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
211.
( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN-US (Equation 211)
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 212)
Simplification of equation 212, solving it for 3I0p and substitution of the result into
equation 211 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 213)
If we finally divide equation 213 with equation 208 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 214)
Section 8 1MRK 504 163-UUS A
Impedance protection
332 Transformer protection RET670 2.2 ANSI
Application manual
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