1D Heat Conduction using explicit Finite Difference Method
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Hello I am trying to write a program to plot the temperature distribution in a insulated rod using the explicit Finite Central Difference Method and 1D Heat equation. The rod is heated on one end at 400k and exposed to ambient temperature on the right end at 300k. I am using a time of 1s, 11 grid points and a .002s time step. When I plot it gives me a crazy curve which isn't right. I think I am messing up my initial and boundary conditions. Here is my code.
L=1;
t=1;
k=.001;
n=11;
nt=500;
dx=L/n;
dt=.002;
alpha=k*dt/dx^2;
T0(1)=400;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha*(T0(i+1)-2*T0(i)+T0(i-1));
end
T0=T1;
end
plot(x,T1)
2 个评论
KSSV
2016-12-15
First place, it is not giving any curve..there is a error in your code. Please recheck your code once.
Derek Shaw
2016-12-15
I am not sure I understand correctly. In the above I wrote this equation to be iterated
With Boundary conditions
and Initial Conditions
with T0=400k and TL=Ti=300k
I am not sure how to set these boundary conditions in the code. Or if there is a curve I need to derive before doing the iterations.
采纳的回答
It seems your initial condition and boundary conditon (x = L) are missing in the code. Try
L=1;
t=1;
k=.001;
n=11;
nt=500;
dx=L/n;
dt=.002;
alpha=k*dt/dx^2;
T0=400*ones(1,n);
T1=300*ones(1,n);
T0(1) = 300;
T0(end) = 300;
for j=1:nt
for i=2:n-1
T1(i)=T0(i)+alpha*(T0(i+1)-2*T0(i)+T0(i-1));
end
T0=T1;
end
plot(T1)
http://jp.mathworks.com/matlabcentral/fileexchange/59916-simple-heat-equation-solver could make a good example for you.
9 个评论
Derek Shaw
2016-12-15
michio
2016-12-15
Glad to know that you figure things out. One additional tip is vectorization instead of for-loop, ie.
for i=2:n-1
T1(i) = T0(i) + alpha*(T0(i+1)-2*T0(i)+T0(i-1));
end
is equivalent to
T1(2:n-1) = T0(2:n-1) + alpha*(T0(3:n)-2*T0(2:n-1)+T0(1:n-2));
The later runs much faster.
Kunpeng Liao
2017-10-13
Are the time step and grid spacing missing? I guess it would be:
if true
T1(2:n-1) = T0(2:n-1) + alpha*dt/dx^2*(T0(3:n)-2*T0(2:n-1)+T0(1:n-2));
end
Torsten
2017-10-16
Look at the definition of "alpha" in the code ...
Best wishes
Torsten.
Siddarth Banerjee
2022-11-1
@Derek Shaw Can you please share the code. I am also struck in the same and cannot figure out the correct code for plotting..
Thanks
L=1;
t=1;
k=.001;
n=11;
nt=10000;
dx=L/n;
dt=.002;
alpha=k*dt/dx^2;
T0=400*ones(1,n+1);
T1=300*ones(1,n+1);
T0(1) = 300;
T0(end) = 300;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha*(T0(i+1)-2*T0(i)+T0(i-1));
end
T0=T1;
end
plot((0:n)*L/n,T1)
If one end should be at 400K and the other end is at 300K.
Why, before the for cycle, is T0 = [300,400,..,400,300]?
For me it should have been T0=[400,300,...,300,300]. If T(0,t)=T0=400 K, then this boundary condition should be fixed during the whole calculation. Can you explain?
Yes, code should be
L=1;
t=1;
k=.001;
n=11;
nt=10000;
dx=L/n;
dt=.002;
alpha=k*dt/dx^2;
T0=300*ones(1,n+1);
T0(1) = 400;
T1 = T0;
for j=1:nt
for i=2:n
T1(i)=T0(i)+alpha*(T0(i+1)-2*T0(i)+T0(i-1));
end
T0=T1;
end
plot((0:n)*L/n,T1)
更多回答（3 个）
youssef aider
2019-2-12
here is one, you can just change the boundaries
clear
clc
clf
% domain descritization
alpha = 0.05;
xmin = 0;
xmax = 0.2;
N = 100;
dx = (xmax-xmin)/(N-1);
x = xmin:dx:xmax;
dt = 4.0812E-5;
tmax = 1;
t = 0:dt:tmax;
% problem initialization
phi0 = ones(1,N)*300;
phiL = 230;
phiR = phiL;
% solving the problem
r = alpha*dt/(dx^2) % for stability, must be 0.5 or less
for j = 2:length(t) % for time steps
phi = phi0;
for i = 1:N % for space steps
if i == 1 || i == N
phi(i) = phiL;
else
phi(i) = phi(i)+r*(phi(i+1)-2*phi(i)+phi(i-1));
end
end
phi0 = phi;
plot(x,phi0)
shg
pause(0.05)
end
Alugunuri
2023-2-8
0 个投票
how to write code with neumqn BCs ?
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