AME 53 Principles of Combustion " Lecture 8 Premixed flames I: Propagation rates Outline" Rankine-Hugoniot relations Hugoniot curves Rayleigh lines Families of solutions Detonations Chapman-Jouget Others Deflagrations
Rankine-Hugoniot relations" What premixed flame propagation speeds are possible in D? Assumptions Ideal gas, steady, D, constant area, constant C P, C v, γ C P /C v Governing equations Equations of state: P =!RT;h! h = C P (T!T ) Mass conservation:!m / A =! u =! u Navier-Stokes, D, no viscosity:!u!u!x +!P!x = 0,!u =!m A = const. "!u + P = const " +! u =P +! u Energy conservation, no work input/output: h + u / + q = h + u / q = heat input per unit mass = fq R if due to combustion Mass, momentum, energy conservation eqns. can be combined yielding Rankine-Hugoniot relations:! " P " % $!'! " P %" $!' " % $ +' = q ; u! u = (u = "!!!# " & # &# " & RT u u " 3 D momentum balance - constant-area duct " Coefficient of friction (C f ) C f " Wall drag force #u $ (Wall area) d(mu)! Forces =! dt! Forces = PA " (P + dp)a " C f (/!u )(Cdx) d(mu)! =!!mu =!mu "!m(u + du) dt Combine: AdP+!mdu + C f (/!u )Cdx = 0 If C f = 0 : AP+!mu = const;!m =!ua # P +!u = const # +! u = P +! u Velocity u pressure P density! duct area A mass flux =!ua Velocity u + du pressure P + dp density! duct area A mass flux =!ua Thickness dx, circumference C, wall area Cdx 4
P / Hugoniot curves" Defines possible end states of the gas (P, ρ, u ) as a function of the initial state (, ρ, u ) & heat input (q/rt ) ( 3 for stoich. hydrocarbon-air, 43 for stoich. H -air) equations for the 3 unknowns (P, ρ, u ) (another unknown T is readily found from ideal gas law P = ρ RT ) For every initial state & heat input there is a whole family of solutions for final state; need one additional constraint how to determine propagation rate u? 00 0 H = 0 H = 35 0. 0. 0! /! P!/ 0 0 3 4 5 6 7 8 9 0! "! 0 8 6 4 0 8 6 4 H = 0 H = 35 5 Hugoniot curves" In reference frame at rest with respect to the unburned gases u = velocity of the flame front Δu = u u = velocity of burned gases Combine mass & momentum: +! u =P +! u ;!u =!m / A = const. (! +!m / A ) ( ) = P +!m / A! P " P # = "!m & % (!!! "! $ A ' Thus, straight lines (called Rayleigh lines) on P vs. /ρ (or specific volume, v) plot correspond to constant mass flow Note Rayleigh lines must have negative slope Entropy another restriction on possible processes (S S ) S! S = C P ln T! Rln P ; P =! RT ;C P = " T! RT "! R " S! S # = ln P! & % (! "! ln P = C P $ ' " " ln P! ln! ;S ) S " P # )! %! $!!!" & ( ' 6 3
Rankine-Hugoniot relations" H = 0: shock Hugoniot only P / > (thus ρ /ρ < ) is possible; P / < would result in decrease in entropy H 0, P Usual weak deflagration branch where!u ="! =" T q < 0; T = T + u! T "R / (" ") = T + fq R = T ad (P=const.) Burned gases at T = T ad move away from front u = S L = laminar burning velocity (if perfectly flat, D, laminar); depends on transport properties and reaction rates For turbulent flames, u = S T (depends additionally on turbulence properties) H 0, P > - detonations!u ="! > 0 u! Can t determine T /T as simply as with weak deflagrations Burned gases move in same direction as front Out of all possible choices, how to determine u? C P 7 Rankine-Hugoniot relations" Above D: strong detonations (M < ) D B: weak detonations (M > ) (point B: mass flow = ) B C: impossible (mass flow imaginary, see Rayleigh line discussion) C E: weak deflagrations (M < ) (point C: mass flow = 0) Below E: strong deflagrations (M > ) 5 4 F H = 0 H =!"#!$% 3 D B A C E 0 0 3!$#!"% 8 4
