Elettronica Santerno SINUS K, User Manual

Page: 34 / 191

INSTALLATION
SINUS K
INSTRUCTIONS
34/ 191
7.2. Air Cooling
Make sure to allow adequate clearance around the inverter for the free circulation of air through the equipment.
The table below shows the min. clearance to leave with respect to other devices installed near the inverter. The
different sizes of the inverter are considered.
Size
A – side clearance
(mm)
B – side clearance
between two
inverters (mm)
C – bottom
clearance
(mm)
D – top clearance
(mm)
S05 20 40 50 100
S10 30 60 60 120
S12 30 60 60 120
S15 30 60 80 150
S20 50 100
100 200
S30 100 200 200 200
S40 100 200 200 300
S50 100 200 200 300
S60 150 300 500 300
Size
Minimum
side
clearance
between two
inverter
modules
(mm)
Maximum
side
clearance
between two
inverter
modules
(mm)
Maximum
side
clearance
between two
supply
modules
(mm)
Maximum side
clearance
between
inverters and
supply
modules
(mm)
Top
clearance
(mm)
Bottom
clearance
(mm)
Clearance
between two
inverter units
(mm)
S65 20 50 50 400 300 500 300
The air circulation through the enclosure must avoid warm air intake. Make sure to provide an adequate air cooling
through the inverter. The technical data related to dissipated power are shown in the ratings table.
The air delivery required may be calculated as follows:
air delivery Q= (Pdiss/
Δ t)*3.5 (m 3 /h)
Pdiss is the sum of the values, expressed in W, of the power dissipated by all components installed in the
enclosure;
Δ
t is the difference between the temperature measured inside the enclosure and the ambient
temperature (temperatures are expressed in degrees centigrade).
Example:
Enclosure with no other component installed, SINUS K 0113.
Total power to be dissipated within the enclosure Pti:
generated by the inverter
Pi 2150 W
generated by other components Pa 0 W
Pti = Pi + Pa = 2150 W
Temperatures:
Max. internal temperature desired
Ti 40 °C
Max. external temperature Te 35 °C
Difference between Ti and Te Δ t 5 °C
Size of the enclosure (meters):
width
L 0.6m
height H 1.8m
depth P 0.6m
Free external surface of the enclosure S:
S = (L x H) + (L x H) + (P x H) + (P x H) + (P x L) = 4.68 m
2
External thermal power dissipated by the enclosure Pte (metallic enclosure only):
Pte = 5.5 x
Δ
t x S = 128 W
Pdiss. left :
Pdiss. = Pti - Pte = 2022 W
To dissipate Pdiss. left, provide a ventilation system with the following air delivery Q:
Q = (Pdiss. /
Δ t) x 3.5 = 1415 m 3 /h
(with reference to ambient temperature of 35°C at 1000m above sea level).