4.1. Methodology for 2D Fluid Structure Interaction
4.1.4. Ghost Solid Fluid Method in 2D
4.1.4.1. Defining the Left and Right Status for the Riemann Solver
As MGFM based Ghost Fluid methods involve the solution of interface Riemann problems at the interface for each medium, it is necessary to identify the left and right status across the interface. We need to find two grid points B (Eulerian) and D′′(Lagrangian) across the interface which lies in the normal direction, and then use the variable values at those grid points. But it is almost impossible to find the nearest grid points to lie exactly on the line normal to the interface and hence, some approximations are necessary. Furthermore, as one of the medium uses Eulerian and the other medium uses Lagrangian grid, one of these left or right grid point will be Eulerian and the other will be Lagrangian. In this subsection, the procedure for finding the left (B) and right ( D′′) grid points for the both fluids (air and water) and the solid Riemann problems is discussed under the assumption that the left medium is fluid and the right medium is solid.
For the fluid:
Interface
A B
C
Normal to the Interface at A
Lagrangian Grid around C D"
Fig. 4.1.2. Identification of the left and right Eulerian nodes B and C for each Eulerian grid node A just left of the interface in a line passing through and parallel to the direction of the normal at the point A. D′′is the Lagrangian grid node nearest to the Eulerian node C.
(The two dotted lines parallel to the interface on its either side determines a band within which the points B and C need to be selected)
The advantage of using the level set is that we know the normal to the interface at every Eulerian grid point by using the following equation:
n ϕ
ϕ
= ∇
∇ r
..………..(4.1.4.1)
For each Eulerian grid point A just left to the interface, the left and right points B and C, respectively, need to be found lying on a line passing through and parallel to the normal to the interface at A. The two dotted lines parallel to the interface on its either side (Fig. 4.1.2.) determine a band within which the points B and C need to be selected. The distance of either of these parallel dotted lines from the interface can be denoted by EPS and we can takeEPS =3 min(∆ ∆x, y)for example.
We want B and C to be the Eulerian grid points but it is difficult to find the B and C
required. We choose the points B and C such that AB and AC both make the minimum angles with the normal with respect to the other neighboring grid points.
The steps for finding B are as follows:
i. Calculate the angle made by the normal to the horizontal line, which we call the reference angle.
ii. Find a grid point B' from all the 8 neighboring grid points of A, for which the ϕ(x y, )>0and for which the difference with the reference angle is the minimum.
iii. If AB' < EPS, set B' as A and search for a new B'.
( )
3 min ,
EPS= ∆ ∆x y for example.
iv. If AB′ −EPS <min(∆ ∆x, y), set B=B'
Similarly the grid point C should be found for which ϕ(x y, )<0and
( )
min ,
AC−EPS < ∆ ∆x y difference between the reference angle and the angle made by AC with the horizontal is the minimum.
Now, point B is selected as the left point for the Riemann solver. But, point C cannot be used as the right node for the Riemann solver as it is an Eulerian grid point. The right node D′′should be a grid node of the Lagrangian grid of the solid medium.
Hence, we need to perform a search within the nearest Lagrangian grid points to C and find one for which AD′′makes the minimum angle with the normal to the interface at A.
Steps to find the point D′′can be described as follows:
i. The indexes ( )i j, are the indexes of the cell that contains one or more Lagrangian grid node(s).
ii. The array which keeps the information of the global indexes of Lagrangian grid nodes that exists in a particular cell, which we name as “Cell”, shall be used to identify the Lagrangian Grid nodes near to Eulerian node C.
The definition of the “Cell” array will be given in section 4.1.8.1.
iii. Find the Lagrangian grid point D′′nearby to C, for which AD′′makes minimum angle with the normal at A.
For the Solid:
As we know the normal direction at each of the Eulerian grid points, we can use bilinear interpolation to calculate the normal direction at each of the Lagrangian grid points.
In order to find a left and right Lagrangian grid node, B′′and C′′, respectively, to a Lagrangian grid point A′′just right to the interface, we have to follow the same procedure for finding B and C in case of fluid medium for the Eulerian grid. Now, C′′is taken as the right node D′′for the Riemann solver as shown in Fig. 4.1.3. But, B′′cannot be used as the left node as it is in Lagrangian frame of reference. The left node B should be a grid node of the Eulerian grid of the fluid medium. We need to perform a search within the nearest Eulerian grid points to B′′ in order to find B (Fig.
4.1.3), for which A B′′ makes the minimum angle with the normal to the interface atA′′.
Interface
A’’
B’’
C’’
Normal to the Interface at A’’
Eulerian Grid around B’’
B
D"=C"
Fig. 4.1.3. Identification of the left and right Lagrangian nodes B′′and C′′for each Lagrangian grid node A′′just right of the interface in a line passing through and parallel to the direction of the normal at the point A′′. B is the Eulerian grid node nearest to the Lagrangian node B′′.
(The two dotted lines parallel to the interface on its either side determines a band within which the points B′′ and C′′need to be selected)
Steps to find the point B can be described as follows:
i. The array “Lag_LocalToGlobal” contains global indexes of each of the Lagrangian grid points. Find the global index for the point B′′.
ii. The arrays “Lag_GlobalToCell_i” and “Lag_GlobalToCell_j” contain the row and column indexes of the “Cell” (for definition of Cell array, see Section 4.1.8.1) which occupy B′′.
iii. Find the Eulerian grid point B nearby to B′′, for which A B′′ makes minimum angle with the normal at A′′.