Siemens siprotec 7SA6 User Manual page 201

7SA6 Manual
C53000-G1176-C133-1
These values may either apply to the entire line length or be based on a per unit of line
length, as the quotients are independent of length. Furthermore it makes no difference
if the quotients are calculated with primary or secondary values.
For overhead lines it is generally possible to calculate with scalar quantities as the
angle of the zero sequence and positive sequence system only differ by an
insignificant amount. With cables however, significant angle differences may exist as
illustrated by the following example.
Calculation example:
110 kV single conductor oil-filled cable 3×185 mm
Ω/km
j73°
Z /s = 0.408·e
1
j18,4°
Z /s = 0.632·e
0
(where s = line length)
The calculation of the earth impedance (residual) compensation factor K
Z
0.632
j(18.4°–73°)
0
⋅
----- - -------------- - e
=
Z 0.408
1
 
Z
1
0
⋅
 
K -- - ----- - 1
= – =
0
3  
Z
1
The magnitude of K0 is therefore
1
2
⋅ ( )
K -- - 0.102
= –
0
3
When determining the angle, the quadrant of the result must be considered. The
following table indicates the quadrant and range of the angle which is determined by
the signs of the calculated real and imaginary part of K
Table 6-1 Quadrants and range of the angle of K
Imaginary
Real part
part
+ +
+ –
– –
In this example the following result is obtained:

1.263
ϕ K ( )
arc tan -------------- -
=

0
0.102
The magnitude and angle of the earth impedance (residual) compensation factors
setting for the first zone Z1 and the remaining zones of the distance protection may be
different. This allows to set the exact values for the protected line, while at the same
time the setting for the back-up zones may be a close approximate even when the
following lines have substantially different earth impedance ratios (e.g. cable after an
overhead line). Accordingly, the settings for the address  . = and 
$QJOH . = are determined with the data of the protected line while the addresses
positive sequence impedance
Ω/km
zero sequence impedance
–j54.6°
1.55 e ⋅
= =
= 0.898 – j1.263
1
⋅ ( )
-- - 0.898 j1.263 1
– –
3
2
( )
1.263 0.42
+ – =
tan ϕ(K0)
Quadrant/Range
+ I
– IV
+ III –90° ... –180°

180° 94.6°
– = –

2
Cu with the following data
⋅ ( )
1.55 0.579 j0.815
–
1
⋅ (
-- - 0.102 j1.263
= – –
3
.
0
0
Rules for calculation
0° ... +90° arctan(|Im|/|Re|)
–90° ... 0° –arctan(|Im|/|Re|)
arctan(|Im|/|Re|) – 180°
Functions
results in:
0
)
6-19