Implicit method in slab

Question

Hi every one,
This is my first code I have ever written with some help. I just would like to make sure is good and right before I submitted to my instructor.

The assignment is to solve a heat diffusion in a slab with two neumann boundaries (convection) and one initial (30 C).

I have attached a file for the main equations.

Thanks in advance


% [u,err,x,t] = heat2(t_0,t_f,M,N)


% function [u,err,x,t] = heat2(t_0,t_f,M,N)

% define the mesh in space

N=101;        %Number of space intervals
M=2000;       %number of time steps
l=1;          %thickness of wall in meter;;
t_0=0;         %initial time
t_f=10; %final time (sec)
dx = l/(N-1);    %size of the mesh
x = 0:dx:N-1;
x = x';      %inverse of mesh

bi=0.77;     %coefficient of terms
bo=0.88;

Te=34;          %exterior wall temperature
k=0.7;          %thermal conductivity of bricks
roh=1600;      %density of bricks (kg/m^3)
Cp=840;           %Cp of bricks (J/kg-K)
T_in=22;         %Inner wall temperature
alfa=Cp*roh/k;

% define the mesh in time
dt = (t_f-t_0)/M;
t = t_0:dt:t_f;

% define the ratio r
 r=alfa*dt/dx^2; %Fourier number
 
Beta=[2*r*bo*Te; zeros(N-2,1); 2*r*bi*T_in];

%initial condition
for i=1:N
  %u(i,1) = cos(x(i));
  u(i,1) = 15;
end 
err(:,1) = u(:,1) - exp(t_0-t(1))*cos(x);


A = zeros(N,N);
A(1,1) = 1+2*r+2*r*bo;
A(1,2) = -2*r;
% A(1,3:N) = 0;
for i=2:N-1
  A(i,i-1) = -r;
  A(i,i) = 1+2*r;
  A(i,i+1) = -r;
end
%A(N,N+1) = 0;
A(N,N-1) = -2*r;
A(N,N) = 1+2*r+2*r*bi;



[L,UU] = LU(A);




%[y] = down_solve(L,u_old);
%[u_new] = up_solve(UU,y);

for j=1:M  %time step
  [y] = down_solve(L,u(:,j)+Beta); 
  u(:,j+1) = up_solve(UU,y);
err(:,j+1) = u(:,j+1) - exp(t_0-t(j+1))*cos(x);


the sub_files:

1)

% [L,U] = LU(A) LU 

function [L,U] = LU(A);

[m,n] = size(A);


for i=1:n
  L(i,i) = 1;
end

for k=1:n-1
  for i=k+1:n
    mult = A(i,k)/A(k,k);
    for j=k+1:n
      A(i,j) = A(i,j) - A(k,j)*mult;
    end
    L(i,k) = mult;
  end
end

for i=1:n
  for j=i:n
    U(i,j) = A(i,j);
  end
end


2)
% [x] = up_solve(U,b) takes an upper triangular matrix U and vector b
% and solves Ux=b

function [x] = up_solve(U,b)

[m,n] = size(U);

x(n) = b(n)/U(n,n);

for k=1:n-1
  x(n-k) = b(n-k);
  for j=n-k+1:n
    x(n-k) = x(n-k) - U(n-k,j)*x(j);
  end
  x(n-k) = x(n-k)/U(n-k,n-k);
end
x = x';

3)

% [x] = down_solve(L,b) takes a lower triangular matrix L and vector b
% and solves Lx=b

function [x] = down_solve(L,b)

[m,n] = size(L);

x(1) = b(1)/L(1,1);

for k=2:n
  x(k) = b(k);
  for j=1:k-1
    x(k) = x(k) - L(k,j)*x(j);
  end
  x(k) = x(k)/L(k,k);
end
x = x';

Implicit method in slab

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Implicit method in slab

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