Differential Equation Problem using ODE45

조회 수: 1 (최근 30일)
NURUL AINA SYAHIRAH
NURUL AINA SYAHIRAH 2024년 1월 2일
댓글: Sam Chak 2024년 2월 5일
In below script, I was using the equation of dF/dz [F=CO2, CO, H2O, H2, DME, CH3OH] which written as A(1) until A(6) to solve the molar flowrate of each component. It was assumed that the feed only contained CO2 and H2 in the inlet flowrate. The mole fraction of CO2 and H2 was assumed to 0.25 and 0.75 with a total flowrate of 0.2667 kmol/s [960 kmol/hr].
In this case, the script was successfully run. However, the mole flowrate obtained for each component was NaN that results to no reaction occured/error. Appreciate if you could help me and give suggestion to improve the script. Thank you in advance.
The script:
function A = test_A(Z,F)
%Fixed parameters
Ptot=50;
T=523;
pc=1900;
%Defining constant
FT=F(1)+F(2)+F(3)+F(4)+F(5)+F(6)+F(7);
RT=523*8.314;
%Feed based on inlet as 0.2667 kmol/s
F(1)=0.25*0.2667;
F(2)=0.75*0.2667;
F(3)=0*0.2667;
F(4)=0*0.2667;
F(5)=0*0.2667;
F(6)=0*0.2667;
F(7)=0*0.2667;
%Partial pressure for each component
pCO2=(F(1)/FT)*Ptot;
pH2=(F(2)/FT)*Ptot;
pCO=(F(3)/FT)*Ptot;
pH2O=(F(4)/FT)*Ptot;
pDME=(F(5)/FT)*Ptot;
pCH3OH=(F(6)/FT)*Ptot
pN2=(F(7)/FT)*Ptot;
%Kinetic parameters
k1=35.45*exp(-1.7069e4/(RT));
k2=7.3976*exp(-2.0436e4/(RT));
k3=8.2894e4*exp(-5.2940e4/(RT));
%Adsorption constant
kH2=0.249*exp(3.4394e4/(RT));
kCO2=1.02*10^-7*exp(6.74*10^4/(RT));
kCO=7.99*10^-7*exp(5.81*10^4/(RT));
%Equilibrium constant
kp1=exp((4213/T)-(5.752*log(T))-(1.707e-3*T)+(2.682e-6*(T^2))-(7.232e-10*(T^3))+17.6);
kp2=exp((2167/T)-(0.5194*log(T))+(1.037e-3*T)-(2.331e-7*(T^2))-1.2777);
kp3=exp((4019/T)+(3.707*log(T))-(2.783e-3*T)+(3.81e-7*(T^2))-(6.561e4/(T^3))-26.64);
%Rate of Reaction
r1=k1*(pCO2*pH2*(1-((pCH3OH*pH2O)/(kp1*pCO2*pH2^3))))/(1+kCO2*pCO2+kCO*pCO+sqrt(kH2*pH2))^3;
r2=k2*((((pCO2*pH2)/(kp2*pCO))-pH2O)/((1+(kCO2*pCO2)+(kCO*pCO)+(sqrt(kH2*pH2)))^3));
r3=k3*((pCH3OH^2/pH2O)-(pDME/kp3)); %MeOH dehydration
%Trans-membrane molar flux
permN2=1.3333e-8;
permH2O=1e-6;
prN2=0*Ptot;
prH2O=0*Ptot;
ppN2=0.79320;
ppH2O=0.20680;
JN2=permN2*(prN2-pN2);
JH2O=permH2O*(prH2O-pH2O);
%Mass balance for each component
vf=0.33;
Dsi=0.0198;
Dmo=14e-3;
%Mass balance for each component
A(1)=(pc*(1-vf)*(r1-r2+r3)*(pi*Dmo^2)-(JH2O*pi*Dmo));
A(2)=(-pc*(1-vf)*(3*r1-r2)*((pi/4)*(Dsi^2-Dmo^2)));
A(3)=(pc*(1-vf)*(r2)*((pi/4)*(Dsi^2-Dmo^2)));
A(4)=(-pc*(1-vf)*(r1-r2)*((pi/4)*(Dsi^2-Dmo^2)));
A(5)=(pc*(1-vf)*(r1-2*r3)*((pi/4)*(Dsi^2-Dmo^2)));
A(6)=(pc*(1-vf)*(r3)*((pi/4)*(Dsi^2-Dmo^2)));
A=A';
  댓글 수: 1
Torsten
Torsten 2024년 1월 2일
편집: Torsten 2024년 1월 2일
You didn't include the script, but only the function to be used by ode45.

