[fede32]

Hi, i'm dealing with a trank drainage problem. I'm using the solver interfoam, and i don't know how to set the boundary conditions in the outlet for p_rgh. Any suggestions?

[Saideep]

Hi;

May be you could just elaborate your condition even better.

But in general the boundary conditions for pressure are at inlet it is a zeroGradient Pressure and at outlet it would be a uniform value fixed to 0.

But this is the most general case. Based on your model setup you can try to vary the B.C's.

Saideep

[fede32]

I am using totalPressure for the boundary conditions for p_rgh at inlet.At inlet Ii have

type totalPressure
p0 uniform 0

where uniform 0 is the atmosphere pressure. Should i set

type fixedValue
value uniform 0

for p_rgh at outlet, if i wanna simulate that it's open to atmosphere? Or should i set

type fixedValue
value uniform 0 - delta

where delta = r*g*(h_inlet-h_outlet)

[Saideep]

Hi;

I guess maybe it is good to try with the first option only that is pressure with fixed value of 0.

May be it would be easy is you could give an overview of your case like the figure. I guess your case is somehow similar to that of the damBreak case.

Like i use interfoam to deal with porous media, micro channel flows and i deal with the range of e-6meters. So, I usually remove the gravitational effect for my simulations and therefore i no longer compute total pressure with the static pressure term.

Saideep

[fede32]

Hi, here i let you an image of the case. The tank is open to atmosphere where p = 0.
https://www.dropbox.com/s/5bi9hb7pze..._prhg.png?dl=0
The top of the tank is at z=1, and the bottom is at z = 0. The endo of the network is at z = -2.7. The level of the fluid is at z=0.8.
So i was thinking that if i set totalPressure = 0 at the top, i must set fixedValue p_rgh= -27000 at the end of the network for the pressure. (So p = 0) Is that ok?

[Andrea_85]

Hi,

p_rgh in interFoam is the pressure minus the hydrostatic contribution. I am not using gravity in my simulation but my understanding of the difference between p and p_rgh is that when you set BC for pressure you don't have to take care about the hydrostatic pressure. This is automatically handled by the code.

If you set p_rgh=-27000 Pa at the outlet i am pretty sure you will get p=-27000*2, which is not what you want. Btw this you can easily check.

If you take for example the capillaryRise test case, there you have p_rgh=0 both at the inlet and outlet and there is gravity.

So if inlet and outlet are open to the atmosphere i would use fixedValue equal to zero for both.

Best,
Andrea

[fede32]

Hi,
i check this and when i set fixedValue p_rgh = 0 Pa at the outlet, then p = 27 000 Pa.
If i set p_rgh = -27 000 Pa, then p = 0 Pa,
thanks for the suggestion!

[Shakiro]

Hi,
I'm facing a similar problem. I try to simulate the non-swirl case of a tank drainage with a geometry according to the paper of Park & C.H Sohn "Experimental and numerical study on air cores for cylindrical tank draining":

A cylindrical tank with a Diameter of 90 mm und a height of 450 mm is drained through a concentric outlet pipe with a diameter of 6 mm and a length of 15 mm. The fluid level in the tank is 350 mm at t = 0. At the inlet (top of the tank) and at the outlet (bottom face of the outlet pipe) I want to set the pressure condition as open to atmosphere.

Therefore, I chose (also based on several papers) for p_rgh:

Inlet
type totalPressure
p0 uniform 0

Outlet
type fixedValue
value uniform 0

However, when evaluating the pressure after a certain time I get:

Inlet
p_rgh = 0,0002 Pa
p = - 16,34 Pa

Outlet
p_rgh = 0 Pa
p = -9588 Pa

As a result also the draining velocity is much higher than expected from the paper.

Do you think this error might result from my pressure boundary condition in p_rgh? But why do I get such a negative p at the outlet? How should I adapt it to obtain an open to atmosphere condition?

[gkarlsen]

Originally Posted by Shakiro

Hi,


However, when evaluating the pressure after a certain time I get:

Inlet
p_rgh = 0,0002 Pa
p = - 16,34 Pa

Outlet
p_rgh = 0 Pa
p = -9588 Pa

It has to do with the static head being negative in the positive z quadrant, so: p_rgh=p-rgh could give you a negative pressure. I find it much less confusing to work with prghPressure and prghTotalPressure boundary conditions instead. Maybe try that? As to the difference in draining rate that you observe it could be related to mesh inaccuracies or turbulence modelling perhaps?

[Shakiro]

Originally Posted by gkarlsen

It has to do with the static head being negative in the positive z quadrant, so: p_rgh=p-rgh could give you a negative pressure. I find it much less confusing to work with prghPressure and prghTotalPressure boundary conditions instead. Maybe try that? As to the difference in draining rate that you observe it could be related to mesh inaccuracies or turbulence modelling perhaps?

Using the prghTotalPressure boundary condition for both inlet and outlet, the simulation finally reproduces the results of the measurement and the paper, respectively. Thank you very much!