Differential Equation Problem using ode45

Question

Luqman Hardiyan 2019-11-25

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/493043-differential-equation-problem-using-ode45

回答： Stephan 2019-11-25

采纳的回答： Stephan

在 MATLAB Online 中打开

I was solving the molar flow rate or dF/dW using this equation, it says that there are error on line 32.

Why there's an error on line 32 saying not enough input arguments? where did I go wrong? thank you

function PBR_Isothermal
  clc; clear;
  %A = Dammitol
  %B = Oxygen
  %C = Valualdehyde
  %D = Carbon Dioxide
  %E = Water
  %I = Nitrogen
  
  %Feed based on inlet as 100 kmol/h
  
  Fa0 = 0.1*100; 
  Fb0 = 0.07*100;
  Fc0 = 0;
  Fd0 = 0;
  Fe0 = 0.02*100;
  Fi0 = 0.81*100;
 
  
  [Fa W] = ode45(FunA,[0 1000],Fa0);
%   [Fb W] = ode45(FunA,[0 1000],Fb0);
%   [Fc W] = ode45(FunA,[0 1000],Fc0);
%   [Fd W] = ode45(FunA,[0 1000],Fd0);
%   [Fe W] = ode45(FunA,[0 1000],Fe0);
end
function A = FunA(F,W)
%Total Pressure
Ptot = 114.7;
%Defining Constant
FT   = F(1)+F(2)+F(3)+F(4)+F(5)+F(6);
RT   = (1/353-1/373)/1.987;
%Partial Pressure of Each
PDT  = (F(1)/FT)*Ptot;
PO2  = (F(2)/FT)*Ptot;
PVA  = (F(3)/FT)*Ptot;
PCO  = (F(4)/FT)*Ptot;
PWA  = (F(5)/FT)*Ptot;
PN2  = (F(6)/FT)*Ptot;
%Constants
k1 = 1.771*10^-3; 
k2 = 23295; 
k3 = 0.5; 
k4 = 1.0; 
k5 = 0.8184; 
k6 = 0.5; 
k7 = 0.2314;
k8 = 1.0; 
k9 = 1.25; 
k10 = 2.0; 
k11 = 2.795*10^-4; 
k12 = 33000; 
k13 = 0.5; 
k14 = 2.0;
%Rate of Reaction
r1 = (k1*exp(-k2*RT)*PO2^k3*PDT^k4)/(1+k5*PO2^k6+k7*PDT^k8+k9*PVA^k10); 
r2 = (k11*exp(-k12*RT)*PO2^k13*PVA^k14)/(1+k5*PO2^k6+k7*PDT^k8+k9*PVA^k10);
%Molar Flow Rate of Each Component
F(1) = -r1;
F(2) = -1/2*r1;
F(3) = r1;
F(4) = 2*r2;
F(5) = r1+2*r2;
F(6) = Fi0;
end

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Answer 1

Stephan 2019-11-25

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/493043-differential-equation-problem-using-ode45#answer_403232

在 MATLAB Online 中打开

[W,Fa] = PBR_Isothermal;
% plot results
subplot(3,2,1)
plot(W,Fa(:,1))
subplot(3,2,2)
plot(W,Fa(:,2))
subplot(3,2,3)
plot(W,Fa(:,3))
subplot(3,2,4)
plot(W,Fa(:,4))
subplot(3,2,5)
plot(W,Fa(:,5))
subplot(3,2,6)
plot(W,Fa(:,6))
function [W,Fa] = PBR_Isothermal
  clc; clear;
  %A = Dammitol
  %B = Oxygen
  %C = Valualdehyde
  %D = Carbon Dioxide
  %E = Water
  %I = Nitrogen
  
  %Feed based on inlet as 100 kmol/h
  
  Fa0 = 0.1*100; 
  Fb0 = 0.07*100;
  Fc0 = 0;
  Fd0 = 0;
  Fe0 = 0.02*100;
  Fi0 = 0.81*100;
 
  
  [W, Fa]  = ode45(@FunA,[0 1000],[Fa0 Fb0 Fc0 Fd0 Fe0 Fi0]);
%   [Fb W] = ode45(FunA,[0 1000],Fb0);
%   [Fc W] = ode45(FunA,[0 1000],Fc0);
%   [Fd W] = ode45(FunA,[0 1000],Fd0);
%   [Fe W] = ode45(FunA,[0 1000],Fe0);
end
function A = FunA(W,F)
%Total Pressure
Ptot = 114.7;
%Defining Constant
FT   = F(1)+F(2)+F(3)+F(4)+F(5)+F(6);
RT   = (1/353-1/373)/1.987;
%Partial Pressure of Each
PDT  = (F(1)/FT)*Ptot;
PO2  = (F(2)/FT)*Ptot;
PVA  = (F(3)/FT)*Ptot;
PCO  = (F(4)/FT)*Ptot;
PWA  = (F(5)/FT)*Ptot;
PN2  = (F(6)/FT)*Ptot;
%Constants
k1 = 1.771*10^-3; 
k2 = 23295; 
k3 = 0.5; 
k4 = 1.0; 
k5 = 0.8184; 
k6 = 0.5; 
k7 = 0.2314;
k8 = 1.0; 
k9 = 1.25; 
k10 = 2.0; 
k11 = 2.795*10^-4; 
k12 = 33000; 
k13 = 0.5; 
k14 = 2.0;
%Rate of Reaction
r1 = (k1*exp(-k2*RT)*PO2^k3*PDT^k4)/(1+k5*PO2^k6+k7*PDT^k8+k9*PVA^k10); 
r2 = (k11*exp(-k12*RT)*PO2^k13*PVA^k14)/(1+k5*PO2^k6+k7*PDT^k8+k9*PVA^k10);
%Molar Flow Rate of Each Component
A(1) = -r1;
A(2) = -1/2*r1;
A(3) = r1;
A(4) = 2*r2;
A(5) = r1+2*r2;
Fi0 = 0.81*100;
A(6) = Fi0;
A=A';
end