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FLuent simulation of taylor couette flow of concentric cylinder geometry. |
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October 26, 2012, 07:44 |
FLuent simulation of taylor couette flow of concentric cylinder geometry.
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#1 |
New Member
rshbhb
Join Date: Oct 2012
Posts: 26
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Hello,
I have a concentric cylinder geometry as shown i the figure, such that, Inner wall is Rotating Outer wall is stationary Upper wall is stationary lower wall is stationary Fluid (liquid) is confined in the annular region (volume). Do I have to use a dynamic mesh or moving mesh OR I can mesh the geometry normally and rotate the inner walls in FLUENT ? (I have used the 2nd option). ------------------------------------------------------ REGARDING SIMULATION (I have referred to "Non-Newtonian Transitional Flow in an Eccentric Annulus") STEP BY STEP ANALYSIS OF MY PROBLEM 1. Models: As I am simulating the taylor couette flow of an incompressible fluid, my problem will go from laminar to turbulent flow pattern. In this case what kind of model should I use :Viscous-> k-epsilon or K-omega ? (I had used standard k-epsilon with standard wall functions). Last edited by rshbhb; October 26, 2012 at 08:12. |
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October 26, 2012, 07:56 |
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#2 |
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rshbhb
Join Date: Oct 2012
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2. Boundary Conditions: For interior rotating wall. I gave MOVING WALL -> Absolute -> Speed (rad/s) -> Rotational axis origin (0,0,0) -> roation axis direction (0,0,1) -> No slip
For Fluid (liquids): Rotation axis direction (0,0,1) & Motion type -> Stationary (IS IT CORRECT or should I use MRF or Moving Mesh???) I HAVE NOT USED periodic conditions. |
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October 26, 2012, 08:24 |
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#3 |
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rshbhb
Join Date: Oct 2012
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3. Solve -> Control -> Solution -> I had used SIMPLE MODEL (Pressure Velocity coupling). Is it correct?
Equations used FLOW & TURBULENCE 4. Solve ->Initialize -> in the compute from list i ha used inner wall. is it correct or should i just keep it empty and press INITIALIZE button?? 5.After Iterating it. Solution did converge and I got a velocity magnitude vector profile like the one shown in figure below (fig.1) but at an angular vel. of 1500 rad/s i expected to get tayloe vortex as shown in fig.2 . I THINK I HAVE GONE WRONG SOME WHERE please help me out with this. Kind Regards, Rishabh. |
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October 26, 2012, 09:00 |
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#4 | ||||
Senior Member
Rick
Join Date: Oct 2010
Posts: 1,016
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Quote:
If you have a turbulent regime just try different turbulence model to find which is approaching the "real" solution. You can use k-epsilon for fully turbulent flow, k-omega for transitional laminar to turbulent flow. Use laminar if you have a laminar regime. Consider that you must verify your y+ values depending on the model you use. Quote:
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Try to change your model (turbulence, transition) depending on the rotational speed and set in your parameters second order upwind schemes. You set the bases as walls; are you sure the second picture is taken by applying the same boundary conditions?Are you sure the domain is not periodic in the second picture? Daniele |
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October 28, 2012, 02:29 |
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#5 | ||
Super Moderator
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Quote:
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October 28, 2012, 03:53 |
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#6 | |
Senior Member
Rick
Join Date: Oct 2010
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Quote:
This is a good question and I don't have an answer. I found this: http://www.cfd-online.com/Forums/flu...ting-wall.html This can explain why we use mrf or sliding mesh for a rushton turbine, which is not a surface of revolution and we can use moving wall in this case (?). I also think that mrf will give the same results. Last edited by ghost82; October 28, 2012 at 05:07. |
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October 28, 2012, 04:15 |
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#7 |
Super Moderator
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I am unable to access the links. Would you like to send the papers (also tutorial you were referring earlier in this post) on my email id turboenginner@gmail.com.
Thank-you for your very informative post. |
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October 28, 2012, 04:31 |
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#8 | |
Senior Member
Rick
Join Date: Oct 2010
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Quote:
Added also a mediafire link. Daniele |
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October 28, 2012, 09:20 |
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#10 |
Super Moderator
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Previous results were with u1 = 20 rad/sec and this one with 5 rad/sec. Other parameters areRefer to this article http://www.compassis.com/downloads/M...tte%20Flow.pdf)
Density = 1 Kg/m3 Kinematic viscosity = 0.1 m2/sec d1 (inner cyclinder dia) = 2 m d2 (outer cylinder dia) = 4 m Length of cylinder = 2 m So what is the Reynolds number for u1= 5 rad/sec, 20 rad/sec ? http://research.ncl.ac.uk/quantum-fl...willis-phd.pdf http://www.cats.rwth-aachen.de:8080/...ette181207.pdf http://www.cats.rwth-aachen.de:8080/...ette181207.pdf |
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October 28, 2012, 11:00 |
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#11 |
Senior Member
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 11:31. |
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October 28, 2012, 11:30 |
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#12 |
Super Moderator
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October 28, 2012, 11:37 |
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#13 | |
New Member
rshbhb
Join Date: Oct 2012
Posts: 26
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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, 11:44 |
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#14 | |
Super Moderator
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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, 11:58 |
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#15 | |
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rshbhb
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For faster communication i have sent you chat request on your gmail id. Please accept it. |
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October 28, 2012, 12:11 |
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#16 |
New Member
rshbhb
Join Date: Oct 2012
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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, 02:27 |
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#17 | |
Senior Member
Rick
Join Date: Oct 2010
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Quote:
Yes, you have to link the faces. See button in the attached image. |
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August 28, 2013, 11:53 |
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#18 |
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behzad
Join Date: Aug 2013
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September 13, 2013, 10:10 |
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#19 |
Super Moderator
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someone indicated that link for dropbox files is not working : Post # 25. Here is new link:
https://dl.dropboxusercontent.com/u/...ute_taylor.rar |
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September 12, 2014, 01:10 |
hi rishabh
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#20 |
New Member
rahul kumar
Join Date: Jun 2014
Posts: 24
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if you want to know your flow near the surface wall of your domain then you can use " k-episilon" and if you want to know far away from the surface wall then use "k-omega
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