# Cfd simulation on a smooth cylinder (drag coefficient)

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 October 12, 2016, 21:03 Cfd simulation on a smooth cylinder (drag coefficient) #1 New Member   Simone Join Date: Oct 2016 Posts: 11 Rep Power: 9 Hi to everyone! Now I will explain my problem: -Software: STARCM+ -Goal: find the right value of drag coefficient on my cilinder -Details: u=45.9 m/s, D=1 m, Re=2.9*10^6 - 2D Domain: 10D over the cylinder, 10D belowe the cylinder, 10D at the left of the cylinder, 20D to the right od the cylinder -Boundary conditions: top and bottom planes are symmetry planes, the right plane is pressure outlet, the left plane is velocity inlet I can't find the right value of the drag coefficient on my cylinder!! I am getting crazy since the beginning of Semptember! I have tried both (steady) k-epsilon and k-omega model, but nothing, I have always oscillations (small or large) on the drag coefficients, around 0.4 (that is the right value if I would have a Reynolds of 1*10^6), but I need a value around 0.7 , 0.8 ... Can anyone help me, please?!

 October 13, 2016, 02:52 #2 Super Moderator     Alex Join Date: Jun 2012 Location: Germany Posts: 3,384 Rep Power: 46 Oscillations in a steady-state solution can be a sign that the flow has an unsteadiness that can not be captured correctly by the steady-state approach. In this case this is definitely true. Consequently, at least an unsteady RANS approach would be necessary to get a converged solution. But even then you will not get the correct value for the drag coefficient without tweaking the turbulence model. The correct approach here is a LES.

 October 13, 2016, 03:43 #3 New Member   Simone Join Date: Oct 2016 Posts: 11 Rep Power: 9 Thank you for your answer! After having seen a lot of scientific articles, I know for sure that the only good available methods are the U-RANS and the LES.. And I need to make a comparison between these two method for my master thesis.. I know that by the Strohual number I am able to decide the suitable time-step for my unsteady simulations, so right now I am performing my first unsteady simulation.. do you know how much I have to wai until I get my statistically steady result? Do you have any suggestion to make some improvements on my simulation? I have already a very fine mesh: 20 prism layers, 1E-6 first prism layer to the wall, 0.01 m total thickness prism layer, a wally+ between 0.01 and 0.22, a mesh refinement around the cylinder and on its wake. Maybe something on the initial condition? I don't know, I am desperate.. Thank you very much for your patience!

 October 13, 2016, 05:20 #4 Super Moderator     Alex Join Date: Jun 2012 Location: Germany Posts: 3,384 Rep Power: 46 Let me throw in a few thoughts: Your Y+ values are rather low. Increase the height of the first layer until the maximum y+ value is around 1. The boundary layer will still be resolved accurately but you will have less trouble with high aspect ratio cells away from the wall. And keeping a low volume jump in the transition between prism layers and the rest of the mesh. Keep in mind that a wall-resolving RANS formulation is necessary with y+ values of 1 and below. Don't get me wrong but I heard the phrase "I have already a very fine mesh" numerous times on this site. If you did not estimate the suitable cell sizes in each region of the flow carefully and performed a sensitivity analysis on the cell size this statement is disputable. Feel free to show some images of the mesh you use. The amount of time necessary to reach the steady state is typically in the order of 100 vortex shedding cycles.But do not rely entirely on an initial guess here, simply check your results when they become statistically stationary. Last edited by flotus1; October 13, 2016 at 09:12.

 October 13, 2016, 10:31 #5 Senior Member   Matt Join Date: Aug 2014 Posts: 947 Rep Power: 17 Are you using the force coefficient report to get cd?

 October 13, 2016, 11:14 #6 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,552 Blog Entries: 6 Rep Power: 53 Check these videos. May be useful to you... I will discuss more in detail after you watch these videos... https://www.youtube.com/watch?v=anTkWfMyEPM https://www.youtube.com/watch?v=TFQ_0HaBXXM

