Modern Engineering Thermodynamics

15.15 Fuel Cells 639
The maximum power _W max and maximum reaction efficiency ðη r Þ max
occur when the device is completely reversible
or _S P = 0: Then,
ðη r Þ max =
∑
in
∑
in
_m i g i −∑
out
_m i h i −∑
out
_m i g i
_m i h i
=
∑
in
∑
in
_n i g i −∑ _n i g i
out
(15.42)
_n i h i −∑ _n i h i
out
Utilizing Eq. (15.33), the molar forms of Eqs. (15.40), (15.41), and (15.42) are
_W = ∑
R
_n i g • i
−∑
P
_n i g • i
− T b
_S P + RT ln ∏
R
ðp i /p° Þ _ni /∏ ðp i /p°
P
Þ _ni
(15.43)
and
η r =
∑
R
_n i g • i −∑
P
_n i g • i − T b
_S P + RT ln ∏
R
∑
R
_n i h i −∑
P
ðp i /p° Þ _ni /∏ ðp i /p°
P
Þ _ni
_n i h i
(15.44)
ðη r Þ maximum
fuel cell
efficiency
=
∑
R
ðn i /n fuel
Þg • i
−∑
P
ðn i /n fuel
∑
R
Þg • i
+ RT ln ∏
R
ðn i /n fuel
Þh i −∑
P
ðp i /p°
ð
Þ ni/n fuel Þ /∏
P
ðp i /p°
ð
Þ ni/n fuelÞ
ðn i /n fuel Þh i
(15.45)
If the reactants or products are unmixed and the pressure of each species (or component) in the reaction is
0.1 MPa, then p i = p° and ∏ P
ðp i /p° Þ ni = ∏ R
ðp i /p° Þ ni = 1:0 in these three equations. Otherwise, if the reactants are
premixed or the product gases are mixed (as they normally are) at a total pressure p m , then the partial pressure of
each component gas must be determined. Further, if all the gases present are ideal gases that obey the Gibbs-
Dalton ideal gas mixture law, then the partial pressures can be expressed in terms of the mole fractions as
p i /p° = ðχ i p m Þ/p°, where x i is the mole fraction of gas i and p°=0.1 MPa. Then,
h i
ln ∏ðp i /p° Þ ni = ln ∏ðχ i p m /p° Þ ni
in Eqs. (15.43), (15.44), and (15.45).
In the case of a fuel cell, the output power appears as an electrical current I flowing through a potential
(voltage) difference ϕ, and using Ohm’s law we can calculate the actual power output of the cell as
_W = ϕI = I 2 R e
where ϕ is the cell voltage, I is its current flow, and R e is the external resistance (recall that work output must be
positive with our sign convention). Combining this equation with Eq. (15.40) gives the entropy production rate
of the fuel cell as
ð _S P Þ fuel
cell
= ∑
P
_n i g i −∑
R
Also, it can be shown that the electrical current I produced by a fuel cell is given by
I = ð_n fuel ÞjF
!
_n i g + ϕI /T b ≥ 0 (15.46)
where I is in amperes when _n fuel is in kgmole/s. In this equation j is the total valence of the fuel ions in kgmole
of electrons per kgmole of fuel, and F is Faraday’s constant, defined as
F = ð6:023 × 10 26 electrons/kgmole electronsÞð1:602 × 10 −19 coulombs/electronÞ
= 96:487 kilocoulombs/kgmole electrons
but since 1 coulomb = 1 joule/volt = 1 J/V, this can be written as
F = 96,487 kJ/ðV . kgmole electronsÞ