[insist96]

hello,
i have a question about the setting of thermal boundary conditions in fluid-solid coupling analysis.

i want to coupled the fluid and solid surface and set a heat flux into the solid surface,but i find that coupled and heat flux couldnt be set at the same time.and if choose coupled,there are options for heat generation rate and wall thickness.what does this mean?

whether shoud i convert heat flux to heat genaration rate by thickness?if its wrong,how can i set a heat flux into solid surface with fluid-solid coupled.

poor my english and thank you very very much.

[MKuhn]

If you choose coupled heat will be transfer from your fluid to your solid, otherwise not. The heat transfer coefficient is calculated by the solver. So you can not define a heat flux as it would be overdeterminated. Heat generation rate means an additional heat source in your solid, like a heating wire. With wall thickness you can define additional layers between your solid and fluid like paint, insulation or heating film.

A coupled boundary is always an internal boundary in your model. You need an outer boundary to set a heat flux to a solid surface (which is not in contact with a fluid)

[AlexanderZ]

whether shoud i convert heat flux to heat genaration rate by thickness?if its wrong,how can i set a heat flux into solid surface with fluid-solid coupled.

that is what you should do.
define material property.
set wall thickness (could be very small)
define heat generation rate [W/m3]
total heat [W] = heat generation rate * wall thickness

[insist96]

Originally Posted by MKuhn

If you choose coupled heat will be transfer from your fluid to your solid, otherwise not. The heat transfer coefficient is calculated by the solver. So you can not define a heat flux as it would be overdeterminated. Heat generation rate means an additional heat source in your solid, like a heating wire. With wall thickness you can define additional layers between your solid and fluid like paint, insulation or heating film.

A coupled boundary is always an internal boundary in your model. You need an outer boundary to set a heat flux to a solid surface (which is not in contact with a fluid)

Thank you for your reply. As you mentioned, the heat generation rate means that there is another heat source in the solid. So what is the difference between defining the heating film during the modeling process and giving a thickness and heat generation rate when defining the coupled interface?

[insist96]

Originally Posted by AlexanderZ

that is what you should do.
define material property.
set wall thickness (could be very small)
define heat generation rate [W/m3]
total heat [W] = heat generation rate * wall thickness

Thank you for your reply. Then how could I set the wall thickness? What is the basis for setting the thickness? Whether the thickness is virtual and it doesn't really exist in the result display?
If i've defined a heating film, so does it means that this heating film transfer heat to both the solid and the fluid surface, which is a little bit different from the heat that I want to just load into the solid? How can I do? Or I'm wrong and I don't have to worry about it?

[MKuhn]

Originally Posted by insist96

As you mentioned, the heat generation rate means that there is another heat source in the solid.

Yes like a heater in a solid block or due to radioactivity. In this case the heat is generated in the cells of your solid.

If you define a wall thickness, then fluent will create some virtual cells. In this case you can add a solid layer to a wall, even if you have only a fluid domain.

Originally Posted by insist96

so does it means that this heating film transfer heat to both the solid and the fluid surface

- Yes, like an electrical film heater

Originally Posted by insist96

which is a little bit different from the heat that I want to just load into the solid?

- depending on the thermal conductivity of your solid the heat will also transfer into your fluid, as the fluid is in contact with your solid (coupled boundary) - only the path is longer.


What is your heat source in reality?

[insist96]

we want to simulate the disc during braking. In this process, the surafce of disc will produce frictional heat, and then heat will flow into the disc body by conduction and dissipate into the air by heat convection. In the traditional thermal calculating model, the coefficient of heat convection was obtained by experical equation and then add them to the disc surface. Now, we want to simulate the heat convection effect by coupling the fluid calculation.

[insist96]

Originally Posted by MKuhn

Yes like a heater in a solid block or due to radioactivity. In this case the heat is generated in the cells of your solid.

If you define a wall thickness, then fluent will create some virtual cells. In this case you can add a solid layer to a wall, even if you have only a fluid domain.

- Yes, like an electrical film heater


- depending on the thermal conductivity of your solid the heat will also transfer into your fluid, as the fluid is in contact with your solid (coupled boundary) - only the path is longer.


What is your heat source in reality?

we want to simulate the disc during braking. In this process, the surafce of disc will produce frictional heat, and then heat will flow into the disc body by conduction and dissipate into the air by heat convection. In the traditional thermal calculating model, the coefficient of heat convection was obtained by experical equation and then add them to the disc surface. Now, we want to simulate the heat convection effect by coupling the fluid calculation.

[MKuhn]

If its necessary to concentrat the heat generation in the surface, than use wall thickness with the thickness of the surface where the heat generation occurs. In this way the heat will transfered to the solid and the fluid. The value in w/m³ is (power of friction heat [W])/(disk surface * wall thickness). If the disk is thin, than you have to subtract the wall thickness from the disk thickness. You can not use the heat transfer coefficient (W/(m² K) as input. You need the total friction heat in Joule and you have to define a time in which the friction heat will disappear. Joule/s = W

[insist96]

Originally Posted by MKuhn

If its necessary to concentrat the heat generation in the surface, than use wall thickness with the thickness of the surface where the heat generation occurs. In this way the heat will transfered to the solid and the fluid. The value in w/m³ is (power of friction heat [W])/(disk surface * wall thickness). If the disk is thin, than you have to subtract the wall thickness from the disk thickness. You can not use the heat transfer coefficient (W/(m² K) as input. You need the total friction heat in Joule and you have to define a time in which the friction heat will disappear. Joule/s = W

thank you very much. And how to define the thickness of the wall? Whether the smaller the value,the better it is?

[MKuhn]

Originally Posted by insist96

thank you very much. And how to define the thickness of the wall? Whether the smaller the value,the better it is?

You wrote that the surface produce the friction heat, so I would recommend this thickness. If you have no idea, take 1/100 of the total thickness of the disk.

[insist96]

Originally Posted by MKuhn

You wrote that the surface produce the friction heat, so I would recommend this thickness. If you have no idea, take 1/100 of the total thickness of the disk.

thank you very very much!