Radiation in FDS

[clem...@tpi.setec.fr]

Hi everybody.
Is someone could help me to understand well how FDS calculate the
radiation.

First :
The emissivity of the material represent the solid capacity to absorb
radiation. 1 if it is black solid or 0 if it is a white solid. There
is a part of the radiation produce by the fire that is absorb by the
solids, and other part that is reflect. So at least, the raditation
emited by a solid is :
RADIATIONtotal = P + R

With : P=E*sigma*S*T^4 and R the part who as been reflected at the
time t.

Is FDS take into account this Reflexion ? and how does he ?

Ex: If i put a fire, that no produce soot, in an insulated room (I
mean there is no opening and all the wall has zero emissivity). If FDS
take the reflexion in account, the temperature inside the room should
increase indefinitely, is that true ?

I try to understand all the parameters of the RADI option. Is someone
could help me with those :

- RADTMP : Real Assumed radiative source temp. (what is it ? what
should change if I changed that ?)

- WIDE_BAND_MODEL Logical Non-gray gas assumption (what is it ? what
should change if I changed that ?)

- CH4_BANDS Logical Include extra fuel bands (what is it ? what
should change if I changed that ?)

- KAPPA0 Real Constant absorption coefficient 1/m (what is it ? what
should change if I changed that ?)

- NMIEANG Integer Number of polar angles (what is it ? what should
change if I changed that ?)

- NUMBER_RADIATION_ANGLES Integer Number of solid angles : Why the
default value is 104 ???

All the suggestion you could do will help me ! It's pretty important
for me.

Thanks for your help !

[dr_jfloyd]

Sections 3.3 and 4.3 of the Theory manual describe the details of the
radiation model. FDS accounts for the emissivity of both surfaces and
gasses. Is there something in that description that is unclear?

A fire in a closed room with adiabatic walls will see continuously
increasing temperature until the fire runs out of oxygen.

The table header for Table 15.9 in the User's Guide refers you to
Section 11.3 where the input keywords you list are described.

Each additional radiation angle costs memory and time. 100 was what
judged to be a reasonable balance between the accuracy of the
radiation solution and the cost.

On Mar 30, 5:37 am, "clemen...@tpi.setec.fr" <clemen...@tpi.setec.fr>
wrote:

[Kevin]

Si vous ne comprennez pas quelquechose que nous avons ecrit dans les
FDS documents, c'est tres important que vous nous disez ou nous avons
peur que des autres ne comprennent pas aussi. Quand vous nous demandez
une question, dites nous le section que vous lisez, ou le formula que
vous ne comprennez pas.

On Mar 30, 5:37 am, "clemen...@tpi.setec.fr" <clemen...@tpi.setec.fr>
wrote:

[clem...@tpi.setec.fr]

> Si vous ne comprennez pas quelquechose que nous avons ecrit dans les

> FDS documents, c'est tres important que vous nous disez.

I don't understand well the section 8.4.2
I will try to explain why :

We have two option to designate a fire : HRRPUA in the SURF line or
HEAT_OF_REACTION in the MATL line.
If I specified HRRPUA there is an ejection of fuel from the surf and
this ejection is regulated to keep HRRPUA constant. Then combustion is
allowed or not allowed depending on the temperature and the amount of
oxygen. At least the reaction release a certain amount of energy.
What happened if I specified HEAT_OF_REACTION ? Section 8.4.2 say that
heat of reaction is the amount of energy consumed to produce something
else. So there is no energy release from the combustion ? The
combustion just consumed energy ? that not possible.
My supposition is that we need to specify Heat of reaction just if
there is any other reaction than combustion (like evaporating water).
If not, we just need to specify HEAT OF COMBUSTION and so FDS will
regulate the amount of fuel ejected.
Considering that, I think that if I specified HEAT OF COMBUSTION in
the MATL LAB, Every surfaces which are formed by this material are
ejecting an adequat amount of fuel to respect the HEAT OF COMBUSTION
we wanted ?

Then I don't understand well Section 9.1.2 :
In which case do I need to specify HEAT OF COMBUSTION in the reac line
AND in the MATL line ? Why do I need to do it and what FDS will do
exactly ?

Thanks a lot for your Help

[Kevin]

I must speak English now. I wish my French were good enough to
describe this sort of thing.

HEAT_OF_REACTION is the amount of energy required to convert solid
fuel (like wood, for example) into fuel gas that can combust. Think of
this like the latent heat of vaporization. The energy used to gasify
the solid fuel is subtracted from the energy that is heating up the
solid.

HEAT_OF_COMBUSTION tells FDS how much energy to release in the gas
phase per unit mass of gasified fuel. It is specified in multiple
places because there can be only one gas phase reaction by default. If
you specify on the REAC line that the fuel is methane (C=1, H=4), the
HEAT_OF_COMBUSTION will be based on CH4. However, if you have
something else burning besides methane, you need to specify its HoC so
that FDS can adjust the mass burning rate accordingly. This is a very
common source of confusion. Understand that in the mixture fraction
formulation, there can only be one fuel gas. If you have a room filled
with different fuels, FDS has adjust the burning rates to produce one
effective fuel gas.

On Mar 31, 5:00 am, "clemen...@tpi.setec.fr" <clemen...@tpi.setec.fr>
wrote:

[dr_jfloyd]

Based on the section numbers you have given, it looks like you have an
older version of the User's Manual. There has been a a lot of
reorganization in the manual with some focus on trying to do a better
job explaining how to use the various pyrolysis models. You may wish
to download the new manual from http://fire.nist.gov/fds/documentation.html

On Mar 31, 5:00 am, "clemen...@tpi.setec.fr" <clemen...@tpi.setec.fr>
wrote: