The polarization curve for a fuel cell at 50 C is given below. The reversible an

Question

The polarization curve for a fuel cell at 50 C is given below. The reversible and thermo- neutral voltages for this fuel cell are 1.2 V and 1.4 V, respectively Calculate the practical efficiency of the cell at 600 mA/cm^2 current density Calculate the voltage loss due to fuel cross-over Calculate the ohmic polarization at 600 mA/cm^2. Assume the voltage drop from 200 to 1000 600 mA/cm^2 is linear. Calculate the exchange current density of the fuel cell at 600 mA/cm^2. Assume the mass transport over voltage is negligible at 600 mA/cm^2 and the Tafel slope for a single cell is 30 mV/decade. The mass transport polarization is significant at 1450 mA/cm^2 current density, and assuming it is solely from oxygen transport, calculate the mass transport polarization at 1450 mA/cm^2 current density.

Explanation / Answer

a). Practical Efficiency at 600 mA/cm2 = {(Voltage at 600 mA/cm2)/Vnlv }x100% = (0.712/1.4)x100% = 50.86%

b). Voltage loss due to fuel cross over = Vnlv - Vocv = (1.4 - 1.2) V= 0.2 V

c). Ohmic polarisationat 600 mA/cm2 ={ (V1000 -V200)/800}x600 = {(0.605 - 0.819)/800} x 600 V = - 0.16V

d). Tafel Slope (A) = 30mV/decade = 300mV =0.3V

From (Vocv -V600) = A ln (i600/io)

or, (1.2- 0.712)V = 0.3V. ln (600/io)

or io = 118 mA/cm2 (Exchange Current density)

e). Mass transport polarisation = 0.3V . ln (1450/118) =0.753 V

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