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  1. Home/
  2. Ravi Shankar Yadav/
  3. Week 5.1 - Mid term project - Solving the steady and unsteady 2D heat conduction problem

Week 5.1 - Mid term project - Solving the steady and unsteady 2D heat conduction problem

Aim: To solve the Steady and Transient state 2D Heat Conduction Equation by using the point iteration techniques for both implicit and explicit schemes. Jacobi Gauss-Seidel Successive Over Relaxation   Theory: The heat equation is a partial differencial equation that describes how the heat is distributed over time…

  • MATLAB

  • Ravi Shankar Yadav

    updated on 30 Dec 2021

Aim:

To solve the Steady and Transient state 2D Heat Conduction Equation by using the point iteration techniques for both implicit and explicit schemes.

  1. Jacobi
  2. Gauss-Seidel
  3. Successive Over Relaxation

 

Theory:

The heat equation is a partial differencial equation that describes how the heat is distributed over time in a solid.

The three dimensional temperature heat conduction equation has both time and spatial derivative.

∂T∂t+α(∂2T∂x2+∂2T∂y2+∂2T∂z2)=0∂T∂t+α(∂2T∂x2+∂2T∂y2+∂2T∂z2)=0

here, Î±Î± is thermal diffusivity.

This equation can be solved in two ways:

1. Steady-State Condition:

In this condition, the rate of change of temperature wrt time is equal to zero.

So, the 2D Steady-state heat equation becomes,

∂2T∂x2+∂2T∂y2=0∂2T∂x2+∂2T∂y2=0

 

2. Transient State Condition:

In this condition, both terms are considered,

So, the 2D Transient state heat equation becomes,

∂T∂t+α(∂2T∂x2+∂2T∂y2)=0∂T∂t+α(∂2T∂x2+∂2T∂y2)=0

 

Now, equations (a) and (b), will be solved by using iterative methods, in 2 different ways.

1. Explicit Method

2. Implicit Method

Note: We will use both these methods to solve the transient state equation.

 

Assumptions:

  1. The domain is a unit square.
  2. nx = ny [Number of points along the x-direction is equal to the number of points along the y-direction.]
  3. Tolerence = 1e-4
  4. Strating error = 9e-9
  5. The Boundary conditions are:
  1. Top Boundary = 600 K
  2. Bottom Boundary = 900 K
  3. Left Boundary = 400 K
  4. Right Boundary = 800 K
  5. Your absolute error criteria is 1e-4

 

A. Solving the steady state equation

∂2T∂x2+∂2T∂y2=0∂2T∂x2+∂2T∂y2=0

Tp=(1K){(TL+TRΔx2)+(TT+TBΔy2)}Tp=(1K){(TL+TRΔx2)+(TT+TBΔy2)}

 

T(i,j)=(1K){(Told(i−1,j)+Told(i+1,j)dx2)+(Told(i,j−1)+Told(i,j+1)dy2)}T(i,j)=(1K){(Told(i-1,j)+Told(i+1,j)dx2)+(Told(i,j-1)+Told(i,j+1)dy2)}

K=2â‹…dx2+dy2dx2â‹…dy2K=2â‹…dx2+dy2dx2â‹…dy2

 

MATLAB Code for solving the equation using 3 iterative methods

Using Jacobi Iteration

% Solving the steady-state equation by using
% Jacobi Method

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
k = 2*((dx^2 + dy^2)/(dx^2 * dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

% Average temperature at corners
T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
iterative_solver = 1;
if iterative_solver == 1
    jacobi_iter_value = 1;
    
while(error > tol)
    for i=2:nx-1
        for j=2:ny-1
            term1 = (Told(i-1,j) + Told(i+1,j))/(k*(dx^2));
            term2 = (Told(i,j-1) + Told(i,j+1))/(k*(dy^2));
            T(i,j) = term1 + term2;
        end
    end
    
    
    error = max(max(abs(Told -T)));
    Told = T;
    jacobi_iter_value = jacobi_iter_value +1;
    
    
end
end
% Plotting
figure(1)
contourf(x, y, T)
clabel(contourf(x, y, T))
colormap(jet)
colorbar
set(gca, 'ydir', 'reverse')
title_text = sprintf('Jacobi iteration number for steady state = %d', jacobi_iter_value);
title(title_text)
xlabel('X axis')
ylabel('Y axis')
fprintf(' Number of Iterations for Jacobi Method for Steady state condition = %d', jacobi_iter_value)

