Siemens 6RA7086-6KS22 Operating Instructions Manual page 102

Connections
where
= −
X X
N K
and
No-load voltage in V at the converter supply connection point
U v0
DC current in A of operating point under investigation
I d
Line frequency in Hz
f N
Inductance in H of commutating reactor used
L D
Impedance of the commutating reactor
X
D
Impedance of the network
X
N
Impedance at the converter terminals
X
K
II.) Armature inductance L a
A separate calculation must be performed if the actual values for short-circuit power S K and/or
armature inductance L a deviate from the values calculated on the basis of the above equations.
Example
Let us assume that a drive has the following data:
U v0 = 400 V
I d = 150 A
f N = 50 Hz
L D = 0.169 mH (4EU2421-7AA10 where I Ln = 125 A)
When
=
X . 0 03536
N
the required system short-circuit power at the converter supply connection point is as follows:
2
400
=
S
K
. 0 0412
and the required motor armature inductance as follows:
=
L . 0 0488
a
The harmonic currents I ν listed in the tables above (where I 1 = g x 0.817 x I d for firing angles α =
20° and α = 60°) apply only to the values S K and L a calculated by the above method. If the
calculated and actual values are not the same, the harmonics must be calculated separately.
For the purpose of dimensioning filters and compensation circuits with reactors, the harmonic
values calculated by these equations can be applied only if the values calculated for S
with the actual values of the drive. If they do not, they must be calculated separately (this is
especially true when using compensated motors as these have a very low armature inductance).
6-16
U
= ×
v 0
X . 0 03536
D
I
d
=
L . 0 0488
a
400
π
× − × ×
2 50 . 0
150
=
. 3 88 MVA
400
× =
. 2 60 mH
×
50 150
( )
π
− × Ω
2 f L
N D
U ( )
×
v 0
H
I ×
f
d
N
× − =
3
169 10 . 0 0412
Ω
Siemens AG
SIMOREG DC Master
05.05
and L tally
K a
6RX1700-0AD76
Operating Instructions