Flow over circular cylinder

anuprdk · February 10, 2014, 16:52

Hi,

I am a beginner to Fluent user trying to model the flow past a circular cylinder at Re=1500 (v=0.01 m/s, diameter=0.15m, fluid - water). The boundary conditions I have put in for the rectangular domain of flow are velocity inlet, pressure outlet and the two sides as walls. The turbulence model chosen is the Standard K-e model. I am trying to simulate the Karman street observed in this kind of Transient flows but I do not observe them being shed.

Is it something wrong with my Solver settings, a snapshot of which has been attached? Also I ran the simulation for 40 sec with a time step of 0.1 sec. My mesh is also relatively fine. Can anyone please drop in a suggestion???

Thank you,

Anup

Velocity profile at v=0.01.jpg

Rect Mesh.png

Solver.jpg

flotus1 · February 10, 2014, 18:38

For low the flow around a circular cylinder at low Reynolds numbers, the standard values for turbulent kinetic energy and dissipation at velocity inlets in fluent are way too high.
This produces excessive turbulent viscosity that prevents vortex shedding.
Actually these values are too high for most external flows.

Switch the turbulence values at the inlet to turbulence intensity and length scale and choose an intensity that better represents the undisturbed upstream flow. Values between 1% and 0.01% should do.
Make sure that the temporal scheme is second order accurate, the first order scheme induces too much damping.
If you still dont get a vortex street, try the k-omega SST turbulence model.

If you are only interested in colorfull pictures and videos you can even run the simulation without a turbulence model.

Forgot to mention: 40 seconds are simply not enough time to develop a vortex street with these settings, no matter what you do with the inlet turbulent quantities.
With a Strouhal number of approximately 0.2, the frequency of the vortex shedding will be 0.013 1/s. That is 75 seconds for one period time.
You should choose a time step size in the order of 75s/40 to simulate a reasonable amount of time.

anuprdk · February 17, 2014, 13:03

Quote:

Originally Posted by flotus1

For low the flow around a circular cylinder at low Reynolds numbers, the standard values for turbulent kinetic energy and dissipation at velocity inlets in fluent are way too high.
This produces excessive turbulent viscosity that prevents vortex shedding.
Actually these values are too high for most external flows.

Switch the turbulence values at the inlet to turbulence intensity and length scale and choose an intensity that better represents the undisturbed upstream flow. Values between 1% and 0.01% should do.
Make sure that the temporal scheme is second order accurate, the first order scheme induces too much damping.
If you still dont get a vortex street, try the k-omega SST turbulence model.

If you are only interested in colorfull pictures and videos you can even run the simulation without a turbulence model.

Forgot to mention: 40 seconds are simply not enough time to develop a vortex street with these settings, no matter what you do with the inlet turbulent quantities.
With a Strouhal number of approximately 0.2, the frequency of the vortex shedding will be 0.013 1/s. That is 75 seconds for one period time.
You should choose a time step size in the order of 75s/40 to simulate a reasonable amount of time.

Hello Sir,

This simulation is for my 3rd year project in Fluid Dynamics. Actually I am supposed to determine the drag force for a 3d cylinder under similar conditions. But to start off I have commenced with a 2d model.

I tried using all possible values of turbulent intensity and turbulent length scale but in vain. Attached herewith is the contour of Vorticity Magnitude for a turbulent intensity of 0.01% and turbulent length scale of 0.0057 for a SST K-omega model.

As per your suggestion I used a time step of 0.075 sec and ran it for a 1000 runs.

Please advise.

thanking you,

Anup Radhakrishnan

V 0 01.jpg

flotus1 · February 17, 2014, 13:52

Again, 1000 time steps with a time step size of 0.075s is only 75 seconds, which is way too short to develop a vortex strees with a period time of 75s.

Use the time step size I recommended (75s/40=1.875s) and run for 1000 time steps.

Barzanoni · February 20, 2014, 07:27

hi
i think that your time step is large, try some lower time scale.

flotus1 · February 20, 2014, 09:55

What estimation brought you to this conclusion?

