1. Mass and molar fractions
yi – mass fraction of species i
xi – molar fraction of species i
mi – mass of species i [kg]
ni – number of moles of species i
Mi – molar mass of species i [kg/kmol]
N – total number of species
Thermochemistry
Combustion

2. Mass and molar concentrations
[Xi] – molar concentration of species i [kmol/m3]
In combustion problems, chemical species may generally be treated as perfect gases satisfying the ideal gas law
p V = n Ro T pi V = ni Ro T Ro= J/kmol K : universal gas constant
p v = R T pi vi = Ri T Ri : gas constant of species i
R = Ro / Mmist Ri = Ro / Mi
p – total pressure [atm]
V – total volume [m3]
pi – partial pressure of species i [N/m2]
v – specific volume [m3/kg]
Thermochemistry
Combustion

3. Mass and molar concentrations
Relations between molar concentrations and molar fractions
Relations between molar concentrations and mass fractions
Mass concentration of species i, ri [kg/m3]
Thermochemistry
Combustion

4. Stoichiometry Thermochemistry
A reactive mixture of a fuel and an oxidant is referred to as stoichiometric when the mass of oxidant is the theoretically necessary and sufficient to fully burn the fuel in an ideal combustion process.
Stoichiometric chemical reaction of an hydrocarbon burning in air:
CxHy + (x + y/4) (O N2)  x CO2 + (y/2) H2O (x + y/4) N2
(The mole fractions of O2 and N2 in air are 21% and 79%, respectively, yielding 79/21=3.76 moles of N2 per mole of O2)
The stoichiometric air/fuel ratio for an hydrocarbon is given by
Thermochemistry
Combustion

5. Stoichiometry Thermochemistry Equivalence ratio:
ox stands for oxidant (typically, the air), rather than oxygen
f < 1 – fuel lean mixture
f = 1 – stoichiometric mixture
f > 1 – fuel rich mixture
When the oxidant is the air, the excess air coefficient is defined as
Excess air (in percentage):
Thermochemistry
Combustion

6. Enthalpy of formation Thermochemistry
Enthalpy of formation of a compound, hf [J/kg], is the net change of enthalpy that occurs in the reaction of formation of a unit of mass of that compound from its constituent elements in their natural states (at given pressure and temperature).
Molar enthalpy of formation of a compound, [J/kmol], is the net change of enthalpy that occurs in the reaction of formation of one mole of that compound from its constituent elements in their natural states (at given pressure and temperature).
The natural state of an element is the most stable state (gas, liquid or solid) of that element, i.e., the state at which that element is found in nature.
Standard reference state conditions: Tref = 25 ºc, pref=po = 1 atm (superscript o denotes atmospheric pressure)
The standard molar enthalpy of formation for several substances is given in Table A1.1 of Appendix 1 of the book.
Thermochemistry
Combustion

7. Enthalpy of formation Thermochemistry
The enthalpy of formation of chemical elements in their natural state is arbitrarily set to zero, e.g., O2(g) , H2(g) , N2(g) , Hg(l) , C(s)
The enthalpy of formation of a compound is negative when the chemical reaction of that compound from its constituent elements releases energy, e.g.,
C (s) + O2 (g)  CO2 (g) (Tref=25 ºC) = kJ/kmol (see Table A1.1 for CO2)
In contrast, the enthalpy of formation is positive when the reaction of formation absorbs energy, e.g.,
O2 (g)  2 O (g) (Tref=25 ºC) = kJ/kmol (see Table A1.1 for O)
A substance with high enthalpy of formation is more chemically reactive than another one with a lower enthalpy of formation
Thermochemistry
Combustion

8. Absolute enthalpy Thermochemistry
The absolute enthalpy (specific or molar) of a substance is defined as:
hs – Sensible enthalpy, defined as the
difference between the enthalpy at T and at Tref
In the case of a perfect gas, the sensible enthalpy (specific or molar) is given by
cp,i [J/kg.K]– specific heat capacity at constant pressure
The specific and the molar enthalpies of an ideal mixture are given by
Thermochemistry
Combustion

9. Enthalpy of reaction and enthalpy of combustion
In systems with constant pressure or constant volume, the change of energy in a chemical reaction depends only on the initial and final states of the system.
In an open system at constant pressure, the change of energy in a chemical reaction is equal to the change of enthalpy of the mixture (neglecting variation of potential and kinetic energy, as well as the work done on the system or by the system, except work due to pressure forces).
When the reactants and the products are at the same temperature and pressure, that change of energy is referred to as enthalpy of reaction and, in the case of combustion reactions, by enthalpy of combustion.
Specific enthalpy of reaction/combustion [J/kg]
Molar enthalpy of reaction/combustion [J/kmol]
Total enthalpy of reaction/combustion [J]
Thermochemistry
Combustion

