Combustion Theory

Section 5

1
 Molecules and Covalent Bonds

Chemical bonds result from a mutual sharing of electrons between atoms,

the shared electrons are in the outermost shell, known as valence electrons

Lewis notation:
Hydrogen Atomic # 1  1 valence electron H

Carbon Atomic # 6  4 valence electrons C

Oxygen Atomic # 8  6 valence electrons O

Atoms like to have electron configuration like noble gas, usually eight valence
electrons, an octet.
H
H 2 H H CH4 H C H
H
Atoms and molecules with unpaired valence electrons are called radicals
e.g. O, H, OH, N, C 2
 Hydrocarbon Fuels (organic compounds)

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

Paraffins (alkanes):
CnH2n+2 n= 1 CH4 methane H H H

n= 2 C2H6 ethane H C H H C C H
n= 3 C3H8 propane H H H
n= 4 C4H10 butane
n= 8 C8H18 octane methane ethane

Olefins (alkenes): H H
H
CnH2n n=2 C2H4 ethene C C C H
n=3 C3H6 propene H
H
Note: n=1 yields CH2 is an unstable molecule propene

Acetylenes (alkynes):
CnH2n-2 n=2 C2H2 acetylene H C C H
n=3 C3H4 propyne acetylene
3
 Alcohols

An alcohol molecule is simply a hydrocarbon molecule with one of the

hydrogen atoms replaced by a hydroxyl molecule (OH)

The main alcohols used as engine fuels are:

Ethanol – ethyl alcohol (C2H5OH) consists of ethane molecule (C2H6) with

OH substitution

Methanol –methyl alcohol (CH3OH) consists of methane molecule (CH4)

with OH substitution

Butanol – (C4H9OH) consists of methane molecule (CH4) with OH substitution

H H H
H C C OH H C OH
H H H

Ethyl alcohol Methyl alcohol 4

 IC Engine Fuels

Crude oil contains a large number of hydrocarbon compounds (25,000).

The purpose of refining is to separate crude oil into various fractions via
a distillation process, and then chemically process the fractions into fuels
and other products.

A still is used to heat a sample, preferentially boiling off lighter

components which are then condensed and recovered.

The group of compounds that boil off between two temperatures are
referred to as fractions.

The order of the fractions as they leave the still are naptha, distillate,
gas oil, and residual oil. These are further subdivided using adjectives
light, middle, and heavy.

The adjectives virgin or straight run are often used to signify that no
chemical processing has been performed to a fraction.
5
Distillation Process

Refining Process

6
 Gasoline

Light virgin (or straight run) naptha can be used as gasoline.

Gasoline fuel is a blend of hydrocarbon distillates with a range of boiling

points between 25 and 225oC (for diesel fuel between 180 and 360oC)

Chemical processing is used to:

• Produce gasoline from a fraction other than light virgin, or

• Upgrade a given fraction (e.g., Alkylation increases the MW and octane

number of fuel: produce isooctane by reacting butylenes with isobutane in
the presence of an acid catalyst.

7
 Octane

The octane molecule is often used to simulate the properties of gasoline

H H H H H H H H

H C C C C C C C C H

H H H H H H H H

n-octane

There are 18 isomers of octane, depending on position of methyl (CH3)

branches which replace hydrogen atoms (eg. a side H is replaced with CH3)

CH3 H CH3

CH3 C C C CH3

CH3 H H

iso-octane

8
 Reformulated Gasoline (RFG)

In order to reduce emissions such as carbon monoxide (CO) and unburned

hydrocarbons (HC) the oxygen content of gasoline is increased to about 3%
by weight (U.S. oxygenated fuels program, winter only).

The US Clean Air Act requires certain large US cities to use RFG year-round
in order to reduce ozone by requiring a minimum oxygen content of 2% by
weight and maximum benzene content of 1%.

The primary oxygenates are MTBE (CH3)OC(CH3)3 and ethanol (C2H5OH)

As part of the reformulated gasoline program sulfur is restricted to 31 ppm

9
 Renewable Fuels
Currently most automotive IC engines use fossil fuels (gasoline or diesel)

Due to the ever increasing cost of oil, due to diminishing oil reserves and
accessibility to the oil reserves, and environmental concerns such as global
warming, alternative fuels have become very attractive. The 2007 US Energy
Bill set a 36 billion gal target for renewable fuels to be used in autos by 2020

Alcohols such as ethanol, methanol, and butanol are receiving a lot of

Attention because they can be synthesized biologically, i.e., bioalcohols or
biofuels.

