# forced to stick of soot particles

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 November 27, 2012, 10:09 forced to stick of soot particles #1 Senior Member   Govindaraju Join Date: Apr 2010 Posts: 206 Rep Power: 8 Dear friends I am trying to simulate a mixture of gas and soot particles . My idea is to trap the soot particle by suitable design of the chamber. Please find the CCL and attached geometry . I could not find any particle movement and no sticking effect in cfx post . Kindly rectify my mistake Your kind help is highly appreciated Thank you kmgraju LIBRARY: MATERIAL: Air Ideal Gas Material Description = Air Ideal Gas (constant Cp) Material Group = Air Data, Calorically Perfect Ideal Gases Option = Pure Substance Thermodynamic State = Gas PROPERTIES: Option = General Material EQUATION OF STATE: Molar Mass = 28.96 [kg kmol^-1] Option = Ideal Gas END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 1.0044E+03 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END REFERENCE STATE: Option = Specified Point Reference Pressure = 1 [atm] Reference Specific Enthalpy = 0. [J/kg] Reference Specific Entropy = 0. [J/kg/K] Reference Temperature = 25 [C] END DYNAMIC VISCOSITY: Dynamic Viscosity = 1.831E-05 [kg m^-1 s^-1] Option = Value END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 2.61E-2 [W m^-1 K^-1] END ABSORPTION COEFFICIENT: Absorption Coefficient = 0.01 [m^-1] Option = Value END SCATTERING COEFFICIENT: Option = Value Scattering Coefficient = 0.0 [m^-1] END REFRACTIVE INDEX: Option = Value Refractive Index = 1.0 [m m^-1] END END END MATERIAL: Soot Material Group = Soot Option = Pure Substance Thermodynamic State = Solid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 2000 [kg m^-3] Molar Mass = 12 [kg kmol^-1] Option = Value END REFERENCE STATE: Option = Automatic END ABSORPTION COEFFICIENT: Absorption Coefficient = 0 [m^-1] Option = Value END END END END FLOW: Flow Analysis 1 SOLUTION UNITS: Angle Units = [rad] Length Units = [m] Mass Units = [kg] Solid Angle Units = [sr] Temperature Units = [K] Time Units = [s] END ANALYSIS TYPE: Option = Steady State EXTERNAL SOLVER COUPLING: Option = None END END DOMAIN: Default Domain Coord Frame = Coord 0 Domain Type = Fluid Location = B40 BOUNDARY: Boundary 1 Boundary Type = INLET Location = F54.40 BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Normal Speed = 10 [m s^-1] Option = Normal Speed END TURBULENCE: Option = Medium Intensity and Eddy Viscosity Ratio END END FLUID: soot BOUNDARY CONDITIONS: MASS AND MOMENTUM: Normal Speed = 10 [m s^-1] Option = Normal Speed END PARTICLE MASS FLOW RATE: Mass Flow Rate = 0.8 [kg s^-1] END PARTICLE POSITION: Option = Uniform Injection NUMBER OF POSITIONS: Number = 2000 Option = Direct Specification END END END END END BOUNDARY: out Boundary Type = OUTLET Location = F48.40 BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Option = Average Static Pressure Pressure Profile Blend = 0.05 Relative Pressure = 0 [Pa] END PRESSURE AVERAGING: Option = Average Over Whole Outlet END END END BOUNDARY: wall Boundary Type = WALL Location = \ F41.40,F42.40,F43.40,F44.40,F45.40,F46.40,F47.40,F 49.40,F50.40,F51.40\ ,F52.40,F53.40 BOUNDARY CONDITIONS: MASS AND MOMENTUM: Option = No Slip Wall END WALL ROUGHNESS: Option = Rough Wall Sand Grain Roughness Height = 0.2 [mm] END END FLUID: soot BOUNDARY CONDITIONS: PARTICLE WALL INTERACTION: Option = Equation Dependent END VELOCITY: Option = Restitution Coefficient Parallel Coefficient of Restitution = 1.0 Perpendicular Coefficient of Restitution = 0.3131 END END END END DOMAIN MODELS: BUOYANCY MODEL: Option = Non Buoyant END DOMAIN MOTION: Option = Stationary END MESH DEFORMATION: Option = None END REFERENCE PRESSURE: Reference Pressure = 1 [atm] END END FLUID DEFINITION: Air Material = Air Ideal Gas Option = Material Library MORPHOLOGY: Option = Continuous Fluid END END FLUID DEFINITION: soot Material = Soot Option = Material Library MORPHOLOGY: Option = Dispersed Particle Transport Solid PARTICLE DIAMETER CHANGE: Option = Mass Equivalent END PARTICLE DIAMETER DISTRIBUTION: Diameter = 1e-10 [m] Option = Specified Diameter END PARTICLE SHAPE FACTORS: Cross Sectional Area Factor = 1.0 END END END FLUID MODELS: COMBUSTION MODEL: Option = None END FLUID: soot EROSION MODEL: Option = None END PARTICLE ROUGH WALL MODEL: Option = None END END HEAT TRANSFER MODEL: Fluid Temperature = 300 [C] Option = Isothermal END THERMAL RADIATION MODEL: Option = None END TURBULENCE MODEL: Option = k epsilon END TURBULENT WALL FUNCTIONS: Option = Scalable END END FLUID PAIR: Air | soot Particle Coupling = One-way Coupling MOMENTUM TRANSFER: DRAG FORCE: Option = Schiller Naumann END PRESSURE GRADIENT FORCE: Option = None END TURBULENT DISPERSION FORCE: Option = None END VIRTUAL MASS FORCE: Option = None END END END END INITIALISATION: Option = Automatic INITIAL CONDITIONS: Velocity Type = Cartesian CARTESIAN VELOCITY COMPONENTS: Option = Automatic END STATIC PRESSURE: Option = Automatic END TURBULENCE INITIAL CONDITIONS: Option = Medium Intensity and Eddy Viscosity Ratio END END END OUTPUT CONTROL: RESULTS: File Compression Level = Default Option = Standard END END SOLVER CONTROL: Turbulence Numerics = First Order ADVECTION SCHEME: Option = High Resolution END CONVERGENCE CONTROL: Length Scale Option = Conservative Maximum Number of Iterations = 100 Minimum Number of Iterations = 1 Timescale Control = Auto Timescale Timescale Factor = 1.0 END CONVERGENCE CRITERIA: Residual Target = 1.E-4 Residual Type = RMS END DYNAMIC MODEL CONTROL: Global Dynamic Model Control = On END PARTICLE CONTROL: PARTICLE INTEGRATION: Option = Forward Euler END END END END COMMAND FILE: Version = 14.0 Results Version = 14.0 END SIMULATION CONTROL: EXECUTION CONTROL: EXECUTABLE SELECTION: Double Precision = Off END INTERPOLATOR STEP CONTROL: Runtime Priority = Standard MEMORY CONTROL: Memory Allocation Factor = 1.0 END END PARALLEL HOST LIBRARY: HOST DEFINITION: mechfran Host Architecture String = winnt-amd64 Installation Root = C:\Program Files\ANSYS Inc\v%v\CFX END END PARTITIONER STEP CONTROL: Multidomain Option = Independent Partitioning Runtime Priority = Standard EXECUTABLE SELECTION: Use Large Problem Partitioner = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARTITIONING TYPE: MeTiS Type = k-way Option = MeTiS Partition Size Rule = Automatic END END RUN DEFINITION: Run Mode = Full Solver Input File = Fluid Flow CFX_001.res END SOLVER STEP CONTROL: Runtime Priority = Standard MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARALLEL ENVIRONMENT: Number of Processes = 1 Start Method = Serial END END END END

