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Heat Conductivity, Copper and Thermal Compound scenarios

  1. Sep 2, 2009 #1
    This is not actual homework. I have not had any course in thermodynamics. I have completed Kinematics and Electrical physics courses at my university. My degree is related to Computer Science, not Engineering or Physics.
    I am asking if my solutions to these questions are relatively correct. Actual values are not needed, just scale.

    1. The problem statement, all variables and given/known data
    Given the following scenarios, wherein an unavoidable gap between a heat pipe and GPU die is bridged by a combination of varying thickness copper shims and thermal grease, which scenarios will yield the highest heat conduction ([tex]\frac{\Delta Q}{\Delta T}[/tex])?

    Single Spaced Shim (layered)
    http://img44.imageshack.us/img44/8633/thermcoppershimspaced.jpg [Broken]

    Direct Contact, Thick Shim
    http://img134.imageshack.us/img134/4513/thermcoppershimdirectco.jpg [Broken]

    Thermal Compound only
    http://img524.imageshack.us/img524/5481/thermtconly.jpg [Broken]

    Three layered shims, direct contact
    http://img522.imageshack.us/img522/5781/thermthreecoppernospace.jpg [Broken]

    Three layered spaced shims
    http://img44.imageshack.us/img44/1750/thermthreecopperspacing.jpg [Broken]

    Thermal Conductivity Constants k:
    Copper 360 W/mK
    Thermal Compound 6.2W/mK

    Assume total distance between the heat pipe and die (Xt) is 5mm= 0.005m.
    Xt=0.005
    Let Y represent the thickness of one shim.

    Assume area is 1. [Only looking for scale between scenarios; surface area doesn't change much.]
    A=1

    2. Relevant equations

    Heat Conduction
    [tex]H= \frac{\Delta Q}{\Delta T} = \frac{kA\Delta T}{x} [/tex]
    H= dQ/dT=kAdT/x

    Layered Heat Conduction
    [tex]H= \frac{A\Delta T}{\frac{x1}{k1}+\frac{x2}{k2}+\frac{x3}{k3}...} [/tex]
    H= AdT/(x1/k1+x2/k2+x3/k3)

    3. The attempt at a solution
    [I will use dT instead of [tex]\Delta T[/tex], since it helps clean up the text.]

    Heat conduction (H) solved for the following:

    Single Spaced Shim (layered)
    [tex]\frac{1*dT}{\frac{\frac{Xt-Y}{2}}{6.2} + \frac{Y}{360} + \frac{\frac{Xt-Y}{2}}{6.2} }[/tex]

    =[tex]\frac{dT}{2*\frac{\frac{Xt-Y}{2}}{6.2} + \frac{Y}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt-Y}{6.2} + \frac{Y}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} - \frac{Y}{6.2} + \frac{Y}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} + \frac{6.2*Y - 360*Y}{2232}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} - \frac{353.8*Y}{2232}}[/tex]

    Plaintext version:
    1*dT/( ((Xt-Y)/2)/6.2 + Y/360 + ((Xt-Y)/2)/6.2 )
    =dT/( 2*((Xt-Y)/2)/6.2 + Y/360 )
    =dT/( (Xt-Y)/6.2 + Y/360 )
    =dT/( Xt/6.2 - Y/6.2 + Y/360 )
    =dT/( Xt/6.2 + (6.2Y - 360Y)/2232 )
    =dT/( Xt/6.2 - 353.8Y/2232 )

    If the shim is the following thicknesses (Percentage of gap): Heat conduction
    Shim 1mm (20%): 1543*dT
    Shim 2mm (40%): 2043*dT
    Shim 3mm (60%): 3021*dT
    Shim 4mm (80%): 5800*dT
    Shim 4.5mm (85%): 10735*dT
    Shim ~5mm (99%): 71999*dT

    Direct Contact, Thick Shim
    Through Thermal Compound
    6.2*.05*dT/(~0)=Inf*dT
    Through shim (Maximum)
    360*1*dT/Y
    =360*dT/Xt
    =72000*dT
    As thickness of thermal grease approaches 0, conductance and conduction of thermal grease approaches Inf, and layered conduction approaches direct contact of shim (maximum).
    However, even if direct contact is made (or very close to that), surface area is not 100%. The pores are however filled with thermal grease.
    Thus, as percentage of surface area of direct contact increases or depth of pores decreases, conduction approaches maximum.
    Theory: If pores were perfect insulators, then conduction would be reduced by the percentage surface area of the pores, which would be the minimum conduction.
    So, actual conduction lies somewhere between stated minimum and maximum, depending on percentage of the surface area consisting of pores and depth of those pores.

    Thermal Compound only
    6.2*1*dT/Xt
    =6.2*dT/0.005
    =1240*dT

    Three layered shims, direct contact
    Same as Direct Contact, Thick Shim, but with increased percentage of pores and pore depth.