Since there is oxygen in the fuel, combustion of alcohols produces no CO

but more greenhouse gas carbon dioxide (CO2) than fossil fuel combustion.

However, since the fuel is derived from plant matter the CO2 produced is
extracted from the atmosphere during the growth of the plant, i.e.CO2 neutral

6CO2 + 6H20 + solar energy → C6H12O6 + 6O2

sugar 10
photosynthesis
 Alcohol Fuels

Ethanol used for fuel is obtained by fermentation. Yeast metabolizes sugar

(C6H12O6) in the absence of oxygen to produce ethanol and carbon dioxide
C6H12O6 → 2C2H5OH +CO2
In Brazil ethanol is derived from sugar cane whereas in the US and Canada
corn is used as the feed stock (sugar cane has 30% more sugar than corn).

Sugars for ethanol fermentation can also be obtained from cellulose

(C5H10O5)n that makes up agricultural byproducts, such as corn cobs, corn
stalks, straw, switch grass, and wood, into renewable energy resources

Alcohol fuels have several drawbacks compared to fossil fuels:

- lower energy density (kJ/m3) than gasoline (10% less for butanol,
27% less for ethanol, 55% less for methanol),
- corrosive to fuel systems (methanol > ethanol > butanol)
- toxicity (methanol)

11
 Ethanol as a Fuel
Ethanol is quickly becoming the alternative fuel of choice for IC engines

An IC engine can run on gasoline with up to 10% ethanol (E10) without any
modifications, the use of higher blends requires changing certain components
in the fuel system, i.e, use stainless steel fuel lines and tank.

In Brazil half of the cars can run on 100% ethanol including flex fuel engines
that can run on all ethanol, all gasoline, or any combination of the two

Gasoline with up to 85% ethanol (E85) is now starting to enter the US and
Canadian markets. Flex fuel engines in this market can run on E85 or all
gasoline, 100% ethanol not yet permitted

12
 Other Alternative Fuels
Natural gas and propane have been used for many years as a fuel for IC
engines, especially in Europe where the price of gasoline has been
historically high. Gaseous state at room temperature therefore have a low
energy density (kJ/m3). Same emissions as gasoline and diesel.

Hydrogen (H2) is the new natural gas with similar issues. The main benefit
is no CO, CO2 and HC emissions. Biggest problem with hydrogen is that
there is no infrastructure to transport and store the fuel at fill stations. Also
hydrogen gas cannot be found in nature it must be manufactured

Biodiesel – manufactured from vegetable oils, waste cooking oil, or animal

fats. It is produced by reacting the oil with an alcohol (usually methanol)
and a catalyst (such as sodium hydroxide). The resulting chemical reaction
produces glycerine and alkyl esters (biodiesel). It is a liquid at RTP and
has 9% lower energy content than regular diesel

Dimethyl ether (C2H6O) – good diesel fuel because of good autoignition

quality. Is a gas at RTP, produced from syngas or methanol.
13
 Other Alternative Fuels

Producer gas (syngas) engines run on the gaseous products from thermal
gasification of biomass such as wood. The carbon reacts with steam, or a
limited amount of air, at high temperature (>700C) to produce a mixture
consisting of roughly 25% CO, 15% H2, and 5% CO2 , 50%N2 ….
Drawback is very low energy content, ten times lower than natural gas.
Benefit CO2 neutral.

14
 Ideal Gas Model

The ideal gas equation of state is:

R 
PV  mRT  m T  nR T
M 

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

Specific internal energy (units: kJ/kg) u (T )   cv (T )dT

Specific enthalpy (units: kJ/kg) h(T )   c p (T )dT

Specific entropy (units: kJ/kg K) s( P, T )  s o (T )  R ln( P Po )

(Po = 1 bar, so entropy at Po)

15
 Ideal Gas Model for Mixtures

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

n
m   mi
i 1

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

mi n
xi  and  xi  1
m i 1

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

n n
U  mu   mi ui H  mh   mi hi
i 1 i 1

The mixture specific internal energy u and enthalpy h is:

U n mi ui n H n mi hi n
u    xi ui h    xi hi
m i 1 m i 1 m i 1 m i 1
16
 Ideal Gas Model for Mixtures

