Selection
Selection examples
12
Network of several drives
12.5
12.5.5
L
12.5-6
EDS82EV903-1.0-11/2002
The preceding example now uses the 934X supply unit:
Drive data
Controller Motor
Drive Type Power Efficiency η
ηη
η
Drive 1 9330 22 kW 0.91
Drive 2 9325 5.5 kW 0.83
Drive 3 E82EV302K4B 3.0 kW 0.81
Drive 4 E82EV152K4B 1.5 kW 0.78
1. Determining DC-power requirements:
Power loss P
V
from the table ”supply power” . (¶ 12.5-4)
P
DC
=
4
i=1
P
M
i
η
+ P
V
i
P
DC
=
45 kW
0.9
+ 1.1 kW +
5.5 kW
0.83
+ 0.261 kW +
3.0 kW
0.81
+ 0.15 kW +
1.5 kW
0.78
+ 0.1 kW = 63.3 kW
2. Determining the required supply unit:
Powers 9341 9342 9343
P
DC
P
V934X
P
DCtotal
63.3 kW
0.1 kW
63.4 kW
63.3 kW
0.2 kW
63.5 kW
63.3 kW
0.4 kW
63.7 kW
1. supply P
DC934X
7.2 kW 14.4 kW 27.0 kW
2. supply P
DC9330
28.8 kW 42.2 kW 49.5 kW
P
DC302K4B
2.6 kW
3.8 kW
4.5 kW
P
DC9325
2.6 kW
3.8 kW
4.4 kW
P
DC152K4B
1.1 kW 1.7 k W 2.0 k W
Max. possible supply
power
42.3 kW 65.9 kW 87.4 kW
– 9342 or 9343 supply units can be used for the DC-bus connection. Since
P
DCtotal
is higher than P
DC934X
, the network requires a second supply.
The required feedback power determines the selection of the regenerative
power supply unit.
3. Determine the second supply terminal:
– Network with 9342: Second supply terminal at 9330, third at
E82EV302K4B, fourth at 9325
– Network with 9343: Second supply terminal at 9330 (better, only two
supply terminals)
Four drives supplied via 934X
regenerative power supply unit
(static power) 68