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Transcript of Lecture 9 - Ignition and Extinction › sites › cefrc › files › Files... · 10!2 100 102 104...
6/26/2011
Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior written permission of the owner, M. Matalon. 1
Lecture 9 Ignition and Extinction
1
x
unburned burned gas uYY
TT
u
uu
0; ;
==
== !!
00/
!
!
YdxdT
S < SL laboratory frame
x
unburned burned gas 0
0/!
!
YdxdTS S uYY
TT
u
uu
==
==
; ;!!
in a frame moving with the wave
Planar flame with volumetric heat loss
Due to heat losses the propagation speed S < SL , and the flame temperatureTf < Ta; both need to be determined.
Heat losses
Heat losses
2
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d(!u)dx
= 0
!ucpdTdx
!"d 2Tdx2
=Q# ! g(T )
!u dYdx
! !D d 2Ydx2
= !#
x
unburned burned gas 0
0/!
!
YdxdTS S uYY
TT
u
uu
==
==
; ;!!
Steady one-dimensional equations
where g(T ) is the volumetric heat loss (J/cm3 s), which is assumed to vanishat T = Tu. For conductive losses, for example, g(T ) = k(T − Tu), while forradiative losses g(T ) = k(T 4 − T 4
u).
3
x
unburned burned gas 0
0/!
!
YdxdT
S S uYY
TT
u
uu
==
==
; ;!!
x=0
Using the asymptotic results derived previously, the mathematical formulationbecomes
!
d("u)dx
= 0
"ucpdTdx
# $d2Tdx 2
= #g(T)
"u dYdx
# "Dd2Ydx 2
= 0
on either side of the sheet conditions across the sheet
[[Y ]] = [[T ]] = 0
Q
��ρDdY
dx
��+
��λdT
dx
��= 0
−ρDdY
dx
����x=0−
=λ/cp√ρD
�2ρB/β2 Yu e
−E/2RTf
= ρuSLYu eE/2RTa−E/2RTf
4
SL =λ/cp√ρbD
�2ρB
ρ2uβ2Yu e
−E/2RTa
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Non-dimensionalization
SL (speed), lf =Dth/SL (length), !u,Tu (density, temperature), (" /cp )QYu /# lf2 (heat loss)
s = S/SL, Ta = T̃a/T̃u, Tf = T̃f/T̃u, β = E(T̃a−T̃u)/RT̃ 2a , q = Q/cpT̃u
Boundary conditions
x → −∞ : ρ = T = 1, Y = Yu, u = s
x → +∞ : dT/dx → 0, Y = 0
sdY
dx− d2Y
dx2= 0
ρu = s
Equations
= −β−1(Ta−1)g(T )sdT
dx− d2T
dx2
[[Y ]] = [[T ]] = 0
qLe−1
��dY
dx
��+
��dT
dx
��= 0
where Ta = 1 + qYu
Jump conditions
−Le−1 dY
dx
���x=0−
= Yu exp
�β
2
Ta
Tf
Tf − Ta
Ta − 1
�
5
integrate from −∞ to 0+, namely through and including the reaction zone
Tf = Ta − β−1(Ta−1)ϕ∗ + . . . ,
drop in flametemperature
We have two equations for the determination of s and ϕ∗. We need, however, todetermine first the solution in the preheat and post-flame zones. In particular,in the post-flame zone we need to carry the analysis to O(β−1).
d
dx[s(T + qY )]− d2
dx2
�T + qLe−1Y
�= −(Ta−1)g(T )
−Le−1 dY
dx
����x=0−
= Yue−ϕ∗/2
6
−β−1sϕ∗ −1
Ta−1
dT
dx
����x=0+
= −β−1
� 0+
−∞g(T )dx
s(Tf − Ta)� �� �−dT
dx
����x=0+
= −β−1(Ta − 1)
� 0+
−∞g(T )dx
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Y ∼�
Yu
�1− eLe sx
�(x < 0)
0 (x > 0)
T ∼
1 + (Ta − 1)esx + . . . (x < 0)
Ta − β−1
�ϕ∗ +
1
s(Ta−1)g(Ta)x
�
� �� �(x > 0)
1
Ta−1
dT
dx
����x=0+
= −β−1
sg(Ta)
reaction zone
preheat zone
O(θ-1)
T
Y adiabatic T = Ta
x
O(β−1)
7
the temperature on the burned side decays to Tu on a much longer scale ~ β
sdT
dx− d2T
dx2= −β−1g(T )
let x = βX
sdT
dX= −g(T )
for g(T ) = k(T − 1), for example,
X = −s
� T
Ta
dT
g(T )
which varies from X = 0 when T = Ta
to X = ∞ when T → 1 (or g(T ) → 0).
