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February 22, 2006, 09:26 
boundary condition

#1 
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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 

February 23, 2006, 01:42 
Re: boundary condition

#2 
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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 i1 i i+1 j+1 *** liquid field       j *** wall location       j1 *** 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 

February 24, 2006, 12:20 
Re: boundary condition

#3 
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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*(11/2i)*T(i1,j,n+1) + a*(1+1/2i)*T(i+1,j,n+1) + b*T(i,j1,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. 

February 24, 2006, 13:31 
Re: boundary condition

#4 
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Hi! uter
Your conditions at the solidfluid 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 onedimemsional domain below. (S=Solid,Q=Heat Source at interface,L=Liquid) SSSSSQLLLLL 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 

February 25, 2006, 09:27 
Re: boundary condition

#5 
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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 

February 25, 2006, 10:33 
Re: boundary condition

#6 
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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 

February 26, 2006, 12:16 
Re: boundary condition

#7 
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Hi:
does the heat element between the solid and the fluid has thickness, if it has , then it is very easy to do with . 

February 27, 2006, 10:21 
Re: boundary condition

#8 
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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 

February 27, 2006, 11:10 
Re: boundary condition

#9 
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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 

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