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Modern Engineering Thermodynamics

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640 CHAPTER 15: Chemical <strong>Thermodynamics</strong><br />

WHAT IS j ?<br />

The letter j represents the number of electrons involved in the cell reaction per kgmole of fuel consumed. For example, for<br />

any hydrocarbon fuel, C n H m , j = nð4Þ + mð1Þ = 4n + m kgmole of electrons per kgmole of C n H m .<br />

When there is no electron flow in the external circuit, there are no losses within the fuel cell and we can write<br />

the maximum (or reversible) power output of a fuel cell as approximately 5<br />

_W max ≈ ϕ o I = ϕ o ð _n fuel Þ jF (15.47)<br />

where ϕ o is the open circuit voltage. Then, the maximum reaction efficiency of the fuel cell becomes<br />

ðη r Þ max<br />

=<br />

fuel cell<br />

∑<br />

R<br />

ðϕ o jFÞ<br />

ðn i /n fuel Þh i −∑ ðn i /n fuel Þh i<br />

P<br />

EXAMPLE 15.18<br />

Determine the maximum theoretical reaction efficiency, open circuit voltage, and maximum theoretical work output per<br />

mole of hydrogen consumed for the hydrogen–oxygen fuel cell operating at 25.0°C and 0.100 MPa, shown in Figure 15.15.<br />

R e<br />

Electrical load<br />

Voltmeter<br />

Electrolyte<br />

25.0 ° C<br />

H 2 (g)<br />

O 2 (g)<br />

25.0 ° C<br />

0.100 MPa<br />

2H +<br />

0.100 MPa<br />

Porous anode<br />

Porous cathode H 2 O(<br />

)<br />

FIGURE 15.15<br />

Example 15.18.<br />

The anode reaction is H 2 ðgÞ !2ðH + Þ + 2ðe − Þ:<br />

The cathode reaction is 0:5½O 2 ðgÞŠ + 2ðH + Þ + 2ðe − Þ!H 2 OðlÞ:<br />

The overall reaction is H 2 ðgÞ + 0:5½O 2 ðgÞŠ ! H 2 OðlÞ:<br />

Also, ðp i /p° Þ H2<br />

= ðp i /p° Þ O2<br />

= 1:0:<br />

Solution<br />

Equation (15.45) can be used to calculate the maximum energy conversion reaction efficiency as<br />

ðη r Þ max<br />

= g• H2ðÞ g<br />

+ ð0:5Þðg• O2ðÞ g<br />

Þ − g• H2OðÞ l<br />

+ RT ln 1:0<br />

h H2 ðÞ+ g ð0:5Þ h O2 ðÞ<br />

g − hH2OðÞ<br />

l<br />

where the g • i<br />

are determined from Eq. (15.34), and the h i are determined from Eq. (15.14). Since the pressure and temperature<br />

given in this problem statement are the standard reference state values, Eqs. (15.14) and (15.34) reduce to h i = h° <br />

<br />

f i<br />

and g • i = g°<br />

f : Then, the maximum theoretical reaction efficiency equation becomes<br />

i<br />

ðη r<br />

Þ max = ðg°Þ f H2ðÞ g<br />

+ ð0:5Þðg°Þ<br />

f O2ðÞ g<br />

− ðg°Þ f H2OðÞ l<br />

+ RT ln 1:0<br />

h° f H2ðÞ g<br />

+ ð0:5Þ<br />

h° f O2ðÞ g<br />

− h° f H2OðÞ<br />

l<br />

5 A small amount of internal irreversibility is generated in producing ϕ o ,soϕ o is slightly less than ϕ rev = ϕ max .

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