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:

Table of contents. This description is structured as follows:

• Rotation-symmetric pressure distribution
  • Problem setup
  • Mathematical model
  • Discretization
• Implementation & Post processing
  • Part 1: Rotation-symmetric one-phase flow simulation setup
  • Part 2: Main program flow
  • Part 3: Post-processing with ParaView

Problem setup

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

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 DuMu^x, 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 DuMu^x 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 DuMu^x, 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

Part 2: Main program flow

Part 3: Post-processing with ParaView