Thermodynamics

776 | ThermodynamicsC77°F, 1 atmO 277°F, 1 atmT 0 = 77°FP 0 = 1 atmCombustionchamberFIGURE 15–32Schematic for Example 15–9.CO 277°F, 1 atmEXAMPLE 15–9Reversible Work Associatedwith a Combustion ProcessOne lbmol of carbon at 77°F and 1 atm is burned steadily with 1 lbmol ofoxygen at the same state as shown in Fig. 15–32. The CO 2 formed duringthe process is then brought to 77°F and 1 atm, the conditions of the surroundings.Assuming the combustion is complete, determine the reversiblework for this process.Solution Carbon is burned steadily with pure oxygen. The reversible workassociated with this process is to be determined.Assumptions 1 Combustion is complete. 2 Steady-flow conditions exist duringcombustion. 3 Oxygen and the combustion gases are ideal gases. 4 Changesin kinetic and potential energies are negligible.Properties The Gibbs function of formation at 77°F and 1 atm is 0 for Cand O 2 , and 169,680 Btu/lbmol for CO 2 . The enthalpy of formation is 0 forC and O 2 , and 169,300 Btu/lbmol for CO 2 . The absolute entropy is 1.36Btu/lbmol · R for C, 49.00 Btu/lbmol · R for O 2 , and 51.07 Btu/lbmol · R forCO 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 referencestate and also the state of the surroundings. Therefore, the reversible work inthis case is simply the difference between the Gibbs function of formation ofthe 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 lbmolof O 2 at 77°F and 1 atm in an environment at the same state. The reversiblework 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 involvingthe Gibbs function by using Eq. 15–24:W rev a N r 1h° f h h° T 0 s 2 r a N p 1h° f h h° T 0 s 2 p a N r 1h° f T 0 s 2 r a N p 1h° f T 0 s 2 p N C 1h° f T 0 s °2 C N O21h° f T 0 s °2 O2 N CO2 1h° f T 0 s °2 CO2Substituting the enthalpy of formation and absolute entropy values, we obtainW rev 11 lbmol C230 1537 R211.36 Btu>lbmol # R24 11 lbmol O 2 230 1537 R2 149.00 Btu>lbmol # R24 11 lbmol CO 2 23169,300 Btu>lbmol 1537 R2151.07 Btu>lbmol # R24 169,680 BtuW rev a N r g ° f,r a N p g ° f,p N C g° f,C¡0 N O2g° f,O2¡0 N CO2 g° f,CO2 N CO2 g° f,CO2 11 lbmol21169,680 Btu>lbmol2 169,680 BtuC O 2 S CO 2which is identical to the result obtained before.