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AME 436 Energy and Propulsion

AME 436 Energy and Propulsion. Lecture 4 Basics of combustion. Outline. Why do we need to study combustion? Types of flames - premixed and nonpremixed Basics of chemical reaction rates Law Of Mass Action (LOMA) Arrhenius form of temperature dependence Premixed flames

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AME 436 Energy and Propulsion

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  1. AME 436Energy and Propulsion Lecture 4 Basics of combustion

  2. Outline • Why do we need to study combustion? • Types of flames - premixed and nonpremixed • Basics of chemical reaction rates • Law Of Mass Action (LOMA) • Arrhenius form of temperature dependence • Premixed flames • Deflagrations - burning velocity, flame thickness, temperature effect • Turbulence effects • Homogeneous reaction • Nonpremixed flames • General characteristics • Droplets • Gas-jet • Turbulence effects AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  3. Why do we need to study combustion? • Chemical thermodynamics only tells us the end states - what happens if we wait “forever and a day” for chemical reaction to occur • We also need to know how fast reactions occur • How fast depends on both the inherent rates of reaction and the rates of heat and mass transport to the reaction zone(s) • Chemical reactions + heat & mass transport = combustion • Some reactions occur too slowly to be of any consequence, e.g. 2 NO  N2 + O2 has an adiabatic flame temperature of 2869K (no dissociation) or 2650K (with dissociation, mostly NO & O) but no one has ever made a flame with NO because reaction rates are too slow! • What do we do with this information in the context of engines? • Determine rates of flame propagation and heat generation • Determine conditions for “knock” in premixed-charge engines • Determine rates of pollutant formation and destruction AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  4. Types of flames • Premixed - reactants are intimately mixed on the molecular scale before combustion is initiated; several flavors • Deflagration • Detonation • Homogeneous reaction • Nonpremixed - reactants mix only at the time of combustion - have to mix first then burn; several flavors • Gas jet (Bic lighter) • Liquid fuel droplet • Liquid fuel jet (e.g. Kuwait oil fire, candle) • Solid (e.g. coal particle, wood) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  5. Premixed flames - deflagration • Propagating, subsonic front sustained by conduction of heat from the hot (burned) gases to the cold (unburned) gases which raises the temperature enough that chemical reaction can occur • Since chemical reaction rates are very sensitive to temperature, most of the reaction is concentrated in a thin zone near the high-temperature side • May be laminar or turbulent Turbulent premixed flame experiment in a fan-stirred chamberhttp://www.mech-eng.leeds.ac.uk/res-group/combustion/activities/Bomb.htm AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  6. Premixed flames - detonation • Supersonic propagating front sustained by heating of gas by a shock wave • After shock front, need time (thus distance = time x velocity) before reaction starts to occur (“induction zone”) • After induction zone, chemical reaction & heat release occur • Pressure & temperature behavior coupled strongly with supersonic/subsonic gasdynamics • Ideally only M3 = 1 “Chapman-Jouget detonation” is stable (M = Mach number = u/c; u = velocity, c = sound speed = (RT)1/2 for ideal gas) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  7. Premixed flames - homogeneous reaction • Model for knock in premixed-charge engines • Fixed mass (control mass) with uniform (in space) T, P and composition • No “propagation” in space but propagation in time • In laboratory, we might heat the chamber to a certain T and see how long it took to react; in engine, compression of mixture (increases P & T, thus reaction rate) will initiate reaction AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  8. “Non-premixed” or “diffusion” flames • Only subsonic • Generally assume “mixed is burned” - mixing slower than chemical reaction AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  9.  ≈ ()1/2 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  10. Law of Mass Action (LoMA) • First we need to describe rates of chemical reaction • The Law of Mass Action (LoMA) states that the rate of reaction is proportional to the number of collisions between the reactant molecules, which in turn is proportional to the concentration of each reactant • For a chemical reaction of the form AA + BB  CC + DD e.g. 1 H2 +1 I2 2 HI A = H2, A = 1, B = I2, B = 1, C = HI, C = 2, D = nothing, D = 0 Loma states that the rate of reaction is given by [ i ] = concentration of molecule i (usually moles per liter) kf = “forward” reaction rate constant • Minus sign on right-hand side since A & B depleted, C & D formed • How to calculate [ i ]? • According to ideal gas law, the total moles of gas per unit volume (all molecules, not just type i) = P/T • Then [ i ] = (Total moles / volume)*(moles i / total moles), thus [ i ] = (P/T)Xi(Xi = mole fraction of i (see lecture 2)) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  11. Collision rate • How to estimate kf? Chemical reaction requires • Collision between molecules • Enough kinetic energy; call this the “activation energy Ea • Luck (“steric factor”, p) – e.g. orientation at time of collision • How many collisions does a molecule undergo per unit time? • Molecule of diameter s & speed c will sweep out an effective volume per unit time = πs2c (Why diameter not radius? Molecules collide when center-to-center spacing is 2r = s) • (Molecules per unit volume) * (Volume swept per unit time) ~ Frequency of collisions = [A]πs2c • Multiply by [A] to get collisions per unit volume per unit time = [A][A]πs2c • Kinetic theory of gases give mean molecular speed (c) & number of collisions per unit volume (V) per unit time for A & B (ZAB): k = Boltzmann’s const. = 1.38 x 10-23 J/K = R/NAvogadro; m = mass of 1 molecule AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

  12. Collision rate • On a mole (rather than molecule) basis • Example: how often do N2 molecules collide at 298K & 1 atm? s = 3.61 x 10-10 m, mass = 4.65 x 10-26 kg (µN2-N2= 2.32 x 10-26kg) • At 2200K, ZN2-N2 = 4.98 x 109/s; typical time for hydrocarbon combustion ≈ 10-3 s, thus each molecule collides ≈ 5 x 106 times! AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

  13. Reaction rate • The reaction rate constant kf is usually of the Arrhenius form Z = pre-exponential factor, a = nameless constant, E = “activation energy” (cal/mole) Working backwards, units of Z are (moles per liter)1-A-vB/(Kasec) • With 3 parameters (Z, n, E) any curve can be fit! • The exponential term causes extreme sensitivity to T for E/ >> T! AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  14. Reaction rate • Boltzman (1800’s) showed that the fraction of molecules in a gas with translational kinetic energy greater than a value E ~ exp(-E/T), thus E is the “energy barrier” that must be overcome for reaction to occur • E is not the same as enthalpy of reaction hf (or heating value QR) and in general the two have no relation to each other - E affects reaction rates whereas hf & QR affect end states (e.g. Tad) - of course hf & QR affect reaction rates indirectly by affecting T “Diary of a collision” AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  15. Comments on LoMA • The full reaction rate expression is then • The H2 + I2 2HI example is one of few where reactants products occurs in a single step; most fuels go through many intermediates during oxidation - even for the simplest hydrocarbon (CH4) the “standard” mechanism (http://www.me.berkeley.edu/gri_mech/) includes 53 species and 325 individual reactions! • The only likely reactions in gases, where the molecules are far apart compared to their size, are 1-body, 2-body or 3-body reactions, i.e. A  products, A + B  products or A + B + C  products • In liquid or solid phases, the close proximity of molecules makes n-body reactions plausible AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  16. Comments on LoMA • Recall that the forward reaction rate is • Similarly, the rate of the reverse reaction can be written as kb = “backward” reaction rate constant • At equilibrium, the forward and reverse rates must be equal, thus This ties reaction rate constants (kf, kb) and equilibrium constants (Ki’s) together AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  17. Deflagrations - burning velocity • Since the burning velocity (SL) << sound speed, the pressure across the front is almost constant • How fast will the flame propagate? Simplest estimate based on the hypothesis that Rate of heat conducted from hot gas to cold gas (i) = Rate at which enthalpy is convected through flame front (ii) = Rate at which enthalpy is produced by chemical reaction (iii) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  18. Deflagrations - burning velocity • Estimate of i Conduction heat transfer rate = -kA(T/) k = gas thermal conductivity, A = cross-sectional area of flame T = temperature rise across front = Tproducts - Treactants = Tad - T∞  = thickness of front (unknown at this point) • Estimate of ii Enthalpy flux through front = (mass flux) x Cp x T Mass flux = VA ( = density of reactants = ∞, V = velocity = SL) Enthalpy flux = ∞CpSLAT • Estimate of iii Enthalpy generated by reaction = QR x (d[F]/dt) x Mfuel x Volume [F] = fuel concentration (moles/volume); Volume = A QR = CPT/f [F]∞ = fuel concentration in the fresh reactants AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  19. Burning velocity, flame thickness • Combine (i) and (ii) • = k/CpSL = /SL ( = flame thickness) where  = k/Cp = thermal diffusivity (units length2/time) • For air at 300K & 1 atm,  ≈ 0.2 cm2/s • For gases  ≈  ≈ D ( = kinematic viscosity; D = mass diffusivity) • For gases  ~ P-1T1.7 since k ~ P0T.7,  ~ P1T-1, Cp ~ P0T0 • For typical stoichiometric hydrocarbon-air flame, SL ≈ 40 cm/s, thus  ≈/SL ≈ 0.005 cm (!) (Actually when properties are temperature-averaged,  ≈ 4/SL ≈ 0.02 cm - still small!) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  20. Burning velocity, flame thickness • Combine (ii) and (iii) SL = {w}1/2 Where w= overall reaction rate = (d[F]/dt)/[F]∞(units 1/s) • With SL ≈ 40 cm/s,  ≈ 0.2 cm2/s, w≈ 1600 s-1 • 1/w= characteristic reaction time = 625 microseconds • Enthalpy release rate per unit volume = (enthalpy flux) / (volume) = (CpSLAT)/(A) = (CpSL/k)(kT)/ = (kT)/2 = (0.07 W/mK)(1900K)/(0.0002 m)2 = 3 x 109 W/m3 !!! • Moral: flames are thin, fast and generate a lot of thermal enthalpy (colloquially, heat) quickly! AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  21. Deflagrations - burning velocity • More rigorous analysis (Zeldovich, 1940) Same functional form as simple estimate (SL ~ {w}1/2, where wis an overall reaction rate) with some additional constants • How does SL vary with pressure? Define order of reaction (n) = A+B; since Thus SL ~ {w}1/2 ~ {P-1Pn-1}1/2 ~ P(n-2)/2 • For “real” hydrocarbons, working backwards from experimental results, we find typically SL ~ P-0.4, thus n ≈ 1.2 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  22. Deflagrations - temperature effect • Since Zeldovich number (b) >> 1 For typical hydrocarbon-air flames, E ≈ 40 kcal/mole  = 1.987 cal/mole, Tad ≈ 2200K b ≈ 10, at T close to Tad, w~ T10 !!!  Thin reaction zone concentrated near highest temperature In Zeldovich (or any) estimate of SL, overall reaction rate wmust be evaluated at Tad, not T∞ or any other temperature • How to estimate E? If reaction rate depends more on E than concentrations [ ], SL ~ {w}1/2 ~ {exp(-E/Tad)}1/2 ~ exp(E/2Tad) - plot of ln(SL) vs. 1/Tad has slope of -E/2 • If bisn’t large, then w(T∞) ≈ w(Tad) and reaction occurs even in the cold gases, so no control over flame is possible! • Since SL ~w1/2, SL ~ (Tadb)1/2 ~ Tad5 typically! AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  23. Deflagrations - summary • These relations show the effect of Tad (depends on fuel & stoichiometry),  (depends on diluent gas (usually N2) & P), w(depends on fuel, T, P) and pressure (engine condition) on laminar burning rates • Re-emphasize: these estimates are based on an overall reaction rate; real flames have 1000’s of individual reactions between 100’s of species - but we can work backwards from experiments or detailed calculations to get these estimates for the overall reaction rate parameters AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  24. Deflagrations Schematic of flame temperatures and laminar burning velocities Real data on SL (Vagelopoulos & Egolfopoulos, 1998) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  25. Deflagrations - example • Using the following experimental data, what are the apparent E, n, Z for lean-to-stoichiometric methane-air flames? AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  26. Deflagrations - example • Using the SL vs. pressure data and determining the best power-law fit we find SL ~ P-0.404, thus (n-2)/2 = -0.404 or n = 1.19 • From the Zeldovich equation • The ln( ) term hardly changes with Tad, so to a first approximation ln(SL) = Const. + (-E/2R)(1/Tad), thus the slope of ln(SL) vs. 1/Tad is -E/2R. From the linear fit the slope is -9273, thus E = -2(-9273)(1.987 cal/moleK) = 36.85 kcal/mole AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  27. Deflagrations - example • Finally estimate Z; for a stoichiometric mixture at 1 atm with n = 1.19, E = 36,850 cal/mole, SL = 40.5 cm/s, Tad = 2226K,  = 0.22 cm2/s = 2.2 x 10-5 m2/s and [F] = XF(P/T) = 0.095(101325 N/m2)/(8.314 J/moleK)(300K) = 3.86 moles/m3 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  28. Turbulent flames - motivation • Almost all flames used in practical combustion devices are turbulent because turbulent mixing increases burning rates, allowing more power/volume • Examples • Premixed turbulent flames • Gasoline-type (spark ignition, premixed-charge) internal combustion engines • Stationary gas turbines (used for power generation, not propulsion) • Nonpremixed flames • Diesel-type (compression ignition, nonpremixed-charge) internal combustion engines • Gas turbines • Most industrial boilers and furnaces AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  29. Basics of turbulence (1) • Good (but old) reference: Tennekes: “A First Course in Turbulence” • Job 1: need a measure of the strength of turbulence • Define turbulence intensity (u’) as rms fluctuation of instantaneous velocity u(t) about mean velocity ( ) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  30. Basics of turbulence (2) • Kinetic energy of turbulence = mass*u’2/2; KE per unit mass (total in all 3 coordinate directions x, y, z) = 3u’2/2 • Note that for a typical u’/SL = 5, with SL = 40 cm/s, u’ = 2 m/s, thus KE = 6 m2/s2 = 6 (kg m2/s2)/kg = 6 J/kg • Is this KE large or small? Typical hydrocarbon: f = 0.062, QR = 4.3 x 107 J/kg, thus for fuel-air MIXTURE, energy/mass ≈ 0.062 * 4.3 x 107 J/kg = 2.7 x 106 J/kg i.e. 444,000 times larger than KE of turbulence • Moral: it pays to be turbulent (but not so turbulent that flame quenching occurs as discussed later) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  31. Basics of turbulence (3) • Job 2: need a measure of the length scale of turbulence • Define integral length scale (LI) as Here the overbars denote spatial (not temporal) averages • LI is a measure of size of largest eddies, i.e. the largest scale over which velocities are correlated • Typically related to size of system (tube or jet diameter, grid spacing, …) • A(r) is the autocorrelation function at some time t • Note A(0) = 1 (fluctuations around the mean are perfectly correlated at a point) • Note A(∞) = 0 (fluctuations around the mean are perfectly uncorrelated if the two points are very distant) • For truly random process, A(r) is an exponentially decaying function A(r) = exp(-r/LI) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  32. Basics of turbulence (4) • In real experiments, typicall can measure u(t) at one point easier than u(x) at one time - can define time autocorrelation function A(x,) and integral time scale I at a point x Here the overbars denote temporal (not spatial) averages • With suitable assumptions LI = (8/p)1/2u’I • Define integral scale Reynolds numberReL u’LI/ ( = kinematic viscosity) • Note generally ReL ≠ Reflow = Vd/; typically u’ ≈ 0.1V, LI ≈ 0.5d, thus ReL ≈ 0.05 Reflow (e.g. in pipe-flow turbulence, grid turbulence, flow behind a cylinder, etc.) