convergenceof natural convection prob. in cfx
hi..
i m mechanical engg student, and try to simulate a natural convection problem thar is tharmal stratification and conjugate heat transfer in straight pipe in ansys cfx. try to generate temperature gradient across the section, i m fresh in cfx.and dont know much. i have no much time to complete my problem because, plz help me. thanks in advance. prob.: i have a horizontal straight bare pipe length 1m ID, 44mm and OD 56 mm *made of stain less steel). initially filled with 50 degree C, at 160 bar operating. ambient temp is 45C, and inlet 300 C water at 0.002 m/s start entering outlet temperature is average . simulation is carried out for steady state, laminar flow analysis. but it is not meeting desired convergence criteria. convergev=nce required up to RMS 10^4 . my cfx pre. inputs and boundary conditions are follows "C:\Program Files\ANSYS Inc\v110\CFX\bin\perllib\cfx5solve.pl" Setting up CFX Solver run ... ++    CFX Command Language for Run  LIBRARY: CEL: EXPRESSIONS: cpC = 0.20592 [J kg^1 K^3] cpD = 1.75442E4 [J kg^1 K^4] cpA = 6550.39577 [J kg^1 K^1] cpB = 81.26306 [J kg^1 K^2] cp = cpA+cpB*T+cpC*T^2+cpD*T^3 denB = 1.36311 [kg m^3 K^1] denC = 0.0027 [kg m^3K^2] denA = 833.93908 [kg m^3 ] den = denA+denB*T+denC*T^2 kB = 0.02039 [W m^1 K^2] kC = 6.1901E5 [W m^1 K^3] kE = 4.99387E11 [W m^1 K^5] kD = 8.74777E8 [W m^1 K^4] kA = 1.87631 [W m^1 K^1] k = kA+kB*T+kC*T^2+kD*T^3+kE*T^4 outletTemp = areaAve(T)@REGION:OUTLET visD = 6.68023E10 [Pa s K^3] visB = 1.67183E4 [Pa s K^1] visC = 5.00487E7 [Pa s K^2] visA = 0.02121 [Pa s] visE = 3.34164E13 [Pa s K^4] vis = visA+visB*T+visC*T^2+visD*T^3+visE*T^4 END END MATERIAL: ASME SA312 type 316 Material Group = User Option = Pure Substance Thermodynamic State = Solid PROPERTIES: Option = General Material EQUATION OF STATE: Density = 8000 [kg m^3] Molar Mass = 55.85 [kg kmol^1] Option = Value END REFERENCE STATE: Option = Specified Point Reference Temperature = 408.15 [K] END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 500 [J kg^1 K^1] END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 16.3 [W m^1 K^1] 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 REFERENCE STATE: Option = Automatic END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 4.34E+02 [J kg^1 K^1] 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 Option = Pure Substance Thermodynamic State = Liquid PROPERTIES: Option = General Material Thermal Expansivity = 0.001208 [K^1] ABSORPTION COEFFICIENT: Absorption Coefficient = 1.0 [m^1] Option = Value END DYNAMIC VISCOSITY: Dynamic Viscosity = vis Option = Value END EQUATION OF STATE: Density = den Molar Mass = 18.02 [kg kmol^1] Option = Value END REFRACTIVE INDEX: Option = Value Refractive Index = 1.0 [m m^1] END SCATTERING COEFFICIENT: Option = Value Scattering Coefficient = 0.0 [m^1] END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = cp Specific Heat Type = Constant Pressure END TABLE GENERATION: Maximum Absolute Pressure = 20 [MPa] Maximum Temperature = 600 [K] Minimum Absolute Pressure = 0 [MPa] Minimum Temperature = 273.15 [K] END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = k END END END END FLOW: SOLUTION UNITS: Angle Units = [rad] Length Units = [m] Mass Units = [kg] Solid Angle Units = [sr] Temperature Units = [K] Time Units = [s] END SIMULATION TYPE: Option = Steady State EXTERNAL SOLVER COUPLING: Option = None END END DOMAIN: wall Domain