Timeline for Heat transfer: boundary conditions with fluid velocity
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2014 at 15:53 | comment | added | Carlo Beenakker | just take $u=u_b$, so no temperature difference between inside and outside at the boundary; you would hope that that then $du/dn=0$ irrespective of $v$, wouldn't you? | |
| Jun 7, 2014 at 9:03 | comment | added | jokersobak | Reasoning against: math.stackexchange.com/questions/822643/… | |
| Jun 7, 2014 at 8:59 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 7, 2014 at 8:56 | comment | added | Carlo Beenakker | the two terms $a\frac{\partial u}{\partial n} - \gamma(u-u_b)(\mathbf v \cdot \mathbf n)$ give the total heat transfer across the boundary, due to diffusion (first term) and convection (second term). The boundary condition says that this heat transfer is linearly proportional to the temperature difference at the boundary (Newton's law of heat transfer). | |
| Jun 7, 2014 at 8:51 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jun 7, 2014 at 6:06 | comment | added | jokersobak | In the case $\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b)$ there can be outgoing heat flow due to convective transfer ($\mathbf v$), can't it? | |
| Jun 7, 2014 at 6:02 | comment | added | jokersobak | In this case we should impose the condition $\beta \geq (\mathbf v \cdot \mathbf n)$. Why is it so (physically)? | |
| Jun 6, 2014 at 12:02 | vote | accept | jokersobak | ||
| Jun 6, 2014 at 11:42 | comment | added | Carlo Beenakker | if the temperature $u$ equals the temperature $u_b$ of the surroundings, there can be no outgoing heat flow, or you'd be violating the second law of thermodynamics... | |
| Jun 6, 2014 at 11:26 | comment | added | jokersobak | Thank you! This was a typo. I meant the condition $a\dfrac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b) = 0$. The heat flux is $\mathbf q = -a\nabla u + \mathbf v u$. Why $(\mathbf v \cdot \mathbf n)(u-u_b)$ instead of $(\mathbf v \cdot \mathbf n)u$? | |
| Jun 6, 2014 at 10:39 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |