Absorption coefficient

[andrej_f]

Hi all,

I have noticed in the User's Guide p. 32 that absorption coefficient
is mentioned. However, no reference is added where the thorough
explanation of this coefficient can be found. Perhaps this is a pure
experimental value?

Andrej

[Bryan Klein]

Hello Andrej,

First of all, if you do not already have it, make sure you have the
latest documentation. Currently that would be found in the RC3 package
for your platform.

On page 30, within section 4.10.3 of the User Guide, there is a short
comment that states:

"Prior to FDS5, the thermal radiation from the gas space was always
absorbed at the surface of the solid material and the emission to the
gas space took place on the surface. Starting in FDS5, the solid
material can be given an ABSORPTION_COEFFIENT (1/m) that allows the
radiation penetrate and absorb into the solid. Correspondingly, the
emission of the material is based on the internal temperatures, not
just the surface."

Then at the bottom of page 32:
"Note also the ABSORPTION_COEFFICIENT for the liquid. This denotes the
absorption in depth of thermal radiation. Liquids do not just absorb
radiation at the surface, but rather over a thin layer near the
surface. Its effect on the burning rate is significant. An example is
given in the Verification Guide under the name ethanol_pan."

Last, take a look at the Technical Guide. Section 3.4, especially
section 3.4.4 on radiation heat transfer to solids. You will see how
absorption coefficient is used in the equations.

I am working on a method for automatic cross references in LaTeX
between the user and technical guides. For now, searching the PDF
files is the best option.

What other questions do you have about the ABSORPTION_COEFFICIENT
parameter?
If you have specific questions we can answer them here, and add more
information in the guides.

-Bryan

[andrej_f]

Hi Bryan

First, thank you for such a quick reply. It is greatly appreciated. It
is nice to see that the Group is alive and working.
Well, I "explained" K with a Figure 5.5 p 167 from the Drysdale's
book. I believe the concept is the same.

I ran the "ethanol_pan" example. I have a problem with understanding
the DEVC lines. For example this one below:

&DEVC
XYZ=0.01,0.01,0.05,QUANTITY='INSIDE_WALL_TEMPERATURE',IOR=3,ID='temp 1
mm' , DEPTH=0.001 /

It is perhaps very straightforward for an experienced user but not for
the beginner. As I understand the line below means
XYZ=0.01,0.01,0.05 The device position.

QUANTITY='INSIDE_WALL_TEMPERATURE' Is it just a name of the device or
have any executive value? If yes, this should be explained in the
User's Guide?

IOR=3 Direction in the positive z axis.

ID='temp 1 mm' Name!?

DEPTH=0.001 Depth to where? To the wall, liquid?

I will use FDS 5 for my Research project so I might later have some
additional questions/suggestions.

Andrej

[Kevin]

Andrej

In the FDS 5 User's Guide, we are adding more detail about certain
output QUANTITY's. I will add more description about
INSIDE_WALL_TEMPERATURE because it is tricky. The coordinates refer
to the surface, but the actual measurement is made at some DEPTH
inside. This can be easily explained.

Thanks

K

[Khalid Moinuddin]

Ethanol_pan simulation is to simulate the fire test conducted by us. I
am currently testing what are the changes made in FDS5 which are
relevant to this simulation and the individual effect of each changes.
Definitely absorption coefficient is on of them. Its default value is
given as 50000. in table 13.9. Is it same for the liquid and solid
material? What should be its equivalent value in FDS4 where it did not
have a provision to specify?

I mean what is the assumed value of the absorption coefficient for
ethanol (or any liquid) fuel in FDS4?

with regards

Khalid

> > > Andrej- Hide quoted text -
>
> - Show quoted text -

[Kevin]

Two major changes, both by Simo Hostikka of VTT, Finland. First,
multiple layers of materials. Now the ethanol pan fire is modeled via
a layer of ethanol on top of a layer of steel on top of a layer of
substrate. No more homogenous solids, like in FDS 4. So we know we
have better heat conduction. Second, the absorption in depth. Instead
of all the fire's radiative heat being absorbed right at the liquid
surface, it is distributed over some finite depth. This turned out to
be VERY important in this case.

There is no equivalent mechanism in FDS 4 for either of these.

On May 23, 9:28 pm, Khalid Moinuddin <Khalid.Moinud...@vu.edu.au>
wrote:

> > - Show quoted text -- Hide quoted text -

[szilagyics]

Hello Everybody,

Could you help me? I don't really understand that the absorption
coefficiens default value is 50000, but in the ethanol pan test case
it is just 40. It's a very big difference. And where can I find these
data from different materials?

Thanks ,

Cs.

