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mass flow in is not equal to mass flow out

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Old   September 17, 2009, 10:08
Exclamation mass flow in is not equal to mass flow out
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Can anyone pls help me with this problem.
My problem is mass flow in is not equal to mass flow out. thank you

This is my model http://img183.imageshack.us/i/48217966.png/

This is my CCL infor:
Code:
# State file created: 2009/09/14 15:18:14
# CFX-12.0.1 build 2009.04.14-23.02
LIBRARY:
CEL:
EXPRESSIONS:
DenH = (DenWater - DenRef)
DenRef = 1.185 [kg m^-3]
DenWater = 997 [kg m^-3]
DownH = 0.68 [m]
DownPres = DenH*g*DownVFWater*(DownH-y)
DownVFAir = step((y-DownH)/1[m])
DownVFWater = 1-DownVFAir
UpH = 0.78 [m]
UpPres = DenH*g*UpVFWater*(UpH-y)
UpVFAir = step((y-UpH)/1[m])
UpVFWater = 1-UpVFAir
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 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: 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 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: 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 = Steady State
EXTERNAL SOLVER COUPLING:
Option = None
END
END
DOMAIN: Domain 1
Coord Frame = Coord 0
Domain Type = Fluid
Location = Assembly
BOUNDARY: Boundary 1
Boundary Type = INLET
Location = F98.104
BOUNDARY CONDITIONS:
FLOW DIRECTION:
Option = Normal to Boundary Condition
END
FLOW REGIME:
Option = Subsonic
END
MASS AND MOMENTUM:
Mass Flow Rate = 3 [kg s^-1]
Option = Bulk Mass Flow Rate
END
END
FLUID: air
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = UpVFAir
END
END
END
FLUID: water
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = UpVFWater
END
END
END
END
BOUNDARY: Boundary 2
Boundary Type = OPENING
Location = F100.104
BOUNDARY CONDITIONS:
FLOW DIRECTION:
Option = Normal to Boundary Condition
END
FLOW REGIME:
Option = Subsonic
END
MASS AND MOMENTUM:
Option = Opening Pressure and Direction
Relative Pressure = DownPres
END
END
FLUID: air
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = 0
END
END
END
FLUID: water
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = 1
END
END
END
END
BOUNDARY: Boundary 3
Boundary Type = OPENING
Location = F105.104,F108.104
BOUNDARY CONDITIONS:
FLOW DIRECTION:
Option = Normal to Boundary Condition
END
FLOW REGIME:
Option = Subsonic
END
MASS AND MOMENTUM:
Option = Opening Pressure and Direction
Relative Pressure = DownPres
END
END
FLUID: air
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = 1
END
END
END
FLUID: water
BOUNDARY CONDITIONS:
VOLUME FRACTION:
Option = Value
Volume Fraction = 0
END
END
END
END
BOUNDARY: Domain 1 Default
Boundary Type = WALL
Location = \
F101.104,F102.104,F103.104,F106.104,F107.104,F109.104,F110.104,F111.10\
4,F112.104,F113.104,F114.104,F83.104,F84.104,F85.104,F86.104,F87.104,F\
88.104,F89.104,F90.104,F91.104,F92.104,F93.104,F94.104,F95.104,F96.104\
,F97.104,F99.104
BOUNDARY CONDITIONS:
MASS AND MOMENTUM:
Option = No Slip Wall
END
END
END
DOMAIN MODELS:
BUOYANCY MODEL:
Buoyancy Reference Density = 1.185 [kg m^-3]
Gravity X Component = 0 [m s^-2]
Gravity Y Component = -9.8 [m s^-2]
Gravity Z Component = 0 [m s^-2]
Option = Buoyant
BUOYANCY REFERENCE LOCATION:
Option = Automatic
END
END
DOMAIN MOTION:
Option = Stationary
END
MESH DEFORMATION:
Option = None
END
REFERENCE PRESSURE:
Reference Pressure = 1 [atm]
END
END
FLUID DEFINITION: air
Material = Air at 25 C
Option = Material Library
MORPHOLOGY:
Option = Continuous Fluid
END
END
FLUID DEFINITION: water
Material = Water
Option = Material Library
MORPHOLOGY:
Option = Continuous Fluid
END
END
FLUID MODELS:
COMBUSTION MODEL:
Option = None
END
FLUID: air
FLUID BUOYANCY MODEL:
Option = Density Difference
END
END
FLUID: water
FLUID BUOYANCY MODEL:
Option = Density Difference
END
END
HEAT TRANSFER MODEL:
Fluid Temperature = 25 [C]
Homogeneous Model = Off
Option = Isothermal
END
THERMAL RADIATION MODEL:
Option = None
END
TURBULENCE MODEL:
Option = Laminar
END
END
FLUID PAIR: air | water
INTERPHASE TRANSFER MODEL:
Option = None
END
MASS TRANSFER:
Option = None
END
SURFACE TENSION MODEL:
Option = None
END
END
INITIALISATION:
Option = Automatic
FLUID: air
INITIAL CONDITIONS:
VOLUME FRACTION:
Option = Automatic with Value
Volume Fraction = UpVFAir
END
END
END
FLUID: water
INITIAL CONDITIONS:
VOLUME FRACTION:
Option = Automatic with Value
Volume Fraction = UpVFWater
END
END
END
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 [Pa]
END
END
END
MULTIPHASE MODELS:
Homogeneous Model = On
FREE SURFACE MODEL:
Option = Standard
END
END
END
MESH ADAPTION:
Activate Adaption = On
Domain Name = Domain 1
Save Intermediate Files = On
Subdomain List = Assembly
ADAPTION ADVANCED OPTIONS:
Node Allocation Parameter = -2
Number of Adaption Levels = 2
END
ADAPTION CONVERGENCE CRITERIA:
Adaption Target Residual = 0.001
Maximum Iterations per Step = 100
Option = RMS Norm for Residuals
END
ADAPTION CRITERIA:
Maximum Number of Adaption Steps = 2
Node Factor = 4
Option = Multiple of Initial Mesh
Variables List = air.Conservative Volume Fraction
END
ADAPTION METHOD:
Minimum Edge Length = 0.0
Option = Solution Variation
END
END
OUTPUT CONTROL:
RESULTS:
File Compression Level = Default
Option = Standard
END
END
SOLVER CONTROL:
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 = Yes
END
END
END
COMMAND FILE:
Version = 12.0.1
END

