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Exemple 10 : 2D flow around a cylinder using the sharp interface method
In this example we study the flow around a cylinder in 2D using the sharp-interface method to represent the cylinder. The geometry of the flow takes the same basic case defined in Exemple 3. The exact dimensions for this example can be found in the following figure. Although in this case, we define the mesh as a simple rectangular mesh and we define the position and radius of the immersed boundary.
This case use
Definition of the mesh :
`#---------------------------------------------------
# Mesh
#---------------------------------------------------
subsection mesh
set type = dealii
set grid type = subdivided_hyper_rectangle
set grid arguments = 2,1: 0,0 : 32 , 16 : true
set initial refinement = 6
end
As for the Exemple 3 we define the boundary condition in order to have a inlet on the left and 2 slip boundary condition on the top and bottom and an outlet on the right of the domain.
# --------------------------------------------------
# Boundary Conditions
#---------------------------------------------------
subsection boundary conditions
set number = 3
subsection bc 0
set id = 0
set type = function
subsection u
set Function expression = 1
end
subsection v
set Function expression = 0
end
subsection w
set Function expression = 0
end
end
subsection bc 1
set id = 2
set type = slip
end
subsection bc 2
set id = 3
set type = slip
end
end
From there the only specific thing we need to define is the immersed boundary. At this point the immersed boundary tool only allow the representation of spherical objects. This mean the immersed boundary can represent a infinitely tine cylinder in 2D and a spherical shell in 3d.
The immersed boundary needs the following information. Some information is general for all immersed boundary as others are specific to a boundary. General information:
- The number of distinct immersed boundaries.
- The order of the approximation used to represent the immersed boundary. (Note: if the order given is equal or higher than 5 the polynomial order used will be the polynomial order used for the velocity cell: Qn multiplied by the number of dimension dim.)
- The number of mesh refinement around and inside the immersed boundary.
- The factor used to define the region refined around the boundary. This factor is a multiplier of the radius of the immersed boundary.
- The factor used to define the region refined inside the boundary. This factor is a multiplier of the radius of the immersed boundary.
- A Boolean variable that defines if the equations assemble inside the immersed boundary is the GLS Navier-Stokes equation. If false, the equation inside the particle is a Laplace equation for each of the variables (u,v,p). By default this variable is false since we are not interested in what is going on inside the particle.
Specific information:
- The position of the center of the immersed boundary and its radius.
- The boundary condition of the immersed boundary. Velocity of the boundary and rotation speed of the boundary.
- The position of the pressure reference point inside the boundary. This position is relative to the center of the immersed boundary. ( Since the boundary is closed, the fluid inside the boundary need a pressure reference.)
In this example we used 1 immersed boundary to represent the cylinder and applied a 3 order of local refinement around the boundary. The force applied on the boundary are evaluated using 1000 evaluation points and the order of the polynomial used is defined by the solver since we used a stencil of order 5. The boundary is stationary and center (0,0) and as a radius of 0.5. this gives the following parameter:
# --------------------------------------------------
# IB particles
#---------------------------------------------------
subsection particles
set number of particles = 1
set stencil order = 5
set initial refinement = 4
set refine mesh inside radius factor=0
set refine mesh outside radius factor=1.2
set integrate motion = false
set alpha =0.5
set assemble Navier-Stokes inside particles = false
subsection particle info 0
set x = 8
set y = 8
set z = 0
set u = 0
set v = 0
set w = 0
set omega x = 0
set omega y = 0
set omega z = 0
set pressure x =0.00001
set pressure y =0.00001
set radius = 0.5
end
end
This lead to the following solution for the velocity and pressure field. Velocity: Pressure: