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
115.
( )
0 0
p 3 3= 1 +
A L ph N Nm p
V Z I K I K I× × + ×
EQUATION1278 V4 EN (Equation 300)
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 (Equation 301)
Simplification of equation 116, solving it for 3I0
p
and substitution of the result into
equation 115 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 (Equation 302)
If we finally divide equation 117 with equation 112 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
(Equation 303)
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%.
Section 8 1MRK 506 369-UUS -
Impedance protection
366 Line distance protection REL670 2.2 ANSI
Application manual
To Next Page
Loading...
