Adiabatic Flame Temperature

The balanced chemical reaction equation is
where
and
HP?TP? 1?4 HR?TR? 1?4
Ni;Phi;P 1?4 Ni;P1?2Dhoi;P ? hsi;P?TP? ii
Adiabatic Flame Temperature
C12H22O11?s? ? 12O2?g? 1?4 12CO2?g? ? 11H2O?liq?
Since the total number of moles of gas is constant (12) in the products and reactants, DN 1?4 0. Therefore, work is zero and the enthalpy of combustion equals the heat transfer: 5,648 kJ/mol.
2.4 Adiabatic Flame Temperature
One of the most important features of a combustion process is the highest temperature of the combustion products that can be achieved. The temperature of the products will be greatest when there are no heat losses to the surrounding environment and all of the energy released from combustion is used to heat the products. In the next two sections, the methodology used to calculate the maximum temperature, or adiabatic flame temperature, will be presented.
2.4.1 Constant-Pressure Combustion Processes
An adiabatic constant-pressure analysis is used here to calculate the adiabatic flame temperature. Under this idealized condition, conservation of energy is:
HP?TP? 1?4 HR?TR?; X^X^^
(2.37)
X^X^^
Ni;Rhi;R 1?4 Ni;R1?2Dhoi;R ? hsi;R?TR? :
ii
Figure 2.3 is a graphic explanation of how the adiabatic flame temperature is determined. At the initial reactant temperature, the enthalpy of the product mixture
32 2 Thermodynamics of Combustion
HR (T)
HR (TR) = Hp (TP) ?x
Energy Release
? HP (TR)
HR (TR)
HP(T)
Reactant Adiabatic Flame Temperature Temperature
Temperature, T
Fig. 2.3 Graphical interpretation of adiabatic flame temperature
is lower than that of the reactant mixture. The energy released from combustion is used to heat up the products such that the condition HP?TP? 1?4 HR?TR? is met.
The task is finding the product temperature given the enthalpy of reactants. Three different methods can be used to obtain TP:
1. Using an average cp value,
2. An iterative enthalpy balance,
3. Finding the equilibrium state using computer software (such as Cantera).
The first two methods can be performed manually if complete combustion is considered and provide only quick estimates. An equilibrium state solver takes into account dissociation of products at high temperature, making it more accurate than the first two methods.
Method 1: Constant, average cp
From conservation of energy, Hp?Tp? 1?4 HR?TR?, which can be expressed as
X^^X^^ Ni;P1?2Dhoi;P ? hsi;P?TP? 1?4 Ni;R1?2Dhoi;R ? hsi;R?TR?
ii
Rearranging yields
X (XXX)X
^^^^
Ni;Phsi;P?TP? 1?4 Ni;PDhoi;P Ni;RDhoi;R ? iiii
Ni;Rhsi;R?TR?
(2.38)
1?4 Q0 ? Nh^?T? rxn;p i;R si;R R
i
Enthalpy
2.4 Adiabatic Flame Temperature with
33
X^X^
Q0rxn;p 1?4 Ni;RDhoi;R Ni;PDhoi;P: (2.39)
ii
Note that water in the products is likely in gas phase due to the high combustion temperature; therefore Q0 1?4 LHV NP M 1?4 LHV m when the fuel is
rxn;p fuel fuel f
completely consumed. The second term, N h^ ?T ?,ininreEpqre. s2e.n3t8s represents
the difference of sensible enthalpy between TR and T0 (25 C) for the reactant mixture. With the assumption that the sensible enthalpy can be approximated by hˆsi,P(TP) % cˆpi (TP T0) with cˆpi % constant, we have
?T T?XN c^ c^?T T?XN 1?4 Q0 ?XN h^ ?T? P 0 i;P pi p P 0 i;P rxn;p i;R si;R R
(2.40)
(2.41)
iii Rearranging the equation one finds TP as
Q0 ?PNh^?T? rxn;p i;R si;R R
TP 1?4 T0 ?
%TR ?PNi;Pc^pi
i
i;R si;R R i
P i
Ni;P c^pi
i Q0rxn;p
i
LHV Nfuel Mfuel
1?4TR? PNc^ ; i;P pi
where the following approximation has been applied4 P N h^ ? T ? P N c^ ? T T ?
i;R si;R R i;R pi;R R 0 i PNi;Pc^pi 1?4 i PNi;Pc^pi
%TR T0
When reactants enter the combustor at the standard conditions, the above
ii
equation reduces to (as sensible enthalpies of reactants are zero at T0) LHV Nfuel Mfuel
TP1?4T0? PNc^ : i;P pi
(2.42)
4 PNi;Rc^pi;R and PNi;pc^pi are assumed to be approximately equal. ii
i

34 2 Thermodynamics of Combustion
The above procedure is general and can be applied to any mixture. Note that the specific heat is a function of temperature, so the accuracy of this approach depends on the value selected for the specific heat cˆp.
