thermal conductive pads 3

[windell747]

Hi, My goal is to fill a 50mil gap between a heatsink and instrument so that the gap is less thermally resistant. I've come up with two options. 1)use two thin thermally conductive pads with a piece of 6061 aluminum in the middle and 2)use a thick thermal pad. I've calculated the equivalent resistances for each case and concluded that the two thin conductive pads with the aluminum piece is less resistant. During this process there were serveral things that seemed odd.

1)The thermal conductivity of the thin thermal pads were around 1.1 W/m-K whereas the thermal conductivity of aluminum is about 169 W/m-K (matweb). Does this make sense?

2)The thermal resistance as taken from the data sheet for the thicker pad is 3 C-in^2/W. I was reading that the thermal conductivity is simply the inverse of the resistance however the units don't make sense when I do this because the units of conductivity is W/m-K and not W/in^2-K. Am I overlooking something here?

Just as an FYI I am using the formula R=L/kA where R is the resistance, L is the thickness of the pad or aluminum, and k is the conductivity of the material.

Thank you in advance for your help!

 

[IRstuff]

Your second item is related to the fact that the thickness is fixed, hence an areal thermal resistance.

As to your first item, you probably need to look some more:

http://vendor.parker.com/Groups/Sea...ERMAL-SELECTOR-GUIDE-SG1004-Rev-A-3-13-07.pdf

TTFN

FAQ731-376

 

[Ursus40]

The thermal resistance of the pads depends on the the area applied, hence the "in^2".
It makes perfect sense that massive aluminum is a factor 100 better that the so called thermal interface materials.

The best thing that you can say about them is that they are better than air! Furthermore there resistance depends on the compression factor as well. You should demand a data sheet with that information as well.

Thermal interfaces have to be as thin as possible so it makes perfact sense that you want to bridge the gap with aluminum as afar as possible.

 

[unclesyd]

This material doesn't have the same capability of Al but is a product we use a lot of for our heating/tracing both steam and electric. Look at the product selection chart for the properties of the different materials.

http://www.thermon.com/us/products.aspx?prodid=79

 

[windell747]

Thank you very much for your valuable input! what would I do without you guys!

 

[EdStainless]

The use of TIM is to reduce the resistance to heat transfer across a poorly mated joint. After all in many systems the interfaces create more resistance than all of the solid materials involved.
There are heat transfer compounds used for the same reason that have much better heat transfer. In general they are a synthetic greases with silver of copper loading. In order to use these you would need a good fit to start with and fairly high clamping forces.

= = = = = = = = = = = = = = = = = = = =
Rust never sleeps
Neither should your protection

http://www.trent-tube.com/contact/Tech_Assist.cfm

 

[MariconEagles]

windell747, I think that thermal resistance is not an inverse of thermal conductivity but rather an inverse of thermal conductance. When you talk about conductivity, your refering to the length on which the heat would transfer from one point to another whereas conductance is simply the heat spreading to the surface area thus you have W/m^2-K as your unit for conductance. Hope this helps...

 

[unclesyd]

You might want to checkout the material Hitherm. I have not checked on the sizes, etc.

I had very good dealings with this company. They do a lot with the anisotropic properties of graphite.

http://www.graftechaet.com/Home/Brands/eGRAF/Products/Thermal-Interface-Material.aspx?bar=1

http://www.graftechaet.com/Home/Brands/eGRAF.aspx

 

[Onno]

Electrical resistance: Voltage = Current * Resistance

Plug in heat transfer quantitities

Analogue: Temperature drop = Heat Flow * Thermal resistance

In units : [K] = [watt]*[K/watt]
So thermal resistance: temperature drop per watt energy transport.

Thermal conduction equation:
Q[watt]=(Labda [watt/m*K]*area[m2]/length[m])*temperature drop[K]
Rewritten to the anlogue Ohms law
temperature drop[K]=Heat flow[watt]*length[m]/(labda[watt/m*K]*area[m2])


temperature drop[K]= [K/watt]*Heat flow [watt]
There is the resistance agian.

Cheers and best regards Onno