T ∼ 1 + (Ta − 1)e−kx/βs
X = −s
� Ta
T
dT
k(T − 1)= − s
kln
Ta − 1
T − 1
reaction zone
preheat zone
O(β-1)
O(β)
T
Y
X = βxx
8
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sϕ∗ =
� 0
−∞g(T )dx+
1
sg(Ta)
=1
s
� Ta
1g(z)dz + g(Ta)
� �� �
total heat loss from thepreheat and post-flamezones, respectively
=1
sqL
−Le−1 dY
dx
����x=0−
= Yue−ϕ∗/2 ϕ∗ = − ln s2
s2 ln s2 = −qL
where
for conductive losses, g(T ) = k(T − 1), the total loss qL = 2k
qL =
� Ta
1g(z)dz + g(Ta)
9
−β−1sϕ∗ −1
Ta−1
dT
dx
����x=0+
= −β−1
� 0+
−∞g(T )dx
(qL)cr = 1/e ≈ 0.37
At extinction the flame speed is nearly 60% of the adiabatic flame speed
s
sext
qL (qL)cr
s2 ln s2 = −qL
sext = e -1/2 ≈ 0.61
unstable
10
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1. Safety lamp devised in 1815 by Humphry Davy (Davy lamp) and used in coal mines. The flame in a lamp is surrounded by metal gauze [used to distribute the heat over a large area and ensure that qL > (qL)cr ]
and will not propagate through, thus preventing explosion.
2. Flame propagation is not possible in narrow tubes, or when r < rcr total conductive losses (through the walls) ~ (2π r)L qL = heat losses per unit volume ~ 2πrL/πr2L → qL ~ 1/r
rcr defines what is known as the “quenching distance”
11
Two-dimensional flames in a channel with conductive losses at the walls
For simplicity, we use here a constant-density model
)(,0 uTTkyT
yY
!=""
=""
Two important parameters: δ = 2a/lf channel’s width (in units of lf ) k a heat loss parameter (k = 0 adiabatic and k → ∞ for cold wall conditions)
burned
Unburned
2a U u
u
TTYY
=
=0
0
=!
!
=!
!
xTxY
x
y
Results of numerical calculations based on a constant density model with finite rate chemistry (β =10) and unity Lewis number are presented below.
12
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reaction rate and temperature contours
δ = 20 δ = 13
the flame survives all values of k flame extinguishes when losses exceed kcr
13
10!2
100
102
104
0.4
0.6
0.8
1
1.2
!=1 2 4
13
15.4
20
40
heat loss
U/SL
Total extinction in narrow channels of a width smaller than approximately 15lf
The quenching diameter dq ~ 15lf is in accord with experimental observations Only partial extinction occurs in wider channels (δ > δ*); the flame persists for all values of k but is confined to the center of the channel. Calculations how that the dead space ~ 6lF, in agreement with general experimental observations.
Propagation speed as a function of heat loss
14
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Ignition
15
Minimum ignition energy
If energy is deposited locally in a homogeneous combustible mixture, in the formof heat (and/or radicals), ignition will occur only if the energy exceeds a certainminimum value Emin - the minimum ignition energy.
At subcritical conditions the energy diffuses sufficiently fast, faster than the rateat which heat is generated by chemical reaction.
Ignition will occur if the energy added raises the temperature of the gas kernelto the adiabatic flame temperature Ta, in a region comparable in size to theflame thickness lf , so that the energy released is sufficiently fast to overcomethe loss by conduction.
Emin ∼ ρul3f · cp(Ta − Tu)
16
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Calcote et al., Ind. Eng. Chem. 1952
The size of the initial kernel has beenoften related to the quenching distance dq,defined as the minimum distance betweentwo plates that a flame cannot pass through(for example, due to excessive heat loss).This distance scales with thelaminar flame thickness,suggesting a relation
Emin ∼ d3q
which has been verified experimentally.
17
Since dq ∼ lf = λ/ρucpSL,
Emin ∼ S−3L
⇒
Ignition 401
is referred to the literature [17] . However, many of the thermal concepts dis-cussed for homogeneous gas-phase ignition will be fruitful in understanding the phenomena that control dust ignition and explosions.
