Combustion - AER 1304

What is Combustion? Combustion is a key element of many of modern society’s critical technologies. Combustion accounts for approximately 85 percent of the world’s energy usage and is vital to our current way of life. Spacecraft and aircraft propulsion, electric power production, home heating, ground transportation, and materials processing all use combustion to convert chemical energy to thermal energy or propulsive force. 0.Introduction 1 AER 1304–ÖLG Examples of combustion applications: Gas turbines and jet engines Rocket propulsion Piston engines Guns and explosives Furnaces and boilers Flame synthesis of materials (fullerenes, nano- materials) Chemical processing (e.g. carbon black produc- tion) Forming of materials Fire hazards and safety 0.Introduction 2 AER 1304–ÖLG Combustion is a complex interaction of: physical processes - fluid dynamics, heat and mass transfer chemical processes - thermodynamics, and chemical kinetics Practical applications of the combustion phenomena also involve applied sciences such as aerodynamics, fuel technology, and mechanical engineering. 0.Introduction 3 AER 1304–ÖLG The transport of energy, mass, and momentum are the physical processes involved in combustion. The conduction of thermal energy, the diffusion of chemical species, and the flow of gases all follow from the release of chemical energy in the exother- mic reaction. The subject areas most relevant to combustion in the fields of thermodynamics, transport phenom- ena, and chemical kinetics can be summarized as follows: 0.Introduction 4 AER 1304–ÖLG Thermodynamics: Stoichiometry Properties of gases and gas mixtures Heat of formation Heat of reaction Equilibrium Adiabatic flame temperature 0.Introduction 5 AER 1304–ÖLG Heat and Mass Transfer: Heat transfer by conduction Heat transfer by convection Heat transfer by radiation Mass transfer Fluid Dynamics: Laminar flows Turbulence Effects of inertia and viscosity Combustion aerodynamics 0.Introduction 6 AER 1304–ÖLG Chemical Kinetics: Application of thermodynamics to a reacting system gives us - equilibrium composition of the combustion products, and - maximum temperature corresponding to this composition, i.e. the adiabatic flame tempera- ture. However, thermodynamics alone is not capable of telling us whether a reactive system will reach equi- librium. 0.Introduction 7 AER 1304–ÖLG Chemical Kinetics (cont’d): If the time scales of chemical reactions involved in a combustion process are comparable to the time scales of physical processes (e.g. diffusion, fluid flow) taking place simultaneously, the system may never reach equilibrium. Then, we need the rate of chemical reactions in- volved in combustion. 0.Introduction 8 AER 1304–ÖLG Primary sources of combustion research literature: 1 Combustion and Flame (journal) 2 Combustion Science and Technology (journal) 3 Combustion Theory and Modelling (journal) 4 Progress in Energy and Combustion Science (re- view journal) 5 Proceedings of the Combustion Institute (Biennial Combustion Symposia (International) proceedings). 6 Combustion, Explosions and Shock Waves (journal translated from Russian) 0.Introduction 9 AER 1304–ÖLG Fundamental Definitions Chemical Reaction: exchange and/or rearrangement of atoms between colliding molecules CO + H2 O → CO2 + H2 , 1 , 1 Reactants → Products The atoms are conserved (C, H, O) On the other hand, molecules are not conserved. H2 + 0.5(O2 + 3.76N2 ) → H2 O + 1.88N2 Reactants → Products 0.Introduction 10 AER 1304–ÖLG Amount of substance or mole numbers (mol): 1 mol of a compound corresponds to 6.023 · 1023 particles (atoms, molecules, or any chemical species). Avogadro’s constant = 6.023 · 1023 Mole fraction χi of species i with mole number of Ni is Ni χi = S j=1 Nj 0.Introduction 11 AER 1304–ÖLG Mass fraction Yi of species i with mass of mi is mi Yi = S j=1 mj Molar or Molecular Mass, Mi (molecular weight is misleading and should not be used) - MCH4 = 16 g/mol - MH2 = 2 g/mol - MO2 = 32 g/mol 0.Introduction 12 AER 1304–ÖLG Mean molar mass, M , of a mixture of species de- notes an average molar mass: 3 M= χi Mi S = number of species in the system Mi Ni Mi χi Yi = S = S j=1 Mj Nj j=1 Mj χj Yi Yi /Mi χi = = S Mi M j=1 Yj /Mj 0.Introduction 13 AER 1304–ÖLG For a system of volume, V : Mass density (density), ρ = m/V (kg/m3 ) Molar density (concentration), c = N/V (kmol/m3 ) Mean molar mass is given by: ρ m = =M c N Chemical kinetics convention: concentrations c of chemical species are usually shown by species symbol in square brackets. cCO2 = [CO2 ] 0.Introduction 14 AER 1304–ÖLG For most conditions involved in combustion, it is satisfactory to use the perfect gas equation of state for the gas phase. P V = N Ro T (Pa)(m3 ) = (mol)(J/molK)(K) Ro = 8.314 J / mol K, universal gas constant P = pressure, Pa T = temperature, K 0.Introduction 15 AER 1304–ÖLG When the gas phase temperatures are near or less than the critical temperatures, or when pressures are near or above the critical pressures, the density or con- centration is not correctly predicted by the perfect gas relationship. Real gas equations should be used. - van der Waals - Peng-Robinson 0.Introduction 16 AER 1304–ÖLG Basic Flame Types: Premixed Flames - Laminar - Turbulent Non-Premixed (Diffusion) Flames - Laminar - Turbulent Partially Premixed Flames - Laminar - Turbulent ♠ triple flames, edge flames,... 0.Introduction 17 AER 1304–ÖLG Laminar (Turbulent) Premixed Flames: Fuel (in gaseous form) and oxidizer are homoge- neously mixed before the combustion event Flow is laminar (turbulent) Turbulent premixed flames: - combustion in gasoline engines - lean-premixed gas turbine combustion 0.Introduction 18 AER 1304–ÖLG Burned Unburned - Cross-section of a gasoline engine combustion chamber. 0.Introduction 19 AER 1304–ÖLG Stoichiometry: A premixed flame is stoichiometric if the premixed reactants contain right amount of oxidizer to con- sume (burn) the fuel completely. If there is an excess of fuel: fuel-rich system If there is an excess of oxygen: fuel-lean system Standard air composition commonly used for com- bustion calculations: O2 + 3.762N2 0.Introduction 20 AER 1304–ÖLG Stoichiometry (cont’d): C3 H8 + 5(O2 + 3.762N2 ) → 4H2 O + 3CO2 + 18.81N2 (A/F )stoich =air-to-fuel ratio (mass)= (mass of air)/(mass of fuel) (A/F )stoich =[5(32+3.762*28)]/(44) = 15.6 Φ = (A/F )stoich /(A/F )actual = Fuel Equivalence Ratio 0.Introduction 21 AER 1304–ÖLG Stoichiometry (cont’d):: Φ = 1: stoichiometric combustion Φ < 1: lean mixture, lean combustion Φ > 1: rich mixture, rich combustion European convention (and to a certain extent Japanese) is to use Air equivalence ratio, λ: λ = 1/Φ In certain industries, excess air ratio, excess oxygen, and similar terminologies are also used. 0.Introduction 22 AER 1304–ÖLG Laminar (Turbulent) Non- Premixed Flames: Fuel (in gaseous form) and oxidizer are mixed/come in to contact during the combustion process A candle flame is a typical laminar non-premixed (diffusion) flame Turbulent non-premixed flames: - hydrogen rocket engine - current aero gas turbines - diesel engines 0.Introduction 23 AER 1304–ÖLG A candle flame. 0.Introduction 24 AER 1304–ÖLG ¾ Air Inlet ¾ Inlet Port Design ¾ Chamber Design ¾ Turbocharge AIR MOTION / TURBULENCE IN THE COMBUSTION CHAMBER FUEL-AIR PARTIALLY MOSTLY EXHAUST MIXING IGNITION "PREMIXED" NON-PREMIXED EMISSIONS PROCESS COMBUSTION COMBUSTION INJECTION AND SPRAY Fuel Properties EGR CHARACTERISTICS y HEAT RELEASE ¾ Injection Timing y RADIATION EXCHANGE BETWEEN ¾ Injection System Design HOT AND COLD POCKETS ¾ Injection Duration y NOX & SOOT FORMATION ¾ Injection Rate y SOOT OXIDATION Processes in the diesel engine combustion. 