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The minimum required heat conducting area is then:
336.2
-------------------------------- -
A
=
e
(
5.5 40 30
(1m = 3.3 ft)
Estimate two of the enclosure dimensions — the height (H) and depth (D), for instance. Calculate the width (W) from:
A
–
e
W
=
------------------------- -
H
+
Inserting H = 2m and D = 0.6m, obtain the minimum width:
7.8
–
W
----------------------------------------------
=
If the enclosure is too large for the space available, it can be made smaller only by attending to one or all of the following:
•
Using a lower PWM switching frequency to reduce the dissipation in the drives
•
Reducing the ambient temperature outside the enclosure, and/or applying forced-air cooling to the outside of the
enclosure
•
Reducing the number of drives in the enclosure
•
Removing other heat-generating equipment
Calculating the air-flow in a ventilated enclosure
The dimensions of the enclosure are required only for accommodating the equipment. The equipment is cooled by the forced
air flow.
Calculate the minimum required volume of ventilating air from:
3kP
------------------------ -
V
=
T
–
T
i
Where:
V
Air-flow in m
Maximum ambient temperature in °C outside the enclosure
T
amb
Maximum ambient temperature in °C inside the enclosure
T
i
P
Power in Watts dissipated by all heat sources in the enclosure
k
Ratio of
Where:
P
0
P
I
Typically use a factor of 1.2 to 1.3, to allow also for pressure-drops in dirty air-filters.
Example
To calculate the air flow for the following:
•
Two SE23400400 models
•
Each drive to operate at 6kHz
•
RFI filter for each drive
•
Braking resistors are to be mounted outside the enclosure
•
Maximum ambient temperature inside the enclosure: 40°C
•
Maximum ambient temperature outside the enclosure: 30°C
Dissipation of each drive: 158W
Dissipation of each RFI filter: 10.1W
Total dissipation: 2 x (158 + 10.1) = 336.2W
Insert the following values:
T
40°C
i
T
30°C
amb
k
1.3
P
2190W
Then:
×
3
1.3
-------------------------------------- -
V
=
40 30
3
(1m
\hr = 0.59ft
Commander SE Advanced User Guide
Issue Number: 4
2
2
(
)
=
6.11m
66.6ft
)
–
2HD
D
(
×
×
)
2 2
0.6
(
2.1m 6ft10in
=
2
+
0.6
amb
3
per hour
P
0
----- -
P
I
is the air pressure at sea level
is the air pressure at the installation
switching frequency
PWM
×
336.2
3
3
(
=
131m
\hr 504ft
–
3
\min)
)
)
\min
115
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