Directional Phase And Residual Overcurrent Protection - ABB REL650 Applications Manual

Line distance protection relion 650 series
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Section 18
Requirements
18.1.6.5
18.1.7
18.1.7.1
496

Directional phase and residual overcurrent protection

If the directional overcurrent function is used the CTs must have a rated equivalent
secondary e.m.f. E
that is larger than or equal to the required equivalent secondary e.m.f.
al
E
below:
alreq
I
×
³
=
×
E
E
I
s n
k ma x
a l
a lre q
I
pn
EQUATION1681 V1 EN
where:
I
Maximum primary fundamental frequency current for close-in forward and reverse faults (A)
kmax
I
The rated primary CT current (A)
pn
I
The rated secondary CT current (A)
sn
I
The rated current of the protection IED (A)
n
R
The secondary resistance of the CT (W)
CT
R
The resistance of the secondary cable and additional load (W). The loop resistance containing the
L
phase and neutral wires, must be used for faults in solidly grounded systems. The resistance of a
single secondary wire should be used for faults in high impedance grounded systems.
S
The burden of an IED current input channel (VA). S
R
channel for I
=5 A
r
Current transformer requirements for CTs according to other
standards
All kinds of conventional magnetic core CTs are possible to use with the IEDs if they
fulfill the requirements corresponding to the above specified expressed as the rated
equivalent limiting secondary e.m.f. E
different standards and available data for relaying applications it is possible to
approximately calculate a secondary e.m.f. of the CT comparable with E
this with the required rated equivalent limiting secondary e.m.f. E
judge if the CT fulfills the requirements. The requirements according to some other
standards are specified below.
Current transformers according to IEC 61869-2, class P, PR
A CT according to IEC 61869-2 is specified by the secondary limiting e.m.f. E
value of the E
is approximately equal to the corresponding E
alf
æ
ö
S
+
+
R
R
R
ç
÷
CT
L
2
I
è
ø
n
according to the IEC 61869-2 standard. From
al
1MRK 506 334-UUS A
=0.010 VA/channel for I
=1 A and S
r
r
. By comparing
al
it is possible to
alreq
. Therefore, the CTs
al
Application manual
(Equation 184)
=0.250 VA/
r
. The
alf

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