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Domain Reference Pressure and mass flow inlet boundary |
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August 1, 2018, 07:34 |
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#21 |
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Glenn Horrocks
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You have a domain with a single boundary (other than walls and symmetry planes). This boundary is a total pressure boundary. As the simulation is steady state, then the steady state answer to this is zero flow velocity everywhere and the pressure equal to the pressure applied at the inlet. You don't need CFD to solve this, I just solved it in my head
Secondly, your applied pressure is a function of aitern. This means the applied pressure changes every iteration, and this means it can never converge because the goal posts keep moving. So you appear to have two fundamental flaws in your simulation setup.
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August 1, 2018, 08:40 |
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#22 | |
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Hi Glenn, thanks for your reply. I can understand when the velocity is zero then pressure everywhere will be equal to pressure applied. But can you please explain why the velocity will be zero? Will it not change w.r.t the position in the domain? Once this is established then should i consider switching the boundary conditions to study the pressure drops? Regarding second point. I am ramping the pressure so that the maximum value i.e. 110 bar is acheived in 4000 iterations and then the simulation will be run at this maximum pressure for further 4000 ietrations. Is there any other way of ramping the pressure in steady state simulations? |
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August 1, 2018, 09:00 |
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#23 |
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You have it set up as a steady state simulation. This means nothing changes with time. If you want to study the effect of pressure changes then you should be doing a transient simulation. Then the simulation you propose makes sense - however you will have to change the pressure from being a function of aitern to a function of time (t).
Regarding your second question - Do not ramp the pressure up for a steady state simulation. If you do the simulation is not steady state, is it? This explains why it does not converge. You must keep the pressure constant.
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August 1, 2018, 09:29 |
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#24 | |
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yes I understand in a steady state simulation nothing will change w.r.t. time because there are no time terms in N-S equations. But will the flow variables donot change w.r.t. the position in a steady state simulation? |
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August 1, 2018, 09:34 |
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#25 |
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In a steady state simulation things can have a gradient with respect to position, but all gradients with respect to time are zero.
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August 1, 2018, 09:49 |
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#26 | |
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Thanks for continued response. |
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August 1, 2018, 19:50 |
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#27 |
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If you have a vessel with only one inlet, and you define the pressure at that boundary then the steady state result of that vessel will be the vessel just gets to the inlet pressure and stays there. And as there are no pressure gradients then there are no velocities. All the flow required to get the vessel to pressure is transient, they all decay down to zero for the steady state result.
Another way of thinking about the steady state result is the result after infinite time - with nothing driving the flow then the flow will dissipate to nothing and the result at infinite time is zero flow. Simple as that.
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August 2, 2018, 04:48 |
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#28 |
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Hi Glenn,
