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lakeat January 9, 2012 11:40

2D vs 3D
 
Dear all,

I found my simulation of an square cylinder, external flow, the 2D simulation always comes with a low Cd value, like 30%. Is this normal?

Any experience?

Martin Hegedus January 9, 2012 15:18

In regards to drag, I always recommend Hoerner's book "Fluid-Dynamic Drag" It is a treasure chest of information.

In general, 3D shapes have lower drag than 2D shapes. However, it sounds like you may be in an area of great uncertainty. The reason is the separation that occurs at the front lip. If your shape was longer or shorter, for example Length/Diameter (L/D, or fineness ratio) of 4 or 0.1, I'd say the 3D shape would have lower drag. However, at L/D of 1, it is very dependent on how, and if, the flow reattaches.

IMO, you are going to need to build confidence in your solution by starting with a shape with a larger fineness (L/D) ratio. You'll also find, unfortunately, that the transition from low drag to high drag as the fineness ratio is reduced can happen over a short span of fineness ratios.

lakeat January 9, 2012 16:03

Thanks for the help.

Im quite frustrated and exhausted. :)

I find "building confidence" is very important for every CFD people.

Actually, I used to get quite GOOD results for channel flow, circular cylinder flow and 3D simulations results of square cylinder of 22000 compares well with the literature,

but these days I just can't find the "bug" (if there is any) in my bc settings or turbulence modeling, etc. so I am totally lost.


In some CFD paper, some guys use a blockage ratio "correction" to take care their insufficient domain size, do you think it is good idea?

Again, for my square cylinder simulation, 2D, it has 5% blockage ratio, 20B of the wake length, B is the width of the square.. I am testing to see if this is the problematic.

Martin Hegedus January 9, 2012 16:48

In general, for production subsonic runs I'll use an outer boundary 50x-100x of max length for 3D cases and 100x-150x of max length for 2D cases. For 2D, the outer boundary can be less but I'm very hesitant to go less than 50x. If you are concerned about your outer boundary, just double it. Your answer may change by a couple of percentage points, but, I don't think that is your issue. I would suggest trying different fineness ratios and see where you are in regards to the drag rise do to the separation bubble on the top and bottom of the square cylinder.

BTW, when you say 3D (in regards to 2D vs. 3D) square cylinder, I am assuming you mean a cylinder where the axis is aligned with the flow vector and your leading and trailing faces are flat. The 2D representation (i.e. side view) of this would be a simple square and the 2D CFD run would be axisymmetric.

Or, do you mean your CFD runs are 2D and you are comparing it to a 3D wind tunnel test? The square tube for the wind tunnel test would then run from one wind tunnel wall to the one on the other side? If this is the case, and you are at a low Reynolds number, I don't think a 2D CFD run can capture the physics.

lakeat January 9, 2012 17:17

Yes, 3D simulation, I mean the mesh is 3D, spanwise mesh is uniform, geometry is 2D from a side view.

Are you suggesting me testing on a refined mesh? That is what I am doing too.

Will let you know, if I find something. Thanks again.

Martin Hegedus January 9, 2012 18:21

Maybe I miss understood the problem from the beginning. For me, a square cylinder is a circular cylinder where the diameter is equal to the length. Ooops.

For your case, which I assume is at low Reynolds numbers, you can not compare a 2D CFD run of a square to a 3D CFD run, unless your turbulence model somehow captures the physics. And I don't know of one, but I don't know of the latest and greatest research in this area either. If you crank up the Reynolds number, then you probably will get better agreement between a 2D and 3D run.

I assume that your runs are unsteady. So, at any given time, the square will have either an upward or downward lift. I wouldn't be surprised to see that the lift histories for the 2D and 3D cases are significantly different, in the sense of max and mins. In which case the comparison between the 2D and 3D cases for the pressure forces on the forward and aft faces of the square will also be significantly different.

Refining the grid will change the answers, but it will not change the physics.

Now, if your runs are steady, then yes, I would be surprised about the 30% difference. Are your runs steady?

Martin Hegedus January 9, 2012 18:36

Quote:

Originally Posted by lakeat (Post 338533)
Are you suggesting me testing on a refined mesh? That is what I am doing too.

Maybe another misunderstanding. What I really suggest doing is comparing 2D and 3D versions of rectangular cylinders. A square cylinders is a unique case of a rectangular cylinder where Length/Height (fineness ration) is equal to 1.0. Try comparing 2D and 3D for a rectangular cylinder with a fineness ratio of 0.1 and a fineness ratio of 4.0. The square is the hardest one!! The reattachment, or the lack thereof, is a pain.

lakeat January 9, 2012 18:42

Thank you, Martin

Okay, I will elaborate on it. I suppose you are testing and trying OF too, for I have seen your post b4.

Ok, it is an unsteady external flow, I am from civil engineering, which means the Re based on square cylinder edge would be around 10^5 as in a wind tunnel case.

2D in geometry is something just like a bridge deck, because it is very long in its span, so we apply the so-called strip theory, so 2D in geometry!

