ABB REL670 2.2 IEC Applications Manual page 306

Section 8
Impedance protection
300
æ
Z0
L
ç
= × + ×
U Z1 I 3I
L ph 0
ph è
×
3 Z1
EQUATION2312 V1 EN-US
By dividing equation 237
write the impedance present to the relay at A side as:
æ
×
3 0 I
= +
Z Z 1 1
ç
L
+
I ph 3 0
è
EQUATION1277 V3 EN-US
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 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-earth 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 U
(
U = × + × +
p Z1 I K 3I
L ph N 0
A
EQUATION2313 V1 EN-US
One can also notice that the following relationship exists between the zero
sequence currents:
3 ⋅ 0 = 3 0 ⋅
I Z I Z
0 L p
EQUATION1279 V3 EN-US
Simplification of equation 240, solving it for 3I0p and substitution of the result into
equation 239 gives that the voltage can be drawn as:
ö
-
Z1 Z0
L m
÷
+
3I
0p
ø
×
3 Z1
L L
by equation 236 and after some simplification we can
ö
KNm
÷
×
I KN
ø
in the faulty phase at A side as in equation 239.
A
)
×
K 3I
Nm 0p
( )
0 2 −
p
L
1MRK 506 369-UEN B
(Equation 237)
(Equation 238)
(Equation 239)
(Equation 240)
Line distance protection REL670 2.2 IEC
Application manual