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AME 513
Principles of Combustion
Lecture 5
Chemical kinetics II – Multistep mechanisms
Outline
 Analytical solution for 1-step reaction
 Irreversible
 Reversible





Steady-state approximation
Pressure effects – Lindemann mechanism
Spreadsheet-level modeling
Partial equilibrium approximation
Online chemical kinetics calculator
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
2
1-step irreversible reaction
 First consider 1 single (irreversible) reaction:
A+BC+D
d [ A]
= -k f [ A] [ B]
dt
(forward reaction, rate constant kf)
But [A] and [B] are not independent, any decrease in [A] results in
an identical decrease in [B] and increase in [C] & [D]; for a
stoichiometric excess of [B], i.e. [B]o > [A]o
[ A] o - [ A] = [ B] o - [ B] Þ [ B] = ([ B]o - [ A] o ) + [ A]
d [ A]
Þ
= -k f [ A ] {( [ B ] o - [ A ] o ) + [ A ]} Þ -k f dt =
dt
d [ A]
[ A] + ([ B]o - [ A]o ) [ A]
2
æ
ö
A]
[
÷ + Constant
Þ -k f t = ln çç
÷
è [ A] + ([ B] o - [ A ]o ) ø
æ
ö
æ [ A] ö
æ [ A] ö
A]
[
o
o÷
÷ - ln ç
÷÷ Þ -k f t = ln ç
t = 0 : [ A ] = [ A ] o Þ Constant = - ln çç
ç
÷
ç
÷
è [ B]o ø
è [ B] o ø
è [ A] + ([ B] o - [ A]o ) ø
A ] e (1- [ A ] o [ B ] o )
[
Þ
=
[ A]o 1- ([ A]o [ B]o ) e-k t
-k f t
f
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
3
1-step reversible reaction
 Now consider 1 single (but reversible) reaction:
A+BC+D
C+DA+B
(forward reaction, rate constant kf)
(reverse reaction, rate constant kb)
d [ A]
= -k f [ A] [ B] + kb [C ] [ D]
dt
[ A] o - [ A ] = [ B] o - [ B] = [C ] - [C ] o = [ D ] - [ D ] o
Þ [ B] = [ B] o - [ A ] o + [ A] ; [C ] = [C ] o + [ A] o - [ A] ; [ D ] = [ D ] o + [ A ]o - [ A]
d [ A]
= -k f [ A ] {[ B ] o - [ A ] o + [ A ]} + kb {[C ] o + [ A ] o - [ A ]} {[ D ] o + [ A ] o - [ A ]}
Þ
dt
d [ A]
d [ A]
1
2
where
= [ A ] + a [ A ] + b Þ ( kb - k f ) dt =
Þ
2
dt
k
k
( b f)
[ A] - a [ A] + b
aº
k f ( [ B ] o - [ A ] o ) + kb ( 2 [ A ] o + [ C ] o + [ D ] o )
(k
b
- kf )
;b º
kb ( [ A ] o + [ C ] o ) ( [ A ] o + [ D ] o )
(k
b
- kf )
æ a 2 + 4b + a + 2 A ö
[ ] ÷ + Constant
ln ç
Þ ( kb - k f ) t = a 2 + 4b çè a 2 + 4b - a - 2 [ A ] ÷ø
1
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
4
1-step reversible reaction
 Initial condition: t = 0, [A]= [A]o
