( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN (Equation 115)
One can also 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 (Equation 116)
Simplification of equation 116, solving it for 3I0p and substitution of the result into
equation 115 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 (Equation 117)
If we finally divide equation 117 with equation 112 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 (Equation 118)
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 affect the operation of a permissive underreaching scheme.
Parallel line out of service and grounded
1MRK 506 369-UUS - Section 8
Impedance protection
Line distance protection REL670 2.2 ANSI 235
Application manual
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