thanks for the explanation. I have understood it now I started the transient simulation with time step of 0.001 and total simulation time of 5s. I used ramping function to slowly increase the inlet pressure but this time ramping function is a function of time. Simulation ran fine for about 3009 time steps (Note: Maximum pressure of 110 bar is acheived at Inlet at 3000 time step). After that it diverged as you can see in the attached figure also. Also what i can i see from the simulation is that the wall has been placed at portion of an Inlet and this percentage portion keeps on osciallting during the simulation. Is ramping function ok now? Any further suggestions please? CEL: LIBRARY: CEL: EXPRESSIONS: Flow000 = 0 [bar] Flow999 = 109 [bar] flowapplied = Flow000 + \ Flow999*t/time*step((time-t)/1[s])+Flow999*step((t-time)/1[s]) time = 3 [s] END END MATERIAL GROUP: Air Data Group Description = Ideal gas and constant property air. Constant \ properties are for dry air at STP (0 C, 1 atm) and 25 C, 1 atm. END MATERIAL GROUP: CHT Solids Group Description = Pure solid substances that can be used for conjugate \ heat transfer. END MATERIAL GROUP: Calorically Perfect Ideal Gases Group Description = Ideal gases with constant specific heat capacity. \ Specific heat is evaluated at STP. END MATERIAL GROUP: Constant Property Gases Group Description = Gaseous substances with constant properties. \ Properties are calculated at STP (0C and 1 atm). Can be combined with \ NASA SP-273 materials for combustion modelling. END MATERIAL GROUP: Constant Property Liquids Group Description = Liquid substances with constant properties. END MATERIAL GROUP: Dry Peng Robinson Group Description = Materials with properties specified using the built \ in Peng Robinson equation of state. Suitable for dry real gas modelling. END MATERIAL GROUP: Dry Redlich Kwong Group Description = Materials with properties specified using the built \ in Redlich Kwong equation of state. Suitable for dry real gas modelling. END MATERIAL GROUP: Dry Soave Redlich Kwong Group Description = Materials with properties specified using the built \ in Soave Redlich Kwong equation of state. Suitable for dry real gas \ modelling. END MATERIAL GROUP: Dry Steam Group Description = Materials with properties specified using the IAPWS \ equation of state. Suitable for dry steam modelling. END MATERIAL GROUP: Gas Phase Combustion Group Description = Ideal gas materials which can be use for gas phase \ combustion. Ideal gas specific heat coefficients are specified using \ the NASA SP-273 format. END MATERIAL GROUP: IAPWS IF97 Group Description = Liquid, vapour and binary mixture materials which use \ the IAPWS IF-97 equation of state. Materials are suitable for \ compressible liquids, phase change calculations and dry steam flows. END MATERIAL GROUP: Interphase Mass Transfer Group Description = Materials with reference properties suitable for \ performing either Eulerian or Lagrangian multiphase mass transfer \ problems. Examples include cavitation, evaporation or condensation. END MATERIAL GROUP: Liquid Phase Combustion Group Description = Liquid and homogenous binary mixture materials which \ can be included with Gas Phase Combustion materials if combustion \ modelling also requires phase change (eg: evaporation) for certain \ components. END MATERIAL GROUP: Particle Solids Group Description = Pure solid substances that can be used for particle \ tracking END MATERIAL GROUP: Peng Robinson Dry Hydrocarbons Group Description = Common hydrocarbons which use the Peng Robinson \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Peng Robinson Dry Refrigerants Group Description = Common refrigerants which use the Peng Robinson \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Peng Robinson Dry Steam Group Description = Water materials which use the Peng Robinson equation \ of state. Suitable for dry steam modelling. END MATERIAL GROUP: Peng Robinson Wet Hydrocarbons Group Description = Common hydrocarbons which use the Peng Robinson \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Peng Robinson Wet Refrigerants Group Description = Common refrigerants which use the Peng Robinson \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Peng Robinson Wet Steam Group Description = Water materials which use the Peng Robinson equation \ of state. Suitable for condensing steam modelling. END MATERIAL GROUP: Real Gas Combustion Group Description = Real gas materials which can be use for gas phase \ combustion. Ideal gas specific heat coefficients are specified using \ the NASA SP-273 format. END MATERIAL GROUP: Redlich Kwong Dry Hydrocarbons Group Description = Common hydrocarbons which use the Redlich Kwong \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Redlich Kwong Dry Refrigerants Group Description = Common refrigerants which use the Redlich Kwong \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Redlich Kwong Dry Steam Group Description = Water materials which use the Redlich Kwong equation \ of state. Suitable for dry steam modelling. END MATERIAL GROUP: Redlich Kwong Wet Hydrocarbons Group Description = Common hydrocarbons which use the Redlich Kwong \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Redlich Kwong Wet Refrigerants Group Description = Common refrigerants which use the Redlich Kwong \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Redlich