sandbox/lopez/transport.c

Concentration transport

This test case intends to help in the understanding and evaluation of the transport of “VOF concentration” fields if the velocity field does not fulfill the incompressible velocity condition (the velocity is not divergence-free). This test case is complementary of advect.c.

#include "advection.h"
#include "src/vof.h"
#include "curvature.h"

ipos(t,a,xo) compute the exact position of a planar interface located initially at x=xo if the velocity field is \mathbf{u} = (x+a) \mathbf{e_x}.

#define ipos(t, a, xo) (a*(exp(t)-1.) + xo*exp(t))

The volume fraction is stored in scalar field f which is listed as an interface for the VOF solver. We do not advect any tracer with the default (diffusive) advection scheme of the advection solver.

scalar f[], rho1[], rho2[];
scalar * interfaces = {f}, * tracers = NULL;
int MAXLEVEL = 5;

#define a 0.1
#define xo 0.

int main()
{

We center the Box of length 2 on the origin.

  L0 = 2.;
  origin (-L0/2., -L0/2.);

The scalar field rho1 and rho2 are “vof concentrations” associated with phase 1 determined by f obeying to the following equations,
\displaystyle D_t \rho_1 = \partial_t \rho_1 + \mathbf{u} \cdot \nabla \rho_1 = 0 \quad \text{and} \quad \partial_t \rho_2 + \nabla \cdot (\mathbf{u} \rho_2) = 0 Note that both expression are not the same if \nabla \cdot \mathbf{u} \neq 0. By default vof.h solves always D_t \rho = 0 unless attribute comp were set to 1 (the default is zero).

  f.tracers = {rho1, rho2};
  rho2.comp = 1;

We then run the simulation for a certain refinement.

  init_grid (1 << MAXLEVEL);
  run();
}

#define T 2.0

Initially the fieldsrho1 and rho2 are homogeneous and equal to 1. The interface is located initially in xo, the velocity field is unidirectional and linear in the x-direction and the timestep must be set considering CFL.

event init (i = 0)
{
  fraction (f, -x + xo);
  foreach() {
    rho1[] = f[];
    rho2[] = f[];
  }

  foreach_face(x)
    u.x[] = (x + a);
   dt = dtnext (timestep (u, DT));
}

We compare the computed time evolution of the interface position with the analytical solution.

event interf_pos (t += 0.1) {
  scalar pos[];
  position (f, pos, {1,0});
  fprintf (stdout, "%g %g %g\n", t, statsf(pos).max, ipos(t, a, xo));
}

At the final instant we compute the error made in the distribution in both fields.

event field (t = T) {

  scalar e[], ec[];
  foreach () {
    e[] = (f[] > 1e-12 ? rho1[]/f[]-1. : 0.);
    ec[] = (f[] > 1e-12 ? rho2[]/f[]-exp(-t) : 0.);
  }
  norm n = normf (e); 
  norm nc = normf (ec); 
  fprintf (stderr, "%d %g %g %g\n", N, n.avg, n.rms, n.max);
  fprintf (stderr, "%d %g %g %g\n", N, nc.avg, nc.rms, nc.max);
  //dump ();
}

set terminal @PNG enhanced size 640,640 font ",12"
set xlabel 't'
set ylabel 'interface position'
plot 'out' u 1:2  t 'vof.h', 'out' u 1:3 lt 2 w l t 'Analytical'

Position of the interface. (script)