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what is the meaning of p_rgh?

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Old   September 23, 2016, 10:36
Default what is the meaning of p_rgh?
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Simone Colucci
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Hi,
I have a simple question. In many solvers (for example in reactingTwoPhaseEulerFoam, OpenFoam 4.0) the PEqn solves an equation for p_rgh, instead of p. p_rgh is defined as:
Code:
p_rgh = p - rho*gh;
where :
Code:
volScalarField gh("gh", (g & mesh.C()) - ghRef);
If I consider, for example, a vertical pipe with the z_bottom (vertical coordinate)=-10 and the z_top=0, rho*g*h has a physical meaning, but if the coordinates are z_bottom = 0 and z_top = 10, what does it mean?

Thank you in advance

Simone
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Old   September 26, 2016, 12:45
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Can you clarify your point little bit more? I mean, I guess h is the difference in height in both cases you mentioned, so isn't it 10?
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Old   September 26, 2016, 22:15
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Pablo Higuera
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Hi Simone,

the truth is that it has no physical meaning, it is just a convenient mathematical artifact. It is explained in my thesis (page 111):

https://repositorio.unican.es/xmlui/...=1&isAllowed=y

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Old   June 14, 2020, 04:07
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Quote:
Originally Posted by Phicau View Post
Hi Simone,

the truth is that it has no physical meaning, it is just a convenient mathematical artifact. It is explained in my thesis (page 111):

https://repositorio.unican.es/xmlui/...=1&isAllowed=y

Best,

Pablo
Hi Pablo,
So, it mean no matter what the co-ordinate system our model is in. the result will be same as long as our BC remain same.
To elaborate my question, if I did simulation for one geometry and now I translate that geometry by 10 cm. Will the result be same if I perform simulation on this new translated geometry?
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Old   June 14, 2020, 16:37
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Hi,
not necessarily. The main advantage is to avoid large round-off errors if numbers are large, as P can be.
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Pablo
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Old   June 14, 2020, 16:55
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Quote:
To elaborate my question, if I did simulation for one geometry and now I translate that geometry by 10 cm. Will the result be same if I perform simulation on this new translated geometry?
- I haven't read the other replies, so if misunderstood, my apologies.
- The governing equations of fluid dynamics are Galilean invariant, which means that making any transformation or rotation on the computational domain within an inertial frame of reference will not change the predictions if the boundary conditions are modified accordingly.
- Therefore, translating the domain by 10 cm in a given direction will not change the numerical predictions if the control volume keeps modelling the same volume.
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Old   June 15, 2020, 08:49
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Thank You for the replies.
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Old   June 15, 2020, 09:07
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Hi HPE,
despite the equations being Galilean invariant, once you discretize them it is another story. The round off errors and the solution will be different. To what extent is a question of the processes and how long you run the simulation for, but "chaos" plays a role here and results may be noticeably different. For example, you can observe this in my simulations when waves break.


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Old   June 15, 2020, 11:18
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Hi,

There are circumstances where Galilean invariance breaks down for the finite volume method in Eulerian frame, true; but from an engineering perspective, the reason in the current context is the acute sensitivity of the governing equations of particular physics on initial/boundary conditions.

What you have observed can also happen for a restarted simulation wherein the restart is not bitwise, which is very common in any simulation of any software.

Personally, I would not be concerned with such ultra academic break downs of Galilean invariance in discretised domains since even the Navier-Stokes equations have only weak solutions, which means there might be some flow behaviour not obeying N-S in some part of universe at some point. But, who cares.

Thank you for your kind remarks.
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Old   November 6, 2020, 21:21
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Quote:
Originally Posted by Phicau View Post
Hi Simone,

the truth is that it has no physical meaning, it is just a convenient mathematical artifact. It is explained in my thesis (page 111):

https://repositorio.unican.es/xmlui/...=1&isAllowed=y

Best,

Pablo
I think this is not correct. It is pseudo-hydrostatic pressure as it is explained in prghPressureFvPatchScalarField.H. It is nothing related to dynamic or pseudo-dynamic pressure to my opinion.
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