Newton's cooling BC

August 20, 2004, 18:14

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, 04:30

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, 12:47

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, 03:59

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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) .

I hope the derivation is correct (please check) and this may help you to validate your code.

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August 23, 2004, 13:11

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!

August 20, 2004, 18:14	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, 04:30	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, 12:47	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, 03:59	Re: Newton's cooling BC	#4
Rami Guest Posts: n/a	<html> <head> <meta http-equiv=Content-Type content="text/html; charset=windows-1255"> <meta name=Generator content="Microsoft Word 10 (filtered)"> <title>Mohsen,</title> <style> </style> </head> <body lang=EN-US> <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(T-T<sub>∞</sub>)</span> <p style='text-align:justify'><span style='color:#333333'>at y=0. For simplicity I use vertex-centered 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>) / (1-Nu)</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, 13:11	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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