Finite Volume Demo

This is a simple implementation of a cell-centered finite volume method in two dimensions,.

Governing Law

We are interested in solving governing equations for ideal compressible and inviscid fluid, known also as Euler equations:

where $\rho$ is density, $\boldsymbol{v} = (v_1, v_2)$ velocity field and $e$ is internal gas energy. Pressure $p$ is defined by the equation of state for ideal gas as

where $\gamma$ is the ratio of specific heats $\gamma = \frac{c_p}{c_v} = 1.4$ for ideal gas.

In order to discretize the system with finite volumes, we rewrite the equations in conservation form

which can be written in conservative variables $\boldsymbol{w} = (w_1, w_2, w_3, w_4)^T$, where $w_1 = \rho$, $w_2 = \rho v_1$, $w_3 = \rho v_2$, $w_4 = e$.

Euler equations in conservative form now read

The constitutive law for ideal gas rewritten in conservative variables becomes

Discretization

The unkown $\boldsymbol{w}$ is from now on discrete and located in cell centers. After discretization of space derivatives, the time-change of $\boldsymbol{w}$ in the mesh cell/element with index $i$ is given by

with the following notation:

• $|C_i|$ is the surface/area of element $i$ in the mesh
• $N(i)$ denotes the set of all indices of neighbours of element $i$, i.e. all elements $j$ that share an edge with element $i$
• $n_{ij}$ is an oriented edge normal perpendicular to edge between elements $i$ and $j$ and pointing from element $i$ towards element $j$
• $l_{ij}$ is the length of edge separating elements $i$ and $j$
• $\mathbf{H}$ is the numerical flux function. The implementation uses AUSM flux splitting scheme

For time discretization, an explicit Euler method was used. The discretization is only first-order accurate in time and space.

Implementation and Numerical Test

The numerical test runs a 2D Riemann problem configuration number 3:

The initial condition in this configuration is

Quadrant	$\rho$	$v_1$	$v_2$	$p$
1	1.5	0.0	0.0	1.5
2	0.5322581	1.2060454	0.0	0.3
3	0.1379928	1.2060454	1.2060454	0.0290323
4	0.5322581	0.0	1.2060454	0.3

The obtained solution at time $T = 0.3$ on a Cartesian mesh with $200 \times 200$ cells: