Question

4.4 A fuel cell operates at 60°C and ambient pressure. Exchange current density, io is 0.005 mA/cm. Assume charge transfer coefficient to be 1.0. a) Calculate the activation voltage losses at 1.5A/cm. b) If the resistive losses at 1.5A/cm are 1/2 of the activation losses, calculate the cell resistance, R. c) Assume no concentration polarization losses and calculate cell voltage at 1.5A/cm, at 1 A/cm', and at 0.5 A/cm.

Answer 1

Ansco- Given that Temperature Iz 60°c z 60+ 273 T= 333k. Exchange current density to 01005 mA /cm transter coefficient : 1 (a) the activation Voltage Losses Pact 2 2 3 2.3RT log (40) LhF 2.3%8314333 01 X 2 X 96485 DOSX 103 2 0103299. log (300x103) 5,47712 = 0.03299 x Mart 2 0:180690 V (b) ohmic Loss Chm = it Nohm 2 Y Mact

1/2 x 0.180 = 1xr X0 180 15 Alcm x r 0.09 115 0.062 cm Tr (C) " cell voltage at li5 A/cm² V = t (total Losses Neone = 0 Nohm = 0.091 nact 01180 V fuel cell E Il 145 V = 11 14-(0.09 +0.180) 0.27 2 1.14 - Tv 2 0.84 v let V2 1 A/cm2 ncone zo hact : - 2.368,316 X 333 2.348.long lanang init) 0.005X103 x 2x96485 Mact a 0.17496

Nohm 2 m 1x 0.06 0.06. Mohm 2 ₂ 2 - (hatt honet nohm) 2 1.14-(0.174 + 0 + 0.06) 1.14. - 0.2349 V = 0.905074 V - at os A/cm² leone 20. Mact 0.03299 log (1008 103) Mart 20116495 Nohm 0.5x0.06 = 0.03 V. Vz2 t - 2 1.14- (nact & Neone Nohm) (0126495 to +0.03) z 1.14 - 0119495 20.94505 V