combustion theory

1

Combustion Theory

2

Ideal Gas Model

The ideal gas equation of state is:

TRnTM

RmmRTPV

where is the Universal Gas Constant (8.314 kJ/kmol K), is the molecular weight and n is the number of moles.

Specific internal energy (units: kJ/kg)

R M

dTTcTh p )()(

dTTcTu v )()(

Specific enthalpy (units: kJ/kg)

Specific entropy (units: kJ/kg K)

)ln()(),( oo PPRTsTPs

where Po is 1 bar and so is corresponds to Po

3

Ideal Gas Model for Mixtures

The mixture internal energy U and enthalpy H (units: kJ) is:

n

iii

n

iii hmmhHummuU

11

The mixture specific internal energy u and enthalpy h is:

n

iii

n

ii

in

iii

n

ii

i hxhm

mhuxu

m

mu

1111

n

iimm

1

The mass fraction, xi, of any given species is defined as:

n

ii

ii x

m

mx

11 and

The mass m of a mixture is equal to the sum of the mass of n components

4

The mole fraction, yi, of any given species is defined as:

n

ii

ii y

n

ny

11 and

The mixture internal energy U and enthalpy H (units: kJ) is:

n

iii

n

iii hnHunU

11

where are molar specific values (units: kJ/kmol)

The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is:

n

iii

n

iii hyhuyu

11

ii hu and

n

iinn

1

The total number of moles in the mixture is:


5


The partial pressure of a component, Pi, in the mixture (units: kPa) is:

PyPP

P

RTPV

RTVP

n

ny ii

ii

ii or

The mixture molecular weight, M, is given by:

i

n

ii

n

i

ii

n

ii

Myn

Mn

n

m

n

mM

11

1

6

Ideal Gas Model

The mass specific mixture entropy (units: kJ/kg K) at P, T is :

o

n

iii

oii PPRPPRsxs ln)/ln(

1

o

n

iii

oii PPRyRsys lnln

1

The molar specific mixture entropy (units: kJ/mol K) is:

Note, if the mixture pressure is at 1 bar the second terms drops out.

7

Composition of Standard Dry Air

• Air is a mixture of gases including oxygen (O2), nitrogen(N2), argon (Ar),carbon dioxide (CO2), water vapour (H20)…. • For combustion dry air is taken to be composed of 21% O2 and 79% N2 by volume (mole fraction).

76.321.0

79.0

2

2

2

2

2

2 O

N

O

tot

tot

N

O

N

y

y

n

n

n

n

n

n

• For every mole of O2 there are 3.76 moles of N2.

• The molecular weight of air is

kg/kmol 84.28)28(79.0)32(21.0

22221

NNOOi

n

iiair MyMyMyM

8

Composition of Standard Moist Air

• Atmospheric air always contains some amount of water vapour.

• The amount of water in moist air at T is specified by the specific humidity, , or the relative humidity, defined as follows:

10 )(

22 TP

P

m

m

sat

OH

air

OH

61.1

62.02

222

2

2air

OHair

OH

airair

OH

air

OH

OHOH n

nn

M

Mn

M

m

M

mn

)(

)(62.062.0

29

18

2

22222

TPP

TP

PP

P

P

P

nM

nM

m

m

sat

sat

OH

OH

air

OH

airair

OHOH

air

OH

• The two are related by:

• The number of moles of water can be calculated knowing or :

9

Hydrocarbon Fuels (organic compounds)

Most common hydrocarbon fuels are Alkyl Compounds and are grouped as:

Paraffins (alkanes): single-bonded, open-chain, saturated (no more hydrogen can be added by breaking a bond)

CnH2n+2 n= 1 CH4 methanen= 2 C2H6 ethanen= 3 C3H8 propanen= 4 C4H10 butanen= 8 C8H18 n-octane and isooctane

C

H

H

H

C

H

H

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

C

H

H

H

H

C

H

H

methane ethane n-octane

H H

H

C

H

There are several isooctanes, depending on position of methyl (CH3) brancheswhich replace hydrogen atoms (eg. a side H is replaced with 3 CH3)

10

Olefins (alkenes): open-chain containing one double-bond, unsaturated (break bond more hydrogen can be added)

CnH2n n=2 C2H4 ethenen=3 C3H6 propene

propene

Acetylenes (alkynes): open-chain containing one C-C triple-bond unsaturated

CnH2n-2 n=2 C2H2 acetylenen=3 C3H4 propyne

HC CH

acetylene

Hydrocarbon Fuels (organic compounds)

For alcohols one hydroxyl (OH) group is substituted for one hydrogene.g. methane becomes methyl alcohol (CH3OH) also known as methanol ethane becomes ethyl alcohol (C2H5OH) also known as ethanol

Note: n=1 yields CH2 is an unstable molecule

H

C

H

HC C

H

H

H

11

Combustion Stoichiometry

If sufficient oxygen is available, a hydrocarbon fuel can be completely oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen is converted to water (H2O).

The overall chemical equation for the complete combustion of one mole of propane (C3H8) with oxygen is:

OcHbCOaOHC 22283

Elements cannot be created or destroyed, soC balance: b= 3H balance: 2c= 8 c= 4O balance: 2b + c = 2a a= 5

Thus the above reaction is:

OHCOOHC 22283 435

# of moles species

12


Air contains molecular nitrogen N2, when the products are low temperature the nitrogen is not significantly affected by the reaction, it is considered inert.

The complete reaction of a general hydrocarbon CH with air is:

22222 )76.3( dNOcHbCONOaHC

The above equation defines the stoichiometric proportions of fuel and air.

Example: For propane (C3H8) = 3 and = 8

2222283 576.343)76.3(5 NOHCONOHC

22222 476.3

2)76.3(

4NOHCONOHC

C balance: = bH balance: = 2c c = /2O balance: 2a = 2b + c a = b + c/2 a = + /4N balance: 2(3.76)a = 2d d = 3.76a/2 d = 3.76( + /4)

13

112

)2876.332(4

1

)/(

1/

ss AF

FA


Substituting the respective molecular weights and dividing top and bottom by one gets the following expression that only depends on the ratio of thenumber of hydrogen atoms to hydrogen atoms () in the fuel.

Note above equation only applies to stoichiometric mixture

For methane (CH4), = 4 (A/F)s = 17.2

For octane (C8H18), = 2.25 (A/F)s = 15.1

The stoichiometric mass based air/fuel ratio for CH fuel is:

HC

NO

fuelii

airii

fuel

airs MM

MM

Mn

Mn

m

mFA

22 476.3

4/

14

Fuel Lean Mixture

• Fuel-air mixtures with more than stoichiometric air (excess air) can burn

• With excess air you have fuel lean combustion

• At low combustion temperatures, the extra air appears in the products in unchanged form:

for a fuel lean mixture have excess air, so > 1

• Above reaction equation has two unknowns (d, e) and we have two atom balance equations (O, N) so can solve for the unknowns

222222 2)76.3)(

4( eOdNOHCONOHC

15

• Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.

• With less than stoichiometric air you have fuel rich combustion, there is insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.

• Get incomplete combustion where carbon monoxide (CO) and molecular hydrogen (H2) also appear in the products.

222222 2)76.3)(

4( fHeCOdNOHCONOHC

where for fuel rich mixture have insufficient air < 1

Fuel Rich Mixture

• Above reaction equation has three unknowns (d, e, f) and we only have two atom balance equations (O, N) so cannot solve for the unknowns unless additional information about the products is given.

16

The equivalence ratio, , is commonly used to indicate if a mixture is stoichiometric, fuel lean, or fuel rich.

s

mixture

mixture

s

AF

AF

FA

FA

/

/

/

/

Off-Stoichiometric Mixtures

stoichiometric = 1 fuel lean < 1 fuel rich > 1

Products)76.3(4 22

NOHC

Stoichiometric mixture:

Off-stoichiometric mixture:

Products)76.3(4

122

NOHC

17

Example: Consider a reaction of octane with 10% excess air, what is ?

