reflection, absorption and transmission through a solid

[Amy Mensch]

I am exposing a solid surface to an external radiative flux, using the
EXTERNAL_FLUX keyword. I would like some of the flux to be reflected
(5%), and some transmitted (6.6%) and the rest absorbed (88.4%) by the
solid surface.
I think I understood from the user guide, last paragraph of 8.2.2, pg.
59, that emissivity and absorptivity are assumed to be the same
value. So to absorb 88.4%, I specified the emissivity to be 0.884.
To specify the transmissivity, I calculated the absorption coefficient
necessary for a 2cm slab to transmit 6.6%. This value came out to
109.5 /m.

&MATL ID='MASK'
CONDUCTIVITY=0.2
EMISSIVITY=0.884
SPECIFIC_HEAT=1.25
DENSITY=1200.
ABSORPTION_COEFFICIENT=109.5/

Is this correct?

[Kevin]

How did you calculate 109.5 /m? I calculate -ln(0.066)/0.02=135.9 /m.
The rest I have to think about.

[Nico]

Hi

Maybe I make a mistake but if you want to consider the reflected part
in your calculation of the absorption coefficient, the formula is:

alpha = [2 ln(1- r)- ln (tau)] / thickness

I found this one in: P.T. Tsilingiris, “Comparative evaluation of the
infrared transmission of polymer films”, Energy Conversion and
Management, 44: 2839-2856 (2003).

In this case I found alpha = 130.78 1/m

regards

Nicolas

[shostikk]

The actual transmittance depends on many things, including the
reflection from the surface, absorption inside the material, emission
inside the material and reflection on the back surface. Due to the
internal emissions, the relative transmittance therefore depends on
the incoming flux level too.

The formula given by Nicolas assumes that the reflection takes place
both on the front and back surface of the layer, using the same
reflectivity. This is indeed the case in FDS, if your surface only
consists of one material. The reflectivity of surface r = 1-e, where e
= emissivity (= absorptivity) of the surface. e is defined using the
keyword EMISSIVITY, in this case EMISSIVITY = 0.95.

The transmission through the layer is thus (1-r)*(1-r)*tau = 0.066. We
get tau = 0.0731.

The formulas given above by Kevin and Nicolas both apply for a line-of-
sight computation. The internal radiation model in FDS considers semi-
infinite slab where most of the photons do not travel in perpendicular
angle and have to travel a distance with is longer than the layer
thickness 0.02 m. This process is modelled using the two-flux model,
which is verified against analytical solution in the Verification
guide. If we estimate the value of the absorption coefficient using
the two-flux model we get much smaller values than the values
mentioned above.

For instance, if the layer is absolutelly cold, the 0.0731
transmission is achieved by absorption coefficient kappa = 68.3 1/m.

If the material is not cold, the estimated value depends on the
material temperature and incoming flux level. Assuming EXTERNAL_FLUX =
50 kW/m2 we get the following values for absorption coefficient:

T = -273.15 C kappa = 68.3 1/m
T = 20 C kappa = 73.1 1/m
T = 200 C kappa = 112.5 1/m

To avoid the the back surface reflectivity, use two layers of
materials with the second layer having EMISSIVITY = 1.

Simo

[Amy Mensch]

That is the equation I used, and I got 135.9 when I did it again. I
must have made a mistake the first time.

> > Is this correct?- Hide quoted text -
>
> - Show quoted text -

[Kevin]

Ask a simple question ...

I created a simple test case that might help. The important stuff is
here:

&MISC H_FIXED=0. /
&MATL ID='MASK'
CONDUCTIVITY=50.2
EMISSIVITY=0.95
SPECIFIC_HEAT=0.25
DENSITY=120.
ABSORPTION_COEFFICIENT=135.9 /
&SURF ID='HOT', THICKNESS=0.02, MATL_ID='MASK', EXTERNAL_FLUX=5.,
COLOR='RED' /

Note that I turned off convective heat transfer (H_FIXED=0) and I
reduced the thermal inertia of the slab so that it would heat up
uniformly and quickly. The EMISSIVITY is 1-r, and the
ABSORPTION_COEFFICIENT is based on my calculation above. The
EXTERNAL_FLUX was chosen arbitrarily. The way I see it, when the slab
heats up to a steady-state temperature, the energy going into it ought
to equal the energy going out:

5*(1-0.066) = 2*0.95*5.67e-11*(T_s^4-293.15^4)

The factor of 2 indicates that heat is lost from both sides of the
slab. The EXTERNAL_FLUX is just a magical heat source that we impose
in FDS to test algorithms. Solving for the slab temperature, T_s, I
get 201.45 C (remember to convert K back to C). FDS predicts 203.32 C.
The difference is related to the approximations that Simo refers to.

The confusing part of this are the definitions. I interprete your
original post to mean that you desire that 6.6% of the external heat
flux pass right through the slab. In which case, 93.4% of your
external heat flux gets absorbed. Note that the EXTERNAL_FLUX in FDS
is added to, and does not replace, the background radiative heat flux.
Thus, the left hand side of my equation above represents the energy
going into the slab. The right is the radiative loss from both sides
to the assumed ambient temperature void on each side of the slab.