Graph:

ex5.1.1

 

Using Gauss-Seidel Iteration

% Solving the steady-state equation by using
% Gauss Seidel

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
k = 2*((dx^2 + dy^2)/(dx^2 * dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

% Average temperature at corners
T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
iterative_solver = 2;
if iterative_solver == 2
    gauss_iter_value = 1;
    
while(error > tol)
    for i=2:nx-1
        for j=2:ny-1
            term1 = (T(i-1,j) + Told(i+1,j))/(k*(dx^2));
            term2 = (T(i,j-1) + Told(i,j+1))/(k*(dy^2));
            T(i,j) = term1 + term2;
        end
    end
    
    
    error = max(max(abs(Told -T)));
    Told = T;
    gauss_iter_value = gauss_iter_value +1;
    
    
end
end
% Plotting
figure(1)
contourf(x, y, T)
clabel(contourf(x, y, T))
colormap(jet)
colorbar
set(gca, 'ydir', 'reverse')
title_text = sprintf('Gauss seidel iteration number for steady state = %d', gauss_iter_value);
title(title_text)
xlabel('X axis')
ylabel('Y axis')
fprintf(' Number of Iterations for Gauss Seidel Method for Steady state condition = %d', gauss_iter_value)

Graph:

ex5.1.2

 

Successive Over Relaxation Iteration

% Solving the steady-state equation by using
% successive over-relaxation method

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
omega = 1.2; 
k = 2*((dx^2 + dy^2)/(dx^2 * dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
iterative_solver = 3;
if iterative_solver == 3
    sor_iter_value = 1;
    
while(error > tol)
    for i=2:nx-1
        for j=2:ny-1
            term1 = (T(i-1,j) + Told(i+1,j))/(k*(dx^2));
            term2 = (T(i,j-1) + Told(i,j+1))/(k*(dy^2));
            T(i,j) = omega*(term1 + term2)+ (1-omega)*Told(i,j);
        end
    end
    
    
    error = max(max(abs(Told -T)));
    Told = T;
    sor_iter_value = sor_iter_value +1;
    
    
end
end
% Plotting
figure(1)
contourf(x, y, T)
clabel(contourf(x, y, T))
colormap(jet)
colorbar
set(gca, 'ydir', 'reverse')
title_text = sprintf('SOR iteration number for steady state = %d', sor_iter_value);
title(title_text)
xlabel('X axis')
ylabel('Y axis')
fprintf(' Number of Iterations for SOR Method for Steady state condition = %d', sor_iter_value)

Graph:

ex5.1.3

 

 

B. Solving the Transient state equation

Using Implicit Scheme

∂T∂t+α(∂2T∂x2+∂2T∂y2)=0∂T∂t+α(∂2T∂x2+∂2T∂y2)=0

Tn+1P=TnP+(αΔt)(TR−2TP+TLΔx2+TT−2TP+TBΔy2)TPn+1=TPn+(αΔt)(TR-2TP+TLΔx2+TT-2TP+TBΔy2)

Tn+1P=(11+2k1+2k2)(TnP+k1(Tn+1L+Tn+1R)+k2(Tn+1T+Tn+1B))TPn+1=(11+2k1+2k2)(TPn+k1(TLn+1+TRn+1)+k2(TTn+1+TBn+1))

 k1=α(ΔtΔx2),k2=α(ΔtΔy2)k1=α(ΔtΔx2),k2=α(ΔtΔy2)

Ti,j=(11+2k1+2k2)(Ti,j+k1(Ti−1,j+Ti+1,j)+k2(Ti,j−1+Ti,j+1))Ti,j=(11+2k1+2k2)(Ti,j+k1(Ti-1,j+Ti+1,j)+k2(Ti,j-1+Ti,j+1))

 

MATLAB Code for solving the equation using 3 iterative methods

 Using Jacobi Iteration

% Solving the steady-state equation by using
% Jacobi method
% Implicit Method

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
nt = 1400;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
dt = 1e-3;
alpha = 1.4
k1 = alpha*(dt/(dx^2));
k2 = alpha*(dt/(dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T = 300*ones(nx,ny);

