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Combustion: Fundamentals and Applications
Suresh K Aggarwal
Department of Mechanical and Industrial Engineering
University of Illinois at Chicago, Illinois, USA
[email protected]
Contents
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Introduction: why study combustion?
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Droplet vaporization and combustion
Thermochemistry: Classical thermodynamics for reacting mixtures
Transport phenomenon; mass transport
Chemical kinetics and Reaction mechanisms
Governing equations for a chemically reacting flow
Simplified reacting systems
Laminar premixed flames
Laminar diffusion flames
Pollutants formation
Reference Material: Stephen R. Turns: An Introduction to Combustion, Third Edition
(1) F.A. Williams: Combustion Theory, (2) C. K. Law: Combustion Physics, (3) D.B.
Spalding: Some Fundamentals of Combustion, (4) Journal and Conference papers
Why study combustion?
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Combustion deals with the conversion of chemical energy to thermal
energy by burning (oxidizing) a gaseous, liquid or solid fuel.
Thermal energy is subsequently harnessed as useful energy in the form
of (1) mechanical energy, such as in an automobile engine, (2) kinetic
and pressure energy used to provide thrust for aircraft and rocket
engines, (3) produce electric power, or used (4) directly for heating,
processing and manufacturing.
It is the most common mode of energy conversion and vital to the world
economy. Combustion applications include:
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Transportation: spark ignition and diesel engines, gas turbine combustors,
rocket engines, etc.
Power generation
Process industry: production of steel, glass, cement, etc.
Household and industrial heating
Fire safety, material synthesis, and pollutant reduction
Other combustion-related applications include fuel reforming, fuel cells, and
energy production from renewable and alternative fuels
It also represents an area offering many challenges and opportunities
for fundamental research; it remains an area of very active research.
Sustainable Energy Future
US EIA: World
Energy Outlook 2013
n 
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Global energy demand is expected to increase 50% by 2030 with fossil
fuels (oil, coal, gas) providing about 80% of the world’s energy.
Climate change (GHG and other pollutants) and diminishing supply of fossil
fuels are the major drivers for combustion research.
Focus is on improving efficiencies, reducing emissions, and developing
renewable fuels.
4
Fuels
§  Solid, liquid, and gaseous fuels
§  Petroleum-derived fuels
§  Most of these fuels are complex mixtures of hundreds of
§ 
compounds, including both aliphatic and aromatic compounds
These are generally produced by the fractional distillation of crude
oil
Gasoline, diesel, aviation fuel, natural gas
§ 
§  Agriculture and biomass-based fuels
§  Significant interest in developing alternative/renewable fuels in a
§ 
sustainable manner
Alcohols (ethanol), bio-butanol, biodiesels, hydrogen, syngas,
Fischer-Tropsch fuels etc.
§  Synthetic fuels
§  Most of these fuels contain straight-chain, branched-chain,
and cyclic-chain hydrocarbons with saturated and
unsaturated molecular structure, as well as aromatics and
other constituents.
Gasoline
§  Gasoline is generally produced by the fractional distillation of crude oil (1
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§ 
barrel (159 L) of crude oil yields about 72 L (19 gallons) of gasoline)
A complex mixture of over 200 hydrocarbons, mostly aliphatic
hydrocarbons, with 4 to 12 carbons.
Generally a mixture of paraffins (alkanes), cycloalkanes, olefins (alkenes)
and aromatics.
25-40% iso-alkanes such as isooctane, 4-8% n-alkanes such as C7H16,
and branched chain, 3-7% cycloalkanes, 5-10% alkenes, and 20-40%
aromatics; a general formula is CnH1.87n(C8.26H15.5)
Iso-octane has been considered a surrogate for gasoline. A more realistic
surrogate is a mixture of n-heptane, iso-octane, toluene and C5-C6 olefins
Important properties (specified by ASTM):
§  Anti-knocking (resistance to autoignition) defined by Octane rating (87)
§  Volatility defined in terms of temperatures for 10, 50, 90, and 100%
evaporation; for example T=1210C for 50% evaporation
§  Regulations in terms of air quality (emissions of VOC, NOx, benzene,
acetaldehyde, etc.)
Diesel Fuel
§  Three grades of diesel as specified by ASTM
§  D1, D2 and D3 as light, medium and heavy distillates; molecular weight
§ 
§ 
§ 
and viscosity increase with the designation number
Important properties (specified by ASTM):
§  Ignitability measured in terms of cetane number; typical values 40-55,
higher CN means shorter ignition delay time measured between start
of injection and start of combustion
§  Cold start, lubricity, viscosity, very low sulfur content
§  Provide proper functionality and wear characteristics for the fuel
injection systems
§  Regulations in terms of emissions
Contains about 75% alkanes (n, iso, and cyclo) and 25% aromatics
(naphthalenes and alkylbenzenes). Average chemical formula is C12H23,
but ranges from C10H20 to C15H28
N-heptane has been considered as a surrogate for diesel fuel. A more
realistic surrogate should be a mixture of n-, iso- and cyclo-alkanes, and
one or two aromatic compounds
Octane Rating and Cetane Number
§  Octane rating of gasoline is measured in a test engine and is defined by
comparison with the mixture of 2,2,4-trimethylpentane (iso-octane) and
heptane that would have the same anti-knocking capacity as the fuel under
test.
§  For example, petrol with the same knocking characteristics as a mixture of
90% iso-octane and 10% heptane would have an octane rating of
90.Important properties (specified by ASTM)
§  The operator of the Cooperative Fuel Research (CFR) engine uses a handwheel to increase the compression ratio of the engine until the time
between fuel injection and ignition is 2.407ms. The resulting cetane
number is then calculated by determining which mixture of cetane
(hexadecane) and isocetane (2,2,4,4,6,8,8-heptamethylnonane) will result
in the same ignition delay.
Aviation Fuels
Jet A (also Jet A-1) or Kerosene
§  Jet A is mostly used in US, while Jet A-1 is used in the rest of the
world. Main difference is that A-1 has lower freezing point
§  Typically contain 6 to 16 carbon, major components include branched
and straight chain alkanes and cyclo-alkanes, aromatics, and less than
5% olefins
JP-8 (Jet Propellant 8) used by US military and NATO allies
§  Similar to Jet A-1 but with anti-corrosive and anti-icing additives
Bio-Fuels and Synthetic Fuels
Biofuels-Alcohols
Ethanol (C2H5OH)
H3 C
C
H2
N-butanol (C4H9OH)
Iso-butanol (i-C4H9OH)
OH
Biodiesels (Fatty acid esters): made from transesterification or by
chemically reacting a lipid (vegetable oil or animal fat) with an alcohol
Synthetic or Synthesized Fuel: Significant interest in producing
biofuels and other renewable from biomass such as cellulose, agriculture
waste, municipal waste. Various fuels include syngas, hydrogen, biogas,
BTL (biomass to liquid) fuels .
Renewable Diesel
§  Biodiesel produced from trans-esterification of vegetable oils and animal
§ 
§ 
§ 
fat, as well as algae, using alcohols (methanol or ethanol)
Note the first diesel engine demonstrated at the 1898 Paris Exhibition by
Rudolph Diesel used peanut oil as its fuel
Most biodiesel fuel in the United States is made from soybeans, while in
Europe, rapeseed or modified canola oil is commonly used
Biodiesel or green diesel is also produced via other processes, namely
biomass-to-liquid (BTL) and gas-to-liquid (GTL):
§  In BTL, biomass is gasified to produce syngas, and Fischer-Tropsch
synthesis is used to convert syngas into diesel and other liquid fuels
§  Similarly, in GTL, natural gas is used to produce syngas, followed by
Fischer-Tropsch synthesis
Straight chain hydrocarbons: alkanes (CnH2n+2), alkenes (CnH2n),
alkynes (CnH2n-2)
H
H
C
H
H
H
Methane-CH4
H
H
C
C
H
H
C
H
C
C
H
H
H
Ethane (C2H6)
H2
C
H3 C
C
H2
CH3
n-butane (n-C4H10)
n-heptane (n-C7H16)
Alkenes (Olifins)
with 1 double bond
Ethene (ethylene)-C2H4
H
H
C
1-heptene (C7H14)
Alkynes--1 triple bond
H
Ethyne (acetylene)-C2H2
3-heptyne–C7H12
Branched Chain Alkanes
CH3
CH
H3C
CH3
Iso-butane (i-C4H10)
Iso-octane (C8H18)
2,2,4-Trimethylpentane
(Iso-octane)
Cycloalkanes form closed ring structure with carbon atoms
Cyclopropane (C3H6)
Cyclohexane (C6H12)
§  N-heptane and iso-octane are considered reference fuels, and
surrogates for diesel fuel and gasoline, respectively.