But the simulation as well as the mesh and turb model could be 2D or 3D.

I think you are very familiar with this field.


As you have said, I noticed, that low-Re-turb-model is an hot topic now, but me either, not sure how it works for a square with Re around 22000. I would like to hear those who have more experience in low-re turb modeling.

What should be defined as low-Re or high :) ??

I am MORE interested in Cd Cl, etc. these integral force coeffs, and I found for bluff body flows, Cl's RMS is quite difficult to generate.

2D-LES generally gives better Cl time history as far as RMS value as concerned, and it looks more like wind tunnel, but 2D-LES doesn't make sense in math point.

lakeat January 9, 2012 18:51

Quote:

Originally Posted by Martin Hegedus (Post 338539)
Maybe another misunderstanding. What I really suggest doing is comparing 2D and 3D versions of rectangular cylinders. A square cylinders is a unique case of a rectangular cylinder where Length/Height (fineness ration) is equal to 1.0. Try comparing 2D and 3D for a rectangular cylinder with a fineness ratio of 0.1 and a fineness ratio of 4.0. The square is the hardest one!! The reattachment, or the lack thereof, is a pain.

Interestingly, it seems from experience and from other publications that Strouhal number is not sensitive to these "local" flow patterns.

I thought mean(Cd) is not sensitive either,
RMS(Cd) and RMS(Cl) would be very sensitive to any settings..

...

OKAY, All I want to ask is :


"How on earth reliable is a 2D(mesh) simulation???"

Martin Hegedus January 9, 2012 20:00

Quote:

Originally Posted by lakeat (Post 338544)
Interestingly, it seems from experience and from other publications that Strouhal number is not sensitive to these "local" flow patterns.

I thought mean(Cd) is not sensitive either,
RMS(Cd) and RMS(Cl) would be very sensitive to any settings..

...

OKAY, All I want to ask is :


"How on earth reliable is a 2D(mesh) simulation???"

I don't think Strouhal number is necessarily linked to CD. But I'm arm waving here, so I'll let others speak up on this.

In regards to mean CD, it is very much sensitive to it. Well, broadly speaking, there are two values, reattached on the top and bottom side, and not reattached. Figure 22 in Hoerner's Fluid Dynamic Drag book shows Cd0 for a rectangular section tested between WT walls vs. c/t where c is the chord (length) and t is the thickness (height). For low c/t values (<1.0), CD0 is about 2.0. Then at around c/t of 1.0 CD0 begins to drop, until it reaches a value of 0.9 to 1.0 somewhere from c/t of 3.0 and 4.0. The maximum rate of change occurs between a c/t of 2.0 and 3.0. Of course this is all dependent of Reynolds number and even possibly the position of the moon and stars.

This reference (http://hdl.handle.net/2060/19930083675), "Low-speed drag of cylinders of various shapes" by Delany, Noel K; Sorensen, Norman E , from ntrs.nasa.gov , shows the Strouhal and CD for rectangles and rounded corner rectangles for fineness rations of 1:2, 1:1, and 2:1.

Given this, I would not trust a 2D simulation for a rectangular cylinder with a value of c/t between 0.5 and 3.5.

lakeat January 9, 2012 20:08

Quote:

Originally Posted by Martin Hegedus (Post 338546)
This reference (http://hdl.handle.net/2060/19930083675), "Low-speed drag of cylinders of various shapes" by Delany, Noel K; Sorensen, Norman E , from ntrs.nasa.gov , shows the Strouhal and CD for rectangles and rounded corner rectangles for fineness rations of 1:2, 1:1, and 2:1.

Given this, I would not trust a 2D simulation for a rectangular cylinder with a value of c/t between 0.5 and 3.5.

Hmm, this makes sense. :D
Within certain Re range, for aspect ratio between 0.5~3.5, 2D-turbulence is far from the truth.

CFD is full of challenges comparing to wind tunnel, when you want to solve some stuff with mesh motion, and solid body motion, it is very difficult to go 3D, it is too consuming. I am getting old to get a reliable FSI result with a 3D simulation.

Martin Hegedus January 9, 2012 21:17

The interesting part is that you are actually trying to model a bridge deck not a rectangular cylinder. There may be cars, girders, guard rails, etc. All this adds turbulence and may cause the flow to reattach earlier than your 3D rectangular cylinder model. Thus the 2D model may actually be closer to the "real" thing. WT of a bridge deck should be investigated and compared to WT data of rectangular cylinders.

Good Luck!

lakeat January 11, 2012 09:52

As Martin has briefly mentioned, I think we have roughly agreed on that:
  1. There ARE many cases where 2D simulation still make SOME sense and are still reliable in SOME way, for example, when the aspect ratio B/H larger than 4.
  2. For the other cases when 2D turbulence prediction is far away from the 3D results, the behavior is unpredictable and uncertain, in other words, it is very risky.

