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Heat conduction in solids - temperature dependent heat transfer coefficient

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Hi,

I would like to solve a simple heat conduction in solids problem with a temperature-dependent heat transfer coefficient at one of the boundaries

h(T) = 1000*(300/T)^1.3

However, the result computed by Comsol is different than I would expect from analytical techniques and an equivalent Comsol model with a solid layer of very low thermal conductivity.

Are there any issues I should be aware of when using temperature-dependent heat transfer coefficients?

Thanks,
Kevin

8 Replies Last Post 2013年2月23日 GMT-5 07:54
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2013年2月5日 GMT-5 01:52
Hi

That should work, provided you have T in Kelvin, I believe you have models in the model library with all sorts of HT equations, anyhow, most material data has T dependent parameters. The only thing your model becomes non-linear and might take longer to solve

--
Good luck
Ivar
Hi That should work, provided you have T in Kelvin, I believe you have models in the model library with all sorts of HT equations, anyhow, most material data has T dependent parameters. The only thing your model becomes non-linear and might take longer to solve -- Good luck Ivar

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Posted: 1 decade ago 2013年2月5日 GMT-5 17:37
Hi

I tried solving this problem and am concerned that Comsol may not be handling the temperature-dependent heat transfer coefficient properly.

The problem that I want to solve is described in the "convectiveBC" files. I would like to compute the temperature of the solid rectangle with temperature-dependent thermal conductivity and heat transfer coefficient at one of the boundaries. To do this, I have include files with constant (temperature-independent) and variable (temperature-dependent) values.

However, I think there is an equivalent conduction problem involving two rectangles and an interfacial resistance with no convection BCs. In this case, the convection BC is replaced by a solid with very high thermal conductivity with some interfacial resistance between it and the solid domain I would like to solve for.

When the properties are constant (temperature-independent), the temperature is almost the same for these two cases. However, when the properties and BCs are temperature-dependent, the peak temperature differs by ~30 K.

Could you check these files?

Thanks,
Kevin
Hi I tried solving this problem and am concerned that Comsol may not be handling the temperature-dependent heat transfer coefficient properly. The problem that I want to solve is described in the "convectiveBC" files. I would like to compute the temperature of the solid rectangle with temperature-dependent thermal conductivity and heat transfer coefficient at one of the boundaries. To do this, I have include files with constant (temperature-independent) and variable (temperature-dependent) values. However, I think there is an equivalent conduction problem involving two rectangles and an interfacial resistance with no convection BCs. In this case, the convection BC is replaced by a solid with very high thermal conductivity with some interfacial resistance between it and the solid domain I would like to solve for. When the properties are constant (temperature-independent), the temperature is almost the same for these two cases. However, when the properties and BCs are temperature-dependent, the peak temperature differs by ~30 K. Could you check these files? Thanks, Kevin


Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2013年2月6日 GMT-5 09:02
Hi

I do not see that big differences, if for your thin resistive layer add a filter with a (dom!=1) Boolean operator you will see that the T profiles, in the domains 2&3 are quite similar for all your models.

The absolute values differ but if you take a simpler model (work in 1D) for the same heat flux you will get different temperature values depending on how you set up the model, and for me the difference in absolute values are due to the simplifications of the model.

Any how Heat flux are not a conservative property, while the total energy yes.

Try it out on 1D, you see too that T does not simply add up independently of your equation

--
Good luck
Ivar
Hi I do not see that big differences, if for your thin resistive layer add a filter with a (dom!=1) Boolean operator you will see that the T profiles, in the domains 2&3 are quite similar for all your models. The absolute values differ but if you take a simpler model (work in 1D) for the same heat flux you will get different temperature values depending on how you set up the model, and for me the difference in absolute values are due to the simplifications of the model. Any how Heat flux are not a conservative property, while the total energy yes. Try it out on 1D, you see too that T does not simply add up independently of your equation -- Good luck Ivar

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Posted: 1 decade ago 2013年2月6日 GMT-5 10:04
Hi,

Thanks for your help. Do you know what temperature Comsol uses to evaluate the heat transfer coefficient if it is temperature-dependent? For h = h(T), which T is used?

Also, when a temperature-dependent relationship ks = ks(T) is used for a thin thermally resistive layer, which temperature is used in evaluating the thermal conductivity?

