0 0
0
0 1
1 3 3
3 1 3 1
m
L L
ph p
ph L
L L
Z
Z Z
V Z I I I
Z Z
æ ö
-
= × + × +
ç ÷
× ×
è ø
EQUATION1276 V4 EN-US (Equation 28)
By dividing equation 28 by equation 27 and after some simplification we can write
the impedance present to the IED at A side as:
3 0
1 1
3 0
æ ö
×
= +
ç ÷
+ ×
è ø
L
I KNm
Z Z
I ph I KN
EQUATION1277 V3 EN-US (Equation 29)
Where:
KNm = Z0m/(3 · Z1L)
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 overreaches. If
the currents have the same direction, the distance protection underreaches.
Maximum overreach occurs if the fault infeed from remote end is weak. If we
consider a single phase-to-earth fault at "p" unit of the line length from A to B on
the parallel 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
30.
( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN-US
(Equation 30)
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 31)
Simplification of equation 31, solving it for 3I0p and substitution of the result into
equation
30 gives that the voltage can be drawn as:
1MRK 506 369-UEN B Section 8
Impedance protection
Line distance protection REL670 2.2 IEC 155
Application manual
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