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Rotation-symmetric pressure distribution

In this example, you will learn how to

  • solve a rotation-symmetric problem one-dimensionally
  • perform a convergence test against an analytical solution
  • apply the Rotational Extrusion filters in ParaView for a two-dimensional visualization of the one-dimensional results

Result. With the Rotational Extrusion and the Warp By Scalar filters in ParaView, the pressure distribution of this example looks as shown in the following picture:

Rotation-symmetric pressure distribution
Fig.1 - Rotation-symmetric pressure distribution on a disc (warped to 3D).

Table of contents. This description is structured as follows:

Problem setup

We consider a single-phase problem that leads to a rotation-symmetric pressure distribution. The following figure illustrates the setup:

Rotation-symmetric setup
Fig.2 - Setup for the rotation-symmetric problem. The pressure boundary conditions are shown by the colored lines and the simulation domain is depicted in grey.

This could, for example, represent a cross section of an injection/extraction well in a homogeneous and isotropic porous medium, where the well with radius

r_1
is cut out and the injection/extraction pressure
p_1
is prescribed as a Dirichlet boundary condition. At the outer radius
r_2
, we set the pressure
p_2
. In the polar coordinates
r
and
\varphi
, the solution to this problem is independent of the angular coordinate
\varphi
and can be reduced to a one-dimensional problem in the radial coordinate
r
. Therefore, in this example, we want to solve the problem on a one-dimensional computational domain as illustrated by the orange line in the above figure.

Mathematical model

In this example we are using the single-phase model of DuMux, which considers Darcy's law to relate the Darcy velocity

\textbf u
to gradients of the pressure
p
. In the case of rotational symmetry, the mass balance equation for the fluid phase can be transformed using polar coordinates:

-\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\varrho k}{\mu} \frac{\partial p}{\partial r} \right) = 0,

where we identify the Darcy velocity in radial direction

u_r = -\frac{k}{\mu} \frac{\partial p}{\partial r}
, and where
k
is the permeability of the porous medium,
\mu
is the dynamic viscosity of the fluid, and
\varrho
is the fluid density.

Discretization

We employ a finite-volume scheme to spatially discretize the mass balance equation shown above. Let us consider a discretization of the one-dimensional domain into control volumes

K_i = \left[ r_i, r_{i+1} \right]
. The discrete equation describing mass conservation inside a control volume
K_i
is obtained by integration and reads:

- 2 \pi r_{i+1} \left( \varrho u_r \right)_{r_{i+1}} + 2 \pi r_i \left( \varrho u_r \right)_{r_i} = 0.

For this type of equation, the implementation of the finite-volume schemes in DuMux is based on the general form:

\sum_{\sigma \in \mathcal{S}_K} | \sigma | \left( \varrho \textbf u \cdot \textbf n \right)_\sigma = 0,

where

\sigma
are the faces of the control volume and where the notation
( \cdot )_\sigma
was used to denote quantities evaluated for a face
\sigma
. The area of a face is denoted with
| \sigma |
. Thus, comparing the two equations we identify
| \sigma | = 2 \pi r_\sigma
for the case of rotational symmetry on a disc. Here,
r_\sigma
refers to the radius at which the face is situated in the one-dimensional discretization.

In DuMux, this is rotational extrusion is approximated by using modified control volume volumes and control volume face areas for the mid-point integration rule.

Implementation & Post processing

Part 1: Rotation-symmetric one-phase flow simulation setup

➡️ Click to continue with part 1 of the documentation

Part 2: Main program flow

➡️ Click to continue with part 2 of the documentation

Part 3: Post-processing with ParaView

➡️ Click to continue with part 3 of the documentation