Thanks,
Kevin
Hi, Thanks for your help. Do you know what temperature Comsol uses to evaluate the heat transfer coefficient if it is temperature-dependent? For h = h(T), which T is used? Also, when a temperature-dependent relationship ks = ks(T) is used for a thin thermally resistive layer, which temperature is used in evaluating the thermal conductivity? Thanks, Kevin

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2013年2月6日 GMT-5 10:30
Hi

careful be sure you understand what "T" means for COMSOL,

its in fact a "field" T(x,y,z,t) in all generality, hence each dx*dy*dz around a position (x0,y0,z0), will take each a different temperature found by interpolating T(x0,y0,z0) with respect to the T grid mapping (mesh element and discretizaion)

--
Good luck
Ivar
Hi careful be sure you understand what "T" means for COMSOL, its in fact a "field" T(x,y,z,t) in all generality, hence each dx*dy*dz around a position (x0,y0,z0), will take each a different temperature found by interpolating T(x0,y0,z0) with respect to the T grid mapping (mesh element and discretizaion) -- Good luck Ivar

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Posted: 1 decade ago 2013年2月22日 GMT-5 13:49
I faced the same problem.

I have a 3D model. I heat parallelepiped by laser and want to know it's temperature depending on the laser power. And I found that with laser power 0.9 mW and constant h = 8000 W/m^2/K I have temperature around 1400K which is consistent with my experimental results. But at laser power 0.17 I want to have temperature 300K. To do this I need to increase h to h = 200000 W/m^2/K. Then for my experiments with laser power in between 0.17 and 0.9 mW I want to extrapolate h to be linearly dependent on temperature. Than I want h to be the next:

h[W/m^2/K] = -219[W/m^2/K^2] * T[K] + 314000[W/m^2/K]

So, if I put temperature T=1400K to this expression h will be h = 7400W/m^2/K which is very close to 8000W/m^2/K

When I put this dependence for h to COMSOL and calculate temperature with laser 0.9mW again I get 344K which is very inconsistent with what I had before.

How should I put this dependence to COMSOL? Am I doing some mistake?

Thanks.
I faced the same problem. I have a 3D model. I heat parallelepiped by laser and want to know it's temperature depending on the laser power. And I found that with laser power 0.9 mW and constant h = 8000 W/m^2/K I have temperature around 1400K which is consistent with my experimental results. But at laser power 0.17 I want to have temperature 300K. To do this I need to increase h to h = 200000 W/m^2/K. Then for my experiments with laser power in between 0.17 and 0.9 mW I want to extrapolate h to be linearly dependent on temperature. Than I want h to be the next: h[W/m^2/K] = -219[W/m^2/K^2] * T[K] + 314000[W/m^2/K] So, if I put temperature T=1400K to this expression h will be h = 7400W/m^2/K which is very close to 8000W/m^2/K When I put this dependence for h to COMSOL and calculate temperature with laser 0.9mW again I get 344K which is very inconsistent with what I had before. How should I put this dependence to COMSOL? Am I doing some mistake? Thanks.

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Posted: 1 decade ago 2013年2月22日 GMT-5 18:04
Hi,

Thanks for your help and advice. I believe I realized why the two models I proposed were different. When the thin thermally resistive layer boundary condition is used,

ks = ks(T)

the temperature, T, used is the average of the temperatures on the upper and lower sides of the interface. Therefore, the interfacial conductance is different from the case when I used

h = h(T)

with the convective boundary condition.

Perhaps this point (that the thin resistive layer thermal conductivity is evaluated at the average of the upper and lower temperatures) should be included in the manual. I didn't see it in there.

Thanks,
Kevin
Hi, Thanks for your help and advice. I believe I realized why the two models I proposed were different. When the thin thermally resistive layer boundary condition is used, ks = ks(T) the temperature, T, used is the average of the temperatures on the upper and lower sides of the interface. Therefore, the interfacial conductance is different from the case when I used h = h(T) with the convective boundary condition. Perhaps this point (that the thin resistive layer thermal conductivity is evaluated at the average of the upper and lower temperatures) should be included in the manual. I didn't see it in there. Thanks, Kevin

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2013年2月23日 GMT-5 07:54
Hi

its corrct that the thin feature provide the "spatial" average value of T, in fact

( up(T) + down(T) ) / 2

You can also plot up(T) and down(T) on a boundary for comparison ;)

--
Good luck
Ivar
Hi its corrct that the thin feature provide the "spatial" average value of T, in fact ( up(T) + down(T) ) / 2 You can also plot up(T) and down(T) on a boundary for comparison ;) -- Good luck Ivar

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