Thermodynamics relates heat to work. Combustion chemistry relates chemical potential energy to heat resulting from chemical reactions. This chapter presents the basic principles of combustion chemistry.

A mole is an Avagadro's number, N_A, of atoms or molecules; N_A »  6.025 ´ 10²³ molecules/mole. In chemical equations, the coefficients are in units of moles. The "molecular weight" or molar mass, , of a pure substance (element or compound, not a mixture) is the mass of a mole of molecules:

                                                                          (2.2-1)

where n is the number of moles and is the molar mass, with typical units of kg/kmol. The mass of a molecule (in grams) is its molar mass divided by N_A. (The unit abbreviation of mole is mol. The use of ~ over a quantity indicates that it is a molar quantity, in amounts per mole.)

"Atomic weights" (molar masses of atomic elements) are commonly found on the periodic table of the elements, and the molar mass of compounds can be calculated from their chemical composition. For example, the molar mass of hydrogen, H, is 1 kg/kmol and oxygen, O, is 16 kg/kmol. Then the molar mass of water, H₂O, is 18kg/kmol. Molar masses for elements deviate slightly from integer values due to isotopes. For hydrogen,  » 1.008 kg/kmol.

An intensive (specific) quantity, x, can be converted to a molar quantity, , by multiplying by its molar mass, :

                                                             (2.2-2)

The state equation of an ideal gas, (1.5-7), contains a gas constant, R, dependent on the gas. Multiplying both sides of (1.5-7) by ,

P× (× v) = (× R)× T

From (2.2-2), and ,

                                                                 (2.2-3)

where is the universal gas constant:

 »  8.3143 kJ/kg× mol× K

applies to most gases (hence is "universal") and is related to the gas constants by :

= × R                                                                 (2.2-4)

Then (2.2-3) can be expressed as

P× V = n× × T                                                         (2.2-5)

A multicomponent system is a mixture of pure substances. Each substance is a component of the system. For a mixture of ideal gases, denoting components by A and B and applying (2.2-5) to each,

                                          (2.2-6)

When gases A and B are combined, the total volume is

V = V_A + V_B                                                             (2.2-7)

and the total number of moles is

n = n_A + n_B                                                             (2.2-8)

By combining equations (2.2-5) and (2.2-6), the fractional volume or mole fraction of components is

                                             (2.2-9)

In general, Amagat’s Rule is:

                                                     (2.2-10)

The V_i are the partial volumes of the gas mixture at the same pressure and temperature. Therefore, when gases are mixed at the same P and T, the total volume is the sum of the individual component volumes.

A similar rule applies to pressures of a mixture of gases in the same volume. Again, for two components,

                                         (2.2-11)

and for mixture pressure, P, and P_A,

                                             (2.2-12)

and P = P_A + P_B. In general, we have Dalton’s Rule:

                                                 (2.2-13)

The P_i are the partial pressures of the individual component gases.

Thermodynamic properties of state for multicomponent systems can be calculated from the number of moles of each component and their individual properties. For U, H, and S:

                 (2.2-14)

Combustion is a kind of chemical reaction. In an engine, it is an exothermic (heat-releasing) reaction between a fuel like gasoline, such as isooctane, C₈H₁₈, one of the paraffin family of hydrocarbons, and oxygen.

A common rocket fuel for both amateurs and professionals is called RP-1. It is similar to kerosene but is formulated with a narrower range of densities and vapor pressures. It is also low in olefins and aromatics, which can cause carbon deposits in fuel plumbing lines. The chemical composition of RP-1 is 1.953 hydrogen atoms per carbon atom, or C_nH_1.953n.. It has a molar mass of about 175 kg/kmol. We will approximate it as C₁₂H₂₄.

Other common fuels used by rocket amateurs are methyl alcohol, CH₃OH, with a molar mass of 32.04 kg/kmol, and ethyl acohol, C₂H₅OH, with 46.07 kg/kmol.

Air is mainly oxygen and nitrogen. An approximation of air is 79% N₂ and 21% O₂. Air has approximately 78% N₂ and 1% of other gases. Atmospheric N₂ is everything in air except oxygen and therefore is not pure nitrogen. Air has an equivalent molar mass of 28.96, or approximately 29. The chemical expression for air is

O₂ + 3.773N₂                                                                     (2.3-1)

The mole ratio is 3.773 moles of atmospheric nitrogen per mole of oxygen. Air is not used as a rocket propellant because nitrogen does not react appreciably with fuel and does not contribute to thrust, yet it occupies volume in the combustion chamber and oxidizer tank. It is "dead weight." Therefore liquid oxygen is used instead of air.

