29052010-1100-introduction à la cinétique.pdf

N. DARABIHA

Samedi 29 Mai 2010, 11h00 – 12h30

Généralités sur la cinétique de combustion

Nasser DARABIHA

nasser.darabiha@em2c.ecp.fr

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

ChemicalKinetics

Thermodynamics

FluidMechanics

Transport

Phenomena

Combustion

Multi-phase flo

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

Why Chemical Kinetics?

Temperature Collisions Reaction

Reacting mixture

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

Mixing of N species inside a volume ν

nk moles of species k , n = nkk=1

nMole fraction

mMass fraction

1 mole = 6,02252*1023 molecules

mass : m = mk = nk Mkk=1

MMixture Molar mass M = Xk Mk

"Mass fraction

VDensity

" = "k

VConcentration

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

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

Any reaction ... :

CH4 + S’( O2 + γ N2) Products

Stoichiometric reaction:

CH4 + S (O2 + γ N2) Products

XO2( )

Stoich

φ =Mixture

equivalence ratio

Mixture equivalence ratio

Any reaction ... :

CH4 + S (O2 + γ N2) Products

CH4 + S (O2 + γ N2) Productsφ

Φ > 1 Rich mixture

Φ < 1 Lean mixture

Φ = 1 Stoichiometric mixture

Any reaction ... :

CH4 + S’ (O2 + γ N2) Products

YO2( )

Stoich

φ =Mixture

equivalence ratio

" S MO2

φ CH4 + S (O2 + γ N2) Products

Air Factor

F = 1 / φ

Any reaction ... :

CH4 + S (O2 + γ N2) Productsφ

CH4 + F . S (O2 + γ N2) Products

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

CONTENTS

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

dU = dQ + dWdW = -PdV

H=U + PV

Constant volume reaction:dU = dQ

Constant pressure reaction:

dU = dQ -PdV

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

dH = dQ

Thermodynamics

dWAdiabatic Constant volumereaction:

dU = 0

Adiabatic Constant pressurereaction:

dH = 0

Thermodynamics

Species enthalpy

Mixture enthalpy

chemical sensible

Thermodynamics

Mixture enthalpy :

chemical sensible

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)

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

Equilibrium burnt gases temperatureadiabatic constant pressure

Adiabatic:

" # kAk

$ % " " # kAk

initial state : 1 final state: 2

The only unknown

" (T1,P) = Y

" (T2,P)

: H = cst

Equilibrium burnt gases temperatureadiabatic constant pressure

Y1k hk

" (T0,P) + Y

1k c pk

" ( # T ,P)T0

$ d # T =

Y2k hk

" (T0,P) + Y

2k c pk

" ( # T ,P)T0

$ d # T

cp2( " T )

Heat released by thecombustion

Burnt gases temperatureconstant pressure

Unknown

cp2( " T )T0

# d " T $ cp1( " T )T0

# d " T =

(Y1k"Y

# (T0,P)

cp1(T,P) = Y1k c pk

" (T,P)

Assume all cpk = cst = cp and not temperature dependent

Burnt gases temperatureconstant pressure

UnknownHeat released by the combustion

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

# (T0)

Calorific value

Calorific value at constant pressure :

(PC)p = (Y1k "Y2k ) hk

# (T0, p

Calorific value at constant volume:

(PC)p = h1(T0, p

0) " h

2(T0, p

(PC)v =U1(T0, p

2(T0, p

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

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.

CONTENTS

Temperature Collisions Reaction

Reacting mixture

Collision frequency

Motion molecular velocity

Volume through which the moleculesweeps in 1s

For one molecule (if all othermolecules are motionless)

Kuo, 2005

Collision frequency

Kuo, 2005

Collision frequency

But all collisions do not lead to reaction

A + A P

A + B P

2A + 3B P2 3

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!

dt= " fc P e

Order of reaction

Overall order of reaction :

" # kAk

$ % " " # kAk

m = " # k

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

)" # k

First order reactions! Valid only forelementary reactions

Second order reactions

Third order reactions

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

Third order reactions

Consecutive reactions

Parallel reactions

Twin reactions

A + Bk1" # " C

A + Bk2" # " D

Parallel reactions

Competitive reactions

A + Bk1" # " C

A + Ek2" # " D

" lnCB0

Opposing (or reverse) reactions

" # k Ak

$k f% & %

kb' % %

" " # k Ak

Opposing (or reverse) reactions

But complex mechanisms: N species, I reactions :

" # k,i Ak

$k f% & %

kb' % %

" " # k,i Ak

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

Each reaction i is characterized by a rate of progress qi

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

Molar reaction rate:

Mass reaction rate

dt= Mk

"kkgm3s

k= " Y

= 0k=1

# =k=1

Conservationof mass

CHEMKINKinetics Scheme

CHEMKINInterpreter

LINK file

ThermodynamicsData Base

Set of CHEMKINRoutines :

CKXTYCKYTXCKXTCCKCPYCKWXPCKWYP

User’s program (x. : equil, ….)

CONTENTS

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 ,.,.,….,.,.,….

