Thermodynamics

778 | Thermodynamics(b) Noting that combustion is adiabatic, the entropy generation during thisprocess is determined from Eq. 15–20:S gen S prod S react a N p s p a N r s rThe CH 4 is at 25°C and 1 atm, and thus its absolute entropy is s – CH 4186.16 kJ/kmol K (Table A–26). The entropy values listed in the ideal-gastables are for 1 atm pressure. Both the air and the product gases are at atotal pressure of 1 atm, but the entropies are to be calculated at the partialpressure of the components, which is equal to P i y i P total , where y i is themole fraction of component i. From Eq. 15–22:S i N i s i 1T, P i 2 N i 3 s ° i 1T, P 0 2 R u ln y i P m 4The entropy calculations can be represented in tabular form as follows:CH 425°C, 1 atmAIR25°C, 1 atmT 0 = 25°CCombustionchamber25°C,1 atmFIGURE 15–34Schematic for Example 15–11.CO 2H 2 OO 2N 2N i y i s – ° i (T, 1 atm) R u ln y i P m N i s – iCH 4 1 1.00 186.16 — 186.16O 2 3 0.21 205.04 12.98 654.06N 2 11.28 0.79 191.61 1.96 2183.47S react 3023.69CO 2 1 0.0654 302.517 22.674 325.19H 2 O 2 0.1309 258.957 16.905 551.72O 2 1 0.0654 264.471 22.674 287.15N 2 11.28 0.7382 247.977 2.524 2825.65S prod 3989.71Thus,S gen S prod S react 13989.71 3023.692 kJ>kmol # K CH 4(c) The exergy destruction or irreversibility associated with this process isdetermined from Eq. 15–23,That is, 288 MJ of work potential is wasted during this combustion processfor each kmol of methane burned. This example shows that even completecombustion processes are highly irreversible.This process involves no actual work. Therefore, the reversible work andexergy destroyed are identical:That is, 288 MJ of work could be done during this process but is not.Instead, the entire work potential is wasted.EXAMPLE 15–11 966.0 kJ/kmol # KX destroyed T 0 S gen 1298 K21966.0 kJ>kmol # K2 288 MJ/kmol CH 4W rev 288 MJ/kmol CH 4Second-Law Analysisof Isothermal CombustionMethane (CH 4 ) gas enters a steady-flow combustion chamber at 25°C and1 atm and is burned with 50 percent excess air, which also enters at 25°Cand 1 atm, as shown in Fig. 15–34. After combustion, the productsare allowed to cool to 25°C. Assuming complete combustion, determine