The stoichiometric reaction is:

Off-Stoichiometric Conditions

22222188 4798)76.3(5.12 NOHCONOHC

10% excess air is:

222222188 98)76.3)(5.12(1.1 bNaOOHCONOHC

91.0

1/)76.4)(5.12(1.1

1/)76.4(5.12

/

/

mixture

s

FA

FA

16 + 9 + 2a = 1.1(12.5)(2) a = 1.25, b = 1.1(12.5)(3.76) = 51.7

Other terminology used to describe how much air is used in combustion:

150% stoichiometric air = 150% theoretical air = 50% excess air

5.1 )76.3)(4

( 2283 NOHC mixture is fuel lean

18

First Law Analysis for Reacting System

Consider a constant pressure process in which nf moles of fuel react withna moles of air to produce np moles of product:

PnAnFn paf

Applying First Law with state 1 being the reactants at P1, T1 and state 2being products at P2, T2:

Reactants Products

ReactantsProducts

Reaction

State 1 State 2

Q

W

)()( 121221 VVPUUQ

WUQ

19

RRii

PpiiRP ThnThnHH

HH

VPUVPU

)()(

)()(

12

111222

HP < HR Q < 0 exothermic reaction

HP > HR Q > 0 endothermic reaction

)()( 1212 VVPUUQ

First Law Analysis for Reacting System

20

Enthalpy of Reaction

Consider the case where the final temperature of the products is the same as the initial temperature of the reactants (e.g., calorimeter is used to measure Q).

ReactionQ

W

P1=P2=Po

T1=T2=To

The heat released under this situation is referred to as the enthalpy of reaction, HR ,

fuel of kmolor kgper kJ :units )()(

)()(

Roii

Poii

RRii

PpiiR

ThnThn

ThnThnH

To

21

The maximum amount of energy is released from a fuel when reacted with astoichiometric amount of air and all the hydrogen and carbon contained in thefuel is converted to CO2 and H2O

This maximum energy is referred to as the heat of combustion or the heating value and it is typically given per mass of fuel

Heat of Combustion

22222 476.3

2)76.3(

4NOHCONOHC

HR(298K)

hydrocarbons

alcohols

22

Heat of Combustion

There are two possible values for the heat of combustion depending on whether the water in the products is taken to be saturated liquid or vapour.

Tp

T

S

hghf

From steam tables: hfg = hg – hf > 0

HR = HP – HR < 0 (exothermic)

The term higher heat of combustion is used when the water in the products is taken to be in the liquid state (hH20 = hf)

The term lower heat of combustion is used when the water in the productsis taken to be in the vapour state (hH20 = hg)

23

Heat of Formation

Consider the following reactions taking place at atmospheric pressure andwith TP = TR = 298K

222

2222

/ 000,394 )()()(

/ 000,286 )()()(2/1

COkmolkJQgCOgOsC

OHkmolkJQlOHgHgO

In these reactions H2O and CO2 are formed from their elements in theirnatural state at standard temperature and pressure (STP) 1 atm and 298K.

Reactions of this type are called formation reactions and the corresponding measured heat release Q is referred to as the standard heat of formationand takes the symbol so:

ofh

kmolkJh

kmolkJh

C

O

/ 000,394

/ 000,286

oO f,

oH f,

2

2

Values for standard heat of formation for different species are tabulated

24

Heat of Formation for Different Fuels

25

Enthalpy Scale for a Reacting System

By international convention, the enthalpy of every element in its natural state(e.g., O2(g), N2(g), H2(g), C(s)) at STP has been set to zero

0)298,1( ofhKatmh

We need to take into account that for a reacting system the working fluid changes molecularly from reactants to products while undergoing a process.

Consider the following identity:

)]298,1(),([)298,1(),( KatmhTPhKatmhTPh

(note the notation convention)

at STP

26

The enthalpy of all other substances at STP is simply the heat of formation of the substance, since it is formed from its elements, for example:

)]298,1(),([ ),( , KatmhTPhhTPh iio

ifi

sensible enthalpychemical enthalpy TK ip dTc298 ,

Therefore, the enthalpy of the i’th component in a mixture is:

Enthalpy Scale for a Reacting System

olOHflOH

olOHfgHgOlOH

hh

hhhhQSTPrecall

lOHgHgO

)(,)(

)(,)()()(

222

22

2222

2/1 @

)()()(2/1

27

-14000

-5000

-9000

298 2800

H2O

CO2

Temperature K

Enthalpy (kJ/kg)

The data is also found in the JANNAF tables provided at course web site

,o

ifh

29

Adiabatic Flame Temperature

ReactionQ

W

Consider the case where the cylinder is perfectly insulated so the process isadiabatic (Q = 0)

)()(

0)()(

1

Rii

Paii

RRii

Ppii

ThnThn

ThnThnQ

For a constant pressure process, the final products temperature, Ta, is known as the adiabatic flame temperature (AFT).

For a given reaction where the ni’s are known for both the reactants and the products, Ta can be calculated explicitly.

