Injected line-to-line voltages
Ö3
150°
U
23
Figure 5.4 – Example of symmetric component calculation using line-to-line voltages
Unscaling the geometric results gives
U
= 100/√3 x 2/3 = 38.5 %
1
U
= 100/√3 x 1/3 = 19.2 %
2
U
/U
= 1/3:2/3 = 50 %
2
1
Example 3 – Two-phase injection with adjustable phase angle
U
= 100 V
N
Voltage measurement mode is "3LN".
Injection:
= 100/√3 V ∠0° = 57.7 V ∠0°
U
= U
a
L1
= 100/√3 V ∠ 1 20° = 57.7 V ∠ 1 20°
U
= U
b
L2
U
= U
= 0 V
c
L3
This is actually identical case with example 2 because the
resulting line-to-line voltages U
U
= U
– U
= U
23
L2
L3
example 2. The only difference is a +30° phase angle difference,
but without any absolute angle reference this phase angle
difference is not seen by the device.
⎡
⎢
⎡
⎤
⎡
⎤
1
1
1
U
⎢
0
⎢
⎥
⎢
⎥
1
100
⎢
=
2
U
1
a
a
⎢
⎥
⎢
⎥
⎢
1
3
⎢
⎥
⎢
⎥
2
⎢
⎣
⎦
⎣
⎦
U
1
a
a
2
⎢
⎣
∠
−
°
⎡
⎤
⎡
100
60
19
⎢
⎥
⎢
1
=
∠
°
=
200
0
⎢
⎥
⎢
3
3
⎢
⎥
⎢
∠
°
⎣
⎦
⎣
100
60
19
U
= 19.2 %
0
U
= 38.5 %
1
U
= 19.2 %
2
U
/U
= 50 %
2
1
2
a U
23
120°
U
12
120°
aU
U
23
23
= U
– U
12
L1
= 100/√3 V∠ 1 20° are the same as in
L2
⎤
100
∠
°
0
⎥
⎡
3
100
⎥
⎢
1
⎥
∠
−
°
=
120
100
⎢
⎥
3
3
3
⎢
⎥
⎣
100
0
⎥
⎦
∠
−
°
⎤
2 .
60
⎥
∠
°
38
5 .
0
⎥
⎥
∠
+
°
⎦
2 .
60
5-11
Measurement Functions
FortescueEx2
Positive sequence
U = 2/3
1
Ö3
U
12
2
U -a U
-a U
23
12
U
12
Negative sequence
-aU
U -aU
23
12
23
Ö3
U
12
U = 1/3
2
= 100 V ∠30° and
L2
∠
°
+
∠
−
°
⎤
0
100
120
⎥
∠
°
+
∠
°
=
0
100
0
⎥
⎥
∠
°
+
∠
+
°
⎦
0
100
120
857-UM001A-EN-P – July 2009
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23