October 13, 2016, 12:13
#7
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Simone
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Quote:
 Originally Posted by flotus1 Let me throw in a few thoughts: Your Y+ values are rather low. Increase the height of the first layer until the maximum y+ value is around 1. The boundary layer will still be resolved accurately but you will have less trouble with high aspect ratio cells away from the wall. And keeping a low volume jump in the transition between prism layers and the rest of the mesh. Keep in mind that a wall-resolving RANS formulation is necessary with y+ values of 1 and below. Don't get me wrong but I heard the phrase "I have already a very fine mesh" numerous times on this site. If you did not estimate the suitable cell sizes in each region of the flow carefully and performed a sensitivity analysis on the cell size this statement is disputable. Feel free to show some images of the mesh you use. The amount of time necessary to reach the steady state is typically in the order of 100 vortex shedding cycles.But do not rely entirely on an initial guess here, simply check your results when they become statistically stationary.
Here I atached some pictures of my general mesh, then the detail about the cylinder and the result about the drag coefficient (it is about 0.38.. and I would need a greater value!)
I have used k-epsilon method, with implicit unsteady of 2nd order in time, and a time-step of 0.001 s.
I think that I have kept a low volume jump in the transition regionbetween the layers and the general mesh, as you can see..
Attached Images
 Screenshot 2016-10-13 09.56.58.png (101.4 KB, 54 views) Screenshot 2016-10-13 09.59.40.png (21.5 KB, 37 views) Screenshot 2016-10-13 10.06.04.png (82.3 KB, 36 views)

October 13, 2016, 12:15
#8
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Simone
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Quote:
 Originally Posted by Far Check these videos. May be useful to you... I will discuss more in detail after you watch these videos... https://www.youtube.com/watch?v=anTkWfMyEPM https://www.youtube.com/watch?v=TFQ_0HaBXXM
Thank you very much for the kindness!
As soon as I can, I will see these videos!

October 13, 2016, 12:19
#9
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Simone
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Quote:
 Originally Posted by MBdonCFD Are you using the force coefficient report to get cd?
Yes, why?
Here I atached some pictures of my general mesh, then the detail about the cylinder and the result about the drag coefficient (it is about 0.38.. and I would need a greater value!)
I have used k-epsilon method, with implicit unsteady of 2nd order in time, and a time-step of 0.001 s.
I think that I have kept a low volume jump in the transition regionbetween the layers and the general mesh, as you can see..
After having seen a lot of scientific articles, I know for sure that the only good available methods are the U-RANS and the LES.. And I need to make a comparison between these two method for my master thesis.. I know that by the Strohual number I am able to decide the suitable time-step for my unsteady simulations, so right now I am performing my first unsteady simulation.. do you know how much I have to wai until I get my statistically steady result?
Do you have any suggestion to make some improvements on my simulation?
I have already a very fine mesh: 20 prism layers, 1E-6 first prism layer to the wall, 0.01 m total thickness prism layer, a wally+ between 0.01 and 0.22, a mesh refinement around the cylinder and on its wake.
Maybe something on the initial condition? I don't know, I am desperate..
Thank you very much for your patience!
Attached Images
 Screenshot 2016-10-13 10.06.04.png (82.3 KB, 9 views) Screenshot 2016-10-13 09.59.40.png (21.5 KB, 6 views) Screenshot 2016-10-13 09.56.58.png (101.4 KB, 8 views)

 October 13, 2016, 12:30 #10 Senior Member   Matt Join Date: Aug 2014 Posts: 947 Rep Power: 17 I have run into issues with this before. There is a bug, or as CD-Adapco would call it a philosophical difference that they won't resolve, related to how the issue of unit depth is handled. The coefficient report is a 3D tool. For 2D analysis it assumes you have a depth of 1 m regardless of unit system so the reference area you input needs to reflect this to get the 'correct' coefficient back out of it. Now that being said, it sounds like you are already in SI units so this may not be the issue you are experiencing. However, given that this bug exists I always shy away from using the coefficient reports and extract force data directly to calculate my own coefficients. It is easier to track down why a coefficient is wrong if you can look and/or control all of the inputs directly. Try calculating Cd yourself from the drag force and see what you get. There is a good chance you will get a different and possibly better answer.

 October 13, 2016, 12:40 #11 New Member   Simone Join Date: Oct 2016 Posts: 11 Rep Power: 9 Yes, I know what you are talking about and I have tried few days ago to calculate the drag coefficient like you said, but I've obtained the same result..

October 13, 2016, 15:20
#12
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Sijal
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Quote:
 Originally Posted by Simone94 Ok, thank you for your advices! Here I atached some pictures of my general mesh, then the detail about the cylinder and the result about the drag coefficient (it is about 0.38.. and I would need a greater value!) I have used k-epsilon method, with implicit unsteady of 2nd order in time, and a time-step of 0.001 s. I think that I have kept a low volume jump in the transition regionbetween the layers and the general mesh, as you can see.. I wait for your answer, thank you very much again!

what reference values you are using for the Cd?

how did you decide this time step? in how many time steps you are resolving one vortex shedding frequency??

keep Yplus at 0.5-0.8...

