# Help me solve a heat conduction/emission transfer problem. Mathematica has failed me

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, $$Q_{in} = Q_{out}$$

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

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

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

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?

• The problem is that you haven't really constructed a heat balance at all. See :stealthskater.com/Personal/Thesis.pdf , page 3. Although for a rod, it can be adapted for a tube quite easily.
– Gert
Mar 8, 2016 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? Mar 8, 2016 at 17:11
• Possibly related: physics.stackexchange.com/q/151209 & physics.stackexchange.com/q/107761 Mar 8, 2016 at 20:03

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: $$kt\frac{\partial^2T}{\partial x^2}=\epsilon \sigma T^4$$

• Which in effect is just differentiating the original equation and being more careful with signs? Mar 8, 2016 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 $U(x) \propto -x^5$. Mar 8, 2016 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. Mar 8, 2016 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. Mar 8, 2016 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 8, 2016 at 18:56