 November 27, 2012, 23:23 #2 Senior Member   Govindaraju Join Date: Apr 2010 Posts: 206 Rep Power: 8 any suggesstion please Thank you kmgraju

 November 28, 2012, 07:14 #3 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,929 Rep Power: 85 Have you done the CFX tutorials? There are several examples of particle flows. Please post an image of your model, and what you expect the flow to do.

 November 28, 2012, 07:20 #4 Senior Member   Govindaraju Join Date: Apr 2010 Posts: 206 Rep Power: 8 Thank you Mr Glenn I have gone through flow through butterfly valve . Any other tutorial?? Thank you Govind

 November 28, 2012, 17:15 #5 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,929 Rep Power: 85 All the tutorials are useful, and there are a few others which use particle tracking. Have a look through the contents page of the tutorial manual.

November 30, 2012, 00:52
#6
Senior Member

Govindaraju
Join Date: Apr 2010
Posts: 206
Rep Power: 8
Quote:
 Originally Posted by ghorrocks All the tutorials are useful, and there are a few others which use particle tracking. Have a look through the contents page of the tutorial manual.
Dear friend

please find the geometry which I have created to trap the soot particle
Kindly let me know my mistakes in the setting . I could not locate my mistakes

Thank you for your kind help

Regards

Govind
Attached Images
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 November 30, 2012, 01:03 #7 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,929 Rep Power: 85 I do not have time to get your simulation working for you. You are going to have to do the debugging yourself. But I can see you set wall restitution factors. This means the particles will bounce when they hit the wall, not stick.

 December 10, 2012, 01:26 #8 Senior Member   Govindaraju Join Date: Apr 2010 Posts: 206 Rep Power: 8 Dear friend I would like to calculate the deposition of the soot particles. how do I know that soot has been deposited on the wall or not? Thank you Govind

 December 10, 2012, 04:57 #9 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,929 Rep Power: 85 First of all you have to decide what the physics of deposition is. Do the soot particles just stick to the wall if they hit it? At the length scale of soot I doubt this is accurate, all sorts of forces become important for tiny particles at small length scales and you will need to know how to deal with this for an accurate model. But you do not care about accuracy and just want imaginary soot sticking to an imaginary wall (with the expected imaginery accuracy) then just put a zero coefficient of restitution onto the walls you want soot to stick to.

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