Detonation velocities - calculation" From Hugoniot relation! " P " % $!'! " P %" $!' " % $ +' = q (Eq. ) ( d P!!# " & # &# " & RT d " " From Rayleigh line P! =!!m " % $ '!!! # A & ( ) =!(! u ) ( ( d P P d (!! ) =!!M (Eq. 4) From mass conservation ( ) ( ) = +! + P!!!! + " (Eq. )!! " P!!!! =!! u =!! u =! u =! M c =! M "RT =!"M (Eq. 3)! RT RT RT ( )! u =! u! (! u ) = (! u )! M P = M (Eq. 5)!! 9 Chapman-Jouget detonation" 5 equations, solve for the 5 unknowns M, M, P /, ρ /ρ and d(p / )/d(ρ /ρ ) at the tangency point where the slopes d(p / )/d(ρ /ρ ) are equal on the Hugoniot curve & Rayleigh line: ( M = + q RT )(! / "!)% ( $ ' + q RT )(! / "!)% $ ' ;M = #! & #! & This is called the Chapman-Jouget detonation (path is A F D) - why is it the most probable detonation speed? Heat release zone Induction zone Temperature M Pressure Shock 3 M = M < M > 0 5
Chapman-Jouget detonation" Consider structure of detonation shock followed by reaction zone, because shock requires only a few collisions to complete whereas reaction requires 0 6 s If subsonic behind reaction zone, expansion waves can catch up to front and weaken shock, slowing it down (why are expansion waves more prevalent than compression waves? To be discussed in class ) which results in smaller M thus larger M Can t achieve weak detonations (M > ) with this structure because you can t transition from M < to M > with heating in a constant-area frictionless duct (Rayleigh flow) So CJ detonation (M = ) is the only stable detonation mostly borne out by experiments Deflagrations - burning velocity" Recall P 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 conducted through flame front (ii) = Rate at which heat is produced by chemical reaction (iii) Reaction zone 000K Product concentration Temperature Direction of propagation Speed relative to unburned gas = S L Reactant concentration 300K Distance from reaction zone Convection-diffusion zone! "/S L = 0.3-6 mm 6
Deflagrations - burning velocity" Estimate of i Conduction heat transfer rate = -ka(δt/δ) k = gas thermal conductivity, A = cross-sectional area of flame ΔT = temperature rise across front = T products - T reactants δ = thickness of front (unknown at this point) Estimate of ii Enthalpy flux through front = (mass flux) x C p x ΔT Mass flux = ρua (ρ = density of reactants = ρ, u = velocity = S L ) Enthalpy flux = ρ C p S L AΔT Estimate of iii Heat generated by reaction = Q R x (d[fuel]/dt) x M fuel x Volume Volume = Aδ Q R = C P ΔT/f f = = Fuel mass (Mass fuel / volume) = Total mass (Mass total / volume) (Moles fuel / volume)(mass fuel / moles fuel) = [F] M " fuel (Mass total / volume) # " [F] = fuel concentration in the cold reactants 3 Deflagrations - burning velocity" Combine (i) and (ii) δ = k/ρc p S L = α/s L (δ = flame thickness) (same as Lecture 7) Recall α = k/ρc p = thermal diffusivity (units length /time) For air at 300K & atm, α 0. cm /s For gases α ν (ν = kinematic viscosity) For gases α ~ P - T.7 since k ~ P 0 T.7, ρ ~ T -, C p ~ P 0 T 0 For typical stoichiometric hydrocarbon-air flame, S L 40 cm/s, thus δ α/s L 0.005 cm (!) (Actually when properties are temperature-averaged, δ 4α/S L 0.0 cm - still small!) Combine (ii) and (iii) S L = (αω) / ω = overall reaction rate = (d[fuel]/dt)/[fuel] (units /s) With S L 40 cm/s, α 0. cm /s, ω 600 s - /ω = characteristic reaction time = 65 microseconds Heat release rate per unit volume = (enthalpy flux) / (volume) = (ρc p S L AΔT)/(Aδ) = ρc p S L /k)(kδt)/δ = (kδt)/δ = (0.07 W/mK)(900K)/(0.000 m) = 3 x 0 9 W/m 3!!! Moral: flames are thin, fast and generate a lot of heat! 4 7
