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.
2.2 Multicomponent systems
A mole is an Avagadro's number, NA, of atoms or molecules; NA » 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:
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 NA. (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, :
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 ,
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)
The symbolR is also used for in the literature.
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,
When gases A and B are combined, the total volume is
V = VA + VB (2.2-7)
and the total number of moles is
n = nA + nB (2.2-8)
By combining equations (2.2-5) and (2.2-6), the fractional volume or mole fraction of components is
In general, Amagat’s Rule is:
The Vi 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,
and for mixture pressure, P, and PA,
and P = PA + PB. In general, we have Dalton’s Rule:
The Pi 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:
2.3 Chemical reactions and combustion stoichiometry
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 CnH1.953n.. 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 as oxidizer for rockets
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 = xC = (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
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
From (2.3-12) and (2.3-13), the fractional flow rates of each component are:
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:
Water has a density of 1000 kg/m3 or 1 g/cm3 (1g/cc). RP-1 has a density of 0.8 g/cm3 and LOX has 1.2 g/cm3.
2.4 The first and second laws for reactions
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, PA (1 atmosphere or 1 bar = 0.987 atm = 100 kPa) and temperature, TA (25 C = 298.15 K) to be zero. The enthalpy of compounds formed as products have an enthalpy of formation hf° at the same temperature and pressure. (The ° symbol indicates standard conditions.) Then hf° is an offset to the relative h function for which h(25 C) = 0. Then
h = h(T - TA) + hf°
The h(D T) term is called the sensible enthalpy and is cp× D T.
When a reaction gives off heat, the enthalpy of the product, HP, must be less than that of the reactants, HR. When elements in their naturally occurring composition at standard state react to form a compound, their HR is zero by definition, and HP is negative. In general,
The first law applied to reactions takes into account the heat released by the reaction as enthalpies of formation:
Qcv + HR = Wcv + HP (2.4-3)
The heat produced by the reaction is the heat of reaction:
The heat of reaction is also called the heating value, QHV, 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, QLVH.
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 cp for h and cv for u. For significant deviation from an ideal gas, tables of can be used directly, or formulas for cp(T) can provide an approximation to cp in the temperature range of the reaction. The first law, taking temperature variation into account for a control volume, becomes
where are molar enthalpies of formation and are the enthalpies under standard conditions. Using (2.4-3), the enthalpy of combustion is
If combustion of ideal gases occurs at constant volume, then the internal energy of combustion is:
When all components are ideal gases, this reduces to
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),
HR = HP
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):
For reactions at temperatures other than TA,,
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
For practical fuels, the difference between heating value and free energy is small.
2.5 Phase and chemical equilibrium
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)
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 Ž dU + PdV - TdS £ 0 (2.5-4)
The left side of (2.5-4) is dG for constant T and P:
In other words, G decreases until it reaches a minimum value at equilibrium. Then
equilibrium Ž minimum G at constant T and P
Or, at equilibrium,
A system in equilibrium is incapable of doing work on itself. In isolation, Wrev = 0. From (2.4-9), GR = GP , 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:
The free energies of two phases of a pure substance in equilibrium are equal. To show this, begin with (1.5-19):
T× ds = dh - vdP
For constant T and P, this reduces to:
T× ds = dh
For the process of vaporization,
hl - T× sl = hg - T× sg Ž gl = gg (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
dgl = dgg (2.5-11)
vl× dP - sl× dT = vg× dP - sg× dT (2.5-12)
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 xi be the mole fraction of liquid for the ith component, and similarly, let yi be the vapor mole fraction. For partial pressures Pi and total pressure P, Raoult's rule is:
Raoult's rule: xi × Pi (sat) = yi× × P = Pi , T constant (2.5-13)
where Pi (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,
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,
The conservation of mass for a closed system requires that the sum of ni, or N, remain constant. For a two-phase system,
nl + ng = N = constant Ž dnl = - dng (2.5-16)
For dni ¹ 0, must equal for equilibrium, or , between phases. This is extended for each component of a multicomponent system.
2.6 Gas equilibrium
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:
where the n i are the coefficients that establish the molar proportions of the components, Ai. The conditions for chemical equilibrium are:
1. thermodynamic equilibrium
2. unchanging amounts of components
Conservation of matter requires that
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),
Then the condition for chemical equilibrium is:
Thus G is minimized, taking the ni into account, and
dGR = - dGP
For ideal gas reactions, the equilibrium condition leads to another important result. Beginning with (1.6-8),
dg = v× dp - s× dT
at a reference temperature, TA . From the ideal gas law, (1.9-3),
where . When, from (2.6-5),
is applied to (2.6-4),
where i covers both products and reactants. Written out, this is:
This reduces to
where the equilibrium constant is defined as
Eqn. (2.6-7) can then be written more compactly as
and is the law of mass action. Kp is a function of temperature only since D GA is a function of temperature only. (For real gases, Kp is also a very weak function of pressure). Kp can be expressed in mole fractions, xi = Pi /P, as
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:
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
Then the mole fractions of products are:
From a table of equilibrium constants, Kp (TH ) is found. Then
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:
HR = HP - Qcv (2.6-16)
For P = 1 atm and with tabulated enthalpies of formation,
Then substituting (2.6-17) and (2.6-18) into (2.6-16), the heat release, Qcv, results. The effect of dissociation on engine performance is small but not negligible.