Edit: another thing I forgot to mention: use at least a second order upwind scheme for the convective fluxes and a second order implicit scheme for the time discretization.
THis is not specific to the problems you are encountering here, but applies to nearly all simulations.

anuprdk · February 24, 2014, 07:20

Hi again,

I have tried out the suggestions you had suggested and yes I am getting a Vortex street. The drag coefficient comes to around 0.4 which does not tally for this Reynolds number.

I read a post about the need for boundary layer settings and grid alternation for these kind of problems. I do not know how to do that. I thought boundary layer would have been self defined with the boundary conditions we input.

Please advise.

flotus1 · February 24, 2014, 07:49

http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F

Dont expect to get good agreement with the standard k-epsilon model.
Getting drag coefficient values that are in agreement with measurements is harder than it looks for a simple cylinder.
I dont really know what grid alternation could mean, but there are quite a few tutorials on this site and all over the web that show how a mesh for a cylinder is created.

anuprdk · February 27, 2014, 22:06

Hi,

Can any one please help me with this drag coefficient!!! Just to make things easier I am running the simulation at Re=100 which is completely laminar. I should be getting a Cd somewhere close to 1 but I still get the value as 0.4. The time step I have chosen is 0.01 and I ran it for 100 seconds but of no avail.

Please I need help in this. It has been a month trying to figure out this problem!!

Thanks,

Anup

flotus1 · February 28, 2014, 04:26

No one can help you with this If you keep ignoring the advice you are given.
I dont know how you lowered the Reynolds number, but no matter how you did it your results still depend more on the initial conditions than on the flow itself.
If you plot the value of the drag coefficient over time (what you should do anyway to calculate a time-averaged value) you will see what I mean.
Just to make sure: You did set the reference values for velocity, density and area in Fluent to match the properties of your flow?

So one last time: estimate the time scale of your flow problem using the Strouhal number, choose a time step size 20-40 times smaller and simulate a sufficient amount of time.
This time has to be long enough to allow the flow become statistically steady, which will take somewhere between 10-100 times the time scale.
Use the plot drag coefficient over time to determine the exact time. After an initial transition it should oscillate around a constant value when the flow is statistically steady.

anuprdk · February 28, 2014, 09:56

Sir,

I had followed your advise with respect to the model I was working on (Re=1500). But since I did not get the expected drag, I reduced my flow speed to get Re=100 (Laminar). There are numerous technical papers available for this Re which would be easy to compare with.

What does reference area mean? I have attached with a snap shot of what is present in my case. I have not played around with this. Please advise.

Thank you so much for all the help,

Anupreference.png

flotus1 · February 28, 2014, 10:18

I just wanted to make sure you changed the reference values to match your simulation setup.
Remember: $c_d = \frac{F_d}{\frac{1}{2} \rho u^2 A}$
Since Fluent cant guess which values you had in mind, you need to provide them as reference values.
Make sure you adjust them correctly to whatever values you have for the new Reynolds number.
If you want to compare your results to some paper, use the same reference values they used.
If they dont give the reference values they used, throw away the paper.

Then go ahead and follow my advice in the last post.

ghobold · February 28, 2014, 10:19

Reference area means the area from which adimensional quantities will be calculated from. Drag coefficient, for example, is defined by

$C_D = \frac{F_D}{\frac{1}{2}\rho U^2 A}$

where $\rho$ is the refence density,

is the reference velocity and

is the reference area. For cylinders, you will find out that most available data use the projected area instead of the total area of the cylinder when calculating the drag coefficient. Still, you have to check the data to see how they use their reference values if you want to do a valuable comparison, so you can use the same references in yours.

Far · March 1, 2014, 05:47

Please have a look at the thread located here : http://www.cfd-online.com/Forums/ans...linders-3.html

anuprdk · March 13, 2014, 12:36

Hi,

Thank you for all the advice. I am getting the Street and the expected Drag coefficient. But what I do not understand is the flow time. For a Re=1500 (v=0.01 m/s) the time step is 1.875 and I need to run the simulation for 1000 time steps. For Re=150 (v=0.001 m/s) I chose a time step of 1.933 and had to run the same for around 5000 time steps to see some shedding.