The chemical equation of RP-1 reacting with liquid oxygen (LOX) is illustrative of stoichiometry: the chemistry of mass balance in chemical reactions. The combustion reaction is:

C₁₂ H₂₄ + aO₂ + ® bCO₂ + cH₂O                                             (2.3-2)

where a, b, and c are the unknown molar amounts of each component. The components on the left side are the reactants and to the right of the arrow (pronounced "yields") are the products of the reaction. Analysis of the combustion products proceeds based on conservation of matter. The amounts of each element must be the same on both sides of the reaction equation:

C: 12 = b

H: 24 = 2× c

O: 2× a = 2× b + c

Solving for the molar amounts relative to 1 mole of RP-1, the resulting reaction is:

C₁₂H₂₄ + 18O₂ ® 12CO₂ + 12H₂O                                         (2.3-3)

By mass, the oxidizer-to-fuel ratio, or O/F, is:

kg oxidizer/kg fuel                 (2.3-4)

The molar mass of O₂ is close to 32 kg/kmol and RP-1, as approximated here, is 168 kg/kmol. This ratio results in complete combustion, with no excess fuel or oxygen, and is "chemically correct" or stoichiometric. In the reaction of (2.3-2), by varying a, O/F can be varied from its stoichiometric value in (2.3-4). Then partial products of combustion appear, such as carbon monoxide, CO. At the temperature of this reaction, other reactions occur, including a reversible reaction between CO₂ and CO. The dissociation of CO₂ to CO requires high energy which reduces thrust and is undesireable. The extent of disocciation varies with temperature. A stoichiometric burn produces high temperatures but does not result in maximum thrust. (See Chapter 3 for details.)

The fractional mass composition or mass fraction of the fuel is calculated for carbon, for example, as:

fractional mass of C = x_C =                      (2.3-11)

The mass fraction of hydrogen must then be 14.3 %.

In a rocket engine, the mass flow rates of propellants times burn time gives the total required mass of each propellant. Consequently, it is more common to express O/F as the mixture ratio

                                                                 (2.3-12)

This is the ratio of oxidizer to fuel mass flow rates. The × above m indicates the rate of m, or dm/dt, in kg/s. From the mixture ratio, the total mass flow rate of propellant is

                                                         (2.3-13)

From (2.3-12) and (2.3-13), the fractional flow rates of each component are:

                                        (2.3-14)

The volumetric flow rates can be calculated from the mass flow rates by dividing by their densities. Sometimes density values are given as specific gravity, the ratio of density to that of a reference substance, usually water:

To apply the first law to a process involving a change in chemical composition, the enthalpy of elements is established at a reference state: at a standard pressure, P_A (1 atmosphere or 1 bar = 0.987 atm = 100 kPa) and temperature, T_A (25 C = 298.15 K) to be zero. The enthalpy of compounds formed as products have an enthalpy of formation h_f° at the same temperature and pressure. (The ° symbol indicates standard conditions.) Then h_f° is an offset to the relative h function for which h(25 C) = 0. Then

h = h(T - T_A) + h_f°

The h(D T) term is called the sensible enthalpy and is c_{p× D} T.

When a reaction gives off heat, the enthalpy of the product, H_P, must be less than that of the reactants, H_R. When elements in their naturally occurring composition at standard state react to form a compound, their H_R is zero by definition, and H_P is negative. In general,

                                                             (2.4-1)

and

                                                             (2.4-2)

The first law applied to reactions takes into account the heat released by the reaction as enthalpies of formation:

Q_cv + H_R = W_cv + H_P                                                 (2.4-3)

The heat produced by the reaction is the heat of reaction:

                                         (2.4-4)

The heat of reaction is also called the heating value, Q_HV, for fuels of unknown chemical composition. For known compositions, reaction energies are calculable from enthalpy or internal-energy data for individual chemical components, as in (2.4-4). When the enthalpy or internal energy of the product as a liquid is taken, it is called the higher heating value, and for a gaseous product, the lower heating value, Q_LVH.

