# Homework Solution: Consider the process flow diagram shown in Fig. 2.14. We can write this out as the system of equation…

Consider the process flow diagram shown in Fig. 2.14. We can write this out as the system of equations m1 = m2 + m3 + m4 + m5 m_2 = m_9 + m_10 + m_11 m_5 = m_8 + m_7+m_6 m_12 = m_4 + m_7 + m_11 m_1 =100 m_5 = 5 ms 0.84 m_12 = m_4 + m_7 0.7 m_1 = m_2 + m_3 0.55 m_1 = m_9 + m_12 0.2 m_9 = m_10 0.85m_2 = m_9 + m_11 3.2 m_6 = m_7 + m_8 The first four equations are the mass balances: the next eight equations are the process specifications. (a) Write a MATLAB program that generates the matrix A and vector b required to solve the problem for this set of equations. (b) Write a MATLAB program that solves this system using naive Gauss elimination. What happens? How do you fix it?

From the given equations A 12x12 matrix is: 1 -1 -1 -1 -1 0 0 0 0 0 0 0

Consider the regularity issue diagram shown in Fig. 2.14. We can transcribe this quenched as the arrangement of equations m1 = m2 + m3 + m4 + m5 m_2 = m_9 + m_10 + m_11 m_5 = m_8 + m_7+m_6 m_12 = m_4 + m_7 + m_11 m_1 =100 m_5 = 5 ms 0.84 m_12 = m_4 + m_7 0.7 m_1 = m_2 + m_3 0.55 m_1 = m_9 + m_12 0.2 m_9 = m_10 0.85m_2 = m_9 + m_11 3.2 m_6 = m_7 + m_8 The original foul-mouthed equations are the bulk balances: the instant prospect equations are the regularity specifications. (a) Transcribe a MATLAB program that generates the matrix A and vector b required to explain the example ce this decideed of equations. (b) Transcribe a MATLAB program that explains this arrangement using easy Gauss encircleation. What happens? How do you decide it?

## Expert Disconnection

From the fond equations A 12×12 matrix is:

1 -1 -1 -1 -1 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 -1 -1 -1 0

0 0 0 0 1 -1 -1 -1 0 0 0 0

0 0 0 -1 0 0 -1 0 0 0 -1 1

1 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 -5 0 0 0 0

0 0 0 -1 0 0 -1 0 0 0 0 0.84

0.70 -1 -1 0 0 0 0 0 0 0 0 0

0.55 0 0 0 0 0 0 0 -1 0 0 -1

0 0 0 0 0 0 0 0 0.20 -1 0 0

0 0.85 0 0 0 0 0 0 -1 0 -1 0

0 0 0 0 0 3.20 -1 -1 0 0 0 0

The b 12×1 matrix is:

0

0

0

0

100

0

0

0

0

0

0

0

Code:

%% Explain rectirectilinear arrangement of eqution Ax=b using Easy Gaussian Encircleation Method

clc; evident all; suspend all;

A = [1,-1,-1,-1,-1,0,0,0,0,0,0,0;…

0,1,0,0,0,0,0,0,-1,-1,-1,0;…

0,0,0,0,1,-1,-1,-1,0,0,0,0;…

0,0,0,-1,0,0,-1,0,0,0,-1,1;…

1,0,0,0,0,0,0,0,0,0,0,0;…

0,0,0,0,1,0,0,-5,0,0,0,0;…

0,0,0,-1,0,0,-1,0,0,0,0,0.84;…

0.7,-1,-1,0,0,0,0,0,0,0,0,0;…

0.55,0,0,0,0,0,0,0,-1,0,0,-1;…

0,0,0,0,0,0,0,0,0.2,-1,0,0;…

0,0.85,0,0,0,0,0,0,-1,0,-1,0;…

0,0,0,0,0,3.2,-1,-1,0,0,0,0];

b = [0;0;0;0;100;0;0;0;0;0;0;0];

disp(‘A matix:’)

disp(A)

disp(‘b matix:’)

disp(b)

[n,~] = extent(A);

%Initialize disconnection X

X = zeros(n,1);

%%Forward Encircleation (convertion Of A to Upeer triangle matrix)

ce i = 1:n-1

while A(i,i)==0

temp =A;

A = [temp(1:(i-1),:);temp(i+1:n,:);temp(i,:)];

tempb = b;

b = [tempb(1:(i-1),:);tempb(i+1:n,:);tempb(i,:)];

end

m = A(i+1:n,i)/A(i,i);

A(i+1:n,:) = A(i+1:n,:)-m*A(i,:);

b(i+1:n,:) = b(i+1:n,:)-m*b(i,:);

end

%%Back Substitution

X(n,:) = b(n,:)/A(n,n);

ce i = n-1:-1:1

X(i,:) = (b(i,:) – A(i,i+1:n)*X(i+1:n,:))/A(i,i);

end

disp(‘Solution:’)

disp(X)

Solution:

100.0000

40.0000

30.0000

9.4565

20.5435

4.8913

11.5435

4.1087

30.0000

6.0000

4.0000

25.0000