0.Introduction 25 AER 1304–ÖLG Spark-ignited gasoline engine TURBULENT Low-NOx stationary gas turbine PREMIXED Flat flame LAMINAR Bunsen flame Aircraft turbine Hydrogen-oxygen rocket motor TURBULENT Diesel engine NON-PREMIXED Pulverized coal combustion (DIFFUSION) Candle flame LAMINAR Radiant burners for heating Wood fire FUEL/OXIDIZER FLUID EXAMPLES MIXING MOTION Examples of combustion systems. 0.Introduction 26 AER 1304–ÖLG 1. Combustion & Thermochemistry This section will cover the following concepts: Basic property relations for ideal gas and ideal gas mixtures. First law of thermodynamics. Enthalpy/heat of reaction; adiabatic flame tempera- ture. Chemical equilibrium. 1. Combustion & Thermochemistry 1 AER 1304–ÖLG Review of Property Relations: Extensive Properties: depends on amount of sub- stance considered. Usually denoted by capital let- ters. Examples are: V for volume, U for total internal energy, H for total enthalpy. Intensive Properties: expressed per unit amount of substance (mass or mole). Its numerical value is independent of the amount of substance present. Usually denoted with lower case letters. Examples are: specific volume v, specific enthalpy h, spe- cific heat cp. 1. Combustion & Thermochemistry 2 AER 1304–ÖLG Intensive Properties (Cont’d): Important excep- tions to this lower case conventions are tempera- ture T and pressure P. Molar based properties will be denoted by an overbar, e.g., h̄ for specific enthalpy per unit mole, J/mol. Extensive properties are related to the intensive ones by the amount of substance present: V = m · v(or N · v̄) (1.1) H = m · h(or N · h̄) 1. Combustion & Thermochemistry 3 AER 1304–ÖLG Equation of State: Provides the relationship among the pressure, tem- perature, and volume. Ideal-gas behaviour: intermolecular forces and volume of molecules are ignored. P V = N Ru T (1.2a) P V = mRT (1.2b) P v = RT (1.2c) P = ρRT (1.2d) 1. Combustion & Thermochemistry 4 AER 1304–ÖLG Equation of State (Cont’d): The specific gas constant R is related to the uni- versal gas constant Ru (or Ro ) by: Ru R= (1.3) MW Ru = 8314.7 J/(kmol K) M W is the molecular weight (or, more precisely, molecular mass). Throughout this course, all gaseous species and gas mixtures will be assumed to be ideal. 1. Combustion & Thermochemistry 5 AER 1304–ÖLG Calorific Equations of State: Relates enthalpy and internal energy to pressure and temperature: u = u(T, v) (1.4a) h = h(T, P ) (1.4a) By differentiating Eqns 1.4a and b: p ∂u Q p ∂u Q du = dT + dv (1.5a) ∂T v ∂v T p ∂h Q p ∂h Q dh = dT + dP (1.5b) ∂T P ∂P T 1. Combustion & Thermochemistry 6 AER 1304–ÖLG Calorific Equations of State (Cont’d): In Eqs 1.5a and b: p ∂u Q cv = (1.6a) ∂T v p ∂h Q cp = (1.6b) ∂T P For an ideal gas: (∂u/∂v)T = 0 and (∂h/∂P )T = 0 1. Combustion & Thermochemistry 7 AER 1304–ÖLG Calorific Equations of State (Cont’d): Integrating Eqs 1.5 and substitute in 1.6 provide 8 T u(T ) − uref = cv dT (1.7a) Tref 8 T h(T ) − href = cp dT (1.7b) Tref We will define an appropriate reference state in a subsequent section. 1. Combustion & Thermochemistry 8 AER 1304–ÖLG Calorific Equations of State (Cont’d): For both real and ideal gases, the specific heats are generally functions of temperature. Internal energy of a molecule: translational, vibra- tional and rotational are temperature dependent. Monatomic species: only translational energy. Diatomic and triatomic: all three, i.e., transla- tional, vibrational and rotational. In general, the more complex the molecule, the greater its molar specific heat. 1. Combustion & Thermochemistry 9 AER 1304–ÖLG Ideal Gas Mixtures: Mole fraction of species i: Ni χi ≡ = Ni /Ntot (1.8) Nj j Mass fraction of species i: mi Yi ≡ = mi /mtot (1.9) mj j 1. Combustion & Thermochemistry 10 AER 1304–ÖLG Ideal Gas Mixtures (Cont’d): By definition the sum of all the mole (and mass) fractions must be unity: 3 3 χi = 1 and Yi = 1 (1.10a&b) i i Relations between χ and Y : Yi = χi M Wi /M Wmix (1.11a) χi = Yi M Wmix /M Wi (1.11b) 1. Combustion & Thermochemistry 11 AER 1304–ÖLG Ideal Gas Mixtures (Cont’d): Mixture molecular weight (mass): 3 M Wmix = χi M Wi (1.12a) i 1 M Wmix = (1.12b) (Yi /M Wi ) i Total pressure is the sum of the partial pressures: 3 P = Pi (1.13) i 1. Combustion & Thermochemistry 12 AER 1304–ÖLG Ideal Gas Mixtures (Cont’d): Partial pressure of the ith species is the pressure of this species if it were isolated from the mixture. Pi = χi P (1.14) For ideal gas mixtures: 3 hmix = Yi hi (1.15a) i 3 h̄mix = χi h̄i (1.15b) i 1. Combustion & Thermochemistry 13 AER 1304–ÖLG Ideal Gas Mixtures (Cont’d): Mixture entropies also is calculated as a weighted sum of the constituents: 3 smix (T, P ) = Yi si (T, P ) (1.16a) i 3 s̄mix (T, P ) = χi s̄i (T, P ) (1.16a) i Pure species entropies depend on the species par- tial pressures as implied in Eqs 1.16. 1. Combustion & Thermochemistry 14 AER 1304–ÖLG Ideal Gas Mixtures (Cont’d): If we take the standard state as 1 atm (Pref = P o = 1 atm), then: Pi si (T, Pi ) = si (T, Pref ) − R · ln (1.17a) Pref Pi s̄i (T, Pi ) = s̄i (T, Pref ) − Ru · ln (1.17b) Pref Appendix A in the textbook lists entropies for sev- eral species. 1. Combustion & Thermochemistry 15 AER 1304–ÖLG Latent Heat (Enthalpy) of Vaporization: hf g is the heat required in a P = const. process to completely vaporize a unit mass of liquid at a given temperature: hf g ≡ hvapor (T, P ) − hliquid (T, P ) (1.18) Clausius-Clapeyron equation: dPsat hf g dT = 2 (1.19) Psat R Tsat 1. Combustion & Thermochemistry 16 AER 1304–ÖLG First Law of Thermodynamics: First Law - Fixed Mass: Conservation of energy is the fundamental prin- ciple in the first law of thermodynamics. For a fixed mass system, energy conservation is ex- pressed for a finite change between two states, 1 and 2, as: 1 Q2 − 1 W2 = ∆E1−2 (1.20) , 1 , 1 , 1 Heat added Work done Change in total to the system by the system energy of system 1. Combustion & Thermochemistry 17 AER 1304–ÖLG First Law - Fixed Mass (Cont’d): Both 1 Q2 and 1 W2 are path functions and occur only at the system boundaries. ∆E1−2 ≡ E2 − E1 is the change in the total energy of the system, i.e. E = m[ ,u 1 + (1/2)v 2 + gz ] , 1 , 1 system internal system potential system kinetic energy energy energy (2.21) The sytem energy is a state variable and does not depend on the path taken. 1. Combustion & Thermochemistry 18 AER 1304–ÖLG First Law - Fixed Mass (Cont’d): We can write Eq 1.20 as a unit mass basis, or ex- pressed to represent an instant in time: 1 q2 −1 w2 = ∆e1−2 = e2 − e1 (1.22) Q̇ − Ẇ 1 , = dE/dt (1.23) , 1 , 1 instantaneous instantaneous instantaneous rate of heat rate of work change of transferred done system energy q̇ − ẇ = de/dt (1.24) 1. Combustion & Thermochemistry 19 AER 1304–ÖLG First Law - Control Volume: Steady-state, steady-flow form of the first law: Q̇cv − Ẇcv = ṁeo , 1 , 1 , 1 Rate of heat Rate of work Rate of energy transferred from done by control flow out of CV surroundings volume excluding flow work − ṁei + ṁ(Po vo − Pi vi ) (1.25) , 1 , 1 Rate of energy Net rate of work flow into CV associated with pressure forces 1. Combustion & Thermochemistry 20 AER 1304–ÖLG First Law - Control Volume (Cont’d): Main assumptions in the previuos equation: - The control volume is fixed relative to the coordinate system. - The properties of the fluid at each point within CV, or on the control surface, do not vary with time. - Fluid properties are uniform over inlet and outlet areas. - There is only one inlet and one outlet stream. 