Dimensioning Cooling Equipment - Siemens SINAMICS S120 Manual

6.11.4

Dimensioning cooling equipment

Manufacturers provide calculation programs for selecting cooling 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.
Formula to calculate the power loss: q = Q - k · A · ΔT
q = thermal power that has to be dissipated by a cooling unit [W]
Q = power loss [W]
k = thermal resistance coefficient [W / (m
(m
2
A = free-standing cabinet surface area [m
∆T = difference between temperatures inside and outside the cabinet [K]
Table 6- 14
Component
CU320-2
Basic Line Filter for AIM / ALM
36 kW
Active Interface Module
Active Line Module 36 kW
Motor Module 18 A
Motor Module 30 A
SMC
SITOP 20
Line contactor
Total:
Assumption:
Free-standing cabinet surface area A = 5 m
Difference between temperature inside and outside the cabinet ∆T = 10 K
q = 2281 W - 5.5 W / (m
Power Units in Booksize C/D-Type Format
Manual, 12/2018, 6SL3097-5AC20-0BP0
K))
Example of power loss calculation for a drive configuration
K) · 5 m
2
K)] (example: Painted sheet steel k = 5.5 W /
2
]
2
Quanti- Total power loss [W]
ty (including electronic losses)
1 24
1 26
1 340
1 666
2 150
3 270
5 10+
1 53
1 12
2
· 10 K = 2006 W
2
Cabinet design
6.11 Notes on control cabinet cooling
Total power loss [W]
24
26
340
666
300
810
50
53
12
2281
133