7.1 Introduction

Source: Reproduced with permission from Rolls-Royce plc

To study the mechanism of heat release in an aircraft gas turbine engine combustion chamber, we need to revisit the thermochemistry principles that we learned in our freshman chemistry class. As engineers, we are interested in maximizing the heat release while minimizing the space requirements for the combustion chamber of an aircraft engine. Besides the mechanism of heat release, we are also interested in the rate at which chemical reactions takes place. We may think of it as one characteristic timescale of the problem. The question of rate is governed by the chemical kinetics, which is a subject in thermochemistry. In addition, we hope to understand the characteristic length scales of the problem involved in flame stability and length, which ultimately are tied to the combustion chamber geometry and sizing. In a combustion chamber of an airbreathing engine, we bring a liquid fuel and air together and an ignition source to start off the chemical reaction. We understand that the fuel has to be first vaporized before any reaction can take place between it and the oxygen in the air. We also recognize that a reaction between the oxygen in the air and the fuel must take place upon a collision between the two molecules. Only very energetic collisions result in the molecular dissociation of the fuel and a reaction with the oxygen. Therefore, not only the collisions are of necessity for a chemical reaction but also the collision energy is of importance. We may want to think of those collisions that lead to a chemical reaction as effective collisions in contrast to those collisions that result in a bounce back of the molecules. The question of whether collision energy is sufficient for a chemical reaction is answered by a chemical parameter known as the activation energy of the reaction. Now that we are relating to chemical reactions as energetic collisions between the oxygen and the fuel molecules, we may relate to the rate of reaction by the frequency of such collisions. We immediately recognize that these parameters, namely the characteristic time and length scales in a combustion chamber, depend on many parameters, including the vaporization rate of the atomized fuel droplets and the turbulence intensity of the mixing process. Finally, as citizen engineers, we are committed to zero pollution, which will prove to be a daunting task in the combustion of a hydrocarbon fuel in air.

Chemical reactions take place at the molecular level and are always expressed in the following form:

(7.1)

where a, b, m, and n are the number of moles of reactants A and B and the products M and N, respectively. The left-hand side of the reaction is known as the reactants, and the right-hand side of the reaction is known as the products. The number of moles of a substance is the ratio of the mass of the substance to its molecular weight, for example,

(7.2)

with the dimensions of

(7.3)

Another interpretation of the mole of a substance is the amount of the substance that contains 6.023 × 10²³ molecules (in a gram. mole), which is known as the Avogadro’s number N_a. Avogadro’s number is defined as the number of carbon atoms in 12 g of ¹²C, which is 6.023 × 10²³ molecules/gmol, therefore,

(7.4)

However, presenting the number of moles of a substance is easiest based on mass and molecular weight rather than the number of individual molecules and the Avogadro’s number. In an engineering combustion problem, we typically introduce a certain mass of fuel and oxidizer in a reaction and not (explicitly) their number of individual molecules.

An important conservation law applicable to chemical reactions is the law of conservation of atomic species. It simply states that the numerical count of atoms of each species on both sides of a reaction has to be the same. The following two examples illustrate this conservation principle.

There are four hydrogen atoms on the reactants side as well as the product side. Two oxygen atoms appear on both sides as well. In the following reaction, methane (CH₄) is reacting with oxygen, with C, H, and O in balance.

In dealing with the problems of reacting gases, such as the type experienced in a combustion chamber of a gas turbine engine, we are faced with a mixture of gases and hence we need to review the laws governing such mixtures. The most fundamental question regarding mixtures is whether we know the mixture properties, as a function of its constituent properties. The mixture properties range from the molecular weight of the mixture to the specific heats of the mixture, the specific enthalpy and entropy of the mixture. In addition, the constituent gases are assumed to obey the perfect gas law, which at sufficiently low densities, we understand that all gases behave as perfect.

7.2 Laws Governing Mixture of Gases

Imagine there is a volume V that contains several gases. The equilibrium mixture of gases has attained a mixture temperature T_m, which we now identify as being shared by all the constituents, that is, all constituents have the same random kinetic energy after numerous collisions that brought them to an equilibrium state, namely,

(7.5)

Before we define the pressure of the mixture p_m, let us first define the partial pressure associated with the constituents of the mixture. A constituent of a mixture that occupies the entire volume of the mixture, V, and is at the mixture temperature, exerts a pressure on the vessel, which is called its partial pressure. Now, Dalton’s law of additive pressures states that the mixture pressure is the sum of all its constituents’ partial pressures (a direct consequence of perfect gas law), namely,

(7.6)

where p_n is the partial pressure of the nth constituent.

On the question of internal energy, enthalpy and the entropy of the mixture of gases, we have the Gibbs–Dalton law that states

(7.7)

(7.8)

(7.9)

where E is the internal energy, H the enthalpy, and S is the entropy. The constituent properties in Equations 7.7 to 7.9 are all based on the assumption that a constituent occupies the entire volume and is at the mixture temperature. We can also express the intrinsic variables of state, namely, the specific internal energy, the specific enthalpy, and the specific entropy based on the above Equations 7.7 to 7.9, as

(7.10)

(7.11)

(7.12)

Now, the mass of the mixture m_m is obviously the sum of the individual masses m_i based on the law of conservation of mass. We can express the mass of each constituent as the product of the number of moles and the molecular weight, of that constituent, based on Equation 7.2, as

(7.13)

Therefore, the mixture molecular weight is now expressible in terms of the individual molecular weights and the mole fraction of each constituent, which we now define as χ

(7.14)

and, therefore,

(7.15)

As an example, air is composed of nitrogen, oxygen, and traces of inert gases, such as argon. Its composition issaid to be nearly 78% N₂, 21% O₂, and 1% Ar, by volume. Therefore, 0.78, 0.21, and 0.01 are the volume fractions of the nitrogen, oxygen, and argon, respectively, in air.

