*
14
M’Ax Installation Guide
www.controltechniques.com
Issue Number: 6
Figure 2-8
Enclosure having front, sides and top panels free to
dissipate heat
Insert the following values:
T
i
55°C
T
amb
30°C
k
5.5
P
1071W
The minimum required heat conducting area is then:
(1m
2
= 10.8 ft
2
)
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:
•
Reducing the ambient temperature outside the enclosure, and/or
applying forced-air cooling to the outside of the enclosure
•
Removing other heat-generating equipment, eg. braking resistors
•
Reducing the number of drives in the enclosure
•
Air circulating fan inside enclosure (see section 2.25 Enclosure
calculations for heat removal on page 13)
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
Ti
Maximum ambient temperature in °C inside the enclosure
P
Power in Watts dissipated by all heat sources in the
enclosure
k
p
Ratio of
Where:
P
0
is the air pressure at sea level
P
1
is the air pressure at the installation
Typically use a factor k
a
of 1.2 to 1.3, to allow also for pressure-
drops in dirty air-filters.
Example
To calculate the required air flow in an enclosure for the following:
•
Three M’Ax 409
•
Each drive is to have an external braking resistor mounted outside
the enclosure
•
Maximum ambient temperature inside the enclosure: 55°C
•
Maximum ambient temperature outside the enclosure: 30°C
•
At sea level (k
p
= 1 for this example)
Dissipation of each drive: 180W (from section 2.11 Output current,
Ambient temperature, Heat dissipation, De-rating on page 7)
Total dissipation: 3 x 180 = 540W
Insert the following values:
T
i
55°C
T
amb
30°C
k
a
1.3
P
540W
Then:
W
H
D
A
e
1071
5.5 55
30
–
(
)
---------------------------------
7.8
2
84 ft
2
(
)
=
=
W
A
e
2HD
–
H
D
+
--------------------------
=
W
7.8
2
2
×
0.6
×
(
)
–
2
0.6
+
----------------------------------------------
2.08m 6ft10in
(
)
=
=
V
3kP
T
i
T
am b
–
-------------------------
=
P
0
P
1
------
V
3
1.3
×
540
×
55
30
–
----------------------------------
84m
3
hr 49
3
min
(
)
=
=