    Three layered spaced shims (Y represents total added thickness of all three shims)
    [tex]\frac{1*dT}{\frac{\frac{Xt-Y}{4}}{6.2} + \frac{\frac{Y}{3}}{360} + \frac{\frac{Xt-Y}{4}}{6.2} + \frac{\frac{Y}{3}}{360} + \frac{\frac{Xt-Y}{4}}{6.2} + \frac{\frac{Y}{3}}{360} + \frac{\frac{Xt-Y}{4}}{6.2} }[/tex]

    =[tex]\frac{dT}{4*\frac{\frac{Xt-Y}{4}}{6.2} + 3*\frac{\frac{Y}{3}}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt-Y}{6.2} + \frac{Y}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} - \frac{Y}{6.2} + \frac{Y}{360}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} + \frac{6.2*Y - 360*Y}{2232}}[/tex]

    =[tex]\frac{dT}{\frac{Xt}{6.2} - \frac{353.8*Y}{2232}}[/tex]

    Plaintext version:
    1*dT/( ((Xt-Y)/4)/6.2 + (Y/3)/360 + ((Xt-Y)/4)/6.2 + (Y/3)/360 + ((Xt-Y)/4)/6.2 + (Y/3)/360 + ((Xt-Y)/4)/6.2 )
    =dT/( 4*((Xt-Y)/4)/6.2 + 3*(Y/3)/360 )
    =dT/( (Xt-Y)/6.2 + Y/360 )
    =dT/( Xt/6.2 - Y/6.2 + Y/360 )
    =dT/( Xt/6.2 + (6.2Y - 360Y)/2232 )
    =dT/( Xt/6.2 - 353.8Y/2232 )

    Same as Single spaced Shim, but with shim thicker, thus still approaching Single Thick shim maximum, but with more thinner individual shims.

    Conclusion, Heat Conduction:
    Direct Contact, Thick Shim > Three layered shims, direct contact > Three layered spaced shims > Single Spaced Shim > Thermal Compound only
    (Wherein the additive thickness of the three layered spaced shims is greater than the single spaced shim)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 3, 2009 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    One problem is that your model assumes a perfect conduction between the joins of different materials - where in fact this is the major source of thermal resistance.

    In general the guideline would be
    1, minimize joins, if you must have a join use the minimum amount of heatsink compound to create a seal between the parts
    2, minimize thickness - get the heatsink as close to the source as possible
    3, use the highest conductivity material possible

    Modelling the conduction of a join is difficult, it scales roughly linearly with pressure and slightly less than linearly with surface flatness.
     
  4. Sep 3, 2009 #3
    Thanks for the feedback.
    Number 2 is impossible in this situation. The gap is unavoidable.
    I am following number 3, wherein only Silver and Diamond have higher conductivity than Copper, and the Thermal Compound has the highest conductivity (Tuniq Tx-3) (and other properties that help, like low bleed).

    Concerning 1 and 3, for the Three layered spaced shims and Single Spaced Shim scenario, am I correct in thinking that the higher the percentage of the gap that is filled with Copper, the greater the conductivity rises? Even when it is layered?
     
  5. Sep 3, 2009 #4

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Then the thick copper shim is the best case.
    Pay attention to the joints the copper and heat pipe should be as flat and smooth as possible, the layer of heatsink compound should be very thin - just use enough to coat the surface into a very thin smear
     
  6. Sep 3, 2009 #5
    Yep. I am aware of the optimal application approach. This was a unique situation for me, since the gap is unavoidable.
    I have had the "Thermal Compound only" application in place for the last week, as it was the only thing I had the resources to do. Originally there was a "thick thermal pad/tape" which is both an adhesive and a thermal agent, but a very poor one. When I removed that and replaced it with only thermal grease (which is why low bleed was important, I can't afford for the grease to evaporate and break the physical connection between the die and heatpipe) temperatures in the die have dropped to all time lows for me, and when stressed (even overclocked) it still maxes out with a slightly lower temperature than the previous idle (minimum). So, "Thermal Compound only" is an acceptable solution for me, as long as it persists in its current behavior, but I was looking for a mathematically better solution (rather than just arguing logic in my head).
    If I cannot acquire a thick copper shim, I hope I have semi-correctly proven to myself that a floating shim is better than none, and as much copper in the gap (thus reducing the thickness of grease) as possible will be an even better solution, but obviously not as much as a thick shim with direct contact and minimal grease (just enough to fill microscopic pores, but not too much to cause the shim to float).
    Basically, the more copper in there the better, yes?
    [Note: I am not only looking for the best solution, but how each compares to the others, in case a non-optimal solution is the only practically usable.]
     
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