Radiating sphere inside spherical shell - heat transfer

[FranzDiCoccio]

Homework Statement

• A sphere of radius $r_s$ is at the center of a spherical shell of inner radius $r_i=10\, r_s$ and thickness $s = 10\, {\rm cm}\ll r_i$.
  
• The sphere has a temperature $T_s=1073\, {\rm K}$ and and an emissivity $e=0.90$.
• The inner surface of the shell has a temperature $T_i = 873 \,{\rm K}$ and it is made of asbestos . The heat transfer coefficient of asbestos is $k_a = 0.090 J/(s K m)$.
• The text says that the temperatures of the sphere and of the inner surface of the shell are "kept constant".
  
• The problem asks to find the temperature $T_e$ of the outer radius of the spherical shell.

Homework Equations

• Stefan-Boltzmann law (radiated heat power)

$$P = \sigma e A T^4$$

• Heat transfer through a rod (should be ok for the spherical shell on account of its thinness)

$$P = k_a \frac{A}{s} \Delta T$$

The Attempt at a Solution

Of course the problem is solved if one can estimate P in the second equation.

I guess it could be a matter of balance of heat radiated by the sphere and (possibly) by the inner surface of the shell. The sphere radiates heat, and, if the shell is not totally reflecting, it is going to absorb part of it.
If that is the case, it should also radiate some heat back.
I'm not sure of how to take into account that "the temperatures of the sphere and of the inner surface of the shell are kept constant".

The power absorbed by the inner shell could be

$$P_i = \sigma e_i A_i (T_s^4-T_i^4)$$

But I'm not sure about at least a couple of things

1. the S-B law applies to a concave surface. Perhaps it works because the shell radius is "large". Anyway, I do not know $e_i$. The problem does not give it.
   
2. it seems to me that the shell absorbs more energy than the sphere radiates (because of the larger surface).

Another possibility is that the inner shell is considered as the (radiating) "surroundings" of the sphere. So the net power emitted by the sphere is

$$P_i = \sigma e_s A_s (T_s^4-T_i^4)$$

and all this power for some reason is trasmitted through the shell.
This does not convince me either.

I think that this problem should be solved using the two equations I wrote (these are the only equations about heat transfer that could be found in the book where I found the problem).

 

[TSny]

I'm not sure about the interpretation of the problem. But, suppose you are meant to assume that the net amount of radiant energy in the volume between the sphere and the shell remains constant in time. Can you deduce the net rate at which radiant energy is absorbed by the shell? Then, can you get the temperature of the outer surface of the shell assuming that the shell is in a steady state of heat conduction?

 

[FranzDiCoccio]

Hi TSny

I'm not sure about the interpretation either.
I think that your second question can be rephrased as: can you get the temperature of the outer surface if you know P in the heat transfer equation?
Of course yes. I would know everything but the external temperature, so that would be easy.

I'm not sure about how to answer your first question.

 

[TSny]

I'm not sure about how to answer your first question.

It seems to me that you can determine the net radiant heat transfer to the shell by assuming that the total amount of radiant energy in the space between the sphere and the shell remains constant.

 

[FranzDiCoccio]

Uhm, I'm not seeing this. The sphere is radiating heat at some rate (S-B law). Possibly the inner shell is also radiating back some energy, at some other rate. Whatever these rates are, wouldn't the radiant energy between the sphere and the shell be time-independent?
Maybe I'm saying a very stupid thing, but I'm very tired now. Here's pretty late, and I have been up for 18 hours now.

I think I'd better go to sleep now. I'll think again about this tomorrow.

Thanks a lot for your help.

 

[FranzDiCoccio]

Maybe I'm getting a little unstuck on this.
Perhaps I can see it this way

1. The net power radiated by the sphere is $P_s = \sigma e_s A_s (T_s^4 - T_i^4)$, where the term with $T_i$ accounts for the power radiated back by the shell. I can use this in the lhs of the heat transfer equation, meaning that whatever power is not scattered back goes out
2. This is not entirely satisfactory, because I'm focusing on the sphere, whereas I need the power through the shell. I'd be happier if I had some equation(s) like the one in 1., but for the shell. I still think that there is something weird, something I cannot put my finger on. The net power radiated by the shell would be something like $P_i = \sigma e_i A_i (T_i^4 - T_s^4)$. Since the sphere is hotter than the shell, this would be negative. Ok, indeed there is some power going out. But then perhaps it should be $P_i =- \sigma e_s A_s T_i^4$. I do not know whether this makes sense. Is this supposed to fix the relation between $e_i$ and $e_s$ for equilibrium at those temperature to occur? Also, probably the equation for $P_i$ should contains some additional term to account for the fact that some of the power radiated by the inner shell does not hit the sphere and goes back to another point of the shell.

Can I use the equation in 1 safely? Am I missing something?

Thanks

 

[TSny]

I might be oversimplifying the problem. I'm considering that we have steady-state conditions. So, the net amount of radiant energy in the space between the sphere and the shell is not changing with time. The net rate at which radiant energy is transferred to this space by the sphere is $P_s = \sigma e_s A_s (T_s^4 - T_i^4)$. Since the total amount of radiant energy in the space between the sphere and the shell must remain constant for steady-state conditions, you can deduce the net rate at which radiant energy is removed from the space by the shell. In order for the shell not to "heat up", it must conduct a certain amount of heat per second to the outside environment.

If I look at it this way, I seem to get a reasonable answer for T_e. The emissivity e_a of the asbestos is not needed.

 

[FranzDiCoccio]

Hi again TSny,

and thanks for your insight. I came to the same conclusion as you, but in a much less clear headed way. I could not wrap my head around this problem because I was kind of obsessing on the spherical shell, which is possibly the worst angle to look at the problem.
I solved the problem according to the above formulas and I got the solution given in the book. So this must be what they had in mind (btw, one of the numbers above is wrong, the shell thickness is 1,0 cm and not 10 cm as I wrote).

Thanks again

 

[TSny]

OK, good.