# Newton's cooling BC

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 August 20, 2004, 17:14 Newton's cooling BC #1 Mohsen Guest   Posts: n/a Hi, I am trying to solve a 2-d steady state heat transfer problem with finite difference (successive under-relaxation). I am trying to apply a convective boundary condition on a horizontal surface with a variable convection heat transfer coeffient. i.e. Heat is to convect outside from a horizontal surface with known but variable (in x, of course) convection heat transfer coefficient, h. I expect the temperature on the surface be variable. But it is not. I use this BC: |K. grad(T)|=h(T-T_infinity) where || means "magnitude," and K is the conductivity. Then I discretize it: backward in x and forward in y (since the convective surface is a bottom surface) and find T and use it as BC in my equations. It really seemed like an easy problem at first. When I use an scheme O(dx^2, dy^2), backward for x and forward in y, it never converges. But O(dx, dy^2) converges however the temperature is constant on the surface which it should not be. I really appreciate any help/idea. Thanks, Mohsen

 August 22, 2004, 03:30 Re: Newton's cooling BC #2 Rami Guest   Posts: n/a Mohsen, I think the BC should relate the NORMAL flux due to conduction with the convection, i.e., K grad(T)*n = h(T-T_infinity) (where *n is the scalar product with the normal unit vector), rather than the BC you use ( |K. grad(T)|=h(T-T_infinity) ).

 August 22, 2004, 11:47 Re: Newton's cooling BC #3 Mohsen Guest   Posts: n/a thanks Rami. I tried that one first. And I think you are right. In that case, I get a uniform temprature on the surface whose heat convection coefficient is non-uniform; which is not correct. It could be a mere coding problem though.

 August 23, 2004, 02:59 Re: Newton's cooling BC #4 Rami Guest   Posts: n/a Mohsen,
Mohsen, Let's apply the BC -k grad(T) · n = h(T-T)

at y=0. For simplicity I use vertex-centered temperature and 1st order discretization for the derivative. After some simple manipulation, the discrete boundary temperature, Ti,0, is Ti,0 *= ( Ti,1 *- Nu T) / (1-Nu)

where Nu, the local Nusselt number is defined as

Nu = h(x) âˆ†y / k(x,0) .

 August 23, 2004, 12:11 Re: Newton's cooling BC #5 Mohsen Guest   Posts: n/a Thanks Rami. Last night I figured all this time the code was giving me correct results! My geometry is super small (micro) therefore the temperature x-gradient is very small. I multiplied the dimensions by 1000 and observed there IS a non-zero x-gradient. I used the same equation as yours for floor and this: k grad(T) ¡¤ n = h(T-T¡Þ) fot ceiling. I used second order backward for x. For ceiling I used second order forward and for floor I used second order backward. And it worked!

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