The second part in the parentheses is the error introduced to the measurement of the
line impedance.
If the current on the parallel line has negative sign compared to the current on the
protected line, that is, the current on the parallel line has an opposite direction
compared to the current on the protected line, the distance function will overreach. If
the currents have the same direction, the distance protection will underreach.
Maximum overreach will occur if the fault current infeed from remote line end is
weak. If considering a single phase-to-ground fault at 'p' unit of the line length from A
to B on the parallel line for the case when the fault current infeed from remote line end
is zero, the voltage
V
A
in the faulty phase at A side as in equation
41.
( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN-US (Equation 41)
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-US (Equation 42)
Simplification of equation 42, solving it for 3I0
p
and substitution of the result into
equation 41 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 43)
If we finally divide equation 43 with equation 38 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 44)
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%.
Section 8 1MRK 502 071-UUS A
Impedance protection
210 Generator protection REG670 2.2 ANSI and Injection equipment REX060, REX061, REX062
Application manual
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