29052010-1100-introduction à la cinétique.pdf

N. DARABIHA

Samedi 29 Mai 2010, 11h00 – 12h30

Généralités sur la cinétique de combustion

1

Nasser DARABIHA

[email protected]

http://www.em2c.ecp.fr

Téléphone : +33 1 41 13 10 72

Télécopie : +33 1 47 02 80 35

Laboratoire d'Energétique Moléculaireet Macroscopique, Combustion

ECOLECENTRALEPARIS

2

ChemicalKinetics

Thermodynamics

FluidMechanics

Transport

Phenomena

Combustion

Multi-phase flo

w

3

• Production and destruction of pollutants• Ignition,• Extinction,• Flame Structure,• ……..

Why Chemical Kinetics?

4

Temperature Collisions Reaction

Reacting mixture

5

CONTENTS

- Some Definitions- Thermodynamics - Chemical kinetics Arrhenius Law Reaction rate Chemkin- Reaction mechanism H2/O2 combustion

- 0-D formulation Auto-ignition PSR

- Pollutants in atmosphere

6

Mixing of N species inside a volume ν

nk moles of species k , n = nkk=1

N

Σ

!

Xk=nk

nMole fraction

!

Yk=m

k

mMass fraction

1 mole = 6,02252*1023 molecules

mass : m = mk = nk Mkk=1

N

Σk=1

N

Σ

!

Yk= X

k

Mk

MMixture Molar mass M = Xk Mk

!

=m

nk=1

N

Σ

7

!

Yk=m

k

m="k

"Mass fraction

!

"k=m

k

VDensity

!

"k

=nkM

k

V= C

kM

k

!

" = "k

k=1

N

#

!

Ck=nk

VConcentration

!

Ck

k=1

N

" =n

V#1

8

mass stoichiometric coefficient

stoichiometric coefficient

F + S O + γ’N2 Products

νFF + νO O + γN2 Products

Stoichiometric reaction

Enough oxidant O to burn all fuel F

9

Stoichiometric reactions:

CnHm + (n+m/4) (O2 + γ N2) nCO2 +(m/2) H2O + (n+m/4) γ N2

CH4 + 2 (O2 + γ N2) CO2 +2 H2O + 2 γ N2

C3H8 + 5 (O2 + γ N2) 3CO2 +4 H2O + 5 γ N2

H2 + 1/2 (O2 + γ N2) H2O + 1/2 γ N2

10

Any reaction ... :

CH4 + S’( O2 + γ N2) Products

Stoichiometric reaction:

CH4 + S (O2 + γ N2) Products

XCH4

XO2

XCH4

XO2( )

Stoich

φ =Mixture

equivalence ratio

!

=

1

" S

1

S

=S

" S

Mixture equivalence ratio

11

Any reaction ... :

CH4 + S (O2 + γ N2) Products

!

S

" S

CH4 + S (O2 + γ N2) Productsφ

Φ > 1 Rich mixture

Φ < 1 Lean mixture

Φ = 1 Stoichiometric mixture

12

Any reaction ... :

CH4 + S’ (O2 + γ N2) Products

YCH4

YO2

YCH4

YO2( )

Stoich

φ =Mixture

equivalence ratio

!

=

MCH 4

" S MO2

MCH 4

S MO2

=S

" S

φ CH4 + S (O2 + γ N2) Products

13

Air Factor

F = 1 / φ

Any reaction ... :

CH4 + S (O2 + γ N2) Productsφ

CH4 + F . S (O2 + γ N2) Products

14

Air volume (excess)

e < 0 Rich mixture

e > 0 Lean mixture

e = 0 Stoichiometric mixture

e = F - 1 = (1 −Φ)/Φ

CH4 + (1+ e ) . S (O2 + γ N2) Products

15

CONTENTS




16

ThermodynamicsFirst principle(no variation of kinetic and potential enery) :

dU = dQ + dWdW = -PdV

H=U + PV

dUdQ

dW

Constant volume reaction:dU = dQ

Constant pressure reaction:

