AME 513 Principles of Combustion Lecture 4 Chemical kinetics I – Basic Concepts

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Outline Law of Mass Action (LOMA) Kinetic theory of gases as applied to chemical reactions Homogeneous reaction Well-stirred reactor Chain branching H2 – O2 reaction AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

Law of Mass Action (LoMA) First we need to describe rates of chemical reaction 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 the Law of Mass Action (LoMA) states that the rate of reaction [ i ] = concentration of molecule i (usually moles per liter) kf = “forward” reaction rate constant 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) Minus sign on d[A]/dt and d[B]/dt since A & B are being depleted Basically 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 AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 10-26 kg) At 2200K, ZN2-N2 = 4.98 x 109/s; typical time for hydrocarbon oxidation ≈ 10-3 s, thus each molecule collides ≈ 5 x 106 times! AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Energy distribution Maxwell-Boltzman distribution: f = # of molecules with KE of E, per unit energy Zero probability of zero or infinite KE (a = E/kT = 0 or ∞); peak probability at a = 0.5 Fraction of molecules with Ea/kT greater than some value b Textbook uses exp(-b) not - similar trends but not technically correct! exp(-a) ∫f(a)d(a) = fraction of molecules with KE > E/kT f(a) AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Energy distribution Ea = “activation energy” representing the “energy barrier” that must be overcome for reaction E is not enthalpy of reaction hf (or heating value QR); in general there is no relation between E & hf - E affects reaction rates whereas hf & QR affect end states (e.g. Tad), though hf & QR affect reaction rates indirectly by affecting T “Diary of a collision” AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Reaction rate So finally the reaction rate expression should be the combination of collision rate (ZAB/V), energy (Ea) and steric “luck” (p) factors AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Reaction rate Compare to book example AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Comments on LoMA In practice, terms other than exp(-b) are treated as expendable; reaction rate constant kf expressed in Arrhenius form; for E per mole not per molecule: Z = pre-exponential factor, n = another (nameless) constant, Ea = activation energy (cal/mole);  = gas constant = NAvogadrok; working backwards, units of Z must be (moles per liter)1-A-vB/(K-nsec) With 3 parameters (Z, n, E) any curve can be fit! The exponential term causes extreme sensitivity to T for E/ >> T! AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Comments on LoMA The full reaction rate expression (omitting over-wiggles) 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 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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-pressure or constant-volume, ideal gas, constant CP and CV Constant mass AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Homogeneous reaction Energy equation (constant volume) - 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 (this simply says that there is a linear relationship between the amount of fuel consumed and the temperature rise) (If constant pressure CV  CP) Since we assumed A = B = 1, where A = fuel, B = oxygen AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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); if constant pressure equation are exactly the same except Tad is the constant-P value AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 homogeneous reaction are very different AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 (only way to break O=O bond directly) 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 faster “runaway” than thermal runaway since 2x > e-a/x for sufficiently large x H can also be removed from the system via H + O2 + M  HO2 + M (M = any molecule) H + wall  H(wall); H(wall) + H(wall)  H2(wall)  H2(gas) AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Homogeneous reaction But at the beginning there is no H – how to get it? 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) What happens to HO2? HO2 + H2  HOOH + H (HOOH = H2O2 = hydrogen peroxide) HO2 + HO2  HOOH + O2 (radical-radical – is it more important?) HOOH + M  2OH (chain branching if H + O2  OH + O ineffective) Which chain-branching route is more important, H or HO2? H + O2  OH + O; d[O2]/dt = -1016.7[H][O2]T-0.8e-16500/RT [ ]: mole/cm3; T: K; R: cal/moleK; t: sec Depends on P2 since [ ] ~ P, strongly dependent on T H + O2 + M  HO2 + M; {M = any molecule} d[O2]/dt = -1015.2[H][O2][M]T0e+1000/RT for M = N2 (higher rate for CO2 and especially for H2O) Depends on P3, nearly independent of T AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Homogeneous reaction Rates equal (“crossover”) when [M] = 101.5T-0.8e-17500/RT Ideal gas law: P = [M]RT thus P = 103.4T0.2e-17500/RT (P in atm) crossover at 950K for 1 atm, higher T for higher P, thus at low T / high P, H + O2  OH + O branching does not occur; relatively inactive HO2 