Domain Reference Pressure and mass flow inlet boundary

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 August 1, 2018, 06:34 #21 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 1, 2018, 07:40
#22
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Quote:
 Originally Posted by ghorrocks 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.

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?

 August 1, 2018, 08:00 #23 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 1, 2018, 08:29
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Quote:
 Originally Posted by ghorrocks 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.
Hi Glenn,

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?

 August 1, 2018, 08:34 #25 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 In a steady state simulation things can have a gradient with respect to position, but all gradients with respect to time are zero. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 1, 2018, 08:49
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Quote:
 Originally Posted by ghorrocks In a steady state simulation things can have a gradient with respect to position, but all gradients with respect to time are zero.
Yes Glenn, that's what I am trying to reason that when I am performaing steady state simulation then pressure and velocity should have gradients w.r.t. position in a domain. But you said in a steady state simulation (in my case) velocity every where will be zero. Why? Am i missing some basic thing or concept?

Thanks for continued response.

 August 1, 2018, 18:50 #27 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 2, 2018, 03:48
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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:
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
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
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
Attached Images
 CFX_transient_Model.jpg (140.6 KB, 9 views)

 August 2, 2018, 07:51 #29 Senior Member   Join Date: Mar 2011 Location: Germany Posts: 552 Rep Power: 20 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

 August 2, 2018, 07:53 #30 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 2, 2018, 07:56
#31
Senior Member

Join Date: Mar 2011
Location: Germany
Posts: 552
Rep Power: 20
Quote:
 Originally Posted by ghorrocks 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.
Thanks Glenn,

i would consider the compressibility of oil but could it cause the divergence of simulation?

What about changing the boundary conditions?

Thanks!!

Regards

 August 2, 2018, 21:24 #32 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 3, 2018, 03:44
#33
Senior Member

Join Date: Mar 2011
Location: Germany
Posts: 552
Rep Power: 20
Quote:
 Originally Posted by ghorrocks 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.
Hi Glenn,

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

 August 3, 2018, 06:32 #34 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 3, 2018, 07:15
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Quote:
 Originally Posted by ghorrocks 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.
Hi Glenn,

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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 CFX_Model.jpg (38.0 KB, 9 views)

August 3, 2018, 07:57
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OK, then

Quote:
 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.
And I should add that it will not converge either.
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August 3, 2018, 08:05
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 Originally Posted by ghorrocks OK, then And I should add that it will not converge either.
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

 August 3, 2018, 19:09 #38 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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? __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

August 4, 2018, 17:13
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Quote:
 Originally Posted by ghorrocks 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?
Yes actually you are right. I thought about it and in incompressible flow without Outflow one cannot solve this probelm as Continuity equation would not be satisfied.

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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 CFX_Model.jpg (46.9 KB, 5 views)

 August 5, 2018, 07:30 #40 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,729 Rep Power: 143 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. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.