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Mass and molar fractions yi mi mtot mi xi N m ni ntot j j 1 N ni 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 N n j j 1 N y x i i 1 i 1 i 1 mi ni M i mtot ntot M mist N M mist m tot ntot n j j 1 ntot Mj N n j 1 nj N Mj tot x j Mj M mist j 1 mtot m N tot ntot nj j 1 yi mi ni M i Mi xi mtot ntot M mist M mist xi M i N j 1 Thermochemistry xj M j xi ni ntot mtot N 1 N M M mj j j 1 mi yi Mi Mi mtot 1 M mist M mist j 1 yj j yi Mi N M j 1 yj j Combustion Mass and molar concentrations X i ni [Xi] – molar concentration of species i [kmol/m3] V 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=8 314.3 J/kmol K : universal gas constant pv=RT 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] p i ni xi p n Thermochemistry Combustion Mass and molar concentrations Relations between molar concentrations and molar fractions X i xi xi X i p Ro T Ro T p X i X N j j 1 Relations between molar concentrations and mass fractions X i p Ro T yi Mi N yj Mj j 1 y i Mi yi X i M i X M N j j j 1 Mass concentration of species i, i [kg/m3] i mi mi mtot yi V V mtot Thermochemistry Combustion Stoichiometry • 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) (O2 + 3.76 N2) x CO2 + (y/2) H2O + 3.76 (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 mar x y 4 M O2 3.76 M N 2 M 4.76 x y 4 ar m M fu M fu fu esteq Thermochemistry Combustion Stoichiometry • Equivalence ratio: m fu mox m fu m ox esteq ox stands for oxidant (typically, the air), rather than oxygen < 1 – fuel lean mixture = 1 – stoichiometric mixture > 1 – fuel rich mixture • When the oxidant is the air, the excess air coefficient is defined as 1 • Excess air (in percentage): Thermochemistry mar m fu mar m fu esteq mar mar m fu m fu 1 esteq e % 100% 100% mar m fu esteq Combustion Enthalpy of formation • 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, h f [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 denotes atmospheric pressure) o • The standard molar enthalpy of formation h fo Tref for several substances is given in Table A1.1 of Appendix 1 of the book. Thermochemistry Combustion Enthalpy of formation • 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) h fo (Tref=25 ºC) = -393 522 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) h fo (Tref=25 ºC) = 249 173 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 Absolute enthalpy • The absolute enthalpy (specific or molar) of a substance is defined as: hi T h fo,i Tref hs ,i T hi T h of ,i Tref hs ,i T • In the case of a perfect gas, the sensible enthalpy (specific or molar) is given by hs ,i T T hs ,i T T Tref Tref • hs – Sensible enthalpy, defined as the difference between the enthalpy at T and at Tref c p,i T * dT * cp,i [J/kg.K]– specific heat capacity at constant pressure c p,i T * dT * The specific and the molar enthalpies of an ideal mixture are given by N hmist y i hi i 1 Thermochemistry N hmist xi hi i 1 Combustion 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. hR T h prod (T ) hreag T Specific enthalpy of reaction/combustion [J/kg] hR T h prod (T ) hreag T Molar enthalpy of reaction/combustion [J/kmol] H R T H prod (T ) H reag T Total enthalpy of reaction/combustion [J] Thermochemistry Combustion 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) hRo = -393 522 kJ/kmol and CO (g) + ½ O2 (g) CO2 (g) hRo = -282 995 kJ/kmol Then C (s) + ½ O2 (g) CO (g) hRo = -393 522 + 282 995 = -110 527 kJ/kmol • The enthalpy of reaction/combustion depends on the temperature Thermochemistry Combustion Enthalpy of reaction and enthalpy of combustion • The enthalpies of combustion at two different temperatures may be related as follows: H R T2 mi T1 i reag H R T2 ni i reag T2 T2 T1 c p,i T dT H R T1 mj T1 j prod c p,i T dT H R T1 T2 nj T2 T1 j prod c p, j T dT c p, j T dT • If T1= 25 ºC and p = 1 atm: H R T2 j prod m j