What exactly is the concept of time steps? Does it have any relation to the time taken for the flow commencing from inlet to reach the outlet?

flotus1 · March 18, 2014, 12:04

What you observed in your simulation is more related to physics than to a numerical concept.
For the unsteady flow around a bluff body, the Strouhal number describes the frequency of the vortex shedding:
$Sr = \frac{f \cdot D}{u}$ .
Here f is the frequency , D is the characteristic length (diameter in our case) and u is the velocity of the undisturbed flow.
The Strouhal number is in the order of 0.2 over a wide range of Reynolds numbers.
Conversely, the period time of the vortex shedding
$T = \frac{1}{f} = \frac{Sr \cdot D}{u}$
will strongly depend on u since Sr is nearly constant.
So if you decrease your Reynolds number by decreasing u, the period time of the vortex shedding will increase.
The time it takes for the vortex street to develop is linked to the period time.
Thats why you need to simulate a physical time that is 10 times higher when you lower the Reynolds number the way you did it (through u).
And that is why i have been estimating time step sizes since the beginning of this thread.
Adjusting the time step size to a fraction of the period time allows you to simulate the flow with somewhat arbitrary Reynolds numbers with the same amount of time steps.

kaabya · April 13, 2014, 00:31

Hey,

It is an interseting discussion about the reference values. I have the same problem but with a more complicated geometry of a parabolic trough. Even I fixed my area to the frontal area of my system, I found a strange results for drag coefficient?

If there is any relation with the Reynolds number and the time of simulation ? How Can I control those parameter? For first time I simulated a laminar steady flow.

Thank you

kazemiakk · September 25, 2014, 17:03

Quote:

Originally Posted by flotus1

Again, 1000 time steps with a time step size of 0.075s is only 75 seconds, which is way too short to develop a vortex strees with a period time of 75s.

Use the time step size I recommended (75s/40=1.875s) and run for 1000 time steps.

Hi,
I modeling the same problem in fluent in Re=120 but I can not capture vortex shedding in my results. I changed time steps and number of time steps, but still have the same problem and everything is steady. I mean it start from zero and Vorticity profile starts to develop but not shed . what should I change?

flotus1 · September 26, 2014, 10:53

Did you read, understand and follow the advice given here carefully?
If so, give more information about what you did so far and the simulation parameters you used.

kazemiakk · September 26, 2014, 18:26

Quote:

Originally Posted by flotus1

Did you read, understand and follow the advice given here carefully?
If so, give more information about what you did so far and the simulation parameters you used.

Yes I Did. but everything in my case is growing the vorticty to fully develop but no vortex shedding. I used the following parameters:
D=0.22 m
Re=1200
V=0.098 m/s
dynamic viscosity: 0.018 kg/ms
time step: 0.3
Number of time step:4000

February 10, 2014, 16:52	Flow over circular cylinder	#1
anuprdk New Member anup radhakrishnan Join Date: Jan 2014 Posts: 24 Rep Power: 12	Hi, I am a beginner to Fluent user trying to model the flow past a circular cylinder at Re=1500 (v=0.01 m/s, diameter=0.15m, fluid - water). The boundary conditions I have put in for the rectangular domain of flow are velocity inlet, pressure outlet and the two sides as walls. The turbulence model chosen is the Standard K-e model. I am trying to simulate the Karman street observed in this kind of Transient flows but I do not observe them being shed. Is it something wrong with my Solver settings, a snapshot of which has been attached? Also I ran the simulation for 40 sec with a time step of 0.1 sec. My mesh is also relatively fine. Can anyone please drop in a suggestion??? Thank you, Anup Velocity profile at v=0.01.jpg Rect Mesh.png Solver.jpg