Aromatics
• 
• 
• 
These species have a ring structure with alternating double and single bonds
between carbon atoms
Monocyclic (benzene) or polycyclic aromatic hydrocarbons (PAH) are present
in fossil fuels, and also formed by incomplete or fuel-rich combustion
PAHs are most widespread pollutants (carcinogens), and precursors for soot
Toluene (C7H8)
Benzene (C6H6)
Naphthalene (C10H8)
Biofuels-Alcohols
Ethanol (C2H5OH)
N-butanol (C4H9OH)
Iso-butanol (i-C4H9OH)
Ether, Aldehydes, Ketones, Esters
Organic compounds which are either constituents in fuels, or formed
during partial oxidation of hydrocarbon combustion, or added in fuel
blends
Ether: commonly used as solvent, contains an oxygen atom
connected to two alkyl or aryl groups
H3C
R: alkyl
O
CH3
Dimethyl ether (C2H6O)
Phenyl: simplest aryl group
Aldehydes: Formed during hydrocarbon combustion, used in fragrances,
contribute to smog, small amounts emitted from engines
H3 C
Formaldehyde (CH2O)
Carbonyl group: C=O
C
H
O
Acetaldehyde (C2H5O)
Ketones: Formed during hydrocarbon combustion, have many applications
in industry, also used in solvents
O
H3C
Ketones contain a carbonyl group bonded
to two other carbon atoms; R and R’ can
be a variety of carbon containing species
C
CH3
Acetone (C3H6O)
Esters: Consist of a carbonyl group adjacent to an ether linkage
O
H3 C
R and R’ are the hydrocarbon parts of the
carboxylic acid and alcohol
C
O
CH3
Methyl acetate (C3H6O2)
Biodiesel Components
methyl butanoate (butyrate)
methyl butenoate
Methyl decanoate (C9H19COOCH3)
Biodiesels can also be produced from biomass and organic
waste
Review of Classical Thermodynamics
§ 
§ 
Review of classical thermodynamics
Thermochemistry: thermodynamic analysis of a gaseous reacting mixture
§ 
§ 
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§ 
§ 
§ 
Ideal gas equation
Thermodynamic system; state properties, extensive and intensive variables
Caloric equation of state, specific heats
First Law and internal energy
Properties of a reacting gas mixture, stoichiometry, equivalence ratio
Enthalpy, enthalpy of formation, sensible enthalpy, enthalpy of reaction,
enthalpy of combustion, fuel heating value
§  Adiabatic flame temperature
§  Second Law, entropy and Gibbs free energy
§  Equilibrium composition and temperature
Review of Thermodynamics
Ideal gas equation:
p = ρRuT / M
p = ρRT pv = RT R = Ru / M
p=pressure (N/m2), ρ=density (kg/m3), T=temperature (K), M=mol. weight (kg/
kmol), Ru=universal gas constant =8314.5 J/(kmol-K), v=specific volume (m3/kg)
and R = Ru/M
€
€
Other€forms: pV = nkBT = NRuT = CRuT
€
V=volume, n=number of molecules, kB=Boltzmann constant=1.3806x10-23 (J/K)
N=number of moles=n/Av, Av=Avogadro constant=6.0221x1023 (1/mol),
C=concentration
Assumptions:
§ 
Dilute gas mixtures at pressures relatively low pressures and high
temperatures compared to critical pressure and temperatures
§ 
§ 
No intermolecular forces except for during collision
Volume occupied by individual molecules (or atoms) is negligible.
