ame 513 principles of combustion lecture 8 premixed flames i: propagation rates
TRANSCRIPT
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
3AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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
4AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
1D momentum balance - constant-area duct
Coefficient of friction (Cf)
5AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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
6AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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
7AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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?
Rankine-Hugoniot relations
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)
Rankine-Hugoniot relations
A
F
9AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
From Hugoniot relation
From Rayleigh line
From mass conservation
Detonation velocities - calculation
10AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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
11AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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
12AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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)
13AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
Deflagrations - burning velocity
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
14AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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!
15AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
Deflagrations - burning velocity
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
16AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
Deflagrations - burning velocity
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
17AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
Deflagrations - burning velocity
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
18AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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!
19AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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)
20AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
Schematic of flame temperatures and laminar burning velocities
Deflagrations – burning velocities
Real data on SL (CH4-air, 1 atm)
21AME 513 - Fall 2012 - Lecture 8 - Premixed flames I
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