Kwong Wet Steam Group Description = Water materials which use the Redlich Kwong equation \ of state. Suitable for condensing steam modelling. END MATERIAL GROUP: Soave Redlich Kwong Dry Hydrocarbons Group Description = Common hydrocarbons which use the Soave Redlich Kwong \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Soave Redlich Kwong Dry Refrigerants Group Description = Common refrigerants which use the Soave Redlich Kwong \ equation of state. Suitable for dry real gas models. END MATERIAL GROUP: Soave Redlich Kwong Dry Steam Group Description = Water materials which use the Soave Redlich Kwong \ equation of state. Suitable for dry steam modelling. END MATERIAL GROUP: Soave Redlich Kwong Wet Hydrocarbons Group Description = Common hydrocarbons which use the Soave Redlich Kwong \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Soave Redlich Kwong Wet Refrigerants Group Description = Common refrigerants which use the Soave Redlich Kwong \ equation of state. Suitable for condensing real gas models. END MATERIAL GROUP: Soave Redlich Kwong Wet Steam Group Description = Water materials which use the Soave Redlich Kwong \ equation of state. Suitable for condensing steam modelling. END MATERIAL GROUP: Soot Group Description = Solid substances that can be used when performing \ soot modelling END MATERIAL GROUP: User Group Description = Materials that are defined by the user END MATERIAL GROUP: Water Data Group Description = Liquid and vapour water materials with constant \ properties. Can be combined with NASA SP-273 materials for combustion \ modelling. END MATERIAL GROUP: Wet Peng Robinson Group Description = Materials with properties specified using the built \ in Peng Robinson equation of state. Suitable for wet real gas modelling. END MATERIAL GROUP: Wet Redlich Kwong Group Description = Materials with properties specified using the built \ in Redlich Kwong equation of state. Suitable for wet real gas modelling. END MATERIAL GROUP: Wet Soave Redlich Kwong Group Description = Materials with properties specified using the built \ in Soave Redlich Kwong equation of state. Suitable for wet real gas \ modelling. END MATERIAL GROUP: Wet Steam Group Description = Materials with properties specified using the IAPWS \ equation of state. Suitable for wet steam modelling. END 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: Air at 25 C Material Description = Air at 25 C and 1 atm (dry) Material Group = Air Data, Constant Property Gases Option = Pure Substance Thermodynamic State = Gas PROPERTIES: Option = General Material EQUATION OF STATE: Density = 1.185 [kg m^-3] Molar Mass = 28.96 [kg kmol^-1] Option = Value 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-02 [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 THERMAL EXPANSIVITY: Option = Value Thermal Expansivity = 0.003356 [K^-1] END END END MATERIAL: Aluminium Material Group = CHT Solids, Particle Solids Option = Pure Substance Thermodynamic State = Solid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 2702 [kg m^-3] Molar Mass = 26.98 [kg kmol^-1] Option = Value END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 9.03E+02 [J kg^-1 K^-1] END REFERENCE STATE: Option = Specified Point Reference Specific Enthalpy = 0 [J/kg] Reference Specific Entropy = 0 [J/kg/K] Reference Temperature = 25 [C] END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 237 [W m^-1 K^-1] END END END MATERIAL: Copper Material Group = CHT Solids, Particle Solids Option = Pure Substance Thermodynamic State = Solid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 8933 [kg m^-3] Molar Mass = 63.55 [kg kmol^-1] Option = Value END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 3.85E+02 [J kg^-1 K^-1] END REFERENCE STATE: Option = Specified Point Reference Specific Enthalpy = 0 [J/kg] Reference Specific Entropy = 0 [J/kg/K] Reference Temperature = 25 [C] END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 401.0 [W m^-1 K^-1] END END END MATERIAL: Oil Material Group = User Option = Pure Substance Thermodynamic State = Liquid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 881 [kg m^-3] Molar Mass = 1.0 [kg kmol^-1] Option = Value END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 1861 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END DYNAMIC VISCOSITY: Dynamic Viscosity = 0.029073 [kg m^-1 s^-1] Option = Value END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 0.14 [W m^-1 K^-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 MATERIAL: Steel Material Group = CHT Solids, Particle Solids Option = Pure Substance Thermodynamic State = Solid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 7854 [kg m^-3] Molar Mass = 55.85 [kg kmol^-1] Option = Value END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 4.34E+02 [J kg^-1 K^-1] END REFERENCE STATE: Option = Specified Point Reference Specific Enthalpy = 0 [J/kg] Reference Specific Entropy = 0 [J/kg/K] Reference Temperature = 