ABB RELION RET670 Applications Manual page 205

1MRK 504 163-UUS A
Transformer protection RET670 2.2 ANSI
Application manual
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 38.
(
V = Z 1 I + K
×
p
A L ph N
EQUATION1278 V4 EN-US
Notice that the following relationship exists between the zero sequence currents:
3 I Z 0 3 0 I Z
⋅ = ⋅
0 L p
EQUATION1279 V3 EN-US
Simplification of equation 39, solving it for 3I0p and substitution of the result into
equation 38 gives that the voltage can be drawn as:
æ
V Z 1 ç I
= × +
p
A L è ph
EQUATION1280 V2 EN-US
If we finally divide equation
to the IED as

I KN
+

ph

Z p ZI
= ⋅
L



EQUATION1379 V3 EN-US
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%.
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 infeed in the line 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 under-reach scheme.
Parallel line out of service and grounded
)
I K I
× + ×
3 3
0 Nm 0 p
0 2 ( p )
−
L
I
3
K I K 0
× + ×
3
N 0 Nm
-
2
40 with equation 35
I p
3
⋅
0
3 I KN
⋅ + ⋅
0 m
p
2
−
I 3 I KN
+ ⋅
ph 0
Impedance protection
ö
×
p
÷
ø
p
we can draw the impedance present






Section 8
(Equation 38)
(Equation 39)
(Equation 40)
(Equation 41)
SEMOD168232-229 v3 SEMOD168232-227 v3
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