Source of a river

In this example we impose a variable flow rate for a “source” located inside the computation domain.

#include "saint-venant.h"
#include "discharge.h"

The domain is 10 metres squared, centered on the origin. Time is in seconds.

#define LEVEL 8

int main()
  size (10.);
  origin (- L0/2., - L0/2.);
  G = 9.81;
  N = 1 << LEVEL;

Boundary conditions

We create a new boundary for the source, with a Neumann condition for the normal velocity (i.e. an inflow).

bid source;
u.n[source] = neumann(0);

The flow rate varies in time and is set by computing the the elevation ηs of the water surface necessary to match this flow rate.

double etas;

event inflow (i++) {
  etas = eta_b (0.1*(1.1 - cos(4.*π*t)), source);
  h[source] = max (etas - zb[], 0.);
  η[source] = max (etas - zb[], 0.) + zb[];

Initial conditions

The river bed is a single valley. The source is created by masking and is a narrow slot located at the head of the valley.

event init (i = 0)
  mask (fabs(x) < 0.5 && fabs(y - 3.5) < Δ/2. ? source : none);

We start with a dry riverbed, so that the problem does not have a natural timescale the Saint-Venant solver can use. We set a maximum timestep to set this timescale.

  DT = 1e-2;

    zb[] = (- cos(x) + y)/2.;
  boundary ({zb});


We compute the time-derivative of the total water volume (i.e. the net flow rate), and make a GIF movie.

event logfile (i++; t <= 2.) {
  static double volo = 0., to = 0.;
  double vol = statsf(h).sum;
  if (i > 0)
    fprintf (ferr, "%g %.6f %g %g\n", t, (vol - volo)/(t - to), vol, etas);
  volo = vol, to = t;

event output (i += 5) {
  output_ppm (h, min = 0, max = 0.05, file = "source.gif");


Evolution of the flow rate

Evolution of the flow rate

Evolution of the water level

Evolution of the water level

See also