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

2. 2 AME 436 - Lecture 4 - Spring 2013 - 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  Deflagrations - burning velocity, flame thickness, temperature effect  Turbulence effects  Homogeneous reaction  Nonpremixed flames  General characteristics  Droplets  Gas-jet  Turbulence effects

3. 3 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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  N 2 + O 2 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

4. 4 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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)

5. 5 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 chamber http://www.mech- eng.leeds.ac.uk/res- group/combustion/activities/Bomb.htm http://www.mech- eng.leeds.ac.uk/res- group/combustion/activities/Bomb.htm

6. 6 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 M 3 = 1 “Chapman-Jouget detonation” is stable (M = Mach number = u/c; u = velocity, c = sound speed = (  RT) 1/2 for ideal gas) c = sound speed = (  RT) 1/2 for ideal gas)

7. 7 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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

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

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

10. 10 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 A A + B B  C C + D D A A + B B  C C + D D e.g. 1 H 2 + 1 I 2  2 HI A = H 2, A = 1, B = I 2, 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) k f = “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)X i (X i = mole fraction of i (see lecture 2)) [ i ] = (P/  T)X i (X i = mole fraction of i (see lecture 2))

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

12. 12 AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Collision rate  On a mole (rather than molecule) basis  Example: how often do N 2 molecules collide at 298K & 1 atm?  3.61 x 10 -10 m, mass = 4.65 x 10 -26 kg (µ N2-N2 = 2.32 x 10 -26 kg)  At 2200K, Z N2-N2 = 4.98 x 10 9 /s; typical time for hydrocarbon combustion ≈ 10 -3 s, thus each molecule collides ≈ 5 x 10 6 times!

13. 13 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion  The reaction rate constant k f 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 /(K a sec)  With 3 parameters (Z, n, E) any curve can be fit!  The exponential term causes extreme sensitivity to T for E/  >> T! Reaction rate

14. 14 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion  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  h f (or heating value Q R ) and in general the two have no relation to each other - E affects reaction rates whereas  h f & Q R affect end states (e.g. T ad ) - of course  h f & Q R affect reaction rates indirectly by affecting T “Diary of a collision” Reaction rate

15. 15 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion  The full reaction rate expression is then  The H 2 + I 2  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 (CH 4 ) the “standard” mechanism (http://www.me.berkeley.edu/gri_mech/) includes 53 species and 325 individual reactions! http://www.me.berkeley.edu/gri_mech/  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 Comments on LoMA

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

17. 17 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Deflagrations - burning velocity  Since the burning velocity (S L ) << 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)

18. 18 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 = T products - T reactants = T ad - T ∞  = thickness of front (unknown at this point)  Estimate of ii Enthalpy flux through front = (mass flux) x C p x  T Mass flux =  VA (  = density of reactants =  ∞, V = velocity = S L ) Enthalpy flux =  ∞ C p S L A  T  Estimate of iii Enthalpy generated by reaction = Q R x (d[F]/dt) x M fuel x Volume [F] = fuel concentration (moles/volume); Volume = A  Q R = C P  T/f [F] ∞ = fuel concentration in the fresh reactants

19. 19 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Burning velocity, flame thickness  Combine (i) and (ii)  = k/  C p S L =  /S L (  = flame thickness) where  = k/  C p = thermal diffusivity (units length 2 /time)  For air at 300K & 1 atm,  ≈ 0.2 cm 2 /s  For gases  ≈ ≈ D ( = kinematic viscosity; D = mass diffusivity)  For gases  ~ P -1 T 1.7 since k ~ P 0 T.7,  ~ P 1 T -1, C p ~ P 0 T 0  For typical stoichiometric hydrocarbon-air flame, S L ≈ 40 cm/s, thus  ≈  /S L ≈ 0.005 cm (!) (Actually when properties are temperature-averaged,  ≈ 4  /S L ≈ 0.02 cm - still small!)

20. 20 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Burning velocity, flame thickness  Combine (ii) and (iii) S L = {  } 1/2 Where  = overall reaction rate = (d[F]/dt)/[F] ∞ (units 1/s)  With S L ≈ 40 cm/s,  ≈ 0.2 cm 2 /s,  ≈ 1600 s -1  1/  = characteristic reaction time = 625 microseconds  Enthalpy release rate per unit volume = (enthalpy flux) / (volume) = (  C p S L A  T)/(A  ) = (  C p S L /k)(k  T)/  = (k  T)/  2 = (0.07 W/mK)(1900K)/(0.0002 m) 2 = 3 x 10 9 W/m 3 !!!  Moral: flames are thin, fast and generate a lot of thermal enthalpy (colloquially, heat) quickly!

