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November 26, 2014, 12:22 |
Vortices in CFX
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#1 |
New Member
Rishikesh Prashant
Join Date: Nov 2014
Posts: 6
Rep Power: 12 |
I am trying run of Taylor-Couette in ANSYS CFX and have been trying to get vortices in the annular gap. My inner cylinder rotates at 10rad/s which is a Reynolds number of approx 1000.
I get 2 vortices - one near the upper stationary end wall and one near the lower wall - also stationary. However I should be seeing counter rotating vortices through entire length of the cylinder. Any help would be appreciated, have been working on this for almost 2 month now. Thanks. Rishi |
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November 26, 2014, 17:23 |
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#2 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,872
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Have you run your simulation for enough physical time? Your flow looks like it has just started up, so it probably just needs to run longer.
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November 27, 2014, 14:26 |
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#3 |
New Member
Rishikesh Prashant
Join Date: Nov 2014
Posts: 6
Rep Power: 12 |
Thanks Ghorrocks for the reply. I should have mentioned that the picture I posted was indeed a transient simulation. I set the time per run to 100s and the timestep per run as 10s.
Do you think it needs to run for a longer period of time? |
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November 27, 2014, 18:11 |
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#4 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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Have a look at how the flow is evolving in time. If it has not reached a steady state or fully developed then it needs to run longer.
In your case I suspect the vorticies grow in from the edges, so you need to run it longer to get the vorticies to grow to fill the entire domain. |
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November 28, 2014, 09:01 |
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#5 |
New Member
Rishikesh Prashant
Join Date: Nov 2014
Posts: 6
Rep Power: 12 |
I have now run a simulation for 800s with timestep of 10s but it still gives me the same result. I also tried running the simulation but with a much higher angular velocity of 90 rad/s and i managed to get the taylor vortices.
But my simulation is for a bioreactor for which I want to remain in the laminar regime. I have read papers which states that you should see the taylor vortices form in laminar region above Re of around 100 |
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November 28, 2014, 11:51 |
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#6 |
Super Moderator
Alex
Join Date: Jun 2012
Location: Germany
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Are you sure that 10 seconds is an appropriate time step size? How did you estimate that?
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November 28, 2014, 13:06 |
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#7 |
New Member
Rishikesh Prashant
Join Date: Nov 2014
Posts: 6
Rep Power: 12 |
I know the timestep should be smaller ideally but otherwise then my simulation takes too long.
I tried running a simulation for a shorter overall period of time but with timestep of 0.1s and I still got the same result. Just out of curiosity I am going into the turbo tab - defining the rotation axis and then clicking calculate velocity components - then plotting the radial velocity on the sliced plane. I believe this is what should show me the vortices |
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November 30, 2014, 17:39 |
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#8 | |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,872
Rep Power: 144 |
Quote:
While you are at it you should check your convergence criteria and mesh density. You need to do a sensitivity analysis to determine what time step size you need. Do not guess the time step size, you will gt it wrong. And definitely do not constrain your time step size by the computing resources you have available. |
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November 30, 2014, 20:15 |
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#9 |
New Member
Rishikesh Prashant
Join Date: Nov 2014
Posts: 6
Rep Power: 12 |
I ran the new simulation with a much smaller timestep of 0.1s which gave me Courant number of around 5 --> I believe this is acceptable from what I have seen. But still no luck. My mesh is 512k nodes and it is quite a small device.
I would like the accurate solution of course! But from everything I have read Taylor vortex flow is not time dependent (wavy vortex flow is) so maybe something else wrong. At the moment I am experimenting with the convergence criteria (default i believe is RMS = 1x10-4). Perhaps this will give me my solution. Many thanks for all your advice I am still learning so making mistakes. Last edited by Rishi91; December 1, 2014 at 11:48. |
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November 30, 2014, 20:47 |
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#10 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,872
Rep Power: 144 |
CFX is an implicit solver so the Courant number is not a good reference value to determine time step size.
The way to determine suitable time step size is by doing a time step sensitivity analysis. Do a simulation at a certain time step and extract the key numerical values which are the result of the simulation (it could be heat transfer, pressure drop, minimum pressure, whatever is important for you). Then repeat the simulation with a time step of half the size. Compare the key numerical values - are they the same within a tolerance you are happy with? If not you have to keep making the time step size smaller (by factors of 2) until they are the same within a tolerance you are happy with. A very common beginner mistake is to guess a time step and justify it by saying it looks really small. The result is an inaccurate simulation as the time step size needs to be adequate to resolve the temporal transient in the simulation and guesses are rarely adequate. Also comparing time step size to other simulations needs to be done carefully as they will use different numerics and different numerics needs different resolutions. |
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December 1, 2014, 13:33 |
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#11 |
Super Moderator
Alex
Join Date: Jun 2012
Location: Germany
Posts: 3,428
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To get an initial guess for the time scale of your problem divide the width of the gap between inner and outer cylinder by the tangential speed of the inner cylinder wall.
Divide this time by ten to get an initial guess for a suitable time step size to start your sensitivity analysis. Multiply it by ten to get the total simulation time. If the solution takes too long use a coarser mesh. Right now your spatial resolution is way better than the temporal resolution. You should try to balance them. If you still dont get the result you expect initialize your transient simulations with the result of a steady-state simulation. |
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