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FLuent simulation of taylor couette flow of concentric cylinder geometry. |
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October 28, 2012, 10:37 |
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#21 |
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rshbhb
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Dear Daniele Sir,
Thank you for the information. I was a bit busy in the previous two days. I'll simulate the system again and will let you know the results or any doubts that I may inccur. I would like to Thank Mr.Far for joining the conversation and sharing his results and doubts with us. |
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October 28, 2012, 11:50 |
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#22 | |
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Quote:
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October 28, 2012, 12:00 |
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#23 |
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Rick
Join Date: Oct 2010
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In my opinion, the Reynolds number, which is dimensionless is:
Re=(U0*d*density)/mu U0 cannot be expressed in rad/s, otherwise the Re is not dimensionless, so it must be in m/s, so I think U0 is the peripheral speed of the rotating cylinder. U0=omega*radius, where omega is the angular velocity in rad/s d is the characteristic length, in my case=(outer radius - inner radius) and it is expressed in m density is the density of the fluid (kg/m3) mu is the dynamic viscosity of the fluid, expressed in Pa*s or, in other words, kg/(s*m) With the dimensional analysis: Re=(m/s)*(m)*(kg/m3)*(s*m/kg) Re is dimensionless. This is how I calculate the Re Referring to your data and your cited article I think Re, to be dimensionless is calculated as: Re=(density*(angular velocity)*radius*radius)/mu where (angular velocity)*radius is the peripheral velocity; the only difference is the characteristic length density in kg/m3 angular velocity in rad/s radius in m (of the rotating cylinder) mu in Pa*s So for u1=5 rad/s, Re (your article)=50 for u2=20 rad/s, Re (your article)=200 PS: I think in page 4 of your cited article there's an error: Re is calculated with radius=1 m, but it is not rotating! PS2: density=1 kg/m3: what type of fluid is it?is it a gas?Is the Taylor-Couette flow valid for gases?? Last edited by ghost82; October 28, 2012 at 12:31. |
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October 28, 2012, 12:02 |
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#24 |
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Rick
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October 28, 2012, 12:16 |
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#25 | |
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What if, we use the slip wall instead of translational periodic boundary condition?
Quote:
https://dl.dropbox.com/u/68746918/Co...nstability.rar If only meshes are needed then click on this link : https://dl.dropbox.com/u/68746918/me...ute_taylor.rar PS: Dropbox will take some time to upload the files. May be few hours. |
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October 28, 2012, 12:30 |
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#26 |
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October 28, 2012, 12:37 |
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#27 | |
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rshbhb
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Quote:
I want to ask you which fluent version are you using? I am using GAMBIT for geometry and FLuent Solver (Fluent 6.3.26). I want to know that will i get such good images with the versoin of Fluent that I have? Sir I'll upload the geometry (in GAMBIT) in 15 mins or so could you please verify if it'll be ok for correct simulation. MY Laptop is not very fast (only 3GB RAM). |
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October 28, 2012, 12:44 |
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#28 | |
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Quote:
Yes you can get the same images in Fluent, there is no change in these features. They have just improved flow schemes and included other turbulnece model and wall treatments. I have also 4GB laptop. But I will be more than happy to look at your mesh. Meanwhile you can compare your mesh with mine or Daniele's. Daniele's opinion will be more valuable in this regard. |
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October 28, 2012, 12:58 |
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#29 | |
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rshbhb
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Quote:
For faster communication i have sent you chat request on your gmail id. Please accept it. |
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October 28, 2012, 13:11 |
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#30 |
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rshbhb
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Sir,
How do I hard link the faces before meshing in GAMBIT? I am just getting the option of linking the face meshes. |
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October 29, 2012, 03:24 |
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#31 | |
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Rick
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Quote:
Grid is the same as the previous simulation. Dimensions are the same as reported in some posts above, Re is 1000 (considering the caracteristic length R2-R1). I had more troubles to get convergence with this case. I also have a laptop with win7 64 bit with only 4 gb ram and a workstation with win7 64 bit and 32 gb ram; on both systems ansys 14 is installed. For these simulations I used the laptop and it took about 1 day to reach a drop in residuals of 1*10^-7. I think you will have no problem using fluent 6.x version. Daniele Last edited by ghost82; October 29, 2012 at 03:41. |
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October 29, 2012, 03:27 |
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#32 | |
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Rick
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Quote:
Yes, you have to link the faces. See button in the attached image. |
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October 29, 2012, 04:07 |
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#33 | ||
Senior Member
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Quote:
In fluent K-epsilon and K-omega both models are equipped with both : wall function and integration to wall. It should be noted that K-epsilon models are by defaults wall function models but modified (coupled with single equation model in boundary layer) to account for the near wall effects while K-omega models are originally designed for intergration to wall approach. So be more careful with Y+ for older versions. Quote:
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October 29, 2012, 12:17 |
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#34 |
New Member
rshbhb
Join Date: Oct 2012
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I have now got somewhere close to getting the result. Here is the result of my simulation done yesterday.