dU = dQ -PdV

dH=dU + VdP + PdV= dQ + dW + PdV

dH = dQ

0

17

Thermodynamics

dUdQ

dWAdiabatic Constant volumereaction:

dU = 0

Adiabatic Constant pressurereaction:

dH = 0

0

18

Thermodynamics

Species enthalpy

Mixture enthalpy

chemical sensible

19

Thermodynamics

Mixture enthalpy :

chemical sensible

with

20

YCO2

T

initial state : 1

final state: 2

Initial state - final state

If time is infinite Final state = Equilibrium

P1T1Y1k

P2T2Y2k

Gibbs energy: G = H - TS

Equilibrium is reached when G is minimal (S is maximal)

21

At the equilibrium state, burnt gases are composed of:

Thermodynamic Equilibrium

CH4 + 2(O2 + 3,76 N2) Final equil. products

CO2 , H2O , N2 , CO, CH2 , CHi , H, H2 , OH , , O , O2 ,.,.,…. , NO , NO2 ,.,.,….

Mathematically, the equilibrium compositionis obtained by minimizing G

22

Equilibrium burnt gases temperatureadiabatic constant pressure

Adiabatic:

!

" # kAk

k=1

N

$ % " " # kAk

k=1

N

$

initial state : 1 final state: 2

The only unknown

!

Y1khk

k=1

N

" (T1,P) = Y

2khk

k=1

N

" (T2,P)

: H = cst

23

Equilibrium burnt gases temperatureadiabatic constant pressure

!

Y1k hk

k=1

N

" (T0,P) + Y

1k c pk

k=1

N

" ( # T ,P)T0

T1

$ d # T =

!

Y2k hk

k=1

N

" (T0,P) + Y

2k c pk

k=1

N

" ( # T ,P)T0

T2

$ d # T

!

cp2( " T )

24

Heat released by thecombustion

Burnt gases temperatureconstant pressure

Unknown

!

cp2( " T )T0

T2

# d " T $ cp1( " T )T0

T1

# d " T =

!

(Y1k"Y

2k) h

k

k=1

N

# (T0,P)

25

CO

CO2!

cp1(T,P) = Y1k c pk

k=1

N

" (T,P)

26

Assume all cpk = cst = cp and not temperature dependent

Burnt gases temperatureconstant pressure

UnknownHeat released by the combustion

!

cp (T2 "T1) = (Y1k "Y2k ) hk

k=1

N

# (T0)

27

Calorific value

Calorific value at constant pressure :

!

(PC)p = (Y1k "Y2k ) hk

k=1

N

# (T0, p

0)

Calorific value at constant volume:

!

(PC)p = h1(T0, p

0) " h

2(T0, p

0)

!

(PC)v =U1(T0, p

0) "U

2(T0, p

0)

It is the quantity of heat that can theoretically be released per unit mass of fuel

28

Calorific value

Calorific value: The quantity of heat released by thecomplete combustion, per unit mass of a fuel, the vaporproduced by the combustion of the gas being assumedto remain as a vapour.

High calorific value: The amount of heat released bycomplete combustion, per unit mass of a fuel, the vaporproduced by the combustion of the gas being assumedto be completely condensed and its latent heat released.

29

CONTENTS




30

Temperature Collisions Reaction

Reacting mixture

31

Collision frequency

Motion molecular velocity

Volume through which the moleculesweeps in 1s

For one molecule (if all othermolecules are motionless)

Kuo, 2005

32

Collision frequency

Kuo, 2005

33

Collision frequency

But all collisions do not lead to reaction

A + A P

A + B P

2A + 3B P2 3

2 3

34

Arrhenius Law

Svante Arrhenius (1859-1927)

Only molecules with E > Ea will react ...Ea =Activation Energy

collision frequencysteric factor (depends on theorientation of the collidingmolecules)

Boltzmann factor

with RR!

dCB

dt= " fc P e

"Ea

RT

35

Order of reaction

Overall order of reaction :

!

