FuelCellSolns

# FuelCellSolns - Design of Alternative Energy Systems Fuel...

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1 ME 417 Design of Alternative Energy Systems Fuel Cell Problems Solutions 1. Determine the ideal voltage, current, and required mass flow rate of hydrogen for an ideal air/hydrogen fuel cell operating at 101 kPa and 350 K that is producing 25 kW. Solution: Our general energy equation is s s T - h h w products j j reactants i i FC products j j reactants i i elec ± ² ³ ´ µ ν - ν ν - ν = For a hydrogen/air fuel cell the balanced chemical reaction equation is given by H 2 + ½O 2 + ½(3.76)N 2 H 2 O + ½(3.76)N 2 which allows use to write our energy equation as { } s ) 88 . 1 ( s s ) 88 . 1 ( s ) 5 . 0 ( s T - h ) 88 . 1 ( h h ) 88 . 1 ( h ) 5 . 0 ( h w p , N O H r , N O H FC p , N O H r , N O H elec 2 2 2 2 2 2 2 2 2 2 - - + + - - + + = We assume that we have an isothermal fuel cell at 350 K, so that T FC = T p = T r = 350 K So we can now evaluate our enthalpies kmole / kJ 1503 ) 8468 9971 ( 0 h h h 2 2 2 H H , f H = - + = Δ + = kmole / kJ 1531 ) 8682 213 , 10 ( 0 h h h 2 2 2 O O , f O = - + = Δ + = kmole / kJ 070 , 240 ) 9904 652 , 11 ( 820 , 241 h h h O H O H , f O H 2 2 2 - = - + - = Δ + = We have not included the enthalpies for the N 2 , since we know that they will cancel out. To evaluate the entropies we need to have our partial pressures. Calculating them we have 0.30 88 . 1 5 . 0 1 1 y 2 H = + + = 0.15 88 . 1 5 . 0 1 5 . 0 y 2 O = + + = 0.55 88 . 1 5 . 0 1 88 . 1 y r , N 2 = + + = 0.35 88 . 1 1 1 y O H 2 = + =

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ME 417 Design of Alternative Energy Systems 2 0.65 88 . 1 1 88 . 1 y p , N 2 = + = Going to the ideal gas tables at 350 K we find ) K kmole /( kJ 21 . 135 s o H 2 = ) K kmole /( kJ 765 . 209 s o O 2 = ) K kmole /( kJ 173 . 196 s o r , N 2 = ) K kmole /( kJ 125 . 194 s o O H 2 = ) K kmole /( kJ 173 . 199 s o p , N 2 = Now substituting }) 65 . 0 ln{ ) 314 . 8 ( (196.173 ) 88 . 1 ( }) 35 . 0 ln{ ) 314 . 8 ( 94.125 1 ( }) 55 . 0 ln{ ) 314 . 8 ( (196.173 ) 88 . 1 ( }) 15 . 0 ln{ ) 314 . 8 ( 09.768 2 ( ) 5 . 0 ( }) 3 . 0 ln{ ) 314 . 8 ( 21 . 135 ( (298) - 240,070) ( ) 531 1 ( ) 5 . 0 ( 503 1 w elec ± ² ³ ´ µ - - - - - + - + - - - + = kJ/kmole 230,360 2) (350)(34.2 - 242,340
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## This note was uploaded on 07/25/2008 for the course ME 417 taught by Professor Somerton during the Spring '07 term at Michigan State University.

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FuelCellSolns - Design of Alternative Energy Systems Fuel...

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