10. Enthalpy of reaction and enthalpy of combustion
The enthalpy of combustion is always negative, since the combustion reactions are exothermic.
At standard reference state conditions (T = 25 ºC and p = 1 atm), we have the standard enthalpy of reaction and the standard enthalpy of combustion.
Hess’s law: the heat released or absorbed in a chemical process is the same whether the process takes place in one or in several steps. Example:
Given C (s) + O2 (g)  CO2 (g) = kJ/kmol
and CO (g) + ½ O2 (g)  CO2 (g) = kJ/kmol
Then C (s) + ½ O2 (g)  CO (g) =
= kJ/kmol
The enthalpy of reaction/combustion depends on the temperature
Thermochemistry
Combustion

11. Enthalpy of reaction and enthalpy of combustion
The enthalpies of combustion at two different temperatures may be related as follows:
If T1= 25 ºC and p = 1 atm:
Thermochemistry
Combustion

12. Internal energy of formation, reaction and combustion
The enthalpies of formation, reaction and combustion are used to determine changes of energy in systems at constant pressure (e.g., combustion chambers, boilers, furnaces, gas turbine combustors)
The internal energies of formation, reaction and combustion are used to determine changes of energy in systems at constant volume (e.g., spark ignition engines)
The enthalpy and the internal energy of reaction may be related as follows:
Thermochemistry
Combustion

13. Heating value of a fuel Thermochemistry
Lower heating value at constant pressure = symmetric of the standard enthalpy of combustion, per unit mass of fuel, when there is water vapour in the combustion products (always positive)
Lower heating value at constant volume = symmetric of the standard internal energy of combustion, per unit mass of fuel, when there is water vapour in the combustion products (always positive)
Higher heating value at constant pressure = symmetric of the standard enthalpy of combustion, per unit mass of fuel, when there is liquid water in the combustion products (always positive)
Higher heating value at constant volume = symmetric of the standard internal energy of combustion, per unit mass of fuel, when there is liquid water in the combustion products (always positive)
The difference between the higher and the lower heating value is equal to product of the mass of water (liquid or gas) in the combustion products by the latent heat of vaporization (hfg for constant pressure or ufg for constant volume)
Thermochemistry
Combustion

14. Adiabatic flame temperature
The adiabatic flame temperature is the temperature of the combustion products in an adiabatic combustion process with negligible variation of the kinetic and potential energy and with no work exchange with the surroundings.
The adiabatic flame temperature is an upper limit of the temperature in a combustion process, the actual temperature of the combustion products being generally lower due to heat losses by conduction, convection or radiation.
The adiabatic flame temperature in a combustion process at constant pressure with the reactants at temperature T1 is obtained from:
Thermochemistry
Combustion

15. Adiabatic flame temperature
Knowledge of the chemical composition of the combustion products is required to determine the adiabatic flame temperature.
At typical adiabatic flame temperatures of hydrocarbons (close to 2300 K), dissociation of the combustion products occurs, and the chemical composition of the products must be determined together with the adiabatic flame temperature. For the time being, dissociation of the combustion products will be ignored.
The specific heat capacities as a function of temperature may be accurately approximated by polynomials, which are given in table A2.1 for several common species. In the case of a manual calculation, it is sufficient to take constant values evaluated at the average temperature (Tref+Tad)/2. Instead, the sensible enthalpies given in tables A2.2 to A2.13 may be used.
The adiabatic flame temperature is a function of the temperature of the reactants and of the equivalence ratio. If there is no dissociation, the adiabatic flame temperature is maximum for a stoichiometric mixture.
Thermochemistry
Combustion

16. Adiabatic flame temperature
The adiabatic flame temperature in a combustion process at constant volume with the reactants at temperature T1 is obtained from:
The ideal gas equation of state must be solved together with the above equation, since the pressure of the combustion products is unknown, and it influences the chemical composition of the combustion products
Thermochemistry
Combustion

17. Dissociation Thermochemistry
Chemical species such as CO2, H2O, N2 and O2 are generally stable up to about 1250 K. At higher temperatures, dissociation occurs, yielding other species such as CO, H2, OH, H, O, etc., which are formed in reactions like
CO2  CO + ½ O2 (symbol  denotes a reversible reaction)
The proportion of CO2, CO and O2 at a particular temperature and pressure adjusts itself such that, after some time, the reaction rates of the direct and inverse reactions are equal, i.e., the number of moles of CO2 that dissociate, according to the direct reaction, is equal to the number of moles that are formed, according to the inverse reaction, per unit of time.
A chemical equilibrium state is then achieved,
with the direct and inverse reactions proceeding
at the same rate.
Mass balances are sufficient to determine the
chemical composition of the combustion products
only if no dissociation occurs.
Thermochemistry
Combustion