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  33. Characteristics of turbulent flames • Most important property: turbulent flame speed (ST) • Most models based on ideas introduced by Damköhler (1940) • Behavior depends on Karlovitz number - ratio of turbulent strain rate to chemical rate; for standard “Kolmogorov” turbulence model • Low Ka: “Huygens propagation,” thin fronts that are wrinkled by turbulence but internal structure is unchanged • High Ka: “Distributed reaction zones,” broad fronts • Note at low u’/SL, ST/SL ~ increases rapidly with increasing u’/SL, but at high enough u’/SL there is almost no increase in ST/SL and in fact flame quenching occurs at sufficiently high u’/SL (thus high Ka ~ (u’/SL)2) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  34. Characteristics of turbulent flames AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  35. Turbulent burning velocity data • Compilation of data from many sources = ST/SL Bradley et al. (1992) = u’/SL AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  36. Turbulent burning velocity • Experimental results shown in Bradley et al. (1992) based on smoothed data from many sources, e.g. fan-stirred chamber • … and we are talking MAJOR smoothing! AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  37. Turbulent premixed flame modeling • “Thin-flame” behavior observed in most practical combustors • Damköhler (1940): in Huygens propagation regime, flame front is wrinkled by turbulence but internal structure and SL are unchanged • Propagation rate ST due only to area increase via wrinkling: ST/SL = AT/AL • Many models, still much controversy about how to model, but most show ST/SL ~ u’/SL… AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  38. Turbulent premixed combustion • Models of premixed turbulent combustion don’t agree with experiments nor each other! • All models are trying to predict the same thing AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  39. Turbulent premixed flame modeling • Low u’/SL: weakly wrinkled flames • ST/SL = 1 + (u’/SL)2 (Clavin & Williams, 1979) - standard for many years • Actually Kerstein and Ashurst (1994) showed this is valid only for periodic flows - for random flows ST/SL - 1 ~ (u’/SL)4/3 • Higher u’/SL: strongly wrinkled flames • Schelkin (1947) - AT/AL estimated from ratio of cone surface area to base area; height of cone ~ u’/SL; result • Yahkot (1988) • Other models based on fractals, probability-density functions, etc.; most predict ST/SL ~ u’/SL at high u’/SL • For this class we’ll usually assume ST/SL ~ u’/SLwith possibility of “bending” or quenching at high Ka~ (u’/SL)2 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  40. Turbulence in engines • How to get high turbulence in engines? • Geometry of intake valves, ports, etc. - cause gas to swirl as it enters combustion chamber • Cup-shaped piston head (“squish”) • Obstacles in flow AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  41. Homogeneous reaction • Given a homogeneous system (T, P, [ ] same everywhere at any instant in time, but may change over time), how long will it take for the mixture to react (explode?) • Model for “knocking” in premixed-charge piston engines • As reaction starts, heat is released, temperature increases, overall reaction rate  increases, heat is released faster, T rises faster,  increases faster, … <BOOM> • Simple analysis - assumptions • Single-step reaction AA + BB  CC + DD • Excess of B (example: “lean” mixture with A = fuel, B = oxygen) • A = B = 1 • Adiabatic, constant-volume, ideal gas, constant Cv • Constant mass AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  42. Homogeneous reaction • Energy equation - if all fuel consumed (Yf = fuel mass fraction) So at any instant in time where Yf(t) is the instantaneous fuel mass fraction (at t = 0, no fuel consumed, T = initial temperature = T∞; at t = ∞, Yf = 0, all fuel consumed, T = Tad); then from page 16 (this simply says that there is a linear relationship between the amount of fuel consumed and the temperature rise) • Since we assumed A = B = 1, where A = fuel, B = oxygen AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  43. Homogeneous reaction • Reaction rate equation (assume n = 0) • Combine Eqs. 1, 2, 3, non-dimensionalize: • Notes on this result •  is the equivalence ratio for our special case A = B = 1; only valid for lean mixtures since we assumed surplus of A = fuel • Get pressure from P(t) = ∞RT(t) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  44. Homogeneous reaction • Equation looks scary but just a 1st order nonlinear ordinary differential equation - integrate to find () (amount of product formed as a function of time) for various (stoichiometry),  (activation energy), initial temp. (T∞), H (heat release) • Initial condition is  = 