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AME 513 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 AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 2 Rankine-Hugoniot relations What premixed flame propagation speeds are possible in 1D? Assumptions Ideal gas, steady, 1D, constant area, constant CP, Cv, CP/Cv Governing equations Equations of state: P = r RT;h2 - h1 = CP (T2 -T1 ) Mass conservation: m / A = r1u1 = r2u2 Navier-Stokes, 1D, no viscosity: ru ¶u ¶P m + = 0, ru = = const. Þ ru2 + P = const Þ P1 +r1u12 =P2 +r2 u 22 ¶x ¶x A Energy conservation, no work input/output: h1 + u12 / 2 + q = h2 + u22 / 2 q = heat input per unit mass = fQR if due to combustion Mass, momentum, energy conservation eqns. can be combined yielding Rankine-Hugoniot relations: g æ P2 r1 ö 1 æ P2 öæ r1 ö q u2 - u1 Du r1 -1÷ - ç -1÷ç +1÷ = ; = = -1 ç g -1 è P1 r2 ø 2 è P1 øè r2 ø RT1 u1 u1 r2 AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 3 1D momentum balance - constant-area duct Coefficient of friction (Cf) Cf º Wall drag force 1 ru 2 × (Wall area) 2 d(mu) dt å Forces = PA - (P + dP)A - C f (1 / 2 ru2 )(Cdx) å Forces = å d(mu) å dt = å mu = mu - m(u + du) Combine: AdP+mdu + C f (1 / 2 ru2 )Cdx = 0 If C f = 0 : AP+mu = const;m = ruA Þ P + ru 2 = const Þ P1 + r1u12 = P2 + r2 u22 AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 4 Hugoniot curves Defines possible end states of the gas (P2, 2, u2) as a function of the initial state (P1, 1, u1) & heat input (q/RT1) (≈ 31 for stoich. hydrocarbon-air, ≈ 43 for stoich. H2-air) 2 equations for the 3 unknowns (P2, 2, u2) (another unknown T2 is readily found from ideal gas law P2 = 2RT2) 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 u1? 20 100 H=0 18 H=0 H = 35 16 H = 35 14 P2/P1 P2/P1 10 1 12 10 8 6 4 2 0.1 0 0.1 1 r1/r2 10 0 1 2 3 4 5 r1/r2 6 AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 7 8 9 10 5 Hugoniot curves In reference frame at rest with respect to the unburned gases u1 = velocity of the flame front u = u2 – u1 = velocity of burned gases Combine mass & momentum: P1 +r1u12 =P2 +r2 u 22 ; ru = m / A = const. Þ P1 + ( m / A) 1 r1 2 = P2 + ( m / A) 1 r2 2 æmö P2 - P1 Þ = -ç ÷ è Aø 1 r2 -1 r1 2 Thus, straight lines (called Rayleigh lines) on P vs. 1/ (or specific volume, v) plot correspond to constant mass flow Note Rayleigh lines must have negative slope Entropy – another restriction on possible processes (S2 ≥ S1) T P P r RT g S2 - S1 = CP ln 2 - R ln 2 ; 2 = 2 2 ;CP = R T1 P1 P1 r1RT1 g -1 -g æ P2 r1 ö g -1 P2 1 P2 S2 - S1 r2 P2 æ r1 ö Þ = ln ç ln = ln - ln ;S2 ³ S1 Þ ³ ç ÷ ÷CP P1 g P1 r1 P1 è r2 ø è P1 r2 ø g AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 6 Rankine-Hugoniot relations H = 0: shock Hugoniot – only P2/P1 > 1 (thus 1/2 < 1) is possible; P2/P1 < 1 would result in decrease in entropy H ≠ 0, P2 ≈ P1 Usual “weak deflagration” branch where Du r T q fQ =1- 1 =1- 2 < 0; T2 = T1 + = T1 + R = Tad (P=const.) u1 r2 T1 g R / (g -1) CP Burned gases at T = Tad move away from front u1 = SL = laminar burning velocity (if perfectly flat, 1D, laminar); depends on transport properties and reaction rates For turbulent flames, u1 = ST (depends additionally on turbulence properties) H ≠ 0, P2 > P1 - detonations Du r =1- 1 > 0 u1 r2 Can’t determine T2/T1 as simply as with weak deflagrations Burned gases move in same direction as front Out of all possible choices, how to determine u1? AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 7 Rankine-Hugoniot relations Above D: