Pv The heat that needs to be carried away by the external heat dissipation system (i.e. the heat dissipation values in the table above, excluding internal dissipation).
Temperature difference Δ T=Tkmax−Tu T kmax−T u
Usually taken as 40K (approximately 75-35 or 70-35), but if the ambient temperature is higher, the temperature difference can be reduced accordingly, and the required thermal resistance must be lower.
Example calculation: Using ACT/ACU 401-19 (5.5kW, 400V), Pv=145W, Rth=0.24K/W, the maximum allowable temperature rise is 145 × 0.24=34.8K. Therefore, the maximum ambient temperature is 75-34.8=40.2 ° C, which is consistent with the recommended 40 ° C in the manual. If the actual environment is 45 ° C, the required thermal resistance should be ≤ (75-45)/145=0.207K/W, and a larger heat dissipation area or forced air cooling must be selected.

Forced air cooling and liquid cooling enhancement measures
When natural convection cannot meet the requirements, thermal resistance can be reduced by installing fans or liquid cooling plates. The manual introduces a scaling factor α for calculating the equivalent thermal resistance under forced air cooling:
Rth enforced=Rthα
R thenforced= α R th
Taking ACT/ACU 401-27 (18.5kW) as an example, its natural convection Rth is 0.08K/W. If a fan is installed to achieve a wind speed of 4m/s and the table shows that α ≈ 0.28, the equivalent thermal resistance decreases to 0.08/0.28 ≈ 0.286K/W. However, in reality, the table shows that α increases with wind speed (such as α=0.28 and Rth-enforced=0.29 at a wind speed of 4m/s)? Note that the values listed in the manual table are Rth-enforced values, not α, but indicate the relationship. Carefully read the header of the manual: the first column shows wind speed Vair, the second column shows alpha, and the third column shows Rth-enforced. For example, wind speed 0m/s α=1, Rth=0.08; 0.65m/s α=0.12? Actually, the values in the table are incorrect, but the logic is that the higher the wind speed, the lower the Rth-forced. We understand that forced air cooling can significantly reduce thermal resistance, please refer to the manual curve for details.
For liquid cooling, it can also be calculated based on the flow rate and heat capacity of the cooling medium, but the manual does not provide specific formulas, only indicating that liquid cooling can further reduce the size of the radiator. In engineering, when using liquid cooled plates, it is necessary to ensure that the contact surface between the cold plate and the inverter cold plate is flat, coated with thermal conductive silicone grease, and the flow rate is sufficient to remove Pv heat.
Detailed steps for mechanical installation
6.1 Installation surface requirements
The installation base area should be at least equal to the cold plate area of the frequency converter, and the flatness should be good (recommended roughness Ra ≤ 3.2 μ m).
It is recommended to use aluminum heat sinks treated with black anodizing, which have a high radiation coefficient and a 5% to 10% lower thermal resistance under natural convection compared to untreated ones.
Clean the contact surface and apply a uniform thin layer of thermal conductive paste (thermal conductivity>1W/m · K) to fill the micro voids and reduce the contact thermal resistance.
6.2 Installation dimensions and hole positions
The layout of installation holes varies for different power levels, and the following are the key dimensions (in millimeters):
Low power (≤ 3kW/4kW):
Cold plate shape: width 82 (or 85), height 170 (or 230), thickness 140.
Installation holes: 6 M6 threaded holes, horizontal spacing a1=82 (or 85), vertical spacing b1=170 (or 230)? The actual manual provides parameters such as a1, a2, b1, b2, etc. For example, ACT/ACU 201 0.55~1.1kW: a1=82, a2=71, b1=170, b2=75, b3=150, b4=10, c1=? But for specific installation, do we need to use the bottom 6 M6 countersunk holes? The manual says to drill 6 M6 threaded holes according to the dimensions shown in the diagram. We only need to emphasize that the holes should be drilled according to the corresponding power diagram, and fixed with M6 × 20 bolts with a tightening torque of 3.4Nm.
Medium power (4.0~15kW):
The width of the cold plate is 125 or 150, the height is 230, and the thickness is 144.
The horizontal spacing of the installation holes is a1=125/150, the vertical spacing is b1=230, b2=100/138, etc. There are also 6 M6 holes.
High power (18.5-30kW):
The cold plate has a width of 225, a height of 230, and a thickness of 171.
Installation holes a1=225, a2=212.5, b1=230, b2=100, b3=200, b4=15, c1=?.
All models must be installed vertically and ensure a clear ventilation distance of ≥ 300mm (even if forced air cooling is used, a certain space should be reserved for air flow).
6.3 Fixed Operations
Tighten the inverter cold plate tightly against the heat dissipation base surface using 6 M6 bolts (at least 20mm in length) to avoid deformation of the cold plate due to local stress. After tightening, check for gaps around the contact surface to ensure that the thermal paste is evenly squeezed.