Wrong calculation of hydrostatic pressure (InterFoam)

Lorena2fdez · October 16, 2017, 07:47

Hi foamers!

Currently, we are working on rivulets. In our case we have an uniform distribution of liquid on a inclined plane. It doesn't matter if you are on the top or on the bottom of the inclined plane, the rho*g*h distribution must be almost the same (only depends on the liquid above), because liquid height is the same.

$HydrostaticPressure ({\bf x}) = \int_L \rho(h) g \, dh$

Where

is the line between $({\bf x} \cdot {\bf g}/g)$ and ${\bf x_0} \cdot {\bf g}/g$ .

Currently, openFoam simplify the problem, It is assumed than $\rho$ doesn't depend on

. And we get:

$HydrostaticPressure ({\bf x}) = hg\, \rho({\bf x}) + P_{ref}$

where $hg ={\bf x} \cdot {\bf g}-{\bf x}_0 \cdot {\bf g}$ .

In the attached image you can see four points. On OpenFoam the hydrostatic pressure is calculated:

$HydrostaticPressure ({\bf x_1}) = h_1 g\, \rho_{air} + P_{ref}$
$HydrostaticPressure ({\bf x_2}) = h_2 g\, \rho_{water} + P_{ref}$
$HydrostaticPressure ({\bf x_3}) = h_3 g\, \rho_{water} + P_{ref}$
$HydrostaticPressure ({\bf x_4}) = h_4 g\, \rho_{water} + P_{ref}$

If my ${\bf x}_0$ is on the interface, (1) and (4) are well calculated. If ${\bf x}_0$ is on the top, (4) is wrong. No matter what ${\bf x}_0$ I selected, (2) and (3) have the same hydrostatic pressure (in OpenFoam), and that is not true.

Does anyone know how to fix this problem?

Greetings,
Lorena

hjasak · October 17, 2017, 00:42

Well, “works for me”...

Joking aside: this must work because if it did not my ships would not float.

Are you sure you are looking at the right thing?

Lorena2fdez · October 17, 2017, 03:33

Hi Jasak,

I have taken a second view at the code.

The wrong calculation is for p as it is assume that p=p_rhg-rhg where is rgh is a wrong aproximation (for my case) of the hydrostatic pressure. However, p is not used on the calculation. It is p_rhg, and this is correct evalueated. My fear was that the wrong value of p could affect the simulation.

On the attached image you can see rho*g*h where rho*g*h = p-p_rgh.
In our case (an uniform distribution of liquid on a inclined plane), if rho*g*h must be the hydrostatic pressure, the estimation is wrong.

I have seen that exist interFoamPressure on Foam-extend, This utility is used for take a good calculation of p.

Thanks and greetings,
Lorena

hjasak · October 17, 2017, 07:54

Hi Lorena,

D0 not worry; the decomposition is fine. Please look for some recent papers by Vukčević and Jasak and you will find a detailed derivation. Sorry, I do not recall,exactly which one it is...

October 17, 2017, 00:42		#2
hjasak Senior Member Hrvoje Jasak Join Date: Mar 2009 Location: London, England Posts: 1,905 Rep Power: 33	Well, “works for me”... Joking aside: this must work because if it did not my ships would not float. Are you sure you are looking at the right thing? __________________ Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk

October 17, 2017, 07:54		#4
hjasak Senior Member Hrvoje Jasak Join Date: Mar 2009 Location: London, England Posts: 1,905 Rep Power: 33	Hi Lorena, D0 not worry; the decomposition is fine. Please look for some recent papers by Vukčević and Jasak and you will find a detailed derivation. Sorry, I do not recall,exactly which one it is... sharonyue likes this. __________________ Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk

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