src/ehd/stress.h

    Electrohydrodynamic stresses

    The EHD force density, \mathbf{f}_e, can be computed as the divergence of the Maxwell stress tensor \mathbf{M}, \displaystyle M_{ij} = \varepsilon (E_i E_j - \frac{E^2}{2}\delta_{ij}) where E_i is the i-component of the electric field, \mathbf{E}=-\nabla \phi and \delta_{ij} is the Kronecker delta.

    We need to add the corresponding acceleration to the Navier–Stokes solver.

    If the acceleration vector a (defined by the Navier–Stokes solver) is constant, we make it variable.

    event defaults (i = 0) {
      if (is_constant (a.x))
        a = new face vector;
    }

    We overload the acceleration event of the Navier–Stokes solver to add the electrohydrodynamics acceleration term.

    event acceleration (i++) {
      assert (dimension <= 2); // not 3D yet
      vector f[];
      foreach_dimension() {
        face vector Mx[];
        foreach_face(x)
          Mx.x[] = epsilon.x[]/2.*(sq(phi[] - phi[-1,0]) - 
                                   sq(phi[0,1] - phi[0,-1] + 
                                      phi[-1,1] - phi[-1,-1])/16.)/sq(Delta);
        foreach_face(y)
          Mx.y[] = epsilon.y[]*(phi[] - phi[0,-1])*
          (phi[1,0] - phi[-1,0] + 
           phi[1,-1] - phi[-1,-1])/sq(2.*Delta);
        boundary_flux ({Mx});

    The electric force is the divergence of the Maxwell stress tensor \mathbf{M}.

        foreach()
          f.x[] = (Mx.x[1,0] - Mx.x[] + Mx.y[0,1] - Mx.y[])/(Delta*cm[]);
      }

    If axisymmetric cylindrical coordinates are used an additional term must be added.

    #if AXI
      foreach()
        f.y[] += (sq(phi[1,0] - phi[-1,0]) +
    	      sq(phi[0,1] - phi[0,-1]))/(8.*cm[]*sq(Delta))
        *(epsilon.x[]/fm.x[] + epsilon.y[]/fm.y[] +
          epsilon.x[1,0]/fm.x[1,0] + epsilon.y[0,1]/fm.y[0,1])/4.;
    #endif
      boundary ((scalar *){f});

    To get the acceleration from the force we need to multiply by the specific volume \alpha.

      face vector av = a;
      foreach_face()
        av.x[] += alpha.x[]/fm.x[]*(f.x[] + f.x[-1])/2.;
    }

    Usage

    Tests