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November 22, 2009, 07:26 
Difference between p in OF1.6.x and pd in OF1.5

#1 
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Suresh kumar Kannan
Join Date: Mar 2009
Location: Luxembourg, Luxembourg, Luxembourg
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Hello everybody,
I have been using OF1.5, interFoam and lesInterFoam for my project work. Recently I also installed the OF1.6.x and ran some preliminary simulations. I find that there is a difference in the output pressure between the two versions of OF1.6.x and OF1.5. In OF1.5 the output is pd and in OF1.6.x it is p. But I still have some problem in understanding pd in OF1.5. 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 OF1.5 and OF1.6.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 

November 22, 2009, 15:33 

#2 
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Niels Nielsen
Join Date: Mar 2009
Location: NJ  Denmark
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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
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November 23, 2009, 04:19 

#3 
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Suresh kumar Kannan
Join Date: Mar 2009
Location: Luxembourg, Luxembourg, Luxembourg
Posts: 129
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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 

November 23, 2009, 04:45 

#4 
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Niels Nielsen
Join Date: Mar 2009
Location: NJ  Denmark
Posts: 462
Rep Power: 15 
Hi
jep then pd=1/2*rho*U^2=normalizedpressure 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.
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