Neumann (zerogradient and open) boundary condition

8 vues (au cours des 30 derniers jours)
Kumaresh Kumaresh
Kumaresh Kumaresh le 16 Sep 2022
Hello all,
In my work of 2D rectangulat channel, right boundary is symmetry and top boundary is open to atmosphere. So, I need to use Neumann boundary condition.
At right side, u(:,Ny) = u(:,Ny-1);
At top side, u(1,:) = u(2,:);
At top and right boundary, the value is chosen from (N-1)th node.
The below code works well separately but when the same boundary condition is implemented in actual problem (involving numerical equations), I'm getting the velocities (u, v) as zeros at right and top boundaries.
Kindly someone share your ideas about it and guide me if I'm wrong anywhere.
Thank you
clear; clc; close all;
W =5; L=0.25;
Nx=20; %row
Ny=10; %column
x = linspace(0,L,Ny);
y = linspace(0,W,Nx);
[X, Y] = meshgrid(x,y);
dx = L/Ny; dy = W/Nx;
rhoo = 0.8; por = 0.3; mu =0.8E-5; dia = 0.003;
u = zeros(Nx, Ny);
v = zeros(Nx, Ny);
P = 101325 * ones(Nx, Ny);
GVecx = zeros(Nx,Ny);
GVecy = zeros(Nx,Ny);
%GVec = zeros(Nx,Ny);
% Boundary conditions
u(:,1) = 0; % left - 1st column @u=0
v(:,1) = 0; % left - 1st column @v=0
u(2,9) = 2;
u(3,9) = 3;
%for i = 2 : Nx-1
u(:,Ny) = u(:,Ny-1); % right - symmetry (Neumann condition)
v(:,Ny) = v(:,Ny-1); % right - symmetry (Neumann condition)
%end
%for j = 2 : Ny
u(1,:) = u(2,:); % top - open to atmosphere (Neumann condition)
v(1,:) = v(2,:); % top - open to atmosphere (Neumann condition)
%end
u(Nx,:) = 0; % bottom
v(Nx,:) = 0; % bottom
u(Nx,1) = 0; % bottom left corner(for left inlet condition @ u =0)
v(Nx,1) = 0; % bottom left corner(for left inlet condition @ u =0)
u(1,1) = 0; % top left corner (for left inlet condition @ v =0)
v(1,1) = 0; % top left corner (for left inlet condition @ v =0)
  3 commentaires
Torsten
Torsten le 18 Sep 2022
Modifié(e) : Torsten le 18 Sep 2022
Maybe in your actual problem code, you initialize the velocities once at the start to 0, but don't update them as
u(:,Ny) = u(:,Ny-1); % right - symmetry (Neumann condition)
v(:,Ny) = v(:,Ny-1); % right - symmetry (Neumann condition)
%end
%for j = 2 : Ny
u(1,:) = u(2,:); % top - open to atmosphere (Neumann condition)
v(1,:) = v(2,:); % top - open to atmosphere (Neumann condition)
during the computation.
If you did update them, u(:,Ny-1), v(:,Ny-1), u(2,:) and v(2,:) also had to be 0 to make u(:,Ny), v(:,Ny), u(1,:) and v(1,:) equal 0 (which is not the case, I guess ?)
Kumaresh Kumaresh
Kumaresh Kumaresh le 18 Sep 2022
Yes well said. I need to update u(:,Ny-1), v(:,Ny-1), u(2,:) and v(2,:) inside my for loop for i and j.
Thank you Mr. Torsten.
That solved my query.

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