Schweitzer Engineering Laboratories SEL-700G Series, Instruction Manual, Page: 124 / 824

4.18
SEL-700G Relay Instruction Manual Date Code 20170814
Protection and Logic Functions
Group Settings (SET Command)
phase sequence will be rotated in a counterclockwise direction when
multiplied by [CTC(m)].
If a balanced set of currents with ACB phase rotation undergoes the same
exercise, the rotations by the [CTC(m)] matrices are in the clockwise
direction. This is because the compensation matrices, when performing phasor
addition or subtraction involving B or C phases, will produce “mirror image”
shifts relative to A-phase, when ACB phase rotation is used instead of ABC.
In ACB phase rotation the three phases still rotate in a counterclockwise
direction, but C-phase is in the 120-degree lagging position and B-phase leads
by 120 degrees, relative to A-phase.
The discussions below assume ABC phase rotation, unless mentioned
otherwise.
The “0” setting value is intended to create no changes at all in the currents and
merely multiplies them by an identity matrix. Thus, for CTCn = 0,
that is,
The “1” setting performs a 30-degree compensation in the counterclockwise
direction, as would a delta CT connection of type DAB (30-degree leading).
The name for this connection comes from the fact that the polarity end of the
A phase CT connects to the nonpolarity end of the B phase CT, and so on, in
forming the delta. Thus, for CTCn = 1, the relay uses the following [CTC(m)]
matrix:
that is,
The “11” setting performs a 330-degree compensation (11 • 30) in the
counterclockwise direction, or a 30-degree compensation in the clockwise
direction, as would a delta CT connection of type DAC (30-degree lagging).
The name for this connection comes from the fact that the polarity end of the
A phase CT connects to the nonpolarity end of the C phase CT, and so on, in
I1nC = IAn
I2nC = IBn
I3nC = ICn
CTC 0
   
1 0 0
0 1 0
0 0 1
=
CTC 1
   
1
3
-------
1 –1 0
0 1 –1
1 0 1 –

=
I1 n C
IA n IB n –
 
3
------------------------------ =
I2 n C
IB n IC n –
 
3
------------------------------ =
I3 n C
IC n IA n –
 
3
------------------------------ =