# couette flow

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 March 29, 2008, 16:15 couette flow #1 thiagobaixo Guest   Posts: n/a Sponsored Links i try to simulate a flow between two parallel plates, the couette flow. i´m using a 2D structured grid. The fact is that i´m not getting the parabolic profile. I try to refine the grid, and it looks fine, so i think that the problem is not concerned with the grid´s quality. My domain is: x = 3 m; y= 1 m; The reynolds number is 10, therefore the laminar condition is OK. My boundary conditions are: 1- The flow into in the domain with a velocity equal 0.0005 m/s; 2- The outlet condition is 0 Pa, relative static pressure; 3- symmetry conditions; 4- no slip condition in y = 0 and y= 1. when i plot the profile velocity in post, i generaly use a line in the Y direction, and plot a chart of velocity u in x axis against Y in y axis. the result is not the parabolic profile. the point where i get the result is far away from the inlet, hence the flow is completely developed. where am i wrong ? please help me !!! best regards !!
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 March 30, 2008, 17:53 Re: couette flow #2 Glenn Horrocks Guest   Posts: n/a Hi, What differencing scheme are you using? You will need an accurate second order scheme to get a good velocity profile. For this model I would recommend the hybrid differencing scheme with a blend factor of 1.0 (therefore entirely second order). Glenn Horrocks

 March 30, 2008, 19:26 Re: couette flow #3 Rogerio Fernandes Brito Guest   Posts: n/a Tiago, dê uma lida no link abaixo sobre os esquemas do Solver do CFX: http://www.padtinc.com/epubs/focus/2..._54.pdf#Page=1 Dear Tiago, give a look on http://www.padtinc.com/epubs/focus/2..._54.pdf#Page=1 to see about the schemes of cfx

 March 31, 2008, 11:48 Re: couette flow #4 thiagobaixo Guest   Posts: n/a i already made a simulations with other advections schemes, from high order to specif blend factor = 1, and with one order, like upwind and blend factor =0...the results is the same...i can´t get the parabolic profile... another suggestions ? regards !

 March 31, 2008, 12:12 Re: couette flow #5 Rogerio Fernandes Brito Guest   Posts: n/a how about increasing the mesh on y direction? Try using tetra mesh with boundary layer close to the 2 walls.

 March 31, 2008, 13:02 Re: couette flow #6 andy2O Guest   Posts: n/a thiagobaixo, I can't spot anything obvious from your comments, so can you provide more information?... What convergence criteria are you using, and what are your final maximum residuals? How many iterations were performed? Did you set the turbulence model to 'laminar'? What *does* the velocity profiles look like? If it's short enough, try and post the CCL data (i.e. the problem description) from the top of your solver output file. Regards, Andy

 March 31, 2008, 13:06 Re: couette flow #7 andy2O Guest   Posts: n/a PS: What is the mesh like in the z-direction? Just 1 cell thick, or more? What is the thickness of the domain (I mean in metres, not cells) in the Z direction? Andy

