13th Step of the 12 steps to Navier-Stokes 😑

     Indeed, the 13th step now exists! This case is called the case of the backward facing step (BFS)! ⬜ This is an unofficial continuation to this. If I get it approved by Dr. Barba, then it will be official. The original series is in Python but I coded this in MATLAB without using many MATLAB specific functions so the code can be translated to other programing languages 🖧 quite easily.

The Code

%% clear and close

close all

clear

clc

%% define spatial and temporal grids

l_square = 1; % length of square

h = l_square/50; % grid spacing

dt = 0.1; % time step

L = 21; % cavity length

D = 2; % cavity depth

Nx = round((L/h)+1); % grid points in x-axis

Ny = round((D/h)+1); % grid points in y-axis

nu = 0.000015111; % kinematic viscosity

Uinf = 0.0060444; % free stream velocity / inlet velocity / lid velocity

cfl = dt*Uinf/h; % % cfl number

travel = 4; % times the disturbance travels entire length of computational domain

TT = travel*L/Uinf; % total time

ns = TT/dt; % number of time steps

Re = l_square*Uinf/nu; % Reynolds number

rho = 1.2047; % fluid density

%% initialize flowfield

u = zeros(Nx,Ny); % x-velocity

v = zeros(Nx,Ny); % y-velocity

p = zeros(Nx,Ny); % pressure

i = 2:Nx-1; % spatial interior nodes in x-axis

j = 2:Ny-1; % spatial interior nodes in y-axis

[X, Y] = meshgrid(0:h:L, 0:h:D); % spatial grid

maxNumCompThreads('automatic'); % select CPU cores

%% solve 2D Navier-Stokes equations

for nt = 1:ns

pn = p;

p(i, j) = (pn(i+1, j)+pn(i-1, j)+pn(i, j+1)+pn(i, j-1))/4 ...

-h*rho/(8*dt)*(u(i+1, j)-u(i-1, j)+v(i, j+1)-v(i, j-1)); % pressure poisson

p(1, :) = p(2, :); % dp/dx = 0 at x = 0

p(Nx, :) = 0; % p = 0 at x = L

p(:, 1) = p(:, 2); % dp/dy = 0 at y = 0

p(:, Ny) = p(:, Ny-1); % dp/dy = 0 at y = D

p(round(1:1*Nx/L), round(1:1*Ny/D)) = 0; % step geometry

p(round(1*Nx/L), round(1:1*Ny/D)) = p(round(1*Nx/L)+1, round(1:1*Ny/D)); % dp/dx = 0 at x = 1 and y = 0 to 1

p(1:round(1*Nx/L), round(1*Ny/D)) = p(1:round(1*Nx/L), round(1*Ny/D)+1); % dp/dy = 0 at x = 0 to 1 and y = 1

p(1:round(1*Nx/L), 1) = p(1:round(1*Nx/L), 2); % dp/dy = 0 at x = 0 to 1 and y = 1

un = u;

vn = v;

u(i, j) = un(i, j)-dt/(2 * h)*(un(i, j).*(un(i+1, j)-un(i-1, j))+vn(i, j).*(un(i, j+1)-un(i, j-1))) ...

-dt/(2*rho*h)*(p(i+1, j)-p(i-1, j)) ...

+nu*dt/h^2*(un(i+1, j)+un(i-1, j)+un(i, j+1)+un(i, j-1)-4*un(i, j)); % x-momentum

u(round(1:1*Nx/L), round(1:1*Ny/D)) = 0; % step geometry

u(1, round(1:1*Ny/D)) = 0; % u = 0 at x = 0 and y = 0 to 1

u(1, round(1*Ny/D:2*Ny/D)) = Uinf; % u = Uinf at x = 0 and y = 1 to 2

u(round(1*Nx/L), round(1:1*Ny/D)) = 0; % u = 0 at x = 1 and y = 0 to 1

u(1:round(1*Nx/L), round(1*Ny/D)) = 0; % u = 0 at x = 0 to 1 and y = 1

u(1:round(1*Nx/L), 1) = 0; % u = 0 at x = 0 to 1 and y = 1

u(Nx, :) = u(Nx-1, :); % du/dx = 0 at x = L

u(:, 1) = 0; % u = 0 at y = 0

u(:, Ny) = 0; % u = 0 at y = D

v(i, j) = vn(i, j)-dt/(2*h)*(un(i, j).*(vn(i+1, j)-vn(i-1, j))+vn(i, j).*(vn(i, j+1)-vn(i, j-1))) ...

-dt/(2*rho*h)*(p(i, j+1)-p(i, j-1)) ...

