Solution_SP2

Solution_SP2 - = = 0175 10 6487 9 926 2 926 229 1 8 8297 50...

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1/26/2008 ME 145 - Solution to Supplemental Problem 2 Power fc = Power rating/DC to AC efficiency = 2x10 6 /0.80 = 2.5 MW(e) DC Assume 0.8 V output (unrealistic but the numbers will work to give a reasonable estimate) Power fc = V x I = 0.8 x I = 2.5x10 6 Watts I=3.125x10 6 amps (C/sec) From pg. 440 of SN 2 = F jn I E V c th fc η From pg. 439, Example 11.4 of SN 2, E=1.229V. Also, η th =0.8297. We know j=2 for a H 2 -O 2 fuel cell and F=9.6487x10 7 C/kg-mole sec mole kg . ) x . )( . ( I n . . . . . E V F jn I c th fc c - = = =
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Unformatted text preview: = = 0175 10 6487 9 926 2 926 229 1 8 8297 50 7 Every 2 kg-mole of e-requires 1 kg-mole of H 2 (MW=2), 0.5 kg-mole O 2 (MW=32) and 1 kg-mole H 2 O (MW=18). Therefore H 2 consumption rate = 1 x 0.0175 x 2 = 0.035 kg/sec O 2 consumption rate = 0.5 x 0.0175 x 32 = 0.280 kg/sec H 2 O production rate = 1 x 0.0175 x 18 = 0.315 kg/sec (=1134 kg/hr=1.13 m 3 /hr)...
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