February 10, 2014, 18:38		#2
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	For low the flow around a circular cylinder at low Reynolds numbers, the standard values for turbulent kinetic energy and dissipation at velocity inlets in fluent are way too high. This produces excessive turbulent viscosity that prevents vortex shedding. Actually these values are too high for most external flows. Switch the turbulence values at the inlet to turbulence intensity and length scale and choose an intensity that better represents the undisturbed upstream flow. Values between 1% and 0.01% should do. Make sure that the temporal scheme is second order accurate, the first order scheme induces too much damping. If you still dont get a vortex street, try the k-omega SST turbulence model. If you are only interested in colorfull pictures and videos you can even run the simulation without a turbulence model. Forgot to mention: 40 seconds are simply not enough time to develop a vortex street with these settings, no matter what you do with the inlet turbulent quantities. With a Strouhal number of approximately 0.2, the frequency of the vortex shedding will be 0.013 1/s. That is 75 seconds for one period time. You should choose a time step size in the order of 75s/40 to simulate a reasonable amount of time. nima_nzm, Chris_321, pakk and 5 others like this. Last edited by flotus1; February 11, 2014 at 04:18.

February 20, 2014, 09:55		#6
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	What estimation brought you to this conclusion? Edit: another thing I forgot to mention: use at least a second order upwind scheme for the convective fluxes and a second order implicit scheme for the time discretization. THis is not specific to the problems you are encountering here, but applies to nearly all simulations. Last edited by flotus1; February 21, 2014 at 12:43.

March 18, 2014, 12:04		#16
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	What you observed in your simulation is more related to physics than to a numerical concept. For the unsteady flow around a bluff body, the Strouhal number describes the frequency of the vortex shedding: $Sr = \frac{f \cdot D}{u}$ . Here f is the frequency , D is the characteristic length (diameter in our case) and u is the velocity of the undisturbed flow. The Strouhal number is in the order of 0.2 over a wide range of Reynolds numbers. Conversely, the period time of the vortex shedding $T = \frac{1}{f} = \frac{Sr \cdot D}{u}$ will strongly depend on u since Sr is nearly constant. So if you decrease your Reynolds number by decreasing u, the period time of the vortex shedding will increase. The time it takes for the vortex street to develop is linked to the period time. Thats why you need to simulate a physical time that is 10 times higher when you lower the Reynolds number the way you did it (through u). And that is why i have been estimating time step sizes since the beginning of this thread. Adjusting the time step size to a fraction of the period time allows you to simulate the flow with somewhat arbitrary Reynolds numbers with the same amount of time steps. Far, DWPQ, karachun and 2 others like this.

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February 17, 2014, 13:52		#4
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	Again, 1000 time steps with a time step size of 0.075s is only 75 seconds, which is way too short to develop a vortex strees with a period time of 75s. Use the time step size I recommended (75s/40=1.875s) and run for 1000 time steps.

February 20, 2014, 07:27		#5
Barzanoni New Member Yaser Join Date: May 2013 Location: I. R. Iran, Tehran Posts: 20 Rep Power: 13	hi i think that your time step is large, try some lower time scale.

February 24, 2014, 07:20		#7
anuprdk New Member anup radhakrishnan Join Date: Jan 2014 Posts: 24 Rep Power: 12	Hi again, I have tried out the suggestions you had suggested and yes I am getting a Vortex street. The drag coefficient comes to around 0.4 which does not tally for this Reynolds number. I read a post about the need for boundary layer settings and grid alternation for these kind of problems. I do not know how to do that. I thought boundary layer would have been self defined with the boundary conditions we input. Please advise.

February 24, 2014, 07:49		#8
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F Dont expect to get good agreement with the standard k-epsilon model. Getting drag coefficient values that are in agreement with measurements is harder than it looks for a simple cylinder. I dont really know what grid alternation could mean, but there are quite a few tutorials on this site and all over the web that show how a mesh for a cylinder is created.

