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March 1, 2020, 02:03 
Heat flux boundary condition

#1 
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Hello everyone,
I am simulating airflow in a room with a block heater using a CFD code. Does anyone know how to implement heat flux B.C of the block heater in a finite difference CFD code? In energy equation (I mean temperature transport equation), temperature B.C. can be easily applied to the code however applying heat flux B.C. is a little tricky. Can I use T(i)=q''/h+T(i1) to implement the boundary condition in the fluid domain (T(i) is the temperature on the surface of the heater and shall be computed, T(i1) is the temperature of adjacent node within the fluid)? If yes, how can I determine the h coefficient? Any help or comment is appropriated. Mike 

March 1, 2020, 04:52 

#2  
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Filippo Maria Denaro
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Quote:
On a boundary (for example a wall) you set the continuity between the conduction in the solid (flux q) and the heat flux in the fluid (k Grad T) along the normal to the boundary. That is k dT/dn = q you see that q must be known from the physics of your problem and k is known from the fluid. Then, you have to discretize dT/dn using an asymmetric FD formula (I strongly suggest to use at least a second order formula) that gives the T at the wall as a function of the internal layer and the condition q. 

March 1, 2020, 05:03 

#3  
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As I mentioned for obtaining the wall temperature, q=h(delta T) equation can be used. However the convection heat transfer coefficient (h) is unknown and that's my problem. 

March 1, 2020, 05:32 

#4  
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Filippo Maria Denaro
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No, that depends on the problem you are considering. If you are studying a CHT problem then yes, the temperature field in fluid and solid must be solved together. Conversely, in a classical heat transfer study the heat flux q applied at a wall is a known condition, that is you do not need at all to prescribe h but only the flux. This is a classical non homogenous Neumann condition that is expressed only in terms of the normal derivative of the temperature on the boundary. 

March 2, 2020, 17:04 

#5  
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Pedro Costa
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Quote:
Since you are doing CFD, it doesn't make sense to lump convective effects into a heat transfer coefficient h. You are already resolving convective effects from 'firstprinciples' in the CFD code, since you solve for a velocity field! So, for instance, if you use central finite differences, at the boundary i=0 you could have something like: (T(1)t(0))/dx = q'' > T(0) = q''dx+T(1). But this BC is not centered at the boundary, so it may be nicer to use secondorder onesided differences as suggested above, and you have for constant grid spacing:T(0) = (2*q''dx+4*T(1)T(2))/3 [google is your friend for confirming this result ]. I hope this helps! 

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cfd, finite difference, heat flux 
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