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We have a single straight pipe with 7 nodes and 6 elements. (The nodes are at each end and between the elements). The following code

We have a single straight pipe with 7 nodes and 6 elements. (The nodes are at each end and between the elements). The following code is Newton's Method applied to this problem. Some statements have been removed. Fix the code and then solve for the paramenters shown. This is a MatLab Code

Element 1 Element 2 Element 3 Element 4 Element 5 Element 6

clear all;

close all;

pbar=3.0; % characteristic pressure

kio=1.0; % Initial flow conductance

nel = 6; %number of elements

nnode = 7; % number of nodes

p=zeros(nnode,1); % Initialize pressure vector

p(1)=1; % boundary condition

rhs=[0 kio 0 0 0 0 0]'; %Initialize RHS vector

residual=ones(7,1); % Initialize residual

toler_r = 0.001;

maxiter=100;

iter=0;

while iter < maxiter && norm(residual)/norm(rhs) > toler_r

iter=iter+1;

amat=zeros(nnode,nnode); % Initialize the matrix.

jacob=zeros(nnode,nnode); % Initialize the Jacobian matrix.

for i = 1:nel % construct the matrix element-by-element

p_el=(p(i)+p(i+1))/2.0;

k_el = kio*(1+p_el/pbar)^4;

dkdp= 4*kio*(1+p_el/pbar)^3/(2*pbar);

amat(i,i)=;

amat(i,i+1)=;

amat(i+1,i)=

amat(i+1,i+1)=;

jacob(i,i)= ;

jacob(i,i+1)= ;

jacob(i+1,i)= ;

jacob(i+1,i+1)= ;

if i==1

rhs(i+1)=k_el;

end

end

jacob

residual=rhs(2:6)-amat(2:6,2:6)*p(2:6)

p_s= amat(2:6, 2:6) \ rhs(2:6);

p(2:6)=p(2:6)+p_s

pause;

end

iter

plot(p);

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