 March 31, 2008, 13:33 Re: couette flow #8 thiagobaixo Guest   Posts: n/a i´m using the rms criteria, (10^-9)... ================================================== ==================== OUTER LOOP ITERATION = 2000 CPU SECONDS = 3.084E+02 ---------------------------------------------------------------------- | Equation | Rate | RMS Res | Max Res | Linear Solution | +----------------------+------+---------+---------+------------------+ | U-Mom | 1.00 | 2.8E-09 | 4.0E-09 | 1.6E-01 ok| | V-Mom | 1.00 | 4.6E-10 | 2.2E-09 | 9.9E-01 ok| | W-Mom | 1.00 | 1.0E-09 | 4.2E-09 | 1.0E+00 F | | P-Mass | 1.00 | 6.8E-11 | 5.2E-10 | 22.9 2.1E-01 ok| +----------------------+------+---------+---------+------------------+ CFD Solver finished: Mon Mar 31 12:57:31 2008 CFD Solver wall clock seconds: 3.0900E+02 Execution terminating: maximum number of time-step iterations, or maximum time has been reached. +--------------------------------------------------------------------+ | V-Mom | +--------------------------------------------------------------------+ Boundary : bottom 5.4917E-04 Boundary : inl 5.8768E-14 Boundary : out -4.8983E-14 Boundary : placa Default 4.0204E-10 Boundary : wall -5.4917E-04 ----------- Domain Imbalance : 1.0920E-13 Domain Imbalance, in %: 0.0000 % +--------------------------------------------------------------------+ | W-Mom | +--------------------------------------------------------------------+ Boundary : inl 6.0706E-08 Boundary : out -6.1018E-08 Boundary : placa Default -5.4917E-01 Boundary : symmetry 5.4917E-01 Boundary : wall 3.1205E-14 ----------- Domain Imbalance : 9.0329E-13 Domain Imbalance, in %: 0.0000 % +--------------------------------------------------------------------+ | P-Mass | +--------------------------------------------------------------------+ Boundary : inl 5.9250E-06 Boundary : out -5.9248E-06 ----------- Domain Imbalance : 2.3097E-10 Domain Imbalance, in %: 0.0039 % in this case for example the simulation not converged, because i used a very low value for the rms, but in another ocasions i used like 400 interactions with a criteria rms 5.10^-5, in this case the solution converged well, but the results related with the profile are the same... the flow is completely laminar, for this case there is no need to use a turbulent model.

 March 31, 2008, 13:46 Re: couette flow #9 thiagobaixo Guest   Posts: n/a the mesh in the z direction is 0.001 m, i extruded the structured mesh whith one layer... my simulation is very simple...make a retangular geometry in ICEM (x = 3m, and y= 1 m), and then build a hexa mesh (trough the block topology), convert to unstructured mesh (only in this way, you can use the extrude mesh for hexa mesh), the extrusion i made with the conditions said above. the i create the parts, like the inlet (the left face), outlet (the right face), wall (the upper face) and bottom (the bottom face). once the mesh is extruded and the parts was created, you can import it into pre cfx. for the inlet i use 0.0005 m/s, for the outlet = 0Pa, for the wall no slip condition, and for the bottom, front face and back face i use symmetry... for the solver, i try to use the high order method for the advection scheme, and the blend factor = 1.0.. for the residual RMS criteria = 5.10^-5, and the number of interactions= 500..the solution finish very quick, because in this case the residuals are quickly obtained. try to run this case, and the in the post, make a line in the y direction, plot the chart of velocity u against y trough that line, in my case the parabolic profile was not obtained... regards !

 March 31, 2008, 13:51 Re: couette flow #10 thiagobaixo Guest   Posts: n/a thanks for the suggestions people, but unfortunately the results are the same ! that sucks !!

 March 31, 2008, 13:52 Re: couette flow #11 thiagobaixo Guest   Posts: n/a thanks for the suggestions people, but unfortunately the results are the same ! that sucks !

 March 31, 2008, 13:58 Re: couette flow #12 andy2O Guest   Posts: n/a "the flow is completely laminar, for this case there is no need to use a turbulent model." I agree 100%! The *default* on CFX is to use the k-epsilon turbulence model. You have to explicity select the 'None (Laminar)' model to model laminar flow - that's what I was asking about. It appears that you did do this correctly since there are no turbulence equation residuals in your output. What about the other convergence control data? What timescale control did you use? What timescale did CFX actually use in the solution? If you would like us to try to help more: 1) What *does* the profile look like? Is it smooth? All we know is that it is not parabolic!! That's not very much information to help you from. 2) What is the mesh like in the z direction? And roughly how many cells in the other directions. 3) Please post the CCL from the def file or the problem setup stage (i.e. beginning) of the solver if possible... there is obviously something slightly wrong - perhaps we can spot it for you. Best wishes, andy

 March 31, 2008, 15:10 Re: couette flow #13 thiagobaixo Guest   Posts: n/a thanks for the advices andy2O ! can i send to you the image of the profile and the CCL by e-mail ? in this case, what´s your e-mail ? you´ll see that´s almost the correct solution. i don´t use a specific time scale, my simulation is steady, and i use the auto timescale option. the auto timescale calculated by cfx is something like 10^-2...

