Siemens SINAMICS DCM Operating Instructions Manual page 65

Control module for variable-speed dc drives
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U
v0
I
is the DC current of the operating point under investigation in A,
d
f
is the line frequency in Hz,
N
L
is the inductance of the line reactor used in H,
D
X
is the line reactor impedance,
D
X
is the network impedance, and
N
X
is the impedance at the unit terminals
K
II.) Armature inductance La
L
= 0.0488 × U
a
If the actual values for the short-circuit power S
the values calculated using the formulae above, a separate calculation procedure will need
to be carried out.
Example:
Let us take a drive with the following data:
U
v0
I
= 150 A
d
f
= 50 Hz
N
L
= 0.169 mH (4EU2421-7AA10 with I
D
Where
X
N
the network short-circuit power at the converter unit terminal is as follows:
S
K
and the armature inductance of the motor is as follows:
L
= 0.0488 × 400 / (50 × 150) = 2.60 mH
a
The harmonic currents I
can be taken from the tables only apply for the values S
method. If the actual values deviate from these, a separate calculation procedure will need to
be carried out.
For the purpose of dimensioning filters and compensation equipment with reactors, it is only
possible to draw on the information provided by the harmonics values calculated in this way
if the values calculated for S
separate calculation procedure needs to be carried out (this particularly applies when using
machines with compensation, as they demonstrate a very low armature inductance level).
SINAMICS DCM Control Module
Operating Instructions, 06.2010, C98130-A7067-A1-02-7619
is the no-load voltage at the converter unit terminal in V,
/ (f
× I
v0
N
d
= 400 V
= 0.03536 × 400 / 150 - 2π × 50 × 0.169 × 10
= 400
/ 0.0412 = 3.88 MVA
2
(where I
ν
6.1 Instructions for EMC-compliant drive installation
) (H)
K
= 125 A)
Ln
= g × 0.817 × I
1
and L
match the actual drive values. In all other cases, a
K
a
and/or armature inductance L
= 0.0412 Ω,
-3
for firing angles α = 20° and α = 60°) that
d
and L
calculated using this
K
a
Connecting
deviate from
a
65

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