æ a 2 + 4b + a + 2 A ö
[ ]o ÷ + Constant
t = 0, [ A ] = [ A ] o Þ ( kb - k f ) (0) = ln ç
a 2 + 4b çè a 2 + 4b - a - 2 [ A ] o ÷ø
æ a 2 + 4b + a + 2 A ö
[ ]o ÷
1
Þ Constant =
ln ç
a 2 + 4b çè a 2 + 4b - a - 2 [ A ] o ÷ø
æ a 2 + 4b + a + 2 A
ö
2
a
+
4b
a
2
A
[
]
1
[
]
o
÷
Þ ( kb - k f ) t =
ln ç
a 2 + 4b çè a 2 + 4b - a - 2 [ A ] o a 2 + 4b + a + 2 [ A ] ÷ø
1
 In general the algebra horrendous but consider special case
[A]o = [B]o, [C]o = [D]o = 0
[ A] o = [ B] o , [C ] o = [ D] o = 0 Þ a = Þ
a 2 + 4b + a + 2 [ A ] o
a 2 + 4b - a - 2 [ A] o
=
2kb [ A ] o
kb - k f
kb [ A ] o
2
;b = -
kb - k f
; a 2 + 4b =
2 kb k f [ A ] o
kb - k f
1- k f kb
1+ k f kb
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
5
1-step reversible reaction
æ a 2 + 4b + a + 2 A
[ ]
Þ ( kb - k f ) t =
ln ç
2 kb k f [ A ] o çè a 2 + 4b - a - 2 [ A ]
é [ A ] k k -1 + [ A ] ù
f
b
1
oú
Þt=
ln ê
2 kb k f [ A ] o ê [ A ] k f kb +1 - [ A ] o ú
ë
û
é [ A]
1
o
Check #1: [ A ] = [ A ] o Þ t =
ln ê
2 kb k f [ A ] o ê [ A ] o
ë
(k
b
- kf )
(
(
)
)
(
(
1- k f kb ö
÷
1+ k f kb ÷ø
)
)
k f kb -1 + [ A ] o ù
1
ú=
ln [1] = 0 OK
ú
2
k
k
A
k f kb +1 - [ A ] o û
b f [ ]o
Check #2: As t ® ¥, need to approach equilbrium concentration [ A ] eq
t ® ¥ Þ [ A]
(
)
k f kb +1 - [ A ] o ® 0 Þ [ A ] eq =
[ A]o
k f kb +1
At equilbrium forward & reverse rates are equal, i.e.
-k f [ A ] [ B ] = -kb [C ] [ D ] Þ -k f [ A ] [ A ] = -kb ( [ A ] o - [ A ]) ( [ A ] o - [ A ])
Þ -k f [ A ] = -kb ( [ A ] o - [ A ]) Þ k f kb [ A ] = [ A ] o - [ A ] Þ [ A ] =
2
2
[ A]o
k f kb +1
= [ A ] eq OK
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
6
1-step reversible reaction
12
kf = 10, kb = 1, [A]o = 1
[A]eq = 0.240
1
0.8
Concentration of A
Concentration of A
1.2
0.6
0.4
0.2
0
10
8
6
kf = 1, kb = 10, [A]o = 10
[A]eq = 7.60
4
2
0
0
0.5
1
Time
1.5
0
0.05
0.1
0.15
0.2
Time
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
7
Steady-state approximation - concept
 These calculations are tedious, need some simplifications to
handle more complex reaction mechanisms!
 Consider a two-step reaction with very reactive radical N
O + N2  NO + N
N + O2  NO + O
(rate constant k1)
(rate constant k2)
 Net production of N: d [ N ] dt = k1 [O] [ N2 ] - k2 [ N ] [O2 ]
 The steady-state approximation assumes d[N]/dt ≈ 0, i.e.
