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Suppose you have a small container, say 2 cubic centimeters (1cm x 1cm x 2cm), positioned so the 2cm side is vertical, and there is a tiny light bulb or heating element at the very top.

Is it the case that the hotter air near the bulb will stay at the top?

Or will the air inside will be approximately the same temperature because it is so small, and the conduction or brownian motion or radiation will surpass the effects of convection?

Is it reasonably feasible to attempt to estimate the difference in temperature between the top and bottom of the container?

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  • $\begingroup$ Can you remark on the wall thickness and material? I'm trying to gauge whether the horizontal air temperature will be uniform or whether the air will be cooler next to the walls. $\endgroup$ Jan 29, 2023 at 18:39
  • $\begingroup$ @Chemomechanics Say it's a 1cm thick foam wall. $\endgroup$ Feb 2, 2023 at 19:08
  • $\begingroup$ A light bulb will produce IR radiation that will not be absorbed by the air well, most of it will be absorbed by the solid walls of the container, which will heat up first. Then conditions for the convection are quite plausible. But magnitude of temperature gradient matters too - if temperature variation across the system are small enough, convection may not occur and all transport will be by radiation and conduction. $\endgroup$ Feb 2, 2023 at 21:46

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Given this description, yes, the hotter air near the heater will likely stay at the top.

Here's the reasoning: There's no source of forced convection such as a fan, and there's no strong driving force for natural convection, as the densest material (i.e., the colder air) is generally already at the bottom. You've indicated that the walls are foam (similar thermal conductivity to air), so large thermal gradients are unlikely to form that might drive small convection cells.

Aside from actually building a prototype or conducting a numerical simulation using finite elements, for example, there are a couple of analytical approaches we might use to estimate the internal air temperature.

Let's assume radiative transfer is negligible for simplicity; this will give us the maximum temperature difference between the top and bottom of the box. If convection is absent inside the box, that leaves conduction as the only relevant heat transfer mechanism.

One analytical approach relies on a happy coincidence: since the foam walls have a similar thermal conductivity to air, let's just forget about the walls and model the spherical conduction of heat emanating from the source. The heat equation at steady state is simply $\nabla^2 T=0$. Working with spherical symmetry, we integrate twice to obtain $T(r)=C_2-\frac{C_1}{r}$, with temperature $T$, radial distance $r$, and constants $C_1$ and $C_2$. At infinite distance, the temperature must be the ambient temperature $T_\infty$, so $C_2=T_\infty$. At any radial distance, Fourier's law gives a radial heat flux of $-k\frac{dT}{dr}=-\frac{C_1k}{r^2}$, and multiplied by the surface area of the sphere at that distance, $4\pi r^2$, we must recover the heater power $J$, so $-4\pi C_1k=J$ and $T(r)=T_\infty+\frac{J}{4\pi kr}$. Thus, at 2 cm, a 1 W heater is predicted to produce a temperature rise of about 150°C at the bottom of the box, increasing with decreasing distance to the heater. One disadvantage of this approach is that the solution diverges at the heater location. Also, the fact that air convectively mixes outside the box is ignored. (For example, 1 cm above the heater, the open air is probably around room temperature, but this equation predicts a temperature increase of 306°C.)

To try to address the boundary condition problem (namely, that the room-temperature boundary condition is not at infinity but several centimeters from the heater), we could set $T_\infty$ at 2 cm, for example. Now the temperature profile is predicted to be $T(r)=T_\infty+\frac{J}{4\pi k}\left(\frac{1}{r}-\frac{1}{2\,\mathrm{cm}}\right)$, and we see a gradient of about 150°C moving in the first centimeter. Hey, that's broadly similar to the earlier prediction: maybe our estimate within the box should be around 150°C per watt per vertical centimeter, for a foam enclosure.

Another analytical approach is to divide the inside air into horizontal slices with heat transfer to the air above and below and heat transfer to the container wall material along the perimeter. The steady-state 1-D heat equation in this case ends up being $$\frac{d^2T}{dx^2}-\frac{k_\mathrm{wall}P}{k_\mathrm{air}At}(T-T_\infty)=0,$$ with distance from the bottom $x$, wall thermal conductivity $k_\mathrm{wall}$, perimeter length $P$, air thermal conductivity $k_\mathrm{air}$, wall thickness $t$, and vertical cross-sectional area $A$. (For the geometry and materials you've specified, that coefficient $\frac{k_\mathrm{wall}P}{k_\mathrm{air}At}$ ends up reducing to approximately $\frac{4}{t^2}$, where $t = 1\,\mathrm{cm}$.) This heat equation is derived from an energy balance on each horizontal slice. The advantages here are that any wall material can be incorporated, and the solution won't diverge at the heater. The boundary equations would be

$$q^{\prime\prime}_\mathrm{bottom}=-k_\mathrm{air}\left.\frac{dT}{dx}\right|_{x=0}=-k_\mathrm{wall}\frac{T(0)-T_\infty}{t};$$

$$q^{\prime\prime}_\mathrm{top}=-k_\mathrm{air}\left.\frac{dT}{dx}\right|_{x=L}=-\frac{J}{A}+k_\mathrm{wall}\frac{T(L)-T_\infty}{t};$$

with interior box height $L$; that is, the heat flowing out of the air at the bottom must equal the heat flow through the bottom of the box, and the heat flowing out of the air at the top must equal the heat flow through the top of the box minus the heater areal power. Now, this approach ignores diagonal heat flow through the corners of the box; unfortunately, the 1-D approach can't capture that nuance, and finite element solution might be needed for better precision.

The solution to the 1-D heat equation above is $T(x)=T_\infty+C_3\exp(\sqrt{h}x)+C_4\exp(-\sqrt{h}x),$ where $h$ is the coefficient $\frac{k_\mathrm{wall}P}{k_\mathrm{air}At}$. This can be inserted into the boundary conditions above to find $C_3$ and $C_4$ and then the temperature difference between the top and bottom of the box

$$T(L)-T(0)=C_3\left[\exp(\sqrt{h}L)-1\right]+C_4\left[\exp(-\sqrt{h}L)-1\right];$$ the process is somewhat tedious, so I won't derive the complete solution here. I will, however, run through the example for this particular set of geometry and materials, for which $\sqrt{h}\approx\frac{2}{t}$ and thus $\sqrt{h}L\approx 4$. Here, we find that $C_3\approx3C_4$ and $C_4\approx\frac{J}{k_\mathrm{air}t(9e^4+e^{-4})}\approx\frac{J}{491k_\mathrm{air}t}$ (check all these calculations!), so that the difference over the 2 cm is predicted to be about 400°C per watt (from 30°C at the bottom to 430°C at the top)—not too far from the earlier estimate of 150°C per watt per centimeter.

So what do we think of these numbers? Can a 1 W heater measuring 1 square centimeter exceed 400°C at the top of a foam enclosure of still air?

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  • $\begingroup$ Thank you, this is very interesting. Although, you made the assumption that radiative transfer is negligible. Is that really so? Maybe the radiation is sufficient to even out the temperature inside? $\endgroup$ Feb 3, 2023 at 16:47
  • $\begingroup$ The radiative transfer depends on the emissivity of the heater and foam. As noted, neglecting radiation gives an upper bound for the top-to-bottom temperature difference. $\endgroup$ Feb 3, 2023 at 17:01

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