Unformatted text preview: rd temperature, i.e., cp;R ¼ cp ðT Þ, where T ¼ ðTR þ T0 Þ=2.
From conservation of energy, Hs,P equals the amount of heat released from
combustion plus the sensible enthalpy of the reactants, Hs,P ¼À Q0 ;p þ Hs;R ¼ mfb
rxn
ÁLHV þ Hs,R, where mfb is the amount of fuel burned. For fb1, mfb ¼ mf since there
is enough air to consume all the fuel in a lean mixture. For rich combustion (f > 1), the
limiting factor is the amount of air available, ma. Therefore, for f>1, the amount of
fuel burned (with air, ma) is mfb ¼ ma fs , where fs is the stoichiometric fuel/air ratio by
mass. Then the adiabatic ﬂame temperature is calculated for a lean mixture as fb1
mf Á LHV þ ðma þ mf Þcp;R ðTR À T0 Þ
ðma þ mf Þcp;P
mf Á LHV
mf =ma Á LHV
% TR þ
¼ TR þ
ðma þ mf Þcp;P
ð1 þ mf =ma Þcp;P
f Á LHV
f Á fs Á LHV
¼ TR þ
¼ TR þ
ð1 þ f Þcp;P
ð1 þ f Á fs Þcp;P TP ﬃ T0 þ (2.43)
where cp;R % cp;P is used in deriving the second line. Similarly, for the rich mixtures
one gets
fr1 Tp ¼ TR þ fs Á LHV
fs Á LHV
¼ TR þ
ð1 þ f Þcp;P
ð1 þ f Á fs Þcp;P (2.44) Note that fs is very small for hydrocarbon fuels (e.g., fs ¼ 0.058 for methane).
As such, the product (ﬂame) temperature increases almost linearly with equivalence ratio, f, for lean combustion as shown in Fig. 2.4. As expected, the ﬂame
temperature peaks at the stoichiometric ratio. In rich combustion, the ﬂame
temperature decreases with f.
Method 2: Iterative enthalpy balance
A more accurate approach is to ﬁnd the ﬂame temperature by iteratively assigning
the ﬂame temperature Tp until Hp(Tp) % HR(TR). The enthalpy of reactants is
assumed given. The enthalpy of products can be expressed in the following form 2.4 Adiabatic Flame Temperature 35 2500 Estimate with constant cp Temperature (K) 2000
Enthalpy balance
1500 Simulated flame 1000 Equilibrium 500 0
0.1 1
Equivalence Ratio, φ 10 Fig. 2.4 Comparison of ﬂame temperatures with different approaches HP ðTP Þ ¼...
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This document was uploaded on 01/20/2014.
 Winter '14
 Physics, Energy, Heat

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