Siemens siprotec 7SA522 User Manual page 288

Functions
6-140
s (length) = 60 km
= 0,19 Ω/km
R /s
1
= 0,42 Ω/km
X /s
1
Short circuit power at the beginning of the line:
S ' = 2,5 GVA
k
current transformer 600 A/5 A
The line impedance Z and source impedance Z
L
follows:
/s = √0.19 2
Z + 0.42
1
= 0.46 Ω/km · 60 km
Z
L
The three phase short circuit current at the end of the line is I
=
I ------------------------------------- -
F end
3
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:
Setting value I>>
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).
2 Ω/km = 0.46 Ω/km
= 27.66 Ω
2 2
110 kV
Z =
----------------------------- -
U
2500 MVA
⋅
1.1 U
1.1 110 kV
N
=
--------------------------------------------------------------- -
⋅ ( ) ⋅ ( 4.84 Ω
Z Z 3
+
V L
2150 A
⋅
1.1 ------------------ - 5 A
=
600 A
are calculated with these values as
U
4.84 Ω
=
:
F end
⋅
= 2150 A
27.66 Ω )
+
⋅
19.7 A
=
C53000-G1176-C119-2
7SA522 Manual