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CH4 þ ? O2 þ N2 ! ?CO2 þ ?H2 O þ ?N2 ;
(2.10)
21
where air consisting of 21% O2 and 79% N2 is assumed.1 The coefﬁcients associated
with each species in the above equation are unknown. By balancing the atomic 18 2 Thermodynamics of Combustion abundance on both the reactant and product sides, one can ﬁnd the coefﬁcient
for each species. For instance, let’s determine the coefﬁcient for CO2: on the reactant
side, we have 1 mol of C atoms; hence the product side should also have 1 mol of
C atoms. The coefﬁcient of CO2 is therefore unity. Using this procedure we can
determine all the coefﬁcients. These coefﬁcients are called the reaction stoichiometric
coefﬁcients. For stoichiometric methane combustion with air, the balanced reaction
equation reads:
CH4 þ 2ðO2 þ 3:76N2 Þ ! 1CO2 þ 2H2 O þ 7:52N2 : (2.11) Note that on the reactant side there are 2·(1 + 3.76) or 9.52 mol of air and
its molecular mass is 28.96 kg/kmol. In this text, the reactions are balanced using
1 mol of fuel. This is done here to simplify the calculations of the heat of reaction
and ﬂame temperature later in the chapter. Combustion stoichiometry for a general
hydrocarbon fuel, Ca Hb Og , with air can be expressed as
bg
b
bg
Ca Hb Og þ a þ À ðO2 þ 3:76N2 Þ!aCO2 þ H2 O þ 3:76 a þ À N2 : (2.12)
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2
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The amount of air required for combusting a stoichiometric mixture is called
stoichiometric or theoretical air. The above formula is for a singlecomponent fuel
and cannot be applied to a fuel consisting of multiple components. There are two
typical approaches for systems with multiple fuels. Examples are given here for a
fuel mixture containing 95% methane and 5% hydrogen. The ﬁrst method develops
the stoichiometry of combustion using the general principle of atomic balance,
making sure that the total number of each type of atom (C, H, N, O) is the same in
the products and the reactants.
0:95CH4 þ 0:05H2 þ 1:925ðO2 þ 3:76N2 Þ !
0:95CO2 þ 1:95H2 O þ 7:238N2 :
The...
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This document was uploaded on 01/20/2014.
 Winter '14
 Physics, Energy, Heat

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