30

Adiabatic Flame Temperature

P R

oifi

oifi

Riii

Piaii

Rii

oifi

Piai

oifi

Rii

Paii

hnhnKhThnKhThn

KhThhnKhThhn

ThnThn

,,1

1,,

1

)298()()298()(

)298()()298()(

)()(

)298()298()()( ,,1 KhnhnhnKhThnThn iP

iP R

oifi

oifi

Riii

Paii

oRH

P R

oifi

oifi

R

Tipi

P

Tipi hnhndTcndTcn ia

,,298 ,298 ,OR

If reactants are at 298K, term =0

31

o

HCfo

OHfo

COf

NaNOHaOHCOaCO

hhh

KhTh.KhThKhTh

10422

222222

,,, 54

)298()(424)298()(5)298()(4

22222104 4.2454)76.3(5.6)( NOHCONOlHC

Consider constant pressure complete combustion of stoichiometric liquid butane-air initially at 298K and 1 atm

P R

oifi

oifi

Riiii

Paii hnhnKhThnKhThn ,,1 )298()()}298()({

Note: first term = 0 since T1 = 298K, also 022 ,, o

Nfo

Of hh

Look up enthalpy values, and iteration gives Ta = 2500K

Adiabatic Flame Temperature, example

32

Ta,

Constant Pressure Adiabatic Flame Temperaturewith products at equilibrium

33

Now consider butane air with 300% excess air

222222104 5.198.9754)76.3)(5.6(4)( ONOHCONOlHC

Look up values and iteration gives Ta = 1215K

Excess air adds 92.8 moles of diatomic molecules (O2 and N2) into the products that does not contribute to heat release just soaks it up.

Adiabatic Flame Temperature, example

o

HCfo

OHfo

COfOaO

NaNOHaOHCOaCO

hhhKhTh

KhThKhThKhTh

1042222

222222

,,, 54 )298()(9.51

)298()(8.97)298()(5)298()(4

34

Constant Pressure Adiabatic Flame Temperaturewith products at equilibrium

nitromethane

octaneethanol

hydrogen

35

Constant Volume AFT

ReactionQ

W

Consider the case where the piston is fixed and the cylinder is perfectly insulated so the process is adiabatic (Q = 0)

)()(

0)()(

1

Rii

Paii

RRii

Ppii

TunTun

TunTunQ

Note h = u + pv = u + RT, so

T) R)(Th(nTRThnR

iiP

aii 1))((

36

p Riai

iP

iP R

oifi

oifi

Riii

Paii

TRnTRn

KhnhnhnKhThnThn

1

,,1

)298()298()()(

Extra term compared to constant pressure AFT ( term > 0)

The AFT for a constant volume process is larger than for a constant pressure process.

The AFT is lower for constant pressure process since there is Pdv work done

Constant Volume AFT

R

iiio

ifiP

iiaio

ifi TRKhThhnTRKhThhn )298()()298()( 1,,

37

i

a

R

p

i

CV

R

p

R

p

R

p

p

pp

R

RR

PR

T

T

n

n

P

P

T

T

n

n

P

P

P

TRn

P

TRn

VV

Constant Volume Combustion Pressure

Assuming ideal gas behavior:

For large hydrocarbons the mole ratio term is close to one

38

Stoichiometric octane-air (HR= 47.9 MJ/kg):

1.15A 4798)76.3(5.12 22222188 /FNOHCONOHC

1.8604.706.1298

2266

5.60

64

i

a

R

p

i

CV

T

T

n

n

P

P

Stoichiometric hydrogen-air (HR= 141.6 MJ/kg):

3.34 88.11)76.3(5.0 22222 A/F NOHNOH

8.60.885.0298

2383

38.3

88.2

i

a

R

p

i

CV

T

T

n

n

P

P

Engine Fuel Comparison

Stoichiometric ethanol-air (HR= 29.7 MJ/kg):

7.16 2.1332)76.3(5.3 2222262 A/F NOHCONOOHC

5.737.702.1298

2197

7.17

2.18

i

a

R

p

i

CV

T

T

n

n

P

P

39

Chemical Equilibrium

• In general the combustion products consist of more than just CO2, H2O O2 and N2

• For rich mixtures CO also exists in the products. At high temperaturesthe molecules dissociate to form H, O, OH, NO via the following reactions:

NONOOHOHOOHH 2 2 2 2 222222

• At equilibrium the rate of the forward reaction equals the rate of the backward reaction.