Near critical conditions, we use a compressibility factor and cubic equation of state
Review of Thermodynamics
§ 
Pressure, density, temperature, internal energy (U), and entropy (S) are
fundamental thermodynamic properties
§ 
Enthalpy (H), Gibbs free energy, etc. are derived thermodynamic properties
used in analysis
§ 
Extensive property is the total property for a given mass or number of moles,
while intensive property is the same property per unit mass (or mole)
Pressure, density and temperature are by definition intensive properties
§ 
§ 
Any extensive property can also be defined as an intensive property; for
example, enthalpy per unit mass (or mole) ==> h=H/m
Caloric equations of state:
u=u(T,v) and h=h(T,p)
h=u + p.v
§ 
For pure gas (or liquid), the state of a given system is defined by two
properties
§ 
For a mixture, we need another property to define the state; usually mass
fraction or mole fraction of each species, u=u(T,v, Yi)
§ 
§ 
For ideal gases, u = u(T) and h = h(T)
For a gas mixture u = u(T,Yi) and h = h(T,Yi)
Specific Heats at Constant Pressure and Volume
# ∂h &
cp = % (
$ ∂T ' p
# ∂u &
cv = % (
$ ∂T ' v
For an ideal gas:
du
cv =
and du = c v (T)dT
dT
€
c p = cv + R
Using the ideal gas equation and h = u +R.T
dh
c p = € and dh = c p (T)dT
dT
€
§ 
γ = c p /c v
R
γR
cv =
and c p =
γ −1 €
γ −1
Specific heat at constant pressure (or volume)€
represents the energy (heat)
required to raise the temperature of 1 kg of material by 1 degree Kelvin
§  Specific
heat depends
€
€ on the temperature and also on the molecular structure,
i.e., whether gas is made up of atoms or molecules, and whether molecules are
monoatomic, diatomic or polyatomic
§ 
For diatomic and polyatomic gases, energy can be stored in translation, rotation,
and vibration modes
§ 
Specific heats are determined experimentally or using statistical thermodynamics
and kinetic theory
Specific Heat at Constant Pressure
c p / Ru = a1 + a2T + a3T 2 + a4T 3 + a5T 4
Extensive data is available for the
values of 5 constants for various
species
Turns: An Introduction to Combustion, 3rd Edition
Properties for an Ideal gas Mixture
Composition of a gas mixture containing Ns number of species is described
by the mass fraction or mole fraction of each species i:
m
n
N
N
Yi = N s i and x i = N s i
By definition:
∑Yi = 1 and ∑ x i = 1
i=1
i=1
∑ mi
∑ ni
s
i=1
i=1
Ns
Ns
∑n M ∑n M
i
Some useful relations:
i
M = i=1€
Ns
i
=
i
i=1
n
∑ ni
€
s
Ns
= ∑ xi M i
i=1
i=1
mi = n i M i and Yi =
ni M i
=
Ns
x i nM i
Ns
∑n M ∑n M
i
i
i=1
i
xi Mi
M
and
∑Y / M
i
Ns
=1/ M
i
Ns
pV = RuT ∑ n i
piV = n i RuT
i=1
Ns
Ns
h = ∑Yi hi and h = ∑ x i hi
€i=1
i
i=1
i=1
Partial pressure of species (i):
Mixture enthalpy
and specific heat
=
Ns
i=1
€
Ns
c p = ∑Yic pi and c p = ∑ x ic pi
i=1
i=1
Stoichiometric Equation
A stoichiometric fuel-oxidizer mixture is defined as the one in which all the
reactants are consumed completely to form products in stable form. For example,
consider the following stoichiometric equation:
CH 4 + 2(O2 + 3.76N 2 ) ⇒ CO2 + 2H 2O + 7.52N 2
€
The reaction involves one mole of methane and 9.52 moles of air, and forms one
mole of CO2 and two moles of H2O. Nitrogen moles are the same on the two sides.
Also the total mass is conserved during the reaction.
Stoichiometric fuel-air ratio:
( F / A) s =
Mf
9.52M air
General stoichiometric equation for any fuel
Cx H y + a(O2 + 3.76N 2€) ⇒ xCO2 + (y /2)H 2O + 3.76aN 2
Mf
( F / A) s =
4.76aM air
€
a = x + y /4
For a given fuel-air mixture
F/A < (F/A)s for
€ fuel lean mixture
€
F/A > (F/A)s for fuel rich mixture
Stoichiometric Equation
Fuel lean or fuel rich mixture condition is better described by using equivalence
ratio:
φ=
F /A
( F / A) s
φ=1: stoichiometric mixture
φ < 1: fuel lean
€
Percentage excess air:
φ > 1: fuel rich
( A /F ) − ( A /F ) s
1−φ
EA =
.100 =
.100
φ
( A /F ) s
In gas turbine combustors, typically φ ≈0.1
€
φ can be included in the stoichiometric equation:
Cx H y + (a / φ )(O2 + 3.76N 2 ) ⇒ xCO2 + (y /2)H 2O + 3.76(a / φ )N 2 + a(1/ φ −1)O2
The above equation is valid for φ < 1. For fuel rich conditions, products also include
CO, H2, and other species in small amounts.
