line for the case when the fault infeed from remote end is zero, we can draw the
voltage V in the faulty phase at A side as in equation 38.
( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN-US (Equation 38)
Notice that the following relationship exists between the zero sequence currents:
3 0 3 0 0 2
0
I Z I Z p
L p L
⋅ = ⋅ −
( )
EQUATION1279 V3 EN-US (Equation 39)
Simplification of equation 39, solving it for 3I0p and substitution of the result into
equation 38 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 40)
If we finally divide equation 40 with equation 35 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 41)
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 infeed in the line 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 under-reach scheme.
Parallel line out of service and grounded
SEMOD168232-227 v3SEMOD168232-229 v3
1MRK 504 163-UUS A Section 8
Impedance protection
Transformer protection RET670 2.2 ANSI 199
Application manual
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