25 [C] END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 60.5 [W m^-1 K^-1] END END END MATERIAL: Water Material Description = Water (liquid) Material Group = Water Data, Constant Property Liquids Option = Pure Substance Thermodynamic State = Liquid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 997.0 [kg m^-3] Molar Mass = 18.02 [kg kmol^-1] Option = Value END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 4181.7 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END REFERENCE STATE: Option = Specified Point Reference Pressure = 1 [atm] Reference Specific Enthalpy = 0.0 [J/kg] Reference Specific Entropy = 0.0 [J/kg/K] Reference Temperature = 25 [C] END DYNAMIC VISCOSITY: Dynamic Viscosity = 8.899E-4 [kg m^-1 s^-1] Option = Value END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 0.6069 [W m^-1 K^-1] END ABSORPTION COEFFICIENT: Absorption Coefficient = 1.0 [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 THERMAL EXPANSIVITY: Option = Value Thermal Expansivity = 2.57E-04 [K^-1] END END END MATERIAL: Water Ideal Gas Material Description = Water Vapour Ideal Gas (100 C and 1 atm) Material Group = Calorically Perfect Ideal Gases, Water Data Option = Pure Substance Thermodynamic State = Gas PROPERTIES: Option = General Material EQUATION OF STATE: Molar Mass = 18.02 [kg kmol^-1] Option = Ideal Gas END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 2080.1 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END REFERENCE STATE: Option = Specified Point Reference Pressure = 1.014 [bar] Reference Specific Enthalpy = 0. [J/kg] Reference Specific Entropy = 0. [J/kg/K] Reference Temperature = 100 [C] END DYNAMIC VISCOSITY: Dynamic Viscosity = 9.4E-06 [kg m^-1 s^-1] Option = Value END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 193E-04 [W m^-1 K^-1] END ABSORPTION COEFFICIENT: Absorption Coefficient = 1.0 [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 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 = Transient EXTERNAL SOLVER COUPLING: Option = None END INITIAL TIME: Option = Automatic with Value Time = 0 [s] END TIME DURATION: Option = Total Time Total Time = 5 [s] END TIME STEPS: Option = Timesteps Timesteps = 0.001 [s] END END DOMAIN: Default Domain Coord Frame = Coord 0 Domain Type = Fluid Location = FLUID BOUNDARY: Inlet Boundary Type = INLET Location = INLET BOUNDARY CONDITIONS: FLOW DIRECTION: Option = Normal to Boundary Condition END FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Option = Total Pressure Relative Pressure = flowapplied END TURBULENCE: Option = Medium Intensity and Eddy Viscosity Ratio END END END BOUNDARY: Symmetry Boundary Type = SYMMETRY Location = Primitive 2D A,Primitive 2D B END BOUNDARY: Walls Boundary Type = WALL Location = GEOM_1 GEOM_OBERFL_CHE_1 BOUNDARY CONDITIONS: MASS AND MOMENTUM: Option = No Slip Wall END WALL ROUGHNESS: Option = Smooth Wall 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 [bar] END END FLUID DEFINITION: Fluid 1 Material = Oil Option = Material Library MORPHOLOGY: Option = Continuous Fluid END END FLUID MODELS: COMBUSTION MODEL: Option = None END HEAT TRANSFER MODEL: Fluid Temperature = 25 [C] Option = Isothermal END THERMAL RADIATION MODEL: Option = None END TURBULENCE MODEL: Option = SST END TURBULENT WALL FUNCTIONS: Option = Automatic END END INITIALISATION: Option = Automatic INITIAL CONDITIONS: Velocity Type = Cartesian CARTESIAN VELOCITY COMPONENTS: Option = Automatic with Value U = 0 [m s^-1] V = 0 [m s^-1] W = 0 [m s^-1] END STATIC PRESSURE: Option = Automatic with Value Relative Pressure = 0 [bar] END TURBULENCE INITIAL CONDITIONS: Option = Medium Intensity and Eddy Viscosity Ratio END END END END OUTPUT CONTROL: BACKUP DATA RETENTION: Option = Delete Old Files END BACKUP RESULTS: Backup Results 1 File Compression Level = Default Option = Standard OUTPUT FREQUENCY: Option = Timestep Interval Timestep Interval = 1000 END END RESULTS: File Compression Level = Default Option = Standard END TRANSIENT RESULTS: Transient Results 1 File Compression Level = Default Option = Standard OUTPUT FREQUENCY: Option = Timestep Interval Timestep Interval = 1000 END END END SOLVER CONTROL: Turbulence Numerics = High Resolution ADVECTION SCHEME: Option = High Resolution END CONVERGENCE CONTROL: Maximum Number of Coefficient Loops = 5 Minimum Number of Coefficient Loops = 1 Timescale Control = Coefficient Loops END CONVERGENCE CRITERIA: Residual Target = 1e-4 Residual Type = RMS END TRANSIENT SCHEME: Option = Second Order Backward Euler TIMESTEP INITIALISATION: Option = Automatic END END END END COMMAND FILE: Version = 19.1 END |
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August 2, 2018, 08:51 |
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#29 |
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Should i consider changing the boundary condition to velocity Inlet or Mass flow Inlet? I am already running the simulation with reduced time step of 1e-4.
