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

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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

##### 1 Comment

小宝 王
on 18 Nov 2021

### Answers (1)

darova
on 1 Sep 2021

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

##### 2 Comments

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