Need some help understanding some thermal physics

[utorstudent]

well the question I have is about black bodies and it says that we have one black body (nonreflecting) at temperature Tu and another black body at temperature T1 (they are parallel and in a vacuum.) The energy flux between them is (Tu^4 - T1^4)*Stefan-Boltzmann's constant and the question wants to know if I place a third plate and allow it to come to steady state temperature what temperature will the middle plate be at and to also show that the flux is exactly half).

The temperature part is easy I think. The temperature of the middle plate should be the average of the two plates. That's the only way I could think of finding this new temperature given how little we know about the problem.

The problem I have though is how would I show that the "net energy flux density" is cut in half. Especially since I'm not really sure what it is in the first place (my book kind of glosses over how to find it and what it is.)

Thanks in advance for any help.

 

[Doc Al]

The key is realizing that the net energy flux in and out of the inserted third plate must be zero. That's all you need to solve for the temperature of the third plate and the net energy flux density. (No, the middle plate temperature is not the average of the two outer plate temperatures.) Assume all plates have perfect emissivity; $\sigma (T_2^4 - T_1^4)$ is the energy flux density.

 

[utorstudent]

EDIT: I talked with my professor a bit about this as well and she said I had the right idea. Thanks a lot for the help. :)

 

[Gokul43201]

EDIT: I talked with my professor a bit about this as well and she said I had the right idea. Thanks a lot for the help. :)

In that case I imagine your idea changed since you last posted here. The temperature of the middle plate will not be the average of the two outer plates.

 

[utorstudent]

In that case I imagine your idea changed since you last posted here. The temperature of the middle plate will not be the average of the two outer plates.

It did. :)

I originally had the whole solution where my post was but I figured I would take it down after I had asked my professor about it and she said it was the right approach.

 

[lalbatros]

About the professor or about thermal radiation ?

It would have been nice to explain to the world that flux conservation easily leads to the halved flux.
It would also have been nice to explain that the temperature of the middle plate is not the average, but that instead the relation is:

$$T^4_{middle} = \frac{T^4_{left} + T^4_{right}}{2}$$

There are also interresting discussions for other geometries and if the emissivities are not 1 and if there is absorption in between the two surfaces.
That's more interresting than homeworks !

 

[Andrew Mason]

It would have been nice to explain to the world that flux conservation easily leads to the halved flux.
It would also have been nice to explain that the temperature of the middle plate is not the average, but that instead the relation is:

$$T^4_{middle} = \frac{T^4_{left} + T^4_{right}}{2}$$

There are also interresting discussions for other geometries and if the emissivities are not 1 and if there is absorption in between the two surfaces.
That's more interresting than homeworks !

I think this needs a bit more explanation (so I can understand it).

If, after reaching a stable temperature, the plate absorbs all the incident radiation it must emit at the same rate. But it will emit equally in both directions. So the energy emitted by the middle plate in each direction is $E_{mid} = A\sigma T_{mid}^4$. Therefore, the total energy emitted is:

$$E_{mid} = 2A\sigma T_{mid}^4$$

which is equal to the absorbed energy from the other bodies

$$2A\sigma T_{mid}^4 = A\sigma T_{left}^4 + A\sigma T_{right}^4$$

which leads to:

$$T_{mid}^4 = \frac{T_{left}^4 + T_{right}^4}{2}$$

So, on the left side of the plate we have:

$$\phi_{left} = \sigma(T_{left}^4 - T_{mid}^4) = \sigma(T_{left}^4 - \frac{T_{left}^4 + T_{right}^4}{2})$$

which results in:

$$\phi_{left} = \sigma(\frac{T_{left}^4 - T_{right}^4}{2})$$

Similarly:

$$\phi_{right} = \sigma(T_{mid}^4 - T_{right}^4) = \sigma(\frac{T_{left}^4 + T_{right}^4}{2} - T_{right}^4)$$

So:

$$\phi_{right} = \sigma(\frac{T_{left}^4 - T_{right}^4}{2})$$

AM