This document appears to be a homework assignment that includes code to model heat transfer using the finite difference method. It contains code to: 1) Define parameters like grid size, velocities, and timestep for the simulation. 2) Calculate Reynolds numbers based on maximum velocities. 3) Initialize temperature matrices and apply boundary conditions. 4) Iterate the simulation over time using an explicit forward-time centered-space scheme. 5) Analyze results by plotting temperature over time at the center and a contour plot at steady state.

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Jared Fertig - ME 541 HW #4
Table of Contents
Problem 2 (dx = dy = .5) ..................................................................................................... 1
Problem 2 (dx = dy = .25) .................................................................................................... 5
Problem 2 (dx = dy = .5)
clear all
close all
clc
%set initial parameters
k = .05;
xmax = 10; % Maximum length (L)
ymax = 5; % Maximum height (H)
tmax = 100; % Maximum time (t)
dx = .5; %delta x
dy = .5; %delta y
dt = .1; %delta t
nx = xmax/dx;
ny = ymax/dy;
nt = tmax/dt;
x = 0:dx:xmax;
y = 0:dy:ymax;
t = 0:dt:tmax;
%coefficients
c = dt/(2*dx);
d = dt/(2*dy);
r = k*dt/(dx^2);
s = k*dt/(dy^2);
%Create velocity matrices
for ii = 1:length(x)
for jj = 1:length(y)
u(ii,jj) = .5*sin(2*pi()*x(ii)/xmax)*sin(pi()*y(jj)/ymax);
v(ii,jj) = 1*sin(2*pi()*x(ii)/xmax)*sin(pi()*y(jj)/ymax);
end
end
%Find maximum cell Reynolds Numbers

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umax = max(max(u));
vmax = max(max(v));
Rcx = umax*dx/k;
disp(['The Maximum Reynolds Number in the x-direction is R_cx =',
num2str(Rcx)])
Rcy = vmax*dy/k;
disp(['The Maximum Reynolds Number in the y-direction is R_cy =',
num2str(Rcy)])
%Create the Temperature Matrix
T = zeros(nx+1,ny+1,nt+1);
%initial conditions
for ii = 1:nx+1
for kk = 2:nt+1
T(ii,1,kk) = 10*sin(pi()*x(ii)/xmax)^2;
T(ii,ny,kk) = 10*sin(pi()*x(ii)/xmax)^2;
end
end
for jj = 1:ny+1
for kk = 2:nt+1
T(1,jj,kk) = 5*sin(2*pi()*y(jj)/ymax)^2;
T(nx,jj,kk) = 5*sin(2*pi()*y(jj)/ymax)^2;
end
end
for ii = 1:nx+1
for jj = 1:ny+1
T(ii,jj,1) = 0;
end
end
%create matrix based on forward time center space
for kk = 1:nt
T(1,1,kk+1) = T(1,1,kk)-...
c*u(1,1)*(T(2,1,kk)-T(nx,1,kk))-...
c*v(1,1)*(T(1,2,kk)-T(1,ny,kk))+...
r*(T(2,1,kk)-2*T(1,1,kk)+T(nx,1,kk))+...
r*(T(1,ny,kk)-2*T(1,1,kk)+T(1,ny,kk));
for ii = 2:nx
for jj = 2:ny
T(ii,jj,kk+1)=T(ii,jj,kk)-...
c*u(ii,jj)*(T(ii+1,jj,kk)-T(ii-1,jj,kk))-...
c*v(ii,jj)*(T(ii,jj+1,kk)-T(ii,jj-1,kk))+...
r*(T(ii+1,jj,kk)-2*T(ii,jj,kk)+T(ii-1,jj,kk))+...
r*(T(ii,jj+1,kk)-2*T(ii,jj,kk)+T(ii,jj-1,kk));
end
for ll = 1:ny
T(nx+1,ll,kk+1)=T(1,ll,kk+1);
end
for mm = 1:nx
T(mm,ny+1,kk+1)=T(mm,1,kk+1);

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end
end
end
%Calculate Temperature Matrix at the Center
T_center = T(nx/2,ny/2,:);
T_cent = T_center(:)';
figure(1)
plot(t,T_cent)
grid on
title('Temperature at Center vs Time')
xlabel('Time (s)')
ylabel('Temperature (deg)')
%Find Steady State Time and Temperature
tol = .0001;
pp = 40;
err = 1;
while err > tol
err = T_cent(pp+1) - T_cent(pp);
pp = pp + 1;
end
disp(['Steady State is Reached at the Center at ', num2str(t(pp)),'
seconds'])
disp(['The Steady State Temperature is ', num2str(T_cent(pp)),'
degrees'])
T_contour = T(:,:,pp);
C = rot90(T_contour);
figure(2)
contour(x,y,C,20);
title('Contour Plot at Steady State (20 Levels)')
xlabel('Distance')
ylabel('Height')
The Maximum Reynolds Number in the x-direction is R_cx =5
The Maximum Reynolds Number in the y-direction is R_cy =10
Steady State is Reached at the Center at 33.2 seconds
The Steady State Temperature is 7.1491 degrees

