Influence of thermal expansion on thermal conductivityby Hologram0110 Tags: conductivity, expansion, influence, thermal 

#1
Nov612, 10:28 AM

P: 158

Hi everyone.
I was wondering if there is a convention for solving heat conduction problems. Specifically should one account for thermal expansion? The SI units for thermal conductivity are W/m/K. Does m include thermal strains already? It seems for basic problems it would be easier to adjust the thermal conductivity. (ie decrease the conductivity with temperature to account for the fact that the geometry would actually be larger due to thermal conductivity). For more complicated problems where there are elastic and plastic strains where you explicitly define a deformed and undeformed frames it seems like you would want thermal conductivity in the deformed frame. I've been working assuming it is in the deformed frame but a colegue has raised concerns. Is there a well known convention? 



#2
Nov612, 12:22 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,351

For many materials, the change in thermal conductivity with temperature is much bigger than the effect of thermal expansion in your question, and the conductivity may either increase or decrease with increasing temperature for different materials. For example compare low alloy and high alloy steels in http://www.kayelaby.npl.co.uk/genera...2_3/2_3_7.html
The definition of thermal conductivity on that web page answers your question (if you read it carefully!), but in practice, the data is often not accurate enough for it to make any practical difference either way. 



#3
Nov612, 12:42 PM

Sci Advisor
PF Gold
P: 2,234

I've never had to take into account expansion of a volume due to thermal expansion for the purpose of a thermal conduction calculation, I typically just use data for thermal conductivity as a function of temperature and call it good.
A quick analysis:  Typical coefficients of thermal expansion are on the order of 10^5 or 10^6; for example Aluminum is 22.2*10^6 m/m*K. Aluminum's thermal conductivity is 205 W/m*K at 298 K.  Imagine analyzing conduction along a 1 m long bar of aluminum with square crosssection of 1 cm x 1 cm. Thermal resistance for conduction is Rt = L/(K*A). So at 298 K, thermal resistance of the bar is 48.780 K/W.  Now increase the temperature of the bar to 398 K; thermal conductivity goes up to 215 W/m*K. If we ignore the volumetric effects this decreases the thermal resistance to 46.512 K/W.  With the increase in temperature the bar also expands 22 microns in length and the crosssection area increases by 4.4*10^9 m^2. If we take into account the increase of thermal conductivity and the volumetric changes to the bar the thermal reistance comes out to 46.511 K/W. So the thermal expansion accounts for a 0.002% difference in the thermal conductivity (my guess well within the margin of error for the physical constants), where as the thermal conductivity increases by 4.9% just due to temperature change. So you would have to be talking about a very large coefficient of thermal expansion or very large temperature changes for it to become a significant effect. 


Register to reply 
Related Discussions  
Thermal Physics question regarding the thermal expansion coefficients  Introductory Physics Homework  1  
coefficeint of thermal expansion of the ice was the same as coeffiecient of thermal c  Introductory Physics Homework  1  
Derivation of electrical conductivity, Seebeck coefficient and thermal conductivity  Advanced Physics Homework  0  
Converting Thermal Conductivity to Thermal Conduction/Resistance  Electrical Engineering  4  
Thermal conductivity: are liquids better thermal conductors?  Introductory Physics Homework  1 