From perfect gas law, written for a constituent as

(7.16)

where is the universal gas constant with a value of

We can sum Equation 7.16 for all the constituents to get

(7.17)

which suggests that the ratio of partial pressure of any constituent to the mixture pressure is equal to the mole fraction of that constituent, namely,

(7.18)

Now, if we consider a mixture of gases, such as air, at a pressure, namely, p_m, and a temperature T_m, we may define a volume occupied by each constituent, V_i, such that

(7.19)

Dividing Equation 7.19 by Equation 7.17 yields

(7.20)

The above equation states that the volume fraction of a gas constituent in a mixture, as defined by Equation 7.19 (i.e., at T_m and p_m), is equal to the constituent’s mole fraction. The same conclusion about the volume fraction and mole fraction (i.e., being equal) may be reached by remembering the Law of Equal Volumes. It states that the volume occupied by one mole of any perfect gas is the same as that of any other perfect gas at a constant pressure and temperature. Consequently, when we describe the air composition as being 0.78, 0.21, and 0.01 for N₂, O₂, and Ar, respectively, by volume, we are stating that the air composition on a mole fraction basis as

(7.21)

with 0.78, 0.21, and 0.01 as mole fractions of the constituents N₂, O₂, and Ar. Now, we can use Equation 7.15 to arrive at the molecular weight of air as

(7.22)

Figure 7.1 shows the meaning of the volume occupied by a constituent gas in a mixture of gases. The specific heats at constant volume and pressure follow the perfect gas law:

(7.23)

Now, combining these results with the mixture laws for the internal energy and enthalpy, Equations 7.10 and 7.11, respectively, we get

(7.24)

(7.25)

The ratio of specific heats for the mixture may now be written as

(7.26)

If instead of working with masses, we prefer to work with the moles and mole fractions, we can replace each constituent mass by the product of its molecular weight and its number of moles. This approach is preferred in chemical analysis where compounds reaction is on a molar basis rather than mass basis. For example, let us express specific heat at constant pressure of the mixture in terms of the mole fractions instead of mass fractions, as given in Equation 7.25,

(7.27)

7.3 Chemical Reaction and Flame Temperature

First, we define a series of terms that describe a chemical reaction, such as exothermic or endothermic. Let us assume the following chemical reaction:

If the reaction proceeds from the reactants A and B to the products C and D and the reaction produces heat as a result, we call such reactions as exothermic. If the opposite occurs, that is, if the reactants would not naturally react without an initial external heat input (stimulus), then such reactions are called endothermic. An example of an exothermic reaction is the combustion of hydrocarbon fuels and air. An example of an endothermic reaction is the phase transformation of water from liquid to vapor that requires a heat input. Our interest in chemical reaction lies primarily in the study of combustion and hence exothermic processes. In gas turbine engines, the reactants are typically hydrocarbon fuels and air. In a liquid propellant chemical rocket, the reactants are typically highly energetic fuels, such as hydrogen, and an oxidizer, such as oxygen. Let us consider octane (C₈H₁₈) as a typical hydrocarbon fuel in a gas turbine engine. Let us also describe air (by lumping argon into the nitrogen molecule) as being approximately composed of

(7.28)

Now, we define a unique mass ratio between the fuel and oxidizer, which results in the complete combustion of the fuel and a stable product composition. This unique ratio is called the stoichiometric ratio and now we apply it to the combustion of octane and air, as an example.

(7.29)

We took a mole of octane and brought in 25/2 moles of O₂ and (3.76) (25/2) moles of N₂. Note that the need for 25/2 came from turning “C₈ ” of octane into “8 CO₂” and the “H₁₈” of octane into “9 H₂O, ” therefore 16 + 9 = 25 oxygen atoms were needed for the complete combustion (which also means complete oxidation) of one mole of octane.

As noted, this unique proportion of fuel to oxidizer results in neither excess oxygen nor any excess fuel. Any more fuel would result in unburned fuel in the products of combustion and any more air would result in excess oxygen in the products of combustion. The nitrogen (molecule) (N₂) is treated as remaining unreacted (or inert in chemical terms) in the combustion process. Now, the stoichiometric fuel-to-air ratio for the combustion of octane in air is

(7.30)

Therefore, we say that the stoichiometric ratio for hydrocarbon fuels and air is about 6.7%. This is a number to remember. If we use proportionately less fuel than the stoichiometric ratio, the combustion is said to be on a fuel lean basis and the opposite is called a fuel rich combustion. A parameter that describes the fuel lean or rich condition of a combustor is called the equivalence ratio , which is defined as

(7.31)

Consequently,

(7.32)

(7.33)

(7.34)

It will be demonstrated in this chapter that the highest combustion temperature is achieved very near the stoichiometric ratio (proportion of the fuel to the oxidizer). This fact, which is depicted in Figure 7.2, should also be stored in memory as a useful concept. For many fuels, the maximum temperature, that is, the adiabatic flame temperature, occurs at the equivalence ratio between 1 and 1.1, since the product-specific heat is reduced with a slightly fuel rich mixture ratio.