1. Combustion & Thermochemistry 21 AER 1304–ÖLG First Law - Control Volume (Cont’d): Specific energy e of the inlet and outlet stream consist of: ,e 1 = ,u 1 + (1/2)v 2 + gz , 1 , 1 Total Internal Kinetic Potential energy per energy per energy per energy per unit mass unit mass unit mass unit mass (1.26) v = velocity where the stream crosses the CV z = elevation where stream crosses the CV g = gravitational acceleration 1. Combustion & Thermochemistry 22 AER 1304–ÖLG First Law - Control Volume (Cont’d): Enthalpy: h ≡ u + P v = u + P/ρ (1.27) Eqns 1.25-1.27 yield: 1 2 Q̇cv − Ẇcv = ṁ[(ho −hi )+ (vo −vi2 )+g(zo −zi )] 2 (1.28) On a mass-specific basis, Eqn 2.28 reads qcv − wcv = (ho − hi ) + (1/2)(vo2 − vi2 ) + g(zo − zi ) (1.29) 1. Combustion & Thermochemistry 23 AER 1304–ÖLG Reactant and Product Mixtures: Stoichiometry: The stoichiometric quantity of oxidizer is just that amount needed to completely burn a quantity of fuel. For a hydrocarbon fuel, Cx Hy , stoichiometric relation is Cx Hy + a(O2 + 3.76N2 ) → xCO2 + (y/2)H2 O + 3.76aN2 (1.30) where a = x + y/4 (1.31) 1. Combustion & Thermochemistry 24 AER 1304–ÖLG Stoichiometry (Cont’d): The stoichiometric air-fuel ratio is X ~ mair 4.76a M Wair (A/F )stoic = = (1.32) mfuel 1 M Wfuel The equivalence ratio is defined as (A/F )stoic (F/A) Φ= = (1.33a) (A/F ) (F/A)stoic 1. Combustion & Thermochemistry 25 AER 1304–ÖLG Stoichiometry (Cont’d): Φ > 1: Fuel-rich mixtures. Φ < 1: Fuel-lean mixtures. Φ = 1: Stoichiometric mixtures. % stoichiometric and % excess air; %stoichiometric air = 100%/Φ (1.33b) %excess air = [(1 − Φ)/Φ] · 100% (1.33c) 1. Combustion & Thermochemistry 26 AER 1304–ÖLG Absolute (or Standardized) Enthalpy and Enthalpy of Formation: For any species, we can define an absolute en- thalpy that is the sum of an enthalpy of formation and sensible enthalpy change: h̄i (T ) = h̄of,i (Tref ) + ∆h̄s,i (Tref ) , 1 , 1 , 1 absolute enthalpy enthalpy of sensible enthalpy at temperature T formation at change in going standard ref from Tref to T state (Tref ,P o ) (1.34) where ∆h̄s,i ≡ h̄i (T ) − h̄of,i (Tref ) 1. Combustion & Thermochemistry 27 AER 1304–ÖLG Absolute (or Standardized) Enthalpy (Cont’d): Define a standard reference state. Tref = 298.15K (25 o C) Pref = P o = 1atm (101.325 kPa) Enthalpies of formation are zero for the elements in their naturally occuring state at the reference temperature and pressure. For example, at 25 o C and 1 atm, oxygen exists as diatomic molecules; then o (h̄f,O2 )298 =0 1. Combustion & Thermochemistry 28 AER 1304–ÖLG Absolute (or Standardized) Enthalpy (Cont’d): To form oxygen atoms at the standard state re- quires breaking of a chemical bond. The bond dissociation energy of O2 at standard state is 498, 390 kJ/kmol. Breaking of the bond creates two O atoms; there- fore the enthalpy of formation of atomic oxygen is (h̄of,O )298 = 249, 195 kJ/kmol 1. Combustion & Thermochemistry 29 AER 1304–ÖLG Enthalpy of Formation (Cont’d): Thus, enthalpies of formation is the net change in enthalpy associated with breaking the chemical bonds of the standard state elements and forming new bonds to create the compound of interest. Examples: (h̄of,N2 )298 = 0 (h̄of,H2 )298 = 0 (h̄of,N )298 = 472, 629 kJ/kmol o (h̄f,H )298 = 217, 977 kJ/kmol 1. Combustion & Thermochemistry 30 AER 1304–ÖLG Abs. enthalpy, heat of formation, and sensible enthalpy 1. Combustion & Thermochemistry 31 AER 1304–ÖLG Enthalpy of Combustion and Heating Values -qcv REACTANTS PRODUCTS Stoichiometric Complete combustion at standard state at standard state Steady-flow reactor 1. Combustion & Thermochemistry 32 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d) Reactants: stoichiometric mixture at standard state conditions. Products: complete combustion, and products are at standard state conditions. For products to exit at the same T as the reac- tants, heat must be removed. The amount of heat removed can be related to the reactant and product absolute enthalpies by apply- ing the steady-flow form of the first law: 1. Combustion & Thermochemistry 33 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d): From eqn 1.29: qcv = ho − hi = hprod − hreac (1.35) The enthalpy of reaction, or the enthalpy of com- bustion, ∆hR , is (per mass of mixture) ∆hR ≡ qcv = hprod − hreac (1.36a) or, in terms of extensive properties ∆HR = Hprod − Hreac (1.36b) 1. Combustion & Thermochemistry 34 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d): Example: - At standard state, the reactants enthalpy of a stoi- chiometric mixture of CH4 and air, where 1 kmol of fuel reacts, is −74, 831 kJ. - At the same conditions, the combustion products have an absolute enthalpy of −877, 236 kJ. ∆HR = −877, 236 − (−74, 831) = −802, 405kJ 1. Combustion & Thermochemistry 35 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d): - This value can be adjusted to a per-mass-of-fuel basis: p kJ Q ∆hR = ∆HR /M Wfuel (1.37) kgfuel - or p kJ Q ∆hR = (−802, 405/16.043) kgfuel = −50, 016 1. Combustion & Thermochemistry 36 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d): - For a per-unit-mass-of-mixture basis: p kJ Q p kJ Q m fuel ∆hR = ∆hR (1.38) kgmix kgfuel mmix where mfuel mfuel 1 = = (1.39) mmix mair + mfuel (A/F ) + 1 ∆hR (kJ/kgmix ) = −50, 016/(17.11 + 1) = −2761.8 1. Combustion & Thermochemistry 37 AER 1304–ÖLG Enthalpy of Combustion and Heating Values (Cont’d): Enthalpy of combustion depends on temperature chosen for its evaluation since enthalpies of reac- tants and products are temperature dependent. The heat of combustion, ∆hc (known also as the heating value or calorific value), is numerically equal to the enthalpy of combustion, but with op- posite sign. HHV: higher heating value (H2 O→ liquid) LHV: lower heating value (H2 O→ vapor) 1. Combustion & Thermochemistry 38 AER 1304–ÖLG Adiabatic Flame Temperatures: We will differentiate between two adiabatic flame temperatures: - constant-pressure combustion - constant-volume combustion Constant-Pressure: If a fuel-air mixture burns adi- abatically at constant pressure, absolute enthalpy of reactants at the initial state (say, T1 = 298 K, P = 1 atm) equals absolute enthalpy of the prod- ucts at final state (T = Tad , P = 1 atm). 1. Combustion & Thermochemistry 39 AER 1304–ÖLG Constant Pressure Tad (Cont’d): Definition of the constant-pressure adiabatic flame temperature is Hreac (Ti , P ) = Hprod (Tad , P ) (1.40a) or, on a per-mass-of-mixture basis, hreac (Ti , P ) = hprod (Tad , P ) (1.40b) Conceptually, adiabatic flame temperature is sim- ple, however, evaluating this quantity requires knowledge of the composition of the combustion products. 1. Combustion & Thermochemistry 40 AER 1304–ÖLG Constant Volume Tad : When we are dealing with constant-pressure com- bustion systems, such as gas turbine combus- tors and rocket engines, the appropriate approach would involve constant-pressure Tad. When we are dealing with constant-volume com- bustion systems, such as enclosed explosions or Otto-cycle (idealized thermodynamic cycle for gasoline engine combustion) analysis, the appro- priate approach would involve constant-volume Tad. 1. Combustion & Thermochemistry 41 AER 1304–ÖLG Constant Volume Tad (Cont’d): Definition: Ureac (Tinit , Pinit ) = Uprod (Tad , Pf ) (1.41) where U is the absolute (or standardized) internal energy of the mixture. Most thermodynamic property compilations and calculations provide H (or h) rather than U (or u). So we consider the fact that: H = U + PV 1. Combustion & Thermochemistry 42 AER 1304–ÖLG Constant Volume Tad (Cont’d): Eqn 1.41: Hreac − Hprod − V (Pinit − Pf ) = 0 (1.42) If we apply the ideal-gas law; 3 Pinit V = Ni Ru Tinit = Nreac Ru Tinit reac 3 Pf V = Ni Ru Tad = Nprod Ru Tad prod 1. Combustion & Thermochemistry 43 AER 1304–ÖLG Constant Volume Tad (Cont’d): Substituting in Eqn 1.42 Hreac − Hprod − Ru (Nreac Tinit − Nprod Tad ) = 0 (1.43) Since mmix /Nreac ≡ M Wreac mmix /Nprod ≡ M Wprod p T T Q init ad hreac − hprod − Ru − =0 M Wreac M Wprod (1.44) 1. Combustion & Thermochemistry 44 AER 1304–ÖLG Chemical Equilibrium: Products of combustion are not a simple mixture of ideal products. We used ideal products approach to determine sto- ichiometry. Ideal combustion products for a hydrocarbon fuel: - Φ = 1: CO2 , H2 O, N2 - Φ < 1: CO2 , H2 O, N2 , O2 - Φ > 1: CO2 , H2 O, N2 , CO, H2 1. Combustion & Thermochemistry 45 AER 1304–ÖLG Real combustion products of a hydrocarbon fuel may include: - CO2 , H2 O, N2 , O2 , H2 , OH, CO, H, O, N, NO,.... Major species, i.e., CO2 , H2 O, N2 , O2 , H2 , CO, dissociate into a host of minor species, i.e., H, N, O, OH, NO. Our objective here is to calculate the mole fraction of all product species. 