This test case solves the Poiseuille flow with immersed boundary conditions.

Configurations

Physical domain.

The global boundaries are defined by the intevales:

\begin{align} [0, 2] &\quad\text{in the flow direction,} & [-1, +1] &\quad\text{otherwise.} \end{align}

The immersed boundary is either a flatchannel of a cylinder:

\begin{align} y = \pm h &\quad\text{(flatchannel),} & x^2 + y^2 = h^2 &\quad\text{(cylinder),} \end{align}

where \(h = \frac{1}{2} \frac{262144}{177147} \), which is half the wolf interval of the Pythagorean scale. This has been chosed to avoid the boundary to match any grid.

Solutions

The velocity has the following parabolic shape: solution:

\begin{align} u(y) = \hat{u} \left(1 - \frac{y^2}{h^2}\right) &\quad\text{(flatchannel),} & u(r) = \hat{u} \left(1 - \frac{r^2}{h^2}\right) &\quad\text{(cylinder).} \end{align}

The flow is pushed by the navier source term, which should be equal to:

\begin{align} f_x = 2 \frac{\mu\hat{u}}{h^2} &\quad\text{(flatchannel),} & f_y = 4 \frac{\mu\hat{u}}{h^2} &\quad\text{(cylinder),} \end{align}

where \( \mu \) is the dynamic viscosity of the fluid. Using the navier source term yields a constant pressure gradient, which allows the domain to be periodic in the flow direction.

The average velocity is computed from the integral:

\begin{align} V \bar{u} = \int_{-h}^{+h}{u(y)\mathrm{d}y} = 2 h \hat{u} \int_0^1 {(1 - z^2) \mathrm{d}z} &\quad\text{(flatchannel),} \\ V \bar{u} = \int_{0}^{2\pi}\int_{0}^{+h}{u(r)r\mathrm{d}r\mathrm{d}\theta} = 2 \pi h^2 \hat{u} \int_0^1 {(1 - z^2) z \mathrm{d}z} &\quad\text{(cylinder),} \end{align}

giving:

\begin{align} \bar{u} = \frac{2}{3}\hat{u} &\quad\text{(flatchannel),} & \bar{u} = \frac{1}{2}\hat{u} &\quad\text{(cylinder).} \end{align}

In the actual test case, the dynamic viscosity of the fluid is set to match a given Reynolds number, which is defined by:

\begin{align} \mathit{Re} = \frac {2h \times \bar{u}} {\mu} \end{align}

We choose to set \(\bar{u}\) and \(\mu\) as follows:

\begin{align} \bar{u} &= \frac {1} {2h} & \mu = \frac {1} {\mathit{Re}} \end{align}

Values are given in the table below.

and \( \mathit{Re} = 20\).

Boundary conditions

Grids

Test case list

Runtime parameters

The following common settings are used:

• The time step is set to 60.
• The MUMPS solver is used by default.

3. Results

poiseuille