October 13, 2016, 16:21
#13
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Simone
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Quote:
 Originally Posted by Far what reference values you are using for the Cd? how did you decide this time step? in how many time steps you are resolving one vortex shedding frequency?? keep Yplus at 0.5-0.8...
I am using as reference values: 101325 Pa, 1,18415 kg/m^3, reference area: 1 m^2 (my diameter is 1 m, so in 2D simulation they coincide), free-stream velocity 45.9 m/s ...
Since the Strohual number should be around 0.25, re-expressing the formula I obtain the inverse of the frequency, that is the period one vortex, and that will be the maximum time-step! However, I resolve in 20, 30 times one vortex shedding..
I will stop the simulation when the solution become statistically steady..
Honestly, I have the y+ quite low (<0.2)..

October 13, 2016, 17:05
#14
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Sijal
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Quote:
 that is the period one vortex, and that will be the maximum time-step!
that can be quit dangerous

Quote:
 I will stop the simulation when the solution become statistically steady..
I guess that is the time when you start to observe key parameters and take the time mean of values such as Cd...

October 13, 2016, 17:23
#15
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Simone
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Quote:
 Originally Posted by Far that can be quit dangerous I guess that is the time when you start to observe key parameters and take the time mean of values such as Cd...

October 13, 2016, 18:42
#16
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Quote:
 Originally Posted by Simone94 I didn'the understand both of your statements.. explain yourself better, please
Quote:
 that is the period one vortex, and that will be the maximum time-step!
put in this way : you should divide your time for one vortex shedding frequency into 20, 40 or 80 intervals and see which one gives you results which does not change with decreasing time step size.

Quote:
 I guess that is the time when you start to observe key parameters and take the time mean of values such as Cd...
You should run your solution up to point when the solution becomes periodic with time. This can be achieved after a long time, so patience. Once this is achieved next thing is to activate the monitors for the lift and drag. This is usually done for 5 cycles, but can be increased to increase accuracy.

Let say for each cycle, you have 20 time steps. so for 5 cycles it is 100 time steps and 100 values for Cd. Just sum all values of Cd and divide sum by 100. that is your time mean drag.

For Cl, practice is different. Normally people quote the root mean square value, because simple mean will give you zero Cl for symmetric bodies such as cylinder.

October 13, 2016, 20:56
#17
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Simone
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Quote:
 Originally Posted by Far put in this way : you should divide your time for one vortex shedding frequency into 20, 40 or 80 intervals and see which one gives you results which does not change with decreasing time step size. You should run your solution up to point when the solution becomes periodic with time. This can be achieved after a long time, so patience. Once this is achieved next thing is to activate the monitors for the lift and drag. This is usually done for 5 cycles, but can be increased to increase accuracy. Let say for each cycle, you have 20 time steps. so for 5 cycles it is 100 time steps and 100 values for Cd. Just sum all values of Cd and divide sum by 100. that is your time mean drag. For Cl, practice is different. Normally people quote the root mean square value, because simple mean will give you zero Cl for symmetric bodies such as cylinder.
Ah ok! Thank you very much, I have understood what you said! The only problem is that the residuals are too high, for example in that simulation stated before.. And I don't know why..
However, I will try to apply your advices and I will let you know

 October 14, 2016, 04:16 #18 Super Moderator     Alex Join Date: Jun 2012 Location: Germany Posts: 3,384 Rep Power: 46 The residuals are too high? This means that the results we are discussing here are not converged? Run more iterations per time step. You should also have a look at these resources: http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F http://www.cfd-online.com/Forums/flu...nvergence.html

 October 14, 2016, 05:25 #19 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,552 Blog Entries: 6 Rep Power: 53 or alternatively reduce the time step by order of magnitude (means divide your current time step by 10, two orders means divide by 100 and so on...) Example : current time step Delta T = 0.05 order of magnitude reduction = 0.05 / 10 = 0.005 Two orders of magnitude reduction = 0.05 / 100 = 0.0005

October 14, 2016, 19:14
#20
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Simone
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Quote:
 Originally Posted by flotus1 The residuals are too high? This means that the results we are discussing here are not converged? Run more iterations per time step. You should also have a look at these resources: http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F http://www.cfd-online.com/Forums/flu...nvergence.html
Ok, thank you
Now I have almost finished my time-step independence study and the residuals are periodically converged, and the same for my drag coefficient.. what I'm worried about is the fact that, smaller the time-step is, smaller the drag coefficient.. I know that I have to do the space grid-indepence study yet, but it is not so comforting..
I would desire my drag coefficient to grow up towards the right value.. XD

 Tags cfd, cylinder, drag coefficient