SI units for Heat transfer equation


by jeff1evesque
Tags: equation, heat, transfer, units
jeff1evesque
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#1
Jul9-09, 03:22 PM
P: 312
1. The problem statement, all variables and given/known data
I am trying to get the units to match up on both sides. [tex]\frac{\partial T}{\partial t} = \frac{\nabla ^{2} T}{k} = \frac{1}{K}(\frac{\partial ^2 T}{\partial x^2} + \frac{\partial ^2 T}{\partial y^2} + \frac{\partial ^2 T}{\partial z^2}})[/tex], where k is the thermal resistivity.

3. The attempt at a solution
By searching all over the web, I've found that thermal resistivity "k" has units of [tex]\frac{Kelvins}{W}[/tex].

Therefore,

[tex]\frac{\partial T}{\partial t} = \frac{Kelvins}{s} \neq (\frac{W}{Kelvins})*(\frac{Kelvins}{m^2})[/tex]. Could someone help me correct the units.


Thanks,


JL
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LowlyPion
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Jul9-09, 03:33 PM
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This may be useful:
http://en.wikipedia.org/wiki/Thermal_resistivity

I think there needs to be a factor of m added to your statement.
jeff1evesque
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Jul9-09, 03:40 PM
P: 312
Quote Quote by LowlyPion View Post
This may be useful:
http://en.wikipedia.org/wiki/Thermal_resistivity

I think there needs to be a factor of m added to your statement.
Oh I see it now, but what about the unit for seconds "s"?

Mapes
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#4
Jul9-09, 04:01 PM
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SI units for Heat transfer equation


k is not the thermal resistivity, it's the inverse thermal diffusivity.
jeff1evesque
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#5
Jul9-09, 04:16 PM
P: 312
Quote Quote by Mapes View Post
k is not the thermal resistivity, it's the inverse thermal diffusivity.
Are you sure? I mean the units do infact check out if you are correct- but according to my class notes they've referred it to thermal resistivity.

Thanks a lot.

JL
Mapes
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#6
Jul9-09, 08:04 PM
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It doesn't matter if I'm sure. Check the literature; ask your teacher for a reference.


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