1 D Unsteady Diffusion Equation by Finite Volume (Fully Implicit) Scheme

Hi, Community,

Need some help to solve 1 D Unsteady Diffusion Equation by Finite Volume (Fully Implicit) Scheme . MATLAB Code is working. When I compare it with Book results, it is significantly different. If it is possible to improve. Attached files are

a -Figure(taken from Book) which show origional example (problem)

b- PDF(taken from Book) shows main scheme to implement.

Your idea will be highly appreciated.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% This code solves the unsteady 1D diffuion equation using FVM for the
% unknown Temperature.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all
close all
clc
tic
%% Sort out Inputs
% Number of Control Volumes
N = 5;
% Domain length
L = 0.02; %[m]
% Grid Spacing (Wall thickness)
h = L/(N);  
% Diffusivity (Thermal conductivity)
k = 10;  %[W/m-K]
% Specific Heat Capacity
Row_c = 10e6;  %[J/m^(3)-K]
% Uniform Heat Generation, A Source Term
q = 1000e3;     %  [W/m^(3)]
% The center of the first cell is at dx/2 & the last cell is at L-dx/2.
x = h/2 : h : L-(h/2);
% Cross sectional area of the 1D domain
A = 1;          %[m^2]  
% Diffusivity (Heat)
alpha = k/Row_c;
% Simulation Time
t_Final = 8;
% Discrete Time Steps
dt = 2;
% Left Surface Temperature
T_a = 0;    %[\circ c] 
% Right Surface Temperature
T_b = 0;    %[\circ c] 
% Parameteric Setup
lambda = (alpha.*dt)/(h^2);
%% Initializing Variable
% Unknowns at time level n
T_Old = zeros(N,1);
% Unknowns at time level n+1
T_New = zeros(N,1);
% Initial Value (Initial Condition)
T_Old(:) = 200; 
% Fully Implicit Scheme to Solve 1D Unsteady Diffusion.
% Time loop
for n = 1:t_Final/dt;
C(N,N) = 0;
D(N) = 0;
for i = 1: N
   if  i > 1 && i < N
      C(i,i) = (1 + 2*lambda).*T_Old(i); 
      C(i,i+1) = -lambda.*T_Old(i);
      C(i,i-1) = -lambda.*T_Old(i); 
      D(i)=T_Old(i); 
   elseif i == 1
    % For 1st boundary node
      C(i,i) = (1 + lambda).*T_Old(i); 
      C(i,i+1) = -lambda.*T_Old(i); 
      D(i) = T_Old(i); 
     else
     % For Last boundary node
       C(i,i) = (1 + 3*lambda).*T_Old(i); 
       C(i,i-1) = -lambda.*T_Old(i); 
       D(i) = 2*T_b*lambda + T_Old(i); 
   end
 end
 T_New = (inv(C)*D').*T_Old;
%        Update calculations
          T_Old = T_New;
end
T_New

[小宝 王]

May I ask your advice on how to learn finite volume method ? and this question comes from that book, can you share it with me,I would be very grateful.

Answer 1

[darova]

Try this simple difference scheme:

lam = 0.5;                  % k*dt/rho/dx^2
n = 20;
T = zeros(n);
T(1,:) = 200;               % t = 0 condition (T=200)
T(:,end) = 0;               % x = L condition (T=0)
h = surf(T);
for i = 1:size(T,1)-1
    for j = 2:size(T,2)-1
        T(i+1,j) = T(i,j) + lam*(T(i,j+1) - 2*T(i,j) + T(i,j-1));
    end
    T(i+1,1) = T(i+1,2);    % x = 0 condition (dT/dx=0)
    set(h,'zdata',T)
    pause(0.3)
end

[Shahid Hasnain]

@darova Thank you for idea, it is forward Euler's scheme on time. As I mentioned im my problem it is fully Implicit. My result are at 8s (t_Final) is correct. I got same coefficient matrix which is on the book but at time 40s. Results are significantly different. I am stuck in time loop which change the results. Also, T_old are changing strangely. Any other idea. Your idea is not aligned, it is very good. I am sorry to say.

[Shahid Hasnain]

@darova

in time n=1, I got the following

C =

225 -25 0 0 0

-25 250 -25 0 0

0 -25 250 -25 0

0 0 -25 250 -25

0 0 0 -25 275

and

D =

200

200

200

200

200

which are perfectly match with book. After that there is something I am missing.

Answer 2

[Nadza Basyuni]