Now, assuming that we know the composition of the products in a chemical reaction, that is, the constituents of the product and their number of moles, the fundamental question that remains is the product’s temperature. How do we arrive at the product’s temperature? If we treat the combustor as a black box, where the reactants enter the box at certain temperature, say T₁, and the products leave the box at another temperature, say T₂, and we can describe the heat transfer through the sides of the box, that is, the heat exchange with the surrounding, as say Q, we can use the law of conservation of energy to write:

(7.35)

where H is the absolute enthalpy and Q is the external heat interaction with the combustion chamber. We show this box schematically in Figure 7.3, as an aid.

Remember absolute enthalpy from thermodynamics that was defined as the sum of heat of formation of the substance at the reference temperature (298.16 K) and what was called the sensible enthalpy, which raised the enthalpy of the substance from the reference temperature to the desired temperature. Absolute enthalpy H may be written as the product of mass and the specific absolute enthalpy for a substance. The mass may be expressed as the product of the number of moles and the molecular weight of the substance. Now, for a perfect gas, if we incorporate all of the above concepts, the absolute enthalpy may be written as

(7.36)

where T_f is the reference temperature, is the standard heat of formation of the substance at the reference temperature per unit mass at the pressure of 1 bar. The subscript “f ” refers to the reference temperature and the superscript “0” refers to the standard state of the chemical compound at the pressure of 1 bar. We define the product of the molecular weight and the specific heat at constant pressure (that appears above) as the molar specific heat, and give it a special symbol, namely, the c_p with a bar, that is, let

(7.37)

The dimensions of the molar specific heat are

(7.38)

We also define the product of the molecular weight and the heat of formation per unit mass of a substance, as the molar heat of formation, and give it a special symbol (with a bar to identify the molar), as

(7.39)

The molar heat of formation is the evolved heat in forming one mole of the substance from its elements in their standard states at constant temperature of T_f = 298.16 K and the constant pressure of p_f = 1 bar. The dimensions of the molar heat of formation at a reference temperature are

(7.40)

Imagine a reaction chamber, where the basic elements of a chemical compound, at their standard state, enter at a reference temperature T_f at the pressure of 1 bar and the compound is then formed in the reaction chamber. Now, in order to maintain the temperature of the compound at the reference temperature, we need to interact with the reaction chamber through heat transfer to or from the box. The heat of formation of a chemical compound is the amount of heat exchange to (positive) or from (negative) a reaction chamber, which forms the chemical compound from its constituents, and maintains a constant (known as reference) temperature. The heat of formation of all naturally occurring elements is then, by definition, zero. The reference temperature is 298.16 K, which corresponds to a standard room temperature of 25°C (77°F) and the reference pressure of 1 bar is the standard atmospheric pressure (also very close to 1 atm). This process can be schematically shown in Figure 7.4.

The exothermic reactions from basic elements to form one mole of a chemical compound then lead to a negative heat of formation, because we need to extract heat from the reaction chamber in order to maintain a constant temperature. The opposite is true for endothermic reactions to form a compound. These concepts are best learned through a series of examples.

The heat of formation of carbon dioxide can be found from the following isothermal reaction at 1 bar:

Therefore, CO₂(g), carbon dioxide (gaseous), has a standard heat of formation of kJ/kmol. We note that in the previous reaction the carbon enters the reaction as a solid and the oxygen as a gas, that is, the standard states of carbon and oxygen, to form one mole of carbon dioxide (in gaseous form).

In forming one mole of carbon monoxide in an isothermal process with T_f = 298 16 K and at a pressure of 1 bar, carbon and oxygen in their standard states have reacted according to the above. We note that a heat extraction of 110, 530 kJ/kmol (same as J/mol) is necessary to maintain isothermal status for the reaction. Hence, we conclude that standard heat of formation of carbon monoxide, CO (g), is kJ/kmol. The following four reactions serve as additional examples on the standard heat of formation.

A blended jet fuel known as JP-4 has an approximate formulation, CH_1.93 and the following isothermal reaction defines its standard heat of formation in liquid form, that is,

The energy balance equation across the burner (Equation 7.35) may now be expressed on a molar basis, to include the complete enthalpy expression. This will account for the reactants entering the combustion chamber at other than reference temperature, namely, T₁, and the products of combustion leave the chamber at another temperature, namely, T₂, as follows:

(7.41)

where “P” and “R” stand for products and reactants, respectively, and the last term defined as

(7.42)

is the difference between the standard heats of formation of all the product constituents and the reactant constituents, at the reference temperature at the pressure of 1 bar. This quantity is also called heat of reaction with a symbol Q_R. Therefore, heat of reaction in a combustion process is defined as

(7.43)

We can specialize the concept of the heat of reaction to combustion of a fuel in pure oxygen O₂. This will be the basis for the definition of the fuel heating value. Assuming that the fuel and the gaseous oxygen, that is, O₂ (g), enter the combustion chamber at the reference temperature of 298.16 K and a complete combustion takes place, the heat release per unit mass of the fuel, which will return the products of combustion to the reference temperature, is called the heating value of the fuel. In simple terms, substitute T_f for T₁ and T₂ in Equation 7.41, which will force the summations over the products and the reactants to zero, and note that

(7.44)

Although the negative sign on the summations of Equation 7.43 or 7.44 shows the necessity of removing a quantity of heat from the combustion chamber in order to maintain an isothermal environment, it is customary to report the fuel heating value as a positive quantity. Hence, we define

(7.45)

Now, depending on the state of water in the products of combustion, that is, as a liquid or vapor, the fuel heating value is referred to as higher heating value (HHV) or lower heating value (LHV), respectively. Since the difference between the higher and lower heating values depends on the energy spent in vaporizing the condensed water in the products of combustion, the following energy balance relates the two heating values:

(7.46)

where the last term h_lg is the latent heat of vaporization for water at 25°C, which is 2443 kJ/kg. Now, let us put these concepts to work and obtain a fuel heating value based on the standard heats of formation and the water/fuel mass fraction and the latent heat of vaporization of water.

