complex numbers as function output

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Sunetra CV
Sunetra CV il 22 Ott 2019
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 Commenti
dpb
dpb il 22 Ott 2019
I'd guess z goes negative...use debugger to step through and watch what happens where to find your logic error...
Sunetra CV
Sunetra CV il 23 Ott 2019
Thank You. I'll look into it.

Accedi per commentare.

Risposte (1)

Hari Krishna Ravuri
Hari Krishna Ravuri il 7 Nov 2019
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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