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Old   February 8, 2014, 14:31
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  #81
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Georgi Angelov
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Thank you a lot. I have figured out why my simulation was returning with code for overflow. This was due to the gambit spliting and meshing. I have splited the volumes in a wrong way and this was the problem. I split the Big volume with the inside cylinder and the big volume with the outside cylinder and now it works.

I have two problems:

1.The first is : As I don't have enough time for running the simulation on server, because I have to wait about 2 weeks for free one and in this period I have to finish my thesis. I want to run the simulation on my computer with a not so fine mesh, but enough to give me some results. This will take several days, so I want to do the simulation on steps. Around 300 [s] for a day from 1200[s] simulation, but after I run the steady state analysis, during the transient analysis[timestep 4, 3.trn files) when I click pause, it stops, writes an res. file and the .trn files. And when I want to continue the simulation from the 3.trn file and timestep 4 it starts from Steady State again. I have read about this problem, set initial value conditions to 3.trn files, and continue history from the initial values, but it starts from the steady state again.
Is there a way to run my simulation after pausing from the current stage?
Attached Images
File Type: jpg Pause.jpg (48.3 KB, 23 views)
File Type: jpg Error.jpg (59.5 KB, 21 views)
File Type: jpg Edit.jpg (85.1 KB, 17 views)
File Type: jpg Define run.jpg (80.5 KB, 23 views)
File Type: jpg Steady State.jpg (58.9 KB, 20 views)
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Old   February 8, 2014, 14:43
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  #82
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Georgi Angelov
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And the second problem is about a Gambit extract:

2.Now I'm running my simulation with a Gambit geometry and mesh, but as I don't like the mesh in Gambit I want to create my model with a Gambit geometry and Ansys Mesh. I have watched some tutorials about meshing and modeling. And I understand some things but I cannot run my simulation with an Ansys mesh. After I extract the geometry from Gambit it writes me 8 parts 9 bodies(solids) and add material to the bodies, and they are solids(picture 1 - Import 1). I click to add frozen, I formed a new part from all of them and it goes like 1 part 9 bodies(fluids) add frozen( picture 2- Geometry 2). I mesh the volumes in mesh tool and after that I continue in CFX. When it reads the model it creates automatically interfaces between the jet fans and the main volume ( picture Interface and Interface 2).I set my expressions and start to do a run but at the begining it crashes with an error( picture ERROR). I don't know what to do with this error the interfaces are wrong or my method in DM is wrong? What are the steps for setting the right conditions and interfaces between the jet fans, the main volume and the fire volume? Thank you in advance.
Attached Images
File Type: jpg Import 1.jpg (89.6 KB, 34 views)
File Type: jpg Geometry 2.jpg (91.6 KB, 27 views)
File Type: jpg Interface.jpg (81.0 KB, 28 views)
File Type: jpg Interface1.jpg (93.1 KB, 36 views)
File Type: jpg Error CFX.jpg (78.1 KB, 26 views)
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Old   May 18, 2017, 01:31
Default FIRE and visibility
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Nirav
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good morning
i am new to fluent i want to predict fire , its smoke distribution and visibilty and time required to clear smoke from a given domain ( basement). can please help do it using fluent ?
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Old   April 30, 2019, 06:40
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This is a great details Mr. CFDST. I have understood most of the details however I would like to understand what does InsideSource, rSource, SourceFlag, FirePower indicates. Can you please explain all the equations which you have cosndiered.


my email: cfdengineer44@gmail.com


Thanks.



Quote:
Originally Posted by CFDST View Post
Inert Fire Modelling
--------------------


#
# Fire Source
#


For inert fire simulations a source of heat is added to a source volume. This type of simplification is appropriate if accurate predictions of the shape of the flame and the near field region of the fire are not important.


The FireEfficiency parameter determines how much of the fire power is added as heat. It can be used to account for the combustion efficiency. In well ventilated fires with clean burning fuels this is likely to be quite high (0.9-1.0).


