Conduction coefficient 1

[cymeryss]

I am working on a heat transfer problem, and need a good estimation of predicting the conduction coefficient for some type of adhesive tape that is used to attach thin,cooper fin strip to an aluminum case. All I know about it is that it is green, crumbles if I try to take it off (probably because of heat), and has a thickness of .01 in. I am thinking that this tape came with the fins which are very delicate and this 'paper type' tape most likely had an adhesive side that attached to the aluminum cylinder to increase its heat disipation properties. I have modeled the forced conv through the fins, but need to include the resistance of the adhesive tape for calcs to be some what accurate. If anybody knows what I am talking about, and has an idea of at least the range o k(W/m deg C), I would greatly appreciate it. A name for the possible manufacturer of these types of fins might also help.

Fins: width (46mm), heigth (11.5), fin thickness (.127 mm) louvered

THanks

Cymeryss

 

[ko99]

Cymeryss,

I don't think this is exactly what you're talking about, but here's a spec on one type of copper "fin" with adhesive attachment:

http://www.aavidthermalloy.com/products/bga/thinfin.pdf

In general, the conductivity of thermal adhesives is fairly poor (typically about 1.5 W/m-C) so it's important to keep it thin (0.01" is on the thick side) and spread the heat evenly over a large area. Calculating resistance due to bulk conductivity is easy:

R = t/KA

R = resistance (C/W)
t = thickness (m)
K = conductivity (W/m-C)
A = heated contact area (m2)

Unfortunately, bulk conductivity does not account for all the thermal loss. There's additional loss due to contact resistance on either side of the adhesive. This depends on the adhesive's ability to penetrate the microscopic nooks and crannies, as well as surface finish, attachment force, etc. It's best to measure it, but lacking better data, a conservative approach is to double the conductive resistance.

Good luck,

ko (

http://www.ecooling.biz)

 

[cymeryss]

ko,

Thank you very much for your response. The value that I was using for my conductivity was 0.7 W/m-C, but then again I was neglecting the contact resistance and was not aware that it was as large as double the magnitude. Thus, in result using 0.7 resulted in similar results as using 1.5 for the value of k, but I will do some more reserach to try to better define this unknown. THanks.

Cymeryss