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Problem 2 (dx = dy = .25)
%set initial parameters
k = .05;
xmax = 10; % Maximum length (L)
ymax = 5; % Maximum height (H)
tmax = 100; % Maximum time (t)
dx = .25; %delta x
dy = .25; %delta y
dt = .1; %delta t
nx = xmax/dx;
ny = ymax/dy;
nt = tmax/dt;
x = 0:dx:xmax;
y = 0:dy:ymax;
t = 0:dt:tmax;
%coefficients
c = dt/(2*dx);
d = dt/(2*dy);
r = k*dt/(dx^2);
s = k*dt/(dy^2);
%Create velocity matrices
for ii = 1:length(x)
for jj = 1:length(y)
u(ii,jj) = .5*sin(2*pi()*x(ii)/xmax)*sin(pi()*y(jj)/ymax);
v(ii,jj) = 1*sin(2*pi()*x(ii)/xmax)*sin(pi()*y(jj)/ymax);
end
end
%Find maximum cell Reynolds Numbers
umax = max(max(u));
vmax = max(max(v));
Rcx = umax*dx/k;
disp(['The Maximum Reynolds Number in the x-direction is R_cx =',
num2str(Rcx)])
Rcy = vmax*dy/k;
disp(['The Maximum Reynolds Number in the y-direction is R_cy =',
num2str(Rcy)])
%Create the Temperature Matrix
T = zeros(nx+1,ny+1,nt+1);
%initial conditions
for ii = 1:nx+1
for kk = 2:nt+1

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T(ii,1,kk) = 10*sin(pi()*x(ii)/xmax)^2;
T(ii,ny,kk) = 10*sin(pi()*x(ii)/xmax)^2;
end
end
for jj = 1:ny+1
for kk = 2:nt+1
T(1,jj,kk) = 5*sin(2*pi()*y(jj)/ymax)^2;
T(nx,jj,kk) = 5*sin(2*pi()*y(jj)/ymax)^2;
end
end
for ii = 1:nx+1
for jj = 1:ny+1
T(ii,jj,1) = 0;
end
end
%create matrix based on forward time center space
for kk = 1:nt
T(1,1,kk+1) = T(1,1,kk)-...
c*u(1,1)*(T(2,1,kk)-T(nx,1,kk))-...
c*v(1,1)*(T(1,2,kk)-T(1,ny,kk))+...
r*(T(2,1,kk)-2*T(1,1,kk)+T(nx,1,kk))+...
r*(T(1,ny,kk)-2*T(1,1,kk)+T(1,ny,kk));
for ii = 2:nx
for jj = 2:ny
T(ii,jj,kk+1)=T(ii,jj,kk)-...
c*u(ii,jj)*(T(ii+1,jj,kk)-T(ii-1,jj,kk))-...
c*v(ii,jj)*(T(ii,jj+1,kk)-T(ii,jj-1,kk))+...
r*(T(ii+1,jj,kk)-2*T(ii,jj,kk)+T(ii-1,jj,kk))+...
r*(T(ii,jj+1,kk)-2*T(ii,jj,kk)+T(ii,jj-1,kk));
end
for ll = 1:ny
T(nx+1,ll,kk+1)=T(1,ll,kk+1);
end
for mm = 1:nx
T(mm,ny+1,kk+1)=T(mm,1,kk+1);
end
end
end
%Calculate Temperature Matrix at the Center
T_center = T(nx/2,ny/2,:);
T_cent = T_center(:)';
figure(3)
plot(t,T_cent)
grid on
title('Temperature at Center vs Time')
xlabel('Time (s)')
ylabel('Temperature (deg)')
%Find Steady State Time and Temperature

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tol = .0001;
pp = 40;
err = 1;
while err > tol
err = T_cent(pp+1) - T_cent(pp);
pp = pp + 1;
end
disp(['Steady State is Reached at the Center at ', num2str(t(pp)),'
seconds'])
disp(['The Steady State Temperature is ', num2str(T_cent(pp)),'
degrees'])
T_contour = T(:,:,pp);
C = rot90(T_contour);
figure(4)
contour(x,y,C,20);
title('Contour Plot at Steady State (20 Levels)')
xlabel('Distance')
ylabel('Height')
The Maximum Reynolds Number in the x-direction is R_cx =2.5
The Maximum Reynolds Number in the y-direction is R_cy =5
Steady State is Reached at the Center at 34.5 seconds
The Steady State Temperature is 6.7535 degrees

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Published with MATLAB® R2015b