1. Combustion & Thermochemistry 46 AER 1304–ÖLG Second-Law Considerations: Second-Law of Thermodynamics → Concept of chemical equilibrium. Consider a V =const., adiabatic reaction vessel in which a fixed mass of reactants form products. As reaction proceeds, T and P rise until a final equilibrium condition is reached. This final state is not governed solely by first-law considerations, but requires invoking the second- law. 1. Combustion & Thermochemistry 47 AER 1304–ÖLG Second-Law Considerations (Cont’d): Consider the following combustion reaction: CO + 0.5O2 → CO2 (1.45) If the final T is high enough, CO2 will dissociate. Assuming products are CO2 , CO, and O2 = CO + 0.5O2 cold → reactants = (1 − α)CO2 + αCO + (α/2)O2 hot (1.46) products 1. Combustion & Thermochemistry 48 AER 1304–ÖLG Second-Law Considerations (Cont’d): α = fraction of CO2 dissociated. It is possible to calculate the adiabatic flame tem- perature as a function of α using Eqn 1.42. α = 1: no heat release; T , P , and χi remain unchanged. α = 0: maximum possible heat release; P and T would be the highest possible allowed by the first-law. Variation of temperature with α: 1. Combustion & Thermochemistry 49 AER 1304–ÖLG Chemical equilibrium for a fixed mass system. 1. Combustion & Thermochemistry 50 AER 1304–ÖLG Second-Law Considerations (Cont’d): Let’s see what constraints second-law imposes on this system Entropy of the mixture: 3 Smix (Tf , P ) = Ni s̄i (Tf , Pi ) i=1 = (1 − α)s̄CO2 + αs̄CO + (α/2)s̄O2 (1.47) 8Tf dT Pi s̄i = s̄oi (Tref ) + c̄p,i − Ru ln o (1.48) T P Tref 1. Combustion & Thermochemistry 51 AER 1304–ÖLG Second-Law Considerations (Cont’d): If we plot mixture entropy versus α, we see that entropy reaches maximum at about α = 0.5. For our choice of conditions, the second law re- quires that entropy change internal to the system dS ≥ 0 (1.49) χi will shift toward the point of maximum entropy when approaching from both sides. Once maximum entropy is reached, no change in χi is allowed. 1. Combustion & Thermochemistry 52 AER 1304–ÖLG Second-Law Considerations (Cont’d): Then, formally the condition for equilibrium is: (dS)U,V,m = 0 (1.50) If we fix the internal energy, U , volume, V , and mass, m, of an isolated system, application of the - first law (Eqn 1.41), - second law (Eqn 1.49) and - equation of state (Eqn 1.2) define the equilibrium T , P , and χi. 1. Combustion & Thermochemistry 53 AER 1304–ÖLG Gibbs Function: For an isolated system of fixed volume, mass, and energy system, the maximum entropy approach demonstrates the role of second law. In most typical sysytems, however, the equilibrium χi are required for a given T , P , and Φ. For such cases, the Gibbs free energy, G replaces entropy as the important thermodynamic property. Gibbs free energy is defined as: G ≡ H − TS (1.51) 1. Combustion & Thermochemistry 54 AER 1304–ÖLG Gibbs Function (Cont’d): Second law in terms of G: (dG)T,P,m ≤ 0 (1.52) which states that Gibbs function always decreases for a spontaneous, isothermal, isobaric change of a fixed-mass system. This allows us to calculate equilibrium χi at a given P , and T. At equilibrium: (dG)T,P,m = 0 (1.53) 1. Combustion & Thermochemistry 55 AER 1304–ÖLG Gibbs Function (Cont’d): For a mixture of ideal gases, the Gibbs function of the ith species o ḡi,T = ḡi,T + Ru T ln(Pi /P o ) (1.54) o ḡi,T is Gibbs function at standard-state pressure, Pi = P o. In reacting systems, Gibbs function of formation is 3 o ḡf,i (T ) ≡ ḡio (T ) − νjI ḡjo (T ) (1.55) j 1. Combustion & Thermochemistry 56 AER 1304–ÖLG Gibbs Function (Cont’d): νjI are the stoichiometric coefficients of the ele- ments required to form one mole of the compound of interest. νOI 2 = 0.5 and νC = 1 for forming one mole of I CO from O2 and C. Similar to enthalpies, ḡf,i o (T ) of the naturally oc- curing elemets are assigned values of zero at the reference state. 1. Combustion & Thermochemistry 57 AER 1304–ÖLG Gibbs Function (Cont’d): Gibbs function for a mixture of ideal gases o Gmix = Ni ḡi,T = 3 o Ni [ḡi,T + Ru T ln(Pi /P o )] (1.56) For a fixed T and P , the equilibrium condition becomes dGmix = 0 (1.57) or o dNi [ḡi,T + Ru T ln(Pi /P o )] 3 o + Ni d[ḡi,T + Ru T ln(Pi /P o )] = 0 (1.58) 1. Combustion & Thermochemistry 58 AER 1304–ÖLG Gibbs Function (Cont’d): Second term in the last equation is zero, because d(lnPi ) = dPi /Pi and dPi = 0 and total pressure is constant. Then, 3 o dGmix = 0 = dNi [ḡi,T + Ru T ln(Pi /P o )] (1.59) For a general system, where aA + bB +..... ⇔ eE + f F +...... (1.60) 1. Combustion & Thermochemistry 59 AER 1304–ÖLG Gibbs Function (Cont’d): The change in the number of each species is pro- portional to its stoichiometric coefficient, dNA = −κa dNB = −κb.=. (1.61) dNE = +κe dNF = +κf.=. 1. Combustion & Thermochemistry 60 AER 1304–ÖLG Gibbs Function (Cont’d): substitute Eqn 1.61 into 1.59 and eliminate κ o − a[ḡA,T + Ru T ln(PA /P o )] o − b[ḡB,T + Ru T ln(PB /P o )] −.. o o (1.62) + e[ḡE,T + Ru T ln(PE /P )] o + f [ḡF,T + Ru T ln(PF /P o )] +.. = 0 or o −(eḡE,T o + f ḡF,T o +.. − aḡA,T o − bḡB,T −..) (PE /P o )e · (PF /P o )f ·.. = Ru T ln (1.63) (PA /P o )a · (PB /P o )b ·.. 1. Combustion & Thermochemistry 61 AER 1304–ÖLG Gibbs Function (Cont’d): Left hand side of Eqn 1.63 is the standard-state Gibbs function change: ∆GoT = (eḡE,T o o + f ḡF,T o +.. − aḡA,T o − bḡB,T −..) (1.64a) ∆GoT ≡ (eḡf,E o o + f ḡf,F o +.. − aḡf,A o − bḡf,B −..)T (1.64b) Argument of the natural logarithm in Eqn 1.63 is defined as the equilibrium constant Kp (PE /P o )e · (PF /P o )f ·.. Kp = (1.65) (PA /P o )a · (PB /P o )b ·.. 1. Combustion & Thermochemistry 62 AER 1304–ÖLG Gibbs Function (Cont’d): Then we have: ∆GoT = −Ru T lnKp (1.66a) Kp = exp[−∆GoT /(Ru T )] (1.66b) Eqns 1.65 and 1.66 give a qualitative indication of whether a particular reaction favors products or reactants at equilibrium: Reactants: If ∆GoT > 0 ⇒ lnKp < 0 ⇒ Kp < 1 Products: If ∆GoT < 0 ⇒ lnKp > 0 ⇒ Kp > 1 1. Combustion & Thermochemistry 63 AER 1304–ÖLG Gibbs Function (Cont’d): Similar physical insight can be obtained by con- sidering the definition of ∆GoT in etrms of en- thalpy and entropy changes: ∆GoT = ∆H o − T ∆S o which can be substituted into Eqn 1.66b Kp = exp[−∆H o /(Ru T )] · exp(∆S o /Ru ) - For Kp > 1, which favors products, ∆H o should be negative (exothermic reaction). Also positive changes in entropy lead to Kp > 1 1. Combustion & Thermochemistry 64 AER 1304–ÖLG Equilibrium Products of Combustion: Full Equilibrium: Calculate the adiabatic flame temperature and de- tailed composition of the products of combustion: - Eqn 1.40 (or 1.41) (1st law) - Eqn 1.66 (Gibbs funct.-Equilibrium const.) - Apropriate atom conservation constants Constant pressure combustion of Propane, C3 H8 , with air, assuming that the products are CO2 , CO, H2 O, H2 , H, OH, O2 , O, NO, N2 , and N: 1. Combustion & Thermochemistry 65 AER 1304–ÖLG Tad and major species distribution. 1. Combustion & Thermochemistry 66 AER 1304–ÖLG Minor species distribution of propane-air combustion. 1. Combustion & Thermochemistry 67 AER 1304–ÖLG Water-Gas Equilibrium: Develop simple relations to calculate ideal prod- ucts of combustion (no dissociation producing mi- nor species) for lean and rich conditions: We employ a single equilibrium reaction (water- gas shift reaction): CO + H2 O ⇔ CO2 + H2 to account for simultaneous presence of CO and H2 , considered as the incomplete combustion products. 