CT ACCURACY 1

[bbgg]

I want to know if the accuracy of a CT ratio 20/5 is affected if the I1 (primary current) is 35 Amp and the RF (Thermal Rating Factor)is 2. The CT has metering purpose. Thanks in advance.

 

[alehman]

Accuracy will be reduced if the secondary voltage exceeds the saturation point. The CT may overheat if high current is allowed to persist.

 

[jbartos]

Suggestion: The CT burden can determine whether the CT VAs are exceeded or within CT rating, thus potentially being a root to inaccuracy when VAs are greater than rated. If the burden is R=0 Ohms, then VA=0x35^2=0VA. If there is VA=0, then the CT cannot be overloaded.

 

[scottf]

The CT will be (is required to be) in accuracy class up to the rating factor, meaning that with a rated primary current of 20A and a rating factor of 2.0, the CT will be accurate up to 40A primary current...assuming the CT has a metering accuracy class. In fact, the accuracy will most likely be better at 35A than 20A primary. The CT's accuracy will improve up to the point that it starts to saturate, which by definition must be above the rating factor.

As for the points on the VA rating...

I don't believe the way is the IEC and other standards define burden in terms of VA is the best. I believe the IEEE/CSA method of defining CT burden as an impedance at a particular power factor is the best. Mainly because folks get confused at believe that the burden in terms of VA must vary based on secondary current. Please note that the VA rating is based on the rated secondary current...i.e. a VA rating of 15 VA with a 1A secondary is a 15 ohm burden.

For the CT originally inquired about, the CT is still accurate up to the rating factor of 2.0. The applied burden is still 15 ohms. It's not proper for CT ratings to say with a 2A secondary (RF of 2.0) that the burden imposed on the CT is 60 VA (2^2 * 15) and therefore the CT is out of accuracy. It's best to think of everything in ohms based on the rated sec current and VA rating.

 

[jbartos]

Comment: Normally, the thermal rating is linked to resistive load, not to impedance load or reactive load. The thermal rating factor would be linked to the resistive burden.

 

[scottf]

Jbartos-

I'm not sure what you're saying. According to IEC, burdens are assumed to have a PF of 0.8 lagging. Under IEEE/CSA, metering burdens are defined with a 0.9 lagging PF and protection burdens are defined with a PF of 0.5 lagging.

The rating factor (full name continuous thermal rating factor) is defined as the multiplier of rated current the unit can run at continuously at a 30 C ambient without exceeding the temperature rise (normally 55C). It is also used as the upper limit for the metering accuracy class.

For instance, a Class 0.2 CT must maintain a <0.2% error from rated current to the rating factor and <0.4% error from 10% rated current to rated current.

The rating factor has nothing to do with PF of the load being measured or the PF of the connected burden.

 

[jbartos]

Comment: The temperature rise is related to RI^2 in Watts.
The burdens have set limits on Power Factors to have limited the magnitude of current |I|; consequently, the heat generation by R|I|^2. Notice that jIm +jIburden is smaller then jIm + Iburden + jIburden. I do not see this aspect addressed in the above posting by scottf.

 

[scottf]

Jbartos-

The temperature rise in a CT is based on the characteristics (resistance and heat transfer) of the primary and secondary windings and the current flowing through it. It has nothing to do with the power factor of the load being measured by the CT or the connected burden.

In insulated CTs, really it's only a function of the primary resistance.

Current is current....power factor is the relationship between the phase of the current and voltage.

 

[bbgg]

Thank you all, in the end I can confirm that the accuracy is not affect if I1 (primay current) doesn't exceed the rating multiply by the RF.