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Emission radiance in participating media

  1. Mar 4, 2015 #1
    This link discusses radiation in participating media. Eq. 9.13 gives a prediction of the changing power of a ray along the path length as:

    [itex] I_\eta(x)=I_\eta(0)\exp(-\tau_\eta)+I_{b\eta}[1-\exp(-\tau_\eta)][/itex]

    where [itex]\tau_\eta[/itex] is the absorption coefficient times the length.

    So, the first term gives energy of the incident ray [itex]I_\eta[/itex] lost to absorption as it propogates, and the second term gives the energy gained from emission.

    With this in mind, I want to know how to figure out what the dimensionless coefficient of emission is. This way I can just multiply [itex]I_{b\eta}[/itex] by that coefficient and know the emission radiance everywhere for that temperature. Is this coefficient the value of [itex](1-\exp(-\tau_\eta))[/itex] at one meter of propogation? after one centimeter? Is it related to the slope of this curve near x=0? How do I find it?



    Also, the above equation is derived from Eq. 9.11 in that reference, where we are told that emission is the absorption coefficient times the radiance of a blackbody ([itex]\kappa_\eta I_{b\eta}[/itex]). I assume that this is not the same as the dimensional absorption coefficient, but I don't know how to get the dimensionless one (that has a range of 0 to 1), from the dimensional one.
     
    Last edited: Mar 4, 2015
  2. jcsd
  3. Mar 4, 2015 #2
    The provided link is not sufficient for understanding the exact situation. After some search, I found a textbook which looks like discussing similar topic. So, my thought is based on the chap. 10.5.2 in http://www.thermalfluidscentral.org/e-books/book-viewer.php?b=37&s=11

    From the product of the dimensionless coefficient of emission[itex] (1-\exp(-\tau_\eta))[/itex] and [itex]I_{b\eta}[/itex], you can only know the emission radiance propagating into a certain direction originated from the concerning volume (see Fig. 10.27 in the textbook). If you want to know the total emission radiance density at some location, this is a different matter. You need to know the shape of the material etc... and need to integrate the whole emission radiance propagating in various directions.

    I think [itex] \kappa_\eta[/itex] is just the dimensional absorption coefficient. What made you think that it should be dimensionless?
     
  4. Mar 5, 2015 #3
    Thanks for the reference, it's really good. What is discussed in 10.5.2 is very close to what I am trying to figure. To clarify, I don't think I need to know the total emission radiance density after radiative transport. I need to know the total emission radiance at the point where radiation originates. It appears that I need to solve something like Eq. 10.120 in the text.

    I want to know the energy of all photons instantaneously emitted inside an arbitrary volume having some absorption coefficient and temperature. I don't care about what happens to the photons once they are emitted (whether it is absorbed, scattered, transmitted, etc.), that is secondary, I just want to know how much energy per volume.

    But, if it is true that this energy is equal to [itex]\kappa_\eta I_{\eta b}(T)dV[/itex], then I don't understand the phenomenology, because my value ought to be linearly dependent on the volume as it is an extensive property. Moreover, the result seems wrong if the absorption coefficient is >1 m^-1, since this means that radiation is greater than blackbody.

    Edit: Rereading the text, I don't think it is discussing what I want. I don't want energy conservation of a propogating ray. I want the net emission within a volume, which is linearly extensive.
     
    Last edited: Mar 5, 2015
  5. Mar 5, 2015 #4
    [itex] \kappa_\eta I_{\eta b}dV[/itex] is proportional to the volume. You have the same [itex]\kappa_\eta[/itex] irrespective of the concerning volume size. [itex]\kappa_\eta[/itex] is the absorption coefficient per unit length. So [itex] \kappa_\eta I_{\eta b}dV[/itex] is an extensive quantity.
     
  6. Mar 6, 2015 #5
    What happens when the absorption coefficient is greater than 1 unit? I must be missing something in the dimensional analysis because it seems that if [itex]\kappa_\eta[/itex]=50 m^-1, the intensity will be 50 times greater than a blackbody, which is impossible.

    Can you show me where I am failing here?
     
  7. Mar 8, 2015 #6
    Sorry, I misunderstood your point.

    eq (10.119) looks like just an approximation (and contains some error). From eq (10.118), we can have a thermal radiation intensity in a given direction from an small element of medium [itex]\delta s[/itex] as
    [tex]\delta I=(1-\exp(-\kappa \delta s)) I_b[/tex].
    If the element size is enough small or [itex]\kappa[/itex] is enough small, we have
    [tex]\delta I=-\kappa \delta s I_b[/tex].
     
  8. Mar 8, 2015 #7
    Actually, I've recognized that the problem is that I am looking at [itex]dI_{\lambda b}[/itex], and not [itex]I_{\lambda b}[/itex]. [itex]dI[/itex] can, of course, be infinite, because it's just describing the change in intensity over the ray path.

    My problem is that I want to know [itex]I_{\lambda,emitted}[/itex], not [itex]dI_{\lambda}[/itex]. That is: the radiation emitted from each point in the volume. So It seems that Eq. 10.118 is more appropriate, or do I need to integrate Eq. 10.119?

    But then it seems like everything radiates like a blackbody, because (1-exp(-KS)) always goes to 1 as S goes to zero...

    I really appreciate the help thusfar.
     
  9. Mar 9, 2015 #8
    Eq. 10.118 is the one you need.

    No. Everything behaves as a blackbody, when the size is infinitely large [itex]S\rightarrow\infty[/itex]. It is intuitively reasonable in this simplified situation. An infinitely large object absorb all the incident energy (in this textbook, surface reflection is neglected). This means that the object is a blackbody.
     
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