1. Combustion & Thermochemistry 68 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): Combustion of an arbitrary hydrocarbon is consid- ered: Cx Hy + a(O2 + 3.76N2 ) → bCO2 + cCO + dH2 O + eH2 + f O2 + 3.76aN2 (1.67a) For Φ ≤ 1 becomes: Cx Hy + a(O2 + 3.76N2 ) → bCO2 + dH2 O + f O2 + 3.76aN2 (1.67b) 1. Combustion & Thermochemistry 69 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): For rich conditions, Φ > 1: Cx Hy + a(O2 + 3.76N2 ) → bCO2 + cCO + dH2 O + eH2 + 3.76aN2 (1.67c) Note that a can be related to Φ: x + y/4 a= (1.68) Φ So for a given fuel and Φ, a is a known quantity. 1. Combustion & Thermochemistry 70 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): For Φ ≤ 1, c and e are zero: b=x d = y/2 (1.69) f = [(1 − Φ)/Φ](x + y/4) Total number of moles of products: p x + y/4 Q NTOT = x + y/2 + (1 − Φ + 3.76) Φ (1.70) 1. Combustion & Thermochemistry 71 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): For Φ ≤ 1 mole fraction of products: χCO2 = x/NTOT χCO =0 χH2 O = (y/2)/NTOT (1.71) χH2 =0 χO2 = [(1 − Φ)/Φ](x + y/4)/NTOT χN2 = 3.76(x + y/4)/(ΦNTOT ) 1. Combustion & Thermochemistry 72 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): For Φ > 1 no oxygen appears, f = 0. To cal- culate the remaining constants, we use the three atomic balances (C, H, and O) and water-gas shift equilibrium: (PCO2 /P o ) · (PH2 /P o ) b·e KP = o o = (1.72) (PCO /P ) · (PH2 O /P ) c·d c=x−b d = 2a − b − x (1.73) e = −2a + b + x + y/2 1. Combustion & Thermochemistry 73 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): Combining Eqn 1.73 with 1.72 yields a quadratic equation in b. Its solution is (negative root se- lected to yield positive values of b): 2a(Kp − 1) + x + y/2 b= 2(Kp − 1) 1 − {[2a(K)p − 1) + x + y/2]2 2(Kp − 1) − 4Kp (Kp − 1)(2ax − x2 )}1/2 (1.74) 1. Combustion & Thermochemistry 74 AER 1304–ÖLG Water-Gas Equilibrium (Cont’d): For Φ > 1 mole fraction of products: NTOT = b + c + d + e + 3.76a = x + y/2 + 3.76a (1.75) χCO2 = b/NTOT χCO = (x − b)/NTOT χH2 O = (2a − b − x)/NTOT (1.76) χH2 = (−2a + b + x + y/2)/NTOT χO2 = 0 χN2 = 3.76a/NTOT 1. Combustion & Thermochemistry 75 AER 1304–ÖLG 2. Chemical Kinetics Introduction: Thermodynamic laws allow determination of the equilibrium state of a chemical reaction system. If one assumes that the chemical reactions are fast compared to the other transport processes like - diffusion, - heat conduction, and - flow, then, thermodynamics describe the system locally. 2. Chemical Kinetics 1 AER 1304–ÖLG Introduction (Cont’d): In most combustion cases, however, chemical reac- tions occur on time scales comparable with that of the flow and the molecular transport processes. Then, information is needed about the rate of chemical reactions. Chemical reaction rates control pollutant forma- tion, ignition, and flame extinction in most com- bustion processes. 2. Chemical Kinetics 2 AER 1304–ÖLG Global & Elementary Reactions An elementary reaction is one that occurs on a molecular level exactly in the way which is de- scribed by the reaction equation. OH + H2 → H2 O + H The equation above is an elementary reaction. On the contrary, the following is not an elementary reaction: 2H2 + O2 → 2H2 O Above reaction is global or overall reaction. 2. Chemical Kinetics 3 AER 1304–ÖLG A general global reaction mechanism involving overall reaction of a moles of oxidizer with one mole of fuel to form b moles of products: F + aOx → bPr (2.1) Experimental observations yield the rate at which fuel is consumed as d[F] = −kG (T )[F]n [Ox]m (2.2) dt 2. Chemical Kinetics 4 AER 1304–ÖLG [X] denotes molar concentration of X, e.g. kmol/m3. kG (T ) is the global rate coefficient. n and m relate to the reaction order. According to Eqn 2.2, reaction is - nth order with respect to fuel, - mth order with respect to oxidant, and - (m + n)th order overall. m and n are determined from experimental data and are not necessarily integers. 2. Chemical Kinetics 5 AER 1304–ÖLG Use of global reactions to express chemistry is usually a black box approach and has limited use in combustion. It does not provide a basis for understanding what is actually happening. Let’s consider the following global reaction: 2H2 + O2 → 2H2 O (2.3) It implies that two moles of hydrogen molecule react with one mole of oxygen to form one mole of water, which is not strictly true. 2. Chemical Kinetics 6 AER 1304–ÖLG In reality many sequential processes occur that involve several intermediate species. Following elementary reactions, among others, are important in conversion of H2 and O2 to water: H2 + O2 → HO2 + H (2.4) H + O2 → OH + O (2.5) OH + H2 → H2 O + H (2.6) H + O2 + M → HO2 + M (2.7) 2. Chemical Kinetics 7 AER 1304–ÖLG Radicals or free radicals or reactive species are reactive molecules, or atoms, that have unpaired electrons. To have a complete picture of the combustion of H2 with O2 , more than 20 elementary reactions can be considered. Reaction mechanism is the collection of elemen- tary reactions to describe the overall reaction. Reaction mechanisms may involve a few steps or as many as several hundred (even thousands). (State-of-the-art). 2. Chemical Kinetics 8 AER 1304–ÖLG Elementary Reaction Rates - Using the concept of elementary reactions has many advantages. - Reaction order is constant and can be experimen- tally determined. - molecularity of the reaction: number of species that form the reaction complex. - Unimolecular - Bimolecular - Trimolecular / Termolecular 2. Chemical Kinetics 9 AER 1304–ÖLG Bimolecular Reactions & Collision Theory Most combustion related elementary reactions are bimolecular: A+B→C+D (2.8) The rate at which the reaction proceeds is d[A] = −kbimolec [A][B] (2.9) dt kbimolec ∝ f (T ) and has a theoretical basis, unlike kG , rate coefficient of a global reaction. 2. Chemical Kinetics 10 AER 1304–ÖLG Collision theory for bimolecular reactions has sev- eral shortcomings. Approach is important for historical reasons and may provide a simple way to visualize bimolecular reactions. Uses the concepts of wall collision frequency, mean molecular speed, and mean free path. The simpler approach is to consider a single molecule of diameter σ travelling at constant speed v and experiencing collisions with identi- cal, but stationary molecules. 2. Chemical Kinetics 11 AER 1304–ÖLG If the distance travelled (mean free path) between colisions is large, then moving molecule sweeps out a cylindrical volume of vπσ 2 ∆t. For random distribution of stationary molecules with number density n/V , number of collisions collisions Z≡ = (n/V )vπσ 2 (2.10) per unit time For Maxwellian velocity distribution for all molecules √ Zc = 2(n/V )πσ 2 v̄ (2.11) 2. Chemical Kinetics 12 AER 1304–ÖLG Eqn.2.11 applies to identical molecules. For dif- ferent molecules, we can use σA + σB ≡ 2σAB √ 2 Zc = 2(nB /V )πσAB v̄A (2.12) which expresses frequency of collisions of a single A molecule with all B molecules. For all A molecules 2 2 2 1/2 ZAB /V = (nA /V )(nB /V )πσAB (v̄A + v̄B ) (2.13) 2. Chemical Kinetics 13 AER 1304–ÖLG If we express mean molecular speed in terms of temperature, 8kB T 1/2 2 ZAB /V = (nA /V )(nB /V )πσAB πµ (2.14) kB = Boltzmann constant. µ = (mA mB )/(mA + mB ) = reduced mass. T = absolute temperature. 