NONOOHOHOOHH 2 2 2 2 222222

• The opposite direction reactions are also possible

222222 2 2 2O 2 NONOOHOHOHH

40


• At equilibrium the relative proportion of the species mole fraction is fixed

• For the general equilibrium reaction

• The equilibrium composition for species CD is given by:

DnCnBnAn DcBA

BADC

BA

DCnnnn

refnB

nA

nD

nC

P

P

XX

XXTK

)(

where K is the equilibrium constant which is tabulated as a function of temperature for different equilibrium reactions, Pref is 1 atm and P is in units of atmospheres.

DCBA

AA nnnn

nXNote

41

222222 )76.3)(4

( fHeCOdNObHaCONOHC


Recall that for a rich mixture (<1) the reaction equation could not be balanced (5 unknowns a, b, d, e, f and only 4 atom balance equations for C,H,O,N ) even if we neglect dissociation (i.e., low product temperature)

If the product species CO2, H2O, CO and H2 are at equilibrium and described by the water-gas reaction:

OHCOHCO 222

The equilibrium constant for this reaction provides the fifth equation :

atm 1 )(22

2

Pfa

be

XX

XXTK

HCO

OHCO

Note K is tabulated as a function of T

42

Chemical Equilibrium, example

1 kmol of CO2, ½ kmol of O2 and ½ kmol of N2 reacts to form a mixtureconsisting of CO2, CO, O2, N2 and NO at 3000K and 1 atm. Determine theequilibrium composition of the product mixture.

222222 2/12/1 eNdOcCObNOaCONOCO

3000K and 1 atm

C 1 = a+c c = 1 - aO 3 = a+b+2c+2d d = 1/2(1 + a - b) N 1 = b+2e e = 1/2(1 - b)

Have 2 unknowns a, b so need 2 equilibrium equations

122203000 2/12/1 .2

327303000 2/1 .1

222

122

.K)(KNONO

.K)(KOCOCO

43

ntot= a+b+c+d+e = a+b+(1-a)+1/2(1+a-b)+1/2(1-b) = (4+a)/2

)(a

ba

a

a

X

XXK

CO

OCO 1 4

1

13273.0

2/12/1

1

2

2

2/)4(

1

2/)4(

)1(2/1

2/)4( 22 a

aX

a

baX

a

aX COOCO


2

2

2/1

1 3273.0CO

OCO

X

XXK

From the equilibrium constant expression

Substituting yields:

44

Solving equations 1 and 2 yields:

a= 0.3745 b= 0.0675

From the atom balance equations get:

c= 0.6255 d= 0.6535 e= 0.4663

Substituting and dividing through by the total number of moles gives:

)(

bba

b

XX

XK

NO

NO 2 )1)(1(

21222.0 2/12/12/12

22


Similarly for the second equilibrium reaction

222 21.030.029.003.017.0 NOCONOCO

45

kNjNOiOHhOgH

fHeCOdOcNObHaCONOHC

)76.3)(4

( 2222222

• If the products are at high temperature (>2000K) minor species will be present due to the dissociation of the major species CO2, H2O, N2 and O2.

• Hand calculations are not practical when many species are involved, one uses a computer program to calculate the product equilibrium composition.

http://www.wiley.com/college/mechs/ferguson356174/wave_s.htmlEquilibrium Combustion Solver Applet

A popular program used for chemical equilibrium calculations is STANJAN

Computer Programs Equilibrium Solvers

http://www.wiley.com/college/mechs/ferguson356174/apps/solver/solver.html

46

Mo

le f

ract

ion

, X

i

Temperature (K)

Equilibrium Composition for Combustion Products of Octane-air

(lean) (rich)

Nitric Oxide (NO)is an air pollutant

(stoich)

47

• One can calculate the AFT for the above stoichiometric reaction where the products are at equilibrium:

• Note dissociation in the products will result in a lower AFT since dissociation reactions are endothermic.

Adiabatic Flame Temperature for Products at Equilibrium

kNjNOiOHhOgH

fHeCOdOcNObHaCONOHC

)76.3)(4

( 2222222

• Again computer programs are used for these calculations: http://www.wiley.com/college/mechs/ferguson356174/wave_s.html

Adiabatic Flame Temperature Applet

and STANJAN

)()( 1R

iiP

aii ThnThn

http://www.wiley.com/college/mechs/ferguson356174/apps/flame/flame.html

48

Chemical Kinetics

First and Second Laws of thermodynamics are used to predict the final equilibrium state of the products after the reaction is complete.

