Rockwell Automation 1336T Wiring and Grounding Guide, (PWM) AC Drives User Manual

Page 44

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Rockwell Automation Publication DRIVES-IN001M-EN-P - March 2014

Chapter 2 Power Distribution

Use the formula below to verify that the impedance of the selected reactor
is more than 0.5% (0.25% for drives with internal inductors) of the
smallest drive in the group. If the impedance is too small, select a reactor
with a larger inductance and same amperage, or regroup the drives into
smaller groups and start over.

EXAMPLE

There are five drives. Each drive is rated 1 Hp, 480V, 2.7 A. These drives do not
have internal inductors.

Total current is 5 x 2.7 A = 13.5 A

125% x Total current is 125% x 13.5 A = 16.9 A

From 1321 Power Conditioning Products Technical Data, publication

1321-

TD001

, we selected the catalog number 1321-3R12-C reactor. This reactor has

a maximum continuous current rating of 18 A and an inductance of 4.2 mH
(0.0042 henries).

1.54% is more than the 0.5% impedance recommended. The catalog number
1321-3R12-C reactor can be used for the five 2.7 A drives in this example.

Z

drive

=

Z

reactor

= L * 2 * 3.14 * f

V

line-line

3 * I

input-rating

L is the inductance of the reactor in henries and f is the AC line frequency.

Z

drive

=

V

line-line

3 * I

input-rating

480V

3 * 2.7

=

= 102.6 Ohms

Z

reactor

= L * (2 * 3.14) * f = 0.0042 * 6.28 * 60 = 1.58 Ohms

Z

reactor

Z

drive

1.58

102.6

= 0.0154 = 1.54%

=

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