Correlated-k method (absorption of radiation)

[hilbert2]

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

 

[DrDu]

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

Isn't this the non-absorbed fraction rather than the absorbed fraction?

 

[hilbert2]

Yeah, sorry, that's of course the non-absorbed fraction. I'm just applying the Lambert-Beer law for non-monochromatic radiation in there.

 

[hilbert2]

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:

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.