G fs 0058 for methane as such the product ame

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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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