Safety
Information
Product
Information
Mechanical
Installation
Electrical
Installation
Getting
Started
Basic
parameters
Running
the motor
Optimization
SMARTCARD
operation
PC tools
Advanced
parameters
Technical
Data
Diagnostics
UL Listing
Information
Affinity User Guide
49
Issue Number: 5 www.controltechniques.com
Dissipation of each drive: 187 W (see section 12-8
Dissipation of each external EMC filter: 9.2 W (max) (see section
12.2.1
Total dissipation: 2 x (187 + 9.2) =392.4 W
The enclosure is to be made from painted 2 mm (0.079 in) sheet steel
having a heat transmission coefficient of 5.5 W/m
2
/
o
C. Only the top,
front, and two sides of the enclosure are free to dissipate heat.
The value of 5.5 W/m
2
/ºC can generally be used with a sheet steel
enclosure (exact values can be obtained by the supplier of the material).
If in any doubt, allow for a greater margin in the temperature rise.
Figure 3-43 Enclosure having front, sides and top panels free to
dissipate heat
Insert the following values:
T
int
40
°
C
T
ext
30
°
C
k
5.5
P
392.4
W
The minimum required heat conducting area is then:
= 7.135 m
2
(77.8 ft
2
)
(1 m
2
= 10.9 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:
=1.821 m (71.7 in)
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 (1 m
3
/hr = 0.59 ft
3
/min)
T
ext
Maximum expected temperature in
°
C
outside
the
enclosure
T
int
Maximum permissible 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 size of an enclosure for the following:
•
Three BA1403 models operating at the Normal Duty rating
•
Each drive to operate at 6kHz PWM switching frequency
•
Schaffner 10A (4200-6118) external EMC 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: 101 W
Dissipation of each external EMC filter: 6.9 W (max)
Total dissipation: 3 x (101 + 6.9) = 323.7 W
Insert the following values:
T
int
40
°
C
T
ext
30
°
C
k
1.3
P
323.7 W
Then:
=
126.2 m
3
/hr (74.5 ft
3
/min)
(1 m
3
/ hr = 0.59 ft
3
/min)
3.7 Enclosure design and drive ambient
temperature
Drive derating is required for operation in high ambient temperatures
Totally enclosing or through panel mounting the drive in either a sealed
cabinet (no airflow) or in a well ventilated cabinet makes a significant
difference on drive cooling.
The chosen method affects the ambient temperature value (T
rate
) which
should be used for any necessary derating to ensure sufficient cooling
for the whole of the drive.
The ambient temperature for the four different combinations is defined
below:
1. Totally enclosed with no air flow (<2 m/s) over the drive
T
rate
= T
int
+ 5°C
2. Totally enclosed with air flow (>2 m/s) over the drive
T
rate
= T
int
3. Through panel mounted with no airflow (<2 m/s) over the drive
T
rate
= the greater of T
ext
+5°C, or T
int
4. Through panel mounted with air flow (>2 m/s) over the drive
T
rate
= the greater of T
ext
or T
int
Where:
T
ext
= Temperature outside the cabinet
T
int
= Temperature inside the cabinet
T
rate
= Temperature used to select current rating from tables in
W
H
D
A
e
392.4
5.5 40 30
–
(
)
---------------------------------
=
W
A
e
2HD
–
H D
+
--------------------------
=
W
7.135
2 2
×
0.6
×
(
)
–
2 0.6
+
-----------------------------------------------------
=
V
3kP
T
int
T
ext
–
---------------------------
=
P
o
P
l
-------
V
3 1.3
×
323.7
×
40 30
–
---------------------------------------
=
Содержание Affinity
Страница 274: ...0474 0000 05 ...