d [N]
d [N]
k1 [O] [ N 2 ]
k1 [O] [ N 2 ] >>
;k2 [ N ] [O2 ] >>
Þ [N] »
dt
dt
k2 [O2 ]
in which case
d [ NO]
k1 [O] [ N 2 ]
= k1 [O] [ N 2 ] + k2 [ N ] [O2 ] = k1 [O] [ N 2 ] + k2
[O2 ] = 2k1 [O] [ N2 ]
dt
k2 [O2 ]
 Does NOT imply [N] = constant, only that [N] varies slowly
compared to its production & consumption rates individually
 When valid? Typically when a rapidly-reacting intermediate (N
in this case) is produced by a slow reaction (O + N2  NO + N)
8
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
Lindemann (1922) - P effects on decomposition
 Unimolecular decomposition of A (e.g. H2O2 + M  2 OH)
A + M  A* + M
A* + M  A + M
A*  B
(rate constant k1f; A* = activated state of A)
(rate constant k1b; k1f/k1b = Kequilibrium = KA/KA*)
(rate constant k2)
 Steady state assumption for A* yields
d éë A* ùû
k [ A] [ M ]
k k [ A] [ M ]
d [ B]
= k1 f [ A] [ M ] - k1b éë A* ùû[ M ] - k2 éë A* ùû » 0 Þ éë A* ùû » 1 f
Þ
= k2 éë A* ùû = 2 1 f
dt
k1b [ M ] + k2
dt
k1b [ M ] + k2
 “Apparent” overall reaction rate: find keff such that d[A]/dt =
-d[B]/dt (i.e. rate of consumption of reactant A = rate of
production of product B):
d éë A* ùû
dt
k [ A] [ M ]
= k1 f [ A ] [ M ] - k1b éë A* ùû[ M ] - k2 éë A* ùû » 0 Þ éë A* ùû » 1 f
k1b [ M ] + k2
k1 f [ M ]
d [ A]
d [ B]
Þ
= -keff [ A] = Þ keff =
dt
dt
1+ ( k1b k2 ) [ M ]
 Note at low P ([M] small): keff  k1f[M] ~ P1;
high P: keff  k1fk2/k1b ~ P0 (falloff in rate at high P)
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
9
Steady-state approx. – chain branching
 Overall reaction A2 + B2  2AB
A2 + M  A + A + M
A + B2  AB + B
A2 + B  AB + A
(rate constant k1, chain initiation)
(rate constant k2, chain branching)
(rate constant k3, chain branching)
A + B + M  AB + M
(rate constant k4, chain termination)
 Steady state assumption for A and B:
d [ A]
2k [ A ] [ M ] + k3 [ A2 ] [ B ]
= 2k1 [ A2 ] [ M ] - k2 [ A ] [ B2 ] + k3 [ A2 ] [ B ] - k4 [ A ] [ B ] [ M ] » 0 Þ [ A ] » 1 2
dt
k2 [ B2 ] + k4 [ B ] [ M ]
d [ B]
k2 [ A ] [ B2 ]
= k2 [ A ] [ B2 ] - k3 [ A2 ] [ B ] - k4 [ A ] [ B ] [ M ] » 0 Þ [ B ] »
dt
k3 [ A2 ] + k4 [ A ] [ M ]
Þ 2k1 [ A2 ] [ M ] - k2 [ A ] [ B2 ] + k3 [ A2 ]
k2 [ A ] [ B2 ]
k2 [ A ] [ B2 ]
- k4 [ A ]
[M ] » 0
k3 [ A2 ] + k4 [ A] [ M ]
k3 [ A2 ] + k4 [ A ] [ M ]
ìï k k [ A ] 2 üï
ìï k [ A ] [ M ] üï
k1 [ A2 ] [ M ] ìï
4k2 k3 [ B2 ] üï
1
2
1 3
2
ý [ A] - í
ý = 0 Þ [ A] =
í1+ 1+
[ A] - í
2 ý
2k2 [ B2 ] îï
ïî k2 [ B2 ] ïþ
k1k4 [ M ] þï
ïî k2 k4 [ B2 ] ïþ
2
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
10
Steady-state approx. – chain branching
 Text states that k1 & k4 should be much smaller than k2 & k3
since k2 & k3 are radical-molecule reactions - what’s wrong
with that statement?
