ABB RELION 670 Series Applications Manual page 319

1MRK 504 152-UEN B
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
the faulty phase at A side as in equation 322.
U = ⋅ p ZI
A
IECEQUATION1278 V2 EN-US
One can also notice that the following relationship exists between the zero sequence currents:
3 ⋅ 0
I Z
0
EQUATION1279 V3 EN-US
Simplification of equation 323, solving it for 3I0
322 gives that the voltage can be drawn as:
= ⋅
U p ZI
A
IECEQUATION1280 V2 EN-US
If we finally divide equation
IED as
Z = ⋅ p ZI
EQUATION1379 V3 EN-US
Calculation for a 400 kV line, where we for simplicity have excluded the resistance, gives with
X1L=0.303 Ω/km, X0L=0.88 Ω/km, 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 earthed
Application manual
(
I + K ⋅ 3 I + K
L ph N 0
( )
= 3 0 ⋅ 0 2 −
I Z p
L p L

+ ⋅ 3 +
I K I K

L ph N 0

324

I + KN ⋅ 3 I + KN

ph 0

L

+ 3 ⋅
I I KN
ph 0


)
⋅ 3 I
Nm 0 p
and substitution of the result into equation
p

3 I ⋅ p
0
⋅

Nm
2 − p

with equation 319 we can draw the impedance present to the

3 I ⋅ p
0
⋅

m
2 − p




Section 7
Impedance protection
in
A
(Equation 322)
(Equation 323)
(Equation 324)
(Equation 325)
GUID-DFB5C5AD-407D-4325-BE46-092F5F5D2F59 v1 GUID-A93E0062-A4E1-447B-A3A3-74B49581AD70 v1
313