Siemens SINAMICS G130 Engineering Manual page 264

General Engineering Information for SINAMICS
Engineering Information
Example of how to calculate heat losses P
A closed cabinet of 2000 mm in height, 600 mm in width and 600 mm in depth is the first cabinet element at the
beginning of a long row of cabinets which are installed with the rear panel against the wall. The interior cabinet
temperature must not exceed 50 °C at an external temperature which is not expected to exceed 30 °C.
Calculation of the permissible power losses P
a) The calculation is based on the assumption that the air inside and outside the cabinet will be still. The applicable
heat transfer coefficient is thus k = 3 W / (m
b) As the cabinet is wall-mounted at the beginning of a row of cabinets, the effective cabinet surface for heat transfer
to the external air comprises only the front, one side panel and the roof of the cabinet. Heat cannot be transferred
to the external air via the rear panel, the side panel to which the adjacent cabinet is joined, or the base, and these
areas cannot therefore be included in the calculation of effective cabinet surface. Taking these installation
conditions into account, the effective surface is determined by the following equation:
c) With a maximum interior temperature of 50°C and a maximum exterior temperature of 30°C, the temperature
J
difference equals = 20 K.
D
Using the values above, the permissible heat losses of the cabinet are finally calculated to be:
This example shows that only relatively low heat losses of less than a few 100 W can be dissipated from a cabinet by
natural convection. This can be sufficient in the case of cabinets which house only electronic components such as
Control Units, Terminal Modules and Sensor Modules, or for cabinets containing only switches, contactors, fuses and
conductor bars.
However, if a cabinet is to accommodate power components such as G130 Power Modules, S120 Line Modules or
S120 Motor Modules in the power range of Chassis format devices, which produce heat losses of several kilowatts,
then cooling by natural convection is no longer an option. For this purpose forced ventilation by fans is required.
Cabinet cooling by means of forced ventilation with fans
The principle of cooling by forced ventilation involves transferring the power losses produced inside the cabinet to the
cooling air forced through the cabinet, which heats up as a result. The power losses P
proportional to both the cooling air circulated through the cabinet by the fans as well as to the temperature difference
between inlet air T and outlet air T
a
The thermal capacity and density of the cooling
air are also included in the calculation. These
variables are in turn dependent on moisture
content and air pressure.
To estimate the power losses P
dissipated from the cabinet, the following quantity
equation can be applied for typical industrial
environments and installation altitudes below
2000 m:
=
P [ W ] 1200
V
where
D J
= T - .
T
c a
Using this equation, it is easy to estimate that
forced ventilation is capable of dissipating heat
losses in the order of 12 kW from the cabinet,
even where the temperature difference is
relatively low at about 10 K and the cooling air
3
flow approximately 1 m /s. However, the cabinet
needs to be fitted with air openings of an
appropriate size to ensure adequate air flow.
This example clearly proves that air-cooled SINAMICS power components, such as G130 Power Modules, S120 Line
Modules or S120 Motor Modules in the power range of Chassis format devices, require forced ventilation and these
power units are therefore equipped as standard with fans.
SINAMICS Engineering Manual – November 2015
264/528
Ó Siemens AG
of a cabinet to be dissipated by natural convection:
v
= k • A •
v
2
• K).
2 2
A = 1.2 m (front) + 1.2 m
J
P = k • A • = 3 W / (m
D
v
.
c
which can be
v
·
J
× 3 D ×
V [ m / s ] [ K ]
J
:
D
2
(one side) + 0.36 m (roof) = 2.76 m
2 2
• K) • 2.76 m • 20 K = 166 W.
Cooling of a cabinet by forced ventilation
2
.
which can be dissipated are
v