Thermodynamics

760 | ThermodynamicsThus,T dp T sat @ 20.88 kPa 60.9°CDiscussion If the combustion process were achieved with dry air instead ofmoist air, the products would contain less moisture, and the dew-point temperaturein this case would be 59.5°C.EXAMPLE 15–4Reverse Combustion AnalysisC 8 H 1810.02% CO 2AIRCombustionchamberFIGURE 15–13Schematic for Example 15–4.5.62% O 20.88% CO83.48% N 2Octane (C 8 H 18 ) is burned with dry air. The volumetric analysis of the productson a dry basis is (Fig. 15–13)CO 2 : 10.02 percentO 2 : 5.62 percentCO: 0.88 percentN 2 : 83.48 percentDetermine (a) the air–fuel ratio, (b) the percentage of theoretical air used,and (c) the amount of H 2 O that condenses as the products are cooled to25°C at 100 kPa.Solution Combustion products whose composition is given are cooled to25°C. The AF, the percent theoretical air used, and the fraction of watervapor that condenses are to be determined.Assumptions Combustion gases are ideal gases.Properties The saturation pressure of water at 25°C is 3.1698 kPa (Table A–4).Analysis Note that we know the relative composition of the products, butwe do not know how much fuel or air is used during the combustion process.However, they can be determined from mass balances. The H 2 O in the combustiongases will start condensing when the temperature drops to the dewpointtemperature.For ideal gases, the volume fractions are equivalent to the mole fractions.Considering 100 kmol of dry products for convenience, the combustionequation can be written asxC 8 H 18 a 1O 2 3.76N 2 2 S 10.02CO 2 0.88CO 5.62O 2 83.48N 2 bH 2 OThe unknown coefficients x, a, and b are determined from mass balances,N 2 :C:H:3.76a 83.48 S a 22.208x 10.02 0.88S x 1.3618x 2bS b 12.24O 2 :a 10.02 0.44 5.62 b S 22.20 22.202The O 2 balance is not necessary, but it can be used to check the valuesobtained from the other mass balances, as we did previously. Substituting,we get1.36C 8 H 18 22.2 1O 2 3.76N 2 2 S10.02CO 2 0.88CO 5.62O 2 83.48N 2 12.24H 2 O