Rockwell Automation Publication 2198-UM005C-EN-P - February 2022 31
Chapter 2 Plan the Kinetix 5300 Drive System Installation
Table 10 - External Shunt Module Specifications
How the Bulletin 2097-Rx and 2198-Rxxx shunts connect to the Kinetix 5300
drive is explained in External Passive-shunt Resistor Connections on page 96
and illustrated with interconnect diagrams in Shunt Resistor Wiring Example
on page 172.
Enclosure Selection
This example is provided to assist you in sizing an enclosure for your
Kinetix 5300 drive system. You need heat dissipation data from all components
planned for your enclosure to calculate the enclosure size (refer to Table 11 on
page 32).
With no active method of heat dissipation (such as fans or air conditioning)
either of the following approximate equations can be used.
If the maximum ambient rating of the Kinetix 5300 drive system is 50 °C
(122 °F) and if the maximum environmental temperature is 20 °C (68 °F), then
T=30. In this example, the total heat dissipation is 416 W (sum of all
components in enclosure). So, in the equation below, T=30 and Q=416.
In this example, the enclosure must have an exterior surface of at least 2.99 m
2
.
If any portion of the enclosure is not able to transfer heat, do not include that
value in the calculation.
Because the minimum cabinet depth to house the Kinetix 5300 system
(selected for this example) is 300 mm (11.8 in.), the cabinet needs to be
approximately 1500 x 700 x 300 mm (59.0 x 27.6 x 11.8 in.) HxWxD.
1.5 x (0.300 x 0.70) + 1.5 x (0.300 x 2.0) + 1.5 x (0.70 x 2.0) = 3.31 m
2
Shunt Module
Cat. No.
Resistance
Ohms
Continuous Power
W
Weight, approx
kg (lb)
2097-R6 75 150 0.3 (0.7)
2097-R7 150 80 0.2 (0.4)
2198-R004 33 400 1.8 (4.0)
2198-R014 9.4 1400 9.1 (20)
2198-R031 33 3100
(1)
(1) The 2198-R031 shunt is limited to 2000 W when used with 2198-C1015-ERS, 2198-C1020-ERS, 2198-C2030-ERS, 2198-C4015-ERS, 2198-
C4020-ERS, 2198-C4030-ERS (frame 2) drives.
16.8 (37)
IMPORTANT To comply with UL requirements, the minimum cabinet size must be 508
mm (20.0 in.), height; 406 mm (16.0 in.), width; and 300 mm (11.8 in.) depth.
Metric Standard English
Where T is temperature difference between inside air and
outside ambient (°C), Q is heat generated in enclosure
(Watts), and A is enclosure surface area (m
2
). The exterior
surface of all six sides of an enclosure is calculated as
Where T is temperature difference between inside air and
outside ambient (°F), Q is heat generated in enclosure
(Watts), and A is enclosure surface area (ft
2)
. The exterior
surface of all six sides of an enclosure is calculated as
A = 2dw + 2dh + 2wh A = (2dw + 2dh + 2wh) /144
Where d (depth), w (width), and h (height) are in meters.
A =
4.08Q
T - 1.1
A =
0.38 (416)
1.8 (30) - 1.1
= 2.99 m
2
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