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jaykobe
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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.
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
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)
[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)
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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