2 . Ignition by Adiabatic Compression and Shock Waves
Ignition by sparks occurs in a very local region and spreads by fl ame char-acteristics throughout the combustible system. If an exoergic system at stand-ard conditions is adiabatically compressed to a higher pressure and hence to a higher temperature, the gas-phase system will explode. There is little likeli-hood that a fl ame will propagate in this situation. Similarly, a shock wave can propagate through the same type of mixture, rapidly compressing and heating the mixture to an explosive condition. As discussed in Chapter 5, a detonation will develop under such conditions only if the test section is suffi ciently long.
Ignition by compression is similar to the conditions that generate knock in a spark-ignited automotive engine. Thus it would indeed appear that compres-sion ignition and knock are chain-initiated explosions. Many have established
Methane Ethane
Hexane
Butane
Propane
Hexane
Cyclohexane
Cyclopropane
ButanePropane
Diethyl ether
Benzene
Heptane
1 At
1 At
Fraction of stoichiometric percentage of combustible in air0 0.5
0.1
0.2
0.40.50.60.8
1
2
34
0.3
0.20 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4
0.40.50.60.8
1
2
34
0.3
1.0 1.5 2.0 0 0.5 1.0 1.5 2.0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
Min
imum
ene
rgy
(mJ)
FIGURE 7.6 Minimum ignition energy of fuel–air mixtures as a function of stoichiometry (from Blanc et al . [14] ) .
CH07-P088573.indd 401CH07-P088573.indd 401 7/24/2008 5:37:35 PM7/24/2008 5:37:35 PM
more reactive mixturesrequire a smaller ignition energy. fraction of stoichiometric percentage of combustible in air
minim
um
energy
(mJ)
Lewis & von Elbe 1986
We can also deduce a dependence of the minimum ignition energy on pressure,using the relations SL ∼ pn/2−1 and lf ∼ p−n/2 derived previously, with n thereaction order. We find that
Emin ∼ p1−3n/2
While reduction in flame thickness facilitates ignition, the increase in the mix-ture density makes it harder.
Note that the phenomenological relation for Emin may depend on chemical ki-netic considerations (reaction mechanism, surface reactions, etc...) and on thesource and geometry of the ignition kernel (spark ignition, electrical wire, etc...,cylindrical or spherical kernel, etc...) 18
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Thermal Explosion
Thermal explosion and ignition theories are based on the same concept. Theyboth depend on the rate of energy release compared to the rate of energy dissi-pation, but in one case the objective may be to prevent explosion, while in theother to enable ignition.
Typical questions in thermal explosion theory:
• criteria that predict whether explosion occurs, or not
• delay time, or thermal induction period before the abrupt transition to aburning state
• time history of the explosive event
The concept is also similar to chain explosions, depending on the rates of chainbranching and chain termination.
19
Semenov theory1. Adiabatic explosion
Consider a homogeneous combustible mixture at a sufficiently low temperaturein a thermally insulated vessel of volume V.
ρcvdT
dt= BQρY e−E/RT
ρdY
dt= −BρY e−E/RT
T (0) = T0, Y (0) = Y0
d
dt(cvT +QY ) = 0
the adiabatic temperature at constant volume Te is largerthan the adiabatic flame temperature Ta (at constant pressure),because cp > cv.
Te = T0 +QY0/cv
⇒ Y =Te − T
Q/cv
20
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Introduce dimensionless variables, using the initial temperature T0 and thechemical reaction time (Be−E/RT0)−1, the problem reduces to
dT
dt= (Te − T )eβ0(T−1)/T
T (0) = 1
Te = 1 + q, q = QY0/cvT0, β0 = E/RT0.
t = e−β0
�Ei(β0
T )−Ei(β0)−eβ0/Te
�Ei(β0(Te−T )
TTe)−Ei(β0(T−1)
Te)��
Ei(z) =
� z
−∞
ex
xdx
is the exponential integral function (the principle value is needed for z > 0).
The exact solution
The implicit form of the solution, and the use of “ special functions” makes itdifficult to interpret the result.
21
T
1
ttign
β � 1
β � 1
Te
Of primary importance is the case β0 � 1, which yields a sharp transition from
the unburned state to equilibrium.