+ nu*dt/h^2*(vn(i+1, j)+vn(i-1, j)+vn(i, j+1)+vn(i, j-1)-4*vn(i, j)); % y-momentum

v(round(1:1*Nx/L), round(1:1*Ny/D)) = 0; % step geometry

v(1, round(1:1*Ny/D)) = 0; % v = 0 at x = 0 and y = 0 to 1

v(1, round(1*Ny/D:2*Ny/D)) = 0; % v = Uinf at x = 0 and y = 1 to 2

v(round(1*Nx/L), round(1:1*Ny/D)) = 0; % v = 0 at x = 1 and y = 0 to 1

v(1:round(1*Nx/L), round(1*Ny/D)) = 0; % v = 0 at x = 0 to 1 and y = 1

v(1:round(1*Nx/L), 1) = 0; % v = 0 at x = 0 to 1 and y = 1

v(Nx, :) = v(Nx-1, :); % dv/dx = 0 at x = L

v(:, 1) = 0; % u = 0 at y = 0

v(:, Ny) = 0; % u = 0 at y = D

end

%% post-processing

velocity_magnitude = sqrt(u.^2 + v.^2); % velocity magnitude

u1 = u; % u-velocity for plotting with box

v1 = v; % v-velocity for plotting with box

p1 = p; % p-velocity for plotting with box

u1(1:round(1*Nx/L) , 1:round(1*Ny/D)) = NaN; % step geometry

v1(1:round(1*Nx/L) , 1:round(1*Ny/D)) = NaN; % step geometry

p1(1:round(1*Nx/L) , 1:round(1*Ny/D)) = NaN; % step geometry

velocity_magnitude1 = sqrt(u1.^2 + v1.^2); % velocity magnitude with box

%% Visualize velocity vectors and pressure contours

hold on

contourf(X, Y, u1', 64, 'LineColor', 'none'); % contour plot

set(gca, 'FontSize', 20)

% skip = 20;

% quiver(X(1:skip:end, 1:skip:end), Y(1:skip:end, 1:skip:end),...

% u1(1:skip:end, 1:skip:end)', v1(1:skip:end, 1:skip:end)', 1, 'k','LineWidth', 0.1); % Velocity vectors

% hh = streamslice(X, Y, u1', v1',2); % Streamlines

% set(hh, 'Color', 'k','LineWidth', 01);

colorbar; % Add color bar

colormap parula % Set color map

axis equal % Set true scale

xlim([0 L]); % Set axis limits

ylim([0 D]);

xticks([0 9 L]) % Set ticks

yticks([0 D]) % Set ticks

xt = [0 21]; % draw top wall

yt = [2 2];

xb = [1 21]; % draw bottom wall

yb = [0 0];

xbox = [0 1 1 0 0]; % draw box

ybox = [0 0 1 1 0];

plot(xbox, ybox, 'k', 'LineWidth', 2)

plot(xt, yt, 'k', 'LineWidth', 2)

plot(xb, yb, 'k', 'LineWidth', 2)

% clim([0 max(velocity_magnitude(:))]) % legend limits

% title('Velocity [m/s]');

xlabel('x [m]');

ylabel('y [m]');

     The results from this code at Re 400 are presented in Fig. 1. The re-attachment length is ~8 m from the trailing-edge of the box, which is same as previously published results, from example in [1].

Fig. 1, post-processes results

     Thank you for reading! If you want to hire me as your next PhD student, please do reach out!

References

 [1] Irisarri, D., Hauke, G. Stabilized virtual element methods for the unsteady incompressible Navier–Stokes equations. Calcolo 56, 38 (2019). https://doi.org/10.1007/s10092-019-0332-5

Open - Source coded 3D Navier–Stokes equations in C++

     Here are 3D Navier–Stokes equations configured for lid-driven cavity flow. The syntax is C++. The results from this code are shown in Fig. 1. From the pressure-Poisson equation, I removed mixed derivative terms to improve solution stability. 😆

p[i][j][k] = ((p[i+1][j][k] + p[i-1][j][k]) * h * h + (p[i][j+1][k] + p[i][j-1][k]) * h * h + (p[i][j][k+1] + p[i][j][k-1]) * h * h) / (2 * (h * h + h * h + h * h)) - h * h * h * h * h * h / (2 * (h * h + h * h + h * h)) * (rho * (1 / dt * ((u(i+1, j, k) - u(i-1, j, k)) / (2 * h) + (v(i, j+1, k) - v(i, j-1, k)) / (2 * h) + (w(i, j, k+1) - w(i, j, k-1)) / (2 * h))));

