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See User Guide |
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Keep in mind that both extinction models in FDS neglect the effect of high strain rate. The extinction method is purely thermodynamic. You may not achieve "blow out" because that would require some accounting for high strain extinction. |
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Hello team,
I'm trying to realize the blow-off of methanol pool fire by airflow. I chose the "extinction 2" as extinction model. Cell size is 0.5 mm. CFT and AIT were set to 1257 ℃ and 200 ℃, respectively.
However, the pool cannot be ignited. When I remove AIT, it can be ignited but cannot be blown out.
The main part of the code is as follows:
&REAC FUEL='METHANOL'
RADIATIVE_FRACTION=0.22
N_SIMPLE_CHEMISTRY_REACTIONS=2
FUEL_C_TO_CO_FRACTION=1.0
HEAT_OF_COMBUSTION = 19000.
CRITICAL_FLAME_TEMPERATURE=1257
AUTO_IGNITION_TEMPERATURE=200/
&COMB EXTINCTION_MODEL='EXTINCTION 2'/
&MATL ID = 'METHANOL LIQUID'
EMISSIVITY = 1.
NU_SPEC = 1.
SPEC_ID = 'METHANOL'
HEAT_OF_REACTION = 1165
CONDUCTIVITY = 0.20
SPECIFIC_HEAT = 2.48
DENSITY = 796
ABSORPTION_COEFFICIENT = 1500
BOILING_TEMPERATURE = 64.65 /
&MATL ID='STEEL', CONDUCTIVITY=54, DENSITY=7800, SPECIFIC_HEAT=0.465, EMISSIVITY=0.8 /
&SURF ID='PAN', COLOR='GRAY 50', MATL_ID='STEEL', THICKNESS=0.003/
&SURF ID='POOL', COLOR='LIGHT BLUE', MATL_ID='METHANOL LIQUID', THICKNESS=0.1, EXTERNAL_FLUX=20, RAMP_EF='rampef' /
&RAMP ID='rampef', T= 0., F=1. /
&RAMP ID='rampef', T=15., F=1. /
&RAMP ID='rampef', T=16., F=0. /
&SURF ID='SUPPLY',
RGB=26,204,26,
VEL=-5.0,
RAMP_V='SUPPLY_RAMP_V'/
&RAMP ID='SUPPLY_RAMP_V', T=0.0, F=0.0/
&RAMP ID='SUPPLY_RAMP_V', T=35, F=0.0/
&RAMP ID='SUPPLY_RAMP_V', T=36, F=1.0/
&RAMP ID='SUPPLY_RAMP_V', T=45, F=1.0/
&RAMP ID='SUPPLY_RAMP_V', T=46, F=0.0/
&VENT MB='ZMIN', SURF_ID='OPEN' /
&VENT MB='ZMAX', SURF_ID='OPEN' /
&VENT MB='YMIN', SURF_ID='OPEN' /
&VENT MB='YMAX', SURF_ID='OPEN' /
&VENT MB='XMIN', SURF_ID='OPEN' /
&VENT ID='XMAX', SURF_ID='OPEN', XB=0.1,0.1,-0.1,0.1,0.15,0.35/
&VENT ID='Vent', SURF_ID='SUPPLY', XB=0.1,0.1,-0.1,0.1,-0.05,0.15, IOR=-1/
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