 Anyway if the 2nd term inside the square root is >> 1 then
[ A] » [ A2 ]
d [ B2 ]
k k k [B ]
k1k3
Þ
» - [ A2 ] 1 2 3 2
k2 k4 [ B2 ]
dt
k4
 Rate of [B2] consumption increases with k1, k2, k3 but
decreases with k4 due to loss of A and B radicals
2
 Note that the requirement 4k2 k3 [ B2 ] k1k4 [ M ] >>1 always breaks
down at sufficiently high P since [B2] ~ P but [M]2 ~ P2, thus
steady-state assumption for this reaction mechanism fails at
high enough P
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
11
Excel spreadsheet model
 OK is this for real? Create Excel spreadsheet model and
compare “exact” solutions using Euler’s method to advance
solution from time step t to t+Dt: f(t+Dt) ≈ f(t) + {df(t)/dt}Dt, e.g.
d [ A]
df (t)
f (t + Dt) » f (t) +
Dt Þ [ A] t+Dt » [ A] t +
Dt
dt
dt
t
d [ A]
= 2k1 [ A2 ] [ M ] - k2 [ A] [ B2 ] + k3 [ A2 ] [ B] - k4 [ A] [ B] [ M ]
dt
Þ [ A ] t+Dt » [ A ] t + D [ A]t = [ A] t + {2k1 [ A2 ] [ M ] - k2 [ A ] [ B2 ] + k3 [ A2 ] [ B] - k4 [ A] [ B] [ M ]} Dt
Euler’s method very simple not very stable or accurate – use
e.g. 4th order Runge-Kutta when Euler fails (actually Euler’s
method gives the first term in the 4th order Runge-Kutta)
 Spreadsheet inputs: k1 – k4; initial concentrations of A2, B2, A,
B, AB, M; Dt
 Outputs: concentrations vs. time; time to reach 90%
completion of reaction to AB; time to peak [A]; time to peak
[B]; validity of steady-state approximation (rates of formation
& destruction of A > 10x net rate of change of [A])
12
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
Excel spreadsheet model – baseline case
 k1 = 0.01, k2 = 2, k3 = 1, k4 = 1
 Time = 0: [A2] = 1, [B2] = 1, [M] = 10, [A] = [B] = [AB] = 0
0.1
1.8
0.09
1.6
0.08
1.4
0.07
1.2
1
A2
B2
A
B
AB
0.06
0.05
0.8
0.04
0.6
0.03
0.4
0.02
0.2
0.01
0
Minor (radical) species (A, B)
Major species (A2, B2, AB)
2
0
0
5
10
15
20
25
30
Time
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
35
13
Validity of steady-state approximation
 Approximation invalid initially when no radical pool exists
 Also invalid when reactants A2 & B2 have been depleted
significantly thus k2[A][B2] & k3[B][A2] are not large
2
1.8
1.6
1.4
Steady state OK? (1 = yes, 0 = no)
1.2
Steady state [A] / Actual [A]
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
Time
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
35
14
Excel spreadsheet model - results
 Radical (A, B) concentrations are much lower than major
(stable) species (A2, B2, AB)
 Radical pool builds quickly to peak, reaches steady-state,
then decreases slowly as reactants are consumed
 Product AB follows mirror-image of reactants A2 & B2,
showing that steady-state exists, radical concentrations shift
quickly as reactant concentrations shift
 How do changes in input parameters affect reaction time
(90% of AB formed)? For 10% increase in parameter (in case
of [A], [B], [AB], added 0.1 to initial mixture), % change in
reaction time:
Property
k1
k2
k3
k4
[M]
[A2]
[B2] [A]
[B]
% change
-6
-4
-2
2
-4
36
-11
-8
-17
k1, k2, k3 increases overall reaction rate, k4 decreases rate
due to loss of radicals; adding [A] decreases rate more than
adding [B] due to bottleneck of initiation A2 + M  2A + M
15
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
Partial equilibrium approximation
 Analogous to steady-state approximation, but applies to a
single (reversible) reaction rather than a single species
 Requires both forward and reverse rates to be fast compared
to net rate, e.g. for A + B2  AB + B
k1 f [ A] [ B2 ] >> k1 f [ A] [ B2 ] - k1b [ AB] [ B] ;k1b [ AB] [ B] >> k1 f [ A] [ B2 ] - k1b [ AB] [ B]
 Example: for reactants A2, B2, product A2B, radicals A, B, AB
A + B2  AB + B
Rate constant k1f
AB + B  A + B2
Rate constant k1b
B + A2  AB + A
Rate constant k2f
AB + A  B + A2
Rate constant k2b
AB + A2  A2B + A
Rate constant k3f
A2B + A  AB + A2
Rate constant k3b
A + AB + M  A2B + M
Rate constant k4 (NOT in partial
equil.)
