CT mismatch in differential protection MiCOM P54x

[Aescobar]

Hello community,

I would like to confirm how the differential protection works on the MiCOM P54x relays in the case of different CT ratios. In the following example, end Ahas a 1400/1 CT ratio and end B has a 2500/1 ratio. 

The manual recommends that the parameter   Ph Ct Corr'tion should be calculated with primary rated current of the smallest CT ratio as a base current:

Ph Ct Corr'tion=CT local/CTmin.

So in this case, should be:

End A: Ph Ct Corr'tion=1400A/1400A=1.00

End B: Ph Ct Corr'tion=2500A/1400A= 1.790

With this configuration, it is okay to set the rest of the differential settings (Phase Is1, Phase Is2 and Phase Is1 CTS) on the same secondary values in both ends of the line?

Looking forward to your reply!

Thank you.

[Gregory_Meunier · Accepted solution]

Hello,

The line differential protection does its internal calculation in per-unit ; consistent per-unit definition must be set at each end.

This means

- For line or cable, the same per-unit values ar each end correspond to the same primary current.

- For transformer in zone application, the same per-unit definition must match the per-unit base of transformer winding.

I only consider the example of the line or application without transformer.

By default, 1 pu corresponds to 1 A for 1 A inputs or 5 A for 5 A inputs.

We can extend the reasoning with the CT primary current rating but ultimately the relay is measuring secondary currents.

If the CT ratios are different at either end, there is no longer one common per-unit definition but two definitions that will cause the relay to trip.

Ph CT Correction parameter is set to adjust the per-unit definition.

Then 1 pu corresponds to 1 A / Ph CT Correction in 1 A inputs or 5 A / Ph CT Correction in 5 A inputs.

Is1, Is2 and Is1 CTS are converted in pu without taking into account the Ph CT Correction.

With 1A rated secondary current, 0.2 A means 0.2 pu.

Example without Phase CT Correction

The line is loaded with 1000 A.

- End A: 1400/1

             Iload(sec) = 1000 A / 1400 A × 1 A = 0.71 A

             Is1 = 0.2 pu => 0.2 A(sec) or 280 A(prim)

             1 pu => 1 A(sec) or 1400 A(prim)

              => Ilocalpu = 0.71 A / 1 A = 0.71 pu

- End B: 2500/1

             Iload(sec) = 1000 A / 2500 A × 1 A = 0.4 A

             Is1 = 0.2 pu => 0.2 A(sec) or 500 A(prim)

             1 pu => 1 A(sec) or 2500 A(prim)

              => Ilocalpu = 0.4 A / 1 A = 0.4 pu

Idiff => 0.71 pu - 0.4 pu  = 0.31 pu

The differential current is due the inconsistency of the pu definition on the both ends, the relay malfunctions by diff trip.

Example with Phase CT Correction

The line is loaded with 1000 A.

- End A: 1400/1 & Ph CT Correction = 1

             1 pu => 1 A(sec) or 1400 A(prim)

             Iload(sec) = 1000 A / 1400 A × 1 A = 0.71 A

              /!\ Is1 set at  0.2 A(sec) - 280 A(prim) => 0.2 pu => Actual threshold at 0.2 A(sec) - 280 A(prim)

              => Ilocalpu = 0.71 A / 1 A = 0.714 pu

- End B: 2500/1 & Ph CT Correction = 2500/1400 = 1.79

             1 pu => 0.559 A(sec) or 1397 A(prim) (same as End A)

             Iload(sec) = 1000 A / 2500 A × 1 A = 0.4 A

             /!\ Is1 set at  0.2 A(sec) - 500A (prim) => 0.2 pu => Actual threshold at 0.112 A(sec) - 280 A(prim)

              => Ilocalpu = Ilocalpu = 0.4 A / (1 A / Ph CT Correction ) = 0.716 pu

Idiff => 0.716 pu - 0.714 pu = 0.02 pu (mismatch due to rounding error)

Consistent pu definition on both ends, no malfunction.

Regarding Is1, Is2 it is recommended to apply the same primary values because the calculated differential and bias currents are the same at all ends.

For Is1 CTS, the same applies, however in the case of three-end line we should consider that the normal load current at all ends which are not the same.