Selection
Selection examples
12
Network of several drives
12.5
12.5.5
L
12.5-5
EDS82EV903-1.0-11/2002
12.5.5 Selection examples
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. Determine first supply terminal:
–P
DC100%
from the table ”supply power” . (¶ 12.5-4)
9330 E82EV302K4B 9325 E82EV152K4B
P
DC100%
60.5 kW 6.1 kW 7.9 kW 3.5 kW
– First supply terminal is 9300 (first controller in line 1).
– I.e. additional supply power required: 63.3 kW - 60.5 kW = 2.8 kW
3. Determine the second supply terminal:
– Find supply power for 9325, E82EV302K4B, E82EV152K4B in column
”9330” in the table ”supply power” .
(¶ 12.5-4)
E82EV302K4B 9325 E82EV152K4B
P
DC2
4.5 kW 4.4 kW 2.0 kW
– The power of 9325 is sufficient.
4. Result:
– The network of drives must be connected to the three-phase AC mains
via the 9330 and 9325 controllers.
Four drives supplied via
controllers (static power) 67