However assuming that the products contain major

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Unformatted text preview: 400 300 fs = 0.05 200 100 0 −100 0 0.1 0.2 0.3 0.4 0.5 Fuel air ratio (mass) 2.2 Combustion Stoichiometry 21 species in addition to the major species (CO2, H2O, N2, O2), and the balance of the stoichiometric equation requires the use of thermodynamic equilibrium relations. However, assuming that the products contain major species only (complete combustion) and excess air, the global equation for lean combustion fb1 is   1 bg a þ À ðO2 þ 3:76N2 Þ ! Ca Hb Og þ f 42      (2.18) b 3:76 bg bg 1 a þ À N2 þ a þ À À 1 O2 aCO2 þ H2 O þ 2 f 42 42f In terms of %EA, we replace f by  100 and the result is %EA þ 100   %EA bg þ 1 a þ À ðO2 þ 3:76N2 Þ ! 100 42      (2.19) b %EA bg b g %EA aCO2 þ H2 O þ 3:76 þ 1 a þ À N2 þ a þ À O2 2 100 42 4 2 100 C a Hb Og þ The amount of excess air can be deduced from measurements of exhaust gases. The ratio of mole fractions between CO2 and O2 is xCO2 a %EA a   ¼ ! ¼ b g %EA b g xCO2 100 xO 2 aþ À aþ À 4 2 100 4 2 xO2 or using Table 2.1 f¼ 100 !f¼ 100 þ %EA 1þ 1 a g aþbÀ2 4  (2.20) xCO2 xO 2 For rich combustion (f>1), the products may contain CO, unburned fuels, and other species formed by the degradation of the fuel. Often additional information on the products is needed for complete balance of the chemical reaction. If the products are assumed to contain only unburned fuel and major combustion products, the corresponding global equation can be written as   1 bg a þ À ðO2 þ 3:76N2 Þ ! f 42     a b 3:76 bg 1 CO2 þ H2 O þ a þ À N2 þ 1 À Ca Hb Og f 2f f 42 f Ca Hb Og þ (2.21) 22 2 Thermodynamics of Combustion Example 2.1 Considering a stoichiometric mixture of isooctane and air, determine: (a) (b) (c) (d) the mole fraction of fuel the fuel-air ratio the mole fraction of H2O in the products the temperature of products below which H2O starts to condense into liquid at 101.3 kPa Solution: The first step is writing and balancing the stoichiometric reaction equat...
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This document was uploaded on 01/20/2014.

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