Feeders With In-Zone Transformers - GE MiCOM P40 Agile Technical Manual

Single breaker current differential (with distance)
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Chapter 6 - Current Differential Protection
10

FEEDERS WITH IN-ZONE TRANSFORMERS

A transformer feeder comprises a transformer directly connected to a transmission circuit without the intervention
of switchgear. Although separate current transformer inputs may be available to allow separate overlapping zones
of protection for the transformer and the feeder this is not always the case and the transformer and the feeder
must be treated and protected as a single item of plant for protection. The integration of in-zone power
transformers into unit feeders causes further complications for current differential protection:
Phase shift and possible imbalance of signals across the transformer
Transfer of earth fault current by the transformer
Magnetising Inrush current
Overfluxing
This product therefore provides facilities to deal with the above issues by providing:
CT compensation for phase shifts and imbalances across the transformer
Zero sequence compensation.
Magnetising Inrush detection and restraint
Second harmonic blocking.
Overfluxing detection and restraint
Fifth harmonic blocking.
To make these available you need to enable the differential protection for an in-zone power transformer. This is
done on a per-setting group basis. The phase differential element (Phase Diff) for the setting group concerned
must be enabled.
To enable the transformer feeder protection you set the Compensation cell in the CURRENT DIFF settings to
Transformer.
10.1
CT PHASE CORRECTION
To compensate for any phase shift between two windings of a transformer, it is necessary to provide phase
correction. Phase correction is provided by specifying the vector group using the Vectorial comp setting in the
CURRENT DIFF column.
The phase compensation options are listed in the table below:
Setting
Yy0
0
Yd1
30 lag
Yy2
60 lag
Yd3
90 lag
Yy4
120 lag
Yd5
150 lag
114
Phase shift
Do nothing
Ia = (IA - IC) / √3
Ib = (IB - IA) / √3
Ic = (IC - IB) / √3
Ia = -IC
Ib = -IA
Ic = -IB
Ia = (IB - IC) / √3
Ib = (IC - IA) / √3
Ic = (IA - IB) / √3
Ia = IB
Ib = IC
Ic = IA
Yd11 and Invert
P543i/P545i
Action
P54x1i-TM-EN-1

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