Chemical kinetics deals with how fast the reaction proceeds.

How fast the fuel is consumed is of interest, the reaction rate ’’’ is definedas:

where [ F ] refers to the fuel concentration (kmol/m3 or kg/m3), negative signdue to the fact that the fuel is consumed.

dt

Fd ]['''

DCAF DCAF

OHOH

NOHCONOHC

222

22222188

2/1

4798)76.3(5.12

Global (or overall) reactions describe the initial and final states:

49

Reaction Mechanism

In reality the reaction proceeds through elementary reactions in a chainprocess known as chain reactions

The global hydrogen-oxygen reaction proceeds via the following elementaryreactions, collectively known as a reaction mechanism:

partner)collision a as actsat present th speciesany is (

MOMOO

MHMHH

MOHMOHH

HOHOH

OOHOH

HOHHOH

OHOHHHO

MHOMOH

2

2

2

2

2

22

222

22

2

M

MHHMH

Chain initiation

Chain propagation

Chain branching

Chain termination

50

Radicals

• Species such as H, O, OH and HO2 are called radicals.

• Radicals have unpaired valence electrons which make them very reactive and short lived, they try to partner with other radicals to form covalent bonds

• In chain branching reactions there is a net production of radicals

• Chain branching reactions lead to rapid production of radicals which causes the overall reaction to proceed extremely fast explosively

• The reaction comes to completion through chain termination reactionswhere the radicals recombine to form the final products.

51

Chemical Reaction Equations

Consider the following general reaction involving n species

nnnn AAAAAAAA "3

"32

"21

"1

'3

'32

'21

'1

Ai = species’ = stoichiometric coefficient for reactants” = stoichiometric coefficients for products

As an example, consider the following elementary reaction

HHHHH 2

For this reaction n = 2 (species are H and H2)

A1 = H 1’ = 3 , 1” = 1

A2 = H2 2’ = 0 , 2” = 1

reactants products

52

Law of mass Action

The law of mass action states that for an elementary reaction the reaction rate is proportional to the product of the concentrations of the reactantspresent raised to a power equal to the corresponding stoichiometriccoefficient. The constant of proportionality is called the reaction rate constant k.

n

jj

n

n

jj

n

jjii

i

nj

j

AAAAA

Akdt

Ad

1

'

3211

1

'"

by given isreaction theoforder overall the

][][][][][ where

][)(][

''3

'2

'1

'

'

For m simultaneous equations:

m

l

n

jjllili

ilj

Akdt

Ad

1 1

',

",

'.

][)(][

53

Law of mass Action

Applying the law of mass action to the reaction HHHHH 2

dt

Hd

dt

Hdk[H]

HkHHkdt

Hd

AAkdt

Ad

HkHHkdt

Hd

AAkdt

Ad

][][

2

1 note

][][][)01(][

][][)(][

][2][][)31(][

][][)(][

23

302

32

22'2

"2

2

302

3

21'1

"1

1

'2

'1

'2

'1

A1 = H

A2 = H2

k

Recall n = 2

54

Reaction Rate Theory

Kinetic theory of gases is used to come up with the following expressionfor the elementary reaction rate constant, first proposed by Arrhenius

TR

EATk ab exp

where A and b are the rate coefficients, Ea is the activation energyand R is the universal gas constant.

Therefore, for the reaction the reaction rate isHHHHH 2

332 ][exp][ HTR

EATHk

dt

]d[H ab

The values of a, b and Ea are tabulated for different reactions.

This expression indicates that molecular hydrogen is produced faster at higher temperature and at higher atomic hydrogen concentrations.

k

55

Rate Coefficients for H2-O2 Reactions

* n is the reaction order

A b Ea Temperature rangeReactions ((cm3/gmol)n-1/s)* (kJ/gmol) (K)

56

Ozone (O3) Decomposition

Global reaction: 23 32 OO

Important elementary reactions include:

MOMOO

OOOO

MOOMO

2

223

23k1

k2

k3

k6

k4

k5

Here m = 6 (3 forward and 3 backward reactions) n =3 (O, O2, O3)

m

l

n

jjllili

ilj

Akdt

Ad

1 1

',

",

'.

