Siemens siprotec 7SA522 User Manual page 199

Resistance Margin
7SA522 Manual
C53000-G1176-C119-2
Accordingly, the reach for any distance zone can be specified as follows:
where
N — is the transformation ratio of the current transformers
CT
N — is the transformation ratio of the voltage transformers
VT
Calculation example:
110 kV overhead line 150 mm
s (length) = 35 km
= 0.19 Ω/km
R /s
1
= 0.42 Ω/km
X /s
1
= 0.53 Ω/km
R /s
0
= 1.19 Ω/km
X /s
0
Current transformers 600 A/5 A
Voltage transformers110 kV/0,1 kV
The line data is calculated with these values as follows:
= 0.19 Ω/km · 35 km = 6.65 Ω
R
L
= 0.42 Ω/km · 35 km = 14.70 Ω
X
L
The first zone should be set to 85 % of the line length; the result is
primary:
= 0.85 · 14.70 Ω = 12.49 Ω
X1 = 0.85 · X
prim L
or secondary:
N
CT
⋅
---------- X1
X1 =
sec prim
N
VT
The resistance setting R allows a margin for fault resistance which appears as an ad-
ditional resistance at the fault location and is added to the impedance of the line con-
ductors. It comprises, for example, the resistance in arcs, the earth distribution resist-
ance of earth points and others. The setting must allow for these fault resistance, but
should at the same time not be larger than necessary. On long heavily loaded lines,
the setting may extend into the load impedance range. Fault detection due to overload
conditions is then prevented with the load trapezoid. Refer to the margin heading
"Load Area" in section 6.2.2.2. The resistance margin setting may be separately set
for the phase–phase faults on the one hand and the phase–earth faults on the other
hand. It is therefore possible to allow for a larger fault resistance for earth faults for
example.
Most important for this setting on overhead lines, is the resistance of the fault arc. In
cables on the other hand, an appreciable arc can not exist, and the resistance of the
cable itself is decisive for this setting. On very short cables, care should however be
taken that an arc fault on the local cable termination is inside the set resistance of the
first zone.
N
CT
⋅
X ---------- X
=
sec prim
N
VT
2
with the following data:
600 A/5 A
------------------------------------- - 12.49 Ω ⋅
=
110 kV/0.1 kV
Functions
1.36 Ω
=
6-51