An Example of a Flux Boundary Condition
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Let us return again to our heat flow test problem and investigate the effect of the Natural boundary condition. As originally posed, we specified Natural(Phi)=0 on both sidewalls. This corresponds to zero flux at the boundary. Alternatively, a convective cooling loss at the boundary would correspond to a flux

Flux = -K*grad(Phi) = Phi – Phi0  

where Phi0 is a reference cooling temperature. With convectively cooled sides, our boundary specification looks like this (assuming Phi0=0):

 
REGION 1 'box'  
START(-1,-1)  
VALUE(Phi)=0      LINE TO (1,-1)  
NATURAL(Phi)=Phi   LINE TO (1,1)  
VALUE(Phi)=1      LINE TO (-1,1)  
NATURAL(Phi)=Phi   LINE TO CLOSE  

The result of this modification is that the isotherms curve upward:

ex701_001