p[0][j][k] = p[1][j][k];

p[num_i - 1][j][k] = p[num_i - 2][j][k];

p[i][0][k] = p[i][1][k];

p[i][num_j - 1][k] = p[i][num_j - 2][k];

p[i][j][0] = p[i][j][1];

p[i][j][num_k - 1] = 0.0;

u[i][j][k] = u[i][j][k] - u[i][j][k] * dt / (2 * h) * (u[i+1][j][k] - u[i-1][j][k]) - v[i][j][k] * dt / (2 * h) * (u[i][j+1][k] - u[i][j-1][k]) - w[i][j][k] * dt / (2 * h) * (u[i][j][k+1] - u[i][j][k-1]) - dt / (2 * rho * h) * (p[i+1][j][k] - p[i-1][j][k]) + nu * (dt / (h * h) * (u[i+1][j][k] - 2 * u[i][j][k] + u[i-1][j][k]) + dt / (h * h) * (u[i][j+1] [k] - 2 * u[i][j][k] + u[i][j-1][k]) + dt / (h * h) * (u[i][j][k+1] - 2 * u[i][j][k] + u[i][j][k-1]));

u[0][j][k] = 0.0;

u[num_i - 1][j][k] = 0.0;

u[i][0][k] = 0.0;

u[i][num_j - 1][k] = 0.0;

u[i][j][0] = 0.0;

u[i][j][num_k - 1] = -Uinf;

v[i][j][k] = v[i][j][k] - u[i][j][k] * dt / (2 * h) * (v[i+1][j][k] - v[i-1][j][k]) - v[i][j][k] * dt / (2 * h) * (v[i][j+1][k] - v[i][j-1][k]) - w[i][j][k] * dt / (2 * h) * (v[i][j][k+1] - v[i][j][k-1]) - dt / (2 * rho * h) * (p[i][j+1][k] - p[i][j-1][k]) + nu * (dt / (h * h) * (v[i+1][j][k] - 2 * v[i][j][k] + v[i-1][j][k]) + dt / (h * h) * (v[i][j+1][k] - 2 * v[i][j][k] + v[i][j-1][k]) + dt / (h * h) * (v[i][j][k+1] - 2 * v[i][j][k] + v[i][j][k-1]));

v[0][j][k] = 0.0;

v[num_i - 1][j][k] = 0.0;

v[i][0][k] = 0.0;

v[i][num_j - 1][k] = 0.0;

v[i][j][0] = 0.0;

v[i][j][num_k - 1] = 0.0;

w[i][j][k] = w[i][j][k] - u[i][j][k] * dt / (2 * h) * (w[i+1][j][k] - w[i-1][j][k]) - v[i][j][k] * dt / (2 * h) * (w[i][j+1][k] - w[i][j-1][k]) - w[i][j][k] * dt / (2 * h) * (w[i][j][k+1] - w[i][j][k-1]) - dt / (2 * rho * h) * (p[i][j][k+1] - p[i][j][k-1]) + nu * (dt / (h * h) * (w[i+1][j][k] - 2 * w[i][j][k] + w[i-1][j][k]) + dt / (h * h) * (w[i][j+1][k] - 2 * w[i][j][k] + w[i][j-1][k]) + dt / (h * h) * (w[i][j][k+1] - 2 * w[i][j][k] + w[i][j][k-1]));

w[0][j][k] = 0.0;

w[num_i - 1][j][k] = 0.0;

w[i][0][k] = 0.0;

w[i][num_j - 1][k] = 0.0;

w[i][j][0] = 0.0;

w[i][j][num_k - 1] = 0.0;

Fig. 1, Velocity and pressure iso-surfaces

     Of course constants need to be defined, such as grid spacing in space and time, density, kinematic viscosity. These equations have been validated, as you might have read and here already! Happy coding!

     If you want to hire me as your PhD student in the research projects related to turbo-machinery, aerodynamics, renewable energy, please reach out. Thank you very much for reading.

