HIGH-EFFICIENCY ENERGY SYSTEMS WITH BIOMASS GASIFIERS AND SOLID OXIDE FUEL CELLS - Biomass as a Sustainable Energy Source for the Future - page 509

In Equation (16.3), R u is the universal gas constant (= 8.3143 J mol −1 K −1 ), T is the
operating temperature (K), and p i

is the partial pressure of the reactants and products

g r is the standard-state Gibbs free energy change for the

(Pa). In Equation (16.4), Δ

reaction (J mol -1 ).
However, the useful voltage output ( V ) under load conditions, i.e., when a current
passes through the cell, is given by

V = E − I × R i − η c − η a ð Eq : 16 : 5 Þ

where n is the number of electrons transferred for each molecule of fuel, R i is the
internal electrical resistance of the SOFC ( Ω ), I is the current passing through the cell

(A), and η a are polarization losses associated with the cathode and anode, respec-

tively. The voltage loss due to internal electrical resistance includes contributions from
the electrodes and the electrolyte, with the highest contribution originating from the
ionic conduction through the solid electrolyte. For a further explanations of cell voltage
losses, readers are referred to O

η c and

Hayre et al. (2009) and Singhal and Kendall (2003).
The current flow is always proportional to the number of oxide ions
reacting with the fuel molecules and thus to the amount of fuel consumed, as
expressed by (16.6)

'

I
n F

n : j = ð Eq : 16 : 6 Þ

Example 16.1
For the following reaction in an SOFC working at 800 C:

6 kJmol −1

H 2 +1 = 2O 2 ! H 2 O with Δ g r = − 188 :

1. Calculate the reversible voltage.
2. Calculate the hydrogen flow (mol s −1 ) required to produce 1 ampere current.

Solution
1. According to Equation (16.4), the reversible voltage is

− −

188 6 × 1, 000

2 × 96,485

:

E = =0 : 977 V

2. According to Equation (16.6), the hydrogen flow rate is

1
2 × 96,485 =5

: H 2 = 18 × 10 −6 mol s −1

: