How do I store Iteration Results

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Isa Isa
Isa Isa 2012-9-3
Hi, Please I need to get values for Z, Za, V, Va at every value of T. The code is as follows:
z=[0.2 0.2 0.2 0.4];
W_PR=0.245;
C=4;
omega=[0.344 0.467 0.578 0.789];
Tc=[600 700 500 570];
Pc=[50 70 58 76];
R=8.314;
P=20;
Z_store = ones(C,C,5);
V_store = ones(5,C);
K_PR=[1.546e-2, 0.456, 1.432e2, 14.32];
iter = 0;
for k=40:10:80
T(k)=273.15+k;
for c=1:C
x_PR(1,c)=z(c)/(1+W_PR*(K_PR(c)-1));
x_PR(2,c)=K_PR(c)*x_PR(1,c);
kappa_PR=0.37464+1.54226.*omega(c)-0.26992.*omega(c).^2;
alpha_PR=(1+kappa_PR.*(1-sqrt(T(k)./Tc(c)))).^2;
a_PR(c,c)=0.45724.*R.^2.*Tc(c).^2./Pc(c).*alpha_PR;
b_PR(c)=0.07780*R.*Tc(c)./Pc(c);
end
for c=2:C
for n=1:(c-1)
a_PR(c,n)=sqrt(a_PR(c,c).*a_PR(n,n));
a_PR(n,c)=a_PR(c,n);
end
end
for c=1:C
A_PR(c,c)=a_PR(c,c).*P./(R.*T(k)).^2;
B_PR(c)=b_PR(c).*P./(R.*T(k));
Z(c,c)=A_PR(c,c)./5;
V(c)=B_PR(c).*6;
end
for c=1:C
Aa_PR(c,c)=a_PR(c,c).*P./(R.*T(k)).^3;
Ba_PR(c)=b_PR(c).*P./(R.*T(k));
Za(c,c)=A_PR(c,c)./8;
Va(c)=B_PR(c).*9;
end
end
Thanks
Isa
[EDITED, Jan, code formatted]
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Isa Isa
Isa Isa 2012-9-3
编辑：Jan 2012-9-11
[EDITED, Jan, formatted code inserted in the question]
Image Analyst
Image Analyst 2012-9-3
You should have done that in your first post, not copied it as a comment.

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回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-9-3
z=[0.2 0.2 0.2 0.4]; W_PR=0.245; C=4;
omega=[0.344 0.467 0.578 0.789];
Tc=[600 700 500 570]; Pc=[50 70 58 76]; R=8.314; P=20;
Z_store = ones(C,C,5); V_store = ones(5,C);
K_PR=[1.546e-2, 0.456, 1.432e2, 14.32]; iter = 0;
for k=40:10:80
T(k)=273.15+k;
for c=1:C
x_PR(1,c)=z(c)/(1+W_PR*(K_PR(c)-1));
x_PR(2,c)=K_PR(c)*x_PR(1,c);
kappa_PR=0.37464+1.54226.*omega(c)-0.26992.*omega(c).^2;
alpha_PR=(1+kappa_PR.*(1-sqrt(T(k)./Tc(c)))).^2;
a_PR(c,c)=0.45724.*R.^2.*Tc(c).^2./Pc(c).*alpha_PR;
b_PR(c)=0.07780*R.*Tc(c)./Pc(c);
end
for c=2:C
for n=1:(c-1)
a_PR(c,n)=sqrt(a_PR(c,c).*a_PR(n,n));
a_PR(n,c)=a_PR(c,n);
end
end
for c=1:C
A_PR(c,c)=a_PR(c,c).*P./(R.*T(k)).^2;
B_PR(c)=b_PR(c).*P./(R.*T(k));
Z(c,c)=A_PR(c,c)./5; V(c)=B_PR(c).*6;
end
for c=1:C
Aa_PR(c,c)=a_PR(c,c).*P./(R.*T(k)).^3;
Ba_PR(c)=b_PR(c).*P./(R.*T(k));
Za(c,c)=A_PR(c,c)./8; Va(c)=B_PR(c).*9;
end
result{k,1}=Z;result{k,2}= Za; result{k,3}= V;result{k,4}= Va
end
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Isa Isa
Isa Isa 2012-9-11
Hi Azzi, As I said before, if I run the code the syntax you put "result{k,1}=Z;result{k,2}= Za; result{k,3}= V;result{k,4}= Va" all give [] as results. That is not correct because the results contain numerical value and zeros. If evaluate using your syntax result{end,1} result{end,2} result{end,3} result{end,3} It gives the results of the last iteration. I need the results of all the iterations.
I will be glad to receive your quick response Thanks Isa
Jan
Jan 2012-9-11
@Isa Isa, I still do not get the problem. Does the code you have posted calculate Z, Za, V and Va correctly? If not, why? Did you set some breakpoints in the code to inspect what's going on? What kind of help do you expect from us now?

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Jan
Jan 2012-9-11
This looks strange:
for c=2:C
for n=1:(c-1)
a_PR(c,n)=sqrt(a_PR(c,c).*a_PR(n,n));
a_PR(n,c)=a_PR(c,n);
end
end
The values of a_Pr are overwritten by its own transposed. Please check if this is really doing what you want.
I do not see a chance that we can find the bugs of your program, because we cannot know what you are wanting to achieve. I suggest to use the debugger to find out, what's going on.
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Jan
Jan 2012-9-11
Hi Isa, and what I ask is, if the posted code is correct, because it is a frequently implemented bug (FIB) to overwrite values in a matrix at an inplace-transpose.
Isa Isa
Isa Isa 2012-9-11
Hi Jan, the code is correct.
Thanks
Isa

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Jan
Jan 2012-9-11
编辑：Jan 2012-9-12
Your code has this problem:
for k = 40:10:80
T(k) = 273.15+k;
...
result{k,1}=Z;result{k,2}= Za; result{k,3}= V;result{k,4}= Va
end
Now the first value written to the index k=40! Improvement:
kList = 40:10:80;
for ik = 1:length(kList)
k = kList(ik); % [EDITED, "iK" -> "ik"]
T(ik) = 273.15 + k; % Not T(k)!
...
result{ik, 1} = Z;
result{ik, 2} = Za;
result{ik, 3} = V;
result{ik, 4} = Va
end
BTW. your code would be cleaner and faster, if it is vectorized:
% Instead of: for c=1:C
x_PR = z ./ (1 + W_PR .* (K_PR.' - 1));
x_PR(2, :) = K_PR .* x_PR.';
kappa_PR = 0.37464+1.54226.*omega-0.26992.*omega.^2;
alpha_PR = (1+kappa_PR.*(1-sqrt(T(ik)./Tc))).^2;
a_PR = diag(0.45724.*R.^2.*Tc .^ 2 ./ Pc .* alpha_PR);
b_PR = 0.07780 * R .* Tc ./ Pc;
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Jan
Jan 2012-9-12
The error message contain "iK" with an uppercase "K", while the variable is called "ik" with a lowercase "k". Is it really too complicated to fix this by your own?
Isa Isa
Isa Isa 2012-9-12
May I ask this? Still your syntax gives Z and Za as 4 by 4 matrix. Instead, the results should be 4 by 4 by 5 matrix. c=1:4 and Z(c,c,ik). That is, the value of Z(c,c) at each T(ik).
Thanks
Isa

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