[dr_jfloyd]

The default value to 50,000 cause the in-depth absorption calc to be
skipped (it is only invoke if less than the default value). So it
really has no physical meaning other than being sufficiently large to
ensure that radiation is absorbed in the first wall cell node.

[Khalid Moinuddin]

Sorry for the asking the same question again. If the default value
50,000 means that radiation is absorbed in the first wall cell node, is
it the equivalent value of the absorption coefficient for ethanol fuel
in FDS4?

With regards

Khalid

[dr_jfloyd]

FDS4 did not have an in-depth model like FDS5. There was no
absorption coefficient in FDS4, therefore, there can be no equivalent
value.

The default value of 50,000 results in the in-depth model not being
used and the net radiative flux is deposited in the first grid cell
which is the behavior of FDS4.

On May 27, 7:27 pm, "Khalid Moinuddin" <Khalid.Moinud...@vu.edu.au>

[Kristopher Overholt]

I would like to resurface this discussion about the absorption coefficient as I cannot seem to grasp how it can be determined. I came across one definition of an absorption coefficient (Kv) being a certain area per unit volume (giving units of 1/m), but I am unable to find a correlation between this Kv and the absorption coefficient used in the ethanol_pan case (40 m^-1) using this method. When 40 was used, the ethanol_pan case matches with the experimental test on the graph shown in the Verification Guide. However, when I try to use the dimensions of the ethanol pool for Kv, I get Kv = ( 0.7m*0.8m) / (0.7m*0.8m*0.009 m) = 111 m^-1.

I suppose what I am asking is, can the absorption coefficient be determined by some relationship of the geometry and depth of the liquid fire (in this ethanol_pan case) or is it an experimentally derived number, or am I completely off with both of these ideas?

Thanks.

Kris

[Hostikka Simo]

Kris,

 

The absorption coefficient is a material property, not something you could calculate based on the case geometry.

 

The coefficient we are using is a spectrally averaged absorption cofficient. When this averagaging is done, the spectral absorption coeffient is integrated over the IR spectrum. In this process, some temperature and possibly some geometrical length scale must be assumed. Spectral absorption coefficients have been determined experimentally for many materials (see e.g. http://webbook.nist.gov/chemistry/). To my knowledge, this data has not been used to get the averaged absorption coefficients we need. My first impression was that for ethanol, for instance, we might end up with a very different number than 40 1/m.

 

In the verification part we use 40 1/m for ethanol pan, but this number was a pure "best fit" value to match the experiment. Since there are still many physical phenomena in pool fires that are not taken into account by FDS, some of these effects may now have been "lumped" to the number 40.

 

Simo

[Kevin]

There are various "absorption coefficients" used in FDS, for solids,
liquids and gases. The basic idea is that some quantity is being
attenuated according to

I = I_0 exp(-K*L)

If the quantity is the Intensity of the radiation, I_0 is what you
start with, and I is what you have after passing a distance L through
the media. The units of 1/m make the argument of the exponential
function dimensionless. Simo's discussion of spectral averaging refers
to the fact that if the quantity being attenuated is thermal
radiation, then K is a function of the wavelength, and a nasty one at
that. In FDS, we can't afford to do transport for all the different
wavelengths (or bands thereof), so we lump all the thermal radiation
into a spectrally-averaged quantity, which requires a spectrally-
averaged K. It is tricky to get K based on measurements done for
particular wavelengths.

On Sep 26, 8:53 pm, "Kristopher Overholt" <koverh...@gmail.com> wrote:
> I would like to resurface this discussion about the absorption coefficient
> as I cannot seem to grasp how it can be determined. I came across one
> definition of an absorption coefficient (Kv) being a certain area per unit
> volume (giving units of 1/m), but I am unable to find a correlation between
> this Kv and the absorption coefficient used in the ethanol_pan case (40
> m^-1) using this method. When 40 was used, the ethanol_pan case matches with
> the experimental test on the graph shown in the Verification Guide. However,
> when I try to use the dimensions of the ethanol pool for Kv, I get Kv = (
> 0.7m*0.8m) / (0.7m*0.8m*0.009 m) = 111 m^-1.
>
> I suppose what I am asking is, can the absorption coefficient be determined
> by some relationship of the geometry and depth of the liquid fire (in this
> ethanol_pan case) or is it an experimentally derived number, or am I
> completely off with both of these ideas?
>
> Thanks.
>
> Kris
>

[Khalid Moinuddin]

Hi Kevin and Simo,

 

I have run ethanol_pan case with three different cell sizes i.e. 50 mm, 25 mm and 12.5 mm. The result is attached to this mail. You can see that the results vary greatly.

 

I apologise that I did not do this during the beta testing. At that time, I thought that the simulation result of fire in open space would be grid independent at 50 mm cell as found with FDS4 which is shown in reference [36] of FDS5 User's guide.