][)(][

The O3 reaction rate is:

224332231

3''' ][]][[]][][[]][[][

3OkOOkMOOkMOk

dt

OdO

57

Relationship Between Rate Constant and Equilibrium Constant

DCBA DCBA kf

kb

DCBA DCkBAkdt

Adbf

][][][][][

At equilibrium:

BA

DC

DCBA

DCBA

BA

DC

k

k

DCkBAk

DCkBAkdt

Ad

b

f

bf

bf

][][

][][

][][][][

0][][][][][

BADC

BA

DCBADC

BA

DCvvvv

refvv

vvvvvv

refvB

vA

vD

vC

P

TR

BA

DC

P

P

XX

XXK

][][

][][Recall:

Note:P

TRC

Vn

Vn

n

nX

total

C

total

CC ][

/

/

58

Relationship Between Rate Constant and Equilibrium Constant

BADC

BADC

vvvv

refb

f

vvvv

refb

f

P

TRK

k

k

P

TR

k

kK

If the equilibrium constant K is available for the elementary Reaction, then you can get kf knowing kb , or get kb knowing kf

For a bimolecular reaction DCBA

Kk

k

b

f

59

Reaction Rate for Global Reaction

Even for the simplest hydrocarbon fuels the chemistry is very complicated.The GRI HC mechanism has 49 species and 227 elementary reactions.

For engineering purposes easier to use global description of reaction:

oductInertOFuel PiOf Pr22

Empirical correlations have been developed for the fuel reaction rate:

lmn InertOFuelTR

EA

dt

Fueld][][][exp

][2

where [ ] in units of gmol/cm3

R = 1.987 cal/gmolKE typically 20 - 40 kcal/gmol

Note this is a correlation so n, m and l don’t have to be integers, typically l= 0

60

Example: Nitric Oxide (NO) Formation

Thermal (Zeldovich) Mechanism:

kJ/mol 62 (2)

kJ/mol 315 (1)

2

2

a

a

EONOON

ENNONO

Forward reaction (1) is endothermic and due to the high Ea requires high temperature (>1800K) to proceed and is very slow.

2/1221 ]][[2

][ON

TR

KPk

dt

NOd reff

From above two reaction using basic arguments one can derive the following expression for the “initial” NO reaction rate for the global reaction:

NOON 222 kG

kG

Global NO formation reaction:

nmG ONk

dt

NOd][][

][22

61

Explosion Limits

Perform an experiment where a fuel-oxygen mixture is injected into a preheated spherical vessel and monitor the vessel pressure.

Fuel-air

VacuumP

T

For a given fuel-oxygen mixture, e.g. stoichiometric, and fill pressure if you vary the temperature of vessel you will find that there is a critical temperature above which an explosion occurs and below which an explosion does not occur (note there is no spark ignition).

An explosion is characterized by a rapid rise in vessel pressure faster than the normal pressure rise due to gas filling.

This critical temperature is referred to as the autoignition or explosion limit temperature

P

time

62

If you repeat the experiment for different fill pressures and plot all the results on a pressure-temperature curve one can define an explosion limit curve.

For H2-O2 the shape of this limit curve can be explained by the temperature and pressure dependencies of the elementary reactions.

Explosion Limits

NO Explosion

Explosion

Explosion Limits ofStoichiometric H2-O2

1 atm

63

Explosion Limits for HC Fuels

Autoignition for hydrocarbon fuels is more complicated than that for hydrogen,different types of behaviour are possible including single- and two-stage ignition.

At 300-400oC one or more combustion waves often appear, accompanied by a faint blue emission, only a small fraction of the reactants react and thetemperature rise is only tens of degrees. These are called cool flames.

A rapid compression machine (RCM) consisting of a piston-cylinder assemblyis used to raise the pressure and temperature of a fuel-air mixture very quicklyto a predetermined value P2, T2.

1

1

2

1

1

1

2

1

2

k

kk

k

rV

V

P

P

T

T

P2 T2

64

Two-stage ignition: first a cool flame propagates through the mixture (D) followed by a “hot flame” or high-temperature explosion (E).

Single-stage ignition: after a certain induction timea “hot flame” or high-temperature explosion (E).

Ignition Process in Isooctane-air Mixtures

After compression the subsequent pressure transient is measured to detect ignition.

