Siemens siprotec SJ62 User Manual page 132

Sensitivity with
High-impedance
Protection
SIPROTEC 4, 7SJ62/63/64 Handbuch
C53000-G1140-C147-A, Edition 07.2015
The voltage across R is then
V = I · ( 2R + R )
R 1 a2 i2
It is assumed that the pickup value of the 7SJ62/63/64 corresponds to half the knee-
point voltage of the current transformers. In the balanced case results
V = V / 2
R KPV
This results in a stability limit I
the scheme remains stable:
Calculation Example:
For the 5-A CT as above with V
longest CT connection lead 22 m (24.06 yd) with 4 mm
= 0.1 Ω
sponds to R
a
that is 15 × rated current or 12 kA primary.
For 1-A CT as above with V
longest CT connection lead 107 m (117.02 yd) with 2.5 mm
= 0.75 Ω
R
a
that is 27 × rated current or 21.6 kA primary.
The voltage present at the CT set is forwarded to the protective relay across a series
resistor R as proportional current for evaluation. The following considerations are rel-
evant for dimensioning the resistor:
As already mentioned, it is desired that the high-impedance protection should pick up
at half the knee-point voltage of the CT's. The resistor R can calculated on this basis.
Since the device measures the current flowing through the resistor, resistor and mea-
suring input of the device must be connected in series. Since, furthermore, the resis-
tance shall be high-resistance (condition: R >> 2R
inherent resistance of the measuring input can be neglected. The resistance is then
calculated from the pickup current I
2.5 Single-Phase Overcurrent Protection
, i.e. the maximum through-fault current below which
SL
= 0.3 Ω
= 75 V and R
KPV i
= 5 Ω
= 350 V and R
KPV i
a2
and the half knee-point voltage:
pu
2
cross-section; this corre-
2
cross-section, results in
+ R , as above mentioned), the
i2
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