[fds-smv] wood pyrolysis

[Lucie Hasalova]

Hi

I have a problem with pyrolysis of wood.

I have a wooden wardrobe in a 10 cm domain. Under the wardrobe there
is a heptane pool set
as a burner releasing approximately 1500 kW. What we would like to do
is to model the fire
spread in the compartment. We want heptane to ignite the wood and we
need the wardorbe to
burn on its own. The problem is, that immediately after we take away
the heptane pool wood
stops burning.
We don´t have our own experimental data on wood, so we used
specification for birch and pine
from the work of Ana Matala - Estimation of solid phase reaction
parameters for fire
simulation - for the one step reaction scheme.
I know that 10 cm grid is quite coarse and 10 cm woodem walls are not
realistic but we
intend to model quite large domain and we can´t use finer grid.
Even though I tried to reduce the grid and the wooden walls to 5 cm
results were pretty much
the same.

When I look on the HRR data from HRR. csv the wood is burning because
there is a change in
HRR.

The wall temperature short before heptane is taken away is 600 °C and
the gas temperature is
1000 °C. When I take away heptane the wall temperature decrease about
200 to 300°C in 10
seconds and wood stops burning.

I would appreciate any help which direction to go and what can be the
problem.

Thank you,
Lucie


&TIME T_END=200.00/
&DUMP DT_RESTART=300.00/

&MESH ID='MESH', IJK=30,30,30, XB=0.00,3.00,0.00,3.00,0.00,3.00/

&REAC ID='HEPTANE',
FYI='NIST NRC FDS5 Validation',
C=7.00,
H=16.00,
O=0.00,
N=0.00,
CO_YIELD=6.0000000E-003,
SOOT_YIELD=0.0150/

&MATL ID='birch_char',
SPECIFIC_HEAT=3.50,
CONDUCTIVITY=0.50,
DENSITY=94.00,
EMISSIVITY=1.00/
&MATL ID='birch',
SPECIFIC_HEAT=2.20,
CONDUCTIVITY=0.2200,
DENSITY=550.00,
EMISSIVITY=1.00,
HEAT_OF_COMBUSTION=1.4500000E004,
N_REACTIONS=1,
HEAT_OF_REACTION=300.00,
NU_FUEL=0.83,
NU_RESIDUE=0.1720,
RESIDUE='birch_char',
N_S=3.12,
A=7.5130000E011,
E=1.6100000E005/

&SURF ID='birch',
RGB=146,202,166,
BURN_AWAY=.TRUE.,
MATL_ID(1,1)='birch',
MATL_MASS_FRACTION(1,1)=1.00,
THICKNESS(1)=0.1000/
&SURF ID='heptan',
COLOR='RED',
HRRPUA=1.6000000E003,
RAMP_Q='heptan_RAMP_Q'/
&RAMP ID='heptan_RAMP_Q', T=0.00, F=0.00/
&RAMP ID='heptan_RAMP_Q', T=1.00, F=1.00/
&RAMP ID='heptan_RAMP_Q', T=99.00, F=1.00/
&RAMP ID='heptan_RAMP_Q', T=100.00, F=0.00/

&OBST XB=1.70,1.80,0.90,2.10,0.2000,2.20, RGB=255,204,102,
SURF_ID='birch'/ zadni deska
&OBST XB=1.20,1.80,0.90,1.00,0.2000,2.20, RGB=255,204,102,
SURF_ID='birch'/ Obstruction[1]
&OBST XB=1.20,1.80,2.00,2.10,0.2000,2.20, RGB=255,204,102,
SURF_ID='birch'/ Obstruction[1]
&OBST XB=1.20,1.30,1.00,2.00,0.3000,2.20, RGB=204,204,255,
SURF_ID='birch'/ predni sklo II
&OBST XB=1.20,1.30,0.90,1.00,0.00,0.2000, RGB=204,102,0,
SURF_ID='birch'/ noha 2
&OBST XB=1.70,1.80,0.90,1.00,0.00,0.2000, RGB=204,102,0,
SURF_ID='birch'/ noha 3
&OBST XB=1.20,1.30,2.00,2.10,0.00,0.2000, RGB=204,102,0,
SURF_ID='birch'/ noha 5
&OBST XB=1.70,1.80,2.00,2.10,0.00,0.2000, RGB=204,102,0,
SURF_ID='birch'/ noha 6
&OBST XB=1.20,1.80,0.90,2.10,2.10,2.20, RGB=255,204,102,
SURF_ID='birch'/ horni deska
&OBST XB=1.20,1.80,0.90,2.10,0.2000,0.3000, RGB=255,204,102,
SURF_ID='birch'/ spodni deska
&OBST XB=1.20,1.80,1.00,2.00,0.00,0.1000, SURF_ID='heptan'/ heptan
zdroj