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

22. 22 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Deflagrations - temperature effect  Since Zeldovich number (  ) >> 1 For typical hydrocarbon-air flames, E ≈ 40 kcal/mole  = 1.987 cal/mole, T ad ≈ 2200K   ≈ 10, at T close to T ad,  ~ T 10 !!!  Thin reaction zone concentrated near highest temperature  Thin reaction zone concentrated near highest temperature  In Zeldovich (or any) estimate of S L, overall reaction rate  must be evaluated at T ad, not T ∞ or any other temperature  How to estimate E? If reaction rate depends more on E than concentrations [ ], S L ~ {   } 1/2 ~ {exp(-E/  T ad )} 1/2 ~ exp(  E/2  T ad ) - plot of ln(S L ) vs. 1/T ad has slope of -E/2   If  isn’t large, then  (T ∞ ) ≈  (T ad ) and reaction occurs even in the cold gases, so no control over flame is possible!  Since S L ~  1/2, S L ~ (T ad  ) 1/2 ~ T ad 5 typically!

23. 23 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Deflagrations - summary  These relations show the effect of T ad (depends on fuel & stoichiometry),  (depends on diluent gas (usually N 2 ) & P),  (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

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

25. 25 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Deflagrations - example  Using the following experimental data, what are the apparent E, n, Z for lean-to-stoichiometric methane-air flames? Pressure = 1 atm Equiv. ratio S L (cm/s) T ad (K) 0.7015.81838 0.7518.21919 0.8024.11997 0.8931.32122 1.0036.52226 Equiv. ratio = 1 Pressure (atm) S L (cm/s) 0.2561.6 0.5051.7 1.0040.5 2.0028.9 3.0022.5

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

27. 27 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Deflagrations - example  Finally estimate Z; for a stoichiometric mixture at 1 atm with n = 1.19, E = 36,850 cal/mole, S L = 40.5 cm/s, T ad = 2226K,  = 0.22 cm 2 /s = 2.2 x 10 -5 m 2 /s and [F] = X F (P/  T) = 0.095(101325 N/m 2 )/(8.314 J/moleK)(300K) = 3.86 moles/m 3 = 3.86 moles/m 3

28. 28 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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

29. 29 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 ( )

30. 30 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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’/S L = 5, with S L = 40 cm/s, u’ = 2 m/s, thus KE = 6 m 2 /s 2 = 6 (kg m 2 /s 2 )/kg = 6 J/kg  Is this KE large or small? Typical hydrocarbon: f = 0.062, Q R = 4.3 x 10 7 J/kg, thus for fuel-air MIXTURE, energy/mass ≈ 0.062 * 4.3 x 10 7 J/kg = 2.7 x 10 6 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)

31. 31 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Basics of turbulence (3)  Job 2: need a measure of the length scale of turbulence  Define integral length scale (L I ) as Here the overbars denote spatial (not temporal) averages  L I 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/L I )

32. 32 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 L I = (8/  ) 1/2 u’  I  Define integral scale Reynolds number Re L  u’L I /  Define integral scale Reynolds number Re L  u’L I / ( = kinematic viscosity)  Note generally Re L ≠ Re flow = V d/ ; typically u’ ≈ 0.1 V, L I ≈ 0.5d, thus Re L ≈ 0.05 Re flow (e.g. in pipe-flow turbulence, grid turbulence, flow behind a cylinder, etc.)

33. 33 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Characteristics of turbulent flames  Most important property: turbulent flame speed (S T )  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’/S L, S T /S L ~ increases rapidly with increasing u’/S L, but at high enough u’/S L there is almost no increase in S T /S L and in fact flame quenching occurs at sufficiently high u’/S L (thus high Ka ~ (u’/S L ) 2 )

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

35. 35 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion = S T /S L Turbulent burning velocity data  Compilation of data from many sources = u’/S L Bradley et al. (1992)

36. 36 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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!

