For Problem 2 in Homework 4 (see below), please further divide the region using 30 x 30 nodes and use Matlab or other programs to obtain the numerical solutions for the temperature values on all nodes. The Matlab codes are provided at the end of the assignment for you to start with. The codes have a lot of errors and bugs to be corrected but the general frame is there. Files to be submitted: your codes and the contour plot of the temperature distribution over the region. Problem 2 in Homework 4: Determine the steady state temperature distribution in the region shown below. It has a thermal conductivity of 10 W/m-K, and the dimension of 1 mx 1 m. The top and bottom surfaces are kept at constant temperature of 30°C and 50°C, respectively. The left surface is subjected to a laser heating, with a heat flux of q" = 10,000 sin(лy) [W/m²], and the right surface is insulated. The system is divided into 3 × 3 points. Node 7 is the original point for both x and y axis ((0,0) point). • Give the temperature equation of the interior point (point 5), ⚫ Give the temperature equation of the points at four boundaries (2, 4, 6, 8), • Give the temperature equation of the points at four corners (1, 3, 7, 9), • Give the matrix equation to solve the temperature distribution. 4.47 Consider the nodal point 0 located on the boundary between materials of thermal conductivity kд and KB. 2 Material A Δν 1 0 3 KA Ax = Ay Material B KB Derive the finite-difference equation, assuming no internal generation.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Codes to start with: (many errors and bugs to be corrected)
m=30;
n=m*m;
x=linspace(0,1,m);
y=linspace(1,0,m);
k=10;
a=zeros(n,n);
b=zeros(n);
c=zeros(n);
d=zeros(m,m);
%interior points
for jj=?
a(jj,jj)=-4;
a(jj,jj-2*m)=1;
a(jj,jj+2*m)=1;
a(jj,jj-2)=1;
a(jj,jj+4)=1;
end
%top bc
for jj=1:m
a(jj,jj)=1;
end
%bottom bc
for jj=?
a(jj,jj)=1;
end
%left bc
for jj=?
a(jj,jj)=-4;
a(jj,jj-m-5)=1;
a(jj,jj+m-2)=1;
a(jj,jj+4)=2;
end
%right bc
for jj=?
a(jj,jj)=-4;
a(jj,jj-m)=1;
a(jj,jj+m)=1;
a(jj,jj-1)=2;
end
for jj=1:m
b(jj)=30;
end
for jj=?
b(jj)=50;
end
for jj=?
b(jj)=-20000/k/m*cos(pi*((m-1)*m-jj+1)/m/m);
end
c=inv(a)*b;
%try to plot c in 2d domain
figure
contourf(x,y,d)
title('Temperature')
xlabel('Distance x')
ylabel('Distance y')