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
jacobi_iter_value = 1;
for k = 1:nt
    error = 9e9;
    
    while(error > tol)
        for i=2:nx-1
            for j=2:ny-1
                term1 = 1/(1 + (2*k1) + (2*k2));
                term2 = k1*term1;
                term3 = k2*term1;
                H = (Told(i-1,j) + Told(i+1,j));
                V = (Told(i,j-1) + Told(i,j+1));
                T(i,j) = (T_initial(i,j)*term1) + H*(term2) + V*(term3);
            end
        end

        error = max(max(abs(Told -T)));
        Told = T;
        jacobi_iter_value = jacobi_iter_value +1;


    end
    % Plotting
    T_initial = T
    figure(1)
    contourf(x, y, T)
    clabel(contourf(x, y, T))
    colormap(jet)
    colorbar
    set(gca, 'ydir', 'reverse')
    title_text = sprintf('Jacobi iteration number for unsteady state = %d', jacobi_iter_value);
    title(title_text)
    xlabel('X axis')
    ylabel('Y axis')
end

fprintf(' Number of Iterations for Implicit Jacobi Method for Transient State Condition = %d', jacobi_iter_value)

Graph:

 ex5.2.1

 

Using Gauss-Seidel Iteration

% Solving the steady-state equation by using
% Gauss-Seidel method
% Implicit Method

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
nt = 1400;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
dt = 1e-3;
alpha = 1.4

k1 = alpha*(dt/(dx^2));
k2 = alpha*(dt/(dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T = 300*ones(nx,ny);

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
Gauss_iter_value = 1;
for k = 1:nt
    error = 9e9;
    
    while(error > tol)
        for i=2:nx-1
            for j=2:ny-1
                term1 = 1/(1 + (2*k1) + (2*k2));
                term2 = k1*term1;
                term3 = k2*term1;
                H = (T(i-1,j) + Told(i+1,j));
                V = (T(i,j-1) + Told(i,j+1));
                T(i,j) = (T_initial(i,j)*term1) + H*(term2) + V*(term3);
            end
        end

        error = max(max(abs(Told -T)));
        Told = T;
        Gauss_iter_value = Gauss_iter_value +1;


    end
    % Plotting
    T_initial = T
    figure(1)
    contourf(x, y, T)
    clabel(contourf(x, y, T))
    colormap(jet)
    colorbar
    set(gca, 'ydir', 'reverse')
    title_text = sprintf('Gauss iteration number for unsteady state = %d', Gauss_iter_value);
    title(title_text)
    xlabel('X axis')
    ylabel('Y axis')
end

fprintf(' Number of Iterations for Implicit Gauss Seidel Method for Transient State Condition = %d', Gauss_iter_value)

Graph:

ex5.2.2

 

Successive Over Relaxation Iteration

% Solving the steady-state equation by using
% Successive Over-Relaxation method
% Implicit Method

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
nt = 1400;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
dt = 1e-3;
omega = 1.2
alpha = 1.4;

k1 = alpha*(dt/(dx^2));
k2 = alpha*(dt/(dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T = 300*ones(nx,ny);

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
T_end = T

% solving
SOR_iter_value = 1;
for k = 1:nt
    error = 9e9;
    
    while(error > tol)
        for i=2:nx-1
            for j=2:ny-1
                term1 = 1/(1 + (2*k1) + (2*k2));
                term2 = k1*term1;
                term3 = k2*term1;
                H = (T(i-1,j) + Told(i+1,j));
                V = (T(i,j-1) + Told(i,j+1));
                T(i,j) = (T_initial(i,j)*term1) + H*(term2) + V*(term3);
                T_end(i,j) = ((1-omega)*Told(i,j))+ (T(i,j)*omega);
            end
        end

        error = max(max(abs(Told -T)));
        Told = T_end;
        SOR_iter_value = SOR_iter_value +1;


    end
    % Plotting
    T_initial = T
    figure(1)
    contourf(x, y, T)
    clabel(contourf(x, y, T))
    colormap(jet)
    colorbar
    set(gca, 'ydir', 'reverse')
    title_text = sprintf('SOR iteration number for unsteady state = %d', SOR_iter_value);
    title(title_text)
    xlabel('X axis')
    ylabel('Y axis')
end

fprintf(' Number of Iterations for Implicit SOR Method for Transient State Condition = %d', SOR_iter_value)

