Analyzing Reaction Mechanisms

School of Aerospace Engineering
Chemical Kinetics:
Analyzing Reaction Mechanisms
Jerry Seitzman
0.2
2500
Mole Fraction
1500
CH4
H2O
HCO x 1000
Temperature
0.1
1000
0.05
500
Temperature (K)
2000
0.15
Methane Flame
0
0
0
0.1
0.2
0.3
Distance (cm)
Kinetics3 -1
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 Combustion
School of Aerospace Engineering
Reaction Mechanisms
• Important combustion problems generally
controlled by a set of chemical reactions
– example, thermal formation of NO in dry air
– controlled by 4 reactions (incl. forward and reverse)
N2  O  NO  N {1} H R  75.1kcal mol
Zeldovich
Mechanism
endothermic
O2  N  NO  O {2} H R  32.1kcal mol
– where did N, O come from?
(initiation reaction)
exothermic
4.1868 kJ/kcal
• Time Scales
– are any of these reactions much slower or faster than
the others?
Kinetics3 -2
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
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Characteristic Reaction Times
• What is characteristic time for a given reaction?
O2  N  NO  O
– e.g., bimolecular {2f}
– can define characteristic time by reaction rate and
change in reactant concentration
N   N 0 d N 
 chem 
 k2 f O2 N 
d N  dt
dt
– assume fractional [O2] change much less than for
[N] and little N initially
chem depends on rate
N   0  1 constant
 chem 
and initial
k2 f O2 0 N  k2 f O2 0 species concentration(s)
AE/ME 6766 Combustion
Kinetics3 -3
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Time Comparisons
k1 f  1.8 1014 e38,370K T
k1r  3.8 1013 e425K T
O2  N  NO  O {2}
k2 f  1.8 1010T e4680K T
k2 r  3.8 109 Te20,820 T
• Assuming initially air (79% N2,
cm3/mol/sec
21%O2) at 1 atm
N2  O  NO  N {1}
1f
2f
1000 K
3107 s
2 s
1500 K
100 s
0.5 s
2000 K
0.2 s
0.2 s
• initially no rev. rxs.
• {2}: fast exothermic,
rapidly follows {1}:
slow endothermic
• Overall NO production limited by{1f} rate , initially
1r
300 s
for 10 ppm NO
Kinetics3 -4
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
400 s
500 s
• {1r} will only compete
with {2f} for high [NO]
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System of Rate Equations
• Net production rates
N2  O  NO  N {1}
O2  N  NO  O {2}
d NO 
 k1 f N 2 O  k1r NO N   k2 f O2 N   k2 r NO O
dt
d N 
 k1 f N 2 O  k1r NO N   k2 f O2 N   k2 r NO O
dt
etc.
• 5 unknowns [N2],[O2],[NO],[N],[O] (+energy/T,…)
– 5 ODE’s (rate equations) to solve (actually 4 + N/O
constraint)
– computer OR simplify analytically
Kinetics3 -5
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 Combustion
School of Aerospace Engineering
Analytic Approximations
• Assume T, [N2], [O2] known
– from energy, other constraints
• Assume [O] given by partial equilibrium with O2
O O2 1 2  Kc, f ,O T   K p, f ,O T  R T 1 2
• Partial Equilibrium Assumption
–
–
–
–
applies to specific reaction(s) (not specific species)
doesn’t require chemical equilibrium of full system
requires some characteristic time to create/maintain
usually associated with fast, energetically neutral
reactions with no faster sources/sinks of species
involved
Kinetics3 -6
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
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Analytic Approximations (con’t)
• So now 3 knowns [N2], [O2], [O]
– only [NO], [N] unknown
• [N] is destroyed almost immediately as it is created
– assume [N] in steady-state d N  dt  0
• Steady-State Assumption
– applies to specific species (not reactions)
– often assumed for minor species
– usually associated with condition where species
concentration determined by instantaneous balance
between formation and destruction rates
– concentration not necessarily invariant in time,
just adapts quickly to other changes
AE/ME 6766 Combustion
Kinetics3 -7
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
N Steady-State
• Apply steady-state to rate equation for [N]
d N  dt  0  k1 f N 2 O  k1r NON   k2 f O2 N   k2r NOO
k1 f N2 O  k1r NO N   k2 f O2 N   k2 r NO O
N 
ss

k1 f N 2 O  k2 r NO O
k1r NO   k2 f O2 
• Define parameters
N 
ss
Kinetics3 -8
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.

RR1  k1 f N 2 O RR2  k2 r NO O
RR1  RR2
k1r NO   k2 f O2 
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Simplified NO Rate
• Use assumptions/definitions in NO rate equation
1
d NO
 k1 f N 2 O  k1r NON   k2 f O2 N   k2 r NOO
dt
d N  dt  0  k1 f N2 O  k1r NO N   k2 f O2 N   k2 r NOO
RR
d NO 
 2RR1  k1r NO N 
dt


1  RR2 RR1

 2 RR1 1 





1

k
O
k
NO
2
f
2
1
r



N * k2 fk1Of 2N2  k2 rk2NO
r NO
*
O kk11r rNO k1kf 2fNO2 2
* 2*
[NO]* is concentration that would occur
if NO in partial equilibrium with N2, O2
N 
ss