The total number of moles in the mixture is:

n
n   ni
i 1

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

n n
yi  i and  yi  1
n i 1

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

n n
U   ni ui H   ni hi
i 1 i 1

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

The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is:
n n
u   yi ui h   yi hi
i 1 i 1
17
 Ideal Gas Model for Mixtures

Mass specific entropy (kJ/kg K) molar specific mixture entropy (kJ/mol K) :

s   xi s  Ri ln( Pi / P)   R lnP Po  s   yi sio  Ri ln yi   R lnP Po 

n n
o
i
i 1 i 1

The mixture molecular weight, M, is given by:

n

m i
mi n
ni M i n
M   1
   yi M i
n n i 1 n i 1

The partial pressure of a component, Pi, in the mixture (units: kPa) is:
PiV
n RT  Pi
yi  i  or Pi  yi P
n PV P
RT

18
 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).
nN 2 nN 2 ntot y N 2 0.79
     3.76
nO2 ntot nO2 yO2 0.21

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

n
• Molecular weight of air is M air   yi M i  yO2  M O2  y N 2  M N 2
i 1

 0.21(32)  0.79(28)  28.84 kg/kmol

• The amount of water in moist air at T is specified by the specific humidity,

w, or the relative humidity, F, defined as follows:
mH 2 O PH 2O
w F 0  F 1 19
mair Psat (T )
 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:
C3 H 8  aO2  bCO2  cH 2O
# of moles species

Elements cannot be created or destroyed, so

C balance: 3=b  b= 3
H balance: 8 = 2c  c= 4
O balance: 2a = 2b + c  a= 5

Thus the above reaction is:

C3 H 8  5O2  3CO2  4 H 2O
20
 Combustion Stoichiometry

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 CaHb with air is:

Ca H b  a(O2  3.76 N 2 )  bCO2  cH 2O  dN 2

C balance: a=b b=a

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

 b b  b
Ca H b   a  (O2  3.76 N 2 )  aCO2  H 2O  3.76 a   N 2
 4 2  4
The above equation defines the stoichiometric proportions of fuel and air.

Example: For octane (C8H18) a= 8 and b= 18

C8 H18  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2 21

 Combustion Stoichiometry

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

 b  b
 a   O2   a  M N2
 ni M i air M 3.76

mair 4  4
 A / F s  
m fuel  ni M i  fuel aM C  b M H

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

 b a  
1  (32  3.76  28)

1 4 
 A / F s 
( F / A) s 12  b a  1

Note above equation only applies to stoichiometric mixture

Example: For octane (C8H18), b/a = 2.25  (A/F)s = 15.1

22
 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:

b b
Ca H b  g (a  )(O2  3.76 N 2 )  aCO2  H 2O  dN 2  eO2
4 2
for a fuel lean mixture have excess air, so g > 1

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

23
 Fuel Rich Mixture
• 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.

b b
Ca H b  g (a  )(O2  3.76 N 2 )  aCO2  H 2O  dN 2  eCO  fH 2
4 2

where for fuel rich mixture have insufficient air  g < 1

• 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.

24
 Off-Stoichiometric Mixtures

The equivalence ratio, f, is commonly used to indicate if a mixture is

stoichiometric, fuel lean, or fuel rich.

 A / F s F / Amixture
f 
 A / F mixture F / As

stoichiometric f = 1
fuel lean f<1
fuel rich f>1

Stoichiometric mixture:
 b
Ca H b   a  (O2  3.76 N 2 )  Products
 4
Off-stoichiometric mixture:

1 b
Ca H b   a  (O2  3.76 N 2 )  Products
f 4 25
 Off-Stoichiometric Conditions
Other terminology used to describe how much air is used in combustion:

110% stoichiometric air = 110% theoretical air = 10% excess air

b
C3 H 8  g (a  )(O2  3.76 N 2 ) g  1.1  mixture is fuel lean
4
Example: Consider a reaction of octane with 10% excess air, what is f?

