I am a little bit hard to understand the reflective wall boundary condition discussed in Fox 2018, JCP, Passalacqua et al. 2010, CES, Patal et al. 2019, JCP... I draw an image which is useful to discuss. For the last case, if we have two velocity nodes flow out of the computational domain, we should have same two velocity nodes with opposite sign flow in. Overall there are four nodes, which is beyond of the nodes number, e.g., three-nodes HyQMOM. How can we handle this problem? or my understanding of the reflective wall condition is misleading? Or we need to assume a uniform velocity (one-node) in the wall?
Last Edit: Mar 24, 2020 20:09:03 GMT -6 by dongyueli
Post by Alberto Passalacqua on Mar 26, 2020 0:54:13 GMT -6
if you think to a fully reflective condition, you change sign to the normal component of the velocity vector. Each node has a velocity, so that should be possible to do for all. Alternatively, and more generally, since this approach works also for a diffuse condition, you can think about moments and fluxes across the wall "face", which leads to the same answer once you do the math.
> if you think to a fully reflective condition, you change sign to the normal component of the velocity vector. Each node has a velocity, so that should be possible to do for all.
Thanks for your reply. Yes I am considering the fully reflective condition and this is where I do not understand. For example, lets take a one-dimensional case. Initially all the particles moves to the left with 1m/s, after they touch the left wall boundary, they reflects to moving right with -1m/s. If I got it right, the key point is that at this time, there are particles moving left and right simultaneously (in each cell, not only the wall adjacent cell). If we say: at the beginning there are only one velocity node (1m/s), now there are two nodes (1m/s and -1m/s). If we have two velocity nodes at the beginning, we end up with four nodes after they collides with the wall.
This is what I draw in the picture. There should be misleading in my thought, but I did not figure out where it is...
Post by Alberto Passalacqua on Mar 30, 2020 17:57:22 GMT -6
The number of nodes does not change. If you have two nodes, you still have two nodes. Each node has its own velocity vector. To understand the implementation, you have to think to a ghost cell behind the wall and enforce the constraint on the flux through the wall face. The velocity in the cell within the domain is unchanged. The velocity in the ghost cell is defined so that the flux through the wall is null, so the normal component of this velocity is equal and opposite to the one you have in the cell on the flow side. See the article from Fox (2008) on a third-order moment method.
Thank you Alberto for your reply. Yes I found Fox 2008's paper. Its consistent with you wrote, which I still did not get it. Another paper I found is that published by Desjardins et al. 2008. In section 4.2.2, it says:
Note that in this example, the particle velocity is bivariate near the impingement point at the wall due to the incoming velocity (1,-1) and the outgoing velocity of (1,1).
Please see the following image, if I got it right, the black region inflow is univariate velocity region (1,-1) (one u node, one v node). Once it touches and is reflected by the wall, the triangle region (I also marked it in red) becomes bivariate: (1,-1) and (1,1) (one u node, two v nodes).
Follow this spirit, if we have bivariate inflow, lets say (1,-1) and (0.5, -0.5), does it mean we end up with four-variate?: (1,-1) (1,1) (0.5, -0.5) (0.5, 0.5). Namely two u nodes, but four v nodes?...
Last Edit: Mar 30, 2020 22:04:18 GMT -6 by dongyueli
I followed the ghost cell spirit and hard coded the velocity nodes. The results look good (see images in below). But the velocity impinging left and right is only one node. I am not sure what will happen if it impinges to the left with three u_x nodes...I did not try it but proceed with other implementation.