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Unformatted text preview: other method of balancing a fuel mixture is to first develop stoichiometry relations for CH4 and H2 individually: CH4 þ 2ðO2 þ 3:76N2 Þ ! CO2 þ 2H2 O þ 2 Á 3:76N2 H2 þ 0:5ðO2 þ 3:76N2 Þ ! H2 O þ 0:5 Á 3:76N2 Then, multiply the individual stoichiometry equations by the mole fractions of the fuel components and add them: 2.2 Combustion Stoichiometry 19 0:95 Á fCH4 þ 2ðO2 þ 3:76N2 Þ ! CO2 þ 2H2 O þ 2 Á 3:76N2 g 0:05 Á fH2 þ 0:5ðO2 þ 3:76N2 Þ ! H2 O þ 0:5 Á 3:76N2 g )0:95CH4 þ 0:05H2 þ 1:925ðO2 þ 3:76N2 Þ ! 0:95CO2 þ 1:95H2 O þ 7:238N2 2.2.1 Methods of Quantifying Fuel and Air Content of Combustible Mixtures In practice, fuels are often combusted with an amount of air different from the stoichiometric ratio. If less air than the stoichiometric amount is used, the mixture is described as fuel rich. If excess air is used, the mixture is described as fuel lean. For this reason, it is convenient to quantify the combustible mixture using one of the following commonly used methods: Fuel-Air Ratio (FAR): The fuel-air ratio, f, is given by f¼ mf ; ma (2.13) where mf and ma are the respective masses of the fuel and the air. For a stoichiometric mixture, Eq. 2.13 becomes  mf  Mf ¼ ; fs ¼  g ma stoichiometric ða þ b À 2Þ Á 4:76 Á Mair 4 (2.14) where Mf and Mair (~28.84 kg/kmol) are the average masses per mole of fuel and air, respectively. The range of f is bounded by zero and 1. Most hydrocarbon fuels have a stoichiometric fuel-air ratio, fs, in the range of 0.05–0.07. The air-fuel ratio (AFR) is also used to describe a combustible mixture and is simply the reciprocal of FAR, as AFR ¼ 1/f. For instance, the stoichiometric AFR of gasoline is about 14.7. For most hydrocarbon fuels, 14–20 kg of air is needed for complete combustion of 1 kg of fuel. Equivalence Ratio: Normalizing the actual fuel-air ratio by the stoichiometric fuelair ratio gives the equivalence ratio, f. f¼ f mas Nas NO2s ¼ ¼ ¼ fs ma Na NO2;a (2.15) The subscript s indicates a value at the...
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