Type = Solid Location = WALL Solids List = Steel BOUNDARY: Domain Interface 1 Side 2 Boundary Type = INTERFACE Location = Primitive 2D A BOUNDARY CONDITIONS: HEAT TRANSFER: Option = Conservative Interface Flux END END END BOUNDARY: wall Default Boundary Type = WALL Location = OUTERSURFACE,Primitive 2D C,Primitive 2D D BOUNDARY CONDITIONS: HEAT TRANSFER: Heat Transfer Coefficient = 10 [W m^2 K^1] Option = Heat Transfer Coefficient Outside Temperature = 320.15 [K] END END END DOMAIN MODELS: DOMAIN MOTION: Option = Stationary END MESH DEFORMATION: Option = None END END INITIALISATION: Option = Automatic INITIAL CONDITIONS: TEMPERATURE: Option = Automatic with Value Temperature = 323.15 [K] END END END SOLID MODELS: HEAT TRANSFER MODEL: Option = Thermal Energy END THERMAL RADIATION MODEL: Option = None END END END DOMAIN: water Coord Frame = Coord 0 Domain Type = Fluid Fluids List = Water Location = WATER BOUNDARY: Domain Interface 1 Side 1 1 Boundary Type = INTERFACE Location = Primitive 2D B BOUNDARY CONDITIONS: HEAT TRANSFER: Option = Conservative Interface Flux END WALL INFLUENCE ON FLOW: Option = No Slip END END END BOUNDARY: INLET Boundary Type = INLET Location = INLET BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END HEAT TRANSFER: Option = Static Temperature Static Temperature = 573.15 [K] END MASS AND MOMENTUM: Normal Speed = 0.002 [m s^1] Option = Normal Speed END END END BOUNDARY: OUTLET Boundary Type = OPENING Location = OUTLET BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END HEAT TRANSFER: Option = Static Temperature Static Temperature = outletTemp END MASS AND MOMENTUM: Option = Static Pressure for Entrainment Relative Pressure = 0 [Pa] END END END DOMAIN MODELS: BUOYANCY MODEL: Buoyancy Reference Density = 891.48 [kg m^3] Buoyancy Reference Temperature = 456 [K] Gravity X Component = 0 [m s^2] Gravity Y Component = 9.81 [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 = 0 [MPa] END END FLUID MODELS: COMBUSTION MODEL: Option = None END HEAT TRANSFER MODEL: Option = Thermal Energy END THERMAL RADIATION MODEL: Option = None END TURBULENCE MODEL: Option = Laminar END END INITIALISATION: Option = Automatic INITIAL CONDITIONS: Velocity Type = Cartesian CARTESIAN VELOCITY COMPONENTS: Option = Automatic END STATIC PRESSURE: Option = Automatic END TEMPERATURE: Option = Automatic with Value Temperature = 323.15 [K] END END END END DOMAIN INTERFACE: Domain Interface 1 Boundary List1 = Domain Interface 1 Side 1 1 Boundary List2 = Domain Interface 1 Side 2 Interface Type = Fluid Solid INTERFACE MODELS: Option = General Connection FRAME CHANGE: Option = None END PITCH CHANGE: Option = None END END MESH CONNECTION: Option = GGI END END OUTPUT CONTROL: RESULTS: File Compression Level = Default Option = Standard END END SOLVER CONTROL: ADVECTION SCHEME: Option = High Resolution END CONVERGENCE CONTROL: Maximum Number of Iterations = 2000 Physical Timescale = 2 [s] Solid Timescale Control = Auto Timescale Timescale Control = Physical Timescale END CONVERGENCE CRITERIA: Residual Target = 0.0001 Residual Type = RMS END DYNAMIC MODEL CONTROL: Global Dynamic Model Control = Off END END END COMMAND FILE: Version = 11.0 Results Version = 11.0 END EXECUTION CONTROL: INTERPOLATOR STEP CONTROL: Runtime Priority = Standard EXECUTABLE SELECTION: Double Precision = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END END PARALLEL HOST LIBRARY: HOST DEFINITION: kewat Host