Stoichiometric : C8 H18  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2

10% excess air is:

C8 H18  1.1(12.5)(O2  3.76 N 2 )  8CO2  9 H 2O  aO2  bN 2

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

 A / F s 12.5(4.76) / 1
f   0.91
 A / F mixture 1.1(12.5)(4.76) / 1 26
 First Law Analysis for Reacting System

Consider a constant pressure process in which nf moles of fuel react with

na moles of air to produce np moles of product:
n f F  na A  n p P

Reactants Products

Reactants
Q Products

State 1 State 2
Reaction

Applying First Law with state 1 being the reactants at P1, T1 and state 2
being products at P2, T2:
Q  U  W

Q12  (U 2  U1 )  P(V2  V1 ) 27
First Law Analysis for Reacting System

Q  (U 2  U1 )  P(V2  V1 )

 (U 2  P2V2 )  (U1  P1V1 )

 H 2  H1

 H P  H R   ni hi (T p )   ni hi (TR )
P R

HP < HR Q<0 exothermic reaction

HP > HR Q>0 endothermic reaction

28
 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).

W
P1=P2=Po
To Reaction
Q
T1=T2=To

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

reaction, HR ,

H R   ni hi (T p )   ni hi (TR )
P R
  ni hi (To )   ni hi (To ) units : kJ per kg or kmol of fuel
P R

29
 Heat of Combustion

The maximum amount of energy is released from a fuel when reacted with a
stoichiometric amount of air and all the hydrogen and carbon contained in the
fuel is converted to CO2 and H2O
 b b  b
Ca H b   a  (O2  3.76 N 2 )  aCO2  H 2O  3.76 a   N 2
 4 2  4
This maximum energy is referred to as the heat of combustion or the heating
value and it is typically given per mass of fuel
HR(298K)

alcohols
30
 Fuel Energy Air- Specific Heat of
density fuel energy vaporizatio
(MJ/L) ratio (MJ/kg air) n
Gasoline and 32 14.6 2.9 0.36 MJ/kg
biogasoline
Butanol fuel 29.2 11.2 3.2 0.43 MJ/kg
Ethanol fuel 19.6 9.0 3.0 0.92 MJ/kg
Methanol 16 6.5 3.1 1.2 MJ/kg

31
 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.
T
hf hg
From steam tables:
Tp 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 products
is taken to be in the vapour state (hH20 = hg)

32
 Heat of Combustion, graphical

Reactants
h(kJ/kg fuel)
0

Products with H2O (g)

hlow Products with H2O (l)

hhigh

hfg,H2O per kg fuel

298K T

33
 Heat of Formation

Consider the following reactions taking place at atmospheric pressure and

with TP = TR = 298K

1 / 2O2 ( g )  H 2 ( g )  H 2O(l ) Q  286,000 kJ / kmol H 2O

C ( s)  O2 ( g )  CO2 ( g ) Q  394,000 kJ / kmol CO2

In these reactions H2O and CO2 are formed from their elements in their
natural 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 formation
and takes the symbol h fo so:
hf,oH 2O  286,000 kJ / kmol
hf,oCO2  394,000 kJ / kmol

Values for standard heat of formation for different species are tabulated
34
Heat of Formation for Different Fuels

35
 Enthalpy Scale for a Reacting System

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:

h ( P, T )  h (1atm,298K )  [h ( P, T )  h (1atm,298K )]

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

i.e, h (1atm,298K )  h f  0
o
(note the notation convention)

at STP

36
 Enthalpy Scale for a Reacting System

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:
1 / 2O2 ( g )  H 2 ( g )  H 2O(l )

recall @ STP Q  hH 2O (l )  1 / 2hO2 ( g )  hH 2 ( g )  h fo, H 2O (l )

 hH 2O (l )  h fo, H 2O (l )

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

hi ( P, T )  h fo,i  [hi ( P, T )  hi (1atm,298K )]

chemical enthalpy sensible enthalpy  298

T
K c p ,i dT

37
 -5000

Enthalpy (kJ/kg)
CO2

-9000
H2O

h of ,i

-14000
298 2800
Temperature K

The data is also found in the JANNAF tables provided at course web site
38
39
 Adiabatic Flame Temperature
Consider the following adiabatic constant pressure process:

Fuel
Reactants Products
Air

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

Q   ni hi (T p )   ni hi (TR )  0
P R
 ni hi (Ta )   ni hi (T1 )
P R

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

40
 Adiabatic Flame Temperature

 ni hi (Ta )   ni hi (T1 )
P R

  
 ni h f ,i  hi (Ta )  hi (298 K )    ni h f ,i  hi (T1 )  hi (298 K ) 
P
o

R
o


 ni hi (Ta )  hi (298K )    ni hi (T1 )  hi (298 K )    ni h f ,i   ni h f ,i 

o o

P R P R 

Sensible heat of products Sensible heat of reactants H Ro

(equal to 0 if T1 = 298K)

