
[Sponsors] 
September 27, 2010, 04:57 
p_rgh in OF 1.7

#1 
Member
Join Date: Nov 2009
Posts: 34
Rep Power: 12 
Dear Foamers,
I'm a little bit confused about the new pressure p_rgh in OpenFOAM. How is it defined? I worry a little bit the boundary conditions I have to define. For example if I would like to have a constant "pressure" at the outflow, how can this be defined ? If I define p_rgh = 0, does that mean that the pressure + (rho)*g*z is fixed to zero in this case? Also I wonder why in some solvers I find a pfile and p_rgh file. Is the pfile only used for postprocessing or is it also possible to define boundary conditions there? Thanks in advance. Stawrogin 

September 27, 2010, 05:20 

#2 
Member
Juho Peltola
Join Date: Mar 2009
Location: Finland
Posts: 89
Rep Power: 13 
In the 1.7.x buoyantPimpleFoam, both the p and p_rgh files are read. The p is used in the thermodynamic model.
In the createFields the p_rgh is defined as: p_rgh = p  rho*gh; So it is the pressure without the hydrostatic pressure and is intialized from the pressure field in the pfile. In the pEqn.H file, the pressure equation is written and solved for the p_rgh, so the boundary conditions important for the pressure solution are the p_rgh conditions. After the pressure solution, the p is calculated with: p = p_rgh + rho*gh; 

September 27, 2010, 05:27 

#3 
Member
Join Date: Nov 2009
Posts: 34
Rep Power: 12 
Dear Juho,
thanks a lot for the help. Now it's clear for me. Stawrogin 

September 30, 2010, 06:22 

#4 
Member
Nico T
Join Date: Aug 2010
Location: Leipzig, Germany
Posts: 39
Rep Power: 11 
Hi Juho,
p_rgh seems to be the dynamic pressure. I work with pimpleFoam and pisoFoam. Is it possible to adjust these codes to obtain p_rgh? Thanks, Nico 

October 5, 2010, 17:52 

#5 
New Member

Hi Nico,
Does p_rgh stand for the dynamic pressure, which means p_rgh=1/2*rho*U^2? I checked the tutorial dam break case, why the value of p_rgh is greater than p? It implies that the rho*g*h is negative. I am so confused. 

October 5, 2010, 19:20 

#6 
Member
Kevin Maki
Join Date: Mar 2009
Location: Ann Arbor, MI, USA
Posts: 43
Rep Power: 13 
Hi Angela,
p_rgh is p  rho*g*h, or, the dynamic pressure. It was called pd in version 1.5, they solved for total pressure in version 1.6, and now are back to solving for only the dynamic pressure in version 1.7. Kevin 

October 6, 2010, 03:07 

#7 
Member
Nico T
Join Date: Aug 2010
Location: Leipzig, Germany
Posts: 39
Rep Power: 11 
Hi Kevin,
but why p_rgh is higher than p? How a negative static pressure (rho*g*h) is possible? Shouldn’t 1/2*rho*U^2 (dynamic pressure def.) leads to p_rgh? PostCalculation of Ufield does not. nico 

October 29, 2010, 15:19 
Why p_rgh is greater than p

#8 
New Member
Nicolás Badano
Join Date: Sep 2010
Posts: 10
Rep Power: 13 
Dear fellows,
Oddly enough, gh is defined in openFoam as: gh = g & mesh.C()In other words, gh is the dot product of the g vector and the cell center position vector. As g is usually defined as (0 0 9.81), gh often results negative in the positive z cuadrant! Finally, as: p_rgh = p  rho * ghp_rgh is greater than p if z > 0 Hope this helps. Best regards Nicolas 

October 29, 2010, 15:26 

#9 
New Member

Thanks. My problem is solved. p_rgh is the dynamic pressure = p rho*g*h


November 17, 2010, 12:01 

#10 
Member
Logan Page
Join Date: Sep 2010
Posts: 38
Rep Power: 11 
one more question regarding this:
Using the buoyantBoussinesqSimpleFoam solver, the units of p and p_rgh is (m^2)/(s^2) [i.e Kinematic Pressure] obviously this is quite simply defined by: (p_rgh / rho) = (p / rho)  g*h My questions is what density is used when dividing through. Is it rhok = 1  beta (T  Tref) ?? Thnks 

December 3, 2010, 12:20 

#11 
Member
abdikerim kurbanaliev
Join Date: Jun 2010
Location: Kyrgyzstan, Osh
Posts: 62
Rep Power: 12 
Hi fellows,
I'd like to continue Nico's idea. It seems to me that p_rgh is not dynamic pressure. I changed endtime to 500 in the laminar dambreak case. After that time initial transition flow will be stationary. So the there no dynamic pressure at t=500. But interfoam gives non zero p_rgh pressure. That is why I assume that p_rgh is not dynamic pressure. Could someone еxplaine me what is going on? Or am I wrong ? Kerim 

