2-16
Power Distribution
Example: There are 5 drives, each is rated 1 HP, 480V, 2.7 amps. These drives do
not have internal inductors.
Total current = 5 * 2.7 amps = 13.5 amps
125% * Total current = 125% * 13.5 amps = 16.9 amps
From publication 1321-2.0, we selected the reactor 1321-3R12-C, which
has a maximum continuous current rating of 18 amps and an inductance of
4.2 mh (0.0042 henries).
1.54% is more than the 0.5% impedance recommended. The 1321-3R12-C
can be used for the (5) 2.7 amp drives in this example.
Ohms
102.6
2.7
3
480
I
3
V
Z
rating
input
line
-
line
drive
=
.
=
.
=
Ohms
58
.
1
60
2
0042
.
0
2
Z
reactor
=
.
. Π
.
=
.
Π
.
=
f
L
%
1.54
0.0154
102.6
1.58
Z
Z
drive
reactor
=
=
=
Содержание Allen-Bradley 1305-AA02A
Страница 1: ...Installation Instructions Wiring and Grounding Guidelines for Pulse Width Modulated PWM AC Drives ...
Страница 4: ...ii Summary of Changes Notes ...
Страница 40: ...2 18 Power Distribution Notes ...
Страница 48: ...3 8 Grounding Notes ...
Страница 68: ...4 20 Practices Notes ...
Страница 78: ...6 8 Electromagnetic Interference Notes ...
Страница 94: ...Glossary 4 UL Underwriters Laboratories ...
Страница 100: ...Index 6 ...
Страница 101: ......