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March 31, 2016, 07:29 |
Cd vs Re cube (FLUENT)
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#1 |
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Hi,
I want to plot the Cd vs Re relation for a cube. The result has to be something like this: (for a cube, psi=0.806) I'am interested in the region 10^4<Re<10^6, because i expect some drop in the Cd value. unfortunatly i haven't found any scientific experiments that involved this region. However, this paper does the same but for spheres, so I use it as a guide... http://www.dtic.mil/get-tr-doc/pdf?AD=ADA494935 For now i have the following simulation results: The laminar region looks ok. The region of 10^4<Re<10^6 however doesn't show the expected drop for both used turbulence models. The Cd also is a bit to low before the expected drop (it has to be around 1.05, i achieve around 0.95). The setup for the LES is a time step size of 1 [s], 50 time steps, max 4 itt/time step. I don't know if this is a suitable setup... If i plot Cd vs Flow Time for Re=10^5 with this setup i get: and for Re=10^6: I don't understand those fluctuations... Can somebody guide me in the right direction about the next steps? |
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March 31, 2016, 14:15 |
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#2 |
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Cees Haringa
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If the correlation predicts 1.05, I don't think 0.95 is a very poor result. Anyway, your timestep size of 1s is very large for LES, too large I'd say (of course, it depends on the physical size of your domain, if your cube is the size of a city, the story is different than for a cube the size of a lego block)
So, how did you estimate the timescale that you need to resolve? For LES, you should take a timescale that is in line with the scale of motions that you resolve. |
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March 31, 2016, 14:45 |
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#3 |
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uh... the 1s timestep was set by default and i wanted to see how that would work out... I don't know how to setup a suitable timestep size. It depends on the expected period, but i dont understand how the Cd would have a period. I just ran a simulation with a timestep of 0.0001 s, and the Cd converges to 1.6068... while i expected it to be around 0.3<cd<0.6...
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March 31, 2016, 14:54 |
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#4 |
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Cees Haringa
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in your setup it should not be periodic I think (of course, it may be unsteady in a turbulent flow, but that's why you take a long-term average). You can base the timetep size on the courant number, or on a physical parameter. For LES, the Taylor microscale may be an instructive timescale to take. In any case, for LES you can stay above the Kolmogorov scale.
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March 31, 2016, 15:10 |
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#5 |
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Thanks for your help,
What would you suggest for a cube of 1 m^3, air at rho=1.225, mu=1.78*10^-5 and V=14.6 m/s? This gives Re=1*10^6. All the scales you named are very new for me (Taylor, Kolmogorov)+ etc). Time to do some more homework. |
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March 31, 2016, 15:23 |
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#6 |
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Cees Haringa
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Depends on your grid size. For LES, the finer the grid, the more scales are resolved and the fewer modeled - smaller resolved lengthscales, means you need smaller resolved timescales. For LES, there is not that much gained when going below the Taylor scale however - since you essentially resolve all energy containing motions (just not the dissipative ones). But before moving to LES, I'd check how steady state RANS simulations perform (k-e, SST), at least they should be close. And they are much less time consuming.
(actually, to start, maybe an axisymmetric sphere is a good test. These can be solved quickly, and benchmarks are easily found) |
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April 1, 2016, 01:45 |
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#7 |
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Lucky
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Does the cube have a faster transition than a sphere? Is it maybe just transitioning at a much lower Re than expected for a sphere? I don't know. For example, putting like rods in a channel can suppress the critical Re from 2300 all the way down to 500.
The LES results are very questionable but it would take a very lengthy discussion to pick it apart. In short, are you sure you did your LES correctly? 4 iterations per time-step is unlikely to be converged, especially with a timestep of 1s, which is too large to resolve anything. This type of simulation would give a result akin to an unsteady laminar simulation. For LES: You should expect a time-dependent drag coefficient in a turbulent flow, whether or not there is any periodic vortex shedding. Of course the time-averaged value is what you should compare. But I would play with simpler turbulence models before doing any LES. It's too much effort, with too many issues, for what you are trying to do right now. |
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April 1, 2016, 02:57 |
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#8 |
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Thank you for your response,
I'am a real beginner in turbulence simulations. Because the lack of a Cd-drop in the more simple turbulence models (k-e, k-w, SST, reynolds 7 eq) in the domain 10^3<Re<10^6, i switched to LES to give it a first try. But until now this hasn't been a succes ;-) |
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April 1, 2016, 03:27 |
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#9 | |
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muhamed
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Quote:
Could you please tell me how to get a drag coefficient for Laminar flows (Re<2000)? I did a simulation to get the same curve (Re versus CD) on a bridge pier into an open channel and I didn't use the turbulent models in ANSYS FLUENT I just kept it Laminar and I used a flow velocity of 0.0001 m/sec for water but the drag force I got is high compared to the velocity value. I tried several things but I got a reverse flow. Could you help me with this problem? |
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April 1, 2016, 03:34 |
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#10 |
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Cees Haringa
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I'm not sure how well the 2-eq models can predict this dip in drag coefficient. In 3 years of TA-ing on a practical course in CFD (where sphere drag is one of the assignments) I've never seen a group resolve it; results of the best groups typically look like the figure below.
I don't know the nature of your assignment/project, but my recommendation would be to first attain the results for say, k-e and SST, to get a good feeling about how they perform and what they can and cannot resolve. You have to take into consideration that the dip in drag coefficient is only a very weak feature, that may easily drown in the error made by the inherent assumptions of the turbulence models you use. Also, the dip occurs in the range of Re where vortex shedding occurs (for a cylinder at least) - meaning that steady RANS may not capture it, but unsteady RANS might (which can still be done on cruder meshes than LES). Do make sure your timestep is sufficiently small to resolve the shedding (the Cornell university FLUENT tutorial page has tutorials on both steady and unsteady cylinder flow - may be a good guideline). Finally, when using RANS, think about your wall resolution. Using wall functions assumes a fully developed flow at the wall, which will affect your wall shear stress and thereby Cd. Using a fine mesh near the wall so that the boundary layers can be resolved may yield better results here. Last edited by CeesH; April 1, 2016 at 03:34. Reason: forgot picture |
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April 7, 2016, 02:51 |
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#11 | |
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muhamed
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Quote:
I just want to ask if it is possible to use the turbulence models provided by ANSYS FLUENT (rather than the laminar model which is exist also in FLUENT) to do a simulation on a water channel flow with "Laminar flow" with a Reynolds number less than 100? |
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April 11, 2016, 12:17 |
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#12 |
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Cees Haringa
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well technically yes, but why would you want to? the results will be less accurate, as turbulence adds additional dissipation which is not physical. So I would not recommend doing so.
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April 12, 2016, 04:24 |
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#13 |
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muhamed
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Well, if I want to study the flow around a cylinder with different Reynolds number, I will use the laminar model for Re< 2000 and turbulence models for higher Reynolds values. So, is it logical to compare between the results taken from different models, i.e. some results will be taken from the laminar model and the other will be taken from the turbulence models? Regards.
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April 12, 2016, 08:25 |
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#14 | |
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Lucky
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Quote:
There are situations when you might want to to use a turbulent model for a laminar flow and a laminar model for a turbulent flow, but there would be a specific need and specific reason for doing so. |
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April 13, 2016, 04:47 |
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#15 | |
Member
muhamed
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Quote:
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