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채택된 답변

Sam Chak
Sam Chak 2024년 1월 2일
First things first, check the values computed by the differential equations. All differential states return either NaN or Inf because the equations contain r2, or r3, or both. In case you don't know, r2 returns Inf due to a division by zero problem (pCO = 0), and r3 returns NaN due to the indeterminate term in pCH3OH^2/pH2O. Fix these two issues and the code should run.
Zspan = [0 1];
F0 = [1 1 1 1 1 1];
[Z, F] = ode45(@test_A, Zspan, F0);
plot(Z, F)
function A = test_A(Z,F)
%Fixed parameters
Ptot=50;
T=523;
pc=1900;
%Feed based on inlet as 0.2667 kmol/s
F(1)=0.25*0.2667;
F(2)=0.75*0.2667;
F(3)=0*0.2667;
F(4)=0*0.2667;
F(5)=0*0.2667;
F(6)=0*0.2667;
F(7)=0*0.2667;
%Defining constant
FT=F(1)+F(2)+F(3)+F(4)+F(5)+F(6)+F(7);
RT=523*8.314;
%Partial pressure for each component
pCO2=(F(1)/FT)*Ptot; % 12.5
pH2=(F(2)/FT)*Ptot; % 37.5
pCO=(F(3)/FT)*Ptot + 0.1; % 0 plus a small non-zero value
pH2O=(F(4)/FT)*Ptot + 0.1; % 0 plus a small non-zero value
pDME=(F(5)/FT)*Ptot; % 0
pCH3OH=(F(6)/FT)*Ptot; % 0
pN2=(F(7)/FT)*Ptot; % 0
%Kinetic parameters
k1=35.45*exp(-1.7069e4/(RT));
k2=7.3976*exp(-2.0436e4/(RT));
k3=8.2894e4*exp(-5.2940e4/(RT));
%Adsorption constant
kH2=0.249*exp(3.4394e4/(RT));
kCO2=1.02*10^-7*exp(6.74*10^4/(RT));
kCO=7.99*10^-7*exp(5.81*10^4/(RT));
%Equilibrium constant
kp1=exp((4213/T)-(5.752*log(T))-(1.707e-3*T)+(2.682e-6*(T^2))-(7.232e-10*(T^3))+17.6);
kp2=exp((2167/T)-(0.5194*log(T))+(1.037e-3*T)-(2.331e-7*(T^2))-1.2777);
kp3=exp((4019/T)+(3.707*log(T))-(2.783e-3*T)+(3.81e-7*(T^2))-(6.561e4/(T^3))-26.64);
%Rate of Reaction
r1=k1*(pCO2*pH2*(1-((pCH3OH*pH2O)/(kp1*pCO2*pH2^3))))/(1+kCO2*pCO2+kCO*pCO+sqrt(kH2*pH2))^3;
r2=k2*( (((pCO2*pH2)/(kp2*pCO)) - pH2O) / ((1 + kCO2*pCO2 + kCO*pCO + sqrt(kH2*pH2))^3) );
r3=k3*((pCH3OH^2/pH2O)-(pDME/kp3)); %MeOH dehydration
%Trans-membrane molar flux
permN2=1.3333e-8;
permH2O=1e-6;
prN2=0*Ptot;
prH2O=0*Ptot;
ppN2=0.79320;
ppH2O=0.20680;
JN2=permN2*(prN2-pN2);
JH2O=permH2O*(prH2O-pH2O);
%Mass balance for each component
vf=0.33;
Dsi=0.0198;
Dmo=14e-3;
%Mass balance for each component
A(1)=( pc*(1-vf)*(r1 - r2 + r3)*(pi*Dmo^2)-(JH2O*pi*Dmo)); % NaN
A(2)=(-pc*(1-vf)*(3*r1 - r2) *((pi/4)*(Dsi^2-Dmo^2))); % Inf
A(3)=( pc*(1-vf)*(r2)*((pi/4) *(Dsi^2-Dmo^2))); % Inf
A(4)=(-pc*(1-vf)*(r1 - r2) *((pi/4)*(Dsi^2-Dmo^2))); % Inf
A(5)=( pc*(1-vf)*(r1 - 2*r3) *((pi/4)*(Dsi^2-Dmo^2))); % NaN
A(6)=( pc*(1-vf)*(r3)*((pi/4) *(Dsi^2-Dmo^2))); % NaN
A=A';
end

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