Most reactions do not maintain constant temperature and pressure, and the enthalpies of formation are consequently affected. If ideal gas behavior is assumed, then h and u are functions of temperature alone and can be found by using c_p for h and c_v for u. For significant deviation from an ideal gas, tables of can be used directly, or formulas for c_p(T) can provide an approximation to c_p in the temperature range of the reaction. The first law, taking temperature variation into account for a control volume, becomes

                         (2.4-5)

where are molar enthalpies of formation and are the enthalpies under standard conditions. Using (2.4-3), the enthalpy of combustion is

                         (2.4-6)

If combustion of ideal gases occurs at constant volume, then the internal energy of combustion is:

                                 (2.4-7)

When all components are ideal gases, this reduces to

                                     (2.4-8)

The maximum combustion temperature is the adiabatic flame temperature. It is achieved when no work or heat transfer occurs during combustion and with complete combustion (stoichiometric mixture). From the first law, (2.4-3),

and the final temperature of the products is found by equating the two terms of (2.4-6) and solving for the temperature. This requires iteration through enthalpy tables or computer solution because h(T) is nonlinear.

The application of the second law to reacting systems can be expressed in terms of the reversible work. From (1.6-3) and (1.6-4), the maximum available work is expressed by the free energy (or Gibbs function):

                                                         (2.4-9)

                         (2.4-10)

For reactions at temperatures other than T_A,,

                     (2.4-11)

The are component enthalpies under standard conditions. This calculation can be simplified by use of the free energy of formation, , given in tables or calculated from and .

The change in free energy is the maximum reversible work available from combustion. Combustion efficiency is defined as the heat released by the reaction relative to the free energy, or

                                     (2.4-12)

For practical fuels, the difference between heating value and free energy is small.

The states of thermodynamic analysis are states of equilibrium; the thermodynamic properties of the system are unchanging. The second law can be applied to make the conditions of equilibrium more explicit. Beginning from eqn. (1.5-3),

G = H - TS = U + PV -  TS                                                 (2.5-1)

Then

dG = dU + PdV + VdP - TdS - SdT                                         (2.5-2)

Heat transfer can be expressed as

d Q = dU + PdV                                                                 (2.5-3)

Applying the second law as the law of increase of entropy, (1.4-12), and substituting (2.5-3):

d Q £ TdS  Þ   dU + PdV - TdS £ 0                                             (2.5-4)

The left side of (2.5-4) is dG for constant T and P:

                                                             (2.5-5)

In other words, G decreases until it reaches a minimum value at equilibrium. Then

equilibrium  Þ   minimum G at constant T and P

Or, at equilibrium,

                                                                 (2.5-6)

A system in equilibrium is incapable of doing work on itself. In isolation, W_rev = 0. From (2.4-9), G_R = G_P , or dG = 0, the same result as (2.5-6). From (2.5-1), we can also conclude that at equilibrium, entropy is maximized in an isolated system. Because H is a function of temperature only, for a given T, G is minimum when S is maximum.

A similar derivation to that involving free energy, but applied to the Helmholtz function, A = U - TS, results in the condition for equilibrium for constant volume and temperature:

                                                                 (2.5-7)

The free energies of two phases of a pure substance in equilibrium are equal. To show this, begin with (1.5-19):

For constant T and P, this reduces to:

For the process of vaporization,

                                         (2.5-8)

or

h_l  - T× s_l  = h_g - T× s_g  Þ  g_l = g_g                                         (2.5-9)

In phase equilibrium, liquid and vapor free energies are the same.

Using (2.5-9), the Clapeyron equation, (1.5-17), can be derived from this equilibrium condition. Beginning with (1.5-8) for a simple compressible substance,

dg = v× dP - s× dT                                                         (2.5-10)

For a pressure change, dP, dT also changes, but dg remains constant, or

dg_l  = dg_g                                                                 (2.5-11)

and substituting,

v_l× dP - s_l× dT = v_g× dP - s_g× dT                                            (2.5-12)

Rearranging,

which is Clapeyron's equation.

In multicomponent, multiphase systems, Raoult's rule, an empirical formula, relates the partial pressures to the mole fractions of the two phases for each component. Let x_i be the mole fraction of liquid for the ith component, and similarly, let y_i be the vapor mole fraction. For partial pressures P_i and total pressure P, Raoult's rule is:

Raoult's rule:  x_i × P_i (sat) = y_{i× ×} P = P_i , T constant                         (2.5-13)

where P_i (sat) are the pressures at saturation. Raoult's rule is linear whereas actual gases are sublinear for ; the partial pressures of each gas in a two-component system will be less than Raoult's rule gives when the difference in their liquid mole fractions is not large.