Deflagrations - burning velocity" More rigorous analysis (Bush & Fendell, 970) using Matched Asymptotic Expansions Convective-diffusive (CD) zone (no reaction) of thickness δ Reactive-diffusive (RD) zone (no convection) of thickness δ/β( ε) where /[β( ε)] is a small parameter T(x) = T 0 (x) + T (x)/[β( ε)] + T (x)/[β( ε)] + Collect terms of same order in small parameter Match T & dt/dx at all orders of β( ε) where CD & RD zones meet!ze!" / " % " S L =.344! 3(!!) % $ ("(!#)) ' $ + ';" ( E,! ( T * # & # "(!!) & )T ad T ad Same form as simple estimate (S L ~ {αω} /, where ω Ze -β is an overall reaction rate, units /s), with additional constants Why β term on reaction rate? Reaction doesn t occur over whole flame thickness δ, only in thin zone of thickness δ/β Reactant concentration isn t at ambient value Y i,, it s at /β of this since temperature is within /β of T ad 5 Deflagrations - burning velocity" What if not a single reactant, or Le? Mitani (980) extended Bush & Fendell for reaction of the form!! A A +! B B! products;! = ZY A! A Y B B exp "E a / #T resulting in # S L =!Ze!" # Y A +# B! # A # B # A % A," $ " ( )! B ("(!#))! A+! B +!! Le A A &!! Le B G ( B ' G ) y! A ( y + a)! B * e!y dy ; a ) "(!#)($!) / Le B ;$ = equivalence ratio 0 ( ) Recall order of reaction (n) = ν A + ν B Still same form as simple estimate, but now β -(n+) term since n may be something other than (as Bush & Fendell assumed) Also have Le A -νa and Le B -νb terms why? For fixed thermal diffusivity (α), for higher Le A, D A is smaller, gradient of Y A must be larger to match with T profile, so concentration of A is higher in reaction zone / 6 8
Deflagrations - burning velocity" How does S L vary with pressure? d[a] =!k f [ A]! A [ B]! B ~P! A dt P! B ~P! A+! B ~P n (for example A = fuel, B = oxidant)!"~ d[a] ~ P! P n ~ P n! [A] " dt Thus S L ~ {αω} / ~ {P - P n- } / ~ P (n-)/ For typical n =, S L independent of pressure For real hydrocarbons, working backwards from experimental results, typically (e.g. stoichiometric CH 4 -air) S L ~ P -0.4, thus n. This suggests more reactions are one-body than two-body, but actually observed n is due to competition between twobody H + O branching vs. 3-body H + O + M which decelerates reaction 7 Deflagrations - temperature effect" Since Zeldovich number (β) >>! T ad "! = E!(T ad ) "T T=Tad "T ad For typical hydrocarbon-air flames, E 40 kcal/mole R =.987 cal/mole, T f 00K if adiabatic β 0, at T close to T f, ω ~ T 0!!! Thin reaction zone concentrated near highest temp. In Zeldovich (or any) estimate of S L, overall reaction rate Ω must be evaluated at T ad, not T How can we estimate E? If reaction rate depends more on E than concentrations [ ], S L ~ {αω} / ~ {exp(-e/rt)} / ~ exp(-e/rt) - Plot of ln(s L ) vs. /T ad has slope of -E/R If β isn t large, then ω(t ) ω(t ad ) and reaction occurs even in the cold gases, so no control over flame is possible! Since S L ~ ω /, S L ~ (T β ) / ~ T 5 typically! 8 9
Burning velocity measurement" Many techniques, all attempt to determine the speed of the unburned gases relative to the flame front or vice versa (since that s the definition of S L ) Counterflow very popular (e.g. Prof. Egolfopoulos) 9 Deflagrations burning velocities" S L or 00K 40 cm/s T ad T ad 500K SL 5 cm/s! 0.5! =! Lean Stoichiometric Rich limit limit Fuel % Schematic of flame temperatures and laminar burning velocities Real data on S L (CH 4 -air, atm) 0 0
Deflagrations - summary" These relations show the effect of T ad (depends on fuel & stoichiometry), α (depends on diluent gas (usually N ) & P), ω (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 000 s of individual reactions between 00 s of species - but we can work backwards from experiments or detailed calculations to get these estimates for the overall reaction rate parameters