[Matheus.Costa]

Hi guys, while checking the results of my simulations I noticed that the pressure has an unfamiliar value, the speed vector seems to have the expected value but the same doesn't apply to the pressure. I'm simulating a pressure drop through an orifice plate inside a pipe and comparing the values with a previous experiment, the pressure at the gages were measured at 1.5 PSI (10.3Kpa) before the orifice plate and 0 PSI after. The calculated pressures for the same experiment were 1.564 PSI and 0.264 PSI respectively, which are close to the experimental result, but when I run the simulation the value for pressure before the plate is 0.764 and -0.267, after researching a little I read that openFOAM uses the pressure normalized by rho (p/rho), so, in order to find the value of the pressure I need to multiply the value that I found and rho? Even after multiplying the value in pascals is too low to be accurate (around 764pa and -267pa)

I set the boundary conditions following the pitzdaily tutorial, maybe I missed something at the inlet boundary condition, I set all the boundaries to be zerogradient for the pressure, except the outlet which is set as fixedvalue: uniform 0. I would be very thankful if any of you guys could help me out.

[olesen]

Pressure is in Pascals for compressible flow solvers, and the"p/rho" for incompressible

[random_ran]

I am just curious that "p/rho" is used in OF.

It seems to have some calculation advantages but I do not know the details

[olesen]

Originally Posted by random_ran

I am just curious that "p/rho" is used in OF.

It seems to have some calculation advantages but I do not know the details

Well if the simulation is incompressible you can either have an implied constant density to drag through all the equations (eg, multiplying by one in various places) or simply do away with density in the equations by more or less factoring it into pressure (p = p/rho, and rho == 1) thus giving the units in question.

[Tobi]

I refer to my book "Mathematics, Numerics, Derivations and OpenFOAM". Here, all equations are described.

As Mark already said, if we have incompressible solvers it is ment that the density is constant and not changing, e.g., by compression or expansion or thermal behavior. As the density is constant, we can take the density out of all derivatives and can cancel via rho. Thus, in all terms including rho, it vanishes and the terms that do not have the density included (such as the pressure term) are divided by the density. Hence, p/rho = p* for incompressible flows. Nevertheless, for incompressible flows the pressure does not have the physical meaning as a "real" pressure as only the pressure gradients are of importantance. Thus, one can see negative pressure values p*.

[random_ran]

I know that, for the incomprensible case, the pressure is usually not
in [Pa] but it is divided by the fluid density (so the unit is
m^2/s^2) in OpenFOAM. In order to get the pressure in the unit [Pa], one
needs to multiply the result from OpenFOAM by the density.

However, this extra step makes me think there might be advantages
in using such a simplification in incompressible solver. Imaging a new
user would expect to set the pressure in the simulation and expect
the calculation result is immediately available to compare to their
experiment tests. In addition, introducing the density into the
incompressible solver might simplify the codebase. I don't know the
OOP details, but just a guess.

"the pressure does not have the physical meaning as a "real"
pressure as only the pressure gradients are of importance."

This perspective is interesting. Is there any other fundamental
measurement that exists that only has importance to their gradient but
not themselves? For example, the second, metre, kilogram, ampere,
kelvin, mole, candela.

[Tobi]

Sure there is a benefit. If you use a density field in the equations, we have to interpolate these quantities to the faces even though it is everywhere the same because we calculate the face fluxes using U and rho in the cell center. Hence, we have to interpolate it to the faces. If we remove the density (as it is a constant) we only need to interpolate U and not both U and rho.

[paran0]

Originally Posted by Tobi

I refer to my book "Mathematics, Numerics, Derivations and OpenFOAM". Here, all equations are described.

As Mark already said, if we have incompressible solvers it is ment that the density is constant and not changing, e.g., by compression or expansion or thermal behavior. As the density is constant, we can take the density out of all derivatives and can cancel via rho. Thus, in all terms including rho, it vanishes and the terms that do not have the density included (such as the pressure term) are divided by the density. Hence, p/rho = p* for incompressible flows. Nevertheless, for incompressible flows the pressure does not have the physical meaning as a "real" pressure as only the pressure gradients are of importantance. Thus, one can see negative pressure values p*.

Hello Tobi,
I am new to CFD and I did not understand "why for incompressible flows the pressure does not have the physical meaning as a "real" pressure?" Could you please explain this or someone else who have seen my text? Thank you.

[AtoHM]

As stated above, the incompressible OpenFOAM solvers solve for the kinematic pressure instead of the physical pressure: https://www.openfoam.com/documentati...-pressure.html
What drives the flow is the pressure gradient, not the physical/absolute value of pressure. It has been described nicely above, I think. All you need to know as a user is that you have to multiply by your fluid density to get the physical pressure value from the result.

