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-   -   coupling vof with single phase flow and gravity term (http://www.cfd-online.com/Forums/main/121675-coupling-vof-single-phase-flow-gravity-term.html)

alame005 August 2, 2013 14:46

coupling vof with single phase flow and gravity term
 
Hey everyone,
currently developing two phase flow solver. I have written a 2D navier stokes solver for a single phase flow and tested it. I also wrote a VOF-plic algorithm and tested that and it works. Now when I couple the two together and add gravity it doesnt work.

Any suggestions on how to treat gravity ?

I have modified my single phase solver to include dimensional quantities and density and rho are used explicitly.
It's based on a predictor corrector step.
When I couple the vof with the single phase I treat the gravity term as a constant and add it to the non-linear and viscous terms in the predictor step.

Is that the way to go about it ? I'm trying to simulate the Rayleigh-Taylor problem.
Your input would be highly appreciated.

rmh26 August 2, 2013 15:17

It would help to know what exactly isn't working. Using the current values of density and viscosity for create an explicit time stepping method should be ok. Not sure what you mean by treating the gravity term as a constant though.

Also you can incorporate gravity as a boundary condition at the interface instead of body force that acts throughout your whole domain.

alame005 August 2, 2013 15:26

Quote:

Originally Posted by rmh26 (Post 443549)
Not sure what you mean by treating the gravity term as a constant though.

Basically your predictor step looks like this for single phase no gravity

u*-u = dt * ( nonlin_terms + visc terms)

what I did was just add gravity to that:

u*-u = dt * ( nonlin_terms + visc terms +g ) where g = -9.8

and it doesnt work. When I plot the velocity vectors I dont get vectors pointing down.

Also how would you apply gravity to boundary conditions ?

rmh26 August 2, 2013 15:56

You should be including the density in your gravity term

u*-u = dt * ( nonlin_terms + visc terms +g*rho(x) )

rho(x) is the density based on the VOF in that cell


You can write the stress condition at the interface as

(P1 − P2)n + (\tau2\tau1 ) n = \frac{1}{Ca} [ \sigman(∇ n) − ∇s\sigma -Bo zn}

Bo = Bond Number and z is the elevation of the interface.

alame005 August 2, 2013 16:20

I had divided my equations by rho basically solving for the non-conservative form.

For the boundary condition at the interface, does it still apply if I don't have the surface stress applied yet?
From what I understand the Rayleigh-Taylor doesnt need the surface stress so can I just eliminate it from the equation you stated ?

Again thanks for your prompt reply.

rmh26 August 2, 2013 16:34

Your density shouldn't be constant throughout the domain so if you are dividing out the density you should be introducing a density ratio term, rho1/rho2. then the equation would look more like

u*-u = dt * ( nonlin_terms + visc terms +g*(x)*rho(x) )/rho2

u*-u = dt * ( nonlin_terms + visc terms)/rho2 +dt*g*(x)*R

where R = rho(x)/rho2

Is you interface perfectly flat, or have you introduced a perturbation to it?
You should examine you pressure field to see if your getting the hydrostatic pressure correct.



The boundary condition will still hold if you set the surface tension to zero and eliminate those terms.

alame005 August 2, 2013 16:42

My density varies it isn't constant. Is rho2 a reference density ?

The way I do it is I say rho(x) = rho1*c + (1-c)*rho2
where c = volume fraction, rho1 = rho_liquid and rho2 = rho_gas

The rho I divide by is rho(x) which is defined as above. So should I be multiplying g by rho1 or rho2 in this case ?

Also my interface is perturbed. Basically for initial conditions, u = v = p = 0 I specify the perturbation by y(x) = 2+0.05*cos(2*pi*x) to give the interface that curve shape. Domain is 1x4

hilllike August 6, 2013 11:02

That depends on the initial situations and your solver, for the boundary condition of the Poisson equation should be consist with the pressure gradient and gravity terms.


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