Differential protection

[Calinic]

Hello,

First of all, excuse my English. I'm a young commisioning engineer and I have a situation on site related to transformer differential protection. The question is reffering to CTs star point connection (both set of CTs). Is that true if the primary position of CTs on HV side is P1 - toward to the breaker P2 - toward to the transformer and on LV side P2 toward to the transformer and P1 toward to the LV busbar, both star point connection must be in oposition? I.e, star point of HV CTs toward to the HV busbar and the star point of LV CTs toward to the LV busbar?
What's happen if both set of CTs are connected as follow: P1 toward to the HV busbar, P2 toward to the transformer for HV side and, on LV side, P1 toward to the transformer and P2 toward to the LV busbar? I guess that the star point in this case can be either toward to the HV side or LV , side, but the most important thing is that both star point have to be arranged in the same direction.

Thanks a lot


Costa

 

[jghrist]

If the differential relay is a microprocessor type, there is considerable flexibility in CT connections. Arranging the CTs in different direction can be compensated by relay settings. Also, if you put P1 toward the transformer and connected the P1 sides together for the neutral, this would be functionally the same as putting P2 toward the transformer and connecting the P2 sides together. If the relay is electromechanical, then unless your transformer is delta-delta, at least one set of CTs must be connected delta, not wye.

It is certainly preferred to connect the CTs as you have indicated, with P1 away from the transformer on each side. Things are easier to understand that way, even if they can be made to work otherwise. However they are connected, make sure that you understand the relative current directions and phase angles when setting the relay.

 

[stevenal]

Usually the CTs are arranged so that the polarity marks face away from the protected transformer, and an internal fault will cause CT secondary current to flow out of the polarity end of the affected CTs. Usually the non-polarity ends of the CT secondaries are connected together to make your star points.

But, there are many connections that will work and even more that won't. Changing the star point to the polarity end on one or both sets of CTs won't make a difference if polarity remains the same. Changing primary polarity (turning the CT around) won't make a difference if secondary polarity is also reversed.

Draw it out. Even if modern relays are used, draw it with imaginary restraint and operate electromechanical coils. A through fault should cause the current to flow only through the restraint coils. Your configuration will work with the proper connection to the relay.

 

[Calinic]

To jhrist:

That's true: the relay is a numerical one, P633 manufactured by Alstom (MiCOM range) with the capability to compensate a possible wrong connection on CTs by settings. The protected transformer has an Yny0yo vector group and even 'inside' of the relay is not neccesary to compensate the phase angle difference, only voltage amplitude (so, no need interposing CTs). I couldn't see any traditional restraint and operating coil in the manual, only dedicated current inputs for three windings applications, the restrain the differential currents being calculeted by the relay itself. I drow out the scheme and I understood now what's the deal with CTs involved in differential schemes.
One more question regarding settings: is I> = 40%Iref to high if we take into account that the power transformer has a tap changer?

Thanks for help

Costa


To Stevenal

I followed your advise to treat the relay as a classical one, with restrain and operating coils and I understood very well the problem. One more question regarding settings: is I> = 40%Iref to high if we take into account that the power transformer has a tap changer?


Thanks for your help,

Costa

 

[jghrist]

I'm not familiar with the Altsom MiCOM. What I are you referring to being > Iref? Is Iref the nominal current, i.e. 1A or 5A?

With a Yny0y0 transformer, you may have to take special consideration of zero sequence currents using wye connected CTs. For the SEL relay, this is done by setting the CT connection compensation for 360° phase difference. Same as 0° as far as phasing goes, but removes zero-sequence components. Check the instruction manual for the MiCOM.

Transformer taps are usually assumed at neutral tap and the slope is set high enough to prevent operation for the maximum ratio difference.

 

[stevenal]

Are you refering to the slope, Iop/Ires? I usually set it 25 to 30% for a transformer with a +/-10% LTC. This is lower than many would set it, but no false trips yet. 40% would be more secure. Good info at selinc.com. Maybe at Alstom also.

 

[Calinic]

I'll give you all details:

The power transformer is: 220/6.6/6.6Kv, 48/28/28MVA, Yny0yo vector group.
CT HV side: 150/5
CT both MV side: 3000/5

The user's guide states: the relay itself calculates Idiff and Irestrain, based on Iref.

I ref, is calculated by the relay using the formula:
Iref, end a (220kv side) = Sref/root3*Vnom, end a = 126 A
Iref, end b (6.6kv side) = Sref/root3*vnom, end b = 4199 A
Iref, end c (6.6kv side) = Sref/root3*vnom, end c = 4199 A

After this figure, the relay calculates the amplitude matching factor.
In this case:
kam, end a = Primary CT rated current, end a/ I ref, end a = 1.191
kam, end b =Primary CT rated current, end b/ I ref, end b = 0.714
kam, end c =Primary CT rated current, end c/ I ref, end c = 0.714

The relay settings are:
Idiff> = 0.34*Iref
Idiff>> = 15*Iref
Idiff>>> = 34*Iref

When I tested the relay, I obtained the following results:

1. 220kv side injection: current injected: 2 A, relay tripped.
Idiff measured by the relay: 0.320Iref;
Irestr measured by the relay: 0.160Iref;

2. 6.6kv side injection: current injected: 3.5 A, relay tripped.
Idiff measured by the relay: 0.320Iref;
Irestr measured by the relay: 0.160Iref;

Thanks

Costa

 

[jbartos]

Suggestion: Visit

http://www.iee.org/OnComms/Branches/UK/England/NorthE/Mersey/Events/docs/MicomProtection.PDF

http://www.alstom.com.cn/en/jv/jv_Products.asp?name=enalstom006

http://www.alstom.com.cn/en/jv/jv_Products2.asp?id=40

http://www.tde.alstom.com/servlet/C...typid=1027433671895&productline=1018253756139

etc. for more info. The more responsible relay settings might be accomplished via tech support and training.

 

[jghrist]

Calinic,

The power transformer is: 220/6.6/6.6Kv, 48/28/28MVA.
You say:
Iref, end b (6.6kv side) = Sref/root3*vnom, end b = 4199 A
but 28000/(1.732·6.6) = 2449

 

[Calinic]

To jhrist:

Yes, but the MiCOM P633 manual states that all Iref currents must be calculated using the same Sref and they recomend also to use as Sref the highest value, in this case 48MVA. Therefore the Iref current on LV side are higher. It's strange for me but this is exactley what the manual states.

Costa