2. Chemical Kinetics 14 AER 1304–ÖLG We can relate ZAB /V to reaction rates No. of collisons Probability that d[A] A and B molecules a collision − = per unit volume · leads to dt per unit time reaction kmol of A · (2.15a) No. of molecules of A or d[A] −1 − = (ZAB /V )PNAV (2.15b) dt 2. Chemical Kinetics 15 AER 1304–ÖLG The probability that a collision will lead to a reac- tion can be expressed as a product of two factors: - an energy factor exp [−EA /(Ru T )] which expresses the fraction of collisions that occur with an energy above the activation energy - a geometrical or steric factor p, that takes into account the geometry of collisions be- tween A and B. 2. Chemical Kinetics 16 AER 1304–ÖLG With substitutions nA /V = [A]NAV and nB /V = [B]NAV , Eqn.2.15b becomes d[A] 8πk T 1/2 2 B − =pNAV σAB · dt µ (2.16) exp [−EA /(Ru T )][A][B] Comparing Eqn. 2.16 with 2.9 8πkB T 1/2 −EA 2 k(T ) = pNAV σAB exp µ Ru T (2.17) 2. Chemical Kinetics 17 AER 1304–ÖLG Collision theory is not capable of providing any means to determine EA or p. More advanced theories do allow calculation of k(T ) from first principles to a limited extent. If the temperature range of interest is not too large, kbimolec can be expressed by the semi- empirical Arrhenius form −EA k(T ) = A exp (2.18) Ru T where A is a constant termed pre-exponential fac- tor or frequency factor. 2. Chemical Kinetics 18 AER 1304–ÖLG Most of the time the experimental values for rate coefficients in Arrhenius form expressed as b −EA k(T ) = AT exp (2.19) Ru T where A, b, and EA are three empirical constants. The standard method for obtaining EA is to graph experimental rate constant data versus inverse of temperature, i.e. log k vs 1/T. The slope gives EA /Ru. 2. Chemical Kinetics 19 AER 1304–ÖLG Unimolecular Reactions: Involves single species A→B (2.20) A→B+C (2.21) - Examples: O2 → O + O; H2 → H + H. - First order at high pressures d[A] = −kuni [A] (2.22) dt 2. Chemical Kinetics 20 AER 1304–ÖLG - At low pressures, the reaction rate may also de- pend a third molecule that may exist within the reaction volume d[A] = −k[A][M] (2.23) dt Termolecular Reactions: A+B+M→C+M (2.24) Termolecular reactions are third order d[A] = −kter [A][B][M] (2.25) dt 2. Chemical Kinetics 21 AER 1304–ÖLG Multistep Mechanisms Net Production Rates Consider some of the reactions in H2 -O2 system kf 1 H2 + O2 kr1 HO2 + H (R.1) kf 2 H + O2 kr2 OH + O (R.2) kf 3 OH + H2 kr3 H2 O + H (R.3) kf 4 H + O2 + M kr4 HO2 + M (R.4) 2. Chemical Kinetics 22 AER 1304–ÖLG The net production rate of any species, say X, involved is the sum of all of the individual ele- mentary rates producing X minus all of the rates destroying X. Net production rate of O2 is then, d[O2 ] =kr1 [HO2 ][H] + kr2 [OH][O] dt (2.26) + kr4 [HO2 ][M] − kf 1 [H2 ][O2 ] − kf 2 [H][O2 ] − kf 4 [H][O2 ][M] 2. Chemical Kinetics 23 AER 1304–ÖLG Net production rate for H atoms: d[H] =kf 1 [H2 ][O2 ] + kr2 [OH][O] dt + kf 3 [OH][H2 ] + kr4 [HO2 ][M] (2.27) − kr1 [HO2 ][H] − kf 2 [H][O2 ] − kr3 [H2 O][H] − kf 4 [H][O2 ][M] d[Xi ](t) = fi {[X1 ](t), [X2 ](t),......[Xn ](t)} dt [Xi ](0) = [Xi ]0 (2.28) 2. Chemical Kinetics 24 AER 1304–ÖLG Compact Notation: Since mechanisms may involve many elementary steps and many species, a generalized compact notation has been developed for the mechanism and the individual species production rates. For the mechanism, N N νji Xj νji Xj for i = 1, 2,...L (2.29) j=1 j=1 where νji and νji are stoichiometric coefficients of reactants and products, respectively. 2. Chemical Kinetics 25 AER 1304–ÖLG N L j Species i Reaction 1 O2 1 R.1 2 H2 2 R.2 3 H2 O 3 R.3 4 HO2 4 R.4 5 O 6 H 7 OH 8 M 2. Chemical Kinetics 26 AER 1304–ÖLG Stoichiometric coefficient matrices:   1 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 νji =   (2.30a) 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1   0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 νji =   (2.30b) 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 2. Chemical Kinetics 27 AER 1304–ÖLG Net production rate of each species in a multistep mechanism: L d[Xj ]/dt ≡ ω̇j = νji qi for j = 1, 2,.....N i=1 (2.31) where νji = (νji − νji ) (2.32) N N I II νji νji qi = kf i [Xj ] − kri [Xj ] (2.33) j=1 j=1 2. Chemical Kinetics 28 AER 1304–ÖLG For example, qi (= q1 ) for reaction R.1 is qi =kf 1 [O2 ]1 [H2 ]1 [H2 O]0 [HO2 ]0 [O]0 [H]0 [OH]0 [M]0 − kr1 [O2 ]0 [H2 ]0 [H2 O]0 (2.34) [HO2 ]1 [O]0 [H]1 [OH]0 [M]0 = kf 1 [O2 ][H2 ] − kr1 [HO2 ][H] Writing similar expressions for i = 2, 3, and 4 and summing completes the total rate expression for ω̇j. 2. Chemical Kinetics 29 AER 1304–ÖLG Rate Coefficients and Equilibrium Constants: At equilibrium forward and reverse reaction rates must be equal. kf A+B kr C+D (2.35) Formation rate of species A: d[A] = −kf [A][B] + kr [C][D] (2.36) dt For equilibrium, time rate of change of [A] must be zero. Same goes for B, C, and D. 2. Chemical Kinetics 30 AER 1304–ÖLG Then, Eqn. 2.36 0 = −kf [A][B] + kr [C][D] (2.37) arranging [C][D] kf (T ) = (2.38) [A][B] kr (T ) Previously we have defined equilibrium con- stant as, (PC /P o )c (PD /P o )d... Kp = (2.39) (PA /P o )a (PB /P o )b... 2. Chemical Kinetics 31 AER 1304–ÖLG Since molar concentrations are related to mol fractions and partial pressures as, [Xi ] = χi P/(Ru T ) = Pi /(Ru T ) (2.40) we can define an equlibrium constant based on molar concentrations, Kc and relate it to Kp , Kp = Kc (Ru T /P o )c+d+...−a−b... (2.41a) or o Σν II −Σν I KP = Kc (Ru T /P ) (2.41b) 2. Chemical Kinetics 32 AER 1304–ÖLG where, Kc is defined as, νiII [Xi ] [C]c [D]d... prod Kc = a b = νiI (2.42) [A] [B]... [Xi ] react So that, kf (T ) = Kc (T ) (2.43) kr (T ) For bimolecular reactions Kc = Kp. 2. Chemical Kinetics 33 AER 1304–ÖLG Steady-State Approximation Analysis of reactive systems can be simplified by applying steady-state approximation to the reactive species or radicals. Steady-state approximation is justified when the reaction forming the intermediate species is slow, while the reaction destroying the interme- diate species is very fast. As a result the concentration of the radical is small in comparison with those of the reactants and products. 2. Chemical Kinetics 34 AER 1304–ÖLG Example (Zeldovich mechanism for NO forma- tion): k1 O + N2 −→ NO + N k2 N + O2 −→ NO + O First reaction is slow (rate limiting); while sec- ond is fast. Net production rate of N atoms, d[N] = k1 [O][N2 ] − k2 [N][O2 ] (2.44) dt 2. Chemical Kinetics 35 AER 1304–ÖLG After a rapid transient allowing buildup of N, d[N]/dt approaches zero. 0 = k1 [O][N2 ] − k2 [N]ss [O2 ] (2.45) k1 [O][N2 ] [N]ss = (2.46) k2 [O2 ] Time rate of change of [N]ss is d[N]ss d k1 [O][N2 ] = (2.47) dt dt k2 [O2 ] 2. Chemical Kinetics 36 AER 1304–ÖLG Mechanism for Unimolecular Reactions Let’s consider a three-step mechanism: ke ∗ A + M −→ A + M (2.48a) ∗ kde A + M −→ A + M (2.48b) ∗ kunim A −→ products (2.48c) In step 1: kinetic energy transferred to A from M; A has increased internal vibrational and rotational energies and becomes an energized A molecule, A∗. 2. Chemical Kinetics 37 AER 1304–ÖLG Two possible scenarios for A∗ : - A∗ may collide with another molecule and goes back to A (2.48b) - A∗ may decompose into products (2.48c) The rate at which products are formed: d[products] = kunim [A∗ ] (2.49) dt Net production rate of A∗ : d[A∗ ] = ke [A][M] − kde [A∗ ][M] − kunim [A∗ ] dt (2.50) 2. Chemical Kinetics 38 AER 1304–ÖLG Steady-state approximation for A∗ , i.e. d[A∗ ]/dt = 0, ∗ ke [A][M] [A ] = (2.51) kde [M] + kunim Substitute Eqn.2.51 into 2.49, d[products] ke [A][M] = (2.52) dt (kde /kunim )[M] + 1 2. Chemical Kinetics 39 AER 1304–ÖLG Another form of writing the overall rate of pro- duction of products for the overall reaction: kapp A −→ products (2.53) d[A] products − = = kapp [A] (2.54) dt dt kapp is an apparent unimolecular rate coef. Equating Eqns.2.52 and 2.54 yields ke [M] kapp = (2.55) (kde /kunim )[M] + 1 2. Chemical Kinetics 40 AER 1304–ÖLG Eqn.2.55 lets us to explain pressure depen- dence of unimolecular reactions: - At high enough pressures (kde [M]/kunim ) >> 1 because [M] increases as the pressure is in- creased; then kapp (P → ∞) = kunim ke /kde (2.56) - At low enough pressures (kde [M]/kunim ) [A]0 , 1 τchem = (2.75) [B]0 kbimolec 2. Chemical Kinetics 55 AER 1304–ÖLG Termolecular Reactions kter A + B + M −→ C + M (2.24) For a simple system at constant T , [M] is con- stant d[A] = (−kter [M])[A][B] (2.9) dt where (−kter [M]) plays the same role as kbimolec does for a bimolecular reaction. 