examples.convection.exponential1DSource.mesh1DΒΆ

Solve the steady-state convection-diffusion equation with a constant source.

Like examples.convection.exponential1D.mesh1D this example solves a steady-state convection-diffusion equation, but adds a constant source, S_0 = 1, such that

\nabla \cdot \left(D \nabla \phi + \vec{u} \phi \right) + S_0 = 0.

>>> diffCoeff = 1.
>>> convCoeff = (10.,)
>>> sourceCoeff = 1.

We define a 1D mesh

>>> from fipy import CellVariable, Grid1D, DiffusionTerm, ExponentialConvectionTerm, DefaultAsymmetricSolver, Viewer
>>> from fipy.tools import numerix

>>> nx = 1000
>>> L = 10.
>>> mesh = Grid1D(dx=L / 1000, nx=nx)

>>> valueLeft = 0.
>>> valueRight = 1.

The solution variable is initialized to valueLeft:

>>> var = CellVariable(name="variable", mesh=mesh)

and impose the boundary conditions

\phi = \begin{cases} 0& \text{at $x = 0$,} \\ 1& \text{at $x = L$,} \end{cases}

with

>>> var.constrain(valueLeft, mesh.facesLeft)
>>> var.constrain(valueRight, mesh.facesRight)

We define the convection-diffusion equation with source

>>> eq = (DiffusionTerm(coeff=diffCoeff)
...       + ExponentialConvectionTerm(coeff=convCoeff)
...       + sourceCoeff)

>>> eq.solve(var=var,
...          solver=DefaultAsymmetricSolver(tolerance=1.e-15, iterations=10000))

and test the solution against the analytical result:

\phi = -\frac{S_0 x}{u_x} + \left(1 + \frac{S_0 x}{u_x}\right)\frac{1 - \exp(-u_x x / D)}{1 - \exp(-u_x L / D)}

or

>>> axis = 0
>>> x = mesh.cellCenters[axis]
>>> AA = -sourceCoeff * x / convCoeff[axis]
>>> BB = 1. + sourceCoeff * L / convCoeff[axis]
>>> CC = 1. - numerix.exp(-convCoeff[axis] * x / diffCoeff)
>>> DD = 1. - numerix.exp(-convCoeff[axis] * L / diffCoeff)
>>> analyticalArray = AA + BB * CC / DD
>>> print(var.allclose(analyticalArray, rtol=1e-4, atol=1e-4))
1

If the problem is run interactively, we can view the result:

>>> if __name__ == '__main__':
...     viewer = Viewer(vars=var)
...     viewer.plot()
Last updated on Jun 27, 2023. Created using Sphinx 6.2.1.