thermal analysis - contact resistance

[tonz1984]

Hello


I'm using Pro Mechanica - Thermal to analyse an LED light fitting. With the LED running at 700mA the Junction Temperature (the areawherethe heatand light is generated/applied)should be around 97degrees, this has been tested in the physical world, and my Mechanica results agree with this, however, the temperature of the rest of the body is about 10degree's different to what it should be!


I've included contact resistance between the different parts (although it's only a guessed value, as I can't find a way of calculating the contact resistance -can anyone can help with this as well?)


I know that the answers will be highly dependant on the guessed 'h' values as well, but i'm confident that the 'h' values are suitable.


tony

 

[stevetto]

Read these suggested techniques in PTC's Knowledge base:

Suggested Technique for Modeling Contact Thermal Resistance For Radial Systems
Suggested Technique for Modeling Thermal Contact Resistance for a Planar Wall
System
Suggested Technique for using Thermal Contact Resistance

 

[tonz1984]

Thanks for the infostevetto, finally getting around to checking those links out, ut the Suggested Technique for Modeling Contact Thermal Resistance For Radial Systems doesn't show the images, which i think are pretty crucial to understanding the concepts.


Is there a chance you'd saved this file somewhere with the images?





thanks

 

[stevetto]

Unfortunately I also can't see the pictures.
But if you can see this document, means that you have ptc support.
So you can indicate this problem to PTC tech support. Or ask them to send you the correct document.

 

[tonz1984]

From reading those files that you suggested (without the images) I think i'm going along the correct lines but I'd be grateful if you could tell me if this is correct.

If had a steel plate joined to an aluminium plate 10mm long * 5mm wide * 2mm thick.

SS surface roughness = ~15*10^-4
AL surface roughness = ~3*10^-4
Therefore fake layer would be SS-AL = 12*10^-4.

If fake layer is 0.2mm thick (L)
Area is 10mm*5mm= 50mm

Conductance= KA/L =(12*10^-4)*(50*10^-3) / (0.2*10^-3)
= 0.3

Therefore <B style="COLOR: #ffa34f">Thermal[/B] Conductivity of this layer would be 0.3 w/mK??

Thanks in advance!