December 3, 2010, 13:29 

#12 
New Member
Nicolás Badano
Join Date: Sep 2010
Posts: 10
Rep Power: 13 
Hi Kerim,
p_rgh is not really dynamic pressure, specially in multiphase flow where rho changes throughout the domain. Is just de difference between real pressure and the rho*g*z field. I think that's one of the reasons the field is no longer called pd, as in 1.5 version of OF. Just to add to the confusion, bear in mind thart rho*g*z is not the real hydrostatic pressure either!! Even if rho is constant it differes from hydrostastic component by a constant (the distance between the z=0 plane and the atmosphere p=0 plane times rho*g). Here's a quick sketch of the relation between p_rgh, rgh and p for a hydrostatic condition (no movement at all). Hope this helps Best regards! Nico 

December 13, 2010, 09:21 
p and p_rgh files

#13 
New Member
Cristiano
Join Date: Jun 2010
Posts: 14
Rep Power: 12 
Hello there,
I'm modelling a burner which has been tested at atmospheric condition and I'm a bit confused . How can I do to set the boundary conditions properly for p and p_rgh? Inlet > Outlet> Wall> Thank you indeed. Cristiano 

January 14, 2011, 10:44 

#14  
Senior Member
Illya Shevchuk
Join Date: Aug 2009
Location: Darmstadt, Germany
Posts: 176
Rep Power: 12 
Quote:
Last edited by linch; January 14, 2011 at 11:01. 

February 10, 2011, 10:55 
Tip

#15 
New Member
Christian Fri
Join Date: Dec 2009
Posts: 7
Rep Power: 12 
Despite this a very simple suggestion it is worth to say it.
In case you have an OpenFOAM 1.6 case that had run successfully and want to run it in OpenFOAM 1.7 remember that you can just rename your p file to p_rgh (and also change some other parameters in your transportProperties and fvSchemes files). Then change the value of g from (0 0 9.81) to (0 0 0). The case should work the same as in OpenFOAM 1.6 and you won't have to worry about if the BC that you are using in OpenFOAM 1.7 are ok if they were ok in OpenFOAM 1.6. Cheers, Christian F. 

May 17, 2011, 16:30 
Bernoulli

#16 
Member
Arnout
Join Date: Nov 2010
Posts: 43
Rep Power: 11 
To understand the different pressures, look at Bernoulli:
Dynamic pressure > 1/2*rho*v^2 Hydraulic pressure> rho*g*h Static pressure > p 1/2*rho*v^2 + rho*g*h + p = Constant From the openFoam site, p_rgh = p  rho*g*h. So, p_rgh is the static pressure minus the hydraulic pressure, based on a arbitrary height. I do not understand where the dynamic pressure came into this discussion. I think it has nothing to do p_rgh. Dynamic pressure is the pressure of the moving fluid and it will convert into static pressure if you bring the velocity of the fluid to zero. Conservation of energy, back to Bernoulli. Good to know: To get my VOF model working, I placed in the fvSolutions file under the PISO solver pRefPoint (0.0 0.0 0.0); pRefValue 1e5; Succes! 

June 24, 2011, 08:39 

#17 
Member
Nico T
Join Date: Aug 2010
Location: Leipzig, Germany
Posts: 39
Rep Power: 11 
Hello,
thanks King, for the clearing explanation. To finish the pressure confusion: The postprocessing utility "ptot" calculates the total pressure (static + dynamic) for every time step: http://www.cfdonline.com/Forums/ope...ldswrite.html Cheers, Nico 

November 19, 2011, 05:52 

#18 
New Member
Fernando Castro
Join Date: Oct 2011
Posts: 1
Rep Power: 0 
I think: P_rgh is the perturbation pressure (from an hydrostatic equilibrium state) that is actually used during the simulations in the momentum equations.


November 29, 2011, 09:11 

#19 
New Member
Howard NJOKU
Join Date: Nov 2010
Location: Nsukka, Nigeria
Posts: 9
Rep Power: 11 
why does p_rgh have units of m^2s^2 in the bouyantBoussinesqPimpleFoam and the bouyantBoussinesqSimpleFoam examples, but has units of kgm^1s^2 under the (multiphase) interFoam examples?


November 29, 2011, 09:23 

#20 
Senior Member
Andrea Ferrari
Join Date: Dec 2010
Posts: 319
Rep Power: 12 
Because it is divided by the density in the first two cases. To have the "real" pressure" you have to multiply by density [kg/m^3].
best andrea 

Thread Tools  Search this Thread 
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
solver for subsonic compressible turbulent flow in OF 1.7  nileshjrane  OpenFOAM Running, Solving & CFD  20  February 13, 2012 05:54 
YPlusRas for InterFoam (Open Foam 1.7)  MrD  OpenFOAM  0  August 11, 2010 15:44 
OpenFOAM 1.7  openSUSE 11.3  gcc 4.5.0  alberto  OpenFOAM  12  July 28, 2010 11:59 
MixSim 1.7 Tutorials  johnnyb  FLUENT  0  August 25, 2003 17:13 