 - 200 VA rated accuracy is about the same as a C800 rated CT.

MOst (but not all) IEC rated CTs have a 1A rated secondary and thus it's a little more involved to convert to IEEE C class definitions, which are all based on a 5A secondary.

 

[mersin33]

Per ANSI, C800 means the error cannot exceed 10% within 1 to 20 times the rated current, provided that the burden does not exceed 8 ohms

Per IEC, 10P20 does not mean anything, because there is no VA rating. But say, it were 30VA, it would mean:

Composite error cannot exceed 10% within 1 to 20 times the rated current, provided that the burden does not exceed 1.2 ohms (30VA/(5Ax5A)).

Maybe it is a completely different standard. Can Charlierod tell us what standard this CT is made to?

 

[charlierod]

Scott:

In the paper cited above, Could you explain why they don't use the same criteria (the one based on X/R ratio of external circuit) for selecting C class of both high and low side CTs.

Does this have to do with the degree of saturation of the CT? here saturation is more critical for high side CTs

 

[stevenal]

The transformer in the paper was evidently fed from the low side, since the high side faults were limited by the transformer impedance. Accuracy is important for the through faults since you want the differential current to be close to zero. For the internal fault on the low voltage winding, accuracy is not important. Even saturated waveforms will show up as differential current, you just want to ensure enough current is available to get above minimum pickup. Sounds like your transformer is the reverse situation.

 

[busbar]

charlierod, my 10-June post was poorly worded. I intended to say that Stan Zocholl has a very good idea of how CTs operate in a transformer protection scheme as he explained it in the paper.

 

[charlierod]

Ok busbar, thanks for your contributions.

I made a mistake in a previous post. The original CT was specified as C800, but in a drawing someone specified it also as 10P20 200 VA maybe using the equivalence mentioned above by Scottf.

Thanks everybody for your contributions