complex numbers as function output

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Sunetra CV
Sunetra CV 2019-10-22
回答： Hari Krishna Ravuri 2019-11-7
The functions used to solve my ODE's using Euler's method give real values in the first iteration ; where as the function returns complex variables from the third iteration. I'm not able to figure out the reason behind this.
function file
function f =f_UP(K)
global M Cp R P z t A tda
P = (R*K(1)*K(2));
% z = (1.4888*(t^0.75)*(Cp^0.25)*(A^0.25)*(K(3)^0.25)*(K(2)^0.25)*((K(2)-tda)^0.25))
% z = 1.4888*(t^0.75)*(Cp^0.25)*(A^0.25)*(K(3)^0.25)*(K(2)^0.25)*((K(2)-tda)^0.25);
f(1) = ((-K(1))*(1.113-2.14*z +2.81*z^2-1.76*z^3+0.396*z^4+ ...
0.08694*z^5-0.0579*z^6+0.0077*z^7))-((K(3))*(0.4235+1.5*z-1.133*z^2+0.3741*z^3-...
0.04537*z^4));
f(2)= (((-1)*R*(K(2)))./(P*M*Cp))*(760.91-1608.08*(z)+2449.22*z^2-...
(2001.61*z^3)+ 904.51*z^4 -212.087*z^5+20.095*z^6);
f(3)= ((-1)/((K(1))*(K(3))))*((1.113-2.14*z+2.81*z^2-1.76*z^3+ ...
0.396*z^4+0.08694*z^5-0.0579*z^6 +0.0077*z^7));
f=f'
Main file
clear all
close all
global M Cp R A tda z t
deltt =0.1;
t_span = 0:1;
x_H2O = 0.5;
x_CO2 = 0.5;
M_CO2 = 44*(10^-3); %kg/mol
M_H2O = 18*(10^-3); %kg/mol
M = (x_CO2*M_CO2) + (x_H2O*M_H2O);
A = 3.14*(0.091)^2;
tda = 300;
R = 8.314; %m3.Pa/mol/K
cp_CO2 = 1168; %J/kg/K
cp_H2O = 2147; %J/kg/K
Cp = (x_CO2*cp_CO2)+(x_H2O*cp_H2O)
K(1,:) = [0.43545,800,1.2]
t = 0;
tend =10;
td = 0:tend;
or i= 1:length(td)
z = 1.4888*(td(i)^0.75)*(Cp^0.25)*(A^0.25)*(K(3)^0.25)*(K(2)^0.25)*((K(2)-tda)^0.25)
K(i+1,:)= K(i,:)' + deltt*f_UP(K(i,:))
g(i)= z
% t = t+1
end
  2 个评论
dpb
dpb 2019-10-22
I'd guess z goes negative...use debugger to step through and watch what happens where to find your logic error...

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回答（1 个）

Hari Krishna Ravuri
You may consider debugging your script using MATLAB Debugger. Please refer https://in.mathworks.com/help/matlab/debugging-code.html for more information on debugging a MATLAB script.

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