Isooctane ignition can either be single or two-stage depending on the post-compression pressure and temperature

65

Isooctane displays different types of ignition, including cool flames and two-stage ignition, whereas methane only displays single-stage ignition.

Explosion Limits for HC Fuels

Two-stage ignition

Single-stage ignition

No ignition

Pressure, atm

Ign

itio

n t

emp

erat

ure

, o C

No ignition

IsooctaneMethane

Single-stage ignition

66

Laminar Premixed Flames

A flame is a thin region in space where chemical reactions convert thefuel-air mixture into combustion products.

A flame can propagate (engine application) or remain stationary (burner application).

For a given P, T, a flame has two basic properties:a) adiabatic flame temperature,Tad

b) laminar burning velocity, Sl

Vu = SlVb

b ,Ta

burntunburnt

Note, Sl is defined in terms of the approaching unburnt gas velocity

Pressure is roughly constant across the flame so ~ 1/T

Stationary flame

Vu = 0u

Vb-Sl

b

Sl

Moving flame

u

67

Structure of Flame

Vu = SL

u

Vb

b

[Fuel][O2]

T

TR

EOF

dt

Fd amn exp][][][

2

Pre-heatZone

ReactionZone

Products zone

[radicals]

Diffusion of heatand mass

Flame thickness Visible part of the flame

68

Laminar Burning velocity

Maillard-LeChatelier theory gives:

)/exp(22ada

nl TREPRRS

Higher flame velocity corresponds to:1) higher unburned gas temperature2) lower pressure (most hydrocarbons)3) higher adiabatic flame temperature (chemical reaction)4) higher thermal diffusivity (= kcond/cp)

1 < n < 2

69

Laminar Burning Velocity Correlation

The following is a correlation developed by Metghalchi and Keck

)1.21(1298, dil

urefll Y

atm

P

K

TSS

where Ydil is the mass fraction of diluent, e.g., residual gas, and

)1(22.016.0

)1(8.018.2

)(2,

MMrefl BBS

Fuel M BM (cm/s) B2 (cm/s)

Methanol 1.11 36.92 -140.51Propane 1.08 34.22 -138.65Isooctane 1.13 26.32 -84.72

is equivalence ratio

70

Flame Velocity

The laminar burning velocity is measured relative to the unburned gas aheadand the flame velocity Vf is measured relative to a fixed observer.

If the flame is propagating in a closed-ended tube the velocity measured is theflame velocity and can be up to 8 times the burning velocity.

This is because the density of the products is lower than the fresh gas so a flow is generated ahead of the flame

Vu Vb=0b

Vf

Moving flame

u

- Vu= SlVb=Vf

b

Vf

Stationary flame

u

Vf

lb

uffblu SVVAS

A

Applying conservation of mass across the flame:

71

Turbulent Flames

Unlike the laminar burning velocity, the turbulent flame velocity is not a property of the gas but instead depends on the details of the flow.

The IC engine in-cylinder flow is always turbulent and the Kolmogorov eddies are typically larger than the laminar flame thickness (1 mm).

Under these conditions the flame is said to display a structure known as wrinkled laminar flame.

A wrinkled laminar flame is characterized by a continuous flame sheet thatis distorted by the eddies passing through the flame.

The turbulent burning velocity depends on the turbulent intensity ut and can be up to 30 times the laminar burning velocity

Laminar flame Turbulent flame

Sl St bltlt SuaSS /1/

72

Flame Thickness and Quenching Distance

A rough estimate of the laminar flame thickness can be obtained by:

mm 122

p

cond

ll c

k

SS

As a flame propagates through a duct heat is lost from the flame to the wall

d

It is found experimentally that if the duct diameter is smaller than somecritical value then the flame will extinguish

This critical value is referred to as the quenching distance dmin and is close in magnitude to the flame thickness.

mind

Local quenching

73

Minimum Ignition Energy and Flammability Limits

A flame is spark-ignited in a flammable mixture only if the spark energy is larger than some critical value known as the minimum ignition energy Eign

It is found experimentally that the ignition energy is inversely proportional to the square of the mixture pressure.

A flame will only propagate in a fuel-air mixture within a composition range known as the flammability limits.

The fuel-lean limit is known as the lower flammability limit and thefuel-rich limit is known as the upper flammability limit.

The flammability limit is affected by both the mixture initial pressure and temperature.

2/1 PEign

combustion theory

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