CFD Wizardry: A 50-Line Python Marvel

     In abundant spare time ⏳, yours truly has implemented the non-conservative and non-dimensional form of the discretized Navier-Stokes 🍃 equations. The code 🖳 in it's simplest form is less than 50 lines including importing libraries and plotting! 😲 For validation, refer here. More examples and free code is available here, here and here. Happy codding!

Code

# Copyright <2025> <FAHAD BUTT>

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import numpy as np

import matplotlib.pyplot as plt

l_square = 1 # length of square

h = l_square / 500 # grid spacing

dt = 0.00002 # time step

L = 1 # domain length

D = 1 # domain depth

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

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

nu = 1 / 100 # kinematic viscosity

Uinf = 1 # free stream velocity / inlet velocity / lid velocity

cfl = dt * Uinf / h # cfl number

travel = 200 # times the disturbance travels entire length of computational domain

TT = travel * L / Uinf # total time

ns = int(TT / dt) # number of time steps

Re = round(l_square * Uinf / nu) # Reynolds number

u = np.zeros((Nx, Ny)) # x-velocity

v = np.zeros((Nx, Ny)) # y-velocity

p = np.zeros((Nx, Ny)) # pressure

for nt in range(ns): # solve 2D Navier-Stokes equations

    pn = p.copy()

    p[1:-1, 1:-1] = (pn[2:, 1:-1] + pn[:-2, 1:-1] + pn[1:-1, 2:] + pn[1:-1, :-2]) / 4 - h / (8 * dt) * (u[2:, 1:-1] - u[:-2, 1:-1] + v[1:-1, 2:] - v[1:-1, :-2]) # pressure

    p[0, :] = p[1, :] # dp/dx = 0 at x = 0

    p[-1, :] = p[-2, :] # dp/dx = 0 at x = L

    p[:, 0] = p[:, 1] # dp/dy = 0 at y = 0

    p[:, -1] = 0 # p = 0 at y = D

    un = u.copy()

    vn = v.copy()

    u[1:-1, 1:-1] = (un[1:-1, 1:-1] - dt / (2 * h) * (un[1:-1, 1:-1] * (un[2:, 1:-1] - un[:-2, 1:-1]) + vn[1:-1, 1:-1] * (un[1:-1, 2:] - un[1:-1, :-2])) - dt / (2 * h) * (p[2:, 1:-1] - p[:-2, 1:-1]) + (1 / Re) * dt / h**2 * (un[2:, 1:-1] + un[:-2, 1:-1] + un[1:-1, 2:] + un[1:-1, :-2] - 4 * un[1:-1, 1:-1])) # x momentum

    u[0, :] = 0 # u = 0 at x = 0

    u[-1, :] = 0 # u = 0 at x = L

    u[:, 0] = 0 # u = 0 at y = 0

    u[:, -1] = Uinf # u = Uinf at y = D

    v[1:-1, 1:-1] = (vn[1:-1, 1:-1] - dt / (2 * h) * (un[1:-1, 1:-1] * (vn[2:, 1:-1] - vn[:-2, 1:-1]) + vn[1:-1, 1:-1] * (vn[1:-1, 2:] - vn[1:-1, :-2])) - dt / (2 * h) * (p[1:-1, 2:] - p[1:-1, :-2]) + (1 / Re) * dt / h**2 * (vn[2:, 1:-1] + vn[:-2, 1:-1] + vn[1:-1, 2:] + vn[1:-1, :-2] - 4 * vn[1:-1, 1:-1])) # y momentum

    v[0, :] = 0 # v = 0 at x = 0

    v[-1, :] = 0 # v = 0 at x = L

    v[:, 0] = 0 # v = 0 at y = 0

    v[:, -1] = 0 # v = 0 at y = D

X, Y = np.meshgrid(np.linspace(0, L, Nx), np.linspace(0, D, Ny)) # spatial grid

plt.figure(dpi = 200)

plt.contourf(X, Y, v.T, 128, cmap = 'jet') # plot contours

plt.colorbar()

plt.streamplot(X, Y, u.T, v.T, color = 'black', cmap = 'jet', density = 2, linewidth = 0.5, arrowstyle='->', arrowsize = 1) # plot streamlines

plt.gca().set_aspect('equal', adjustable='box')

plt.xticks([0, L])

plt.yticks([0, D])

plt.xlabel('x [m]')

plt.ylabel('y [m]')

plt.show()

Lid-Driven Cavity

     The case of lid-driven cavity in the turbulent flow regime can now be solved in reasonable amount of time. The results are shown in Fig. 1. I stopped the code while the flow is still developing as you are reading a blog and not a Q1 journal. 😆 Within Fig. 1, streamlines, v and u component of velocity and pressure are shown going from left to right and top to bottom. At the center of Fig. 1, the velocity magnitude is superimposed. As this is DNS, the smallest spatial scale resolved is ~8e-3 m [8 mm]. While, the smallest time-scale ⌛ resolved is ~8e-4 s [0.8 ms].

