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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.

#1 
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rshbhb
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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 "NonNewtonian 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> kepsilon or Komega ? (I had used standard kepsilon with standard wall functions). Last edited by rshbhb; October 26, 2012 at 08:12. 

October 26, 2012, 07:56 

#2 
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rshbhb
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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. 

October 26, 2012, 08:24 

#3 
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rshbhb
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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. 

October 26, 2012, 09:00 

#4  
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Daniele
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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 kepsilon for fully turbulent flow, komega 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:
Quote:
Quote:
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 

October 26, 2012, 09:13 

#5  
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rshbhb
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I'll do the simulations again and let you know the results. Rishabh. 

October 26, 2012, 14:30 

#6 
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Daniele
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Just for fun I tried a laminar regime with Re=1000; you can see Taylor vortices.
Instead of wall for bases I set periodic condition (in gambit you have to link the faces before meshing and apply a periodic boundary for all the bases). Daniele 

October 26, 2012, 15:40 

#7 
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rshbhb
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Dear Daniele Sir/Ma'am,
I tried the simulation again with the new parameters as you described but still i m getting the same results as shown in the figure below (even with glycerine as the liquid). Its good to know that you got taylor vortices. I want to ask you, 1.what did you use as fluid? 2.Which faces do I have to link ? (do you mean bottom faces together & upper tw faces together?) 3. I have simulated with kw and ke models, should I use laminar model to get taylor vortex? 4.Should the periodic conditions be given as described in the FLUENT tutaorial guide "Non newtonian flow..."? 5.how do you get this image? (I only know to get contours and vectors) Do you use any isosuraface ? 6. How many number of nodes did you have in your mesh ? 7. Let me know If I am able to convey my doubts properly to you ? Rishabh. Last edited by rshbhb; October 26, 2012 at 16:03. 

October 27, 2012, 03:12 

#8  
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Quote:
first of all you need to evaluate the fuid dynamic regime; you need to evaluate the Reynolds number to know if your regime is laminar, transitional or turbulent. In the free article "Direct numerical simulation of turbulent Taylor–Couette flow" you can read that the authors obtain a laminar regime for Re=1000 and a turbulent regime for Re=3000, Re=5000 and Re=8000. When you evaluate the Reynolds number take in mind that U0 is expressed in m/s and it's omega*Rinternal (omega in rad/s and Rinternal in m). Once you know what regime you have you can start your simulations. I imposed the periodic boundary because I read in this article "Consider the incompressible flow between two infinitely long concentric cylinders". 1. I simulated Re=1000 with water as fluid, so my regime is laminar. 2. Yes, you have to link bottom and upper faces in gambit, so when you will mesh the bottom faces, the upper ones will be automatically meshed; also you have to set one periodic boundary contidion for all these 4 faces. Read here: Hardlinking faces in Gambit (periodics) If you will not link the faces, or if you will not set one periodic boundary gambit will not let you export the mesh. 3. It depends on your regime as written above. 4. All you have to do, once you have imported the mesh into fluent, is to set translational periodic condition in the boundary condition panel (you can choose between rotational and translational). I imposed a 0 pressure gradient. 5. the image is a plot contour of velocity magnitude; then from display>scene I checked overlays and plot the vector graph. For the vector graph I choosed from vector options "in plane", black color, and a scale factor of 3. 6. I'm on a mac now so I don't remember exact value but for the laminar regime I choosed a mean of 23 mm between the nodes; take in mind that the distance between nodes is a function of the turbulence model you will use, check the fluent guide for y+ values. 7. All the bests, Daniele P.S. I'm a Sir P.S.2 Can you upload your image in the post above? I can't see it Last edited by ghost82; October 27, 2012 at 03:35. 