Tables of thermochemical data (Table 7.1) that list the standard heat of formation of numerous compounds are found in compilations, such as JANAF or the Handbook of Chemistry and Physics. Here, only a partial listing is reproduced for the purposes of illustration and problem solving involving chemical reactions.

Chemical			(kJ/gmol)	(kJ/gm)
symbol	Name	State	[per mole basis]	[per mass basis]
C	Carbon	Solid	0	0
C	Carbon	Gas	716.67	59.72
CO2	Carbon dioxide	Gas	−393.522	−8.944
CO	Carbon monoxide	Gas	−110.53	−3.947
H2	Hydrogen	Gas	0	0
H	Hydrogen atom	Gas	217.999	217.999
OH	Hydroxyl radical	Gas	39.463	2.321
H2O	Water	Gas	−241.827	−13.435
H2O2	Hydrogen peroxide	Gas	−136.106	−4.003
N2	Nitrogen	Gas	0	0
N	Nitrogen atom	Gas	472.68	33.763
NO	Nitric oxide	Gas	90.291	3.010
NO2	Nitrogen dioxide	Gas	33.10	0.7196
N2O	Nitrous oxide	Gas	82.05	1.8648
O2	Oxygen	Gas	0	0
O	Oxygen atom	Gas	59.56	3.723
O3	Ozone	Gas	34.00	0.708
CH4	Methane	Gas	−74.873	−4.6796
CH3OH	Methyl alcohol	Gas	−201.07	−6.2834
CH3OH	Methyl alcohol	Liquid	−238.66	−7.4581
C2H5OH	Ethyl alcohol	Gas	−235.00	−5.108
C2H5OH	Ethyl alcohol	Liquid	−277.20	−6.026
C3H8	Propane	Gas	−103.90	−2.3614
C4H10	Butane	Gas	−126.148	−2.175
C8H18	Octane	Gas	−208.447	−1.8285
C8H18	Octane	Liquid	−249.93	−2.1924
CH1.553	JP-3	Liquid		−1.11
CH1.93	JP-4	Liquid		−1.77

Some useful conversion factors are 1 kcal = 4.1868 kJ = 3.9684 BTU = 3, 088 ft · Ibf kcal/gmol = 4186.8 kJ/kmol and kJ/kmol = 1000 kJ/gmol

Applying Equation 7.41, to a chemical reaction with a known product composition and a known heat exchange characteristics through the walls of the combustor, Q_external, will result in the calculation of the exit temperature of the products of combustion, T₂. A reference boundary condition on the combustion chamber is the adiabatic assumption of zero heat exchange with the environment, that is, Q_external ≡ 0, which leads to a unique exit temperature value, known as the adiabatic flame temperature T_af. In addition, there are polynomial expressions that are developed for the molar specific heats of various chemical compounds, such as the sample produced in Table 7.2, which provide the remaining input to the enthalpy balance equation across the combustor, that is, Equation 7.41. As the polynomial expressions for specific heat are in temperature, and we do not a priori know the temperature of the products of combustion, the process of selecting a suitable molar specific heat is iterative. Otherwise an average specific heat may be assumed, over the expected temperature range. Also note that in a fuel-rich combustion, , the heat of reaction remains the same as that of stoichiometric reaction, whereas the flame temperature will be reduced as the heat of reaction is now used to raise the temperature of additional fuel.

Gases at low pressures = kJ/kmol · K θ = T(K)/100
Gas		Range K	Max. error (%)
N2	= 39.060 − 512.79θ-1.5 + 1072.7θ−2 − 820.40θ−3	300–3500	0.43
O2	= 37.432 + 0.020102θ1.5 − 178.57θ−1.5 + 236.88θ−2	300–3500	0.30
H2	= 56.505 − 702.74θ−0.75 + 1165.0θ−1 − 560.70θ−1.5	300–3500	0.60
CO	= 69.145 − 0.70463θ 0.75 − 200.77θ-0.5 + 176.76θ−0.75	300–3500	0.42
OH	= 81.564 − 59.350θ 0.25 + 17.329θ0.75−− 4.2660θ	300–3500	0.43
HO	= 59.283 − 1.7096θ 0.5 − 70.613θ−0.5 + 74.889θ−1.5	300–3500	0.34
H2O	= 143.05 − 183.54θ0.25 + 82.751θ0.5 − 3.6989θ	300–3500	0.43
CO2	= −3.7357 + 30.529θ0.5 – 4.1034θ + 0.024 198θ2	300–3500	0.19
NO2	= 46.045 + 216.10θ−0.5 − 363.66θ−0.75 + 232.550θ−2	300–3500	0.26
CH4	= 672.87 + 439.74θ0.25 − 24.875θ0.75 + 323.88θ−0.5	300–2000	0.15
C2H4	= 95.395 + 123.15θ0.5 − 35.641θ0.75 + 182.77θ−3	300–2000	0.07
C2H6	= 6.895 + 17.26θ − 0.6402θ2 + 0.00728θ3	300–1500	0.83
C3H8	= −4.042 + 30.46θ − 1.571θ2 + 0.03171θ3	300–1500	0.40
C4H10	= 3.954 + 37.12θ − 1.833θ2 + 0.03498 θ3	300–1500	0.54

Source: Adapted from Van Wylen and Sonntag 1985.