Architecture String = intel_p3.sse_winnt5.1 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 = kway Option = MeTiS Partition Size Rule = Automatic END END RUN DEFINITION: Definition File = D:\M.Tech\trail 3rd_at anb45c and \ Highresuation\trail_3rd.def Interpolate Initial Values = Off Run Mode = Full END SOLVER STEP CONTROL: Runtime Priority = Standard EXECUTABLE SELECTION: Double Precision = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARALLEL ENVIRONMENT: Number of Processes = 1 Start Method = Serial END END END ++    Solver    ++ ++    ANSYS CFX Solver 11.0     Version 2007.01.1519.20 Mon Jan 15 19:24:21 GMTST 2007     Executable Attributes     singleint3232bitnovc6optimisedsupfortnoprofnospaglcomp     Copyright 19962007 ANSYS Europe Ltd.  ++ ++  Job Information  ++ Run mode: serial run Host computer: KEWAT Job started: Fri Jul 05 16:29:28 2013 ++  Memory Allocated for Run (Actual usage may be less)  ++ Data Type Kwords Words/Node Words/Elem Kbytes Bytes/Node Real 101526.5 211.91 222.18 396587.9 847.63 Integer 51398.6 107.28 112.48 200775.9 429.12 Character 2864.8 5.98 6.27 2797.7 5.98 Logical 65.0 0.14 0.14 253.9 0.54 Double 1208.0 2.52 2.64 9437.5 20.17 ++  ****** Notice ******   Wall Heat Transfer Coefficient written to the results file uses   "Wall Adjacent Temperature" for the bulk temperature. If you want   to override the bulk temperature then set the expert parameter   "tbulk for htc = <value>"  ++ ++  Mesh Statistics  ++ Domain Name : water Total Number of Nodes = 398177 Total Number of Elements = 389760 Total Number of Hexahedrons = 389760 Total Number of Faces = 16224 Minimum Orthogonality Angle [degrees] = 71.6 OK Maximum Aspect Ratio = 54.4 OK Maximum Mesh Expansion Factor = 1.5 OK Domain Name : wall Total Number of Nodes = 80928 Total Number of Elements = 67200 Total Number of Hexahedrons = 67200 Total Number of Faces = 27360 Minimum Orthogonality Angle [degrees] = 86.9 OK Maximum Aspect Ratio = 4.4 OK Maximum Mesh Expansion Factor = 1.2 OK Global Statistics : Global Number of Nodes = 479105 Global Number of Elements = 456960 Total Number of Hexahedrons = 456960 Global Number of Faces = 43584 Minimum Orthogonality Angle [degrees] = 71.6 OK Maximum Aspect Ratio = 54.4 OK Maximum Mesh Expansion Factor = 1.5 OK Domain Interface Name : Domain Interface 1 Nonoverlap area fraction on side 1 = 0.00E+00 Nonoverlap area fraction on side 2 = 0.00E+00 ++  Buoyancy Reference Information  ++ Domain Group: water Buoyancy has been activated. The absolute pressure will include hydrostatic pressure contribution, using the following reference coordinates: (1.71162E11,3.77106E12,1.00000E+00). ++  Average Scale Information  ++ Domain Name : water Global Length = 1.1488E01 Minimum Extent = 4.4000E02 Maximum Extent = 1.0000E+00 Density = 9.9248E+02 Dynamic Viscosity = 5.5002E04 Velocity = 2.0000E03 Advection Time = 5.7441E+01 Reynolds Number = 4.1460E+02 Thermal Conductivity = 6.5603E01 Specific Heat Capacity at Constant Pressure = 4.1267E+03 Prandtl Number = 3.4598E+00 Domain Name : wall Global Length = 9.7951E02 Minimum Extent = 5.6000E02 Maximum Extent = 1.0000E+00 Density = 7.8540E+03 Thermal Conductivity = 6.0500E+01 Specific Heat Capacity at Constant Pressure = 4.3400E+02 Thermal Diffusivity = 1.7749E05 Average Timescale = 5.4056E+02 Minimum Timescale = 1.7669E+02 Maximum Timescale = 5.6341E+04 I need urs expert comment. waiting for the reply.. plz help again thanx.. 