OR  ni 298
Ta
c p ,i dT  ni 298
Ti
c p ,i dT   ni h fo,i   ni h fo,i 
P R  P R 

41
 Adiabatic Flame Temperature, example

C8 H18 (l )  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2

Consider constant pressure complete combustion of stoichiometric liquid

butane-air initially at 298K and 1 atm

 ni {hi (Ta )  hi (298K )}   ni hi (T1 )  hi (298K )    ni h f ,i   ni h f ,i 

o o

P R  P R 

Note: first term = 0 since T1 = 298K, also h fo,O2  h fo, N2  0

8hCO (Ta )  hCO (298K ) 9hH O (Ta )  hH O (298K ) 47hN (Ta )  hN (298K )
2 2 2 2 2 2


  8h fo,CO  9h fo, H O  h fo,C H
2 2 8 18

Look up enthalpy values, and iteration gives Ta = ????K

42
Constant Pressure Adiabatic Flame Temperature
with products at equilibrium

Ta,

43
 Adiabatic Flame Temperature, example

Now consider octane air with 10% excess air

C8 H18  1.1(12.5)(O2  3.76 N 2 )  8CO2  9H 2O  1.25O2  51.7 N 2

8hCO (Ta )  hCO (298K ) 9hH O (Ta )  hH O (298K ) 51.7hN (Ta )  hN (298K )
2 2 2 2 2 2

2

 13.8hO (Ta )  hO (298K )   8h fo,CO  9h fo, H O  h fo,C H
2 2 2 8 18


Look up values and iteration gives Ta = ???? K

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.

44
Constant Pressure Adiabatic Flame Temperature
with products at equilibrium

nitromethane

hydrogen
octane
ethanol

45
 Constant Volume AFT

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

Reaction
Q

Q   ni ui (Tp )   ni ui (TR )  0
P R

 ni ui (Ta )   ni ui (T1 )
P R

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

 ni (hi (Ta )  R T )   ni(hi(T1 )  R T)

P R

46
 Constant Volume AFT

  
 ni h f ,i  hi (Ta )  hi (298K )   R Ti   ni h f ,i  hi (T1 )  hi (298K )   R Ti
P
o

R
o


 ni hi (Ta )   ni hi (T1 )  hi (298K )    ni h f ,i   ni h f ,i    ni hi (298K )
o o

P R  P R  P 
  ni R Ta   ni R T1
p R

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

47
 Constant Volume Combustion Pressure

Assuming ideal gas behaviour:

VR  VP
nR R TR n p R T p

PR Pp
Pp  n p  T p  PCV  n p  Ta 
        
PR  nR  TR  Pi  nR  Ti 

For large hydrocarbons like octane the mole ratio term is close to one

C8 H18 (l )  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2

np 8  9  47 64
   1.06
nR 1  12.5(4.76) 60.5

48
 Engine Fuel Comparison

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

C8 H18  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2 A/F  15.1

PCV  n p  Ta   64  2266 
         1.06  7.604  8.1
Pi  nR  Ti   60.5  298 

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

H 2  0.5(O2  3.76 N 2 )  1H 2O  1.88N 2 A/F  34.3

PCV  n p  Ta   2.88  2383 
         0.85  8.0  6.8
Pi  nR  Ti   3.38  298 

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

C2 H 6O  3.5(O2  3.76 N 2 )  2CO2  3H 2O  13.2 N 2 A/F  16.7

PCV  n p  Ta   18.2  2197 
         1.02  7.37  7.5
Pi  R  i  
n T 17.7  298  49
 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 temperatures
the molecules dissociate to form H, O, OH, NO via the following reactions:

