Setting Outlets with gravity..

Javier Larrondo

Hi, I have been doing the mathematical background when gravity is present. And need some confirmation and explanation in some terms of the equation.

I'm using the reduced pressure formulation or density_difference model to simulate gravity, this will introduce in the momentum equation a reference pressure p_ref, which satisfies the hydrostatic-equilibrium equation:

grad p_ref + rho_ref * G = 0

and if we define G = - grad(f) (**), where f is the gravitational potential, then you have

grad ( p_ref + rho_ref*f ) = 0

grad(p_ref) + grad(rho_ref*f) = 0

the momentum equation

rho_ref*DU/Dt = - grad(p) + h*grad^2(U) + F (*)

where F is the bouyancy-source term, which is defined like:

F = rho*G

defining

delta_rho = rho-rho_ref,

then

F = (rho_ref + delta_rho )*G using (**)

F = delta_rho*G - grad(rho_ref*f)

using this in the momentum equation

rho_ref*DU/Dt = - grad(p) + h*grad^2(U) + delta_rho*G - grad(rho_ref*f)

adding cero to the equation

rho_ref*DU/Dt = - grad(p) + h*grad^2(U) + delta_rho*G - grad(rho_ref*f) + grad(p_ref)+grad(rho_ref*f)

reordering and reducing terms, we have

rho_ref*DU/Dt = grad(p-p_ref) + h*grad^2(U) + (rho-rho_ref)*G

So, the question is :

How do I set the value of the pressure in an outlet if I know that his hydrostatic value p is -100 Pa? Do I have to set in the outlet now (p-p_ref)? The pressure ploted in VR-Viewer is p or (p-p_ref)? Now is this calculation correct?

if gravity in the -Z direccion (G=-g*k) then the equation of p_ref, reduces to

d(p_ref)/dz +rho_ref*g =0, so

p_ref=rho_ref*g*z

which value of z have to use? z=0 or z= height of the domain? is it constant? or depends on z? that means that I have know a new field p_ref?

thanks, for any comments

Javier