" # kAk

k=1

N

$ % " " # kAk

k=1

N

$

!

m = " # k

k=1

N

$!

dCM k

dt= ( " " # k $ " # k ) k f (CM k

)" # k

k=1

N

%

36

First order reactions! Valid only forelementary reactions

37

Second order reactions

AB

38

Third order reactions

--> Becomes a 2nd orderreaction if CM is cst

Third order reactions

39

Consecutive reactions

k1

k2

40

Parallel reactions

Twin reactions

!

A + Bk1" # " C

!

A + Bk2" # " D

!

dCC

dt= k

1CACB

!

dCD

dt= k

2CACB

!

"CC

CD

=k1

k2

41

Parallel reactions

Competitive reactions

!

A + Bk1" # " C

!

A + Ek2" # " D

!

dCC

dt= k

1CACB

!

dCD

dt= k

2CACE

!

" lnCB0

CB0

#CC

=k1

k2

lnCE0

CE0

#CD

42

Opposing (or reverse) reactions

!

" # k Ak

k=1

N

$k f% & %

kb' % %

" " # k Ak

k=1

N

$

43

Opposing (or reverse) reactions

But complex mechanisms: N species, I reactions :

!

" # k,i Ak

k=1

N

$k f% & %

kb' % %

" " # k,i Ak

k=1

N

$ for i =1,.....,I

Each reaction i is characterized by a rate of progress qi

44

Reaction rate of species k in the ith reaction is:

Reaction rate of species k for the overall mechanism is:

General expression of reaction rates

.

.

45

as

Molar reaction rate:


Mass reaction rate

!

dCk

dt=

•

"k

mole

m3s

#

$ %

&

' (

!

dCkMk

dt= Mk

•

"kkgm3s

#

$ %

&

' (

!

CkM

k= "

k= " Y

k

!

d"Yk

dt= M

k

•

#k

46

Then:


!

Mk

•

"k

= 0k=1

N

#

!

d"Yk

dt= M

k

•

#k

!

d"Yk

dtk=1

N

# =k=1

N

# Mk

•

$k

Conservationof mass

47

CHEMKINKinetics Scheme

CHEMKINInterpreter

LINK file

ThermodynamicsData Base

Set of CHEMKINRoutines :

CKXTYCKYTXCKXTCCKCPYCKWXPCKWYP

.

.

User’s program (x. : equil, ….)

48

CONTENTS




49

DETAILED CHEMISTRY

CH4 + 2O2 CO2 +2 H2O

H2 + 1/2 O2 H2O

Even at the equilibrium state, burnt gases are composed of:

CH4 + 2 O2 Products

CO2 , H2O , CO, CH2 , CHi , H, H2 , OH , , O , O2 ,.,.,….,.,.,….

50

Chain reactions• Series of competitives, consecutives and opposing reactions• Apparition of intermediate species• Importance of free radicals

Radicals: highly reactive molecules or atoms

Chain-initiation reaction (A2 has alower Ea than B2)

!

A2

+ M" 2A + M

Chain-carrying reactions:

radical -> radical (fast propagation)

!

A + B2" AB + B

!

B + A2" AB + A

!

A + AB" A2

+ B

!

B + AB" B2

+ A

Chain terminating reaction

!

2A + M" A2

+ M

!

2B + M" B2

+ M

51

H2 / O2 combustion

52

DETAILED CHEMISTRY(H2 - O2 combustion)

INITIATION : free - radical production

Temperature Collisions

O2 + M 2 O + M

H2 + M 2 H + M

M denotes all other species as third body (O2, H2O, …)Delivering its energy to O2 and H2 helping them to dissociate

Chain-Initiatingreactions

53

Chain-branching reaction : Radicals are more produced thanconsumed

O + H2 H + OH

H + O2 O + OH

DETAILED CHEMISTRY

Very fast reactions

54

Example :

A 1 cm3 container with n =1019 moleculesand fc = 108 collisions / s

1 free radical per cm3

★ Chain-carrying: time for all molecules to react = 1019/108 = 30,000 years

55

Chain reactions

Time to react = 64 / 108 ≈ 1 µs

★ Chain-branching:

➡ 1 + 2 + 22 + … + 2L= (2L+1-1)/(2-1) = 1019 molecules after L=64 generations

Extremely fast reaction ....