18. Chemical equilibrium criteria
The chemical equilibrium condition is obtained by combining the 1st and 2nd laws of thermodynamics:
dU = dQ – p dV (1st law for a closed system, assuming that there is only work done by pressure forces)
(2nd law)
Therefore,
It can be shown that for a reactive system in chemical
equilibrium the reactions are reversible. It follows that:
The equilibrium condition for a process with U and V constants is the maximization of the entropy of the system;
The equilibrium condition for a process with S and V constants is the minimization of the internal energy of the system.
Thermochemistry
Combustion

19. Chemical equilibrium criteria
Processes with U and V constant or with S and V constants do not generally occur in combustion systems. In these systems, the Gibbs free energy, G, and the Helmholtz function, F, are more common:
G = H –T S
F = U –T S
The equilibrium condition for a process with p and T constants is the minimization of the Gibbs free energy of the system.
The equilibrium condition for a process with V and T constants is the minimization of the Helhmoltz function of the system.
All chemical equilibrium criteria are equivalent, i.e., all criteria yield the same solution, i.e., the same chemical equilibrium composition of a reactive system.
Thermochemistry
Combustion

20. Chemical equilibrium Thermochemistry
Free Gibbs energy for an isothermal system in chemical equilibrium
Go [J] -free Gibbs energy at po=1 atm
[J/kmol] - molar Gibbs energy at T and po
(Tref) [J/kmol] –standard free Gibbs energy of formation
[J/kmol] – sensible molar Gibbs free energy
Thermochemistry
Combustion

21. Chemical equilibrium Thermochemistry
The Gibbs free energy of formation is equal to zero for chemical elements in their natural state
The Gibbs free energy of formation of several species is given in Tables A2.2 to A2.13
Differentiation of the free Gibbs energy yields
The last term is zero for T and p constant (why?)
Thermochemistry
Combustion

22. Chemical equilibrium Thermochemistry
Hence, the chemical equilibrium condition may be written as
Let’s consider the following general reaction in chemical equilibrium
a A + b B  e E + f F +...
A, B, ..., E, F, ... – chemical species
a, b, …, e, f, … – stoichiometric coefficients
Thermochemistry
Combustion

23. Chemical equilibrium Thermochemistry Chemical equilibrium condition:
Equilibrium constant:
(with pA, pB, …, pE, pF, … in atm)
The equilibrium constants for the reactions of formation of several species from their constituent elements in their natural state are given in table A4.1
The equilibrium constants of other reactions may be determined from the tabulated equilibrium constants by means of algebraic manipulations.
Thermochemistry
Combustion

24. Equilibrium constants
Example: calculate the equilibrium constant of reaction
H2O  H + OH
The following equilibrium constants are tabulated:
H2 + ½ O2  H2O
½ H2  H
½ H2 + ½ O2  OH
The equilibrium constant of the reaction given above is then
Thermochemistry
Combustion

25. Equilibrium constants
The equilibrium constant may be explicitly obtained from as
The equilibrium constant, as well as the change of the Gibbs free energy, indicates the trend of conversion of reactants into products or, equivalently, the conversion of products into reactants.
If Kp<1, then DGo(T)>0  there are more reactants than products in equilibrium.
If Kp>1, then DGo(T)<0  there are more products than reactants in equilibrium.
Accordingly, the larger Kp, the smaller DGo(T), and the greater the amount of products relatively to reactants under equilibrium conditions.
The equilibrium constant Kp is independent of the total pressure, being only a function of the temperature
Thermochemistry
Combustion

26. Equilibrium constants
The change of the Gibbs free energy may be expressed as
yielding
Assuming that DHo(T) and DSo(T) change little with the temperature (this is a good approximation, particularly for small or moderate temperature changes), then ln(Kp) changes linearly with the inverse of temperature.
The relation between the equilibrium constants at two different temperatures may then be expressed as
Exothermic reaction: DHo(T) < 0, T   Kp 
Endothermic reaction: DHo(T) > 0, T   Kp 
Thermochemistry
Combustion

27. Equilibrium constants
The equilibrium constant may also be expressed in terms of mass fractions, molar fractions, molar concentrations and total number of moles of every species, e.g.
In contrast with Kp, the equilibrium constants Kc and Kn depend on both temperature and total number of moles, i.e., total pressure.
Thermochemistry
Combustion

28. Le Châtelier´s principle
If a system in chemical equilibrium is disturbed (e.g., by changes in temperature, pressure, chemical composition), the system will tend to shift its equilibrium state so as to minimize or counteract the effect of the disturbance.
If there is an increase of pressure, keeping the temperature unchanged, the system will reduce the pressure by producing fewer moles.
Example: CO2  CO + ½ O2
An increase of pressure will favour the inverse reaction.
The system will shift to an equilibrium state with lower CO and O2 and higher CO2 mole fractions than originally.
Thermochemistry
Combustion