strong detonations (M2 < 1) D – B: weak detonations (M2 > 1) (point B: mass flow = ∞) B – C: impossible (mass flow imaginary, see Rayleigh line discussion) C – E: weak deflagrations (M2 < 1) (point C: mass flow = 0) Below E: strong deflagrations (M2 > 1) 5 F H=0 H=2 4 P2/P1 D 3 B 2 C 1 E A 0 0 1 2 r1/r2 AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 3 8 Detonation velocities - calculation From Hugoniot relation g +1 P2 d ( P2 P1 ) g æ P2 r1 ö 1 æ P2 öæ r1 ö q g -1 P1 -1÷ - ç -1÷ç +1÷ = (Eq. 1) Þ = (Eq. 2) ç g +1 r g -1 è P1 r2 ø 2 è P1 øè r2 ø RT1 d ( r1 r2 ) 11 g -1 r2 From Rayleigh line 2 1+ æmö P2 - P1 2 = - ç ÷ = - ( r1u1 ) Þ è Aø 1 r2 -1 r1 r1u1 ) P2 P1 -1 r1 2 u12 M12 c12 M12g RT1 ( == - u1 = === -g M12 (Eq. 3) r1 r2 -1 P1r1 P1 RT1 RT1 RT1 2 d ( P2 P1 ) Þ = -g M12 (Eq. 4) d ( r1 r2 ) From mass conservation r1u1 = r2u2 Þ ( r1u1 ) = ( r2u2 ) Þ M12 = M 22 2 2 P2 P1 r1 r2 (Eq. 5) AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 9 Chapman-Jouget detonation 5 equations, solve for the 5 unknowns M1, M2, P2/P1, 1/2 and d(P2/P1)/d(1/2) at the tangency point where the slopes d(P2/P1)/d(1/2) are equal on the Hugoniot curve & Rayleigh line: 1/2 1/2 é ( q RT1 ) (g 2 -1) ù é ( q RT1 ) (g 2 -1) ù M1 = ê1+ ú +ê ú ; M 2 =1 2g 2g ë û ë û This is called the Chapman-Jouget detonation (path is A F D) - why is it the most probable detonation speed? AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 10 Chapman-Jouget detonation Consider structure of detonation – shock followed by reaction zone, because shock requires only a few collisions to complete whereas reaction requires 106s 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 M1 thus larger M2 Can’t achieve weak detonations (M2 > 1) with this structure because you can’t transition from M < 1 to M > 1 with heating in a constant-area frictionless duct (Rayleigh flow) So CJ detonation (M2 = 1) is the only stable detonation – mostly borne out by experiments AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 11 Deflagrations - burning velocity Recall P2 ≈ P1 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) AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 12 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 = thickness of front (unknown at this point) Estimate of ii Enthalpy flux through front = (mass flux) x Cp x T Mass flux = uA ( = density of reactants = ∞, u = velocity = SL) Enthalpy flux = ∞CpSLAT Estimate of iii Heat generated by reaction = QR x (d[fuel]/dt) x Mfuel x Volume Volume = A QR = CPT/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) r¥ [F]∞ = fuel concentration in the cold reactants AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 13 Deflagrations - burning velocity Combine (i) and (ii) = k/CpSL = /SL ( = flame thickness) (same as Lecture 7) Recall = k/Cp = thermal diffusivity (units length2/time) For air at 300K & 1 atm, ≈ 0.2 cm2/s For gases ≈ ( = kinematic viscosity) 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!) Combine (ii) and (iii) SL = (w)1/2 w = overall reaction rate = (d[fuel]/dt)/[fuel]∞ (units 1/s) With SL ≈ 40 cm/s, ≈ 0.2 cm2/s, w ≈ 1600 s-1 1/w = characteristic reaction time = 625 microseconds Heat release rate per unit volume = (enthalpy flux) / (volume) = (CpSLAT)/(A) = CpSL/k)(kT)/ = (kT)/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 heat! AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 14 Deflagrations - burning velocity More rigorous analysis (Bush & Fendell, 1970) using Matched Asymptotic Expansions