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Modelling a heat sink with COMSOL

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Hi,

I'd like to model a certain Aavid heat sink that has a profile that can be seen in the picture I'll attach. I'm new to Comsol so I don't have much experience with creating objects of certain profile. I'd know how to create a heat sink with normal fins but this one is a little bit tricky for me as it has a bit inclined and curved fins. I also considered doing the profile in Solidworks and importing it to Comsol, but I'd like to stick to Comsol for the moment.

Thanks!


4 Replies Last Post Mar 13, 2015, 5:41 a.m. EDT
Edgar J. Kaiser Certified Consultant

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Posted: 10 years ago Feb 24, 2015, 3:10 p.m. EST
Marko,

I would suggest to draw one fin as a polygonal line with a trapezoid shape, apply fillets to the edges and corners, multiply it with an array node and add the end and bottom portion. If you need it in 3D you can extrude this shape from a workplane.

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Marko, I would suggest to draw one fin as a polygonal line with a trapezoid shape, apply fillets to the edges and corners, multiply it with an array node and add the end and bottom portion. If you need it in 3D you can extrude this shape from a workplane. Cheers Edgar -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com

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Posted: 10 years ago Feb 27, 2015, 6:11 a.m. EST
Thank you for your reply.

I still have few questions regarding the modelling of the heat sink. Since I am modelling a heat sink for power electronics applications, I have the value for thermal resistances between the junction and case, case and sink and between sink and air. I'd like to keep the temperature of the junction below 125 degC and I know that in the worst case the losses are 55W. Now, I'm not really how to define the physics of my setup. I tried very simple approach, to define the case as a copper plate and to make it to a heat source of 55W but this gives me around 10,000K which is ridiculous. Another try is with the boundary heat source where I defined the Q to be 166 W/m2, which would correspond to the losses over the case surface. Is this the right approach to it or is there really a proper approach to design and simulate this heat sink?

Thank you
Thank you for your reply. I still have few questions regarding the modelling of the heat sink. Since I am modelling a heat sink for power electronics applications, I have the value for thermal resistances between the junction and case, case and sink and between sink and air. I'd like to keep the temperature of the junction below 125 degC and I know that in the worst case the losses are 55W. Now, I'm not really how to define the physics of my setup. I tried very simple approach, to define the case as a copper plate and to make it to a heat source of 55W but this gives me around 10,000K which is ridiculous. Another try is with the boundary heat source where I defined the Q to be 166 W/m2, which would correspond to the losses over the case surface. Is this the right approach to it or is there really a proper approach to design and simulate this heat sink? Thank you

Edgar J. Kaiser Certified Consultant

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Posted: 10 years ago Feb 27, 2015, 7:10 a.m. EST
Marko,

I would recommend to study some examples in the model library (you can update it online to see more examples), check the knowledge base and the model gallery.

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Marko, I would recommend to study some examples in the model library (you can update it online to see more examples), check the knowledge base and the model gallery. Cheers Edgar -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com

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Posted: 10 years ago Mar 13, 2015, 5:41 a.m. EDT
Hello,

Thank you for your reply. I have managed to simulate the behaviour I need.

Now, I have one more question. It concerns the thermal conductivity of a material I use. Namely, I use Aluminium alloy 6060 which has by default thermal conductivity k = 200W/mK. My problem is that I have a heat sink with a certain thermal resistance, namely 0.5K/W at 200mm length. How can I convert this resistance to conductivity and should I do that at all?

I tried multiplying Rth by the length and inverting it which gives me a conductivity of 13.3W/mK but when this is simulated I got that the temperature at the bottom of my heatsink is 200degC which is unacceptable.

Could you explain me how to take the thermal resistance of the heat sink into profile?

Thank you,

Marko
Hello, Thank you for your reply. I have managed to simulate the behaviour I need. Now, I have one more question. It concerns the thermal conductivity of a material I use. Namely, I use Aluminium alloy 6060 which has by default thermal conductivity k = 200W/mK. My problem is that I have a heat sink with a certain thermal resistance, namely 0.5K/W at 200mm length. How can I convert this resistance to conductivity and should I do that at all? I tried multiplying Rth by the length and inverting it which gives me a conductivity of 13.3W/mK but when this is simulated I got that the temperature at the bottom of my heatsink is 200degC which is unacceptable. Could you explain me how to take the thermal resistance of the heat sink into profile? Thank you, Marko

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