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-   -   convergenceof natural convection prob. in cfx (http://www.cfd-online.com/Forums/cfx/120410-convergenceof-natural-convection-prob-cfx.html)

 cpkewat July 6, 2013 09:02

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 = 1.75442E-4 [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 = -6.1901E-5 [W m^-1 K^-3]
kE = -4.99387E-11 [W m^-1 K^-5]
kD = 8.74777E-8 [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 = -6.68023E-10 [Pa s K^-3]
visB = -1.67183E-4 [Pa s K^-1]
visC = 5.00487E-7 [Pa s K^-2]
visA = 0.02121 [Pa s]
visE = 3.34164E-13 [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:
Length Units = [m]
Mass Units = [kg]
Solid Angle Units = [sr]
Temperature Units = [K]
Time Units = [s]
END
SIMULATION TYPE:
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
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
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:
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 = k-way
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 2007.01.15-19.20 Mon Jan 15 19:24:21 GMTST 2007 |
| |
| Executable Attributes |
| |
| single-int32-32bit-novc6-optimised-supfort-noprof-nospag-lcomp |
| |
| Copyright 1996-2007 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

Non-overlap area fraction on side 1 = 0.00E+00
Non-overlap 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: (-1.71162E-11,-3.77106E-12,-1.00000E+00).

+--------------------------------------------------------------------+
| Average Scale Information |
+--------------------------------------------------------------------+

Domain Name : water
Global Length = 1.1488E-01
Minimum Extent = 4.4000E-02
Maximum Extent = 1.0000E+00
Density = 9.9248E+02
Dynamic Viscosity = 5.5002E-04
Velocity = 2.0000E-03
Reynolds Number = 4.1460E+02
Thermal Conductivity = 6.5603E-01
Specific Heat Capacity at Constant Pressure = 4.1267E+03
Prandtl Number = 3.4598E+00

Domain Name : wall
Global Length = 9.7951E-02
Minimum Extent = 5.6000E-02
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 = 1.7749E-05
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..

 ghorrocks July 7, 2013 06:15

The lack of convergence is an FAQ: http://www.cfd-online.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.

 cpkewat July 7, 2013 06:58

file attech of my pipe model.

1 Attachment(s)
Quote:
 Originally Posted by ghorrocks (Post 438237) The lack of convergence is an FAQ: http://www.cfd-online.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.

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.

 ghorrocks July 7, 2013 07:32

Can you post an image of the temperature cross section you are getting, and what you expect to get?

 cpkewat July 7, 2013 07:58

temp profile

1 Attachment(s)
sir , i attached temperature profile, what i got.

 ghorrocks July 7, 2013 08:06

OK, I see.

 cpkewat July 9, 2013 05:47

respected sir, i tried with increased the physical timescale, but still not conversed . up to RMS 10^-3.

 ghorrocks July 13, 2013 07:22

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?

 Benfa July 14, 2013 04:07

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 1e-4! You have to judge if the simulation is converged and accurate eneough for your needs.

 cpkewat January 31, 2014 03:10

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 K-epsilon model for this case ( Re<1000 and Ra> 10^9) orany other best suited model.

Regards
cpkewat.

 ghorrocks January 31, 2014 04:41

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 k-e has this, SST might have it as well).

 cpkewat January 31, 2014 05:56

turbulence model.

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.

 ghorrocks January 31, 2014 06:12

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.

 cpkewat January 31, 2014 07:08

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.

 ghorrocks January 31, 2014 07:14

So isn't your examiner saying you should model it with a turbulence model? That seems to obvious next step.

 cpkewat January 31, 2014 07:29

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.

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