Now, let us proceed with two example problems that assume the molar composition of the products of combustion.

7.4 Chemical Equilibrium and Chemical Composition

A chemical reaction normally continues until no more changes in molar concentration of the products occur. At that point, we say the reaction has reached an equilibrium state. We take advantage of the state of equilibrium in a chemical reaction and then try to establish the chemical composition of the products of combustion in that state. Take the combustion of one mole of hydrogen and one half mole of oxygen, as an example.

(7.47)

First, note that forward and reverse arrows in Equation 7.47 have replaced the forward reaction arrow of our earlier chemical reaction expressions. This simply means that a fraction of the products convert back into the reactants and as the combustion temperature increases other products may be formed and eventually a state of equilibrium is reached among all constituents. The word “eventually” in the previous sentence signifies a period of time needed to achieve the equilibrium state in a chemical reaction. Therefore, it is entirely reasonable to talk about a rate of reaction, and in particular the rate of formation of chemical species, which belong to the field of chemical kinetics. In the above equilibrium reaction of hydrogen and oxygen, we have expected/identified some water formation, some “leftover” hydrogen, some “leftover” oxygen, some hydroxyl (OH) formation, and possibly some dissociation of oxygen and hydrogen to form the atomic oxygen and hydrogen. We did not have to stop, however! Could the temperature of combustion be so high as for the oxygen and hydrogen atoms to ionize? And if so, what will the product concentrations be? These are all legitimate (and tough) questions. Now, let us stay at the level of Equation 7.47, which identifies six constituents of unknown molar concentrations. The total number of moles of products at equilibrium is n_m, where

(7.48)

The molar concentration of any species is simply the ratio of the number of moles of that species to the total number of moles of the product, that is, the same as the mole fraction of any given species

We shall apply the law of conservation of atomic species to the above reaction and conclude that

Now, we have two equations and six unknowns. We need to produce four additional equations involving the six unknowns before we have a chance of solving for the six (unknown) molar concentrations.

In order to produce additional equations to assist with the unknown molar concentrations, we introduce a new law, which is called the law of mass action.

7.4.1 The Law of Mass Action

Consider a stoichiometric reaction of the type:

where a, b, c, and d are the (stoichiometric) number of moles in the reaction. The law of mass action states that the rate of disappearance of chemical species in the reactants is proportional to the concentrations of the reactants each raised to their respective stoichiometric exponent. Applied to the forward rate of reaction r_f is proportional to the products of the concentrations of the reactants raised to their stoichiometric exponents, that is,

(7.49)

The bracketed terms are molar concentrations of reactants A and B. The proportionality constant in Equation 7.49 is called the forward reaction rate coefficient k_f, that is,

(7.50)

The law of mass action applied to a reverse reaction then relates the rate of formation of the products to the product of concentration of the products raised to their stoichiometric exponents, namely,

(7.51)

An equilibrium state is reached among the reactants and the product species, when the forward and reverse reaction rates are equal, namely,

(7.52)

Therefore the ratio of forward to reverse reaction rate coefficients in an equilibrium reaction is called the equilibrium constant K

(7.53)

It is customary to express the concentrations of species by their partial pressures in the equilibrium mixture and the equilibrium constant is then referred to as K_P since it is based on partial pressures. Hence, in terms of the partial pressures, the law of mass action is written as

(7.54)

We may equivalently use the mole fractions (or molar concentrations) for species concentration in the law of mass action to arrive at an equivalent expression, namely,

(7.55)

The two equilibrium constants are related to each other, as we may relate the ratio of partial pressures and the mixture pressure to the mole fraction (according to Equation 7.18), hence,

(7.56)

Among the two equilibrium constants, K_P and K_n, the former is a function of temperature alone and hence independent of pressure, whereas the latter is in general a function of both temperature and pressure, that is,

(7.57)

(7.58)

Based on a single parameter functional dependence of K_p, tables and charts of this equilibrium constant are produced as a function of temperature and used for product composition calculations. It is also interesting to note that the molar concentrations in a stoichiometric mixture are a function of the mixture pressure (if c + d − a − b ≠ 0, according to Equation 7.56), whereas the equilibrium constant K_P is only a function of the mixture temperature. The tables of values for the equilibrium constant K_P, as a function of temperature for numerous reactions of interest to combustion, are used to setup a series of auxiliary equations for the unknown molar concentrations. Some of these tables and graphical depictions of equilibrium constant K_P for many reactions are shown at the end of this section. The following example shows a simple application of this principle.

7.4.2 Equilibrium Constant K_P

A first attempt in studying the molar concentrations of the products of combustion will require knowledge of the equilibrium constants (as a function of temperature) for at least the following subreactions:

• Dissociation of oxygen, nitrogen, and hydrogen into the atoms of O, N, and H, respectively
• Formation of OH, NO, NO₂, CO, O₃, NH₃ from various stoichiometric reactions
• Dissociation of water into elements, O, H, and/or OH, dissociation of CO₂.

