EXAMPLE 15–9 Reversible Work Associated
with a Combustion ProcessOne lbmol of carbon at 77°F and 1 atm is burned steadily with 1 lbmol of
oxygen at the same state as shown in Fig. 15–32. The CO 2 formed during
the process is then brought to 77°F and 1 atm, the conditions of the sur-
roundings. Assuming the combustion is complete, determine the reversible
work for this process.Solution Carbon is burned steadily with pure oxygen. The reversible work
associated with this process is to be determined.
Assumptions 1 Combustion is complete. 2 Steady-flow conditions exist during
combustion. 3 Oxygen and the combustion gases are ideal gases. 4 Changes
in kinetic and potential energies are negligible.
Properties The Gibbs function of formation at 77°F and 1 atm is 0 for C
and O 2 , and 169,680 Btu/lbmol for CO 2. The enthalpy of formation is 0 for
C and O 2 , and 169,300 Btu/lbmol for CO 2. The absolute entropy is 1.36
Btu/lbmol · R for C, 49.00 Btu/lbmol · R for O 2 , and 51.07 Btu/lbmol · R for
CO 2 (Table A–26E).
Analysis The combustion equation isThe C, O 2 , and CO 2 are at 77°F and 1 atm, which is the standard reference
state and also the state of the surroundings. Therefore, the reversible work in
this case is simply the difference between the Gibbs function of formation of
the reactants and that of the products (Eq. 15–27):since the g–f° of stable elements at 77°F and 1 atm is zero. Therefore,
169,680 Btu of work could be done as 1 lbmol of C is burned with 1 lbmol
of O 2 at 77°F and 1 atm in an environment at the same state. The reversible
work in this case represents the exergy of the reactants since the product
(the CO 2 ) is at the state of the surroundings.
Discussion We could also determine the reversible work without involving
the Gibbs function by using Eq. 15–24:Substituting the enthalpy of formation and absolute entropy values, we obtainwhich is identical to the result obtained before.169,680 Btu¬ 1 1 lbmol CO 223 169,300 Btu>lbmol 1 537 R 21 51.07 Btu>lbmol#R 24
¬ 1 1 lbmol O 2230 1 537 R 21 49.00 Btu>lbmol#R 24
Wrev 1 1 lbmol C 230 1 537 R 21 1.36 Btu>lbmol#R 24
NC 1 h°fT 0 s° (^2) CNO 21 h°fT 0 s° (^2) O 2 NCO 21 h°fT 0 s° (^2) CO 2
aNr 1 h°fT 0 s (^2) raNp 1 h°fT 0 s (^2) p
WrevaNr 1 h°fhh°T 0 s (^2) raNp 1 h°fhh°T 0 s (^2) p
169,680 Btu
1 1 lbmol 21 169,680 Btu>lbmol 2
NCg°f,CNO 2 g°f,O 2 NCO 2 g°f,CO 2 NCO 2 g°f,CO 2
WrevaNrg°f,raNpg°f,p
CO 2 SCO 2
776 | Thermodynamics
Combustion
chamber
C
77 °F, 1 atm
CO 2
T 0 = 77°F
O 2
77 °F, 1 atm
77 °F, 1 atm
P 0 = 1 atm
FIGURE 15–32
Schematic for Example 15–9.
0
¡
0
¡