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Converting Thermal Conductivity to Thermal Conduction/Resistance

 
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Aug12-09, 09:12 AM   #1
 

Converting Thermal Conductivity to Thermal Conduction/Resistance


Hi all

I want to convert thermal conductivity to thermal conductance. The problem is with the units.

for thermal conductivity the units are [W/(m.K)] and for thermal conductance the units are [W/K]

For E.g for copper thermal conductivity is 400[W/m.K]
how can I convert it to thermal conductance?Can it be simply by multiplying it my the length of the copper wire I am using?

Any help with be really appreciated.

Thanks and Regards
Hamza
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Aug12-09, 10:29 AM   #2
 
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Hi Hamza!
Quote by hamzaaaa View Post
for thermal conductivity the units are [W/(m.K)] and for thermal conductance the units are [W/K]

For E.g for copper thermal conductivity is 400[W/m.K]
how can I convert it to thermal conductance?Can it be simply by multiplying it my the length of the copper wire I am using?
Yup!

You can always rely on the units …

just multiply by the length (in metres) …
"/m" means exactly what it says
Aug12-09, 06:16 PM   #3
 
hmm.. I think you need to multiply by the cross-sectional area and then divide by the length. The conductance should decrease with increased length and increase with larger area.

Consider if you add a second wire of equal size. You would expect the conductance to increase by a factor of 2 (and resistance drop by a factor of 2).
Aug13-09, 02:20 AM   #4
 
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Converting Thermal Conductivity to Thermal Conduction/Resistance


Quote by davidrit View Post
hmm.. I think you need to multiply by the cross-sectional area and then divide by the length. The conductance should decrease with increased length and increase with larger area.

Consider if you add a second wire of equal size. You would expect the conductance to increase by a factor of 2 (and resistance drop by a factor of 2).
oops! davidrit is right!

(I thought the thermal conductivity was of the wire, but I see now it's of the material … see http://en.wikipedia.org/wiki/Thermal_conductivity)
Aug13-09, 09:23 AM   #5
 
Thanks all...
very much appreciated.
I have a further query now which I have posted separately.

hxxp://www.physicsforums.com/showthread.php?t=330491
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