AME513-F12-lecture8

Report
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
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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
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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
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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
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8
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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
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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?
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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
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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)
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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?
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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
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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)
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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 = ∞CpSLAT
 Estimate of iii
Heat generated by reaction = QR x (d[fuel]/dt) x Mfuel x Volume
Volume = A
QR = CPT/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
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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)
= (CpSLAT)/(A) = CpSL/k)(kT)/ = (kT)/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!
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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
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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
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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
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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/2T) - 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!
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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)
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Deflagrations – burning velocities
Schematic of flame temperatures and
laminar burning velocities
Real data on SL (CH4-air, 1 atm)
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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
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