Commander SE Advanced User Guide
115
Issue Number: 4
The minimum required heat conducting area is then:
(1m = 3.3 ft)
Estimate two of the enclosure dimensions — the height (H) and depth (D), for instance. Calculate the width (W) from:
Inserting
H
= 2m and
D
= 0.6m, obtain the minimum width:
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:
Where:
V
Air-flow in m
3
per hour
T
amb
Maximum ambient temperature in
°
C outside the enclosure
T
i
Maximum ambient temperature in
°
C inside the enclosure
P
Power in Watts dissipated by
all
heat sources in the enclosure
k
Ratio of
Where:
P
0
is the air pressure at sea level
P
I
is the air pressure at the installation
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
PWM
switching frequency
•
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
i
40
°
C
T
amb
30
°
C
k
1.3
P
2190W
Then:
(1m
3
\hr = 0.59ft
3
\min)
A
e
336.2
5.5 40 30
–
(
)
---------------------------------
6.11m
2
66.6ft
2
(
)
=
=
W
A
e
2HD
–
H D
+
--------------------------
=
W
7.8
2 2
×
0.6
×
(
)
–
2 0.6
+
----------------------------------------------
2.1m 6ft10in
(
)
=
=
V
3kP
T
i
T
amb
–
-------------------------
=
P
0
P
I
------
V
3 1.3
×
336.2
×
40 30
–
---------------------------------------
131m
3
\hr 504ft
3
\min
(
)
=
=
Содержание COMMANDER SE
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Страница 93: ...Commander SE Advanced User Guide 93 Issue Number 4...
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Страница 142: ...142 Commander SE Advanced User Guide Issue Number 4 Figure 6 9 Input line reactor 4400 0241 All dimensions in mm...
Страница 144: ...144 Commander SE Advanced User Guide Issue Number 4 Figure 7 3 Commander SE Size 4 All dimensions in mm...
