[HectorRedal]

Hi

I would like to know how the artificial compressiblity may affect the accurary of a transient solution.

The objective I am pursuing is simulating an imcompressible flow. I have two options:
First option is assuming fluid is completely incompressible, which means that the sound velocity in the fluid is infinite (c= inf, M_p = 0).
Second option is assuming fluid is not completely incompressible, which means that the sound velocity is finite (c!=inf, M_p!=0), but not low (i.e., Mach = 0.01)

Will the solution obtained vary?

I am simulating a flow past a circular cylinder and this is the behaviour I am observing:
The flow oscilates behind the cilinder at Re=100.
Assuming total incompresibility, I am obtaining Strouhal = 0.175
Assuming certain artificial incompressiblity, I am obtaining Strouhal = 0.1645, which matches the value appearing in the references.

It appears that when using artificial incompressiblity the algorithm behaves better than when assuming total incompressiblity.

Any comment / opinion is welcome.

Thanks,
Hector.

[FMDenaro]

By a physical point of view, M=0.01 is practically corresponding to the incompressible model (at least if you do not have an acceleration along the cylinder such to have a greater local Mach number).

However, the compressible and incompressible numerical models produce difference in the description of the pressure field (sound wave). In the latter, there is no description of sound waves travelling during the transient, the spread istantaneously. The compressible model at such a low Mach is stiff, but if you adopt suitable technique you can describe, with small time steps, the acosutic waves travelling. In any case, the velocity field solved by the two methods should be practically equal.

My opinion is that the artifical compressibility technique is more adequate to a steady solution than to a description of a transient.

To give you more help you should provide some plots of the fields, the Strouhal number is a global parameter that cannot provide more detailed infos about the solutions.

[HectorRedal]

I am uploading some pictures of the velocity(x component) and pressure at different instants (0, T/4, 2T/4, 3T/4, 4T/4).
I hope this will help.

I would like to comment also that when I double the Reynolds number (from 100 to 200), the dicrepancy observed in the incompressiblity limit decreases. I mean, it is aligned with the value obtained by references.
The error I got for Re200 is approx 1-2%, whereas at Re100 the error is 7-8%.

It appears as the viscosity decreases, the error obtained is fewer.

This behaviour is not observed in a stationary simulation, where the error obtained (less than 0.1%) is independent of the Reynolds number.

[HectorRedal]

More pictures for velocity x component

[FMDenaro]

1) you have clearly a problem in the outflow
2) I am not able to understand the comparison between the two cases
3) Are you sure that the time and space discretizations are of the same type and accuracy?

[HectorRedal]

Originally Posted by FMDenaro

1) you have clearly a problem in the outflow
2) I am not able to understand the comparison between the two cases
3) Are you sure that the time and space discretizations are of the same type and accuracy?

Why do you think there is a problem at the outflow? Which trace in the flow makes you be suspicious of a problem in the outlet?

The time and space discretization are of second order, O(delta t ^2) and O(delta x ^ 2).

[FMDenaro]

Originally Posted by HectorRedal

Why do you think there is a problem at the outflow? Which trace in the flow makes you be suspicious of a problem in the outlet?

The time and space discretization are of second order, O(delta t ^2) and O(delta x ^ 2).

From the oscillations in the pressure field, at upper and lower corners of the right edge of the domain.

The type of discretization in time and space is the same?

[HectorRedal]

Originally Posted by FMDenaro

From the oscillations in the pressure field, at upper and lower corners of the right edge of the domain.

The type of discretization in time and space is the same?

When you state "type of discretization", what are you understanding.
Sorry for asking about this, it is only to give you a proper answer.

I am using a an implicit scheme for both pressure and velocity.
I am evaluating velocity and pressure at nodal points.
So, for a tria element that I am using, I have three nodes (vertices of the triangle) where the velocity and pressure are evaluated.

Is this what you are referring to?

[FMDenaro]

If you are using FE, I mean if the same shape functions are used and if the time advancement is performed in the same way and same time step.

But before that, check for the error at outflow

[HectorRedal]

Originally Posted by FMDenaro

If you are using FE, I mean if the same shape functions are used and if the time advancement is performed in the same way and same time step.

But before that, check for the error at outflow

Ok, now I understand your question.
The same shape functions are used for velocity and presure (Linear interpolation), and the same time step is used in both equations (momentum and pressure).

Regarding your second point (error in the outflow), when I analyzed the outflow pattern (a couple of weeks ago) I saw an oscilating pattern in the pressure field, but I understood that in the same way the velocity oscilates (both along and transversal velocity oscilate), this should apply to the pressure field.

For coming into this conclusion, I am thought about the Bernoulli equation: the sum of energy should remain constant. So if the velocity diminishes, the pressure should increase (and other way around).

I have integrated the flow at the outlet, for several time step, and it has the same value as for the inflow (in order to double check there is not any loss of mass). I mean, althouth the velocity oscilates, the sum of velocity at all points at the exit should remain constant.

[FMDenaro]

Bernoulli does not apply for viscous and rotational flows due to the dissipative effects.
It has also a different form for unsteady velocity but only provided that the velocity depends on a potential.

[HectorRedal]

Originally Posted by FMDenaro

Bernoulli does not apply for viscous and rotational flows due to the dissipative effects.
It has also a different form for unsteady velocity but only provided that the velocity depends on a potential.

So, what would be the expected behaviour at the exit?
A constant pressure field at all the points of the domain?

I would expect a pressure drop between the front and rear part of the cylinder, which may oscilate.
But at the exit, constant independent of the velocity field?

[FMDenaro]

You should get a regular pressure field without those spots at corners

[HectorRedal]

Originally Posted by FMDenaro

You should get a regular pressure field without those spots at corners

I get your point, you mean regular = constant presure value at the exit (along the vertical line)

[FMDenaro]

almost constant, if the outflow is not located quite far from the body you have some effects due to the vortex shedding but along the centerline, there is no physical reason to have two pressure spots at the corners

[FMDenaro]

see this example
https://www.google.it/imgres?imgurl=...act=mrc&uact=8

[HectorRedal]

Originally Posted by FMDenaro

see this example
https://www.google.it/imgres?imgurl=...act=mrc&uact=8

I have taken a look at the reference you have provided me with, and now I understand what you are referring to. I will investigate a bit what's going on, but clearly the problem is located at the right boundary exit, as you pointed out.
Thanks for your support.