h of , j Tref T 2 ref H Ro Tref mj j prod Thermochemistry T T2 Tref c p , j T dT c p , j T dT i reag mi h of ,i Tref T 2 ref i Tref mi T2 T c p ,i T dT c p ,i T dT reag Combustion 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: H R T U R T pV U R T RoT n prod nreag U R T U prod T U reag T j prod m j u of , j Tref T ref T m j T ref T U Ro Tref j prod Thermochemistry cv , j T dT cv, j T dT i reag mi u of ,i Tref T ref i Tref mi T T cv ,i T dT cv ,i T dT reag Combustion Heating value of a fuel • 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 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: H prod Tad H reag T1 o o m j h f , j Tref Trefad c p , j T dT mi h f ,i Tref Tref1 c p ,i T dT j prod T i reag T H Ro Tref m j Trefad c p , j T dT mi Tref1 c p ,i T dT T j prod Thermochemistry T i reag Combustion 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 Adiabatic flame temperature • The adiabatic flame temperature in a combustion process at constant volume with the reactants at temperature T1 is obtained from: U prod Tad U reag T1 H prod Tad H reag T1 U prod Tad U reag T1 V p 2 p1 H prod Tad H reag T1 Ro n prod T prod nreag Treag • 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 Dissociation • 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 Chemical equilibrium criteria • The chemical equilibrium condition is obtained by combining the 1st and 2nd laws of thermodynamics: dU = dQ – p dV dQ dS T rev • Therefore, (1st law for a closed system, assuming that there is only work done by pressure forces) dQ dS T irrev dS U ,V 0 (2nd law) dU S ,V 0 • 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 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 dG p,T dF V ,T 0 0 • 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 Chemical equilibrium dG dH TdS SdT dU pdV Vdp TdS SdT dQ Vdp TdS SdT • Free Gibbs energy for an isothermal system in chemical equilibrium dG V dp n Ro T dp n Ro T d ln p p GT , p G o T n Ro T ln p p o GT , p N n i g i T , p i 1 n g N i o i Go [J] -free Gibbs energy at po=1 atm g io T [J/kmol] - molar Gibbs energy at T and po T Ro T ln pi po i 1 g io T g io, f Tref g io,s T gio, s T d gio T Tref Thermochemistry g io, f (Tref) [J/kmol] –standard free Gibbs energy of formation g io, s [J/kmol] – sensible molar Gibbs free energy Combustion Chemical equilibrium • 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 dGT , p N dni g io i 1 T Ro T ln pi n d g N p o i o i T Ro T ln pi po i 1 The last term is zero for T and p constant (why?) Thermochemistry Combustion Chemical equilibrium Hence, the chemical equilibrium condition may be written as dGT , p dn g N i o i T Ro T ln pi po 0 i 1 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 dn A dn dn dn B E F a b e f e g Eo T Ro T ln pE p o f g Fo T Ro T ln pF p o 0 o T Ro T ln p A p o b g Bo T Ro T ln pB p o a gA Thermochemistry Combustion Chemical equilibrium Chemical equilibrium condition: p po o o o o a g A T b g B T e g E T f g F T Ro T ln E p A p o p p e F po B po a f b G o T e g Eo T f g Fo T a g Ao T b g Bo T Equilibrium constant: pEe pFf Kp a b p A pB (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 Equilibrium constants Example: calculate the equilibrium constant of reaction H2O H + OH The following equilibrium constants are tabulated: K p1 H2 + ½ O2 H2O ½ H2 H K p2 ½ H2 + ½ O2 OH K p3 p H 2O pH2 pO2 pH pH2 pOH pH2 pO2 The equilibrium constant of the reaction given above is then K p4 K p2 p H pOH p H 2O Thermochemistry pH2 K p3 pH2 K p1 p H 2 pO2 pO2 K p2 K p3 K p1 Combustion Equilibrium constants • The equilibrium constant may be explicitly obtained from G T Ro T ln K p o as G o T K p exp Ro T • 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 Go(T)>0 there are more reactants than products in equilibrium. If Kp>1, then Go(T)<0 there are more products than reactants in equilibrium. Accordingly, the