Ignition is an asymptotic concept
22
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For β0 � 1 the temperature evolves on the “slow time” ∼ β0t , which suggeststhat the “real” characteristic time for nondimensionalization should have been(Bβ0e−β0)
−1. This rescaling yields
dT
dt= β−1
0 (1 + q − T )eβ0(T−1)/T
T = 1 + β−10 ϕ(t) + . . .
dϕ
dt= qeϕ ϕ(0) = 0
ϕ = − ln (1− qt)
tign =cvT0
QY0Bβ0e−β0, β0 = E/RT0
10
T
t tign
T0
T ∼ 1− β−10 ln (1− qt)
23
The ignition time decreases with increasingQ, Y0 and the reaction rate Bβ0e−β0 .
2. Nonadiabatic explosion
Semenov theory
ρcvdT
dt= BQρY e−E/RT − S
Vk(T − T0)
We now allow for conductive heat losses at the surface S of the vessel.
We assume that during the induction period, there is no significant reactantdepletion and the density of the mixture remains constant.
Using the conduction time tc = cvρV/kS as a unit time,
dT
dt= De−β0/T − (T − 1)
heat releaseparameter
D = δeβ0/β0,δ =
QY
cvT0� �� �× cvρV/kS
(Bβ0e−β0)−1
� �� �conduction time
reaction time24
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Steady statesintersection of the two curves
e−β0/T = 1D (T − 1)
e−β0/T
TT1 Tc
Dc(T
− 1)/D
exist only when D < Dc
For D1 < Dc, a steady state exists with T = T1.
Here energy is dissipated faster than produced,
and the system will reach the steady state T1 as t → ∞.
For D > Dc, the rate of energy generated is faster than the energy dissipationand explosion occurs at a finite time.
The critical conditions are obtained from the tangency conditions
Tc =12 (β0 ±
�β20 − 4β0)
Dc = (Tc − 1)eβ0/Tc
De−β0/T = T − 1
Dβ0
T 2e−β0/T = 1
⇒
25
e−β0/T
TT1 Tc
Dc(T
− 1)/D
Of interest here is the solution with the lower
value of T and, in particular, when β0 � 1. Then
(ρY )c =k(S/V)RT 2
0
QBEeE/RT0−1
pc =k(S/V)R2T 3
0
QY BWEeE/RT0−1critical pressure
minimum massneeded for explosion
QY
cvT0=
k(S/V)B−1eβ0−1
β0cvρ
at criticality
Tc = 1 + β0−1
Dc = β0−1eβ0−1 or δc = e−1
26
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For β0 � 1
T = 1 + β−10 ϕ(t) + . . .
ϕ(0) = 0
dϕ
dt= δeϕ − ϕ
and explosion occurs for δ > δc = e−1
then, ϕ → ∞ when t → texp
texp =
� ∞
0
dϕ
δeϕ − ϕ
An approximation, near the critical conditions, yields
which must be done numerically.
note that texp → ∞ when δ → e−1 as it should.
texp ≈ π
�2e−1
δ − e−1
Estimate of the induction time
⇒
27
Frank-Kamenetski theory
We allow now for spatial temperature variations inside the vessel, but retain theassumptions of no significant reactant depletion (and constant density).
ρcv∂T
∂t− λ∇2T = BQρY e−E/RT
T��walls
= T0
We nondimensionalize the equation as follows:the characteristic size a of the vessel for the spatial coordinate,the conduction time cvρa2/λ for timethe temperature of the walls T0 for temperature.
∂T
∂t−∇2T =
δ
β0eβ0(1−1/T )
T��walls
= 1
δ =QY
cvT0× cvρa2/λ
(Bβ0e−β0)−1=
conduction time
reaction time28
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Steady states
−∇2T =δ
β0eβ0(1−1/T )
The solution T (x, t) would give the temperature distribution in the vessel as afunction of time. The question we will address here is whether steady statesexist, and under what conditions.
For β0 � 1
T = 1 + β−10 ϕ(t) + . . .
⇒∇2ϕ = −δeϕ
ϕ��walls
= 0
Frank-Kamenetskii equation
Solutions were obtained analytically for a slab and a cylinder, and numericallyfor a sphere. In all cases, it was found that solutions exist only for δ < δc.
The time dependent problem will reach a steady state only when δ < δc; other-wise explosion occurs.
29
QY
cvT0
a2Bβ0e−β0
λ/ρcv=
0.88 slab2.0 cyinder2.0 sphere
critical conditions
1 2 3
2
4
6
8
10
ϕmax
δ
30