Hello all,

I have a question on how to handle a boundary (interface) condition in the following scheme. One has a solid and on top of a liquid, with each their properties (k,alpha). On the interface there is a heating element. It is known that the total flux generated is Q (part goes into solid, part into liquid). Furthermore the temperature on the interface has to be continuous offcourse.

I was wondering how one takes care of this interface condition in an implicit finite difference scheme?

thanks

Hi uter

the boundary condtion can be simply implimented by an energy balance of the cell at the interface

consider the cell havin the interface of liquid on top, solid wall, and the heating element qiving in the heat flux of Qf at temp Tf

i-1 i i+1 j+1 *------*------* liquid field

| | |

| | | j *------*------* wall location

| | |

| | | j-1 *------*------* solid field

consider the energy balance at the node (i,j) consider heat flux at (i,j) is Qf

energy content of liquid cells above (i,j) would be

El(i,j) = rho*Cp*(dT/dt) + k*A1*(dT/dx) + k*A2*(dT/dy)

at (i,j) for liquid

Es(i,j) = rho*Cp*(dT/dt) + k*A1*(dT/dx) + k*A2*(dT/dy)

+ Qf*As

at (i,j) for solid

Taking respective Cp and k and other properties according to liquid and solid properly and differentiate them accordingly

I feel this would work well If anything thing is wrong please tell me

all the best bye Ravi Kiran

Hi Ravi, thanks for your answer, but I'm not sure I understand. I'm new to this computational stuff, I'm even new in the field of fluid mechanics, so let me explain the situation a bit better.

I split the temperature fields of the solid and the liquid regions. The situation is cylindrical and we have rotational symmetry, so only the radial and axial components are taken into account. Now, for internal nodes (in liquid as well as solid) the equations are written as:

T(i,j,n+1) = (T(i,j,n) + a*(1-1/2i)*T(i-1,j,n+1) + a*(1+1/2i)*T(i+1,j,n+1) + b*T(i,j-1,n+1) + b*T(i,j+1,n+1))/A

where a = D*dt/dr^2 b = D*dt/dz^2 A = 1 + 2*a + 2*b

The coefficients are different for the liquid and solid in that the diffusivity parameter is different and the axial grid spacing, dz, is different.

Now, on the last row of nodes of the solid (N,j) and the first of the liquid (0,j) the following should hold:

-k_liq*dT_liq/dz_liq + k_sol*dT_sol/dz_sol = Q where Q is the total heat flux, so into the liquid and solid. and

T_liq = T_sol

The problem is that I don't know how the internal node equation is modified to include this condition on the boundary.

What I figure is that the flux can be split into two parts. The part that goes into the solid is (k_solid*Q)/(k_solid+k_liquid) likewise for the liquid.

then equate this flux to the temperature gradient in the respective media. Incorporate these into the internal node equation, and take the average at the interface. This however is not working, but I don't know where the flaw in my reasoning lies.

Hi! uter

Your conditions at the solid-fluid interface are correct. But I think that the method of splitting heat flux may be wrong. (From your posting "The part that goes into the solid is (k_solid*Q)/(k_solid+k_liquid) likewise for the liquid.")

Let's consider one-dimemsional domain below. (S=Solid,Q=Heat Source at interface,L=Liquid)

S-S-S-S-S-Q-L-L-L-L-L

Assuming that the temperature at the left end is fixed at 0. And the temperature at the right end are (1) 0, and (2) 100.

Can you expect that the heat fluxes go into solid and fluid domain are the same values for both cases?

Best regards

worasit

Hi Worasit,

I'm sorry, I didn't mean to say that the two heat fluxes were the same. I meant that the heat flux for the solid is as stated before and for the liquid it is k_liquid*Q/(k_solid+k_liquid), where Q is the total heat flux. So, the ratio of the heatfluxes into the two media is equal to the ratio of there heat conduction parameters Q_solid/Q_liquid = k_solid/k_liquid.

I hope this is more clear.

greetings uter

Hi! uter

I known what you mean from your second posting.

What I want to say is 'the ratio of heat fluxes go into solid and fluid domain is not necessary to be k_solid/k_liquid because it depends on the boundary conditions of your problem and you must left it as an unknown'.

Cheer

worasit

Hi:

does the heat element between the solid and the fluid has thickness, if it has , then it is very easy to do with .

You have a point there Worasit. Would you perhaps have any idea how to do it then, because I'm lost. About the boundary condition: the temperature at the boundaries stay unchanged, the calculational space is chosen such that the energy doesn't reach the boundaries during the time that it is calculated, and so the temperature is the same as the initial temperature.

greets uter

Hi! uter

Because you have two conditions at the interface. You should find the additional equation to solve this problem more easier.

Find the paper about conjugate heat transfer problems, many researchers use "ghost node" to cope the conditions at the interface. I have never use FVM to solve the problems of this type but I'm using FEM to solve them. So, I can comment this point.

After you correct this point, you must check other steps carefully.

Best regards

worasit