# Total radiation power emitted in a semi-transparent volume

1. Jun 18, 2015

### Hypatio

(note I am designing this question myself, in order to understand this phenomenon by example)
1. The problem statement, all variables and given/known data
Find the total radiative power emitted within a 0.1 m2 semi-transparent medium, having an absorption coefficient of 5 m-1 over all spectra, and a temperature of 1000 K. Note that total emitted power is needed, do not consider absorption of power, and do not find the power emitted from the surface of the medium.

2. Relevant equations
We should be able to use the equation

$4\pi \kappa_\lambda I_{\lambda b}(s)dVd\lambda=4\kappa_\lambda E_{\lambda b}(s)dVd\lambda$

where $\kappa$ is the absorption coefficient, $\lambda$ is the wavelength, V is volume, I is the radiative intensity, E is emissive power, and s is a distance. This equation should give the spectral emission by an isothermal volume element.

The Planck function is

$I_{\lambda b}=\frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_bT}}-1}$

3. The attempt at a solution
To get the total power emitted by a blackbody (change units from W/m3/sr to W/m3) I multiply by $4\pi$:

$I^{net}_{\lambda b}=\frac{8\pi hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_bT}}-1}$

Numerically integrating over all the spectra for T=1000K to get the total black body intensity, I get about 18 kW/m2. So if my medium had an infinite absorption coefficient (black body), my rounded answer for emission in a 0.1 m2 volume would be 1.8 kW. So far, so good, I think. Now I need to figure out how this value changes with a different absorption coefficient.

I get 5*1.8=9 kW. But this is impossible because 1.8 kW should be the maximum possible. I don't see how to calculate the correct value.

Thanks in advance for any help.

Last edited: Jun 18, 2015
2. Jun 18, 2015

### Staff: Mentor

0.1 m2 is not a volume.
The first equation looks nice, where did you use it?

3. Jun 19, 2015

### Hypatio

Sorry, I think typing m2 can be treated as a typo. I am trying to figure out how to use an equation like that posted but I am unsure. I am following 'Thermal Radiation Heat Transfer', fifth edition, by Howell et al. Today I found a statement that is more directly useful:

"The total energy emitted by a volume element dV is given by Equation 1.54 [in my original post] integrated over all $\lambda$ to obtain

$4dV\int_0^\infty\kappa_\lambda E_{\lambda b}d\lambda$"

From the initial equation I posted this must also be

$4\pi dV\int_0^\infty\kappa_\lambda I_{\lambda b}d\lambda$

so now by my logic I can do the following:

$4\pi dV\int_0^\infty\kappa_\lambda I_{\lambda b}d\lambda=4\pi 0.1m^3\int_0^\infty\frac{5}{m}I_{\lambda b}d\lambda=0.1m^3\frac{5}{m}18\frac{kW}{m^2}(m)=9kW$

This is what I got before. This shouldn't be right because it would mean that the radiative intensity is proportional to the absorption coefficient. I don't get it.

4. Jun 19, 2015

### Staff: Mentor

It leads to mismatching units if changed to m3.

I don't see the problem with this. More absorption also means more emission.

5. Jun 20, 2015

### chrgrose

I am the thread starter -- new account made since I cant access my former account.

The problem is that radiative intensity cannot be linearly proportional to the absorption coefficient, otherwise a body with an absorption coefficient greater than 1/m will emit more power than a blackbody. In my example, the volume emits 5 times more energy than a blackbody. Something is wrong with my solution, but I don't know what.

6. Jun 20, 2015

### Staff: Mentor

You compare quantities with different units here. That's like saying a second is more than a meter, it does not work.
But even if you fix that: it will not, most of its emitted power is directly absorbed internally again. A blackbody is the limit of "infinite" absorption coefficient and "infinite" emitted radiation, but you only see the surface, not the volume.

Your units are wrong, and that is not just a typo.

7. Jun 20, 2015

### chrgrose

Well then what is the correct answer and how do the units follow. I want you to be right but I don't see it.

By the way I don't care about reabsorption. This should be dealt with by separate terms.

On the other hand your last statement suggests that the internal volume of a black body actually does emit an infinite power which is very surprising to me. I understand that you are suggesting that it's reasonable because we only see the emissions from an infinitesimally thick surface, but I am still suprised by the thermodynamic implications.. in fact If this is true then I think it is the source of my misunderstandings... can you direct me to literature which discusses this aspect in particular? Thanks so much.

8. Jun 20, 2015

### Staff: Mentor

That's not how it works. Units are not some nasty detail, things without units are completely meaningless. Is "4" the right answer? Well, maybe 4 days is, but 4 seconds is not, and if it would be 4 meters something is really wrong.

Fix the units in your calculation, and the issue will go away.

The thermodynamic implication is the nonexistence of such a perfect blackbody. You cannot get absorption below the length scale of atoms, which keeps everything finite. A more realistic length scales is the wavelength of the frequency considered (if larger than an atom).

9. Jun 20, 2015

### chrgrose

I'm not implying that units are irrelevant. I arrived at an answer for my particular problem of 9 kW, or 9 kJ/s. This is my calculation of the power within a 0.1 m^3 body of material at 1000 K and $\kappa_\lambda$=5/m. It seems that the units are just fine. I only disputed the quantity, as it seemed to be unphysical. It is seeming to me that I am just mistaken, and such a quantity is correct for a medium with those properties.

If that is the case, then OK. Although I am uncertain of how to then interpret the usual description of the power of a ray moving along a path length s:

$I_{\lambda}(s)=I_\lambda(0)\exp(-\kappa_\lambda s)+I_{b\lambda}[1-\exp(-\kappa_\lambda s)]$

where the first term gives the energy of the incident ray $I_\lambda$ lost to absorption as it propagates, and the second gives the energy gained from emission. In this description, the power of the ray can never be more than $I_{b\lambda}$ if the incident ray is zero. Previously I interpreted this as indicating that the power must always be finite, even if the absorption coefficient were infinite. With your help it seems that I should interpret this as follows: For a medium with a high absorption coefficient, huge amounts of energy really are emitted along the ray path, becoming infinite for a blackbody. However, the energy of the ray $I_\lambda$ can never become infinite, or even larger than $I_{\lambda b}$, because of the complementary rate of absorption of the emitted energy.

I may be understanding now...

10. Jun 20, 2015

Right.