Correlated-k method (absorption of radiation)

  • #1
hilbert2
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I've been recently studying the correlated-k method of calculating the absorption of EM radiation when passing through a sample of given thickness. I'm not sure if anyone here has experience on the same subject, but in case there is I have some questions...

Suppose I have a material sample that has only one Gaussian-shaped absorption spectral line, with the absorption coefficient ##\kappa## given as a function of wavenumber ##\eta## as

##k(\eta ) = Ae^{-b(\eta - \eta_0 )^2}##.

Now, I guess that if I have a beam of light/IR radiation that has a constant spectral intensity on a given wavenumber range ##[\eta_1 ,\eta_2 ]## and zero intensity outside that range, the fraction of total radiative energy absorbed when passing through a sample of thickness ##\Delta x## is

##\tau = \frac{\int\limits_{\eta_1}^{\eta_2}\exp\left[-\kappa (\eta )\Delta x\right] d\eta}{\eta_2 - \eta_1}## (*)

(or is there a weighting with ##\eta## inside the integral in numerator? If the wavelength range ##[\eta_1 , \eta_2 ]## is narrow, this doesn't matter though...) The correlated-k method is based on the assumption, that if I define a function ##f(k)## with the relation ##d\eta = (\eta_2 -\eta_1 )f(k)dk##, and form the functions

##g(k) = \int\limits_{k(\eta_1 )}^{k}f(k')dk'##,
##k(g) = g^{-1}(g)##,

then the absorbed fraction can be calculated by

##\tau = \int\limits_{0}^{1}e^{-k(g)\Delta x}dg## (**).

This result can supposedly be shown to be exact. Now, the function ##f(k)## seems to give the relative length of a small wavenumber interval ##d\eta## corresponding to a small absorption coefficient interval ##dk##. If I choose a small subinterval ##dk## from the set ##[0,A]##, there are two wavenumber intervals (on both sides of ##\eta_0##), that are "equivalent" to this. For the Gaussian spectral line, it seems to be that

##f(k) = \frac{\exp\left|\log{A/k}\right|}{(\eta_2 - \eta_1 )A\sqrt{b\log(A/k)}}##.

And the functions ##g(k)## and ##k(g)## are easy to deduce from this by numerical integration and function inversion in Mathematica. The graph of function ##f(k)## for ##k\in [0,A]## looks a bit similar to how the gamma function ##\Gamma (x)## behaves on some intervals between negative integer values of ##x##, but I'm not sure if this has any relevance.

The problem is, that if I choose some values for the parameters ##A,b,\eta_1 ,\eta_2 ,\eta_0## and calculate the absorbed fraction for several values of ##\Delta x##, I will not get the same result from (*) and (**)... Does anyone here see any obvious mistake in the calculations I have made? The correlated-k method has been described in this article: http://heattransfer.asmedigitalcollection.asme.org/article.aspx?articleid=1445408 (not sure if there's a free full text available somewhere).

Thanks,
Hilbert2
 

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  • #2
DrDu
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##\tau = \frac{\int\limits_{\eta_1}^{\eta_2}\exp\left[-\kappa (\eta )\Delta x\right] d\eta}{\eta_2 - \eta_1}## (*)
Isn't this the non-absorbed fraction rather than the absorbed fraction?
 
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  • #3
hilbert2
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Yeah, sorry, that's of course the non-absorbed fraction. I'm just applying the Lambert-Beer law for non-monochromatic radiation in there.
 
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  • #4
hilbert2
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Just mentioning that I solved this problem. When the spectral line is defined as that kind of a Gaussian function, the function ##f(k)## looks something like this:

2hro29s.jpg


And then ##g(k)## is difficult to evaluate numerically because ##f(k)\rightarrow\infty## as ##k\rightarrow 0##. The solution is to shift the Gaussian downwards by some small number:

##k(\eta ) = Ae^{-b(\eta - \eta_0 )^2} - \epsilon##

and then consider only the interval where ##k(\eta )\geq 0##. That way there's no infinities in the ##f(k)## as the graph of ##k(\eta )## meets the horizontal axis with a nonzero angle of incidence. The k-distribution result seem to be very close to the exact transmittance when calculating it like this.
 
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