# examples.convection.robin¶

Solve an advection-diffusion equation with a Robin boundary condition.

This example demonstrates how to apply a Robin boundary condition to an advection-diffusion equation. The equation we wish to solve is given by,

The analytical solution for this equation is given by,

where

>>> from fipy import CellVariable, FaceVariable, Grid1D, DiffusionTerm, PowerLawConvectionTerm, ImplicitSourceTerm, Viewer
>>> from fipy.tools import numerix
>>> nx = 100
>>> dx = 1.0 / nx

>>> mesh = Grid1D(nx=nx, dx=dx)
>>> C = CellVariable(mesh=mesh)

>>> D = 2.0
>>> P = 3.0

>>> C.faceGrad.constrain([0], mesh.facesRight)


We note that the Robin condition exactly defines the flux on the left, so we introduce a corresponding divergence source to the equation.

Note

Zeroing out the coefficients of the equation at this boundary is probably not necessary due to the default no-flux boundary conditions of cell-centered finite volume, but it’s a safe precaution.

>>> convectionCoeff = FaceVariable(mesh=mesh, value=[P])
>>> convectionCoeff[..., mesh.facesLeft.value] = 0.
>>> diffusionCoeff = FaceVariable(mesh=mesh, value=1.)
>>> diffusionCoeff[..., mesh.facesLeft.value] = 0.

>>> eq = (PowerLawConvectionTerm(coeff=convectionCoeff)
...       == DiffusionTerm(coeff=diffusionCoeff) - ImplicitSourceTerm(coeff=D)
...       - (P * mesh.facesLeft).divergence)

>>> A = numerix.sqrt(P**2 + 4 * D)

>>> x = mesh.cellCenters[0]
>>> CAnalytical = CellVariable(mesh=mesh)
>>> CAnalytical.setValue(2 * P * numerix.exp(P * x / 2)
...                      * ((P + A) * numerix.exp(A / 2 * (1 - x))
...                         - (P - A) * numerix.exp(-A / 2 *(1 - x)))
...                      / ((P + A)**2*numerix.exp(A / 2)
...                         - (P - A)**2 * numerix.exp(-A / 2)))

>>> if __name__ == '__main__':
...     C.name = '$C$'
...     CAnalytical.name = '$C_{analytical}$'
...     viewer = Viewer(vars=(C, CAnalytical))

>>> if __name__ == '__main__':
...     restol = 1e-5
...     anstol = 1e-3
... else:
...     restol = 0.5
...     anstol = 0.15

>>> res = 1e+10

>>> while res > restol:
...     res = eq.sweep(var=C)
...     if __name__ == '__main__':
...         viewer.plot()

>>> print(C.allclose(CAnalytical, rtol=anstol, atol=anstol))
True

Last updated on Jun 15, 2022. Created using Sphinx 5.0.1.