DFG flow around cylinder, laminar case (Re=20)

General information and references

The DFG-benchmark in general simulates flow in a pipe around a fixed circular object. Different inflow speeds decide about the behaviour of the flow. The benchmark is set up in 2D.

Please consider also the following literature concerning this benchmark:

[1] On High Order Methods for the Stationary Incompressible Navier-Stokes Equations; University of Heidelberg; Preprint 42/98, 1998; Download
[2] Higher-order finite element discretizations in a benchmark problem for incompressible flows; John, V.; Matthies, G.; Int. J. Num. Meth. Fluids 2001; 37: 885-903

Geometry and flow configuration:

The geometry of the benchmark consists of a simple channel of length 2.2 and height 0.41. At (x,y)=(0.2,0.2) a cylinder with diameter L=0.1 is placed.

The fluid is characterized by a viscosity-parameter of ν=0.001 and a density of ρ=1. The problem to solve is the stationary Navier-Stokes equation

\begin{displaymath} -\nu\Delta \vec u + \vec u\nabla \vec u+ \nabla p = 0, \quad \nabla\cdot \vec u=0.\end{displaymath}

On the upper wall, the lower wall and on the cylinger, Dirichlet-0-boundary with velocity

$\vec u=(u,v)=(0,0)$ is imposed. The left wall is set to a parabolic inflow profile with maximum inflow velocity U=0.3:

\begin{displaymath}(u,v)=(\frac{4Uy(0.41-y)}{0.41^2},0) \textrm{\quad for all\quad}(x,y)\in[0]\times[0,0.41]\end{displaymath}

The boundary conditions on the right wall are Neumann boundary conditions (i.e. do-nothing boundary conditions).

The defined benchmark setup results in a Reynolds number of

with the characteristic velocity
of the parabolic profile.

Verification of the results by the picture norm:

To verify the results visually, check the following pictures, which show the velocity, the pressure and the streamfunction in the domain:

Verification of the results by numerical data:

The DFG benchmark is knwon for many years and had been extensively analyzed. In the following, we describe how to check the computed results by numerical numbers, independent of the above pictures.

In the stationary case, the correctness of the results are measured by three numbers:

  • The drag coefficient
    $C_D = \frac{2 F_D}{\rho U^2_{\textrm{mean}} L}$
  • The lift coefficient
    $C_L = \frac{2 F_L}{\rho U^2_{\textrm{mean}} L}$
  • The pressure drop
    $\Delta p = p(A)-p(B)$ with A=(0.15,0.2) and B=(0.25,0.2)
    being the leftmost and rightmost point of the cylinder, respectively.

The surface of the cylinder is denoted by S. Let $\vec t$, $\vec n$ denote the tangential and normal vector of S, respectively. Then, the forces are calculated using the formula

\begin{displaymath}\left(\begin{array}{c} F_D \\ F_L \end{array}\right) = \int_S \sigma \vec n\ dS,\qquad \end{displaymath}

The definition of the stress tensor σ in this situation is

$\sigma := -pI + \rho\nu ( \nabla \vec u ).$

In this situation, the stress tensor can also be defined as the following surface integral, which gives usually better values in line integration along the surface S:

\begin{eqnarray*} \sigma &:=& -pI + \rho\nu ( Du_t/D\vec n ) \vec n \\ &=& -pI + \rho\nu <D\vec u*\vec n,\vec t> \vec t \end{eqnarray*}

This leads to the following formula for calculating the forces:

\begin{eqnarray*} F_D &=& \int_S (\rho\nu\frac{\partial u_t}{\partial \vec n} n_y-pn_x) dS \\ F_L &=& -\int_S (\rho\nu\frac{\partial u_t}{\partial \vec n} n_x+pn_y) dS \end{eqnarray*}

The reference values for this situation were calculated in [1] as follows:

  • CD=5.57953523384
  • CL=0.010618948146
  • Δp=0.11752016697