&VENT SURF_ID='OPEN', XB=0.00,0.00,0.00,3.00,0.00,3.00/ Vent
&VENT SURF_ID='OPEN', XB=3.00,3.00,0.00,3.00,0.00,3.00/ Vent[1]

&BNDF QUANTITY='GAS TEMPERATURE'/
&BNDF QUANTITY='WALL TEMPERATURE'/

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[TimoK]

The usual misunderstanding: MESH-line IJK+XB ==> 0.1 m cells
and:

&SURF ID='birch',
RGB=146,202,166,
BURN_AWAY=.TRUE.,
MATL_ID(1,1)='birch',
MATL_MASS_FRACTION(1,1)=1.00,
THICKNESS(1)=0.1000/

BUT: THICKNESS(1) has nothing to do with the IJK+XB cell
sizes. The solid state calculation is completely different
calculation than the gas phase calculation. So the two
different calculations can have completely different
thicknesses. So try to put:

THICKNESS(1)=0.018 /

So you have 18 mm thick wall made of plank wood. If you have
THICKNESS(1)=0.1 you have log walls and log walls do not
burn easily. The log is a heat sink. The default back
side boundary condition is "VOID" and for a wooden wall
this is good, because behind the wood you have the studs
and air between the studs. If you have there mineral
wool, you could try "INSULATED" backside boundary condition,
or make the material explicitely two layer material: first
(the fire side) layer is wood and the second layer is
mineral wool. Befind the mineral wool can be what so
ever, if there is enough mineral wool (it is a good
insulator of heat).

If it still does not burn, put THICKNESS(1)=0.005.
5 mm thick wood should burn easily.

TimoK

[Lucie Hasalova]

Thank you for a very fast answer.

I´ve read posts about MATL and SURF THICKNESS here in the discussion
to understanf better what you wrote and consequently I tried what you
suggest.
but it´s still not working.
I simplified the geometry and use just a wooden board

&OBST XB=1.40,1.50,0.90,2.10,0.2000,2.20, SURF_ID='pine'/ zadni deska

in the same domain.

&MESH ID='MESH', IJK=30,30,30, XB=0.00,3.00,0.00,3.00,0.00,3.00/

I went down to 5 mm surface thickness

&SURF ID='pine',
RGB=255,153,0,
BURN_AWAY=.TRUE.,
MATL_ID(1,1)='pine',
MATL_MASS_FRACTION(1,1)=1.00,
THICKNESS(1)=5.0000000E-003/

First time I set the simulation time to 120 s and the heptane pool was
taken away at 80 s but already at 60 s the wooden board started to
burn away and at 80 seconds there was almost nothing to burn just on
sides where is quite low temperature.
So I tried the same once more and set the time when the heptane stops
burning to 60 s. The wall temperature was the same as in the previous
case at the time 60 s but after the heptane stop burning wood stopped
burning as well.

In this case wouldn´t be the back side boundary condition BACKING
EXPOSED better? And the same in case I would like to do - wooden
wardrobe?
And in connection to this question arises. I understand that solid
state and gas state calculations can both work on different meshes but
whatt I don´t understand is how they exchange information when the
SURF and MATL thickness is not the same. For example if I a have a 10
cm thick wall defined by MATL thickness and I apply a SURF on the wall
1 cm thick and use BACKING EXPOSED. Then I say to FDS that I want the
heat to be transfered through 1 cm thick wall and this heat is given
to the gas phase on the other side of the wall. But where? Where the
MATL thickness ends - at 10 cm distance? 9 cm in between are simply
ignored by the solid phase solver and information is given to the
closest mesh boundary of the gas phase solver?
And the same problem arises with the BURN AWAY function. When the surf
thickness is less than the MATL thickness, BURN AWAY is governed by
the SURF thickness. When the energy from the material of the surf
thickness is exhausted then it burns away but physically disapperas
the volume given by MATL thickness.
Do I understand this correct

Thanks for your help,
Lucie

On 20 dub, 16:25, TimoK <tkorh...@gmail.com> wrote:
> The usual misunderstanding: MESH-line IJK+XB ==> 0.1 m cells
> and:
>
> &SURF ID='birch',
>       RGB=146,202,166,
>       BURN_AWAY=.TRUE.,
>       MATL_ID(1,1)='birch',
>       MATL_MASS_FRACTION(1,1)=1.00,
>       THICKNESS(1)=0.1000/
>
> BUT: THICKNESS(1) has nothing to do with the IJK+XB cell
> sizes. The solid state calculation is completely different
> calculation than the gas phase calculation. So the two
> different calculations can have completely different
> thicknesses. So try to put:
>
>       THICKNESS(1)=0.018 /
>

> So you have 18 mm thick wall made of plankwood. If you have

> THICKNESS(1)=0.1 you have log walls and log walls do not
> burn easily. The log is a heat sink. The default back
> side boundary condition is "VOID" and for a wooden wall

> this is good, because behind thewoodyou have the studs

> and air between the studs. If you have there mineral
> wool, you could try "INSULATED" backside boundary condition,
> or make the material explicitely two layer material: first

> (the fire side) layer iswoodand the second layer is

> mineral wool. Befind the mineral wool can be what so
> ever, if there is enough mineral wool (it is a good
> insulator of heat).
>
> If it still does not burn, put THICKNESS(1)=0.005.

> 5 mm thickwoodshould burn easily.

[R_Webster]

Lucie

I guess you can say that 9 cm are ignored if it helps you to visualize
it better. However, it is more appropriate to say that the geometry of
the gas phase cell has nothing to do with solid phase heat transfer. I
haven't looked at your input file in great detail, but it is possible
to use the "right" input parameters, and still not produce a
reasonable flame spread rate.

You must also consider the assumptions you are making. Is it possible
to use a parameter set optimized for 1-dimensional pyrolysis and apply
it (1) to combustion, (2) to a material with different boundary
conditions, (3) to a material in a different orientation, and (4) to 3-
dimensional flame spread, with minimal error? Maybe you're a little
more optimistic than I am.

Rob Webster

> For more options, visit this group athttp://groups.google.com/group/fds-smv?hl=en.- Hide quoted text -
>
> - Show quoted text -

[TimoK]

Lucie,

If you are thinking to model a burning wardrope, you
should be using much finer mesh for the gas phase
than 10 cm cells. And same is true for a wall made of
plank wood, if you want to model "burn away" features.
This means that you want to model the gas flow (and
radiation and...) behind the wood in the void between
the studs. So you need a reasonable number of gas cells
there.

One test you could do, is burning the 5 mm thick wood
wall with burn_away=.false. Then you should see if
you get an upward flame spread. Put BNDF outputs for
surface temperature and heat fluxes. Note that the
gas phase cell size has an impact on how the flame
spreads. So in principle, you need different material
parameters for different gas phase cell size calculations.
The finer the gas cell size the "hotter" fire you get.
The grid cell temperature is the "average" temperature
of the gas at the volume 10cm x 10cm x 10cm in your case.

[RidhaDjebali]

Hi,

I am trying to do a pyrolysis exercice of wood in a box and I am converting now the fds4 code inputs to the fds 6.0.1 (compile date Tue 26 Nov 2013). I find it difficult du to numerous changed input functions.

The code is here enclosed. If possible, would like to help me know the correct FDS version 6.0.1 input file ?  

Many thanks for you in advance.

[Randy McDermott]

I suggest you read the FDS 6 user guide and work the pyrolysis examples in the verification suite.

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