A more elaborate analysis will require additional subreactions, which go beyond the scope of our treatment. Now, let us examine some of the more useful subreactions we encounter more often in our analysis, namel, y

Reaction	Equilibrium Constant
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.

The equilibrium constants are depicted in graphical form in Figure 7.5 (adapted from Hill and Peterson, 1992). The equilibrium constants for additional dissociation reactions of interest are tabulated in Table 7.3 (from Strehlow, 1984).

Temperature	Temperature
K	°R
600	1080	−18.574	−16.336	−38.081	18.633	34.405	14.318
700	1260	−15.449	−13.599	−32.177	15.583	29.506	12.946
800	1440	−13.101	−11.539	−27.744	13.289	25.830	11.914
900	1620	−11.272	−9.934	−24.292	11.498	22.970	11.108
1000	1800	−9.807	−8.646	−21.528	10.062	20.680	10.459
1100	1980	−8.606	−7.589	−19.265	8.883	18.806	9.926
1200	2160	−7.604	−6.707	−17.377	7.899	17.243	9.479
1300	2340	−6.755	−5.958	−15.778	7.064	15.920	9.099
1400	2520	−6.027	−5.315	−14.406	6.347	14.785	8.771
1500	2700	−5.395	−4.756	−13.217	5.725	13.801	8.485
1600	2880	−4.842	−4.266	−12.175	5.180	12.940	8.234
1700	3060	−4.353	−3.833	−11.256	4.699	12.180	8.011
1800	3240	−3.918	−3.448	−10.437	4.270	11.504	7.811
1900	3420	−3.529	−3.102	−9.705	3.886	10.898	7.631
2000	3600	−3.178	−2.790	−9.046	3.540	10.353	7.469
2100	3780	−2.860	−2.508	−8.449	3.227	9.860	7.321
2200	3960	−2.571	−2.251	−7.905	2.942	9.411	7.185
2300	4140	−2.307	−2.016	−7.409	2.682	9.001	7.061
2400	4320	−2.065	−1.800	−6.954	2.443	8.625	6.946
2500	4500	−1.842	−1.601	−6.535	2.224	8.280	6.840
2600	4680	−1.636	−1.417	−6.149	2.021	7.960	6.741
2700	4860	−1.446	−1.247	−5.790	1.833	7.664	6.649
2800	5040	−1.268	−1.089	−5.457	1.658	7.388	6.563
2900	5220	−1.103	−0.941	−5.147	1.495	7.132	6.483
3000	5400	−0.949	−0.803	−4.858	1.343	6.892	6.407
3100	5580	−0.805	−0.674	−4.587	1.201	6.668	6.336
3200	5760	−0.670	−0.553	−4.332	1.067	6.458	6.269
3300	5940	−0.543	−0.439	−4.093	0.942	6.260	6.206
3400	6120	−0.423	−0.332	−3.868	0.824	6.074	6.145
3500	6300	−0.310	−0.231	−3.656	0.712	5.898	6.088
3600	6480	−0.204	−0.135	−3.455	0.607	5.732	6.034
3700	6660	−0.103	−0.044	−3.265	0.507	5.574	5.982
3800	6840	−0.007	0.042	−3.086	0.413	5.425	5.933
3900	7020	0.084	0.123	−2.915	0.323	5.283	5.886
4000	7200	0.170	0.201	−2.752	0.238	5.149	5.841

Temperature	Temperature
K	°R
600	1080	−7.210	−6.111	−2.568	−11.040	2.001	−13.212
700	1260	−6.086	−5.714	−2.085	−10.021	0.951	−11.458
800	1440	−5.243	−5.417	−1.724	−9.253	0.146	−10.152
900	1620	−4.587	−5.185	−1.444	−8.654	−0.493	−9.145
1000	1800	−4.062	−5.000	−1.222	−8.171	−1.011	−8.344
1100	1980	−3.633	−4.848	−1.041	−7.774	−1.440	−7.693
1200	2160	−3.275	−4.721	−0.890	−7.442	−1.801	−7.153
1300	2340	−2.972	−4.612	−0.764	−7.158	−2.107	−6.698
1400	2520	−2.712	−4.519	−0.656	−6.914	−2.372	−6.309
1500	2700	−2.487	−4.438	−0.563	−6.701	−2.602	−5.974
1600	2880	−2.290	−4.367	−0.482	−6.514	−2.803	−5.681
1700	3060	−2.116	−4.304	−0.410	−6.347	−2.981	−5.423
1800	3240	−1.962	−4.248	−0.347	−6.198	−3.139	−5.195
1900	3420	−1.823	−4.198	−0.291	−6.065	−3.281	−4.991
2000	3600	−1.699	−4.152	−0.240	−5.943	−3.408	−4.808
2100	3780	−1.586	−4.111	−0.195	−5.833	−3.523	−4.642
2200	3960	−1.484	−4.074	−0.153	−5.732	−3.627	−4.492
2300	4140	−1.391	−4.040	−0.116	−5.639	−3.722	−4.355
2400	4320	−1.305	−4.008	−0.082	−5.554	−3.809	−4.230
2500	4500	−1.227	−3.979	−0.050	−5.475	−3.889	−4.115
2600	4680	−1.154	−3.953	−0.021	−5.401	−3.962	−4.009
2700	4860	−1.087	−3.928	0.005	−5.333	−4.030	−3.911
2800	5040	−1.025	−3.905	0.030	−5.270	−4.093	−3.820
2900	5220	−0.967	−3.884	0.053	−5.210	−4.152	−3.736
3000	5400	−0.913	−3.864	0.074	−5.154	−4.206	−3.659
3100	5580	−0.863	−3.846	0.094	−5.102	−4.257	−3.584
3200	5760	−0.815	−3.828	0.112	−5.052	−4.304	−3.515
3300	5940	−0.771	−3.812	0.129	−5.006	−4.349	−3.451
3400	6120	−0.729	−3.797	0.145	−4.962	−4.391	−3.391
3500	6300	−0.690	−3.783	0.160	−4.920	−4.430	−3.334
3600	6480	−0.653	−3.770	0.174	−4.881	−4.467	−3.280
3700	6660	−0.618	−3.757	0.188	−4.843	−4.503	−3.230
3800	6840	−0.585	−3.746	0.200	−4.807	−4.536	−3.182
3900	7020	−0.554	−3.734	0.212	−4.773	−4.568	−3.137
4000	7200	−0.524	−3.724	0.223	−4.741	−4.598	−3.095

