Modern Engineering Thermodynamics

642 CHAPTER 15: Chemical Thermodynamics
transport rate due to the chemical species crossing the system boundary, it represents the net chemical flow
availability of the system, or
_A flow = ∑ _n i ½ða f Þ
chemical
i Š chemical −∑ _n i ½ða f Þ i Š chemical
net R
P
2 3
∏ ðp i /p° Þ _ni
(15.49)
= ∑ _n i g .
i −∑ _n i g . 6 R 7
i + RT ln 4 5
R
P
ðp i /p° Þ _ni
∏
P
and the specific molar flow availability of chemical species i is
½ða f Þ i Š chemical = g i °+RT ln
p
i
p°
(15.50)
The following example illustrates the use of this material.
EXAMPLE 15.19
Determine the net molar specific flow availability of the hydrogen–oxygen fuel cell operating at 25°C and 0.1 MPa analyzed
in Example 15.18. Assume that the ground state is the standard reference state, SRS (25°C and 0.1 MPa).
Solution
Recall that the reaction is H 2 + 0.5 O 2 → H 2 O. The fuel cell has three flow streams (H 2 ,O 2 ,andH 2 O), all at the SRS pressure p i = p°. Also,
since the Gibbs function at the SRS reduces to the Gibbs function of formation, Eq. (15.50) gives ½ða f Þ i Š chemical = g i °+RT ln 1 = ðg°Þ f i .
Consequently, ½ða f Þ H2
Š chemical
= ½ða f Þ O2
Š chemical
and ½ða f Þ Š H2O chemical = ðg°Þ f H2OðlÞ
= −237:178 MJ=kgmole.
Then, the net molar specific flow availability is given by
_a flow
chemical
!
net
=
_A flow
chemical net
n f
= ∑
R
ðn i /n f Þ½ða f Þ i
Š chemical
−∑
P
ðn i /n f Þ½ða f Þ i
Š chemical
= ðn H2 /n H2 Þ½ða f Þ H2
Š chemical
+ ðn O2 /n H2 Þ½ða f Þ O2
Š chemical
− ðn H2O/n H2 Þ½ða f Þ H2O Š chemical
= 0 + 0 − ð1Þð−237:178Þ = 237:178 MJ/kgmole H 2
Exercises
55. Determine the net chemical flow availability of the fuel cell described in Example 15.19 when hydrogen is consumed at
a rate of 2.00 kgmole/min. Answer: [_A( flow ) net ] chemical = 474.4 MJ/min.
56. Determine the molar specific chemical flow availability of oxygen in air at a total pressure of 3.50 MPa and the SRS
temperature of 25.0°C. Assume air consists of a mixture of 21.0% oxygen and 79.0% nitrogen on a molar basis. Answer:
[ _a( flow ) O2 ] chemical = 4940 kJ/kgmole O 2 .
57. If all the flow streams entering and exiting the hydrogen–oxygen fuel cell discussed in Example 15.19 are at 1.00 MPa
instead of 0.100 MPa, determine the net chemical flow availability of the fuel cell per kgmole of hydrogen consumed.
Answer: [_a( flow ) net ] chemical = 3090 kJ/kgmole H 2 .
SUMMARY
In this chapter, we deal with the fundamental elements of chemical thermodynamics. Chemistry has its roots in
thousands of years of alchemy; its accurate mathematical notation is relatively recent. The 19th century stoichiometric
mass balance and the basic concepts of stereochemistry provide a framework on which an accurate combustion
analysis of organic fuels can be built. Concepts such as percent of theoretical air, fuel modeling, heat of
formation, and the standard reference state plus the first law of thermodynamics applied to a chemical reaction
lead to a useful understanding of the heat of combustion of a chemical compound. The adiabatic flame temperature
and maximum explosion pressure calculations provide conservative upper bounds for real combustion processes.
The introduction of the third law of thermodynamics provides the basis on which to build an absolute
entropy scale that can be used to determine chemical reaction irreversibilities via the entropy balance. Also, the
Gibbs function from the combined first and second laws is found to be a controlling factor in chemical reactions,
chemical equilibrium, and dissociation reactions. Finally, fuel cell analysis provides a means of investigating the
maximum possible work that can be produced directly from a chemical reaction.