 

Regards,

[Kevin]

I need to look at this case and get the integrated heat flux for the
various mesh sizes. Thanks for the info. K

On Nov 1, 11:31 pm, "Khalid Moinuddin" <Khalid.Moinud...@vu.edu.au>
wrote:
> Hi Kevin and Simo,...
>
> read more »
>
> ethanol_pan_grid.doc
> 78KDownload

> > > > > > > > > On May 6, 6:11 am,- Hide quoted text -

[Kevin]

Khalid

I have done some work on this problem last week. The ethanol_pan case
is difficult to work with because there are so many parameters. So I
decided to just look at a simple "burner" in FDS. Nothing more than a
fixed HRRPUA, with no specified material properties. I had to add a
new output to FDS in order to integrate the HEAT_FLUX to the burner
surface. I noticed that for cases with grids of 10 cm, 5 cm and 2.5
cm, the radiative loss (which is roughly proportional to the
integrated heat flux to the burner) varied from 35% to 42%. The 35% is
understandable in the 10 cm case, because FDS is just using the given
RADIATIVE_FRACTION. In the 2.5 cm case, the sigma*T^4 term is also
playing a role. In the ethanol_pan case, RADIATIVE_FRACTION=0, and the
radiative loss is calculated purely from temperature and gas/soot
composition. It appears that this is the leading cause of grid
dependence in your case and in mine.

Our goal is to eliminate as much as possible grid dependence in the
gas phase calculation. We have had good success for total HRR and
flame height. Now we need to turn our attention to the radiative loss.
If we succeed in this, then we can be confident that the overall
spatial distribution of the fire's energy is independent of the grid,
and we can then turn our attention to the details of the pyrolysis. We
are trying to decouple gas and solid phase phenomena. Results such as
yours are a combination of the two. By looking at the burner only, we
focus on the gas. On the other hand, we have simple examples of just
the solid (or liquid) phase. Grid dependence is mainly in the gas
phase. For the solid, we have as fine a grid as we need.

I might bump this thread out into a separate one, to keep open the
lines of communication, and to let others on the team know what we're
up to. K

On Nov 1, 11:31 pm, "Khalid Moinuddin" <Khalid.Moinud...@vu.edu.au>
wrote:

> Hi Kevin and Simo,...
>
> read more »
>
> ethanol_pan_grid.doc
> 78KDownload
>

> > > > > > > > > On May 6, 6:11 am,- Hide quoted text -

[zpe...@gmail.com]

hi,Kevin

I want to know that the fds6 can predict the pool fire burning rate  accuratly now?

now there is any nother one who  gain the liquid fuel  absorption coefficients experimently?

I want to simulate the pool fire accuratly, is it possible?

In the user guide ,about pool fire ,there is only the thermal conduction model in fds,so it is impossible to achieve now ?

• 
  

在 2007年11月12日星期一UTC+8下午10时05分23秒，Kevin写道：

[shostikk]

Check the FDS6 nightly build validation guide, and the section on liquid pool burning rates. There is some discussion on the absorption coefficients.

Please note that using a single value (i.e. gray assumption) for the liquid absorption coefficient is very strong approximation and inherently leads to
some uncertainty in the computation because many of the liquids have very strong peaks in their spectra at the infrared region.

[pei zhu]

Thank you very much,shostikk!

how can I download the FDS6 build validation guide? is that in the 

Editing:  svn/  trunk/ FDS/ trunk/ Manuals/ FDS_Validation_Guide/ Burning_Rate_Chapter.tex  ?

2013/9/9 shostikk <simo.h...@vtt.fi>

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[Gregor]

see:
"As part of the FDS 6 beta release, you can access nightly versions of the FDS-SMV documentation/manuals. "

[Chris]

I couldn't find the Absorption Coefficient of Steel. What value would you recommand as I don't think the Standard-Value of 50'000 is right.

Thanks.

Chris

[zpe...@gmail.com]

Sorry, Chris, I only considered the Absorption Coefficient of fuel before, but I never focused on the materical of steel.

---

zpe...@gmail.com

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[Chris]

Thanks. Maybe someone else :-)

[Khalid Moinuddin]

Absorption coefficient represents the depth from the surface to where radiation can go/be absorbed. I don't think for steel it happens to any depth, rather at the surface. So the default value should be ok.

---

From: fds...@googlegroups.com [fds...@googlegroups.com] on behalf of Chris [chr...@hotmail.com]
Sent: Wednesday, 3 September 2014 1:13 AM
To: fds...@googlegroups.com
Subject: Re: Re: [fds-smv] Re: Absorption coefficient

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