Last edited by wyldckat; September 3, 2015 at 17:09. Reason: Added [CODE][/CODE] markers and disabled embedded image
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Old   September 18, 2009, 08:22
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ghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the rough
Your simulation is not converged. Have you looked here: http://www.cfd-online.com/Wiki/Ansys...gence_criteria

Also output some backup files as the simulation progresses. You can load them in CFD-Post and it may help you diagnose what is wrong.
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Old   September 18, 2009, 09:07
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thank you.
which tutorial should i look through, if i want to add lio into the flow?
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Old   March 19, 2018, 01:06
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D.R
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Hey, I am in a same boat right now. Just in case if you know, please let me know - where can I find the function TO FIND the mass flow rate in ANSYS Fluent?

Secondly, I want to know whether the fluid flow is distributed equally or not to my two outlets from one inlet pipe. If you know how to find the fluid distribution (that function in ANSYS fluent), please let me know. Thanks!
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Old   March 19, 2018, 04:14
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I would suggest to use CFX. This gives many more options in monitoring, like changing range and adding monitors on the fly.. Also, it automatically saves your monitors in your res files, whereas Fluent doesn't. Very frustrating in my opinion.

But if you want to use Fluent and get the values at the end of your run, just go to Results > Reports > Fluxes > MassFlowRate at your boundaries.
If you want to monitor them during your run, go to Report definitions and create the samething there and make sure you create a monitor.
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Old   March 19, 2018, 05:16
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D.R
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Quote:
Originally Posted by Gert-Jan View Post
I would suggest to use CFX. This gives many more options in monitoring, like changing range and adding monitors on the fly.. Also, it automatically saves your monitors in your res files, whereas Fluent doesn't. Very frustrating in my opinion.

But if you want to use Fluent and get the values at the end of your run, just go to Results > Reports > Fluxes > MassFlowRate at your boundaries.
If you want to monitor them during your run, go to Report definitions and create the samething there and make sure you create a monitor.
Hi Gert, Thank you for your swift reply. I will certainly try in CFX. To be honest, I am bit new with CFD and hence, I am learning. I followed your instruction and indeed it was helpful. I can at least find the mass flow rate now - i.e. under Solution in ANSYS Fluent. However, the result for Mass Flow Rate that I got for my model is; At Inlet: 1.006 Kg/s and at outlet 1: -0.4291 kg/s and at outlet 2; -0.5767 kg/s. It seems, in my case I can see Mas in = Mass out