The lack of convergence is an FAQ: http://www.cfdonline.com/Wiki/Ansys...gence_criteria
Why are you modelling this as laminar? Do you know it is laminar? How? Please post an image of what you are modelling, and the results you are getting so far. 
file attech of my pipe model.
1 Attachment(s)
Quote:
thank you sir for the reply. i attached a file.is a 1meter straight bare pipe of ID 44mm, OD 56mm. initialy filled with water temp 50C , and 300C water at 0.002 m/s start flowing through inlet, i calculated Re. no. it is wel below and in laminor range. plz reply. thanks alote. 
Can you post an image of the temperature cross section you are getting, and what you expect to get?

temp profile
1 Attachment(s)
sir , i attached temperature profile, what i got.
thanx for reply again. 
OK, I see.
Have you read the FAQ link I posted? This talks about the exact question you are asking. 
respected sir, i tried with increased the physical timescale, but still not conversed . up to RMS 10^3.

The link I posted talked about a lot more than just larger time steps. Have you read the other things it talks about as well?

Hello,
natural convection often does not converge well. Mostly you have to simulate everything in full 3D and not using symmetries. Another important point is to run your simulation transient. mostly the natural convec. is transient and only converges well using time step developement. And by the way ther is no rule that your simulation has to converge lower than 1e4! You have to judge if the simulation is converged and accurate eneough for your needs. 
turbulence model.
Dear sir,
I am solving a thermal stratification problem in horizontal surge pipe, where Reynolds number is < 1000 but Rayleigh number goes beyond the 10^9, Reynolds decides the flow is Laminar, since it is the case of natural convection so Rayleigh number decides it as a turbulent flow. can i use Kepsilon model for this case ( Re<1000 and Ra> 10^9) orany other best suited model. Please reply. and thank in advance, Regards cpkewat. 
The best thing to do is simply run a few turbulence models and see which works the best. Also try a few of the options, such as the buoyant flows turbulence options (I think ke has this, SST might have it as well).

turbulence model.
sir thanks for the reply.,
I am new in CFD, my Reynolds number is low less than 1000 which is within the laminar regime but Rayleigh number is greater than 10^9. and this is thermal stratification problem simple straight pipe. is this problem is Laminar or Turbulent case. 
It is probably turbulent due to the Rayleigh number. But you should check that as well, you might have a laminar core and turbulent boundary layers.

turbulence model.
Sir
i m mechanical engg student, and try to simulate a natural convection problem thar is tharmal stratification and conjugate heat transfer in straight pipe in ansys cfx. try to generate temperature gradient across the section, i m fresh in cfx.and dont know much. i have no much time to complete my problem. i have a horizontal straight bare pipe length 1m ID, 44mm and OD 56 mm *made of stain less steel). initially filled with 50 degree C, at 160 bar operating. ambient temp is 45C, and inlet 300 C water at 0.002 m/s start entering, just because of buoyancy stratification is expected. I calculated Re ( 158 <Re<750) but same time Ra > 10^9. and solved the problem with Laminar model. just not because of proper guidance i wasted my most of the time. and my work has rejected by examiner b/c of this turbulent case and solved with laminar case. 
So isn't your examiner saying you should model it with a turbulence model? That seems to obvious next step.

turbulence model.
sir thanx for replying silly doubts..
he didn't directly say. he was saying with his opinion this problem falls in turbulent regime, even Re is very low, and i have no experimental setups for validating my results. that's become more problematic. 
All times are GMT 4. The time now is 22:18. 