I think I've not done the iterations for sufficient no. of times.( Right?) I had used k-omega SST model for this. Angular Velcity 6 rad/s. Also please could you explain what do you mean by 'Y+' (is the residual y values you are talking about or something else and how can I control the Y+ values) Thank you "SIRs" |
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October 29, 2012, 12:27 |
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#35 | |
Senior Member
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Quote:
Y+ is non dimensional parameters which is due to friction velocity, first mesh point distance from wall and kinematic viscosity. http://www.cfd-online.com/Wiki/Dimen...tance_(y_plus) You must ensure two things in mesh: 1. First cell height (Expansion ratio should be less than 1.2) 2. Total height of boundary layer. you can get estimate from this link (for flat plate but good enough for initial estimate) http://www.see.ed.ac.uk/~johnc/teach...dboundary.html Also try this link: http://www.ae.metu.edu.tr/~ae244/doc...07/node12.html The mesh, which is shared in post, has y= < 1, so it is good for SST-KW model. |
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October 29, 2012, 13:09 |
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#36 | |
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Rick
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Quote:
to see if your solution is converged you can monitor some important value during iterations, for example: integral on the volume of the turbulent kinetic energy; or/and Area weight surface integral of velocity magnitude on different planes (for example at some planes at z=cost) etc. When your monitor value doesn't change during successive iterations your solution is converged. See my picture in posts above; as you can see in my example convergence is reached for residuals<1*10^-7!! Fixing only minimum values for residuals is not a good idea to evaluate if solution is converged; you should monitor some other values! For the y+ subject Far is very expert as you can read on some other threads on this forum. PS: you can plot the y+ value; in plot postprocessing you will find "yplus" under "turbulence"; plot yplus values selecting wall surfaces and see if these values respect the limits. |
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October 29, 2012, 13:13 |
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#38 |
New Member
rshbhb
Join Date: Oct 2012
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In the mesh you all had made, did you use (or when shall i use) the reverse orientation option while hard linking the mesh (in the 'link mesh faces' option).
Do I use it for 1. the top and bottom periodic faces, OR 2. the vertical periodic side faces, OR 3. should I use this option at all. Daniele Sir had provided me with this link: http://www.cfd-online.com/Forums/flu...periodics.html in which there is a mention that we have to take the direction of the arrows same for both the faces but it not coming out to be same whatever I do. (Is it wrongly written?). Here is the picture for better understanding. THanks !! |
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October 29, 2012, 13:19 |
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#39 |
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Rick
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It computes the integral of kinetic energy on the domain; monitoring some important values is very important to evaluate convergence; in fact in the mixing tank tutorial it is reported:
The default convergence criteria are not sufficient to get the correct flow features in a mixing tank. To judge the convergence, some of the integrated quantities needs to be monitored along with velocity magnitude around the turbine. In this problem, monitor the velocity magnitude on a surface just above and below the turbine. Also monitor the volume integral of kinetic energy in the tank. |
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October 29, 2012, 13:23 |
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#40 |
New Member
rshbhb
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Here is the image.
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