2. Chemical Kinetics 56 AER 1304–ÖLG Then, the characteristic time for termolecular reactions is, ln [e + (1 − e)([A]0 /[B]0 )] τchem = (2.77) ([B]0 − [A]0 )kter [M] And, when [B]0 >> [A]0 , 1 τchem = (2.78) [B]0 (kter [M]) 2. Chemical Kinetics 57 AER 1304–ÖLG 3. Combustion Chemistry - For many kinetically controlled systems, the num- ber of elementary reaction steps is so large that simple analytical solutions are not possible. - Most of the time, a large number of chemical species are involved. - Even in combustion of simple hydrocarbons, num- ber of species involved is quite high. - To illustrate how the fundamental principles of chemical kinetics apply to real-world problems, we will consider some examples. 3. Combustion Chemistry 1 AER 1304 – ÖLG Hydrogen-Oxygen Reaction The reaction between hydrogen and oxygen is a good example of a multicomponent kinetic system. To de- scribe the sysytem properly, we should consider eight major species and at least 16 reactions. - The overall reaction is: 2H2 + O2 → 2H2 O This reaction is exothermic; but mixtures of gaseous hydrogen and oxygen are quite stable at atmospheric conditions. Any conceivable direct reaction between the two gases is zero. 3. Combustion Chemistry 2 AER 1304 – ÖLG - The reaction half-time at atmospheric conditions has been estimated to be much larger than the age of the universe. - If the reaction is initiated by some free-radical species, then the reaction proceeds very rapidly and violently. - The radicals are typically H and O atoms pro- duced from the dissociation of H2 and O2 , re- spectively. - We will consider a simplified mechanism that rep- resents the gross features of H2 + O2 at low P. 3. Combustion Chemistry 3 AER 1304 – ÖLG - Initiation ko H2 −→ 2H (H.0a) ko O2 −→ 2O (H.0b) - Chain branching k2 O2 + H −→ OH + O (H.2) k3 H2 + O −→ OH + H (H.3) - Chain propagation 3. Combustion Chemistry 4 AER 1304 – ÖLG k1 H2 + OH −→ H2 O + H (H.1) - Chain termination k4 H + wall −→ (1/2)H2 (H.4) k5 H + O2 + M −→ HO2 + M (H.5) - Initiation step is the dissociation of some amount of molecular species by a spark, flame, electric discharge, or some other means. 3. Combustion Chemistry 5 AER 1304 – ÖLG - Bond energies: O = O : 5.1 eV H − H : 4.5 eV O − H : 4.4 eV - Reaction (2) is endothermic by 0.7 eV (about 70 kJ/mol) and progresses slowly. - Reactions (3) and (1) are endothermic by 0.1 eV, and these reactions are relatively fast. - The OH and O radicals are therefore rapidly con- sumed, and the principal chain carrier is H atoms. 3. Combustion Chemistry 6 AER 1304 – ÖLG - We will find the rate equation for the free-radical density n∗ , which is taken to be the same as that for [H] atoms: dn∗ d[H] = dt dt = ko [H2 ] − k2 [H][O2 ] + k3 [O][H2 ] + k1 [OH][H2 ] −k4 [H] − k5 [H][O2 ][M] (3.1) - For other free-radical species: d[OH] = k2 [H][O2 ] + k3 [O][H2 ] − k1 [OH][H2 ] dt (3.2) 3. Combustion Chemistry 7 AER 1304 – ÖLG d[O] = ko [O2 ] + k2 [H][O2 ] − k3 [O][H2 ] (3.3) dt - Since [O] and [OH] are both much lower than [H], we can assume that both of these species are at steady-state: d[OH] d[O] = =0 (3.4) dt dt that yields ko [O2 ] + k2 [H][O2 ] [O]ss = (3.5) k3 [H2 ] 3. Combustion Chemistry 8 AER 1304 – ÖLG and for [OH], k2 [H][O2 ] + k3 [O][H2 ] [OH]ss = (3.6) k1 [H2 ] Substituting for [O]ss 2k2 [H][O2 ] + ko [O2 ] [OH]ss = (3.7) k1 [H2 ] - We substitute the steady-state values of [OH] and [O] radicals in the rate equation for free-radical density 3. Combustion Chemistry 9 AER 1304 – ÖLG dn∗ k2 [H][O2 ] + ko [O2 ] = ko [H2 ]−k2 [H][O2 ]+k3 [H2 ] dt k3 [H2 ] 2k2 [H][O2 ] + ko [O2 ] +k1 [H2 ] − k4 [H] − k5 [H][O2 ][M] k1 [H2 ] = ko ([H2 ] + 2[O2 ]) +(2k2 [O2 ] − k4 − k5 [O2 ][M])n∗ , 1 , 1 , 1 wo f g (3.8) so that dn∗ = wo + (f − g)n∗ (3.9) dt 3. Combustion Chemistry 10 AER 1304 – ÖLG - f : chain branching - g: chain termination - Two classes of solutions are possible: 1. g > f : termination exceeds branching 2. g < f : branching exceeds termination - Solution 1 implies that k4 > (2k2 − k5 [M])[O2 ] (3.10) so that g > f is assured at sufficiently low O2 pressures. 3. Combustion Chemistry 11 AER 1304 – ÖLG ωo/(g-f) Free-radical density, n* n* ~ ωo t Time, t 3. Combustion Chemistry 12 AER 1304 – ÖLG - The solution is then ∗ wo n = {1 − e−(g−f )t } (3.11) g−f - At short times, n∗ increases almost linearly with slope ∼ wo t and reaches a steady-state value of n∗ss = wo /(g − f ) (3.12) - At higher O2 pressures g < f , then ∗ wo n = {e(f −g)t − 1} (3.13) f −g 3. Combustion Chemistry 13 AER 1304 – ÖLG Free-radical density, n* Time, t 3. Combustion Chemistry 14 AER 1304 – ÖLG - In this case, the free-radical concentration in- creases exponentially, and, since the overall rate depends on the radical concentration, the reaction velocity increases rapidly. This is usually termed as an explosion. - The hydrogen-oxygen reaction behaves quite dif- ferently in different pressure regimes. - It is possible to construct an explosion boundary for the hydrogen-oxygen reaction as a function of temperature and pressure. 3. Combustion Chemistry 15 AER 1304 – ÖLG I II III IV Total Rate Chain-reaction Thermal Slow reaction Atmospheric flame Explosion Explosion Total Pressure (H2 + O2) P1 * P 2* P 3* 3. Combustion Chemistry 16 AER 1304 – ÖLG - Region I: The reaction is wall-recombination lim- ited, and proceeds to a steady-state. - Region II: f begins to exceed g, the branched- chain reaction takes over and explosion ensues. - Region III: As the pressure is further increased, the explosion is quenched and another regime of steady-state is encountered. In III, kinetics are dominated by relatively unreac- tive HO2 (hydroperoxyl) radicals. Other reactions, in addition to those we considered above, are im- portant and include the following: 3. Combustion Chemistry 17 AER 1304 – ÖLG 2HO2 −→ H2 O2 + O2 (H.6) H2 + HO2 −→ H2 O2 + H (H.7) H + HO2 −→ 2OH (H.8) H + HO2 −→ H2 + O2 (H.9) H + HO2 −→ H2 O + O (H.10) H2 O2 + M −→ 2OH + M (H.11) H2 O2 + H −→ H2 O + OH (H.12) H2 O2 + H −→ H2 + O2 (H.13) H2 O2 + OH −→ H2 O + HO2 (H.14) H + OH + M −→ H2 O + M (H.15) H + H + M −→ H2 + M (H.16) 3. Combustion Chemistry 18 AER 1304 – ÖLG - Region IV: At a still high pressure, the amount of heat liberated in the exothermic steps of the mech- anism becomes larger than can be dissipated by conduction and other thermal transport processes, and the temperature rises. This, in turn increases the rates of initiation and provides heat for en- dothermic chain-branching reactions leading to more heat release, resulting in a thermal explosion. Experimentally observed (P − T ) boundaries of the H2 -O2 reaction in a closed vessel is shown below. 3. Combustion Chemistry 19 AER 1304 – ÖLG 3. Combustion Chemistry 20 AER 1304 – ÖLG Carbon Monoxide Oxidation Oxidation of CO is important in hydrocarbon com- bustion. From a very simplistic point-of-view, hydrocarbon combustion (related to C content) can be charac- terized as a two-step process: - breakdown of fuel to CO. - oxidation of CO to CO2. CO oxidation is extremely slow in the absence of small amounts of H2 or H2 O. 3. Combustion Chemistry 21 AER 1304 – ÖLG If the H2 O is the primary hydrogen-containing species, the CO oxidation can be described by: CO + O2 −→ CO2 + O (CO.1) O + H2 O −→ OH + OH (CO.2) CO + OH −→ CO2 + H (CO.3) H + O2 −→ OH + O (CO.4) - (CO.1) is slow; not much contribution to CO2 formation, but chain inititation reaction. - (CO.3) is the actual CO oxidation step; also chain- propagation step producing H atoms. 