 April 1, 2008, 04:10 Re: couette flow #14 andy2O Guest   Posts: n/a thiagobaixo, Please could you just post to the forum, rather than emailing me... that way other people can help you, as my time is very limited and others have more experience, and other people can learn too. It sounds like the problem must be quite subtle if the results are as close as you say, so I certainly can't promise to spot the cause, but I'm happy to try. You can post pictures to the forum using the method described in the FAQ. See the link: http://www.cfd-online.com/Wiki/Ansys...ible_answer.3F Regards, andy.

 April 1, 2008, 11:02 Re: couette flow #15 thiagobaixo Guest   Posts: n/a The profile that i had obtained is in the following andress: http://reg.imageshack.us/content.php.../perfildw8.png did have anyone made a simple simulation about the flow between to parallels plate in rest ?

 April 1, 2008, 12:32 Re: couette flow #16 andy2O Guest   Posts: n/a It appears that the velocity your figure shows is more than 10 times bigger than the inlet velocity you have told us about (0.0005 m/s). That should not be for parabolic flow between 2 flat plates! So, if you have given us the correct input velocity, it appears something is badly wrong with the setup. 1) Could you add some "symbols" to the graph - so we can see the mesh? (E.g. how many nodes on each section of the graph) 2) Can you post a screen shot of the actual mesh from e.g. CFX Pre? 3) In your original post you said you use no-slip at y=0 and at y=1. In another post you said you use symmetry on 3 boundaries. Which is correct? I cannot see the effect of a no-slip boundary at y=0 or 1 on the plot. but most of all: 3) *Please*, if you would like help, try and post the CCL for the problem - that's the text output you see in CFX solver right at the beginning of the simulation, before the iterations start. Regards, andy

 April 1, 2008, 12:59 Re: couette flow #17 thiagobaixo Guest   Posts: n/a ok Andy ! follow below the ccl related with the simulation, in this case it refers to my last one, i input a timestep of 2 s. about the question 3, my original post was wrong, in fact the condition wall applies only at the upper face of retangule. the mesh is: http://img358.imageshack.us/my.php?image=malhamr1.jpg and the ccl: LIBRARY: 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 Thermal Expansivity = 0.003356 [K^-1] ABSORPTION COEFFICIENT: Absorption Coefficient = 0.01 [m^-1] Option = Value END DYNAMIC VISCOSITY: Dynamic Viscosity = 1.831E-05 [kg m^-1 s^-1] Option = Value END EQUATION OF STATE: Density = 1.185 [kg m^-3] Molar Mass = 28.96 [kg kmol^-1] Option = Value 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 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 = 1.0044E+03 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 2.61E-02 [W m^-1 K^-1] 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: placa Coord Frame = Coord 0 Domain Type = Fluid Fluids List = Air at 25 C Location = PLACA BOUNDARY: bottom Boundary Type = SYMMETRY Location = BOTTOM END BOUNDARY: inl Boundary Type = INLET Location = INL BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Option = Cartesian Velocity Components U = 0.005 [m s^-1] V = 0 [m s^-1] W = 0 [m s^-1] END END END BOUNDARY: out Boundary Type = OUTLET Location = OUT BOUNDARY CONDITIONS: FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Option = Average Static Pressure Relative Pressure = 0 [Pa] END PRESSURE AVERAGING: Option = Average Over Whole Outlet END END END BOUNDARY: placa Default Boundary Type = WALL Location = SYMMETRY BOUNDARY CONDITIONS: WALL INFLUENCE ON FLOW: Option = No Slip END END END BOUNDARY: symmetry Boundary Type = SYMMETRY Location = LIVE END BOUNDARY: wall Boundary Type = WALL Location = WALL BOUNDARY CONDITIONS: WALL INFLUENCE ON FLOW: Option = No Slip 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 [atm] END END FLUID MODELS: COMBUSTION MODEL: Option = None END HEAT TRANSFER MODEL: Option = None END THERMAL RADIATION MODEL: Option = None END TURBULENCE MODEL: Option = Laminar END END END OUTPUT CONTROL: RESULTS: File Compression Level = Default Option = Standard END END SOLVER CONTROL: ADVECTION SCHEME: Blend Factor = 1.0 Option = Specified Blend Factor END CONVERGENCE CONTROL: Maximum Number of Iterations = 10000 Physical Timescale = 2 [s] Timescale Control = Physical Timescale CONVERGENCE CRITERIA: Residual Target = 1.e-10 Residual Type = RMS END DYNAMIC MODEL CONTROL: Global Dynamic Model Control = On 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: ln119 Installation Root = C:\Arquivos de programas\Ansys Inc\v%v\CFX Host Architecture String = intel_xeon64.sse2_winnt5.1 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 = C:/Documents and \ Settings/thiagosp/Desktop/ENCIT/CFX/couette/A.def Interpolate Initial Values = Off Run Mode = Full END SOLVER STEP CONTROL: Runtime Priority = Standard EXECUTABLE SELECTION: Double Precision = On END MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARALLEL ENVIRONMENT: Number of Processes = 1 Start Method = Serial END END END thanks !

