Information About Calculating Power Loss - Siemens SINAMICS S120 Equipment Manual

Cabinet Configuration and EMC Booksize
6.7 Information about heat dissipation
6.7.4

Information about calculating power loss

Cabinet manufacturers provide calculation programs for selecting climate control equipment.
It is always necessary to know the power loss of the components and equipment installed in
the cabinet.
The physical relationship is shown in the following example.
Figure 6-12
q = power that has to be dissipated through a cooling unit [W / K]
Q = power loss [W]
∆T = temperature difference between the room and cabinet interior [K]
k = thermal resistance value, e.g. sheet-steel, painted 5.5 [W / (m
A = free-standing cabinet surface area [m
Table 6-15 Example of a power loss calculation
Component
CU320
Active line module 36 kW
Motor module 18 A
Motor module 30 A
SMC20
SITOP 20
Main contactor
Total:
Assumption:
Free-standing cabinet surface area A = 5 m
Temperature difference between the room and cabinet interior ∆T = 20 K
q = (1970 [W] / 20 [Κ]) - 5.5 [W / (m
6-32
q =
(
Example of dimensioning climate control equipment
Number
1
1
2
3
10
1
1
Q
−
)
∆ T
]
2
Power loss [W]
20
580
165
290
10
53
12
2
* K)] * 5 [m ] = 71 [W/K]
2 2
Equipment Manual, (GH2), 04.2004 Edition, 6SL3097-2AC00-0BP0
k ∗ A
* K)]
2
Total power loss [W]
20
580
330
870
100
53
12
1970
Booksize Power Sections