ame 513 principles of combustion lecture 8 premixed flames i: propagation rates

AME 513

Principles of Combustion

Lecture 8Premixed flames I: Propagation rates

2AME 513 - Fall 2012 - Lecture 8 - Premixed flames I

Outline

Rankine-Hugoniot relations Hugoniot curves Rayleigh lines Families of solutions Detonations

Chapman-Jouget Others

Deflagrations


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: Mass conservation: Navier-Stokes, 1D, no viscosity:

Energy conservation, no work input/output:

q = heat input per unit mass = fQR if due to combustion Mass, momentum, energy conservation eqns. can be

combined yielding Rankine-Hugoniot relations:

Rankine-Hugoniot relations


1D momentum balance - constant-area duct

Coefficient of friction (Cf)


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?

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:

Thus, straight lines (called Rayleigh lines) on P vs. 1/r (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)

Hugoniot curves


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

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

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?


8

D

B

CE

AME 513 - Fall 2012 - Lecture 8 - Premixed flames I

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)


A

F


From Hugoniot relation

From Rayleigh line

From mass conservation

Detonation velocities - calculation


5 equations, solve for the 5 unknowns M1, M2, P2/P1, r1/r2 and d(P2/P1)/d(r1/r2) at the tangency point where the slopes d(P2/P1)/d(r1/r2) are equal on the Hugoniot curve & Rayleigh line:

This is called the Chapman-Jouget detonation (path is A F D) - why is it the most probable detonation speed?

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

Chapman-Jouget detonation


Deflagrations - burning velocity

Recall P2 ≈ P1

How fast will the flame propagate? Simplest estimate based on the hypothesis thatRate 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)



Estimate of iConduction heat transfer rate = -kA(T/)k = gas thermal conductivity, A = cross-sectional area of flameT = 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 TMass flux = uA ( = density of reactants = ∞, u = velocity = SL) Enthalpy flux = ∞CpSLAT

Estimate of iiiHeat generated by reaction = QR x (d[fuel]/dt) x Mfuel x VolumeVolume = AQR = CPT/f

[F]∞ = fuel concentration in the cold reactants


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!



More rigorous analysis (Bush & Fendell, 1970) using Matched Asymptotic Expansions Convective-diffusive (CD) zone (no reaction) of thickness d Reactive-diffusive (RD) zone (no convection) of thickness / (1-d b

) e where 1/[ (1- )]b e is a small parameter T(x) = T0(x) + T1(x)/[ (1- )]b e + T2(x)/[ (1- )]b e 2 + … Collect terms of same order in small parameter Match T & dT/dx at all orders of (1- ) b e where CD & RD zones

meet

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? Reaction doesn’t occur over whole flame thickness d, only in thin

zone of thickness /d b Reactant concentration isn’t at ambient value Yi,∞, it’s at 1/ b of

this since temperature is within 1/ b of Tad



What if not a single reactant, or Le ≠ 1? Mitani (1980) extended Bush & Fendell for reaction of the form

resulting in

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-nA and LeB

-nB terms – why? For fixed thermal diffusivity (a), 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



How does SL vary with pressure?

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 two-body H + O2 branching vs. 3-body H + O2 + M which decelerates reaction


Deflagrations - temperature effect

Since Zeldovich number () >> 1

For typical hydrocarbon-air flames, E ≈ 40 kcal/mole = 1.987 cal/mole, Tf ≈ 2200K if adiabatic ≈ 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∞ Þ 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 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 ~ (T)1/2 ~ T5 typically!


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)


Schematic of flame temperatures and laminar burning velocities

Deflagrations – burning velocities

Real data on SL (CH4-air, 1 atm)


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 principles of combustion lecture 8 premixed flames i: propagation rates

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