Emission radiance in participating media

[Hypatio]

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:

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

where \tau_\eta is the absorption coefficient times the length.

So, the first term gives energy of the incident ray I_\eta 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 I_{b\eta} by that coefficient and know the emission radiance everywhere for that temperature. Is this coefficient the value of (1-\exp(-\tau_\eta)) at one meter of propagation? 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 (\kappa_\eta I_{b\eta}). 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.

 

[optophotophys]

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(1-\exp(-\tau_\eta)) and I_{b\eta}, 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 \kappa_\eta is just the dimensional absorption coefficient. What made you think that it should be dimensionless?

 

[Hypatio]

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 \kappa_\eta I_{\eta b}(T)dV, 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 propagating ray. I want the net emission within a volume, which is linearly extensive.

 

[optophotophys]

\kappa_\eta I_{\eta b}dV is proportional to the volume. You have the same \kappa_\eta irrespective of the concerning volume size. \kappa_\eta is the absorption coefficient per unit length. So \kappa_\eta I_{\eta b}dV is an extensive quantity.

 

[Hypatio]

\kappa_\eta I_{\eta b}dV is proportional to the volume. You have the same \kappa_\eta irrespective of the concerning volume size. \kappa_\eta is the absorption coefficient per unit length. So \kappa_\eta I_{\eta b}dV is an extensive quantity.

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 \kappa_\eta=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?

 

[optophotophys]

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 \delta s as
\delta I=(1-\exp(-\kappa \delta s)) I_b.
If the element size is enough small or \kappa is enough small, we have
\delta I=-\kappa \delta s I_b.

 

[Hypatio]

Actually, I've recognized that the problem is that I am looking at dI_{\lambda b}, and not I_{\lambda b}. dI 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 I_{\lambda,emitted}, not dI_{\lambda}. 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.

 

[optophotophys]

Eq. 10.118 is the one you need.

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

No. Everything behaves as a blackbody, when the size is infinitely large S\rightarrow\infty. 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.