★ If 1% of the reactions are chain-branching:➡ t ≈ 40 µs

Assume chain-branching reactions

1 radical --> 2 radicals

56

Chain-carrying : Radicals and final products

OH + H2 H2O + H

O + H2O 2 OH

DETAILED CHEMISTRY

57

Chain-terminating steps : recombination mechanism

H + H + M H2 + M

O + O + M O2 + M

H + O + M OH + M

H + OH + M H2O + M

DETAILED CHEMISTRY

58

H2 + O2 2 OHH2 + OH H2O + HO + OH H + O2O + H2 OH + HH + O2 +M HO2 + MOH + HO2 H2O + O2H + HO2 2 OHO + HO2 O2 + OHOH + OH O + H2OH + H + M H2 + MH + H + H2 H2 + H2H + H + H2O H2 + H2OH + OH + M H2O + M….….

COMBUSTION of H2 / O2

>10 species>40 reactions

59

C2H4

C2H5

C2H6

CH4

CH3

CH2O

HCO

CO

C2H3

C2H2

CH2

CH

CO, CO2

C3...

C3...

COMBUSTION of CH4 / O2

60

OXYDATION of CO

CO + OH CO2 + H

CO + O2 CO2 + O

O + H2O 2 OH

H + O2 OH + O

CO + HO2 CO2 + OH

Chain-terminating steps:

Main OH production reactions:

At the flame front:

61

CONTENTS




62

Application of spontaneous combustion

• Hypersonic combustion (SCRAMJET)• Diesel engines

To be avoided in :

• Safety characteristic length of fuel tanks Semenov theory

• Spark ignition engines (avoid pinking)

63

Adiabatic, constant pressure

Spontaneous combustion

64

Initial conditions :

Auto-ignition of a reactive mixture Adiabatic, constant pressure

65

Auto-ignition of a reactive mixture

T0 > Ti

τi

Auto-ignitingdelay time

66

Auto-ignition delay time

τi decreases exponentially with T0

τi changes as P01-n

Influence of equivalence ratio and dilutant (YN2)

H2

C2H2

COKerosene

CH4

T °C1100 1000 900 800 700 600

τi ms

40

10

1

0.5

67


• T = ambient: metastable state, reaction rate isalmost null

• When T increases:• If T > Ti Beginning of exothermic oxidationreaction:

production of enough radicals to ignite

• If T < Ti Heat released is not sufficient toincrease the temperature, because endothermicreactions absorb the heat to crack the fuel:

not enough radicals are produced to ignite

68

Container with constant volume withisothermal wall (Tw) and a wall surface S


V T

YF YO

STw

Container balance energy equation ….

Heat losses Q

69

!

d"VYk

dt=V M

k#

k

•

!

d"VU

dt= # K S (T #T

w)

!

U =k=1

N

" YkU

k

!

(" VYk

dUk

dt+U

k" V

dYk

dt)

k=1

N

# = $ K S (T $Tw)

!

(" VYk

dUk

dt+U

kVM

k#

k

•

)k=1

N

$ = % K S (T %Tw)


70

!

" V cv

dT

dt= # (V M

kU

k$

k

•

)k=1

N

% # K S (T #Tw)


!

" V cvdT

dt=V #F

•

Qv0 $ K S (T $Tw )

Production Heat loss

Fuel consumption rate modeling:

71

Auto-ignition conditions

Semenov theory

Production > Heat loss

!

"F

•

Qv0

# KS

V(T $T

w)

!

y1T( )

!

y2T( )

72

y(T)

T

Y1(T)

Production(chemical)

!

"F

•

Qv0

73

T

Heat loss

Tw

Y2 (T)Y (T)

!