February 27, 2014, 22:06		#9
anuprdk New Member anup radhakrishnan Join Date: Jan 2014 Posts: 24 Rep Power: 12	Hi, Can any one please help me with this drag coefficient!!! Just to make things easier I am running the simulation at Re=100 which is completely laminar. I should be getting a Cd somewhere close to 1 but I still get the value as 0.4. The time step I have chosen is 0.01 and I ran it for 100 seconds but of no avail. Please I need help in this. It has been a month trying to figure out this problem!! Thanks, Anup

February 28, 2014, 04:26		#10
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	No one can help you with this If you keep ignoring the advice you are given. I dont know how you lowered the Reynolds number, but no matter how you did it your results still depend more on the initial conditions than on the flow itself. If you plot the value of the drag coefficient over time (what you should do anyway to calculate a time-averaged value) you will see what I mean. Just to make sure: You did set the reference values for velocity, density and area in Fluent to match the properties of your flow? So one last time: estimate the time scale of your flow problem using the Strouhal number, choose a time step size 20-40 times smaller and simulate a sufficient amount of time. This time has to be long enough to allow the flow become statistically steady, which will take somewhere between 10-100 times the time scale. Use the plot drag coefficient over time to determine the exact time. After an initial transition it should oscillate around a constant value when the flow is statistically steady.

February 28, 2014, 09:56		#11
anuprdk New Member anup radhakrishnan Join Date: Jan 2014 Posts: 24 Rep Power: 12	Sir, I had followed your advise with respect to the model I was working on (Re=1500). But since I did not get the expected drag, I reduced my flow speed to get Re=100 (Laminar). There are numerous technical papers available for this Re which would be easy to compare with. What does reference area mean? I have attached with a snap shot of what is present in my case. I have not played around with this. Please advise. Thank you so much for all the help, Anupreference.png

February 28, 2014, 10:18		#12
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	I just wanted to make sure you changed the reference values to match your simulation setup. Remember: $c_d = \frac{F_d}{\frac{1}{2} \rho u^2 A}$ Since Fluent cant guess which values you had in mind, you need to provide them as reference values. Make sure you adjust them correctly to whatever values you have for the new Reynolds number. If you want to compare your results to some paper, use the same reference values they used. If they dont give the reference values they used, throw away the paper. Then go ahead and follow my advice in the last post.

February 28, 2014, 10:19		#13
ghobold New Member Gustavo Join Date: Jun 2013 Posts: 26 Rep Power: 13	Reference area means the area from which adimensional quantities will be calculated from. Drag coefficient, for example, is defined by $C_D = \frac{F_D}{\frac{1}{2}\rho U^2 A}$ where $\rho$ is the refence density, is the reference velocity and is the reference area. For cylinders, you will find out that most available data use the projected area instead of the total area of the cylinder when calculating the drag coefficient. Still, you have to check the data to see how they use their reference values if you want to do a valuable comparison, so you can use the same references in yours.

March 1, 2014, 05:47		#14
Far Senior Member Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,558 Blog Entries: 6 Rep Power: 54	Please have a look at the thread located here : http://www.cfd-online.com/Forums/ans...linders-3.html

March 13, 2014, 12:36		#15
anuprdk New Member anup radhakrishnan Join Date: Jan 2014 Posts: 24 Rep Power: 12	Hi, Thank you for all the advice. I am getting the Street and the expected Drag coefficient. But what I do not understand is the flow time. For a Re=1500 (v=0.01 m/s) the time step is 1.875 and I need to run the simulation for 1000 time steps. For Re=150 (v=0.001 m/s) I chose a time step of 1.933 and had to run the same for around 5000 time steps to see some shedding. What exactly is the concept of time steps? Does it have any relation to the time taken for the flow commencing from inlet to reach the outlet?

April 13, 2014, 00:31		#17
kaabya New Member bassem Join Date: Apr 2014 Posts: 6 Rep Power: 12	Hey, It is an interseting discussion about the reference values. I have the same problem but with a more complicated geometry of a parabolic trough. Even I fixed my area to the frontal area of my system, I found a strange results for drag coefficient? If there is any relation with the Reynolds number and the time of simulation ? How Can I control those parameter? For first time I simulated a laminar steady flow. Thank you

September 26, 2014, 10:53		#19
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,427 Rep Power: 49	Did you read, understand and follow the advice given here carefully? If so, give more information about what you did so far and the simulation parameters you used.