Example
This example uses a class 0.2 voltage transformer with a rated burden of 45 VA. The following data was taken
from the measuring report.
Table 6-18 For Phase A
V/Vn S
b
= S
br
S
b
= S
br
/4
ε
v
[%] δ
v
[min] ε
v
[%] δ
v
[min]
0.8 -0.14 0.31 0.16 -0.34
1 -0.15 0.43 0.15 -0.24
1.2 -0.16 0.68 0.14 -0.06
The values resulting for the phases B and C are almost identical.
Furthermore, a termination with a rated burden (other measuring devices are still connected) is assumed so
that an angle error of +0.43 min (rated-voltage value) is used as the calculation variable.
A current transformer of type 5PR is used here. This current transformer features a gap that limits the rema-
nence to 10 %. However, this gap results in larger angle errors. The following tables show excerpts from the
test reports.
Table 6-19 Phase A
I/I
n
[%] Sb = S
br
(cos β = 0.8)
ε
i
[%] δ
i
[min]
100 0.314 46.40
Table 6-20 Phase B
I/I
n
[%] Sb = S
br
(cos β = 0.8)
ε
i
[%] δ
i
[min]
100 0.247 35.10
Table 6-21 Phase C
I/I
n
[%] Sb = S
br
(cos β = 0.8)
ε
i
[%] δ
i
[min]
100 0.702 41.10
Since the positive-sequence system power is evaluated in the device, take the sign into account when you add
the angle errors per phase, and divide the result by 3.
The following value results in this example:
The resulting correction angle is:
Set the parameter (_:2311:101) Angle correction = -0.67 °.
Protection and Automation Functions
6.38 Reverse-Power Protection
812 SIPROTEC 5, Overcurrent Protection, Manual
C53000-G5040-C017-8, Edition 07.2017
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