37. 37 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 S L are unchanged  Propagation rate S T due only to area increase via wrinkling: S T /S L = A T /A L  Many models, still much controversy about how to model, but most show S T /S L ~ u’/S L …

38. 38 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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

39. 39 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Turbulent premixed flame modeling  Low u’/S L : weakly wrinkled flames  S T /S L = 1 + (u’/S L ) 2 (Clavin & Williams, 1979) - standard for many years  Actually Kerstein and Ashurst (1994) showed this is valid only for periodic flows - for random flows S T /S L - 1 ~ (u’/S L ) 4/3  Higher u’/S L : strongly wrinkled flames  Schelkin (1947) - A T /A L estimated from ratio of cone surface area to base area; height of cone ~ u’/S L ; result  Yahkot (1988)  Other models based on fractals, probability-density functions, etc.; most predict S T /S L ~ u’/S L at high u’/S L  For this class we’ll usually assume S T /S L ~ u’/S L with possibility of “bending” or quenching at high Ka ~ (u’/S L ) 2

40. 40  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 Turbulence in engines

41. 41 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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, …  As reaction starts, heat is released, temperature increases, overall reaction rate  increases, heat is released faster, T rises faster,  increases faster, …  Simple analysis - assumptions  Single-step reaction A A + B B  C C + D D  Excess of B (example: “lean” mixture with A = fuel, B = oxygen)  A = B = 1  Adiabatic, constant-volume, ideal gas, constant C v  Constant mass

42. 42 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Homogeneous reaction  Energy equation - if all fuel consumed (Y f = fuel mass fraction) So at any instant in time where Y f (t) is the instantaneous fuel mass fraction (at t = 0, no fuel consumed, T = initial temperature = T ∞ ; at t = ∞, Y f = 0, all fuel consumed, T = T ad ); 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

43. 43 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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)

44. 44 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 T ad since most of the time needed to react is before self-acceleration occurs  Very different from propagating flames where S L depends mostly on T ad not T ∞ because in for flames there is a source of high T (burned gases) to raise gas T to near T ad 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

45. 45 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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

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

47. 47 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion 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 + O 2  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!)

48. 48 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion “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 S L ~ [  ] 1/2 )  No inherent thickness  (unlike premixed flames where thickness ~  /S L ) - 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/u for a jet, where d = diameter u = velocity, or L I /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  ~ exp(-E/RT) is very sensitive to temperature (like premixed flames)

49. 49 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion “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.

50. 50 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Droplet combustion  Heat from flame is conducted to fuel surface, vaporizes fuel, fuel convects/diffuses to flame front, O 2 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

51. 51 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Droplet combustion  How fast does droplet burn? Spherically-symmetric model (Godsave, Spalding 1953), assuming “mixed is burned,” affectionately called the “dee-squared law” d(0) = droplet diameter at time = 0 d(t) = droplet diameter at some later time t K = droplet burning rate constant (m 2 /s) k = gas thermal conductivity (W/mK)  l = droplet density (kg/m 3 ) C P = gas specific heat at constant pressure (J/kgK) Q R = fuel heating value per unit mass (J/kg) f = stoichiometric fuel mass fraction T ∞ = ambient air temperature (K) T d = droplet vaporization temperature (K) L v = latent heat of vaporization of fuel (J/kg)  B (“Transfer number”) ≈ ratio of heat generation by chemical reaction to heat needed to vaporize fuel; typical values: methanol 3, most hydrocarbons 8 - 10  Diameter of flame surrounding droplet (d flame )

52. 52 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Droplet combustion  The d 2 -law assumes no buoyant or forced convection, but in engines there is likely to be a lot of flow; one relation for the effect of flow on burning rate is Re d = Droplet Reynolds number = ud(t)/ Re d = Droplet Reynolds number = ud(t)/ Nu = Nusselt number based on droplet diameter u = droplet velocity relative to gas Pr = Prandtl number = /  = kinematic viscosity = kinematic viscosity  = thermal diffusivity = k g /  g C p,g  Note that this result reduces to the previous one for u = 0 (thus Re = 0)

53. 53 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Droplet combustion  Note all the heat release (Q R ), heat of vaporization (L v ), etc. is tied up in B which appears only inside a ln( ), thus changing these properties hardly affects burning rate at all  Why? The more you vaporize fuel, the more rapidly the fuel vapor blows out, thus the harder it is for heat to be conducted to the fuel surface Marchese et al. (1999), space experiments, heptane in O 2 -He

54. 54 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Nonpremixed-gas flames - laminar gas-jet flames  Flame height (L f ) determined by equating diffusion time (d j 2 /D, d j = jet diameter, D = oxygen diffusivity) to convection time (L f /u) (u = jet exit velocity) d j 2 /D ~ L f /u  L f ~ ud j 2 /D or L f /d j ~ ud j /D Gases: D ≈  L f /d j ~ ud j / = Re d  Consistent with more rigorous models

55. 55 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Nonpremixed turbulent jet flames  Turbulent (Hottel and Hawthorne, 1949)  For turbulent flows D is not constant but rather D ~ u’L I  U’ ~ u; L I ~ d j L f ~ d j (independent of Re)  U’ ~ u; L I ~ d j  L f ~ d j (independent of Re)  High u u’ Ka large - flame “lifts off” near base  High u  high u’  Ka large - flame “lifts off” near base  Still higher - more of flame lifted  When lift-off height = flame height, flame “blows off” (completely extinguished) Lifted flame (green = fuel; blue = flame)