29. Le Châtelier´s principle
If there is an increase of temperature, keeping the pressure unchanged, the system will reduce the temperature by favouring the endothermic reaction. Similarly, a decrease of temperature favours the exothermic reaction.
p = 1 atm
p=10 atm
T =1500 K
xCO2
0.9994
0.9997
xCO
3.601×10-4
1.672×10-4
xO2
1.801×10-4
8.357×10-5
T = 2000 K
0.9777
0.9895
0.0149
6.96×10-3
0.0074
3.48×10-3
Thermochemistry
Combustion

30. Calculation of chemical equilibrium composition
Whenever dissociation is present, the calculation of the adiabatic flame temperature and the determination of the chemical equilibrium composition require the specification of all the chemical species
At least the following species are present in the combustion of hydrocarbons:
CO2, H2O, O2, CO, H2, OH, O, H, and for combustion in air, N2, N and NO. Other species are also generally present, such as HO2, H2O2, N2O, NO2, C(s), HCN, CH4 or C2H2.
Formulation of the chemical equilibrium problem for a system at constant pressure:
N species  N+1 equations (to determine molar fractions and adiabatic flame temperature)
1 equation for energy conservation:
K equations for atomic mass balances (typically O, H, C, N)
N-K linearly independent equations to be found from chemical equilibrium
Thermochemistry
Combustion

31. Calculation of chemical equilibrium composition
K independent species out of the N species are selected in such a way that none of those independent species may be transformed in another one of them by means of chemical reaction. Moreover, the remaining N-K species should be obtained by chemical reaction from the selected K independent species.
The N-K equations are determined from the condition of chemical equilibrium of the reactions of formation of the N-K species from the selected K independent species.
Example: combustion of methane in air with N = 11 species present under equilibrium - CO2, H2O, O2, CO, H2, OH, O, H, N2, N and NO.
and nar are given, so that there are N = 11 unknown mole fractions
Thermochemistry
Combustion

32. Calculation of chemical equilibrium composition
4 equations may be written for the atomic mass balances:
The following 4 species are selected as independent: O2, H2, N2 and CO
The remaining 7 equations are obtained from the condition of chemical equilibrium of the reactions of formation of the other 7 species from the 4 independent species
Thermochemistry
Combustion

33. Calculation of chemical equilibrium composition
CO + ½ O2  CO2
H2 + ½ O2  H2O
½ H2 + ½ O2  OH
½ O2  O
½ H2  H
½ N2  N
½ N2 + ½ O2  NO
Thermochemistry
Combustion

34. Chemical equilibrium software
PER (Olikara and Borman, 1975) – Calculates the chemical equilibrium composition and the adiabatic flame temperature for the combustion of CxHyOzNw in air considering the 11 species of the previous example (a modified version of this code is available in the book by Turns)
The determination of the equilibrium composition using the reaction constants has a few disadvantages (e.g., the need to specify the chemical reactions in equilibrium, numerical difficulties when there are very small concentrations, difficulty in dealing with problems with solid or liquid phases)
CEA (Gordon and McBride, 1994) – code developed by NASA and based on the minimization of the Gibbs free energy using Lagrange multipliers. See:
STANJAN (Reynolds, 1986) – code based on the method of chemical potentials of the elements. Online version available at
Thermochemistry
Combustion

35. Examples of application of chemical equilibrium software
DADOS DE ENTRADA
Número de átomos de carbono no combustível 3
Número de átomos de hidrogénio no combustível 8
Número de átomos de oxigénio no combustível 0
Número de átomos de azoto no combustível 0
Razão de equivalência 1.000
Temperatura (K) – estimativa inicial
Pressão (Pa)
Entalpia dos reagentes (kJ/kmol combustível)
DADOS DE SAÍDA
Temperatura da chama e propriedades dos produtos da combustão
Temperatura adiabática de chama [K] =
Entalpia da mistura [J/kg] = E+06
Calor específico da mistura cp [J/kg-K] = E+04
Razão entre os calores específicos, cp/cv =
Massa molar da mistura [kg/kmol] =
Número de moles de combustível por mole de produtos =
Fracções molares das espécies em condições de equilíbrio
H: O: N:
H2: OH: CO:
NO: O2: H2O:
CO2: N2:
Example of application of the PER code
Thermochemistry
Combustion

36. Examples of application of chemical equilibrium software
Thermochemistry
Combustion

37. Examples of application of chemical equilibrium software
Thermochemistry
Combustion

38. Examples of application of chemical equilibrium software
Thermochemistry
Combustion