Temperature	Temperature
K	°R
600	1080	−30.678	−45.842	−16.687	−7.652	−1.377	−18.326	−31.732
700	1260	−25.898	−38.448	−13.882	−7.114	−2.023	−15.996	−27.049
800	1440	−22.319	−32.905	−11.784	−6.728	−2.518	−14.255	−23.537
900	1620	−19.540	−28.597	−10.155	−6.438	−2.910	−12.905	−20.806
1000	1800	−17.321	−25.152	−8.856	−6.213	−3.228	−11.827	−18.621
1100	1980	−15.508	−22.336	−7.795	−6.034	−3.490	−10.948	−16.834
1200	2160	−14.000	−19.991	−6.913	−5.889	−3.710	−10.216	−15.345
1300	2340	−12.726	−18.008	−6.168	−5.766	−3.897	−9.598	−14.084
1400	2520	−11.635	−16.310	−5.531	−5.664	−4.058	−9.069	−13.004
1500	2700	−10.691	−14.838	−4.979	−5.575	−4.197	−8.610	−12.068
1600	2880	−9.866	−13.551	−4.497	−5.497	−4.319	−8.210	−11.249
1700	3060	−9.139	−12.417	−4.072	−5.430	−4.426	−7.856	−10.526
1800	3240	−8.493	−11.409	−3.695	−5.369	−4.521	−7.542	−9.883
1900	3420	−7.916	−10.507	−3.358	−5.316	−4.605	−7.261	−9.308
2000	3600	−7.397	−9.696	−3.055	−5.267	−4.681	−7.009	−8.790
2100	3780	−6.929	−8.963	−2.782	−5.223	−4.749	−6.780	−8.322
2200	3960	−6.503	−8.296	−2.532	−5.183	−4.810	−4.572	−7.896
2300	4140	−6.115	−7.687	−2.306	−5.146	−4.866	−6.382	−7.507
2400	4320	−5.760–	−7.130	−2.098	−5.113	−4.916	−6.208	−7.151
2500	4500	−5.433	−6.617	−1.506	−5.081	−4.963	−6.048	−6.823
2600	4680	−5.133	−6.144	−1.730	−5.052	−5.005	−5.899	−6.520
2700	4860	−4.854	−5.706	−1.566	−5.025	−5.044	−5.762	−6.240
2800	5040	−4.596	−5.300	−1.415	−5.000	−5.079	−5.635	−5.979
2900	5220	−4.356	−4.922	−1.274	−4.977	−5.112	−5.516	−5.737
3000	5400	−4.132	−4.569	−1.142	−4.955	−5.143	−5.405	−5.511
3100	5580	−3.923	−4.239	−1.019	−4.934	−5.171	−5.300	−5.299
3200	5760	−3.728	−3.930	−0.903	−4.915	−5.197	−5.203	−5.100
3300	5940	−3.544	−3.639	−0.795	−4.897	−5.221	−5.111	−4.914
3400	6120	−3.372	−3.366	−0.693	−4.880	−5.244	−5.024	−4.738
3500	6300	−3.210	−3.108	−0.597	−4.864	−5.265	−4.942	−4.572
3600	6480	−3.056	−2.865	−0.506	−4.848	−5.285	−4.865	−4.416
3700	6660	−2.912	−2.636	−0.420	−4.834	−5.304	−4.791	−4.267
3800	6840	−2.775	−2.418	−0.339	−4.821	−5.321	−4.721	−4.127
3900	7020	−2.646	−2.212	−0.262	−4.808	−5.338	−4.655	−3.994
4000	7200	−2.523	−2.016	−0.189	−4.796	−5.353	−4.592	−3.867

Let us work with these equilibrium constants to deduce equilibrium constants for other reactions that may be of interest. For example, how does the equilibrium constant of the following two Stoichiometric reactions relate to each other?

Equilibrium Constant, KP, for Various Stoichiometric Reactions

We note that the reaction on the right depicts one half of the reaction on the left, on a molar basis, there fore the equilibrium constant on the right is the square root of the reaction on the left.

As another example, let us look at a reaction where the reactants and products are switched. The first reaction dissociates water into molecular hydrogen and oxygen, such as

And the second reaction is the reverse of the first, namely,

The equilibrium constant for the first reaction is

The second reaction has

as its equilibrium constant.