I suppose there is no problem in initial conditions or mesh because the first unsteady simulation ran for 309 time steps. Isn't it? Thanks in advance. Regards |
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August 2, 2018, 08:53 |
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#30 |
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Glenn Horrocks
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This question is very similar to this FAQ: https://cfd-online.com/Wiki/Ansys_FA...do_about_it.3F
In short: double precision, smaller time steps, better mesh quality. Also: Are you sure compressibility is not significant here? Oil with 100 bar pressure behind it might have enough density change that this is important.
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August 2, 2018, 08:56 |
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#31 | |
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Quote:
i would consider the compressibility of oil but could it cause the divergence of simulation? What about changing the boundary conditions? Thanks!! Regards |
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August 2, 2018, 22:24 |
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#32 |
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Before you think about numerical issues you need to think about whether it is significant or not. If the system has significant compressibility then you need to model it regardless of whether it is easy or hard. But you are correct in that if you model compressibility then a whole range of new issues are created and convergence may be more difficult.
The boundary condition needs to match the physical condition. Whether that is the case depends on what you are doing and only you can assess that.
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August 3, 2018, 04:44 |
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#33 | |
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Quote:
i also know the mass flow rate of oil at the Inlet. My point is, can mass flow boundary condition alone (as there is no outlet in the domain) be used to study the pressure drops in the system? Thanks!! Regards |
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August 3, 2018, 07:32 |
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#34 |
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I have just gone back over this thread to remind myself of the history (I am doing many threads at once, I cannot recall the history of all of them).
Are you modelling an object with a single non-wall boundary, where the mesh is stationary and the fluid incompressible? In this case this simulation is badly posed for both the steady state and transient cases. No net fluid can go in or out, so this configuration will generate no flow anywhere.
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August 3, 2018, 08:15 |
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#35 | |
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no the domain do have walls. Infact there is only one Inlet boundary and rest are walls (ofcourse there are symmetry because of 2D simulation) but the mesh is stationary and the fluid is incompressible. You can see the domain again in the attached image. Thanks |
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August 3, 2018, 08:57 |
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#36 | |
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OK, then
Quote:
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August 3, 2018, 09:05 |
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#37 |
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what's the reason for that?
Previously you said the simulation with One Inlet boundary in steady state will not converge as there will be no flow. But why it will not converge even in transient? What can I do if I have to study pressure drop in such a system? Regards |
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August 3, 2018, 20:09 |
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#38 |
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Think about it - the fluid is incompressible and you have only one entry/exit. Nothing can happen.
You say you want to study pressure drop - but pressure drop from where to where? You only have one obvious point to take the pressure at. You need two points to get a pressure drop. What is the other point?
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August 4, 2018, 18:13 |
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#39 | |
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Quote:
By pressure drop I mean pressure losses. You can see the figure in the attachment. I labeled the regions where I want to know the pressure losses. I will explain the system again so that you know what i am trying to model. This is actually 2D sectional view of a Hydraulic system where the flow is coming from a pump (pump flow not modelled). The high pressure oil enters the system and generates the pressure to lift one surface of a system which in turn lifts some other external load. So i want to know that when oil at 110 bar enters this hydraulic system, how much pressure loss is there till it reaches the lifting surface and sealing rings of the system. I hope I made myself clear. Is there any other way to model this system to study pressure losses? Thanks a lot. |
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August 5, 2018, 08:30 |
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#40 |
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Your description of the system appears to suggest that there should be an outlet in your system to a hydraulic actuator, or one of the walls in your modelled system should move as it is the actuator. If this is correct then you have an inlet and somewhere for the fluid to go and your model makes sense.
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