Charging-current compensation in Differential protection-understanding

[markmarko]

Hi all...
I have found some difficulties understanding charging-current compensation (Ic), in my case when working with line differential protection.
I have a relay with setting Ic=1, and charging current of 400A, and differential setting of 480A.
How this Ic has influence on pickup value of differential protection? What if the Ic setting is 2?
How does voltage influence on pickup value of diff protection....
I was reading 7SD86 manual, but it's still not clear for me how to calculate all this situations (Ic=OFF, with applied voltage,
two or three end lines...)
Can anybody give me some explanation or link to some more detail manuals.....
thank you in advance...

 

[Kiribanda]

You have mentioned a charging current of 400A!! Is it possible?
What is the length of the cable & its voltage rating?
Also what is Ic?

 

[markmarko]

Thanks for response....
It's around 50km of cable, 220kV.
Ic is setting (according manual) ''Is stabilization/ Ic rated. Parameter which should increase stabilization when we lost voltage.
For example, if differential setting is 480A (when voltage is present), what would be the pickup value when there is no voltage on the line? Should be higher...but, how much?
Also...this parameter should '' increase sensitivity of diff protecten...'' but how? I mean for which value? How to calculate this values?

 

[RRaghunath]

Charging current compensation (Ic) is optional. If the charging currents are already considered while deciding Idiff> pickup threshold setting, there is no need for Ic.
If more sensitive setting is required (for Idiff>), then it is advised to set Ic- capacitive charging current compensation.

 

[markmarko]

Thanks foe explanation...this seems very clear...
I would just like to know how to calculate Idiff pickup for different situations and how does voltage influence on the calculations.
Here are some Example settings, and test situations:
charging current= 400A
Idiff>480A (1.2*400)
CT 1000/1A
Ic=1
Test situations:
1. Ic=OFF, no voltage applied --> Idiff pick up= 0.51A (with CT error of 3%, this is clear)
2. Ic=ON, no voltage applied --> Idiff pick up= 0.775A
3. Ic=ON, voltage applied on one side of the line--> Idiff pickup = 0.45A
4. Ic=ON, voltage applied on both side of the line -->Idiff pickup = 0.29A
What would be results if Ic=2? (I would like to know the calculation/relationship for different situations)..