how to I solve 2 eqns with 2 unknowns and plot 5 plots according to these solutions?

1 Ansicht (letzte 30 Tage)
clc;
clear;
deltaGrxn_std_rxn1 = -24800; %(J/mol)
deltaHrxn_std_rxn1 = -90200; %(J/mol)
deltaSrxn_std_rxn1 = -219;
deltaGrxn_std_rxn2 = -28600;
deltaHrxn_std_rxn2 = -41200;
deltaSrxn_std_rxn2 = -42;
R = 8.314;
T = 298:10:800;
for j = 1:length(T)
deltaGrxn1_funcT = (deltaHrxn_std_rxn1) - (T(j) .* deltaSrxn_std_rxn1);
Ka_rxn1(j) = exp((-deltaGrxn1_funcT)./(R.*T(j)));
deltaGrxn2_funcT = (deltaHrxn_std_rxn2) - (T(j) .* deltaSrxn_std_rxn2);
Ka_rxn2(j) = exp((-deltaGrxn2_funcT)./(R.*T(j)));
syms ksi1
syms ksi2
y_CO = (1-ksi1-ksi2)./(3-2.*ksi1);
y_H2 = (2-2.*ksi1+ksi2)./(3-2.*ksi1);
y_CH3OH = (ksi1)./(3-2.*ksi1);
y_H2O = (1-ksi2)./(3-2.*ksi1);
y_CO2 = (ksi2)./(3-2.*ksi1);
eqn1 = ((y_CH3OH)./((y_H2O.^2).*(y_CO))) - Ka_rxn1(j) == 0
eqn2 = ((y_CO2.*y_H2)./(y_CO.*y_H2O)) - Ka_rxn2(j) ==0
E = [eqn1, eqn2];
S = vpasolve(E, ksi1, ksi2, [0 1]);
end
plot(T,y_CO)
hold on
plot(T,y_H2)
hold on
plot(T,y_CH3OH)
hold on
plot(T,y_H2O)
hold on
plot(T,y_CO2)

Antworten (1)

Alan Stevens
Alan Stevens am 20 Jun. 2021
Here's a possible way. Only you will know if the results make sense
deltaGrxn_std_rxn1 = -24800; %(J/mol)
deltaHrxn_std_rxn1 = -90200; %(J/mol)
deltaSrxn_std_rxn1 = -219;
deltaGrxn_std_rxn2 = -28600;
deltaHrxn_std_rxn2 = -41200;
deltaSrxn_std_rxn2 = -42;
R = 8.314;
T = 298:10:800;
deltaGrxn1_funcT = (deltaHrxn_std_rxn1) - (T.* deltaSrxn_std_rxn1);
Ka_rxn1 = exp((-deltaGrxn1_funcT)./(R.*T));
deltaGrxn2_funcT = (deltaHrxn_std_rxn2) - (T.* deltaSrxn_std_rxn2);
Ka_rxn2 = exp((-deltaGrxn2_funcT)./(R.*T));
y_CO = @(ksi1,ksi2) (1-ksi1-ksi2)./(3-2.*ksi1);
y_H2 = @(ksi1,ksi2) (2-2.*ksi1+ksi2)./(3-2.*ksi1);
y_CH3OH = @(ksi1,ksi2) (ksi1)./(3-2.*ksi1);
y_H2O = @(ksi1,ksi2) (1-ksi2)./(3-2.*ksi1);
y_CO2 = @(ksi1,ksi2) ksi2./(3-2.*ksi1);
yco = zeros(size(T));
yh2 = zeros(size(T));
ych3oh = zeros(size(T));
yh2o = zeros(size(T));
yco2 = zeros(size(T));
K = [0 1]; % Initial guesses
for j = 1:length(T)
K0 = K; % Update initial guesses
F = @(K) norm(((y_CH3OH(K(1),K(2)))./((y_H2O(K(1),K(2)).^2).*(y_CO(K(1),K(2))))) - Ka_rxn1(j))...
+ norm(((y_CO2(K(1),K(2)).*y_H2(K(1),K(2)))./(y_CO(K(1),K(2)).*y_H2O(K(1),K(2)))) - Ka_rxn2(j));
K = fminsearch(F,K0);
yco(j) = y_CO(K(1),K(2));
yh2(j) = y_H2(K(1),K(2));
ych3oh(j) = y_CH3OH(K(1),K(2));
yh2o(j) = y_H2O(K(1),K(2));
yco2(j) = y_CO2(K(1),K(2));
end
figure(1)
plot(T,yh2,T,yco2),grid
xlabel('T'), ylabel('y')
legend('H2','CO2')
figure(2)
plot(T,yco,T,yh2o),grid
xlabel('T'),ylabel('y')
legend('CO','H2O')
figure(3)
plot(T,ych3oh),grid
xlabel('T'),ylabel('y_CH3OH')
CO and H2O are numerically too close to separate graphically.

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