Fig. 1, The results at Reynolds number of 10,000

Free-Jet

          The case of free jets in the turbulent flow regime can now be solved in reasonable amount of time. The results are shown in Fig. 2. I stopped the code while the flow is still developing. Once again, I remind you that you are reading a blog and not a Q1 journal. 😆 Within Fig. 2, streamlines and species are shown. As this is DNS, the smallest spatial scale resolved is ~0.02 m [2 cm]. While, the smallest time-scale ⌛ resolved is ~4e-4 s [0.4 ms]. The code for implementing species, in this case temperature using the energy equation is available on the previous post.

Fig. 2, Free jet at Reynolds number 10000

Heated Room

     The benchmark case of mixed convection in an open room in the turbulent flow regime can now be solved in reasonable amount of time as well. The results are shown in Fig. 3. I stopped the code while the flow field stopped showing any changes. 😆 As this is DNS, the smallest spatial scale resolved is ~0.0144 m [1.44 cm]. While, the smallest time-scale ⌛ resolved is ~1e-4 s [1 ms]. The code for implementing species, in this case temperature using the energy equation is available on the previous post. In the previous post, the momentum equation has no changes as the gravity vector is at 0 m/s2. For this example, Boussinesq assumption is used.

Fig. 3, Flow inside a heated room at Reynolds number of 5000

Backward - Facing Step (BFS)

     Another benchmark case of flow around a backwards facing step can now be solved in reasonable amount of time as well. The flow is fully turbulent. The results are shown in Fig. 4. I stopped the code while the flow field is still developing. 😆 As this is DNS, the smallest spatial scale resolved is ~0.01 m [1 cm]. While, the smallest time-scale ⌛ resolved is ~1e-4 s [1 ms]. As can be seen from Fig. 4, there are no abnormalities in the flow field.

Fig. 4, Flow around a backwards facing step at Reynolds number of 10000

PS: I fully understand, there is no such thing as 2D turbulence 🍃. Just don't kill the vibe please 💫.

     Thank you for reading! If you want to hire me as your next shinning post-doc, do let reach out!

1D Laplace's equation using Finite Difference Method

This post is about FDM for Laplace Equation with various boundary conditions.

MATLAB Code (1D, Dirichlet Boundary Conditions)

%% initialize the workspace, clear the command window

clear; clc

%% finite difference 1D laplace dirichlet boundary conditions %% d2u/dx2 = 0 %% u(o) = 10, u(L) = 4 %% Ax=b%%

N = 4 ; %number of grid points

L = 1; %length of domain

dx = L/(N-1); %element size

%% initialize variables %%

l = linspace(0,L,N); %independent

u=zeros(1,N); %dependent

%% boundary conditions %%

u(1)=10;

u(end)=4;

%% b vector %%

b=zeros(N-2,1);

b(1) = b(1) - u(1);

b(end) = b(end) - u(end);

%% A matrix

A = -2*eye(N-2,N-2);

for i=1:N-2

    if i<N-2

        A(i,i+1)=1;

    end

    if i>1

        A(i,i-1)=1;

    end

end

%% solve for unknowns %%

x = A\b;

%% fill the u vector with unknowns %%

u(2:end-1) = x;

%% plot the results %%

hold on; grid on; box on, grid minor

set(gca,'FontSize',40)

set(gca, 'FontName', 'Times New Roman')

ylabel('u','FontSize',44)

xlabel('l','FontSize',44)

plot(l,u,'-o','color',[0 0 0],'LineWidth',2,'MarkerSize',20)

MATLAB Code (1D, Mixed Boundary Conditions)

%% initialize the workspace, clear the command window

clear; clc

%% finite difference 1D laplace mixed boundary conditions %% d2u/dx2 = 0 %% u(o) = 10, du/dx(L) = 4 %% Ax=b%%

N = 5; %number of grid points

L = 1; %length of domain

dx = L/(N-1); %element size

a = 4;

%% initialize variables %%

l = linspace(0,L,N); %independent

u=zeros(1,N); %dependent

%% dirichlet boundary condition %%

u(1) = 10;

%% b vector %%

b=zeros(N-1,1);

b(1) = b(1) - u(1);

b(end) = b(end) + dx*a; %neumann boundary condition added to b vector

%% A matrix

A = -2*eye(N-1,N-1);

for i=1:N-1

    if i<N-1

        A(i,i+1)=1;

    end

    if i>1

        A(i,i-1)=1;

    end

end

A(N-1,N-1) = -1; %neumann boundary condition added to A matrix

%% solve for unknowns %%

x = A\b;

% fill the u vector with unknowns %%

u(2:end) = x;

%% plot the results %%

hold on; grid on; box on, grid minor

set(gca,'FontSize',40)

set(gca, 'FontName', 'Times New Roman')

ylabel('u','FontSize',44)

xlabel('l','FontSize',44)

plot(l,u','-o','color',[0 0 0],'LineWidth',2,'MarkerSize',20)

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