AB
[ ] [ B] » K AB K B º K ; [ AB] [ A] » K AB K A º K ; [ A2 B] [ A] » K A2B K A º K
eq,1
eq,2
eq,3
A
B
K
K
B
A
K
K
AB
A
K
K
[ ][ 2 ]
[ ][ 2 ]
[ ] [ 2 ] AB A2
A B2
B A2
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
16
Partial equilibrium approximation
 Combining these results to solve for radical species A, B, AB
Keq,1Keq,2 [ A2 ] [ B2 ] Keq,3
3
[ A] =
[ A2 B]
K eq,1K eq,3 [ A2 ] [ B2 ]
; [ B] =
; [ AB] = Keq,1Keq,2 [ A2 ] [ B2 ] ;
[ A2 B]
 Then the product formation rate is
K eq,1K eq,2 [ A2 ] [ B2 ] K eq,3
d [ A2 B]
= k4 [ A ] [ AB ] [ M ] = k4
K eq,1K eq,2 [ A2 ] [ B2 ] [ M ]
dt
[ A2 B]
3
K eq,1K eq,2 K eq,3 [ A2 ] [ B2 ] [ M ]
2
= k4
[ A2 B]
 Obviously cannot apply at t = 0 since [A2B] = 0, but early on,
before [A2] and [B2] depletion is significant,
[ A2 B] » 2k4 Keq,1Keq,2 Keq,3 [ A2 ] [ B2 ] [ M ] t
2
t1/2 behavior
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
17
Online chemical kinetics calculator
 What if the mechanism is more complex or less stable?
http://navier.engr.colostate.edu/~dandy/code/code-5/index.html
 Chemical mechanisms for H2-O2, CH4-O2 and a few others
 Input page simple except for need to choose by trial & error:
 Total integration time (long enough to see fuel consumption &
product formation)
 Time interval (Dt) (small enough to resolve transients)
 Output format hard to feed into to Excel!
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
18
Online chemical kinetics calculator
 Typical example - stoich. H2-O2, T = 850K, P = 1 atm, near 3rd
explosion limit, above 2nd limit so mole fractions of H, O, OH
very small, ≈ 10-9
 Similar to schematic A2 + B2 results
0.007
0.9
0.006
0.8
0.005
0.7
0.6
H2
0.5
H2O
0.004
HO2 x 100
0.4
0.003
H2O2
0.3
0.002
0.2
0.001
0.1
0
Mole fraction HO2 and H2O2
Mole fraction H2 and H2O
1
0
0
20
40
60
80
100
Time (seconds)
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
19
Summary
 Chemical kinetic systems for realistic fuels are comprised of
many individual reactions, resulting in coupled Ordinary
Differential Equations in terms of the concentrations of
reactants [A], [B], [C], [D], … of the form
d ì[A] ü d ì[B] ü
d ì[C] ü
d ì[D] ü
n
n
í ý = í ý = - í ý = - í ý = -k f [ A] [ B]
dt î n A þ dt î n B þ
dt î n C þ
dt î n D þ
A
B
 Analytical solution of these equations is impossible for all but
the simplest cases
 Techniques exist for simplifying these large sets of equations
 Steady-state (for one species)
 Partial equilibrium (for one reversible reaction)
 Many others…
 Simplifying methods exploit the fact that some reactions (e.g.
radical + stable species) are typically faster than others (e.g.
chain initiation, 3-body recombination, radical + radical) and
so adjust to changing concentrations on a shorter time scale
20
AME 513 - Fall 2012 - Lecture 5 - Chemical Kinetics II
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