H 2  2H O2  2O H 2  O2  2OH O2  N 2  2 NO

• The opposite direction reactions are also possible

2H  H 2 2O  O2 2OH  H 2  O2 2 NO  O2  N 2

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

H 2  2H O2  2O H 2  O2  2OH O2  N 2  2 NO
50
 Chemical Equilibrium

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

• For the general equilibrium reaction

n A A  nB B  ncC  nD D

• The equilibrium composition for species A, B, C, D is given by:

nC  nD  n A  nB
X nC
X nD
 P 
K (T )  C D  
X nA
X nB P 
A B  ref 

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.
nA
Note XA 
n A  nB  nC  nD
51
 Chemical Equilibrium
Recall that for a rich mixture (g<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)

b
Ca H b  g (a  )(O2  3.76 N 2 )  aCO2  bH 2O  dN 2  eCO  fH 2
4

If the product species CO2, H2O, CO and H2 are at equilibrium, an additional

equation can be obtained from the water-gas reaction:
CO2  H 2  CO  H 2O

The equilibrium constant for this reaction provides the fifth equation :
X CO  X H 2O eb
K (T )   P  1 atm
X CO2  X H 2 a f

Note K is tabulated as a function of T

52
 Chemical Equilibrium, example

1 kmol of CO2, ½ kmol of O2 and ½ kmol of N2 reacts to form a mixture

consisting of CO2, CO, O2, N2 and NO at 3000K and 1 atm. Determine the
equilibrium composition of the product mixture.

CO2  1/ 2O2  1/ 2 N 2  aCO  bNO  cCO2  dO2  eN2

3000K and 1 atm

C 1 = a+c c=1-a
O 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

1. CO2  CO  1 / 2O2 K1( 3000 K)  0.3273

2. 1 / 2O2  1 / 2 N 2  NO K 2( 3000 K)  0.1222

53
 Chemical Equilibrium, example

From the equilibrium constant expression

X CO  X O1 / 2
K1  0.3273  2

X CO 2

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

a 1 / 2(1  a  b) 1 a
X CO  X O2  X CO2 
(4  a) / 2 (4  a) / 2 (4  a) / 2

Substituting yields:

X CO  X O1 / 2 a 1  a  b 
1/ 2
K1  0.3273  2
   (1 )
X CO 2
1 a  4  a 

54
 Chemical Equilibrium, example

Similarly for the second equilibrium reaction

X NO 2b
K 2  0.1222   (2)
X O1 /22  X 1N/22 (1  a  b)(1  b)1 / 2

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:

0.17CO  0.03NO  0.29CO2  0.30O2  0.21N 2
55
 Computer Programs Equilibrium Solvers

• 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.

b
Ca H b  (a  )(O2  3.76 N 2 )  aCO2  bH 2O  cN 2  dO2  eCO  fH 2
4
 gH  hO  iOH  jNO  kN  

• 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.html
Equilibrium Combustion Solver Applet

A popular program used for chemical equilibrium calculations is STANJAN

56
Equilibrium Composition for Combustion Products of Octane-air

(lean) (stoich) (rich)

Mole fraction, Xi

Nitric Oxide (NO)

is an air pollutant

57
Temperature (K)
 Adiabatic Flame Temperature for Products at Equilibrium

b
Ca H b  (a  )(O2  3.76 N 2 )  aCO2  bH 2O  cN 2  dO2  eCO  fH 2
4
 gH  hO  iOH  jNO  kN  

• One can calculate the AFT for the above stoichiometric reaction where the
products are at equilibrium:  ni hi (Ta )   ni hi (T1 )
P R

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

• Again computer programs are used for these calculations:

http://www.wiley.com/college/mechs/ferguson356174/wave_s.html
Adiabatic Flame Temperature Applet

and STANJAN
58
 Chemical Kinetics

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

C8 H18  12.5(O2  3.76 N 2 )  8CO2  9 H 2O  47 N 2

H 2  1 / 2O2  H 2O
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.

 F F  A A   C C  D D

How fast the fuel is consumed is of interest, the reaction rate w’’’ is defined
as:
d[ F ]
w'''  
dt
where [ F ] refers to the fuel concentration (kmol/m3 or kg/m3), negative sign
due to the fact that the fuel is consumed. 59
 Reaction Mechanism
In reality the reaction proceeds through elementary reactions in a chain
process known as chain reactions

The global hydrogen-oxygen reaction proceeds via the following elementary

reactions, collectively known as a reaction mechanism:
H2  M  H  H  M Chain initiation

H  O 2  M  HO 2  M
HO 2  H 2  H 2 O  OH Chain propagation
OH  H 2  H 2 O  H

H  O 2  OH  O
Chain branching
H 2  O  OH  H

H  OH  M  H 2 O  M
H  H  M  H2  M Chain termination
O  O  M  O2  M
60
( M is any species present th at acts as a collision partner)
 Chain Branching

• In chain initiation reactions radicals are formed

• In chain branching reactions there is no net production or destruction of

free radicals

• 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 reactions

where the radicals recombine to form the final products.