56. 56 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Example Ronney Oil and Gas Company claims to have developed a chemical called PDR TM, that, when an infinitesimal amount is added to air, increases the thermal conductivity (k) of air by 10%. No other properties of air or fuels are affected. With this additive, how would each of the following be affected? In particular, state whether each of the above will increase or decrease or remain constant, and by less than 10%, more than, or exactly 10%? (a) (a)Heating value of CH 4. Unaffected, heating values and all thermodynamic properties are unrelated to thermal conductivity. (b) (b)S L of a stoichiometric CH 4 -air flame. S L ~ [  ] 1/2 and  = k/  C P ~ k, so a 10% increase in k will lead to a [1.1 1/2 - 1.0 1/2 ]/1.0 1/2 ≈ 0.05 = 5% increase in S L (c) (c)S T of a stoichiometric CH 4 -air flame in the “thin-flame” (low Ka) regime (same u’). To a first approximation, S T /S L ~ u’/S L, thus S T ~ u’, which is independent of S L and thus k. (d) (d)S T of a stoichiometric CH 4 -air flame at high Ka (same u’). Under these conditions S T /S L increases more slowly than linearly with u’/S L, for example we might say S T /S L ~ (u’/S L ) 0.5. Thus if S L increased 5%, (u’/S L ) 0.5 would decrease by about 2.5%, thus S T /S L would decrease 2.5%, thus S T = S L (S T /S L ) would increase about 2.5%. (e) (e)Time for homogeneous explosion in a stoichiometric CH 4 -air mixture. Unaffected since homogeneous reactions are unaffected by transport processes. (f) (f)Time for a liquid fuel droplet to burn out completely. Since the burning rate constant K ~ k and no terms in the expression for B are affected by k, the burning time decreases by exactly 10% (g) (g)Length of a laminar gas-jet flame. Simple answer: no effect. Better answer:  = k/  C P ~ k and since ,, and D are similar for gases, if  increases, D will probably increase by about the same amount, thus L would decrease by exactly 10% (h) (h)Length of a turbulent gas-jet flame. No effect since L ~ d, independent of transport properties like D

57. 57 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Summary - combustion  Combustion is the combination of chemical reaction with convective and diffusive transport of thermal energy and chemical species  The most important distinction between flames is premixed vs. non- premixed, i.e. whether the reactants are mixed before combustion  Chemical reactions relevant to combustion are generally VERY complicated but can often be approximated (roughly) by a one step overall reaction  Chemical reactions relevant to combustion generally have high activation energy (more precisely, high Zeldovich number  ) and thus are more sensitive to temperature than any other property  Premixed flames  Deflagrations - subsonic - burning velocity S L ~ (  ) 1/2 (  = reaction rate at T ad )  Detonations - supersonic wave  Homogeneous reaction - time of reaction depends mainly on T ∞ not T ad  Nonpremixed flames  To a first approximation “Mixed is burned” - burning rates/times NOT affected by chemical reaction rates  Turbulence increases the rates of combustion by increasing surface area (premixed) or mixing rates (nonpremixed)

58. 58 AME 436 - Lecture 4 - Spring 2013 - Basics of Combustion Summary - combustion  There are 4 types of properties we use to describe combustion systems  Thermodynamic properties & combinations of thermodynamic properties »Temperature (T), pressure (P), volume (V), internal energy (U), enthalpy (H), entropy (S), specific heat (C P, C v,  = C P /C v ), Gibbs free energy (G), equilibrium constants (K), density (  ), mole fraction (X), fuel mass fraction (f), transfer number (B), molecular weight (M), gas constant (  or R), sound speed (c), heating value (Q R ), latent heat of vaporization (L V )  Transport properties »Thermal conductivity (k), thermal diffusivity (  = k/  C P ), mass diffusivity (D), momentum diffusivity aka kinematic viscosity ( )  Chemical rate properties »Activation energy (E), pre-exponential factor (Z), order of reaction (n), exponent of temperature dependence outside Arrhenius term (a)  Fluid mechanical properties »Velocity (u), turbulence intensity (u’), integral length scale (L I )  From these we determine other properties, e.g.  Thermodynamic - T ad, X i at equilibrium, P for const-V combustion, work in/out due to compression/expansion  Combustion - S L, S T, ignition time, droplet burning rate constant (K), jet flame length (L f )