Emerson SP1201 - Calculating the Air-Flow in a Ventilated Enclosure

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Unidrive SP Regen Installation Guide 63
Issue Number: 2 www.controltechniques.com
Example
To calculate the size of a non-ventilated enclosure for the following:
• Two SP 1405 (1 x Regen and 1 x motoring drive) models
operating at the Normal Duty rating
• Each drive to operate at 6kHz PWM switching frequency
• Schaffner 16 A (4200-6119) external EMC filter for each drive
• Maximum ambient temperature inside the enclosure: 40°C
• Maximum ambient temperature outside the enclosure: 30°C
Dissipation of each drive: 147 W (see Chapter 12 Technical Data in the
Unidrive SP User Guide)
Dissipation of external EMC filter: 9.2 W (max) (see Chapter 12
Technical Data in the Unidrive SP User Guide)
Dissipation of each external regen inductor: 125 W x 1 (see section
10.4.1 Regen inductors on page 196)
Dissipation of external switching frequency filter: 28 W x 1 (see Chapter
10 Technical data on page 183)
Total dissipation: ((147 x 2) + 9.2 + 125 + 28) = 456.2 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
cubicle (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 5-30 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 456.2 W
The minimum required heat conducting area is then:
= 8.294 m
2
(90.36 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:
=3.2 m (126.02 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:
• Two SP1406 (1 x Regen and 1 x motoring drive) models
operating at the Normal Duty rating
• Each drive to operate at 6kHz PWM switching frequency
• Schaffner 16A (4200-6119) external EMC filter for each drive
• Maximum ambient temperature inside the enclosure: 40°C
• Maximum ambient temperature outside the enclosure: 30°C
Dissipation of each drive: 147 W (see Chapter 12 Technical D ata in the
Unidrive SP User Guide)
Dissipation of external EMC filter: 9.2 W (max) (see Chapter 12
Technical Data in the Unidrive SP User Guide)
Dissipation of external regen inductor: 125 W x 1 (see section
10.4.1 Regen inductors on page 196)
Dissipation of external switching frequency filter: 28 W x 1 (see Chapter
10 Technical data on page 183)
Total dissipation: ((147 x 2) + (9.2 + 125 + 28) = 456.2 W
Insert the following values:
T
int
40°C
T
ext
30°C
k 1.3
P 456.2 W
Then:
= 118.6 m
3
/hr (70.05 ft
3
/min) (1 m
3
/ hr = 0.59 ft
3
/min)
W
H
D
A
e
456.2
5.5 40 30–()
---------------------------------=
W
A
e
2HD–
HD+
--------------------------=
W
10.72 2 2× 0.6×()–
20.6+
-----------------------------------------------------=
V
3kP
T
int
T
ext
–
---------------------------=
P
o
P
l
-------
V
21.3× 456.2×
40 30–
---------------------------------------=