61
 Global Reaction Rate

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:

 f Fuel   O O2  i Inert   P Pr oduct 

2

Empirical correlations have been developed for the fuel reaction rate:

d [ Fuel ]  E 
  A exp     [ Fuel ] [O2 ] [ Inert]
n m l
dt  RT 
where [ ] in units of gmol/cm3
R = 1.987 cal/gmolK
E typically 20 - 40 kcal/gmol

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

62
 Explosion Limits

Perform an experiment where a fuel-oxygen mixture is injected into a

preheated evacuated vessel and monitor the vessel pressure.

P
Fuel-air
P
Vacuum Pfill

time
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

63
limit temperature
 Explosion Limits
If you repeat the experiment for different fill pressures and plot all the results on a
pressure-temperature graph 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.

1 atm

NO Explosion

Explosion Limits of
Stoichiometric H2-O2
Explosion

64
 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 the
temperature rise is only tens of degrees. These are called cool flames.

A rapid compression machine (RCM) consisting of a piston-cylinder

assembly is used to raise the pressure and temperature of a fuel-air
mixture very quickly to a predetermined value P2, T2.

P2 T2
k 1
k 1
T2  P2  k V 
     1   r k 1
T1  P1   V2 

65
 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

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 time

a “hot flame” or high-temperature explosion (E).

66
 Explosion Limits for HC Fuels

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

Isooctane
Methane
Ignition temperature, oC

Single-stage ignition

Single-stage ignition No ignition

Two-stage ignition

No ignition

Pressure, atm
67
 Laminar Premixed Flames

A flame is a thin region in space where chemical reactions convert the

fuel-air mixture into combustion products.

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

application).

For a given P, T, f a flame has two basic properties:

a) adiabatic flame temperature,Tad
b) laminar burning velocity, Sl

Vu = Sl Vb-Sl Sl Vu = 0
burnt
Vb
Rb ,Ta ru unburnt
rb ru

Stationary flame Moving flame

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

Pressure is roughly constant across the flame so r ~ 1/T 68

 Structure of Flame

Vb Vu = SL
rb ru

[Fuel]
T
[O2]
d[ F ]   Ea 
 [ F ]n [O2 ]m exp 
dt  RT  [radicals]

Products Reaction Pre-heat

zone Zone Zone

Flame
Visible part of the flame thickness d
Diffusion of heat
and mass 69
 Laminar Burning velocity

Maillard-LeChatelier theory gives:

Sl2  a  RR  a  P n  2 exp(  Ea / R Tad ) 1<n<2

Higher flame velocity corresponds to:

1) higher unburned gas temperature
2) lower pressure (most hydrocarbons)
3) higher adiabatic flame temperature (chemical reaction)
4) higher thermal diffusivity a (= kcond/rcp)

70
 Laminar Burning Velocity Correlation

The following is a correlation developed by Metghalchi and Keck

g b
 T   P 
Sl  Sl ,ref  u    (1  2.1Ydil )
 298K   1atm 

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

Sl ,ref  BM  B2 (f  fM ) f is equivalence ratio

g  2.18  0.8(f  1)
b  0.16  0.22(f  1)

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

Methanol 1.11 36.92 -140.51

Propane 1.08 34.22 -138.65
Isooctane 1.13 26.32 -84.72

71
 Flame Velocity

The laminar burning velocity is measured relative to the unburned gas ahead
and 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 the

flame 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

Vb=0 Vf Vu Vf Vb=Vf Vf - Vu= Sl

rb ru rb ru
Moving flame Stationary flame

Applying conservation of mass across the flame:

r 
r u  S l  A  r b  V f  A  V f   u   S l
 rb  72
 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 that

is 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

St / Sl  1  aut / Sl 
b
Sl St

Laminar flame Turbulent flame 73

 Flame Thickness and Quenching Distance

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

2a 2  kcond 
d 
 
 1 mm
Sl Sl  r  c p 

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

Local quenching d

It is found experimentally that if the duct diameter is smaller than some

critical 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.
d min  d
74
 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.
Eign  1 / P 2

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 the
fuel-rich limit is known as the upper flammability limit.

The flammability limit is affected by both the mixture initial pressure and
temperature.
75
76