RR1  RR2
k1r NO   k2 f O2 
*
*
k1 f

NO  N 
 K p1 
N 2 O
k1r
*
k2 f

NO  O
 K p2 
O2 N *
k2 r
AE/ME 6766 Combustion
Kinetics3 -9
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Simplified NO Rate (con’t)


d NO 
1  RR2 RR1
k O  k NO 

 2 RR1 1 






dt
1

k
O
k
NO
k
k N 
2f
2
1r




1  RR2 RR1

 2 RR1 1 
2
*


RR2 k 2 r NOO
 1  NO  NO  k2 r NO  k1 f N 2 

RR1 k1 f N 2 O


1  RR2 RR1

 2 RR1 1 
2
 1  NO * NO  RR RR 

2
1 
*
*
  NO  NO  RR2  k2 r NO  O RR1,RR2,[NO]*=f(T,[N2],[O2])
*2
2f
2
1r




2r
1f
2
 okay for hot air,
d
1 d NO  2 RR1 
1 2

 lean combustion;


*
NO *  1  RR1 RR2  [O] partial equil?
dt NO  dt
Kinetics3 -10
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 Combustion
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School of Aerospace Engineering
Simplified NO Formation Result
• Solution for
– T, [N2], [O2] constant
– [NO]0=0
• Maximum relative
change at t=0

d 2 RR1 
1 2



* 
dt NO   1  RR1 RR2 
d 2 RR1
  1 

*
dt NO 

1
– not true; time required
to reach [N]ss
d NO 
 RR1  k2 f O2 N 
dt
• Asymptotes to
equilibrium NO
  NO  NO 
*
RR1  k1 f N 2 O
[N]ss reached
– d/dt=0 @ =1
t
AE/ME 6766 Combustion
Kinetics3 -11
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
NO Formation
• Recall earlier time constant
calculations (1atm, air)
1f
2f
1000 K
3107 s
2 s
1500 K
100 s
0.5 s
N2  O  NO  N {1}
O2  N  NO  O {2}
2000 K
0.2 s
0.2 s
• Formation of NO from Zeldovich (thermal NOx)
will be relatively slow except at very high
temperatures
Kinetics3 -12
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
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School of Aerospace Engineering
NO Destruction
• Consider case when NO initially high
– [NO] >> [NO]*
– for example, sudden cooling of hot products
• Now look at results for >>1
(… and RR1/RR2 O(1))
 2 RR1    2   2RR2
d
2 RR1 
1





* 
dt removal NO   1  RR1 RR2  NO *  RR1 RR2 
NO *
2
• Now limited by {2r}
Compare to
d
dt

formation
2 RR1
NO *
– endothermic
– slower than formation NO  N  N2  O {1r} 75.1kcal mol
NO  O  O2  N {2r} 32.1kcal mol
(2 minor species)
AE/ME 6766 Combustion
Kinetics3 -13
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
NO Destruction (con’t)
• Characteristic NO destruction times
 NO,destr 
NO   NO   NO   1
d NO  dt 2RR 2k NO O 2k O
2
2r
2r
1atm, air, O in parital equilibrium with O2
T(K)
2600
2200
NO, destr. O(ms) O(0.1s)
2000
1500
1000
O(s)
O(hr)
1000’s years
• So if NO produced and gases cooled, approach
chemically frozen flow
– NO levels may be frozen at superequilibrium values
Kinetics3 -14
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
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Comments on Computer Solutions
• Recall time constants for Zeldovich mechanism
1atm, air
1f
2f
–
–
–
–
1000 K
3107 s
2 s
1500 K
100 s
0.5 s
2000 K
0.2 s
0.2 s
N2  O  NO  N {1}
O2  N  NO  O {2}
NO production relatively slow, limited by {1}
N rates relatively fast (steady-state assumption)
makes computer solution “difficult” or expensive
such large variation in time-constants leads to set of
stiff ODEs; need stiff ODE solver (found in
Chemkin, Cantera, …)
Kinetics3 -15
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 Combustion
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Net Mass Production Rates
• Sometimes need to have net production rate in mass
units
w NO 
dwNO
dY
d NO 
  NO  WNO
dt
dt
dt
• Strictly, this is chemical source term w NO,chem
• Total change in mass density is
w NO  w NO,chem  w NO,density  
Kinetics3 -16
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
dYNO
d
 YNO
dt
dt
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Chain Branching
• Many reaction mechanisms depend on rapid growth
of radical population
– to create fast overall reaction
– to turn reactants into products
• Radicals
– typically unpaired electrons
– highly reactive
– abstraction reactions usually faster than dissociation
reactions
AB  R  A  BR
Kinetics3 -17
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 Combustion
School of Aerospace Engineering
Types of Radical Reactions
• Chain initiation reactions
R
H
+2
>>0
N2  O  N  NO
• Chain branching reactions
0
0
H  O2  OH  O
• Chain terminating reactions
+1
<0
O  O  M  O2  M
-2
<<0
– create radical without radical reactant
O2  M  O  O  M
• Chain propagating reactions
– use and produce same # of radicals
– net production of radicals
– net destruction of radicals
Kinetics3 -18
Copyright © 2004-2005, 2014 by Jerry M. Seitzman. All rights reserved.
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