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As pointed out in the Process Description, the water-gas
Chapter 13, Problem 13.5(choose chapter or problem)
As pointed out in the Process Description, the water-gas shift reaction (Equation 13.2) occurs in the reformer along with the reforming reaction (Equation 13.1). It too is controlled by chemical equilibrium.
\(K_{p_{13.2}}=\frac{y_{\mathrm{CO}_{2}} y_{\mathrm{H}_{2}}}{y_{\mathrm{CO}} y_{\mathrm{H}_{2} \mathrm{O}}}\)
\(\log_{10}K_{p_{13.2}}=\frac{1197.8}{T(\mathrm{K})}-1.6485\)
where the nomenclature is analogous to that in the preceding problem.
(a) Taking into account the occurrence of reactions given by both Equations 13.1 and 13.2, estimate the composition of the product gas leaving the reformer and the conversion of \(\mathrm{CH}_{4}\), assuming the product stream leaving the reformer has achieved chemical equilibrium at \(855^{\circ} \mathrm{C}\) and 1.6 MPa. What is the total flow rate of this stream in both kmol/h and kg/h? What effect does the water-gas shift reaction have on the production of CO at the reformer conditions?
(b) The ratio of CO to \(\mathrm{H}_{2}\) can be an important variable in efficient use of raw materials. In this case study a 3:1 steam-to-methane molar ratio of feed streams was specified. Determine how this feed ratio affects the ratio of CO to \(\mathrm{H}_{2}\) in the product from the reformer assuming the reaction products are in chemical equilibrium at \(855^{\circ} \mathrm{C}\) and 1.6 MPa.
Questions & Answers
(2 Reviews)
QUESTION:
As pointed out in the Process Description, the water-gas shift reaction (Equation 13.2) occurs in the reformer along with the reforming reaction (Equation 13.1). It too is controlled by chemical equilibrium.
\(K_{p_{13.2}}=\frac{y_{\mathrm{CO}_{2}} y_{\mathrm{H}_{2}}}{y_{\mathrm{CO}} y_{\mathrm{H}_{2} \mathrm{O}}}\)
\(\log_{10}K_{p_{13.2}}=\frac{1197.8}{T(\mathrm{K})}-1.6485\)
where the nomenclature is analogous to that in the preceding problem.
(a) Taking into account the occurrence of reactions given by both Equations 13.1 and 13.2, estimate the composition of the product gas leaving the reformer and the conversion of \(\mathrm{CH}_{4}\), assuming the product stream leaving the reformer has achieved chemical equilibrium at \(855^{\circ} \mathrm{C}\) and 1.6 MPa. What is the total flow rate of this stream in both kmol/h and kg/h? What effect does the water-gas shift reaction have on the production of CO at the reformer conditions?
(b) The ratio of CO to \(\mathrm{H}_{2}\) can be an important variable in efficient use of raw materials. In this case study a 3:1 steam-to-methane molar ratio of feed streams was specified. Determine how this feed ratio affects the ratio of CO to \(\mathrm{H}_{2}\) in the product from the reformer assuming the reaction products are in chemical equilibrium at \(855^{\circ} \mathrm{C}\) and 1.6 MPa.
ANSWER:Step 1 of 10
The given data are as follows:
Reformer temperature is \(855^{\circ}\mathrm{C}=1128\mathrm{\ K}\)
Reformer pressure is 15.79 atm = 1.6 MPa
The expression for the chemical equilibrium constant given is as follows,
\(K_{p 13.1}=\frac{y_{C O} y_{H_{2}}^{3}}{y_{\mathrm{CH}_{4}} y_{H_{2} \mathrm{O}}} P^{2}\)
Where, \(\log _{10} K_{p 13.1}=\frac{-11769}{T(K)}+13.1927\)
\(\begin{aligned} K_{p 13.2} & =\frac{y_{\mathrm{CO}_{2}} y_{\mathrm{H}_{2}}}{y_{\mathrm{CO}} y_{\mathrm{H}_{2} \mathrm{O}}} \\ \log _{10} K_{p 13.2} & =\frac{1197.8}{T(K)}-1.6485 \end{aligned}\)
The reaction that is taking place in the reactor is as follows:
\(\begin{aligned} \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\ \longrightarrow & \mathrm{CO}+3 \mathrm{H}_{2} \\ \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\ \longrightarrow & \mathrm{CO}_{2}+\mathrm{H}_{2} \end{aligned}\)
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Review this written solution for 209354) viewed: 265 isbn: 9780471687573 | Elementary Principles Of Chemical Processes - 3 Edition - Chapter 13 - Problem 13.5
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