Is this result sufficient enough to tell me that the fluid flow is not distributed equally at outlet 1 and outlet 2 from single inlet pipe? Can you please explain? Your response will be highly appreciated. Thank you!
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Old   March 19, 2018, 05:22
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Fluent says so, so you might believe it. But It depends on your case, the accuracy and validity of your setup, the convergence, etc.
Impossible for me to say based on the input you provided.
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Old   March 19, 2018, 05:30
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D.R
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Quote:
Originally Posted by D.R View Post
Hi Gert, Thank you for your swift reply. I will certainly try in CFX. To be honest, I am bit new with CFD and hence, I am learning. I followed your instruction and indeed it was helpful. I can at least find the mass flow rate now - i.e. under Solution in ANSYS Fluent. However, the result for Mass Flow Rate that I got for my model is; At Inlet: 1.006 Kg/s and at outlet 1: -0.4291 kg/s and at outlet 2; -0.5767 kg/s. It seems, in my case I can see Mas in = Mass out

Is this result sufficient enough to tell me that the fluid flow is not distributed equally at outlet 1 and outlet 2 from single inlet pipe? Can you please explain? Your response will be highly appreciated. Thank you!
In addition to my above statement - I mean I wanted to say;

Mass Flow Rate (M.F.R.) Results can tell me that whether the fluid flow is distributed equally or not at Outlets 1 and 2?

Because my M.F.R. result for outlet 1 = -0.429 kg/s
outlet 2 = -0.5767 kg/s

Does this value tells me that my fluid is not distributed equally?
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Old   March 19, 2018, 05:37
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D.R
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Quote:
Originally Posted by Gert-Jan View Post
Fluent says so, so you might believe it. But It depends on your case, the accuracy and validity of your setup, the convergence, etc.
Impossible for me to say based on the input you provided.
Thank you Gert for your patience as I might sound silly. But like I said, I am a new learner.

Well, I do understand that it truly depends upon my Boundary Conditions, my convergence, and so on. I totally agree with you on that front.

But let's assume, that my BC, convergence, and other criteria are perfect. So, then Mass Flow Rate will be sufficient enough to tell me whether my flow is distributed equally or not? As I mentioned above, in my case, I found two different values at outlets 1 and 2. So, should I consider that the flow is not distributed equally?

Secondly, Mass flow rate is the only way in ANSYS Fluent which can tell me the fluid flow is distributed properly or not?

Thank you so much Gert!
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Old   March 19, 2018, 05:39
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Yes of course. It tells you that it is not equally distributed.
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Old   March 19, 2018, 06:00
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Thank you Gert, I have been breaking my head for whole day to find this. Watched many tutorials but couldn't find what I have been looking for. You save my time.

One last question, with respect to the same enquiry!

In order to find fluid distribution (i.e. equal or non-equal) in two or more pipes from one single pipe, I have set up my boundary condition which is as below:

At Inlet, Velocity = 26 m/s
At outlet 1, Outflow = 1
At outlet 2, outflow = 1 (flow rate weighing)

So, I believe if I will keep outflow same, i.e outflow =1 for both outlets, then ANSYS Fluent will only figure out for the values at Outlet 1 and 2. (Likewise we keep pressure value = 0 at Outlet and then ANSYS Fluent tells us the pressure distribution throughout the body including at outlet - same concept I was thinking here to keep outflow = 1 for both. So, ANSYS will tell me whether my flow is going to be equal at outlets or not?)

Please correct me if I am wrong! Just in case if you know, please let me know on this.


Thank you so much!
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Old   March 19, 2018, 06:12
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Remember, this is CFX Forum, not the Fluent Forum.

I know only a bit of Fluent. I don't know what "Outflow = 1" means.
But it looks like a fraction to me. So I would use 0.5 for both outlets. Not?

I always use pressure outlets or massflow outlets.
If you use 2 massflow outlets you can also distribute the flow easily if you set the 1.006kg/s at your inlet.
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Old   March 19, 2018, 06:21
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Quote:
Originally Posted by Gert-Jan View Post
Remember, this is CFX Forum, not the Fluent Forum.

I know only a bit of Fluent. I don't know what "Outflow = 1" means.
But it looks like a fraction to me. So I would use 0.5 for both outlets. Not?

I always use pressure outlets or massflow outlets.
If you use 2 massflow outlets you can also distribute the flow easily if you set the 1.006kg/s at your inlet.
Oops I did not know that I am on a different platform and asking something else. Although you answered me for my enquiry, so, I truly appreciate you for your support mate. I am working on my Uni project and I am close to my deadline. Your information indeed saved my time. Thank you Gert!
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