The conditions for equilibrium for a multicomponent, multiphase system are that the partial molar free energies of each component, , be the same for all phases. The more common name for is the chemical potential,

                                                 (2.5-14)

D m between phases of a component causes mass transfer between them (just as an electrical potential difference causes charge transfer). Applying (2.5-9) to multiple components, each must be in phase equilibrium for the system to be in equilibrium. From (2.5-6), and for constant T and P,

                                         (2.5-15)

The conservation of mass for a closed system requires that the sum of n_i, or N, remain constant. For a two-phase system,

n_l + n_g = N = constant  Þ    dn_l = - dn_g                                 (2.5-16)

For dn_i ¹  0, must equal for equilibrium, or , between phases. This is extended for each component of a multicomponent system.

We now consider chemical equilibrium of ideal gases. Many reactions are reversible: they can proceed backwards, from products to reactants, as well as forwards. The general reversible reaction is:

                                             (2.6-1)

where the n _i are the coefficients that establish the molar proportions of the components, A_i. The conditions for chemical equilibrium are:

1. thermodynamic equilibrium
2. unchanging amounts of components

Conservation of matter requires that

                                             (2.6-2)

In other words, the changes in molar amounts of the components must vary together to maintain mass balance. If there are two moles of reactant 2 for each mole of reactant 1 (n ₂ = 2× n ₁), then any change in the moles of 1 must be half that of 2 (dn₁ = ¹/_2× dn₂). The coefficients of components in chemical equations can be scaled by the same proportionality factor, and this also applies to changes in amounts of components. In reversible reactions, a positive change (or gain) in mass on one side of the equation is a negative change (or loss) to the other side.

Substituting for all but dn₁ into (2.5-15) from (2.6-2),

Simplifying,

                                         (2.6-3)

Then the condition for chemical equilibrium is:

                                                         (2.6-4)

Thus G is minimized, taking the n_i into account, and

dG_R = - dG_P

For ideal gas reactions, the equilibrium condition leads to another important result. Beginning with (1.6-8),

at a reference temperature, T_A . From the ideal gas law, (1.9-3),

                                         (2.6-5)

where . When, from (2.6-5),

is applied to (2.6-4),

where i covers both products and reactants. Written out, this is:

                                         (2.6-6)

This reduces to

                                             (2.6-7)

where the equilibrium constant is defined as

                                                     (2.6-8)

Eqn. (2.6-7) can then be written more compactly as

                                                     (2.6-9)

and is the law of mass action. K_p is a function of temperature only since D G_A is a function of temperature only. (For real gases, K_p is also a very weak function of pressure). K_p can be expressed in mole fractions, x_i = P_i /P, as

                                     (2.6-10)

Equilibrium applies to engine combustion at high temperatures, where significant dissociation of products occurs. As temperature increases, gas molecules increase in kinetic energy, including vibrations between atoms within a molecule. At sufficiently high temperatures, a significant number of atoms break free, or dissociate, from their molecules. A particular reaction involving dissociation is:

                                                     (2.6-11)

The energy released by the combustion reaction

C + O₂  ® CO₂

heats the CO product, causing dissociation. The two processes are combustion followed by heating and dissociation of CO₂. The heat of combustion is transferred to the dissociation process so that CO₂ leaves the first process at ambient temperature:

The overall reaction is:

C + O₂ ® (1 - z)CO₂ + zCO + ^z/₂O₂                                     (2.6-12)

z is the number of moles of dissociated CO₂ per mole of CO. The total number of moles per mole of CO at equilibrium is

                                                 (2.6-13)

Then the mole fractions of products are:

                             (2.6-14)

From a table of equilibrium constants, K_p (T_H ) is found. Then

or

                                                     (2.6-15)

Next, solve for z subject to the constraint

0 £ z £ 1

Then substitute z to find the n values.

The heat transfer from the combustion chamber to the surroundings can be calculated by applying the first law:

H_R = H_P - Q_cv                                                         (2.6-16)

For P = 1 atm and with tabulated enthalpies of formation,

,                                                 (2.6-17)

         (2.6-18)

Then substituting (2.6-17) and (2.6-18) into (2.6-16), the heat release, Q_cv, results. The effect of dissociation on engine performance is small but not negligible.