我有下面的代码。
for k=40:10:80
T(k)=273.15+k;
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];
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);
end
for c=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));
end
for c=1:C
Z(c,c)=A_PR(c,c)./5;
V(c)=B_PR(c).*6;
end
end
每次运行代码时,我都想要每个 T(k) 处的 Z 和 V 的结果。 该代码只给出一个 T 值的结果,尽管我希望它运行循环并给出所有 T(k) 的 Z 和 V 的结果。
最佳答案
您可能希望在主循环之外定义一些参数,至少预先分配存储空间。
Z_store = ones(C,C,5); % if you want to use a 3d matrix
和
V_store = ones(5,C);
新功能:
function [Z_store V_store] = SO_test1()
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
iter = iter + 1;
Z_store(:,:,iter) = Z;
V_store(iter,:) = V;
end
end
关于matlab - 在 Matlab 中存储迭代结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12234048/