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Strange flow partern (Reverse Flow) in fluid past circular cylinder problem at exit

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Old   June 6, 2016, 17:27
Default Strange flow partern (Reverse Flow) in fluid past circular cylinder problem at exit
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Hector Redal
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Hi,

I would like to comment some strange pattern I am observing when simulating a flow past a circular cylinder.
I have detected that at the exit I see that the flow enters the domain at the upper side of the exit and exits at the lower side of the exit. I would expect that the fluid exits the domain accross the exit.

I am attaching an image where this strange pattern happens.

The boundary conditions I am using at the exit are du/dx = 0, dv/dx = 0 and dp/dx=0.

Additionally I am using:
Reynolds number = 10
Richardson number = 1.0
gravitiy force downwards (negative value of y-direction).
Fluid is air.
The cylinder is hotter than the air.
I am trying to analyze the bouyancy effects caused by the hot cylinder.

If needed I can provide more details of the simulation.

Best regards,
Hector.
Attached Images
File Type: jpg ReverseFlowPastCylinder.jpg (71.8 KB, 30 views)
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Old   June 6, 2016, 17:32
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Filippo Maria Denaro
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The flow pattern appears too strange everywhere, not only at the exit...

Are you using a 2D case? Because if you fix du/dx=0 at the exit, then the continuity constraint at the exit becomes dv/dy=0, therefore the v component must be constant (zero if it is zero at some points)
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Old   June 6, 2016, 17:54
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Hector Redal
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Hi Filippo

You are right. That's a good point.
I will check it.

Thanks,
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Old   June 6, 2016, 18:05
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Hector Redal
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By the way, two questions:

Are the boundary conditions du/dx = dv/dy = dp/dx = 0 the correct ones to impose at the exit?
Or should I impose different boundary conditions?
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Old   June 7, 2016, 00:30
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I would move the outlet further away. You are imposing a boundary condition near a region where there is a lot of fluid interaction.
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Old   June 7, 2016, 17:34
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Hector Redal
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I have a question in regards, with the boundary constrains at the exit.
In this type of simulations, the velocity vy at the exit usually shows a periodic pattern, which oscilates around a mean value which is zero.
The usually constraint imposed at the exit is du/dx = dv / dx = 0.
According to your comment if du/dx= 0, due to the continuity equation, the dv/dy must be also 0, which means constant value. But this is not the case since the velocity profile at the exit oscilates. How can this be?
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Old   June 7, 2016, 17:48
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Quote:
Originally Posted by HectorRedal View Post
I have a question in regards, with the boundary constrains at the exit.
In this type of simulations, the velocity vy at the exit usually shows a periodic pattern, which oscilates around a mean value which is zero.
The usually constraint imposed at the exit is du/dx = dv / dx = 0.
According to your comment if du/dx= 0, due to the continuity equation, the dv/dy must be also 0, which means constant value. But this is not the case since the velocity profile at the exit oscilates. How can this be?

but the fixed condition du/dx= 0 (that drives to dv/dy=0) is your choice, that is an arbitrary boundary condition, therefore if you have at the exit an oscillating flow condition you cannot use du/dx=0. However, I suppose that if you increease the lenght of the domain sufficiently, the physical flow oscillations will be damped and the condition du/dx=0 could be acceptable. Note that the consequence that v=constant along y let such constant to vary in time.
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Old   June 9, 2016, 17:04
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Hector Redal
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Yes, you are right, the boundary condition at the exit is my choice. But it is the usually boundary conditions imposed in this kind of problems (Flow past circular cylinder).

What I think is that one thing is dv/dy = 0 and other different thing is dv/dt = 0.

In my opinion, dv/dt is different of zero because v oscilates among a mid point. But, if dv/dy = 0, I understand that all the points at the exit should have the same v value, but oscillating all them together.

Could it be possible? I think so.
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Old   June 9, 2016, 17:24
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Quote:
Originally Posted by HectorRedal View Post
Yes, you are right, the boundary condition at the exit is my choice. But it is the usually boundary conditions imposed in this kind of problems (Flow past circular cylinder).

What I think is that one thing is dv/dy = 0 and other different thing is dv/dt = 0.

In my opinion, dv/dt is different of zero because v oscilates among a mid point. But, if dv/dy = 0, I understand that all the points at the exit should have the same v value, but oscillating all them together.

Could it be possible? I think so.
Yes, this is the result that must follow from your bc
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Old   June 9, 2016, 18:14
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You didn't say if you're using a commercial code or one that you've developed. In either case, but especially if you're using your own code, I strongly recommend that you start with a simpler problem and work your way up as you gain confidence that the code works ok and/or your boundary conditions are correct. So:

(1a) I agree with others that your exit boundary is too close to the object and needs to be pushed back. For that matter, you will find out upon parametric investigation that your other boundaries are potentially too close to the object as well.

(1b) I would initially start with the "less difficult" problem and just run a Re=1 case. In this Stokes flow case your exist boundary is probably already far enough. So, if you're getting garbage here the problem is most likely _not_ the domain size. Neither is it necessarily the boundary condition (it could be a coding issue). For Re=1, you can safely set v=0 on all boundaries and see what happens. If this works, then move to a higher Re and a longer domain.

(2) You are solving a heat transfer problem, which is significantly more difficult than a "cold flow" simulation. So, I strongly recommend that you begin with the simpler incompressible flow problem with the energy equation turned off. Unless you get acceptable results for this simpler case there is no reason you should go to the next step. The next step should probably ignore gravity and involve simpler temperature boundary conditions. Again, unless this step produces good result your attempt at including gravity is a waste of valuable time. So, apply step (1) to incompressible flow first.

(3) You're solving a heat transfer problem but you didn't mention anything about your temperature boundary condition.

adrin
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