October 27, 2012, 03:49 

#9 
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Daniele
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Also,
if you need to refine your grid and you have not a "powerful pc" you can simulate only a slice (for example 60 degrees; the angle must be a divisor of 360 degrees) of the 2 concentric cylinders (see the attached image I found on google, just for reference). In this case you will have to set an additional rotational periodic boundary on the 2 vertical faces; as for the horizontal faces the 2 vertical faces have to be linked before meshing them. 

October 27, 2012, 12:00 

#10 
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Sijal Ahmed Memon (turboenginner@gmail.com)
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At Re = 8000, flow is still considered as taylor couette flow?
http://www.compassis.com/downloads/M...tte%20Flow.pdf 

October 27, 2012, 16:45 

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Sijal Ahmed Memon (turboenginner@gmail.com)
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I want to try this flow. Could you please tell me the following:
1. speed in rad/s 2. density and viscosity 3. length of cylinder 4. inner and outer dia 5. How you define the Reynolds number 6. Flow is laminar or turbulent and what are basis to consider flow laminar or turbulent. I have tried with arbitary values and this is what I get. Last edited by Far; October 27, 2012 at 17:26. 

October 28, 2012, 01:23 

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Quote:
Edit : Not possible in ICEM, should be defined in Fluent using command define>boundaryconditions>modifyzone>makeperiodic Some quick results from the sector cut simulation are given below with same parameters but refined mesh in the axial direction, radial and tangential direction due to only 45 deg geometry. u1 = 20 rad/sec (arbitrary value) After more iterations: Last edited by Far; October 28, 2012 at 02:01. Reason: given in post 

October 28, 2012, 02:29 

#13  
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Quote:
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October 28, 2012, 04:49 

#14  
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these were my parameters: 1. 2.5 rad/s 2. Water: density: 1000 kg/m3, viscosity: 0,001003 Pa*s (default) 3. referring to this last simulation: 0.12 m 4. Inner radius: 0.02 m, outer radius: 0.04 m 5. Re=density*U0*d/viscosity, where: U0 is omega*Inner radius, omega is the angular velocity in rad/s, d is outer radius minus () inner radius. See here: http://www.math.purdue.edu/~sdong/pd...ettejfm07.pdf or here: http://www.mediafire.com/?4varocdijzhfm5g 6. The article above states that Re=1000 is laminar, Re=3000, 5000 and 8000 are turbulent; so I suppose that between Re=1000 and 3000 there's transition regime (?) I don't know if when Re=8000 flow is still considered TaylorCouette, from the article it seems is still considered TaylorCouette; I only tried one simulation with basic scientific literature research. In this last simulation I consider a slice (60 degrees) with a very fine grid (a mean of 0.5 mm with a total of 576000 hexa cells), with a double periodic boundary (rotational around z axis and translational in z direction). I obtained a converged solution when residuals drop under 1*10^7 (monitor some z iso surface to see if converged solution is reached!!) However I have problems in post processing with the fluent built in; I receive this error when:  plot contour  Display>Views>Periodic repeats>define Error: CAR: invalid argument [1]: wrong type [not a pair] Error Object: #f I think this is related with the double periodic boundary (rotational and translational). No problems in post processing as you can see from the pictures below in CFDPOST and Tecplot. Herein below some pictures: Last edited by ghost82; October 28, 2012 at 05:42. 

October 28, 2012, 04:50 

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Daniele
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Some other pictures:


October 28, 2012, 04:53 

#16  
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Daniele
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This is a good question and I don't have an answer. I found this: Fundamental Question about Rotating wall! 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 06:07. 

October 28, 2012, 05:15 

#17 
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Sijal Ahmed Memon (turboenginner@gmail.com)
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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.
Thankyou for your very informative post. 

October 28, 2012, 05:31 

#18  
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Added also a mediafire link. Daniele 

October 28, 2012, 10:20 

#20 
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Sijal Ahmed Memon (turboenginner@gmail.com)
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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/quantumfl...willisphd.pdf http://www.cats.rwthaachen.de:8080/...ette181207.pdf http://www.cats.rwthaachen.de:8080/...ette181207.pdf 

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