After these basic manipulations, we may want to add and subtract elementary reactions to form other reactions of interest. As an example, let us consider the following (water-gas) reaction in equilibrium:

Now, if we subtract

from the sum of the following two reactions

we get

All terms on the reactant side vanish except H₂ and we may bring the negative CO₂ from the product side to the reactant side, to get the desired reaction

Therefore, the equilibrium constant for this reaction is the product of the reactions that we added divided by the reaction that we subtracted to form the desired reaction, namely,

The adiabatic flame temperature as a function of equivalence ratio for the Jet A fuel combustion in air at 800 K initial temperature and 25 atm initial pressure is shown in Figure 7.6 (from Blazowski, 1985). There are several noteworthy features in Figure 7.6 that are instructive to our study. First, the initial temperature and pressure were chosen to represent the combustion chamber inlet conditions in a modern aircraft gas turbine engine. Second, the actual flame temperature that allows for dissociation falls below the theoretical predictions (of a complete combustion). The difference between the actual curve and theoretical predictions starts around an equivalence ratio of ∼0.4–0.5, which points to a flame temperature of ∼1650–1750 K, where dissociation reactions begin. Interestingly, a combustion chamber in a modern aircraft gas turbine engine operates in the 0.4–0.5 range of the equivalence ratio, which corresponds to a fuel-to-air ratio of ∼2.5–3.5%. Third, although the figure caption identifies the initial temperature at 800 K, we could have discerned this fact from the flame temperature at zero equivalence ratio on the graph, that is, no combustion. Fourth, as noted earlier, the maximum flame temperature occurs at an equivalence ratio slightly greater than 1 (∼1. 1).

Also, note that the burner exit (or flame) temperature is a double-valued function of equivalence ratio, or fuel-to-air ratio, that is, we can reach a burner exit temperature from either the fuel-lean or the fuel-rich side of the stoichiometric point.

The effect of initial air temperature on the stoichiometric flame temperature of a Jet A fuel combustion in air is shown in Figure 7.7a. Comparing the slope of the flame temperature rise with inlet temperature to a slope = 1.0 line drawn in Figure 7.7a for reference, we note that only ∼1/2 of the inlet temperature rise appears as the rise in the final flame temperature. This behavior corresponds to higher molar concentrations of dissociated species that appear with higher temperatures. Consequently, a 100 K increase in the inlet air temperature for this reaction results in ∼50 K final flame temperature rise. The effect of the combustion pressure on the stoichiometric flame temperature of Jet A fuel is shown in Figure 7.7b. The increasing pressure leads to a rise in the stoichiometric flame temperature. The increase is due to a suppressed dissociation levels with pressure, as shown in Example 7.4. We observed that a lower mixture pressure promoted nitrogen dissociation and it was suppressed with an increasing mixture pressure in Example 7.4.

It is instructive to examine the effect of oxidizer, as in pure oxygen or air, on the combustion of various fuels and the impact of the oxidizer on the stoichiometric flame temperature. For this purpose, Table 7.4 is reproduced from Glassman (1987) that examines the maximum temperature at equivalence ratios near stoichiometric of various fuels reacting with air or pure oxygen. Moreover, the effect of pressure on combustion temperature is reported for methane, similar to Figure 7.7b. Combustion of hydrogen in air results in a stoichiometric flame temperature of ∼2400 K at 1 atm pressure. The flame temperature for the combustion of hydrogen with pure oxygen is ∼3080 K, at 1 atm pressure. Air contains nitrogen (∼78% by volume), which for the most part is inert and does not react with the fuel or oxidizer. Also, the fuel heating value remains the same whether the oxidizer is oxygen or air. For the same energy release (per unit mass of the fuel), there are more reactants in air that do not participate in the combustion but need to be heated to the final temperature. Consequently, the flame temperature will be lower when a fuel reacts with air as opposed to pure oxygen. This behavior is observed for all (five) fuels listed in Table 7.4. Now to the effect of pressure on combustion, we note that methane reacts with air at 1 atm pressure and reaches a maximum flame temperature of 2210 K. If we increase the combustion pressure to 20 atm, the flame temperature increases to ∼2270 K. This is a rather minor increase in temperature, which is caused by a decrease in the dissociation levels of the products of combustion with pressure. At this temperature, the dissociation levels were small and hence a suppressed level of dissociation did not cause a pronounced effect in final temperature. To ascertain this point, we examine the combustion of methane with pure oxygen, which reaches a flame temperature of ∼3030 K at 1 atm pressure. The flame temperature rises to ∼3460 K when methane reacts with pure oxygen at 20 atm pressure. At the temperature of 3030 K, the level of dissociation is much higher, hence the effect of pressure on suppressing the product dissociation is more pronounced and leads to a flame temperature of ∼3460 K.

Approximate flame temperatures of various stoichiometric mixtures,
critical temperature 298 K
Fuel	Oxidizer	Pressure (atm)	T(K)
Acetylene	Air	1	2600a
Acetylene	Oxygen	1	3410b
Carbon monoxide	Air	1	2400
Carbon monoxide	Oxygen	1	3220
Heptane	Air	1	2290
Heptane	Oxygen	1	3100
Hydrogen	Air	1	2400
Hydrogen	Oxygen	1	3080
Methane	Air	1	2210
Methane	Air	20	2270
Methane	Oxygen	1	3030
Methane	Oxygen	20	3460

Source: Glassman 1987. Reproduced with permission from Elsevier

^aThis maximum exists at ø = 1.3.

^bThis maximum exists at ø = 1.7.