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16 Nov 2019
The synthesis of methanol from CO and H_2 is carried out in a catalytic reactor. The reactor is fed with a gas stream at 220 degree C consisting of 5 mol% CH_4, 25% CO, 5% CO_2 and the remainder H_2. The reactor and feed stream are at 7.5 MPa. Analysis reveals that three reactions can take place: CO + 2H_2 doubleheadarrow CH_3OH Reaction 1 k_(1) CO_2 + 3H_2 doubleheadarrow CH_3OH + H_2O Reaction 2 K_(2) CO_2 + H_2 doubleheadarrow CO + H_2O Reaction 3 k_(3) a). Show that only two of the three reactions are independent. b) If the temperature and pressure of the equilibrated product stream are 250 degree C and 7.5 MPa, use Matlab to determine the composition (mole fractions) of the product stream and the percentage conversion of CO and H_2. The equilibrium constants (K_(1) and K_(3)) for the first and third reactions are given below. K_(1) = Y_CH_3OH/Y_COy^2H_2P^2 = exp(21.225 + 9143.6/T - 7.492 ln T + 4.076 times 10^-3T - 7.161 times 10^-8 T^2) K_(3) = Y_COY_H_ O/Y_CO_2Y_H_2 = exp(13.148 -5639.5/T - 1.077 ln T - 5.44 times 10^-4 T + 1.125 times 10^-7 T^2 + 49170/T^2)
The synthesis of methanol from CO and H_2 is carried out in a catalytic reactor. The reactor is fed with a gas stream at 220 degree C consisting of 5 mol% CH_4, 25% CO, 5% CO_2 and the remainder H_2. The reactor and feed stream are at 7.5 MPa. Analysis reveals that three reactions can take place: CO + 2H_2 doubleheadarrow CH_3OH Reaction 1 k_(1) CO_2 + 3H_2 doubleheadarrow CH_3OH + H_2O Reaction 2 K_(2) CO_2 + H_2 doubleheadarrow CO + H_2O Reaction 3 k_(3) a). Show that only two of the three reactions are independent. b) If the temperature and pressure of the equilibrated product stream are 250 degree C and 7.5 MPa, use Matlab to determine the composition (mole fractions) of the product stream and the percentage conversion of CO and H_2. The equilibrium constants (K_(1) and K_(3)) for the first and third reactions are given below. K_(1) = Y_CH_3OH/Y_COy^2H_2P^2 = exp(21.225 + 9143.6/T - 7.492 ln T + 4.076 times 10^-3T - 7.161 times 10^-8 T^2) K_(3) = Y_COY_H_ O/Y_CO_2Y_H_2 = exp(13.148 -5639.5/T - 1.077 ln T - 5.44 times 10^-4 T + 1.125 times 10^-7 T^2 + 49170/T^2)