I need to change the boundary condition in the code below
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% Heat diffusion in one dimensional wire within the explicit FTCS method
clear;
% Parameters to define the heat equation and the range in space and time
D = 1*10^-4; % Heat Conductivity parameter in m2/s
K = 1*10^-4; % Heat Conductivity parameter in s^-1
L = 5.; % Total length in m
T = 43200.; % Final Time
% Parameters needed to solve the equation within the explicit method
Nt = 7200; % Number of time steps
Dt = T/Nt; % Time step
Nx = 51; % Number of space steps in m
Dx = L/(Nx-1); % Space step
b = 1-(2*D*Dt/Dx^2)- K*Dt; % beta parameter in the finite- difference implementation
% Remember that the FTCS method is stable for b=<1
disp(b);
% The initial condition: the initial temperature of the pipe
for i = 1:Nx+1
x(i) = (i-1)*Dx; % we also define vector x, due to the space discretization
u(i,1) = sin (pi*x(i));
end
% Boundary conditions: the temperature of the pipe at the boundaries at any time
for k =1:Nt+1
u(1,k) = 1.;
u(Nx+1,k) = 1.;
t(k) = (k-1)*Dt; % we also define vector t, due to the time discretization
end
% Implementation of the explicit FTCS method
for k =1:Nt % Time Loop
for i=2:Nx; % Space Loop
u(i,k+1) =u(i,k) + D*Dt/(Dx)^2*(u(i+1,k)-2*u(i,k)+u(i-1,k)); % Implementation of the FTCS Method
end
end
%Graphical representation of the temperature at different selected times
figure(1)
plot(x,u(:,1),'-b',x,u(:,round(Nt/100)),'--g',x,u(:,round(Nt/10)),':b',x,u(:,Nt),'-.r')
Title('Temperature of the pipe within the explicit FTCS method')
xlabel('X')
ylabel('T')
figure(2)
mesh(t,x,u)
title('temperature of the pipe within the explicit FTCS method')
xlabel('t')
ylabel('x')
Boundary and Initial Conditions: 𝐶|𝑥=0 = 1 mg/L, 𝑑𝐶 𝑑𝑥 |𝑥=𝐿 = 0, 𝐶|𝑡=0 = 1 mg/L
3 个评论
Torsten
2022-4-4
If you want to solve
dC/dt = D * d^2C/dx^2
C(0,t) = 1
dc/dx(L,t) = 0
C(x,0) = 1
then the solution is C(x,t) = 1.
No need to simulate anything.
回答(1 个)
Kayemba Luwaga
2023-7-22
if true % code % Heat diffusion in one dimensional wire within the explicit FTCS method clear; % Parameters to define the heat equation and the range in space and time D = 1*10^-4; % Heat Conductivity parameter in m2/s K = 1*10^-4; % Heat Conductivity parameter in s^-1 L = 5.; % Total length in m T = 43200.; % Final Time % Parameters needed to solve the equation within the explicit method Nt = 7200; % Number of time steps Dt = T/Nt; % Time step Nx = 51; % Number of space steps in m Dx = L/(Nx-1); % Space step b = 1-(2*D*Dt/Dx^2)- K*Dt; % beta parameter in the finite- difference implementation % Remember that the FTCS method is stable for b=<1 disp(b); % The initial condition: the initial temperature of the pipe for i = 1:Nx+1 x(i) = (i-1)*Dx; % we also define vector x, due to the space discretization u(i,1) = sin (pi*x(i)); end % Boundary conditions: the temperature of the pipe at the boundaries at any time for k =1:Nt+1 u(1,k) = 1.; %% Edit: Lower BC at Temp = 1 u(Nx+1,k) = 100.; %% Edit: Upper BC at Temp = 100 t(k) = (k-1)*Dt; % we also define vector t, due to the time discretization end % Implementation of the explicit FTCS method for k =1:Nt % Time Loop for i=2:Nx; % Space Loop u(i,k+1) =u(i,k) + D*Dt/(Dx)^2*(u(i+1,k)-2*u(i,k)+u(i-1,k)); % Implementation of the FTCS Method end end %Graphical representation of the temperature at different selected times figure(1) plot(x,u(:,1),'-b',x,u(:,round(Nt/100)),'--g',x,u(:,round(Nt/10)),':b',x,u(:,Nt),'-.r') title('Temperature of the pipe within the explicit FTCS method') xlabel('X') ylabel('T') figure(2) mesh(t,x,u) % plot(t,u) title('temperature of the pipe within the explicit FTCS method') xlabel('t') ylabel('x')
end
1 个评论
Kayemba Luwaga
2023-7-22
Check lines 23 and 24 for editing the temperature boundary conditions/values, Hope it's what you want Cheers!
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