1 at  = 0 • What do we expect? • Since reaction rate is slowest at low T, reaction starts slowly then accelerates • “Induction time” (e.g. time to reach 90% completion of reaction,  = 0.1) should depend mostly on initial temperature T∞, not final temperature Tad since most of the time needed to react is before self-acceleration occurs • Very different from propagating flames where SL depends mostly on Tad not T∞ because in for flames there is a source of high T (burned gases) to raise gas T to near Tad before reaction started; in the homogeneous case no such source exists • This means that the factors that affect flame propagation and knock are very different AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  45. Homogeneous reaction • Double-click chart to edit or change parameters • Case shown:  = 0.7,  = 10, H = 6 • Note profile and time to “ignite” depend strongly on , much less on  and H AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  46. Homogeneous reaction • In case of “real” chemistry, besides the thermal acceleration mechanism there is also a chemical or “chain branching” acceleration mechanism, e.g. for H2-O2 H + O2 OH + O H2 + OH  H + H2O O + H2  OH + H etc. where 1 “radical” (H, OH, O) leads to 2, then 4, then 8, … radicals • In the case above, the “net” reaction would be 2 H2 + O2  H + OH + H2O which shows the increase in the radical “pool” • This “chain branching” mechanism leads to even faster “runaway” than thermal runaway since 2x > e-a/x for sufficiently large x • What if no H to start with? H2 + O2 HO2 + H (mostly this: h = 55 kcal/mole ) H2 + M  H + H + M (slower since h = 104 kcal/mole - too big) O2 + M  O + O + M (h = 119 kcal/mole - even worse) (M = any molecule) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  47. Homogeneous reaction • Example of ignition time for “real” fuels at engine-like T & P • Low T (1000/T > 1): VERY different times for different fuels, dominated by slow breakdown rate of fuel molecule • High T: Similar times because H + O2  OH + O branching rather than fuel molecule breakdown (except for toluene which is hard to “crack”) • Note ignition time increases with increasing T for 750 < T < 900K (negative effective activation energy!) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  48. “Non-premixed” or “diffusion” flames • Simplest approach to determining properties: “mixed is burned” - chemical reaction rates faster than mixing rates • No inherent propagation rate (unlike premixed flames where SL ~ [w]1/2) • No inherent thickness  (unlike premixed flames where thickness ~ /SL) - in nonpremixed flames, determined by equating convection time scale = /u=  to diffusion time scale 2/   ~ ()1/2 where  is a characteristic flow time scale (e.g. d/ufor a jet, where d = diameteru= velocity, or LI/u’ for turbulent flow, etc.) • Burning must occur near stoichiometric contour where reactant fluxes are in stoichiometric proportions (otherwise surplus of one reactant) • Burning must occur near highest T since w~ exp(-E/RT) is very sensitive to temperature (like premixed flames) AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  49. “Non-premixed” or “diffusion” flames • We’ll look at two examples of non-premixed flames which represent opposite extremes of what might happen in a Diesel engine • Droplet combustion - vaporization of droplets is slow, so droplets burn as individuals • Gas-jet flame - vaporization of droplets is so fast, there is effectively a jet of fuel vapor rather than individual droplets • Reality is in between, but in Diesels usually closer to the gas jet “with extras” Flynn, P.F, R.P. Durrett, G.L. Hunter, A.O. zur Loye, O.C. Akinyemi, J.E. Dec, C.K. Westbrook, SAE Paper No. 1999-01-0509. AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

  50. Droplet combustion • Heat from flame is conducted to fuel surface, vaporizes fuel, fuel convects/diffuses to flame front, O2 diffuses to flame front from outside, burning occurs at stoichiometric location • As fuel burns, droplet diameter d(t) decreases until d = 0 or droplet may extinguish before reaching d = 0 • Experiments typically show d(0)2 - d(t)2 ≈ Kt • Model for droplets in Diesel engine combustion AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion

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