[Tobi]

Quick addition

First: add multiply by a reference density and then second: add a reference pressure

[Tobi]

Not sure what you mean but for incompressible solver in OpenFoam:

p_real = (p_cfd*rho)+p_ref

[paran0]

Hi,
I am running a propeller transient case using pimplefoam for my thesis. But, I want to run it now as a steady state case so that I can understand immediately how it looks like when the transient simulation converges to steady state. Is there any one who can help to change my case setting so that it runs as a steady state simulation? Thanks!

[vv123]

Originally Posted by Tobi

Not sure what you mean but for incompressible solver in OpenFoam:

p_real = (p_cfd*rho)+p_ref

Hi Tobi,

I don't understand what the P_ref value is. I am using the solver pimpleFoam and imply all BC for p as zeroGradient, p_out =0 and p_top = slip. I don't specify a reference pressure anywhere, so how can you know what this value is?

Some background;

I am simulating a river flow, I would like to transform the pressure calculations from OpenFoam to a water level from the rigid lid.

Thanks!!

[unilord]

Originally Posted by vv123

Hi Tobi,

I don't understand what the P_ref value is. I am using the solver pimpleFoam and imply all BC for p as zeroGradient, p_out =0 and p_top = slip. I don't specify a reference pressure anywhere, so how can you know what this value is?

Some background;

I am simulating a river flow, I would like to transform the pressure calculations from OpenFoam to a water level from the rigid lid.

Thanks!!

Hey vv,

I know I am not Tobias, but I think I can answer your question. In incompressible solver, the pressure that you choose as inlet our outlet doesn't really matter. Pressure in itself has no real representation. However, a difference of pressure does.

You can choose your outlet pressure = 0 [Pa/(kg.m^-3)], just like you mentioned. However, that won't happen in real life, as a pressure of 0 will not be a real representation of the physics of your problem. In your case, I assume your p_ref would be the environment pressure (101.325 [P/(kg.m^-3)], if it is at sea level.

[vv123]

Originally Posted by unilord

Hey vv,

I know I am not Tobias, but I think I can answer your question. In incompressible solver, the pressure that you choose as inlet our outlet doesn't really matter. Pressure in itself has no real representation. However, a difference of pressure does.

You can choose your outlet pressure = 0 [Pa/(kg.m^-3)], just like you mentioned. However, that won't happen in real life, as a pressure of 0 will not be a real representation of the physics of your problem. In your case, I assume your p_ref would be the environment pressure (101.325 [P/(kg.m^-3)], if it is at sea level.

Hi Pedro,

Thanks for your answer. That makes sense. However I still have an issue. In my case I have a rigid lid, where there is actually a free surface. The changes from the rigid lid are assumed to not be super large (so that is why this choice was made). However, I would like to retrieve a deviation in the water level from the rigid lid based on the pressure levels.

The problem I am facing is that the pressure is always bigger upstream, than downstream. Therefore, after a correction for velocity it still doesn't work. Also I am not sure about my derivation of pressure to water level. I added a word document with my derivation.

Another way (that seems to work, but I didn't manage to get a derivation) is:
h = p_k/g - 0.5u^2/g

could you maybe help with this?

[unilord]

Originally Posted by vv123

Hi Pedro,

Thanks for your answer. That makes sense. However I still have an issue. In my case I have a rigid lid, where there is actually a free surface. The changes from the rigid lid are assumed to not be super large (so that is why this choice was made). However, I would like to retrieve a deviation in the water level from the rigid lid based on the pressure levels.

The problem I am facing is that the pressure is always bigger upstream, than downstream. Therefore, after a correction for velocity it still doesn't work. Also I am not sure about my derivation of pressure to water level. I added a word document with my derivation.

Another way (that seems to work, but I didn't manage to get a derivation) is:
h = p_k/g - 0.5u^2/g

could you maybe help with this?

Hey vv,

Sorry but I am not able to understand the problem you are facing. Why is it a problem that the pressure is always bigger upstream? And what do you mean by "correction for velocity"? Maybe with some pictures of your domain it would be clearer.

[vv123]

Explanation problem.png

Originally Posted by unilord

Hey vv,

Sorry but I am not able to understand the problem you are facing. Why is it a problem that the pressure is always bigger upstream? And what do you mean by "correction for velocity"? Maybe with some pictures of your domain it would be clearer.

Hi Pedro,

As I want to convert the calculated pressure to the water level. I have a rigid lid on top, but I also have the measured water levels. I made rigid lid smoother than the actual water levels. However if I calculate the water level from the pressure, I am facing a water level setup upstream. I am thinking that this might be due to the pressure that is always bigger upstream as well.