 April 1, 2008, 14:09 Re: couette flow #18 andy2O Guest   Posts: n/a ... but according to the CCL, you have: 1) SYMMETRY boundary called "bottom" at location BOTTOM 2) INLET boundary called "inl" at location INL 3) OULET boundary called "out" at location OUT 4) WALL boundary called "placa Default" at location SYMMETRY ?????!!!!!?????!!!! 5) SYMMETRY boundary called "symmetry" at location LIVE 6) WALL boundary called "wall" at location WALL Number 4 looks very strange, and is not consistent with what you told us. There are 2 wall boundaries in this simulation. Is this a mistake? Also: You have a tiny velocity and a very big domain. It is normally a good idea for you to converge the solution for enough 'time' for the flow to pass several times through the domain. For your case it takes a time of 3 [m] / 0.0005 [m s^-1] = 6000 [s] for the flow to pass through the domain. You have 500 iterations, each of 10^-2 [s] timescale according to your earlier post - which is just 5 [s] in total - so this is perhaps not enough! Try using a bigger timestep! CFX is obviously picking up the small thickness of the 2D mesh when it is choosing the automatic timescale - I'm surprised by this - I thought it would ignore this dimension for a 2D simulation, but perhaps this is to do with boundary condition 4. Because you have a timestep which is so very very small compared to the characteristic time it is easy to get small residuals even though the flow is not converged. Why are you simulating such a slow flow in such a big domain?! I guess it is a test calculation? Finally, the mesh thickness in the Z direction should be comparable to the size of your smallest cell. From your picture this is 3 [m] / (about 100 cells) = 0.03 [m], so your 0.001 [m] is a bit thin. Nothing else jumps out straight away. I hope this helps. I'll have another glance later during another coffee break. andy

 April 1, 2008, 14:44 Re: couette flow #19 thiagobaixo Guest   Posts: n/a thank´s man ! i´ll try your suggestions, i agree about the critic in number 4, the fact is that when i created the domain in pre, automatically the boundary condition "domain default domain" is done ! take your coffe, when you came back you´ll see what´s going on ! thanks !

 April 1, 2008, 15:21 Re: couette flow #20 thiagobaixo Guest   Posts: n/a YES ! you are right !! after your suggestions, i FINALLY get the desired profile !! look that : http://img385.imageshack.us/my.php?image=profilekn9.png Thanks man !! here in brazil, we would say: " you are the man !" thank´s a lot !! regards !

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