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Which pressure OpenFOAM use for incompressible flow? P/rho or (P-101325)/rho ? |
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December 15, 2009, 04:59 |
Which pressure OpenFOAM use for incompressible flow? P/rho or (P-101325)/rho ?
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#1 |
Senior Member
Jiang
Join Date: Oct 2009
Location: Japan
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Dear Foamers:
I am a little confused about pressure for imcompressible flow. I want to know which pressure OpenFOAM uses. Because in outflow boundary, pressure value is fixed. So if you give zero, the pressure result will be very small. If you give 101325/rho, the pressure result will be large. in all the tutorials of OpenFOAM , in case/0 ,pressure is set to zero. If that means , the pressure set in P file is (Real Pressure - 101325)/rho ? Thank you very much! |
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December 15, 2009, 15:23 |
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#2 |
New Member
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Isn't the pressure already divided by rho?
So the real pressure would be p*rho and then if you have 0 at outlet add atmospheric pressure so: p*rho + 101325 But you should see my post about pressure and knowbody has answered yet. http://www.cfd-online.com/Forums/ope...cell-size.html I suspect pressure is not being calculated correctly because it varies too much with cell size. If so you can't use the pressure value. |
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October 11, 2013, 11:37 |
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#3 | |
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Quote:
thanks |
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October 11, 2013, 22:00 |
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#4 |
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Fumiya Nozaki
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You can get the real pressure value (that is not divided by rho) using the formula
that jugghead wrote and you can find the rho value by consulting the physical property books at your simulation condition. Hope this helps, Fumiya |
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October 12, 2013, 03:53 |
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#5 | |
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Quote:
nu=mu/rho So how can i just get the value from the book. Isnt the simplefoam or other incompressible solver default value used? So what is the default value then? i have done the solver, with nu 1.5exp-5 (from motorbike tutorial) Now i need the rho value based on that nu or based on the motorbike tutorial. Thanks |
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October 12, 2013, 05:22 |
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#6 |
Senior Member
Fumiya Nozaki
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Hi nash,
If you look at the table(http://www.engineeringtoolbox.com/ai...ies-d_156.html), you can find that the kinematic viscosity(nu) value is nearly equal to 1.5e-5 at 20 degrees Celsius and the density(rho) value is 1.205 kg/m^3 at this temperature. So, the motor bike tutorial solves the flow on these conditions if the working fluid is air. If you try to do another simulation at different condition(different temperature or fluid etc.), you can find the nu and rho value from books and set nu value in the transportProperties dictionary. When your simulation finishes, you can get the real pressure value using the formula that jugghead wrote and the density value you will have found. |
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October 12, 2013, 05:46 |
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#7 | |
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Quote:
Now i would like to ask, if i want to get the exact pressure direct from the simulation, i plan to set the rho to 1. So i need to set nu. But i dont know the mu. Any idea? Temperature is at 20 degree celcius. Isnt okay if i do so? Thanks again for your help |
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October 29, 2014, 06:56 |
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#9 |
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Hello together,
I know this thread is older but I am a little confused at the moment concerning the pressure. If I use the incompressible solver interFoam and set the pressure to 0 in 0/p, I sometimes get a negative pressure p. Is it also true in this case that I calculate the "real" static pressure with p + 101325? Thanks a lot for your help idefix |
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November 21, 2014, 01:26 |
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#10 |
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Artem Shaklein
Join Date: Feb 2010
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Hello, idefix.
In incompressible limit , so only affects a solution. It doesn't matter whether absolute static pressure p_abs[0] = 1e5 or p_abs[0] = 2e5 (by the way, p_abs[0] = 0 can be used as well, but it is nonsense, because in vacuum just small amount of matter is presented and it can't be modeled by continuum mechanics theory). If density is constant, it's useful to divide all equations by density, so to recover abs pressure one has to do the math . In this case pressure (0/p) has dimensions [Pa/(kg/m^3)]. But this approach can be used in general case as well (just consider a number of digits to store: 1.013250001e5 vs 1.0e-4). To recover pressure one needs next . In this case pressure (0/p) has dimensions [Pa]. I looked inside interFoam case and found that 0/p has [Pa] dimensions. So you should go with . |
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February 5, 2015, 10:53 |
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#11 |
Senior Member
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Just a question for confirmation, as I'm getting confuse with my results, in order to see how to change my BC.
In 0/U I defined pressureInletvelocity and in 0/p I defined total pressure for inlet and set it equal to 0, so I defined: total = static + dynamic --> 0 = 0 + rho*U^2/2 (value for U= (0 0 0) --> all is zero) When I plot a slice on paraview, what pressure I get? static pressure divided by rho? total pressure divided by rho? or in other way to ask, is pressure calculated by openfoam the static one or the total pressure? thanks a lot. Bye |
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January 5, 2016, 11:17 |
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#12 | |
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Quote:
Thanks! |
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April 22, 2016, 04:38 |
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#13 | |
New Member
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Quote:
Notations - p = static pressure ptot = total pressure rho = density to my knowledge, for incompressible flows, OF solves for p/rho. You can find total pressure (ptot = p + 1/2 * rho *U^2) using the ptot utility. |
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August 10, 2018, 12:55 |
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#14 |
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Chayanit Nigaltia
Join Date: Jan 2018
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If I define my own rho and rhoInf, the will the static pressure be normalised by my value of rho?
Please help. |
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August 13, 2018, 05:22 |
it is simple
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#15 |
Member
ijaz fazil
Join Date: Apr 2013
Location: Singapore
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You have to provide boundary value
pressure-(density*g*height) For example if you want to specify pressure to be zero then enter the value 0-(density*g*height) Make sure that your g value specified correctly in constant folder. |
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August 14, 2018, 04:40 |
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#16 |
New Member
Chayanit Nigaltia
Join Date: Jan 2018
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Thanks Ijaz
Since the pressure value is already normalised by density in openfoam( I think they have used a value of 1.2) So now do I have to change the source code and then compile.Or by simply stating my free stream density and acceleration due to gravity, I will have my calculations correct. |
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August 14, 2018, 04:57 |
for interfoam pressure is not normalised
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#17 |
Member
ijaz fazil
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Location: Singapore
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Hi
But for interFoam alone pressure should not be normalised with density, might be because you have two different fluids with different density, so we have to use the actual value of pressure. For example if atmospheric pressure is zero, then use that value to define the boundary. |
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