larger Kp, the smaller Go(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 Equilibrium constants • The change of the Gibbs free energy may be expressed as G o T H o T T S o T yielding H o T S o T exp K p exp Ro T Ro • Assuming that Ho(T) and So(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 o K p T2 T 1 1 H ln K T Ro T2 T1 p 1 Thermochemistry Exothermic reaction: Ho(T) < 0, T Kp Endothermic reaction: Ho(T) > 0, T Kp Combustion 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. e f E F Kc Aa Bb nEe nFf Kn a b nA nB Ro T K p K c o p e f a b p K n o ntot p e f a b • 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 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 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 xCO2 0.9994 0.9997 xCO 3.601×10-4 1.672×10-4 xO2 1.801×10-4 8.357×10-5 xCO2 0.9777 0.9895 T = 2000 K xCO 0.0149 6.96×10-3 xO2 0.0074 3.48×10-3 T =1500 K Thermochemistry Combustion 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: H prod Tad H reag T1 K equations for atomic mass balances (typically O, H, C, N) N-K linearly independent equations to be found from chemical equilibrium Thermochemistry Combustion 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. nCH 4 CH4 nar O2 3.76 N 2 nCO 2 CO2 nH 2 O H 2O nO 2 O2 nCO CO nH 2 H 2 nOH OH nO O nH H nN 2 N 2 nN N nNO NO n CH 4 and nar are given, so that there are N = 11 unknown mole fractions Thermochemistry Combustion Calculation of chemical equilibrium composition nCH 4 CH4 nar O2 3.76 N 2 nCO 2 CO2 nH 2 O H 2O nO 2 O2 nCO CO nH 2 H 2 nOH OH nO O nH H nN 2 N 2 nN N nNO NO 4 equations may be written for the atomic mass balances: nCH 4 nCO 2 nCO 4 nCH 4 2 nH 2 O 2 nH 2 nOH nH 2 nar 2 nCO 2 nH 2O 2 nO 2 nCO nOH nO n NO 3.76 2 nar 2 nN 2 nN nNO 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 Calculation of chemical equilibrium composition nCH 4 CH4 nar O2 3.76 N 2 nCO 2 CO2 nH 2 O H 2O nO 2 O2 nCO CO nH 2 H 2 nOH OH nO O nH H nN 2 N 2 nN N nNO NO 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 Kp p o p ntot nCO 2 nCO nO 2 1 2 Kp nOH Kp nH 2 nO 2 Kp nH nH 2 p po n tot n H 2O nH 2 Kp nO Kp nN nO 2 nO 2 12 Kp nN2 p po n tot 1 2 12 p po n tot p po n tot 12 nNO nN 2 nO 2 Combustion 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: http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm • STANJAN (Reynolds, 1986) – code based on the method of chemical potentials of the elements. Online version available at http://navier.engr.colostate.edu/~dandy/code/code-4/index.html Thermochemistry Combustion Examples of application of chemical equilibrium software Example of application of the PER code DADOS DE ENTRADA Número de átomos de carbono no combustível Número de átomos de hidrogénio no combustível Número de átomos de oxigénio no combustível Número de átomos de azoto no combustível Razão de equivalência Temperatura (K) – estimativa inicial Pressão (Pa) Entalpia dos reagentes (kJ/kmol combustível) 3 8 0 0 1.000 2000.0 101325.0 -103847.0 DADOS DE SAÍDA Temperatura da chama e propriedades dos produtos da combustão Temperatura adiabática de chama [K] = 2266.78 Entalpia da mistura [J/kg] = -0.1420E+06 Calor específico da mistura cp [J/kg-K] = 0.227230E+04 Razão entre os calores específicos, cp/cv = 1.1655 Massa molar da mistura [kg/kmol] = 28.0638 Número de moles de combustível por mole de produtos = 0.03838622 Fracções molares das espécies em condições de equilíbrio H: 0.00046610 O: H2: 0.00331509 OH: NO: 0.00247177 O2: CO2: 0.10273075 N2: Thermochemistry 0.00031115 0.00320690 0.00589086 0.72078613 N: 0.00000002 CO: 0.01242792 H2O: 0.14839331 Combustion Examples of application of chemical equilibrium software Thermochemistry Combustion Examples of application of chemical equilibrium software Thermochemistry Combustion Examples of application of chemical equilibrium software Thermochemistry Combustion