Example
The nominal voltage of the drive in this calculation is 690 V.
Motor
U
=
3300 V
n
I
=
106 A
n
f
=
50 Hz
n
(
ϕ
)
cos
=
0,82
n
------------------- - Ω 17.97 Ω
Z
=
m
106
⇒
17,97 0,82 Ω 14,74 Ω
R
=
m
X
=
17,97
m
Cable
Length = 5 km
Ω
------- -
R
=
0,27
c
km
mH
-------- -
L
=
0,33
c
km
Secondary side
(
Z
=
1,35
+
1
Current for the secondary side:
I
=
I
=
106 A
2
n
Thus, the minimum voltage for the secondary side equals
3 106 A 19.38 Ω 3558 V
⋅
U
=
2
The secondary voltage of the transformer is thus chosen as follows
U
=
3560 V
N2
Primary side
The primary voltage of the transformer is
⋅
U
=
0,9 690 V
N1
3300
≈
⋅
3
⋅
≈
2
⋅
Ω 10,29 Ω
1 0,82
–
⇒
5 0,27 Ω
⋅
R
=
c
2 π 50 0,33 10
⋅
⇒
L
=
c
2
)
(
14,74
+
0,52
+
10,29
⋅
≈
≈
620 V
≈
≈
1,35 Ω
3 –
⋅
⋅
⋅
≈
0,52 Ω
2
)
Ω 19,38 Ω
≈
Appendix – Step-up applications
37