Graph:

ex5.2.3

 

 

Using Explicit Scheme

∂T∂t+α(∂2T∂x2+∂2T∂y2)=0∂T∂t+α(∂2T∂x2+∂2T∂y2)=0

Tn+1P=TnP+(αΔt)(TR−2TP+TLΔx2+TT−2TP+TBΔy2)TPn+1=TPn+(αΔt)(TR-2TP+TLΔx2+TT-2TP+TBΔy2)

Tn+1p=(Tnp+k1(TnL−2TnP+TnR)+k2(TnT−2TnP+TnB))Tpn+1=(Tpn+k1(TLn-2TPn+TRn)+k2(TTn-2TPn+TBn))

 k1=α(ΔtΔx2),k2=α(ΔtΔy2)k1=α(ΔtΔx2),k2=α(ΔtΔy2)

Ti,j=(Ti,j+k1(Ti−1,j−2Ti,j+Ti+1,j)+k2(Ti,j−1−2Ti,j+Ti,j+1))Ti,j=(Ti,j+k1(Ti-1,j-2Ti,j+Ti+1,j)+k2(Ti,j-1-2Ti,j+Ti,j+1))

 

MATLAB Code for solving the equation

 Using Jacobi Iteration

% Solving the steady-state equation by using
% Jacobi method
% Explicit scheme

clear all
close all
clc

% Inputs
% nx = number of grids
nx = 10;
ny = nx;
nt = 1400;
% Domain length
l = 1;
x = linspace(0, l, nx)
y = linspace(0, l, ny)
dx = x(2) - x(1)
dy = y(2) - y(1)

% Tolerance error, start error
tol = 1e-4;
error = 9e9;
dt = 1e-3;
alpha = 1.4

k1 = alpha*(dt/(dx^2));
k2 = alpha*(dt/(dy^2));

% Boundary conditions
T_L = 400;
T_R = 800;
T_T = 600;
T_B = 900;

T = 300*ones(nx,ny);

T(2:ny-1, 1) = T_L;
T(2:ny-1, nx) = T_R;
T(1, 2:nx-1) = T_T;
T(ny, 2:nx-1) = T_B;

T(1,1) = (T_T + T_L)/2;
T(nx,ny) = (T_R + T_B)/2;
T(1,ny) = (T_T + T_R)/2;
T(nx,1) = (T_L + T_B)/2;

T_initial = T;
Told = T;
% solving
jacobi_iter_value = 1;
for k = 1:nt
    error = 9e9;
    
    while(error > tol)
        for i=2:nx-1
            for j=2:ny-1
                term1 = Told(i,j);
                term2 = k1*(Told(i-1,j) - 2*Told(i,j) + Told(i+1,j));
                term3 = k2*(Told(i,j-1) - 2*Told(i,j) + Told(i,j+1));
                T(i,j) = term1 + term2 + term3;
            end
        end

        error = max(max(abs(Told -T)));
        Told = T;
        jacobi_iter_value = jacobi_iter_value +1;


    end
    % Plotting
    figure(1)
    contourf(x, y, T)
    clabel(contourf(x, y, T))
    colormap(jet)
    colorbar
    set(gca, 'ydir', 'reverse')
    title_text = sprintf('Jacobi iteration number for unsteady state = %d', jacobi_iter_value);
    title(title_text)
    xlabel('X axis')
    ylabel('Y axis')
end

fprintf(' Number of Iterations for Explicit Jacobi for Transient method = %d', jacobi_iter_value)

Graph:

ex5.3.1

 

Conclusion:

Type of scheme Methods used No. of Iterations
Implicit Steady State

1. Jacobi

 

2. Gauss Seidel

 

3. SOR

1. 217

 

2. 117

 

3.  78
Implicit Transient State  

1. Jacobi

 

2. Gauss Seidel

 

3. SOR

  

1. 3666 

 

2. 3117

 

3. 2791
Explicit Transient State 1. Jacobi 1. 1833

 

  1. In the Steady-state equation, the SOR method takes the least number of iteration for solving the equation.
  2. In the case of Transient equations, the time required to solve the equation is more when compared to the steady-state condition in solving the problem.
  3. In Transient state cases, explicit methods take less time than implicit methods.

 

 

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