If the heating value of a fuel is given, a mass-based analysis for the same control volume can be conducted. The initial mixture consists of fuel and air with mf and ma, respectively. By mass conservation, the products have a total mass of mf + ma. The sensible enthalpy of the products is approximated by Hs,P 1?4 (ma + mf ) c"p;P (TP – T0), where c"p;P is an average value of specific heat evaluated at the average temperature of the reactants and products, i.e., c"p;P 1?4 cp?T"?; where T" 1?4 ?Tp ? TR?=2. Similarly, the sensible enthalpy of the reactants is estimated by Hs,R 1?4 (ma + mf ) c"p;R (TR – T0), where c"p;R is an average value of specific heat evaluated at the average temperature of reactants and the standard temperature, i.e., c"p;R 1?4 cp?T"?, where T" 1?4 ?TR ? T0?=2. From conservation of energy, Hs,P equals the amount of heat released from combustion plus the sensible enthalpy of the reactants, Hs,P 1?4 Q0rxn;p ? Hs;R 1?4 mfb LHV ? Hs,R, where mfb is the amount of fuel burned. For fb1, mfb 1?4 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 1?4 mafs, where fs is the stoichiometric fuel/air ratio by mass. Then the adiabatic flame temperature is calculated for a lean mixture as fb1
TP ffiT0 ?mf LHV??ma ?mf?c"p;R?TR T0? ?ma ? mf ?c"p;P
%TR? mf LHV 1?4TR? mf=ma LHV ?ma ? mf ?c"p;P ?1 ? mf =ma?c"p;P
1?4TR ? f LHV 1?4TR ? f fs LHV ?1?f?c"p;P ?1?f fs?c"p;P
(2.43)
where c"p;R % c"p;P is used in deriving the second line. Similarly, for the rich mixtures one gets
fr1 Tp1?4TR? fs LHV 1?4TR? fs LHV (2.44) ?1 ? f?c"p;P ?1 ? f fs?c"p;P
Note that fs is very small for hydrocarbon fuels (e.g., fs 1?4 0.058 for methane). As such, the product (flame) temperature increases almost linearly with equiva- lence 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
2000
1500
1000
500
Estimate with constant cp
Enthalpy balance Simulated flame
Equilibrium
0
0.1 1 10
Equivalence Ratio, ?
Fig. 2.4 Comparison of flame temperatures with different approaches
X^X^^ X^ HP?TP? 1?4 Ni;Phi;P 1?4 Ni;P1?2Dhoi;P ? hsi;P?TP? 1?4 HR?TR? 1?4 Ni;Rhi;R
iii
Next, we rearrange the above equation to find an expression for the sensible enthalpy of the products as
X^X^ X^X^ Ni;PDhoi;P ? Ni;Phsi;P?TP? 1?4 Ni;RDhoi;R ? Ni;Rhsi;R?TR?
iiii
X^ X^X^X^
Ni;Phsi;P?TP? 1?4 Ni;RDhoi;R Ni;PDhoi;P ? Ni;Rhsi;R?TR? (2.45)
iiii XN h^ ?T?1?4 Q0 ?XN h^ ?T?:
i;P si;P P rxn;p i;P si;R R ii
With an initial guess of flame temperature, Tp1, one evaluates Hp(Tp1) from tables such as those in Appendix 3. If Hp(Tp1) < HR(TR), we guess a higher flame temper- ature, Tp2. One repeats this process until the two closest temperatures are found such that Hp(Tf1) < HR(TR) < Hp(Tf2). The product temperature can be estimated by linear interpolation. This method, although more accurate, still assumes complete combustion to the major products.
Method 3: Equilibrium State (Free software: Cantera; Commercial software: Chemkin)
Dissociation5 of products at high temperature (T > 1,500 K at ambient pressure) can take a significant portion of energy from combustion and hence the product
5 Dissociation is the separation of larger molecules into smaller molecules. For example, 2H2O ?2H2 + O2.
Temperature (K)
36 2 Thermodynamics of Combustion
temperature is lower than that calculated with only major components as products. The equilibrium state determines the species concentrations and temperature under certain constraints such as constant enthalpy, pressure, or temperature. The equilibrium flame temperature is expected to be lower than the temperatures estimated with Method 1 or Method 2. In addition, the chemical equilibrium state is often used in combustion engineering as a reference point for chemical kinetics (the subject of Chap. 3) if infinite time is available for chemical reactions. At this ideal state, forward and backward reaction rates of any chemical reaction steps are balanced. By constraining certain variables such as constant pressure and enthalpy, the chemical equilibrium state can be determined by minimizing the Gibbs free energy, even without knowledge of the chemical kinetics. Computer programs (such as STANJAN, Chemkin, Cantera) are preferred for this task, as hand calculations are time consuming.
2.4.2 Comparison of Adiabatic Flame Temperature Calculation Methods
The presented methods of estimating adiabatic flame temperature will produce different values from each other. Predicted adiabatic flame temperatures of a methane/air mixture at ambient pressure using these methods are compared in Fig. 2.4 for a range of equivalence ratios. Also included are the results from a flame calculation using a detailed, non-equilibrium flame model. On the lean side, the results agree reasonably well among all methods, as the major products are CO2, H2O, unburned O2, and N2. Visible deviations arise near stoichiometric conditions and become larger in richer mixtures. One reason for the deviation is the assump- tions made about product species in the rich mixtures. For rich mixtures at the equilibrium state, CO is preferred over CO2 due to the deficiency in O2. Because the conversion of CO into CO2 releases a large amount of energy, the rich mixture equilibrium temperatures are lower than those from the flame cal