Radiation should always be modelled. Radiative heat fluxes from the fire zone can be significant. The total fire power will be split between the radiative and convective heat fluxes from the fire zone. If radiation is not being modelled then the efficiency needs to be reduced to account for the fact that all the power is dissipated by convective/diffusive heat fluxes only.


It is important to use a sensible size for the fire volume which will produce sensible temperatures. A useful strategy is to use CEL to set up a source region so that the size of the source can be easily changed, which allows for rapid testing of different sizes without having to generate a new mesh each time. This strategy becomes essential for modelling fires (or any source regions) that grow in time and is described in ANSYS solution number 541877. The mesh needs to be sufficiently fine in the fire region to accommodate any likely changes in size. It is then a simple matter to change the fire size and restart the calculation until an acceptable fire temperature is obtained. Once a suitable size has been obtained the mesh can be modified if desired.


The example below is for a fire in which the heat source is added to a cylindrical volume. The minVolume and minRadius parameters are used to constrain the minimum source volume to be similar to the smallest mesh volume.


#
# CCL Example
#
LIBRARY:
CEL:
EXPRESSIONS:
xFire = 5 [m]
yFire = 2.5 [m]
zFire = 0 [m]
FireRadius = max(0.25 [m], minRadius)
FireHeight = FireRadius
FireEfficiency = 0.9
FirePower = 0.5 [MW] * FireEfficiency
rSource = sqrt((x-xFire)^2 + (y-yFire)^2)
hSource = z-zFire
insideSource = step((FireRadius-rSource)/unitL)*step((FireHeight-hSource)/unitL)
SourceVolume = volumeInt(SourceFlag)@Fire Subdomain
heatSource = insideSource*FirePower/max(SourceVolume, minVolume)
minVolume = minVal(Volume of Finite Volumes)@Fire Subdomain
minRadius = (minVolume/pi)^(1/3)
unitL = 1 [m]
END
END
ADDITIONAL VARIABLE: SourceFlag
Option = Definition
Tensor Type = SCALAR
Units = [ ]
Variable Type = Unspecified
END
END


FLOW: Flow Analysis 1
.
DOMAIN: Default Domain
.
FLUID MODELS:
ADDITIONAL VARIABLE: SourceFlag
Additional Variable Value = insideSource
Option = Algebraic Equation
END
END
.
SUBDOMAIN: Subdomain 1
SOURCES:
EQUATION SOURCE: energy
Option = Source
Source = heatSource
END
END
END
.
END
.
END


Cylindrical source volumes tend to produce flame heights that are quite a bit higher than the source height. A conical source region will produce a flame height closer to the height of the source region, eg:


LIBRARY:
CEL:
EXPRESSIONS:
BaseRadius = max(0.25 [m], minRadius)
FireHeight = 2*BaseRadius
FireRadius = BaseRadius*(1-hSource/FireHeight)
END
END
END


#
# Soot modelling
#


Soot is modelled as an additional variable having a specified volumetric source in the fire region. The soot yield is specified as a fraction of the fuel mass source. Values for soot yields for various fuel types are available in the literature and are dependent on how cleanly the fuel burns.


#
# CCL Example
#
LIBRARY:
CEL:
EXPRESSIONS:
sootYield = 0.05
FuelHeat = 15 [MJ kg^-1]
FuelMassSource = FirePower/FuelHeat
massSource = insideSource*FuelMassSource/max(SourceVolume, minVolume)
sootSource = sootYield * massSource
END
END
ADDITIONAL VARIABLE: soot
Option = Definition
Tensor Type = SCALAR
Units = [ ]
Variable Type = Specific
END
END


FLOW: Flow Analysis 1
.
DOMAIN: Default Domain
.
FLUID MODELS:
ADDITIONAL VARIABLE: soot
Kinematic Diffusivity = 1.0E-5 [m^2 s^-1]
Option = Transport Equation
END
END
.
SUBDOMAIN: Subdomain 1
SOURCES:
EQUATION SOURCE: soot
Option = Source
Source = sootSource
END
END
END
.
END
.
END


The radiation field can be coupled to the gas temperatures via the soot mass fraction by setting the fluid absorption coefficient equal to:
GasAbs = 0.01 [m^-1] + max(1.5*density*soot/(smokeDensity*smokeDiameter), 0.0 [m^-1]).
smokeDensity = 2000 [kg m^-3]
smokeDiameter = 1.0E-6 [m]


The smoke diameter is the effective mean diameter of smoke particles and is essentially a free parameter that determines the degree of coupling between the gas and the radiation field. The formula above is derived from the emission from black, non-interacting spheres at the same temperature as the gas they are immersed in.




