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Functions
6-20
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 insignifi-
cant amount. With cables however, significant angle differences may exist as illustrat-
ed 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) matching factor K
Z
0.632
j(18.4°–73°)
0
⋅
----- -
-------------- - e
=
Z
0.408
1
æ
ö
Z
1
0
⋅
ç
÷
K
-- -
----- - 1
=
–
=
0
3
è
Z
ø
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 fol-
lowing 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:
æ
1.263
ϕ K
(
)
arc tan
-------------- -
=
è
0
0.102
The magnitude and angle of the earth impedance (residual) matching factors setting
for the first zone Z1 and the remaining zones of the distance protection may be differ-
ent. 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 1120 K0 (Z1) and 1121 Angle
K0(Z1) are determined with the data of the protected line while the addresses 1122
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°
ö
180°
94.6°
–
=
–
ø
2
Cu with the following data
results in:
0
⋅
(
)
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°
C53000-G1176-C156-2
)
7SA6 Manual
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