Discussion Forum

The gold sphere is on a block of wood and above the block of wood is a cube of air? If so, there's two obvious choices for boundary conditions: reflecting (symmetric) will block the transfer of heat, or isothermal which will permit the flow of heat but maintain a constant temperature.

One thing I'd worry about is if the sphere is balanced on the block of wood then the boundaries will touch at only one point, and will be tangential at that point. This may make for some ugly meshing. It might be better to have the block extend slightly into the wood by some amount, or to put a small cylinder of wood extending up under the sphere... anything to get rid of that singularity.

As an aside I think if your box is L = 20 R, that's still too small to emulate infinite. The 3-dimensional spreading resistance from a sphere of radius R to infinity is proportional to 1/R. So if I truncate space at distance 10R, I get (1/R - 1/10R) = 0.9 / R -- I'm making an approximate 10% error in thermal resistance.

Hi Daniel,

Thank you for your reply !

To be clearer, my sphere is enclosed in the box fulled with air. The sides of this box are made of wood and the system {sphere + box} is embedded in air (infinite medium). According to experiments, the finite size of the box is to be taken into account in the calculations of the cooling of the sphere and all of the physical phenomena relating to the cooling of a hot body (convection, conduction...).

The box represents a wooden house...The design can be seen in attached file.

Than you for your help.

KB

Hi

This reminds me of chemical engineering course problems :)

You need the overall heat transfer coefficient of the wood wall, k_tot:

1/k_tot = 2/k_(a/w) + h/l_w

where k_(a/w) is the heat transfer coefficient between air and wood (I do not know where to find it), l_w is the heat conducivity of wood and h = 30 cm (thickness).

Next, what is the contact area between the gold sphere and wood? Can be a point.

Provided that these constants are known, the problem is solved. Use the Newton's cooling law for the heat flux between interior and exterior air (the latter is constant): Q = k_tot*(T_out - T_in).

At the contact of gold and wood yet another heat transfer coefficient is needed. If no BC is given here, the temperature profile is continuous, T_gold = T_wood.

The contact gold-air in the box can be treated in various ways, perhaps a thin barrier could be the simplest solution.

I wish this was helpful.

br
Lasse

Is this what you wanted?

br
Lasse

Hi Lasse,

Thank you for your reply and nice job. I will go perform the model by using your advices. Unfortunately, I use Comsol 4.2 so that I cannot open your file.

I'll let you know !

Best

KB

Also, note the only thing not axially symmetric about the problem is the house, so if you can approximate it as a circular hut instead, then you can solve the problem in 2D (with axial symmetry) which is faster.

For example, there are 4 walls in a square box enclosing area L^2. A circle of the same area will have diameter L sqrt(4/π). If you get this working well then you can switch to the more detailed 3D model.

PDF report.

Hi Lasse,

Please, sorry for this late post reply. Thank you for this complete report ! I have some questions about the setting of comsol to solve from your report :

You write :

*Heat transfer in fluids
- Heat flux 1 : k_tot*(Text-T) (boundaries 4,8,12)
- Heat flux 2 : k2*(T2-T) (boundaries 20,21)

*Heat transfer in solids 2
- Heat flux 1 : k2*(T2-T) (boundaries 20,21)
- Heat flux 2 : k*(T-T2) (boundaries 4,8,12)
- Heat flux 3 : -k*(T2-Text) (boundaries 2,9,11,16-19)

1) I understand that "T" is the temperature of air and "T2" is the temperature of the sphere. In the sections "Heat transfer in fluids", the "Heat flux 1" and "Heat flux 2" correspond to the heat transfers by convection : sphere-air and medium exterior-air respectively...Is it exact ?

2) In the section, "Heat transfer in solids 2", I do not understand the expression of q0 that you used...These expressions correspond to a heat transfer by convection as in the previous section ?

Why this egality : Heat flux 1 (Heat transfer in solids 2) = Heat flux 2 (Heat transfer in fluids) = k2*(T2-T) ?

I do not understand the physic represented by these last expressions...I feel that there are redundancies ?

3) A last think...In your report, there is not expression correspond to the heat transfer by conduction between the sphere and the surface on which it is placed...There is a contact (point) between theses two materials...Why ?

Hi

I used Newton's cooling law everywhere at the phase boundaries:

heat flux = k*(Ta - Tb)

Whether it is called convective or general expression does not matter because that is the equation used in the simulation. There is no convection in the model.

Because we have two physics nodes which meet at the phase boundary, the same boundary equation must be presented for both of them, only the sign is changed. When the solid wood receives heat, q = k*(T2 - T), air inside the box looses the same amount q = -k*(T2 - T) = k*(T2 - T). But I made a short cut: I expressed the heat loss of air inside the box with the overall heat transfer coefficient, as I described earlier in this thread.

T is the temperature in solids, both wood and gold, the only difference being in their material properties; T2 is the temperature in the air inside the box. In the Heat Transfer in Solids node, Heat Flux 1 is the cooling law between the outer surface of the cube and the exterior, Heat Flux 2, describes heat transfer between the inner surface of the cube and inner air, Heat Flux 3 heat transfer between the gold sphere and air.

In the Heat Transfer in Fluids node, Heat Flux 1 is the loss of heat across the wooden walls, as explained above, and Heat Flux 2 heat transfer from gold to air; this is equal to Heat Flux 3 in the solid node, only with the sign reversed. Maybe this confuses you that T is the temperature variable for both wood and gold.

I just put the sphere resting on the bottom of the box, giving a contact point, because I did not know where to put it. Is it hanging in the air?

I wish I was clear enough :)

BR
Lasse

Hi,

Thank you for your reply. Now I completely understood your report.

In my problem the object is in fact a half sphere placed on the ground at the center of the room. From your model, I just add a heat flux node to take into account the heat transfer by conduction (q = -k dT/dx) between the sphere and the ground (the surface of the transfer is pi*R^2) but, I feel that this latter term (Fourier's equation) is not necessary to be added...

K