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

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