# Two materials heated in vacuum

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1. Jul 27, 2017

### cc94

1. The problem statement, all variables and given/known data
This isn't a homework question but something I'm working on that I thought should be simple. Two disks (area $A$ and thickness $d$) are joined together and placed under a radiation heater in vacuum, so that one side of the top disk is heated with a constant power. Assume heat is only lost out of the two-disk system by radiation. 1. Is there a temperature gradient in either of the materials? 2. What is the steady-state temperature of the bottom disk/bottom surface?

2. Relevant equations
For conduction between the two materials, I found this equation on Wikipedia: https://en.wikipedia.org/wiki/Thermal_contact_conductance#Thermal_boundary_conductance.

For radiation, we have the Stefan-Boltzmann law,

$$P = \epsilon\sigma AT^4$$

3. The attempt at a solution
I'm not sure what all I'm supposed to consider between these two bodies. My thinking is, first, we have some amount of heat $Q_{in}$ coming in. The top disk absorbs this as

$$Q_{abs} = \epsilon_{top} Q_{in}$$

Now here's where I was confused. I know the top disk emits radiation out of the top face, but do I also include radiation emitted from the back face in contact with the bottom disk? Or does the conduction from Fourier's law actually include the fact that the disk is radiating? If we don't include a radiation term, then we have two equations for the top and bottom disks:

$$\epsilon_{top} Q_{in} = \epsilon_{top}\sigma AT_{top}^4 + \frac{T_{top}-T_{bot}}{d/(k_{top}A) + 1/(hA) + d/(k_{bot}A)}$$
$$\frac{T_{top}-T_{bot}}{d/(k_{top}A) + 1/(hA) + d/(k_{bot}A)} = \epsilon_{bot}\sigma AT_{bot}^4$$

where $k$ is the thermal conductivity and $1/h$ is the thermal contact resistance. If we include radiation, I should just add those terms to each equation. But then I have two more unknowns since I think there's 2 different temperatures at the interface between the disks.

2. Jul 27, 2017

### jambaugh

Treat radiation and absorption separately. If the disks have a finite thickness then there will be a separate question of their thermal conductivity and each side will have a different temperature. If so then at equilibrium the temperature will vary linearly between the surface temps with a slope inversely proportional to the conductivity. If you want to assume a high thermal conductivity and thin disks you can assume the temperature of both sides is constant.

The trick to determining the amount of energy absorbed by each disk is to do "ray tracing" of sorts. Start with your flat surface at temperature T within a large sphere also at temperature T. Then the amount of heat absorbed by the surface must equal the amount radiated since it must be at equilibrium since the temperatures are the same. Since this does not depend on the radius of the sphere it must only depend on the amount of spherical angle the surface sees which is one hemisphere (its sky) or $2\pi$ steradians.

You can then infer that the amount of energy absorbed by a surface from a distant thermal source is the amount of energy it would emit at that temperature times the proportion of the total hemisphere the surface sees as having that temperature. If the source object is close so that different points on the surface see different spherical angles then you will have to integrate an energy absorption density across the surface.

However as I re-read your post it would seem to me that your intent is that the disks be quite close together and wide relative to their thickness. You can treat that case and further simplify by assuming mirror sides in the tube containing them, and then the problem becomes 1-dimensional. For infinite thermal conductivity of the disks they will heat up until both are the same temperature all are radiating the same amount of energy out of each side, that being (for each side) the total energy coming from the heater. With finite thermal conductivity there will be a temperature drop within the disks but each face will have equal temperature to that of the one it faces across the vacuum... since with my added side mirrors each sees the other as covering the entire "sky".

There is also a slight potential for ambiguity in this question because one could interpret the "amount of energy emitted by the radiation heater" two ways. If it is emitting heat it can also absorb it and thence re-radiate it. Here configuration will become important. If you assume a point source then the question arises as to what happens to the heat emitted by the disk facing the heater. Is it reflected back or absorbed by an external heat sink. If the heater is not a point source then it will absorb energy from the disk it heats and the question is how do you figure the amount of energy emitted. Is it Net? or Gross? It is easy enough to resolve. One just has to be a wee bit more specific in describing how the heater is configured. I would suggest for example allowing it to be a similar disk which is heated to a temperature so that is face emits the specified energy as given by the Stefan-Boltzmann law.

To clarify on this last point, if you had a heater inside a mirrored sphere evacuated to vacuum, if you pump in 1 watt of energy then the heater will continue to increase in temperature (no way for it to leave the sphere) emitting a higher and higher gross amount of energy as the temperature goes up. But the net radiated energy will still be 1 watt, the 1 watt more than it absorbs due to the external input. The gross goes up because all the energy radiated earlier is returning to be reabsorbed.

3. Jul 27, 2017

### Staff: Mentor

In my judgment, your formulation is correct as it is written.