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CH4, anode + 2H2O, anode → 4H2, anode + CO2, anode 4H2, anode + 2O2, cathode → 4H2O, anode CH4, anode + 2O2, cathode → 2H2O, anode + CO2, anode (Steam Reforming Reaction) (Fuel Cell Reaction) (Combined Reforming and FC Reactions) For ease of calculation, assume a 100 lb/hr basis for the methane. nfuel,supplied =⎜100 ⎝ 4 ⎛ lbCH ⎞⎛ 1lbmolCH ⎞ 4 ⎟⎜ 4 ⎟=6.23 lbmolCH hr hr ⎠⎝16.043lbCH4 ⎠ Thus, 85 percent, or 5.30 lb mol CH4 /hr, will be reformed and consumed by the fuel cell. The remainder will be reformed but not consumed by the fuel cell reaction. These changes are summarized in the following table: Spent Fuel Effluent Calculation Gas CH4 CO CO2 H2 H2O Total mol percent FC inlet 100.0 0.0 0.0 0.0 0.0 100.0 FC inlet 6.23 0.00 0.00 0.00 0.00 6.23 lb mol/hr Ref / FC rxn Reforming -5.30 -0.93 0.00 0.00 5.30 0.93 0.00 3.74 10.60 -1.87 10.60 1.87 FC outlet 0.00 0.00 6.23 3.74 8.73 18.70 mol percent FC outlet 0.00 0.00 33.33 20.00 46.67 100.00 This intermediate solution reflects only two out of three reactions. Now apply the water gas shift reaction to determine the true exit composition. Use the quadratic equation listed in Example 9-5 to determine how far the reaction will proceed, where x is the extent of the reaction in the forward direction as written: x 2a The equilibrium constant, K, at 1800 °F (1255 °K) is K = e (4276/1255-3.961) = 0.574 a= (1-K)= (1-0.574) =0.426 b = {[C O 2 ] + [H 2 ] + K ([C O ] + [H 2 O ])} = 0 . 3 3 3 3 + 0 . 2 0 0 0 + 0 . 5 7 4 * ( 0 . 0 0 + 0 . 4 6 6 7 ) = 0 . 8 0 1 2 c={[CO2][H2]−[CO][H2O]K}= (0.3333)(0.20) - (0.00)(0.4667)(0.574)= 0.0666 CO+H2O←⎯→CO2 +H2 x=−b± b2−4ac 9-11PDF Image | Fuel Cell Handbook (Seventh Edition)
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