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9.1.2 FuelProcessingCalculations Example 9-8 Methane Reforming - Determine the Reformate Composition Given a steam reformer operating at 1400 oF, 3 atmospheres, pure methane feed stock, and a steam to carbon ratio of 2 (2 lb mol H2O to 1 lb mol CH4), (a) List the relevant reactions; (b) Determine the concentration assuming the effluent exits the reactor in equilibrium at 1400 oF; (c) Determine the heats of reaction for the reformer's reactions; (d) Determine the reformer's heat requirement assuming the feed stocks are preheated to 1400 oF; (e) Considering LeChâtelier's Principle, indicate whether the reforming reaction will be enhanced or hindered by an elevated operating temperature; (f) Considering LeChâtelier's Principle, indicate whether increased pressure will tend to promote or prevent the reforming reaction. Solution: (a) The relevant reactions for the steam reformer are presented below: CH4 + H2O ⇔ 3H2 + CO (Steam Reforming Reaction) CO + H2O ⇔ CO2 + H2 (Water Gas Shift Reaction) A third reaction is presented below; this reaction is simply a combination of the other two. Of the three reactions, any two can be used as an independent set of reactions for analysis, and can be chosen for the user's convenience. CH4 + 2H2O ⇔ 4H2 + CO2 (Composite Steam Reforming Reaction) (b) The determination of the equilibrium concentrations is a rather involved problem, requiring significant background in chemical thermodynamics, and will not be solved here. One aspect that makes this problem more difficult than Example 9-6, which accounted for the steam reforming reaction within the fuel cell, is that the reforming reaction cannot be assumed to proceed to completion as in the former example. In Example 9-6, hydrogen was consumed within the fuel cell, thus driving the reforming reaction to completion. Without being able to assume the reforming reaction goes to completion, two independent equilibrium reactions must be solved simultaneously. The solution to this problem is most easily accomplished with chemical process simulation programs using a technique known as the minimization of Gibbs free energy. To solve this problem by hand is an arduous, time- consuming task. The ASPENTM computer solution to this problem is provided below: Inlet Composition (lb mols/hr) CH4 100 CO0 CO2 0 H20 H2O 200 T otal 300 Effluent Composition (lb mols/hr) 11.7441 64.7756 23.4801 288.2478 88.2639 476.5115 Effluent Composition (mol fraction) 2.47 13.59 4.93 60.49 18.52 100.00 9-13PDF Image | Fuel Cell Handbook (Seventh Edition)
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