forms: H + O2 + M  HO2 + M HO2 + wall  HO2(wall) 2 HO2(wall)  H2(wall) + 2 O2(wall) H2(wall) + 2 O2(wall)  H2 + 2 O2 Causes “2nd explosion limit” of H2 – O2 system in vessel with constant-T walls AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Homogeneous reaction What causes 2 other limits??? Rules: H, OH, O more reactive than HO2 When radicals hit wall, they may adsorb and be converted to stable species Radical concentrations small, thus radical-radical reactions unlikely (but maybe HO2 + HO2 … homework problem!) Reaction rates vary with pressure (P) Collisions (of any molecule) with wall ~ P1 (P = pressure) Bimolecular reactions (e.g. H + O2  OH + O) ~ P2 Trimolecular reactions (e.g. H + O2 + M  HO2 + M ) ~ P3 First limit H + O2  OH + O dominates H + O2 + M  HO2 + M As P, molecule-wall collisions increase more slowly than molecule-molecule collisions, thus probability of radicals reaching wall and being converted to stable species decreases – chain branching can occur - explosion As T, branching rates (e.g. H + O2  OH + O) increase (high Ea) thus impact of loss at walls decreases – explosion AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Homogeneous reaction Third limit HO2 is not a stable species, just much less active a radical compared to H, OH, O Again, as P, molecule-wall collisions increase more slowly than molecule-molecule collisions, thus probability of HO2 reacting in gas phase to produce HOOH before reaching wall and being converted to stable species decreases – chain branching can occur – explosion HO2 + H2  HOOH + H or HO2 + HO2  HOOH + O2 HOOH + M  2 OH OH + H2  HOH + H H + O2 + M  HO2 + M and the cycle continues AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I 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 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Well-stirred reactor The homogeneous-reaction example was for time-dependent behavior of a fixed mass with no flow; what about the opposite (steady, fixed volume, with flow in/out)? Assume reactants and products are perfectly stirred, i.e. as soon as reactants flow into the reactor, they are completely mixed with products (constant T, P and composition throughout reactor) Again will have low fuel + high T or high fuel + low T in reactor Energy balance; ( )R = in reactor AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Well-stirred reactor But how to find YA,R and YB,R? Mass balance on A Mass balance on B: since one B molecule is consumed for every A AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Well-stirred reactor Finally (really finally this time!) combining Eqs. 1, 2, 3 Equation for reactor temperature TR as a function mass flow rate for varying values of the properties in the reference mass flow, heat release parameter (e) and dimensionless activation energy (b) AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Well-stirred reactor Very low flow – plenty of time, nearly complete reaction, TR ≈ Tad Very high flow – not enough time, very little reaction, TR ≈ T∞ Results show classical Z-shaped response with hysteresis Upper – “extinction” branch – when mass flow increased too much, reactor “extinguishes” due to insufficient residence time – TR drops to lowest branch Lower – “ignition” branch – when mass flow decreased enough, reactor “ignites” due to adequate residence time, TR rises to highest branch b = 3, e = 0.2, f = 1 b = 5, e = 0.2, f = 1 AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Well-stirred reactor Middle branch generally unstable Small increase in mass flow for momentarily fixed TR (thus reaction rate) will increase reactant leakage, thus decrease T, moving reaction rate still lower, eventually dropping to lower branch Opposite for small decrease in mass flow (jump to upper branch) Upper and lower branches stable – e.g. increase in mass flow decreases TR and takes you back to same branch Results very sensitive to b As b increases, extinction point rises to higher T, i.e. closer to Tad At realistic values of b, curve spans many decades of mass flow – never looks like textbook plots b of 3 and 5 plotted – not realistic, should be 10 or more for hydrocarbon oxidation Weak effects of e and f Who cares about this? Can use this apparatus to study kinetics and obtain estimates of reaction rate parameters Z and E, especially by measuring mass flow at extinction AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I

AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I Summary Chemical reaction require a collision; collision rates determined by kinetic theory of gases Reaction requires adequate kinetic energy (exp(-Ea/RT) term) and steric (luck) factor, resulting in ZTnexp(-Ea/RT) expression with 3 adjustable parameters Z, n, Ea Chemical kinetic systems for realistic fuels are comprised of 100s or 1000s of individual reactions, but some general rules (e.g. regarding T and P effects) expected Much information can be deduced from well-characterized experiments (e.g. constant P, T vessel) Reaction rates + thermodynamics can be used to describe model systems, e.g. (time-dependent) homogeneous reaction or (steady-state) Well-Stirred Reactor AME 513 - Fall 2012 - Lecture 4 - Chemical Kinetics I