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Siemens siprotec 7SA522 User Manual

Siemens siprotec 7SA522
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Functions
6-140 7SA522 Manual
C53000-G1176-C119-2
s (length) = 60 km
R
1
/s = 0,19 /km
X
1
/s = 0,42 /km
Short circuit power at the beginning of the line:
S
k
= 2,5 GVA
current transformer 600 A/5 A
The line impedance Z
L
and source impedance Z
U
are calculated with these values as
follows:
Z
1
/s = 0.19
2
+ 0.42
2
/km = 0.46 /km
Z
L
=0.46 /km · 60 km = 27.66
The three phase short circuit current at the end of the line is I
F end
:
With a safety margin of 10 % the resultant primary setting value is:
Set value I>> = 1.1 · 2150 A = 2365 A
or the secondary setting value:
i.e. if the short circuit current is greater than 2365 A (primary) or 19.7 A (secondary)
the fault is definitely on the protected feeder. This fault may be cleared immediately by
the overcurrent protection.
Comment
: The calculation was carried out with scalar quantities which is sufficient for
overhead lines. If there is a large difference in the angle of the source and line imped-
ance, the calculation must be done with complex values.
An analogous calculation can be done for earth faults, whereby the maximum earth
fault current that flows during an earth fault at the end of the line is decisive.
The set time delays are pure additional delays, which do not include the operating time
(measuring time).
Z
U
110
2
kV
2
2500 MVA
------------------------------
4.84
==
I
F end
1.1 U
N
3Z
V
(
Z
L
)
+
--------------------------------------
1.1 110 kV
3 4.84
(
27.66
Ω)
+
----------------------------------------------------------------
2150 A
== =
Setting value I>> 1.1
=
2150 A
600 A
-------------------
5A
⋅⋅
19.7 A
=

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Siemens siprotec 7SA522 Specifications

General IconGeneral
BrandSiemens
Modelsiprotec 7SA522
CategoryProtection Device
LanguageEnglish

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