CT Resistance calculation 2

[Arevaengineer]

Gentlemen
Can anybody point out the method to calculate the CT resistance for example 100/1/1,Cl 1,5P10.
Thanks to anyone for helping.
Regards

 

[Arevaengineer]

Additional Info:

6.6kV, 60Hz

 

[scottf]

There is no accurate way to calculate the secondary resistance from the ratings of a CT.

 

[Alex68]

Starting from a list of CT data, I believe to have found a formula:

if Isec = 5A then a=0.1
elseif Isec = 1A then a=1

Rct= a* 8.4 * (1-VA/50) * (Iprim/1000)^0.925

Any suggestion?

 

[scottf]

Alex68-

I'm not in the office now to test to see if this works, but I don't see how it could consistently, for the following reasons:

1) You don't know the secondary wire size used in the CT.
2) For an insulated CT, you don't know the number of primary turns used, hence, you don't know the number of secondary turns used.
3) You don't know the core size. Although you may be able to guess at it, it can make a big difference.
4) Your formula doesn't take rating factor into consideration. This really ties back to point (3) and (1), as it would dictate the size of the core and wire used.

 

[Alex68]

Dear Scottf
sorry but I disagree with you. The formula I found is an approximation, based on the data of a lot of CTs.

1) the factor "a" depends directly on the rated secondary current "Isec"

2) the number of turns is taken into account because the formula contains the "Iprim", i.e. the primary rated current, while the factor "a" is related to "Isec"

3-4) the factor "VA" is the rated burden of the CT

I believe that the formula works with a good approximation
Please try to apply it to a real CT and then tell me what you will obtain.
On the other hand, if you are a CT designer, please tell me the formula you use.

 

[scottf]

Alex-

I do work for an instrument transformer manufacturer and we do not use a specific formula that would be easy to apply without knowing all of the design details. The calculation method we use is based on the total length of secondary wire, based on the size wire used.

Your formula takes into account Iprim and Isec. However, unless this is a window-type CT, you have no way of knowing the number of secondary turns used. For example...let's say you have a 15 kV CT with a ratio of 100:1A. That design most likely uses 10 primary turns or so, meaning that there would be 1000 secondary turns.

Now, for a CT with 1000 secondary turns, the geometry/size of the core matters a great deal to the secondary winding resistance. You have no real way to know that core size for an insulated CT.

You also have no way of knowing the secondary winding cross sectional area. For a 1A rated secondary, a differt wire size may be used for a CT with an RF of 1.2 versus and RF of 3.0, as the unit with an RF of 3.0 has to carry 3A continuously.

The bottom line is that if you need to know the secondary winding resistance of a CT with a reasonable certainty, you need to get this from the manufacturer. This data should be on the original test reports, or you can call the manufacturer with the serial number. If all else fails, you can measure the secondary resistance directly.

 

[Arevaengineer]

Thanks ScottF and Alex68 for the valuable info.I have measured the secondary CT resistance directly.As per the formula given by Alex68 it holds true for the core which is class 5P10 but for the metering core which is class 1 the formula does not apply. I have noticed that the resistance is lower for metering core than the protection core.
I just wanted to compare the measured resistance against the designed as the name plate does not mention the CT secondary resistance.

 

[scottf]

The metering core is probably much smaller than the protection core and that leads to a smaller winding resistance, as mentioned above.

 

[Alex68]

The formula has been obtain for protection transformers and not for metering ones. I do not have data for metering transformers.

Generally the problem is to avoid the saturation of protection CTs and validate their sizing. During the first phase of a plant design I must size the protection CTs with few data. So I developed the formula to foresee the secondary winding resistance.
Perhaps it seems a bit "strange" and without a physical meaning, but it is the result of the interpolation of a lot of data and... it works.

 

[scottf]

Alex68-

Your formula may work some of the time, however, I checked it versus a design which uses multiple primary turns and, as I explained above, it does not work correctly. Your formula assumes that only 1 primary turn is used.

 

[Alex68]

Mr. Scottf, a two stars expert, has spoken.
I surrender.

Official comunication: People do not use my formula!