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My problem: A thin-walled tube (length $L$, diameter $D$ and wall thickness $t \ll D$) is in a vacuum. It is held on one end (at $x=0$) by a heat source at constant temperature $T(0)=T_0$. The only way it can dissipates heat is radiatively. I am assuming emission only occurs from the outer surface of the tube. The conductivity of the tube is $k$ in $[W/mK]$ and the emissivity $\epsilon$. What is the equilibrium temperature profile $T(x)$ in the tube? (a numerical approximation will do).

My attempt:

In a steady state, \begin{equation} Q_{in} = Q_{out} \end{equation}

From Fourier's law of thermal conduction, the heat entering through the end section is \begin{equation} Q_{in} = -k \frac{dT}{dx}\Big|_{x=0} \times \pi Dt \end{equation}

From the Stefan-Boltzmann law of Black-body radiation, the heat dissipated through the outer surface of the tube is given by \begin{equation} Q_{out} = \int_0^L \epsilon \sigma T^4 \mathrm{d}x \times \pi D \end{equation}

Equating the two, the problem becomes \begin{equation} -\frac{kt}{\epsilon \sigma} \frac{dT}{dx}\Big|_{x=0} = \int_0^L T^4 \mathrm{d}x,\ \ \ T(0) = T_0 \end{equation}

Trying to solve this in Mathematica is hopeless. Am I doing something wrong? How can I find a local differential form of the equation? Can I simplify it further?

Thanks for your help.

ACuriousMind
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RegencyAndCo
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  • The problem is that you haven't really constructed a heat balance at all. See :http://www.stealthskater.com/Personal/Thesis.pdf , page 3. Although for a rod, it can be adapted for a tube quite easily. – Gert Mar 08 '16 at 16:47
  • Thanks for the reference, it is extremely relevant. However, how have I not constructed a heat balance? How is $Q_{in} = Q_{out}$ not the condition for a steady state? – RegencyAndCo Mar 08 '16 at 17:11
  • Possibly related: http://physics.stackexchange.com/q/151209/ & http://physics.stackexchange.com/q/107761/ – Kyle Kanos Mar 08 '16 at 20:03

1 Answers1

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You need to do a differential heat balance on a small segment of the tube between x and x + $\Delta x$.

Heat in at x = $-\pi Dtk\left(\frac{\partial T}{\partial x}\right)_x$

Heat in at x + $\Delta x$ = $+\pi Dtk\left(\frac{\partial T}{\partial x}\right)_{x+\Delta x}$

Heat lost due to radiation = $\pi D\Delta x\epsilon \sigma T^4$

Heat balance equation:$$+\pi Dtk\left(\frac{\partial T}{\partial x}\right)_{x+\Delta x}-\pi Dtk\left(\frac{\partial T}{\partial x}\right)_x=\pi D\Delta x\epsilon \sigma T^4$$

Dividing by $\Delta x$ and taking the limit as $\Delta x$ approaches zero gives: $$kt\frac{\partial^2T}{\partial x^2}=\epsilon \sigma T^4$$

Chet Miller
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  • Which in effect is just differentiating the original equation and being more careful with signs? – Farcher Mar 08 '16 at 17:00
  • Exact forms for the solution to this equation probably don't exist, but for physical intuition, this will be equivalent to a particle moving under the influence of a potential $U(x) \propto -x^5$. – Michael Seifert Mar 08 '16 at 17:22
  • Thanks @Farcher . I edited the words heat out to read heat in at x + $\Delta x$. I believe this was the only sign issue. I stand by the rest of the signs in the equations. – Chet Miller Mar 08 '16 at 18:17
  • @ChesterMiller In no way was I criticising your derivation. All I was pointing out was the original derivation has a sign difference from yours. – Farcher Mar 08 '16 at 18:31
  • @MichaelSeifert my intuition seems to be a bit rusty, can you explain that a little more (how a particle's trajecory in U(x) = -ax^5 potential is equivalent to the solution?) Thanks – uhoh Mar 08 '16 at 18:56
  • @uhoh: If a particle was moving in such a potential, then it would obey the equation of motion $m \ddot{x} = - dU/dx = 5 a x^4$. Replace $x \to T$, $t \to x$, $kt \to m$, and $5a \to \epsilon \sigma$ in this equation, and you end up with the same differential equation. – Michael Seifert Mar 08 '16 at 19:23
  • @Farcher I don't see any inconstancy between the signs in the original derivation and in my final result. – Chet Miller Mar 08 '16 at 19:34
  • @MichaelSeifert isn't the solution to the problem the time evolution of a temperature distribution:T(x,t) which hopefully will have an assymtotically constant equilibrium shape as t goes to infinity? How does a single particle accelerating rapidly to the left x = f(t) help here? – uhoh Mar 08 '16 at 19:40