Rate of conduction heat transfer with different hot-side and cold-side areas

  • #1
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When we measure 'the rate of conduction heat transfer'=Q , we assume that the hot side and the cold side's area are same. But if the both side's area is different to each other, how can i know the rate of conduction heat transfer?
like below figure.
1599003995251.png

1599004185709.png

Would you like to help me?? Thanks.
 
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  • #2
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Summary:: When we measure a thermal conductivity, we assume that the hot side and the cold side's area are same. But if the both side's area is different to each other, how can i know the thermal conductivity?

When we measure a thermal conductivity, we assume that the hot side and the cold side's area are same. But if the both side's area is different to each other, how can i know the thermal conductivity?
Would you like to help me?? Thanks.
Please provide a diagram illustrating the kind of geometric arrangement you are referring to.
 
  • #3
Lnewqban
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Summary:: When we measure a thermal conductivity, we assume that the hot side and the cold side's area are same. But if the both side's area is different to each other, how can i know the thermal conductivity?
...
Thermal conductivity remains the same for same material.
In your case, what changes is the heat flux, which dependds on cross section or area.

Please, see:
https://en.m.wikipedia.org/wiki/Heat_flux
 
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  • #4
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Please provide a diagram illustrating the kind of geometric arrangement you are referring to.
I edited it.
 
  • #5
rude man
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As said, conductivity is a property of the material, not iys geometry.
So you prob. meant conductance or resistance.
Just calculate the reisitance of a differential cross-sectional volume & integrate.
 
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  • #6
jbriggs444
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As said, conductivity is a property of the material, not iys geometry.
So you prob. meant conductance or resistance.
Just calculate the reisitance of a differential cross-sectional volume & integrate.
@Chestermiller is the subject matter expert here, but it seems to me that you would need to make some assumptions of uniformity and directionality for the heat flow before a scalar figure for resistance across a cross-section becomes meaningful.

For a reasonably long and thin object, such assumptions may be reasonable, but it is good to be aware that one is making them.
 
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  • #7
rude man
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@Chestermiller is the subject matter expert here, but it seems to me that you would need to make some assumptions of uniformity and directionality for the heat flow before a scalar figure for resistance across a cross-section becomes meaningful.

For a reasonably long and thin object, such assumptions may be reasonable, but it is good to be aware that one is making them.
For the figure the OP provided, what i suggested would be appropriate.

Of course, you can be infInitely picky, such as heat lost/gained along the figure - but this is supposed to be introductory physIcs. Sturm-Liouville is not yet encountered, for example.
 
  • #8
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Oh, I made a mistake. What i meant was 'not' a conductivity but 'the rate of conduction heat transfer'
Sorry :(
I'll edit my subject.
 
  • #9
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I agree with @jbriggs444. If the cross sectional area were changing very gradually (like toward the right side of the figure, but not like toward the left side of the figure in post #1), you could write $$Q=kA(x)\frac{dT}{dx}$$where Q is the rate of heat flow (independent of x) and A(x) is the local cross sectional area. This would integrate to $$Q=k\frac{\Delta T}{\int_0^L{\frac{dx}{A(x)}}}$$

If the geometry does not vary gradually, one would have to solve the 2D or 3D heat conduction equation, a partial differential equation.
 

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