KS

V(T "T

w)

74

y

T

y1 y2

Tw

Case 1: T0 high,

Auto-ignition

75

y

T

y1 y2

Tw

If T < Tc, The temperature increases up to T=Tc

Tc

Case 2: two curves are tangent

If T > Tc, Auto-ignition

dT/dt =0 (unstable)T = Tc + ε , Ignition

c

76

y

T

y1

y2

TwIf T < Ta , the temperature increases up to Ta and remains stable

a

b

Case 3:

If Ta < T < Tb , the temperature decreases to Ta and remains stable

If T > Tb, Auto-ignition

77

Varying K or (S/V)

The critical temperatureTc is not an intrinsicmixture property.

y

TTw

K(S/V)

Indeed, it is functionof K and (S/V)

S / V = 1 / LIf L goes up, S/V decreases and it Increasesthe risk of auto-ignition

If the system is well insulated, K decreases,and it increases the risk of auto-ignition

78

Varying P

There exists a critical pressure Pc above whichthere is always mixture auto-ignition.

w

79

Determination of ignition limits (Pi, Ti)y

T

y1 y2

Tw Ti

C

At the critical point C:

!

Y1

="F

•

Qv0

!

Y2

= KS

V(T "T

w)

80

One shows that:

81

Equivalence ratio

Equivalence ratio

Determination of ignition limits (Pi, Ti)

Ignition

Ignition

Lean limit Rich limit

Lean limit Rich limit

no combustion

no combustion

82

no combustion

Determination of ignition limits (Pi, Ti)

Ignition

83

Hydrocarbons ignition limits

84

Influence of T0 on ignition (C7H16-air)

P0 =30 bars, Φ=0.5

Time (s)

Tem

pera

ture

(K)

85

Influence of T0 on ignition (C7H16-air)

P0 =30 bars, Φ=0.5

Igni

tion

dela

y (s

)

Cold flamePrincipal ignition

Initial temperature (K)

86

CONTENTS




Perfectly stirred reactor (PSR)

Feed at the inlet at the state: X0

Homogeneous combustion

Burnt gases at the outlet

INLET OUTLET

, Yk0 , hk0, T0

Volume V

Yk , hk, T

Yk , hk, T

!

m

•

τ = Residence time

88

!

" VdY

k

dt= m

•

Yk0# m

•

Yk

+$kM

kV k =1,...,N

.

IN OUT PRODUCTION.

Species mass balance equation :

.

89

with

!

" Vdh

dt= m

•

Yk0hk0k=1

N

# $ m•

Ykhkk=1

N

# $Qh

!

h = Yk

k=1

N

" hk

!

dh

dt= Y

k

dhk

dt

"

# $

%

& '

k=1

N

( + hk

dYk

dt

"

# $

%

& '

k=1

N

(

!

Cp = Ykk=1

N

" Cpk

!

dhk

dt= Cpk

dT

dt

!

dh

dt= Cp

dT

dt+ hk

k=1

N

"dYk

dt

Enthalpy balance:

90

!

" V Cp

dT

dt= m

•

Yk0hk0k=1

N

# $ m•

Ykhkk=1

N

# $Qh

$ hkk=1

N

# " VdYk

dt

%

& '

(

) *

!

" VdY

k

dt= m

•

Yk0# m

•

Yk

+$k

•

MkV

!

" V Cp

dT

dt= m

•

Yk0hk0k=1

N

# $m•

Ykhkk=1

N

# $Qh

$ m•

Yk0hkk=1

N

# + m•

Ykhkk=1

N

# + $ V hkk=1

N

# %•

k Mk

&

' (

)

* +

Enthalpy balance:

91

Balance Equations

!

" VdY

k

dt= m

•

Yk0# m

•

Yk

+$k

•

MkV k =1,...,N

!

" V Cp

dT

dt= m

•

Yk0hk0k=1

N

# $ m•

Yk0hkk=1

N

# $Qh +V $ hkk=1

N

# %•

k Mk

&

' (

)

* +

Heat release rate

!

h

•

92

ATTENTION

!

" V Cp

dT

dt= m

•

Yk0hk0k=1

N

# $ m•

Ykhkk=1

N

# $Qh + h•

!

" Vdh

dt= m

•

Yk0hk0k=1

N

# $ m•

Ykhkk=1

N

# $Qh

!