Convective-diffusive (CD) zone (no reaction) of thickness Reactive-diffusive (RD) zone (no convection) of thickness /b(1-e) where 1/[b(1-e)] is a small parameter T(x) = T0(x) + T1(x)/[b(1-e)] + T2(x)/[b(1-e)]2 + … Collect terms of same order in small parameter Match T & dT/dx at all orders of b(1-e) where CD & RD zones meet æ 2a Ze- b ö æ 1.344 - 3(1- e ) ö E T¥ ÷ SL = çç 1+ ; b º , e º ÷ 2÷ ç b (1- e ) ø ÂTad Tad è ( b (1- e )) ø è Same form as simple estimate (SL ~ {w}1/2, where w ~ Ze-b is an overall reaction rate, units 1/s), with additional constants Why b-2 term on reaction rate? 1/2 Reaction doesn’t occur over whole flame thickness , only in thin zone of thickness /b Reactant concentration isn’t at ambient value Yi,∞, it’s at 1/b of this since temperature is within 1/b of Tad AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 15 Deflagrations - burning velocity What if not a single reactant, or Le ≠ 1? Mitani (1980) extended Bush & Fendell for reaction of the form n A A + n B B ® products;w = ZYAn AYBn B exp (-Ea / ÂT ) resulting in nB æ ö n n n 1 1 - b n A +n B -1 A( B A) SL = çç 2a Ze YA,¥ G ÷÷ n A +n B +1 -n A -n B LeA LeB ( b (1- e )) è ø 1/2 Gº ¥ ò n yn A ( y + a) B e- y dy ; a º b (1- e )(f -1) / LeB ; f = equivalence ratio 0 Recall order of reaction (n) = A+ B Still same form as simple estimate, but now b-(n+1) term since n may be something other than 1 (as Bush & Fendell assumed) Also have LeA-A and LeB-B terms – why? For fixed thermal diffusivity (), for higher LeA, DA is smaller, gradient of YA must be larger to match with T profile, so concentration of A is higher in reaction zone AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 16 Deflagrations - burning velocity How does SL vary with pressure? d[A] n n = -k f [ A ] [ B] ~Pn Pn ~Pn +n ~P n (for example A = fuel, B = oxidant) dt 1 d[A] w~ ~ P -1P n ~ P n-1 [A]¥ dt Thus SL ~ {w}1/2 ~ {P-1Pn-1}1/2 ~ P(n-2)/2 For typical n = 2, SL independent of pressure For “ real ” hydrocarbons, working backwards from experimental results, typically (e.g. stoichiometric CH4-air) SL ~ P-0.4, thus n ≈ 1.2 This suggests more reactions are one-body than two-body, but actually observed n is due to competition between twobody H + O2 branching vs. 3-body H + O2 + M which decelerates reaction A B A B A B AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 17 Deflagrations - temperature effect Since Zeldovich number (b) >> 1 Tad ¶w E bº = w (Tad ) ¶T T =Tad ÂTad For typical hydrocarbon-air flames, E ≈ 40 kcal/mole = 1.987 cal/mole, Tf ≈ 2200K if adiabatic b ≈ 10, at T close to Tf, w ~ T10 !!! Thin reaction zone concentrated near highest temp. In Zeldovich (or any) estimate of SL, overall reaction rate must be evaluated at Tad, not T∞ How can we estimate E? If reaction rate depends more on E than concentrations [ ], SL ~ {w}1/2 ~ {exp(-E/T)}1/2 ~ exp(-E/2T) - Plot of ln(SL) vs. 1/Tad has slope of -E/2 If b isn’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 ~ (Tb)1/2 ~ T5 typically! AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 18 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 SL) Counterflow very popular (e.g. Prof. Egolfopoulos) AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 19 Deflagrations – burning velocities Schematic of flame temperatures and laminar burning velocities Real data on SL (CH4-air, 1 atm) AME 513 - Fall 2012 - Lecture 8 - Premixed flames I 20 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 513 - Fall 2012 - Lecture 8 - Premixed flames I 21