Modern Engineering Thermodynamics

624 CHAPTER 15: Chemical Thermodynamics
In the case of mixtures of ideal gases that obey the Gibbs-Dalton ideal gas mixture law, the absolute molar specific
entropy of chemical species i, s i , needed for the entropy balance of Eq. (15.20) is given by
s i = s i °+
Z T
and, when c pi is constant (or averaged) over T° to T,
where p i is the partial pressure of chemical species i in the mixture.
T°
c pi ðdT/TÞ − R ln ðp i /p° Þ (15.25)
s i = s i °+c pi ln ðT/T°
Þ− R ln ðp i /p° Þ (15.26)
Accurate integrated values can be found for s i through the use of Table C.16c in Thermodynamic Tables to accompany
Modern Engineering Thermodynamics. Note that, in this table, we use the condensed notation
Z
ϕ i = s i °+ c pi ðdT/TÞ
so that Eq. (15.25) reduces for use with Table C.16c to
s i = ϕ i − R lnðp i /p° Þ (15.27)
Alternatively, a computer program could be easily written to calculate accurate integrated values for s i :
The partial pressure p i of component i in the mixture is determined from the mixture composition via Eq.
(12.23) as
p i = χ i p m = w i ðM m /M i Þp m
where χ i is the mole fraction, w i is the mass (or weight) fraction, M i is the molecular mass of chemical species i
in the mixture, M m is the equivalent molecular mass of the mixture, and p m is the total pressure of the mixture.
EXAMPLE 15.12
Calculate the entropy produced per mole of fuel when methane is burned with 100.% theoretical air. The reactants are premixed
at 25.0°C at a mixture total pressure of 0.100 MPa, and the products are at 200.°C at a total pressure of 0.100 MPa. The
molar heating value of methane under these conditions with the water of combustion in the vapor phase is −134.158 MJ per
kgmole of methane. Assume constant specific heat ideal gas behavior for all the combustion components.
Solution
Equation (15.20) gives the required entropy production as
ðs P
Þ r = ∑
P
ðn i /n fuel Þs i −∑ ðn i /n fuel Þs i − q r /T b
wherewearegivenq r = −134:158 MJ/kgmole and we assume that T b = 200. + 273.15 = 473 K. The reaction equation for
100.% theoretical air is
CH 4 + 2½O 2 + 3:76ðN 2 ÞŠ ! CO 2 + 2ðH 2 OÞ + 7:52ðN 2 Þ
The partial pressures of the reactants can then be found from Eq. (12.23) as
and the partial pressures of the products are
p CH4 = ðn CH4 /n R Þp m = 1/ð1 + 2 + 7:52Þ ð0:100Þ = ð1/10:52Þð0:100Þ
= 9:51 kPa
p O2 = ð2/10:52Þð0:100Þ = 19:0 kPa
p N2 = ð7:52/10:52Þð0:100Þ = 71:5 kPa
p CO2
= ðn CO2 /n P ÞP m = ð1/10:52Þð0:100Þ = 9:51 kPa
p H2O = ð2/10:52Þð0:100Þ = 19:0 kPa
p N2 = ð7:52/10:52Þð0:100Þ = 71:5 kPa
R
Now,
∑
R
ðn i /n fuel
Þs i = s CH4 + 2:00ðs O2
Þ+ 7:52ðs N2 Þ
where, from Eq. (15.26) with T = 298 K, and using Table 15.7,