#
# Visibility
#


Visibility can be calculated on a local basis using the light-reflecting visibility for an infinite homogenous medium having a given local soot concentration [ref = DiNenno P.J., Handbook of Fire Protection Engineering, NFPA, Quincy, MA, USA, 1990.]


#
# CCL Example
#
LIBRARY:
CEL:
EXPRESSIONS:
c1 = 3.95E-4 [kg m^-2]
visibility = min((c1/max(soot*density, 1e-10[kg m^-3])), 1e3 [m])
END
END
ADDITIONAL VARIABLE: Visibility
Option = Definition
Tensor Type = SCALAR
Units = [m]
Variable Type = Unspecified
END
END


FLOW: Flow Analysis 1
.
DOMAIN: Default Domain
.
FLUID MODELS:
ADDITIONAL VARIABLE: Visibility
Additional Variable Value = visibility
Option = Algebraic Equation
END
END
.
END
.
END


#
# Time stepping
#


Buoyant flows can be difficult for the steady state solver to converge. It is often beneficial to use the transient solver even for a steady state solution. Time steps of around 1s may be possible for an inert fire model. The steady state solver can be run initially with auto time stepping. The time steps reported by the auto time stepping can by used as a guide for setting a transient time step.




#
# Transient fires
#
The power output of a transient fire tends to be described by a quadratic growth phase (P = a t^2) up to some constant maximum power (Pmax) achieved at some time (tg) which is the fire growth time. The size of the quadratic coefficient (a) determines the rate of fire growth.


National Fire Protection Association (NFPA) standardized fire growth curves:
Fire growth rate a (W/s^2)
Slow 2.9
Medium 12
Fast 47.


The fire size for the maximum power can be determined using the procedure above for a steady state fire. The power of a real fire tends to grow proportional to the cross sectional (floor) area of the fire. However, the constant of proportionality is different for differing fuels, and environmental conditions (wind, outdoor/indoor, ventilation, confinement, etc) hence the need to determine a size which gives a sensible temperature as described above. Also, real fires are typically fuelled by evaporation from a surface area, whereas the inert model described here distributes the fire power over a volume. To be consistent with the source modelling, the fire power should be set to be proportional to the source volume:
FireRadius = max((FirePower/MaxPower)^(1/3) * MaxRadius, minRadius)


If the fire power was to be set to be proportional to the area, the power density would be inversely proportional to the radius and grow as the radius shrinks, leading to spurious higher flame temperatures at smaller radii and the danger of extremely high power densities and temperatures at the start of the simulation (the power density would tend to infinity as the radius tends to zero).


The inert fire approximation is not expected to produce accurate solutions for the flame shape or the near field region, so the particular growth law for the radius is not important. What matters is maintaining sensible flame temperatures. If more accuracy than this is required you should use a combustion model.
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Old   April 30, 2019, 08:01
Default arrhenius equation
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Abdullah Arslan
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Hi

When I try to simulate combustion/catalytic reaction (using finite chemistry), I have to set up required reaction. The reaction can be modelled using arrhenius equation or by expression. By assigning a random expression the model works fine but when I use Arrhenius equation parameters, I receive this error which shouldn't be at first place.

Error processing expression 'Pre Exponential Factor'.
The expression assigned to 'Pre Exponential Factor' does not resolve to correct dimensions.
Expected: 'm^9 s^-1 K^-1500 mol^-3'
Actual: '<dimensionless>'.
Error processing expression: Pre Exponential Factor = 2.3e-6
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