CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > OpenFOAM Running, Solving & CFD

Difference between p in OF-1.6.x and pd in OF-1.5

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   November 22, 2009, 07:26
Default Difference between p in OF-1.6.x and pd in OF-1.5
  #1
Senior Member
 
Suresh kumar Kannan
Join Date: Mar 2009
Location: Luxembourg, Luxembourg, Luxembourg
Posts: 129
Rep Power: 9
kumar is on a distinguished road
Hello everybody,
I have been using OF-1.5, interFoam and lesInterFoam for my project work. Recently I also installed the OF-1.6.x and ran some preliminary simulations.

I find that there is a difference in the output pressure between the two versions of OF-1.6.x and OF-1.5.
In OF-1.5 the output is pd and in OF-1.6.x it is p.
But I still have some problem in understanding pd in OF-1.5. WHAT exactly is the formulation for this pd. And also as OF uses relative pressure, where do I specify this reference pressure. And id i dont specify this reference pressure, what value does it take by default. I want to make this clear because when i run the damBreak case of OF-1.5 and OF-1.6.x and then i plot pd aand p respectively. I see negative pressures in both of them, which may be correct given the reference pressure is atmospheric pressure.

Or may be i am not understanding it properly.
Can anybody explain the exact formulation of the p and pd that is written in the solution file after a interFoam calculation.
bye
with regards
K.Suresh kumar
kumar is offline   Reply With Quote

Old   November 22, 2009, 15:33
Default
  #2
Senior Member
 
linnemann's Avatar
 
Niels Nielsen
Join Date: Mar 2009
Location: NJ - Denmark
Posts: 472
Rep Power: 16
linnemann will become famous soon enough
with a single fluid incompressible solver the pressure is normalized so just multiply with the density, the kinematic viscosity is sufficient for the shear forces calculations, as I've understand.

Since it is a multiphase solver you use I don't know if the same applies. The two fluids can be mixed and what to do here?

Best
__________________
Linnemann

PS. I do not do personal support, so please post in the forums.
linnemann is offline   Reply With Quote

Old   November 23, 2009, 04:19
Default
  #3
Senior Member
 
Suresh kumar Kannan
Join Date: Mar 2009
Location: Luxembourg, Luxembourg, Luxembourg
Posts: 129
Rep Power: 9
kumar is on a distinguished road
Hello Linneman,
Thankyou very much for your explanation. Digging further into the source of InterFoam, I found that there is a definition of pd :

p = pd + rho*gh

Then I have a question if i set g as 0 in my problem then the term rho*gh goes to zero and then p = pd.
So if pd is the dynamic pressure then it should be equal to (pd = 1/2*rho*U^2).
Please correct me if i am wrong.

And regarding your comment about the normalizaiton do you mean the pressure is normalized by density and kinematic viscosity, but in that case there is a problem with the dimensions. If iam understanding it wrong correct me.
thankyou
regards
K.Suresh kumar
kumar is offline   Reply With Quote

Old   November 23, 2009, 04:45
Default
  #4
Senior Member
 
linnemann's Avatar
 
Niels Nielsen
Join Date: Mar 2009
Location: NJ - Denmark
Posts: 472
Rep Power: 16
linnemann will become famous soon enough
Hi

jep then pd=1/2*rho*U^2=normalized-pressure

So for incompressible solvers the density is always 1 and the shear forces depend on the kinematic viscosity only.

The reason you only need the kinematic viscosity in the incompressible solvers is that the wall shear forces are calculated using nu_t (turbulent kinetic viscosity) and in incompressible with mu_t (turbulent dynamic viscosity) since it depends on the density.

Using for example water with density 1000kg/m^3 and using paraview you can create a calculator and take the scalar p and multiply with 1000 to get the pressure. Again the is for single phase flow don't know about multiphase.
__________________
Linnemann

PS. I do not do personal support, so please post in the forums.
linnemann is offline   Reply With Quote

Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On



All times are GMT -4. The time now is 18:28.