Heat Conductivity, Copper and Thermal Compound scenarios

In summary, the best scenario for heat conduction in this situation is using a thick copper shim. It is important to minimize joints and use a thin layer of heatsink compound to increase heat conduction efficiency. The conductivity can also be improved by filling a higher percentage of the gap with copper.
  • #1
jaykobe
3
0
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.

Homework Statement


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

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

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

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

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

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

Homework 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)

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)
 
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  • #2
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.
 
  • #3
mgb_phys said:
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.

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?
 
  • #4
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
 
  • #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.]
 

1. What is heat conductivity?

Heat conductivity refers to the ability of a material to transfer heat. It is a measure of how well a material can conduct heat through it, and is typically denoted by the symbol "k". Higher values of heat conductivity indicate that a material is more efficient in transferring heat.

2. How does copper compare to other materials in terms of heat conductivity?

Copper is known for its high heat conductivity, making it one of the most efficient materials for transferring heat. It has a heat conductivity value of 401 W/mK, which is significantly higher than other common metals such as iron (80 W/mK) and aluminum (237 W/mK).

3. What is the role of thermal compound in heat conductivity scenarios?

Thermal compound, also known as thermal paste, is a substance used to improve the heat transfer between two surfaces. In heat conductivity scenarios, thermal compound is applied between the CPU and heatsink to fill any microscopic gaps and improve the contact between the two surfaces, thus increasing the efficiency of heat transfer.

4. Is there a difference between thermal conductivity and thermal resistance?

Yes, there is a difference between thermal conductivity and thermal resistance. Thermal conductivity (k) is a measure of how well a material conducts heat, while thermal resistance (R) is a measure of how much a material resists the flow of heat. They are inversely proportional to each other, with the relationship being R = 1/k. This means that materials with higher heat conductivity will have lower thermal resistance.

5. How can I calculate the heat transfer rate in a copper wire?

The heat transfer rate in a copper wire can be calculated using the formula Q = (k x A x ΔT)/L, where Q is the heat transfer rate, k is the heat conductivity of copper, A is the cross-sectional area of the wire, ΔT is the temperature difference between the two ends of the wire, and L is the length of the wire. This formula can also be used to calculate heat transfer rates in other materials, as long as their respective heat conductivity values are used.

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