How do I store Iteration Results

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]

[Jan]

Please format your code. The topic code formatting has been discussed in this forum such frequently, that I'm starting to get strange and wild emotions when I see the next ugly posting.

[Isa Isa]

[EDITED, Jan, formatted code inserted in the question]

[Image Analyst]

You should have done that in your first post, not copied it as a comment.

Answer 1

[Azzi Abdelmalek]

 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

[Isa Isa]

Thanks Azzi, I ran the code but the result{k,1}=Z;result{k,2}= Za; result{k,3}= V;result{k,4}= Va only gives ones. There is no numerical value for Za, Z, Va, V in the result. Please check it

Thanks Isa

[Azzi Abdelmalek]

i agree, there are many []. but that's the result of your code

[Azzi Abdelmalek]

but there are some results non empty

result{end,1}
result{end,2}
result{end,3}
result{end,3}

[Isa Isa]

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]

@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?

Answer 2

[Jan]

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.

[Isa Isa]

Hi Jan, What I need is to retrieve the numerical value of Z, Za, V, Va after the iterations completed. Number of iterations is determined by the number of T: for k=40:10:80 T(k)=273.15+k; So I need the value for Z, Za, V, Va at k=40,50,60,70 and 80 Thanks Isa

[Jan]

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]

Hi Jan, the code is correct.

Thanks

Isa

Answer 3

[Jan]

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;

[Isa Isa]

Hi Jan, I used the syntax below as you have advised

kList = 40:10:80; for ik = 1:length(kList) k = kList(iK); T(ik) = 273.15 + k;

But I received this error "Undefined function or variable 'iK'.

Regards Isa

[Jan]

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]

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