Simplification of equation 86, solving it for 3I0
p
and substitution of the result into
equation 85 gives that the voltage can be drawn as:
0
0
3 p
p 3
2 p
1
A L ph N Nm
I
V Z I K I K
×
= × + × + ×
-
æ ö
ç ÷
è ø
EQUATION1280 V2 EN-US (Equation 87)
If we finally divide equation 87 with equation 82 we can draw the impedance present
to the IED as
Z p ZI
I KN I KN
I p
p
I I KN
L
ph m
ph
= ⋅
+ ⋅ + ⋅
⋅
−
+ ⋅
3
3
2
3
0
0
0
EQUATION1379 V3 EN-US (Equation 88)
Calculation for a 400 kV line, where we for simplicity have excluded the resistance,
gives with X1L=0.48 Ohm/Mile, X0L=1.4Ohms/Mile, zone 1 reach is set to 90% of
the line reactance p=71% that is, the protection is underreaching with approximately
20%.
The zero sequence mutual coupling can reduce the reach of distance protection on the
protected circuit when the parallel line is in normal operation. The reduction of the
reach is most pronounced with no current infeed in the IED closest to the fault. This
reach reduction is normally less than 15%. But when the reach is reduced at one line
end, it is proportionally increased at the opposite line end. So this 15% reach reduction
does not significantly af
fect the operation of a permissive underreaching scheme.
Parallel line out of service and grounded
GUID-DFB5C5AD-407D-4325-BE46-092F5F5D2F59 v1GUID-A93E0062-A4E1-447B-A3A3-74B49581AD70 v1
Z0m
A B
21
21
en05000222_ansi.vsd
CLOSED
CLOSED
OPEN OPEN
ANSI05000222 V1 EN-US
Figure 101: The parallel line is out of service and grounded
1MRK 502 071-UUS A Section 8
Impedance protection
Generator protection REG670 2.2 ANSI and Injection equipment REX060, REX061, REX062 239
Application manual
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