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Stefan–Boltzmann Law - Radiation in a Sphere

  1. Oct 5, 2012 #1
    Hi,

    I searched the forum and the web and can't seem to find my answer, nevertheless I will continue looking.

    There might be some obvious answer to this and I'm probably missing something, but in any case this is my question:

    I have a sphere inside another sphere with a cavity of air between the spherical planes.

    The radius of the outer sphere is about 4 times larger than the inner sphere.

    I have a temperature gradient between the two spherical planes, (let's say 100K).

    Given my temperature gradient, ratio of the areas, and Boltzmann's law:

    Irradiance=[{σ*ε*Area_small*((T_hot)^4)}-{σ*ε*Area_large*((T_cold)^4)}]

    my net energy due to radiation is a negative number. The ratio of the area is dominating my difference.

    I guess what I'm asking is, what does this negative quantity mean? Energy in this case should still be moving from the inner sphere to the outer sphere directed along the normal path I think?

    This is not a homework problem, I'm designing a vessel and need to know what this means as far as heat transfer.

    Thanks in advance.


    Update:

    This is assuming equal emissivity. Although both bodies will be aluminum so I am wondering if emissivity changes much with a temperature difference of 100K?
     
    Last edited: Oct 5, 2012
  2. jcsd
  3. Oct 5, 2012 #2

    Ken G

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    It's not completely clear what you mean by "irradiance", but it sounds like what you are asking is, what is the rate of energy transfer from the hotter, small sphere to the cooler, larger sphere. Assuming that is your question, you cannot use the expression you have, because it assumes the entire energy irradiated by the outer sphere into the cavity is absorbed by the inner sphere. That's not true, only the fraction that hits the smaller sphere. The easy way to do that is to note that if the outer sphere was at the same T as the inner sphere, that amount of energy would have to balance the rate the inner sphere is emitting light (which is one part of your expression). So just take that amount, and scale it up by the T ratio, to the 4th power. Here you will note that saying one is 100 K more than the other is not enough information, you need the actual T so you can find the ratio. Anyway, the answer is, the net energy transfer is your expression, except with the small area in both terms. The sign just depends on your convention, you know the net transfer is from hotter to cooler, and the same amount of energy goes both ways, just opposite sign.
     
  4. Oct 5, 2012 #3
    Thank you for your response! Yes, Irradiance can be found here: http://en.wikipedia.org/wiki/Irradiance
    although not always the best source wiki defines it well.

    So what you are saying is that my outer area emitting energy inward from the term: σ*ε*T_cold, should be considered as radiation emitted from a surface the same size as my inner area?
     
  5. Oct 5, 2012 #4

    Ken G

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    Gold Member

    Yes. Here's another way to see that. Just consider the surface of the smaller sphere, and ask what is the net rate it is losing energy. That's outward irradiance minus inward irradiance, all at that same sphere. The outward flux just depends on the Area and T of the inner sphere, using the expression you have already. The inward flux is the exact same expression, same area (because it is the same sphere, just looking at the inward flux), but a different T-- it is the T corresponding to the outer sphere (because every inward ray is coming from that outer sphere, it makes no difference at all how large that outer sphere is). The outer sphere could be the 2.7 K cosmic background radiation coming from tens of billions of light years away, for all the difference the size would make-- all that matters is the outer temperature.

    Now, in case what you really want is the net energy transferred to the outer sphere from the inner sphere, rather than the other way around, realize that this has the exact same numerical value, expect with opposite sign. That is because the speed of light is so fast that no energy is "piling up" in between the spheres-- to have energy conservation, the negative of the net energy that is going from the small sphere to the big sphere is going the opposite way.
     
  6. Oct 5, 2012 #5
    All your equation descibes the difference in emmissive power between the 2 surfaces.
    You need to consider that the inner sphere radiates all of its energy to the outer sphere.
    The outer sphere radiates some of its energy to itself and some to the inner sphere, and you would have to determine the ratio.

    I would just like to expand on what Ken G is saying in more concrete terms:

    What is used is called the shape factor as seen at:
    http://en.wikipedia.org/wiki/View_factor#Summation_of_view_factors

    The external link gives a list of view factors ( same thing different name ) for configurations of various radiating shapes.
    http://www.engr.uky.edu/rtl/Catalog/
    Not that a shape factor is not intinsic to a specific body but will take into account the body(ies) to which is exchanges radiation.

    Anyways, for a sphere within a sphere, the shape factor F of outer sphere(2) to inner sphere(1) is the ratio of the radii of inner sphere to outer sphere squared.
    In your case that is the square of 1/4 = 1/16. Lets call that F21.

    The inner sphere to the outer sphere shape factor F12=1. - all radiation from inner sphere is emmitted to outer sphere.
    F11 = 0 - inner sphere does not radiate to itself
    F22 = 1-F21 - outer sphere radiates and some strikes the inner sphere

    Plugging these into your the equation yields the case where both spheres being black bodies:
    Q1 = -Q2
    Q1 = A1 F12( e2 - e1 )
    Q2 = A2 F21 (e1 - e2 )

    If you can get your head around all that then feel free to condider that both your spheres are grey bodies with an emissivity not equal to 1.0. And you would have to take into account how much radiation is reflected or absorbed by each body.
     
  7. Oct 10, 2012 #6
    Thank you 256bits!

    I understand the view factor now, that helped so much! and I do know I must account for grey body differences.

    But I am wondering about the equations you provided, are these complete or just an idea presenting what my view factor(s) will be?

    What is 'e' in the equations, emissivity? What is 'Q' in the equation, Joules/sec ?

    I very appreciate your time in explaining this but it would help me to know what these variables mean, especially with the (T^4) term not there.
     
  8. Oct 10, 2012 #7
    Also, what I am calculating is the steady-state temperature between my two bodies at equilibrium, is the view factor needed for this?
     
  9. Oct 10, 2012 #8
    I am also confused by your use of A1 and A2, are these the small and large surface areas respectively?

    If so, when I am now working out the math theoretically I am getting back to my original question, adding F22 and F21, that sum being the factor for my radiation coming from my outer cold body is resulting as my outer area being the factor for my cold body term.

    I am sorry but with the lack of factors and explanation of variables in those last three equations I am now more confused
     
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