Simo -- why should kappa be temperature dependent?

[shostikk]

> Simo -- why should kappa be temperature dependent?

The radiation leaving the material consists of what is transmitted
through (the external flux + ambient transmitted) and what is emitted
by the material itself. The emission term depends on the material
temperature. So, the result depends if the 6.6 % transmission is
desired at room temp or at the final temperature.

If the original post wants 6.6 % to be transmitted and 5 % to be
reflected, the material should not absorb 100-6.6 % but only 100-5-6.6
%.

[Kevin]

Amy -- as you can see from this discussion, there is sometimes
confusion about definitions. I suggest you read all this, but then do
a few simple calcs to ensure that what you desire is what you get.

[Amy Mensch]

Thanks for your help. Ok I think I am following. First, I'm changing
the emissivity to 0.95.
I'm not sure exactly how Simo found the kappa values for the various
temperatures, but I don't think I need to. The information I have
about the layer material is the absorption coefficient, kappa, at
different wavelength bands (and not that the transmissivity must be
6.6%). I calculated transmissivity=e^-(kappa*L) for each band. Then
I came up with an average transmissivity, weighted by the contribution
of each wavelength band (f factors), which came out to 6.6%. However
I agree that value is not the actual transmissivity, because of
reflections etc., but I think I can use the 6.6% transmissivity to
calculate the average absorption coefficient for this material to be
135.9 /m. Then if I use that value, it sounds like FDS should take
care of the calculations for the actual amount of radiation that is
transmitted.

On Dec 2, 9:59 am, shostikk <simo.hosti...@vtt.fi> wrote:

[Kevin]

Amy, you're a mensch. Simo, learn a little Yiddish:

http://en.wikipedia.org/wiki/Mensch

You're a mensch, too!

> > %.- Hide quoted text -

[Antony]

From Kevin test case
emissivity = 0.95
reflectivity = 0.5
reflectivity = 0.066
material thickness = 0.02
In FDS absorptivity = emissivity = 0.95

reflectivity + reflectivity + absorptivity = 1
But 0.95+0.5+0.066 > 1
I am confuse on this, could anyone explain please?

[shostikk]

Antony, I'm sorry but I don't understand what you are trying to say.
In any case, for EMISSIVITY = 0.95, reflectivity would be 1-0.95 =
0.05.

Amy, your approach seems quite reasonable. As Kevin suggested, do some
testing to ensure you get something you would expect. The "f factors"
you mention - are these the Planck function integrals inside the
bands? I would be very interested how and for what kind of material
you have obtained such data. Can you share any of your results?

I got my kappa values using a stand-alone version of the two-flux
model, and iterating the kappa value untill I got the desired
transmittance. Even more complexity would come from the multiple
reflections, which I assumed would be very small.

[Antony]

Sorry, I wrote the word wrongly, it should be

emissivity = 0.95
reflectivity = 0.5

transmissivity = 0.066
I don't understand why reflectivity is 1-0.95 but not 1-emissivity-
transmissivity
Could you explain this?

[shostikk]

Something like this:

radiation hits surface
fraction r (reflectivity) is reflected back
fraction (1-r) is absorbed. -> SURFACE absorptivity a = (1-r)
emissivity e = absorbtivity a.
fraction (1-r) = e penetrates inside the solid.
fraction t of that gets transmitted through. (t = transmissivity)
fraction (1-t) is absorbed by the material at length of layer
thickness.
intensity reaching the back surface is e*(1-t)

[Kuldeep]

The absorption coefficient should be derived as follows

q_L = (1-r)q_o exp(-alpha * L)
q_L/q_o=tau

Therefore
alpha = [ln(1- r)- ln (tau)] / thickness

This is slightly different from the formula that comes P.T.
Tsilingiris???

Kuldeep

[Amy Mensch]

Yes, I believe "f factors" are the Planck function integrals. I
wasn't aware of that name for them. f is a function of refractive
index, wavelength, and temperature of the radiation source, that gives
the fraction of radiative intensity from zero to the wavelength of
interest. I am trying to simulate the heat transfer through a
polycarbonate layer, from a radiant panel. I have the equivalent
blackbody temperature of the radiant panel = temperature of the
radiation source. I arbitrarily divided the spectrum into 6
wavelength bands. Then I estimated polycarbonate's absorption
coefficient for each band from some papers on polycarbonate optical
properties. The main two were
- Progelhof, Franey, & Haas, Journal of Applied Polymer Science, 15,
1971.
- Wang, Abe, Matsuura, Miyagi, & Uyama, Applied Optics, 37, 1998.
I can share results, just let me know what you are interested in.

> > I am confuse on this, could anyone explain please?- Hide quoted text -

[Amy Mensch]

I think the difference is that it has an extra (1-r), for reflection
on the back face. See Simo's post #4. That will give you the factor
of 2 you are missing.

> > > > Is this correct?- Hide quoted text -