1D Transient Diffusion (MATLAB code)

clear; clc;% clear the screen and memory

Nx=500; %number of space nodes

Nt=10000; %number of time nodes

Lx=0.3; %length of space (m)

Lt=10; %length of physical time (s)

dx=Lx/(Nx-1); %grid spacing

dt=Lt/(Nt-1); %time step

c=0.000097; %speed of wave (constant)

a=c*dt/dx.^2;

u=100*ones(Nx,1); %initialization of matrix for the property under investigation

x=zeros(Nx,1); %initialization of space

for i=1:Nx-1 %space loop

    x(i+1)=x(i)+dx;

end

u(1)=-20; %boundary condition

u(Nx)=50; %boundary condition

for t=0:dt:Lt %time loop

    un=u; %u(i,t)

    for i=2:Nx-1 %solution loop, backward in space forward in time

        u(i)=((1-2*a)*un(i))+a*(un(i+1)+un(i-1)); %discretized equation, u(i,t+1)

    end

    plot(x,u) %plotting

    title({t;'Time Elaplsed (s)'},'FontSize',38.5);

    pause(0.001)

end

 

 

1D non-Linear Convection (MATLAB code)

clear; clc;% clear the screen and memory

Nx=41; %number of space nodes

Nt=82; %number of time nodes

Lx=2; %length of space (m)

Lt=1; %length of time (s)

dx=Lx/(Nx-1); %grid spacing

dt=Lt/(Nt-1); %time step

a=dt/dx;

u=ones(Nx,1); %initialization of solution matrix

x=zeros(Nx,1); %initialization of space

hold on

for i=1:Nx-1 %space loop

    x(i+1)=x(i)+dx;

end

for i=0.5/dx:1/dx %initial conditions

    u(i)=2;

end

for t=1:Nt %time loop

    un=u; %u(i,t)

    for i=2:Nx %solution loop, backward in space forward in time

        u(i)=un(i)-u(i)*a*(un(i)-un(i-1)); %discretized equation, u(i,t+1)

    end

    plot(x,u,'k') %plotting

    pause(0.1)

end

1D Linear Convection (MATLAB code) (Updated with CFL condition)

clear; clc;% clear the screen and memory

Nx=41; %number of space nodes

Nt=82; %number of time nodes

Lx=2; %length of space (m)

Lt=1; %length of time (s)

dx=Lx/(Nx-1); %grid spacing

dt=Lt/(Nt-1); %time step

c=2; %speed of wave (constant)

a=c*dt/dx;

u=ones(Nx,1); %initialization of solution matrix

x=zeros(Nx,1); %initialization of space

hold on

for i=1:Nx-1 %space loop

    x(i+1)=x(i)+dx;

end

for i=0.5/dx:1/dx %initial conditions

    u(i)=2;

end

for t=1:Nt %time loop

    un=u; %u(i,t)

    for i=2:Nx %solution loop, backward in space forward in time

        u(i)=un(i)-a*(un(i)-un(i-1)); %discretized equation, u(i,t+1)

    end

    plot(x,u,'k') %plotting

    pause(0.1)

end



New Code, with the CFL condition satisfied.

clear; clc;% clear the screen and memory

Nx=13; %number of space nodes

Lx=2; %length of space (m)

Lt=1; %length of time (s)

dx=Lx/(Nx-1); %grid spacing

c=2; %speed of wave (constant)

dt=dx*0.1/c; %time step, Courant number = 0.1

Nt=(Lt/dt)+1; %number of time nodes

a=c*dt/dx;

u=ones(Nx,1); %initialization of solution matrix

x=zeros(Nx,1); %initialization of space

t=zeros(fix(Nt),1); %initialization of space

hold on

for i=1:Nt-1

    t(i+1)=t(i)+dt;

end

for i=1:Nx-1 %space loop

    x(i+1)=x(i)+dx;

end

for i=0.5/dx:1/dx %initial conditions

    u(i)=2;

end

for j=0:dt:Lt %time loop

    un=u; %u(i,t)

    for i=2:Nx %solution loop, backward in space forward in time

        u(i)=un(i)-a*(un(i)-un(i-1)); %discretized equation, u(i,t+1)

    end

    plot(x,u,'k') %plotting

    pause(0.1)

end

Solve PDEs using MATLAB (Code)

This is a post about a code I wrote to solve  1-D Poisson's equations. The method employed is called the finite difference method. For 1-D Laplace equation, the RHS of the equation mentioned below goes to zero. Hence all the terms in the B(i) loop equal zero. Also, B(1) and B(Nx) terms equal to the boundary conditions only. I hope this code helps you in learning something.

The example equation is mentioned below.
d²y/dx² = x²+1 and y(0)=10, y(1)=20

The result of the code is a plot between the dependent and the independent variables, as shown in Fig. 1. The accuracy of the solutions lie around 1.5 % of the analytical solution.