" V Cp

dT

dt= m

•

Yk0hk0k=1

N

# $ m•

Yk0hkk=1

N

# $Qh + h•

93

Application example

at t=0 , Yk(0)=Yk0 , T(0) = Ti

!

dYk

dt=1

"(Y

k0#Y

k) +

$k

•

Mk

%k =1,...,N

!

Cp

dT

dt=1

"Yk0hk0

k=1

N

# $1

"Yk0hk

k=1

N

# $Qh

%V+h•

%

T0 = 300 KH2/Air mixture

Inlet Volume V

Outlet

Pressure= 1 atm

Yk (t), hk (t), T(t)

94

Ti = 4000K

τ =0,01 s

Stoic. H2/AirT0 = 300 KPressure= 1 atm

95

Influence of τTi = 4000K

Time (s)

96


Time (s)

97


Residence time (s)

98

Influence of Tiτ =0,01 s

Time (s)

99

Influence of Tiτ =0,01 s

Time (s)

100

CONTENTS




101

• CO2, H2O • CO

• CH4• CHx• Soot

• NOx

• COV (combustion)• SOx• Cl…• Br….• …..• …..

(vegetations, combustion)

(Rice, Ruminants…., combustion)

(Lightning, combustion)

102

Impact of Pollutantson the atmosphere

Troposphere : Altitude < 10-15 km,T 220K

k2 very fast

k1 (hν)day

night! NO2 NO + O

O2 + O +M O3 + M

k1

k2

Production of O3 If steady state for NO:

!

O3

C =k1 NO

2C

k3 NOC

=> Fragile equilibrium

>

k3NO + O3 NO2 + O2

If NO decreases and NO2 increases : Increase of O3

104

Effect of CO

CO + OH CO2 + H

H + O2 + M OH2 + M

•Importance of OH :Initiation of theprocess

105

• If [O3] / [NO] < 5000

HO2 + NO NO2 + OH

Formation of O3

• If [O3] / [NO] > 5000

HO2 + O3 OH + 2 O2

Destruction of O3

Effect of CO

NO2 + hν NO + O (λ > 420 nm)O2+O+M O3 +M

Balance :CO + O3 CO2 + O2

Balance : CO +2 O2 CO2 + O3

106

Effect of CH4

CH4 + OH CH3 + H2O

CH3 + O2 + M CH3O2 + M

2 HO2 + 2 NO 2 NO2 + 2 OH

CH3O2 + NO CH3O + NO2

CH3O + O2 CH2O + HO2

CH2O + O2 HCO + HO2

HCO + OH CO + H2O3NO2 + hν 3NO + 3O (λ > 420 nm)3O2+3O+3M 3O3 +3MBalance :

CH4+ 6 O2 CO + 2 H2O + 3 O320% - 50% of COin atmosphere

107

NO - NO2 Photochemical Cycle

λ < 420nm

NO2NO O

O3

O2O2

• Day time– Solar radiation allows

formation of O– Formation of O3 (O + O2)– Produced NO reacts with O3

and forms O2 et NO2

– Establishment of equilibrium• Night

– no production of O– Consumption of O3 by NO

emitted by cars (unless thewind has moved O3 away)

108

NO - NO2 Photochemical Cyclein polluted atmosphere

λ < 420nm

NO2NO O

O3

O2O2

RO2+RH

• Day time– New mechanism of NO2 /

RH formation– No consumption of O3

– Continuous increase of O3

• Night– no production of O– Consumption of O3 by NO

emitted by cars (unless thewind has moved O3 away)

NO2 + OH HNO3

acid rain

Troposphere : < 10-15 km, T 220K

Stratosphere : > 10-15 km, T 270K

Destruction of O3 by X : NO, OH, Cl, Br, .....

X + O3 XO + O2

XO + O X + O2

O3 + O 2 O2

k1 (hν)O2 2 O

O + O2 O3

k1Ozone layer :

111

Direct Effects of NOx onatmospheric equilibrium

• NOx participate in the formation of ozonein the troposphere : Smog

• NOx participate in the destruction of theozone layer in the stratosphere

29052010-1100-introduction à la cinétique.pdf

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