How do I store Iteration Results
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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]
3 Kommentare
Image Analyst
am 3 Sep. 2012
You should have done that in your first post, not copied it as a comment.
Antworten (3)
Azzi Abdelmalek
am 3 Sep. 2012
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
5 Kommentare
Jan
am 11 Sep. 2012
@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?
Jan
am 11 Sep. 2012
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.
3 Kommentare
Jan
am 11 Sep. 2012
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.
Jan
am 11 Sep. 2012
Bearbeitet: Jan
am 12 Sep. 2012
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;
3 Kommentare
Jan
am 12 Sep. 2012
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?
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