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Which pressure OpenFOAM use for incompressible flow? P/rho or (P-101325)/rho ?

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Old   December 15, 2009, 04:59
Default Which pressure OpenFOAM use for incompressible flow? P/rho or (P-101325)/rho ?
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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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Old   December 15, 2009, 15:23
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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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Old   October 11, 2013, 11:37
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Quote:
Originally Posted by jugghead View Post
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.
may i know which default value of rho is used by simpleFoam (incompressible solver)? Because i need to have the right value for my pressure since i need to do performance curve of my fan simulation.

thanks
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Old   October 11, 2013, 22:00
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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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Old   October 12, 2013, 03:53
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Quote:
Originally Posted by fumiya View Post
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
basically one defines the nu value in transport properties. The nu value is given by this equation

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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Old   October 12, 2013, 05:22
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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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Old   October 12, 2013, 05:46
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Quote:
Originally Posted by fumiya View Post
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.
Thanks for the explanation.

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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Old   October 12, 2013, 10:26
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I think it's not possible and it is easier to multiply the result by rho after calculation.

Fumiya
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Old   October 29, 2014, 06:56
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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
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Old   November 21, 2014, 01:26
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Hello, idefix.
In incompressible limit \rho \ne f (p), so only \mathrm{grad} p 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 p_{stat}^{abs} = p_{stat}^{relative} \rho + p_{stat}^{reference}. 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 p_{stat}^{abs} = p_{stat}^{relative} + p_{stat}^{reference}. 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 p_{stat}^{abs} = p_{stat}^{relative} + p_{stat}^{reference}.
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Old   February 5, 2015, 10:53
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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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Old   January 5, 2016, 11:17
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Quote:
Originally Posted by student666 View Post
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
Did you find the answer for your question as I'm troubling in this, too?

Thanks!
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Old   April 22, 2016, 04:38
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Quote:
Originally Posted by student666 View Post
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
Hi student666,

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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Old   August 10, 2018, 12:55
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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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Old   August 13, 2018, 05:22
Default it is simple
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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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Old   August 14, 2018, 04:40
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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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Old   August 14, 2018, 04:57
Default for interfoam pressure is not normalised
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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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