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August 20, 2004, 17:14 
Newton's cooling BC

#1 
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Hi, I am trying to solve a 2d steady state heat transfer problem with finite difference (successive underrelaxation). 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(TT_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

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Mohsen,
I think the BC should relate the NORMAL flux due to conduction with the convection, i.e., K grad(T)*n = h(TT_infinity) (where *n is the scalar product with the normal unit vector), rather than the BC you use ( K. grad(T)=h(TT_infinity) ). 

August 22, 2004, 11:47 
Re: Newton's cooling BC

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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 nonuniform; which is not correct. It could be a mere coding problem though.


August 23, 2004, 02:59 
Re: Newton's cooling BC

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<html>
<head> <meta httpequiv=ContentType content="text/html; charset=windows1255"> <meta name=Generator content="Microsoft Word 10 (filtered)"> <title>Mohsen,</title> <style> </style> </head> <body lang=ENUS> <div class=Section1> <span style='color:#333333'>Mohsen,</span> <span style='color:#333333'>Let's apply the BC</span> <span style='color:#333333'>k grad(T) · n = h(TT<sub>∞</sub>)</span> <p style='textalign:justify'><span style='color:#333333'>at y=0. For simplicity I use vertexcentered temperature and 1<sup>st</sup> order discretization for the derivative. After some simple manipulation, the discrete boundary temperature, T<sub>i,0</sub>, is</span> <span style='color:#333333'>T<sub>i,0</sub> <sub>*</sub>= ( T<sub>i,1 </sub>* Nu T<sub>∞</sub>) / (1Nu)</span> <p class=MsoNormal><span style='color:#333333'>where Nu, the local Nusselt number is defined as</span> <p class=MsoNormal><span style='color:#333333'>Nu = h(x) âˆ†y / k(x,0) .</span> <p class=MsoNormal><span style='color:#333333'>I hope the derivation is correct (please check) and this may help you to validate your code.</span> </div> </body> </html> 

August 23, 2004, 12:11 
Re: Newton's cooling BC

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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 xgradient is very small. I multiplied the dimensions by 1000 and observed there IS a nonzero xgradient. I used the same equation as yours for floor and this:
k grad(T) ¡¤ n = h(TT¡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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