MATLAB Answers

Results are NaN or absurd if a certain parameter changes.

조회 수: 1(최근 30일)
Furkan Erdem
Furkan Erdem 2021년 6월 10일
답변: darova 2021년 8월 5일
Hello everyone,
The code I have given below is the explicit 1-D finite difference solution of a heat transfer question. The problem I'm having is that depending on the number of nodes (n), the results differ significantly. Between 2 and 46 nodes, the results are reasonable where accuracy, if my solution is correct, improves. However, between nodes 47 and 50, the results are erratic and I can't make any sense of them. After 50, results are given as NaN.
I'm asuuming that beyond 50, "dx" becomes too small, however I don't know what happens between 46 and 50. I can say that lowering "dx" results in a similar behaviour.
I was wondering if there was a problem with my poorly written code or am I missing something else?
clear, clc;
n = 6; %number of nodes
L = 0.05;
dx = L/(n-1);
k = 28;
h = 60;
q = 6*10^5;
alp = 12.5*10^-6;
dt = .1;
t = 0:dt:100;
Ta = 30;
Ti = 200;
tau = alp*dt/dx^2;
c = zeros(length(t),1);
for j = 1:length(t)
if j == 1
T(1) = tau*(q*dx^2/(2*k) + Ti) + Ti*(1-tau);
T(n) = tau*(Ti + q*dx^2/(2*k) + h*Ta*dx/k) + Ti*(1-tau-tau*h*dx/k);
else
T(1) = tau*(q*dx^2/(2*k) + Ti) + T(1)*(1-tau);
T(n) = tau*(T(n-1) + q*dx^2/(2*k) + h*Ta*dx/k) + T(n)*(1-tau-tau*h*dx/k);
end
for i = 2:1:n-1
if j == 1
T(i) = tau*(Ti+Ti) + (1 - 2*tau)*Ti + (tau*q*dx^2)/(2*k);
else
T(i) = tau*(T(i-1)+T(i+1)) + (1 - 2*tau)*T(i) + (tau*q*dx^2)/(2*k);
end
end
end
m = (1:n);
mt = transpose(m);
Tt = transpose(T);
table(mt,Tt)
figure
plot(mt,T)
  댓글 수: 2
Furkan Erdem
Furkan Erdem 2021년 6월 10일
I wouldn't ever have noticed that, I should be more orgnaised. Unfortunately, in this instance, the code is for a class so I'm going to have to leave it.

댓글을 달려면 로그인하십시오.

답변(2개)

Tanmay Das
Tanmay Das 2021년 8월 4일
Hi,
By default, all numeric values are stored as double data type. Once the number of nodes cross 50, the values in T array exceed the bounds of double data type. MATLAB constructs the double data type according to IEEE® Standard 754 for double precision. The range for a negative number of type double is between -1.79769 x 10^308 and -2.22507 x 10^-308, and the range for positive numbers is between 2.22507 x 10^-308 and 1.79769 x 10^308. You can look into the double Documentation for further information.

darova
darova 2021년 8월 5일
I suggest you to visualize the data in 3D. See how values change each step
T = zeros(20);
a = 0.04; % constant
t = linspace(0,pi,20);
T(1,:) = sin(t); % initial condition
h = surf(T);
for i = 1:19
for j = 2:19
T(i+1,j) = T(i,j) - a*(T(i,j+1)-T(i,j-1)+2*T(i,j));
end
set(h,'zdata',T)
pause(0.5)
end

제품


릴리스

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by