My question is this,
  1. Will 2D URANS simulation make more sense than 2D LES simulation?
  2. Will 2D URANS simulation be more accurate than 2D LES simulation?

JBeilke January 11, 2012 11:11

You want to know if pest is better than cholera.

Both types of calculation are unreliable because you can not say in advance if they give reasonable results for your geometry.

lakeat January 11, 2012 11:20

Then why I found journals and proposal reviewers tend to refuse all 2D-LES, while they seems to be MORE tolerable on allowing the 2D-URANS to join the "comparison" games.

Or I was misinformed?

pete January 11, 2012 11:56

2D LES is physically incorrect. Trying to model the large edies in 2D is not possible since their behaviour and development is very much three dimensional. 2D RANS is not physically incorrect in the same way, but requires certain criteria or assumptions that are often not fulfilled

lakeat January 11, 2012 12:11

Yet see how many papers were published with 2D-LES...


I am more interested in the lift coefficient RMS values than in the Cd.

But I got impression that ALL URANS would give sine wave time history, and low RMS(Cl) values.

Is there any better solution within the scope of 2D (2D-mesh) simulations? That what I am looking hard for.

jola January 11, 2012 12:33

This is not at all my field, but I remember that the Kato-Launder modification was designed to predict this kind of problems for square cylinders. A bit of Googling turned up this reference which looks interesting and is by good authors:

http://www.sciencedirect.com/science...42727X02002102

Martin Hegedus January 11, 2012 13:13

On aspect of this question is how much does one believe in 3D URANS? I don't do a lot of unsteady calculations so I can't really comment. However, IMO, RANS equations throw so much eddy viscosity at a problem I don't know how a vortex would be able to break apart by twisting on itself. For example, take the wake behind a truck or axially aligned circular cylinder, I assume the vortex coming off (for a URANS calc) is going to want to be a ring vortex, and stay that way. Yes, it will be unsteady in the down stream direction, but probably not around the axis. In other words, URANS would be unsteady in a 2D sense only. Again, these are conjectures on my part.

Take rotating helicopter blades for example. I've seen many times where the grid around the blade (these are Chimera grids) use the RANS equations and then there is a cartesian outer grid which is run with Euler (i.e. zero viscosity) One advertised reason they do this is to increase the turn around time of the solution. However, I wouldn't be surprised to learn if there are other benefits, i.e. good bye eddy viscosity. Unfortunately, I never asked them.

Question, the original 3D runs of the square cylinder to which you were comparing your 2D runs to, were they LES or URANS? Sorry, I just assumed they were LES. If they were URANS, I would expect the 2D and 3D URANS runs to be similar. :o

Maybe another modeling choice for you would be to do it like the helicopter people. Create an inner grid and use RANS/URANS on it and create a cartesian outer grid which uses Euler. This would have to be 3D. But, maybe, it runs faster than LES.

lakeat January 11, 2012 13:27

Quote:

Originally Posted by Martin Hegedus (Post 338851)
On aspect of this question is how much does one believe in 3D URANS? I don't do a lot of unsteady calculations so I can't really comment. However, IMO, RANS equations throw so much eddy viscosity at a problem I don't know how a vortex would be able to break apart by twisting on itself. For example, take the wake behind a truck or axially aligned circular cylinder, I assume the vortex coming off (for a URANS calc) is going to want to be a ring vortex, and stay that way. Yes, it will be unsteady in the down stream direction, but probably not around the axis. In other words, URANS would be unsteady in a 2D sense only. Again, these are conjectures on my part.

Take rotating helicopter blades for example. I've seen many times where the grid around the blade (these are Chimera grids) use the RANS equations and then there is a cartesian outer grid which is run with Euler (i.e. zero viscosity) One advertised reason they do this is to increase the turn around time of the solution. However, I wouldn't be surprised to learn if there are other benefits, i.e. good bye eddy viscosity. Unfortunately, I never asked them.

Question, the original 3D runs of the square cylinder to which you were comparing your 2D runs to, were they LES or URANS? Sorry, I just assumed they were LES. If they were URANS, I would expect the 2D and 3D URANS runs to be similar. :o

Maybe another modeling choice for you would be to do it like the helicopter people. Create an inner grid and use RANS/URANS on it and create a cartesian outer grid which uses Euler. This would have to be 3D. But, maybe, it runs faster than LES.



Thank you guys,

Hybrid would be definitely better than (U)RANS, but it is still too costly for large Re. Many hybrid still require a low y_1st_plus be well set, around unity or similar.

Low-Re turb model is very promising, but I am not sure which one is more stable. :) I am testing.

Yes, the 3D simulation I mentioned is LES, I will never do 3D URANS, I don;t think it worth the effort.

And btw, in 3DURANS, the vortices do break in the wake a little bit, but 2D rings persist for a quite long distance, this is what I've seen from many published papers.








These days, I am so interested in 2D URANS' performance, is that I am doing some project, where we have thousands of cases to run, or we have body motion dynamics mesh etc., all of these demand the single run to be as quickly as possible.


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