%solution to 1D Poisson's equations with dirichlet boundary conditions, 2nd order accurate, uniform grid
clear; clc;
Nx=101; %Enter the number of grid points
clc;
A=zeros(Nx,Nx); %Stiffness matrix
B=zeros(Nx,1); %Dependent variable vector, also used for boundary conditions and other known terms
x=zeros(Nx,1); %Independent variable vector
x(Nx)=1; %Domain size
dx=x(Nx)/(Nx-1); %Grid spacing
B(1)=(((x(1).^2)+1)*(dx.^2))-10; %Enter the function the PDE equals to multiplied by the square of the grid spacing minus the first boundary condition
B(Nx)=(((x(Nx).^2)+1)*(dx.^2))-20; %Enter the function the PDE equals to multiplied by the square of the grid spacing minus the second boundary condition
for i=2:Nx-1 %Off boundary values for the dependent variable vector
    B(i)=((x(i).^2)+1)*(dx.^2); %Enter the function the PDE equals to multiplied by the the square of the grid spacing
end
for i=1:Nx-1 %Loop for independent variable
    x(i+1)=(i)*dx;
end
for i=1:Nx %Diagonal terms for the stiffness matrix
    A(i,i)=-2;
end
for i=1:Nx-1 %Off diagonal terms for the stiffness matrix
    A(i,i+1)=1;
    A(i+1,i)=1;
end
X=A\B;
hold on
set(gca,'FontSize',28)
xlabel('Independent Variable','FontSize',38.5)
ylabel('Dependent Variable','FontSize',38.5)
title('Dependent VS Independent Variable (d^2y/dx^2 = x^2+1)','FontSize',38.5);
plot(x,X)

Fig. 1. Plot from the code

Solve ODEs using MATLAB (Code)

I have decided to teach myself some CFD coding using my own intuition and some free online tutorials. This is a post about code I wrote to solve a first order ordinary differential equation. The method employed is called the finite difference method. I will start by coding equations using this method and will slowly move to finite volume method. I hope this code helps you in learning something.

The equation in question is mentioned below.
dy/dx = x+4 and y(0)=3

A uniform grid was employed, as I have not yet figured out how to make non uniform grid. When I figure it out, I will update it here, which will be soon.

The code, mentioned below, starts with a variable called Nx, which represents the total number of grid points in which the domain is divided in to. A rule of thumb, the more points in the domain, more accurate the solution becomes at the cost of computation resources, mainly CPU time and RAM and storage.

x and y are the vectors in which the known and the unknown values are stored. x has the known values, like the length of a pipe through which a fluid flow or a rod through which heat flows etc. y stores the unknown values like temperature of the rod or velocity of the fluid etc. The values are stored at nodes.

y(1) means first value in the vector y. which also happens to be our boundary condition. y is the value of the function at x=0. x=0 is the first value in in the x vector, written as x(1).

x(Nx-1) represents the domain size. Minimum domain size is usually zero e.g. starting point of a rod or a pipe and maximum can be any real number.

To calculate the grid spacing, represented by dx, the domain size was divided by the number of elements. In this example, we have four nodes, two at corners and two in-between, forming a total of three elements.

First order finite difference method was used to discretize the given equation. And a for loop was used for the solution. Using 4 grid points, the solution has an accuracy of 2.22% against the analytical solution of the equation which can be verified by hand calculations. Thank you for reading.

%solution to dy/dx=x+4 y(0)=3
%uniform grid
clear; clc;
Nx=4; %number of grid points
x=zeros(Nx,1); %x vector, known, goes from some known value to maximum value of the domain
y=zeros(Nx,1); %y vector, one value known due to boundary condition rest unknown
y(1)=3; %boundary condition
x(Nx-1)=1; %domain size
dx=x(Nx-1)/(Nx-1); %grid spacing
for i=1:Nx-1
    x(i+1)=(i)*dx;
    y(i+1)=(dx.*(x(i)+4))+y(i); %discretized differential equation
end

LU Decomposition MATLAB

%Please don't mess around with my code
clc;
R=input('How many variables are there in your system of linear equations? ');



clc;
disp ('Please enter the elements of the Coefficient matrix "A" and the Constant/Known matrix "B" starting from first equation');



A=zeros(R,R);

B=zeros(R,1);
for H=1:R

for i=1:R

fprintf ('Coefficient Matrix Column \b'), disp(i), fprintf ('\b\b Row \b'), disp(H);

I=input(' ');



A(H,i)=A(H,i)+I;
end

fprintf ('Known Matrix Row \b'), disp(H);

J=input(' ');



B(H,1)=B(H,1)+J;
end
clc;

[L,U] = lu(A);

Y = L\B;

X = U\Y;
fprintf ('The Coefficient Matrix "A" is \n'), disp(A);

fprintf ('The Constant/Known Matrix "B" is \n'), disp(B);

fprintf ('The Solution Matrix "X" is \n'), disp(X);