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)

+ M" 2A + M

Chain-carrying reactions:

radical -> radical (fast propagation)

A + B2" AB + B

B + A2" AB + A

A + AB" A2

B + AB" B2

Chain terminating reaction

2A + M" A2

2B + M" B2

H2 / O2 combustion

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

Chain-branching reaction : Radicals are more produced thanconsumed

O + H2 H + OH

H + O2 O + OH

DETAILED CHEMISTRY

Very fast reactions

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

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

Chain-carrying : Radicals and final products

OH + H2 H2O + H

O + H2O 2 OH

DETAILED CHEMISTRY

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

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

CO, CO2

COMBUSTION of CH4 / O2

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:

CONTENTS

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)

Adiabatic, constant pressure

Spontaneous combustion

Initial conditions :

Auto-ignition of a reactive mixture Adiabatic, constant pressure

Auto-ignition of a reactive mixture

T0 > Ti

Auto-ignitingdelay time

Auto-ignition delay time

τi decreases exponentially with T0

τi changes as P01-n

Influence of equivalence ratio and dilutant (YN2)

COKerosene

T °C1100 1000 900 800 700 600

τi ms

• 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

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

Container balance energy equation ….

Heat losses Q

dt=V M

dt= # K S (T #T

U =k=1

(" VYk

# = $ K S (T $Tw)

(" VYk

$ = % K S (T %Tw)

" V cv

dt= # (V M

% # K S (T #Tw)

" V cvdT

dt=V #F

Qv0 $ K S (T $Tw )

Production Heat loss

Fuel consumption rate modeling:

Auto-ignition conditions

Semenov theory

Production > Heat loss

V(T $T

y1T( )

y2T( )

Production(chemical)

Heat loss

Y2 (T)Y (T)

V(T "T

Case 1: T0 high,

Auto-ignition

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

Case 2: two curves are tangent

If T > Tc, Auto-ignition

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

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

Case 3:

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

If T > Tb, Auto-ignition

Varying K or (S/V)

The critical temperatureTc is not an intrinsicmixture property.

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

Varying P

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

Determination of ignition limits (Pi, Ti)y

At the critical point C:

V(T "T

One shows that:

Equivalence ratio

Determination of ignition limits (Pi, Ti)

Ignition

Lean limit Rich limit

no combustion

Determination of ignition limits (Pi, Ti)

Ignition

Hydrocarbons ignition limits

Influence of T0 on ignition (C7H16-air)

P0 =30 bars, Φ=0.5

Time (s)

Influence of T0 on ignition (C7H16-air)

P0 =30 bars, Φ=0.5

Cold flamePrincipal ignition

Initial temperature (K)

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

τ = Residence time

Yk0# m

kV k =1,...,N

IN OUT PRODUCTION.

Species mass balance equation :

Yk0hk0k=1

# $ m•

Ykhkk=1

h = Yk

( + hk

Cp = Ykk=1

dt= Cpk

dt= Cp

dt+ hk

Enthalpy balance:

" V Cp

Yk0hk0k=1

# $ m•

Ykhkk=1

$ hkk=1

# " VdYk

Yk0# m

" V Cp

Yk0hk0k=1

# $m•

Ykhkk=1

$ m•

Yk0hkk=1

# + m•

Ykhkk=1

# + $ V hkk=1

# %•

Enthalpy balance:

Balance Equations

Yk0# m

MkV k =1,...,N

" V Cp

Yk0hk0k=1

# $ m•

Yk0hkk=1

# $Qh +V $ hkk=1

# %•

Heat release rate

ATTENTION

" V Cp

Yk0hk0k=1

# $ m•

Ykhkk=1

# $Qh + h•

Yk0hk0k=1

# $ m•

Ykhkk=1

" V Cp

Yk0hk0k=1

# $ m•

Yk0hkk=1

# $Qh + h•

Application example

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

%k =1,...,N

"Yk0hk0

"Yk0hk

%V+h•

T0 = 300 KH2/Air mixture

Inlet Volume V

Outlet

Pressure= 1 atm

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

Ti = 4000K

τ =0,01 s

Stoic. H2/AirT0 = 300 KPressure= 1 atm

Influence of τTi = 4000K

Time (s)

Residence time (s)

Influence of Tiτ =0,01 s

Time (s)

Influence of Tiτ =0,01 s

Time (s)

CONTENTS

• CO2, H2O • CO

• CH4• CHx• Soot

• NOx

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

(vegetations, combustion)

(Rice, Ruminants…., combustion)

(Lightning, combustion)

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

Production of O3 If steady state for NO:

C =k1 NO

k3 NOC

=> Fragile equilibrium

k3NO + O3 NO2 + O2

If NO decreases and NO2 increases : Increase of O3

Effect of CO

CO + OH CO2 + H

H + O2 + M OH2 + M

•Importance of OH :Initiation of theprocess

• 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

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

NO - NO2 Photochemical Cycle

λ < 420nm

NO2NO O

• 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)

NO - NO2 Photochemical Cyclein polluted atmosphere

λ < 420nm

NO2NO O

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 :

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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