3. Combustion Chemistry 22 AER 1304 – ÖLG - (CO.2) and (CO.4) are chain-branching reactions producing OH, and OH and H, respectively. - (CO.3) reaction is the key step in CO oxidation. If H2 is present, then following steps are involved: O + H2 −→ OH + H (CO.5) OH + H2 −→ H2 O + H (CO.6) CO + HO2 −→ CO2 + OH (CO.7) - In the presence of H2 , the entire H2 -O2 reaction system should be included to describe CO oxida- tion. 3. Combustion Chemistry 23 AER 1304 – ÖLG Oxidation of Higher Paraffins General Scheme: Alkanes=Paraffins: saturated, straight chain or branched-chain, single-bonded hydrocarbons. General formula: Cn H2n+2. Generic oxidation discussion will be for n > 2. Methane (and ethane) display some unique charac- teristics not common with higher alkanes. Overview of the key points of alkane oxidation. 3. Combustion Chemistry 24 AER 1304 – ÖLG Three sequential processes: I. Fuel is attacked by O and H; breaks down to H2 and olefins (double-bonded straight hydrocarbons). H2 oxidizes to H2 O. II. Unsaturated olefins form CO and H2. Almost all H2 converts to water. III. CO burns to CO2 releasing almost all of the heat associated with combustion: CO + OH −→ CO2 + H (CO.3) 3. Combustion Chemistry 25 AER 1304 – ÖLG These three processes can be further detailed as (with the example of propane, C3 H8 ): Step#1. A C-C bond is broken in the original fuel molecule. A C-C bond is weaker than an H- C bond. C3 H8 + M −→ C2 H5 + CH3 + M (P.1) Step#2. Two resulting hydrocarbon radicals break down further to olefins: H-atom abstraction. C2 H5 + M −→ C2 H4 + H + M (P.2a) CH3 + M −→ CH2 + H + M (P.2b) 3. Combustion Chemistry 26 AER 1304 – ÖLG Step#3. H atoms from Step#2 starts a radical pool: H + O2 −→ O + OH (P.3) Step#4. With the development of a radical pool, at- tack on the fuel molecule intensifies. C3 H8 + OH −→ C3 H7 + H2 O (P.4a) C3 H8 + H −→ C3 H7 + H2 (P.4b) C3 H8 + O −→ C3 H7 + OH (P.4c) 3. Combustion Chemistry 27 AER 1304 – ÖLG Step#5. Hydrocarbon radicals decay to olefins and H atoms via H-atom abstraction. C3 H7 + M −→ C3 H6 + H + M (P.5) This process obeys the β-scission rule, which states that C-C or C-H bond broken will be the one that is one place removed from the radical site. C3 H6 + H + M C3 H7 + M (P.6) C2 H4 + CH3 + M 3. Combustion Chemistry 28 AER 1304 – ÖLG 3. Combustion Chemistry 29 AER 1304 – ÖLG Step#6. Oxidation of olefins created in Steps#2 and 5 by O that produces formyl radicals (HCO) and formaldehyde (H2 CO). C3 H6 + O −→ C2 H5 + HCO (P.7a) C3 H6 + O −→ C2 H4 + H2 CO (P.7b) Step#7. Methyl radicals (CH3 ), formaldeydhe (H2 CO), and methylene (CH2 ) oxidize. Step#8. Carbon monoxide oxidizes following the CO mechanism discussed previously. 3. Combustion Chemistry 30 AER 1304 – ÖLG Hierarchy in the reaction mechanism describing alkane combustion. 3. Combustion Chemistry 31 AER 1304 – ÖLG Global Mechanisms: Global models do not capture all the features of hydrocarbon combustion, but they may be useful in simple engineering approximations as long as their limitations are recognized. A single-step expression: kG Cx Hy + (x + y/4)O2 −→ xCO2 + (y/2)H2 O (G.1) d[Cx Hy ] = −A exp [−Ea /(Ru T )][Cx Hy ]m [O2 ]n dt (G.2) 3. Combustion Chemistry 32 AER 1304 – ÖLG A multi-step global mechanism: Cn H2n+2 → (n/2)C2 H4 + H2 (M.1) C2 H4 + O2 → 2CO + 2H2 (M.2) CO + (1/2)O2 → CO2 (M.3) H2 + (1/2)O2 → H2 O (M.4) which assumes that the intermediate hydrocarbon is ethylene (C2 H4 ). 3. Combustion Chemistry 33 AER 1304 – ÖLG Methane oxidation mechanism 3. Combustion Chemistry 34 AER 1304 – ÖLG Nitrogen Oxide Kinetics - Combustion products contain NO at levels of sev- eral hundred to several thousand parts per million (ppm) and NO2 levels in tens of ppm. - In the atmosphere, in the presence of ultraviolet sunlight, an equilibrium is established: hν NO2 + O2 NO + O3 , 1 ozone 3. Combustion Chemistry 35 AER 1304 – ÖLG - Presence of certain hydrocarbons (e.g., unburned hydrocarbons from combustion, methane from var- ious sources) slowly unbalances the above reac- tion. - NO contributes to destruction of ozone in strato- sphere. - NO contributes to production of ground level ozone. - NO is involved in photochemical smog and haze. 3. Combustion Chemistry 36 AER 1304 – ÖLG The main sources of nitrogen oxide, NOx , emissions from combustion are: - Thermal NO: oxidation of molecular nitrogen in the postflame zone. - Prompt NO: formation of NO in the flame zone (Fenimore mechanism). - N2 O-intermediate mechanism. - Fuel NO: oxidation of nitrogen-containing com- pounds in the fuel. Relative importance of these three are dependent on the operating conditions and fuel. In most practical combustion devices the thermal NO is the main source. 3. Combustion Chemistry 37 AER 1304 – ÖLG The basic mechanism for thermal NO production is given by six reactions known as extended Zeldovich mechanism: k1f O + N2 k1r NO + N (N.1) k2f N + O2 k2r NO + O (N.2) k3f N + OH k3r NO + H (N.3) 3. Combustion Chemistry 38 AER 1304 – ÖLG - The contribution of reaction 3 is small for lean mixtures, but for rich mixtures it should be con- sidered. Forward reaction 1 controls the system, but it is slow at low temperatures (high activation energy). Thus it is effective in post-flame zone where temperature is high and the time is avail- able. - Concentrations of 1000 to 4000 ppm are typically observed in uncontrolled combustion systems. - From reactions 1-3, the rate of formation of ther- mal NO can be calculated: 3. Combustion Chemistry 39 AER 1304 – ÖLG d[NO] = k1f [O][N2 ] − k1r [NO][N] + k2f [N][O2 ] dt −k2r [NO][O] + k3f [N][OH] − k3r [NO][H] (3.14) - To calculate the NO formation rate, we need the concentrations of O, N, OH, and H. - In detailed calculations, these are computed using detailed kinetic mechanisms for the fuel used. - For very approximate calculations, these may be assumed to be in chemical equilibrium. 3. Combustion Chemistry 40 AER 1304 – ÖLG - At moderately high temperatures N does not stay at thermodynamic equilibrium. A better approxi- mation could be to assume N to be at steady-state. - From reactions 1-3, we have d[N] = k1f [O][N2 ] − k1r [NO][N] − k2f [N][O2 ] dt +k2r [NO][O] − k3f [N][OH] + k3r [NO][H] = 0 k1f [O][N2 ] + k2r [NO][O] + k3r [NO][H] [N]ss = k1r [NO] + k2f [O2 ] + k3f [OH] (3.15) 3. Combustion Chemistry 41 AER 1304 – ÖLG - The reaction rate constants, in [m3 / kmol s], for 1-3 are as follows: k1f = 1.8 · 1011 exp (−38, 370/T ) k1r = 3.8 · 1010 exp (−425/T ) k2f = 1.8 · 107 T exp (−4680/T ) 6 (5.16) k2r = 3.8 · 10 T exp (−20, 820/T ) k3f = 7.1 · 1010 exp (−450/T ) 11 k3r = 1.7 · 10 exp (−24, 560/T ) 3. Combustion Chemistry 42 AER 1304 – ÖLG N2 O-intermediate mechanism is important in very- lean combustion (Φ < 0.8). This mechanism can be represented by: O + N2 + M N2 O + M (N.4) H + N2 O NO + NH (N.5) O + N2 O NO + NO (N.6) - This mechanism is important in NO control strate- gies in lean-premixed gas turbine combustion ap- plications. 3. Combustion Chemistry 43 AER 1304 – ÖLG It has been shown that some NO is rapidly pro- duced in the flame zone long before there would be time to form NO by the thermal mechanism. This is also known as the Fenimore mechanism: - The general scheme is that hydrocarbon radicals form CN and HCN CH + N2 HCN + N (N.7) C + N2 CN + N (N.8) 3. Combustion Chemistry 44 AER 1304 – ÖLG - The conversion of hydrogen cyanide, HCN, to form NO is as follows HCN + O NCO + H (N.9) NCO + H NH + CO (N.10) NH + H N + H2 (N.11) N + OH NO + H (N.3) - For equivalence ratios higher than 1.2, chemistry becomes more complex and it couples with the thermal mechanism. 3. Combustion Chemistry 45 AER 1304 – ÖLG

Combustion - AER 1304

Document Details

Tags

Related

Summary

Full Transcript

Upgrade to continue