Φ is one of the key parameters in the design and operation of combustion systems
An Example
Stoichiometric Equation: C7 H16 +11(O2 + 3.76N 2 ) ⇒ 7CO2 + 8H 2O + 41.36N 2
State 1: Reactants at
p=101325 n/m2, and T=298 K
ni = 1, 11, 41.36
n=53.36 moles
Ns
xi = 0.01874, 0.206, 0.775 mass = mr = ∑ ni M i
State is defined by p, T, and ni:
Ns
M=1.61 Kg
i=1
pV = RuT ∑ ni
V=1.304 m3
i=1
State 2: Products at p=101325
n/m2, and T=2086 K
xi = 0.124, 0.142, 0.734
ni = 7, 8, 41.36
n=56.36 moles
Ns
m p = ∑ ni M i
m=1.61 Kg
V=9.648 m3
i=1
Total number of moles and volume change during this process, but total mass
does not.
In real cases, there is dissociation and the temperature and composition at
state 2 will be different due to dissociation and other factors; T=1980 K
C7 H16 +11(O2 + 3.76N 2 ) ⇒ 6CO2 + 7.2H 2O + CO + 0.8H 2 + 41.36N 2
First Law (Conservation of Energy)
The first and second law of thermodynamics can be used to determine the
equilibrium properties (i.e., temperature and composition) of a reacting mixture.
The first law is used to determine the mixture temperature while the second law
is used to determine the composition.
The first law relates changes in internal energy of a system to heat supplied to
the system and work done by the system.
dU = δQ − δW
or
€
€
U 2 − U1 = Q1−2 − W1−2
(1)
state 1
state 2
(2)
Here U is a state property, while Q and W are path dependent. This relation
is applicable even if there is a change in composition going from state 1 to
state 2.
Sign convention:
W > 0: work done by the system, Q > 0 heat transferred to the system
W < 0: work done on the system, Q < 0 heat transferred from the system
€
First Law and Internal Energy
For studying equilibrium properties of a reacting system, only the pdV (or
flow work) is important, i.e., δW=pdv. Then for a constant volume process
Eq. (1) becomes
du = δQ
or
Using h=u+pv, Eq. 1 yields
u2 − u1 = Q
(3)
dh = δQ + Vdp
For a constant pressure process, this becomes
€
dh = δQ
€
or
h2 − h1 = Q
(4)
Eqs. 3 and 4 provide very
€ useful fundamental information:
Ø Eq. (3): For a constant volume process, heat added (Q) to the system
equals the change in internal energy of the system
Ø Eq. (4): For a constant
€ pressure process, it equals the change in enthalpy.
Ø Thus, u is the natural variable for constant volume process, and h is the
natural variable for constant pressure process.
First Law for a Control Volume
For a reacting flow, a control volume is generally used, and the
equivalent form of the above equation at steady state is
m˙ (h2 − h1 ) = Q˙
Ø  Here
(1)
(2)
m˙ is mass flow rate and h is enthalpy per unit mass
˙ is the rate of heat supplied to the control volume.
Ø  Q
€
€
(5)
€
Ø  The system form of this equation (Eq. 4) can be obtained by
dividing by the mass flow rate.
Ø  Eq. 4 can be written on mass basis or mole basis.
Ø  Eq. 5 is used in several different forms. This will be discussed in
later chapters.
Important Notes and Examples
Equation (3), (4) or (5) representing the first law of thermodynamics provide a very
useful relation to calculate the properties of a reacting system at equilibrium.
Notes and Observations:
Ø For an adiabatic process under constant pressure, enthalpy of the system does not
change; h2=h1.
Ø Similarly the enthalpy of a mixture flowing through the control volume remains the
same. As discussed later, this provides the equation for calculating the adiabatic flame
temperature.
Ø For an endothermic process (i.e., boiling), heat is supplied to the system (or flowing
mixture), and h2>h1.
Ø Similarly for an exothermic process (i.e., condensation, combustion), heat is taken
from the system ( h2<h1)
Ø Example: Consider 1 kmol of O2 contained in a system at a temperature of 298K and
1000 kJ of heat is supplied to the system. Then we can calculate the increase in
temperature of O2 or temperature corresponding to state 2 using Eq. 4, i.e.,
T
h2 = h1 + Q
or
∫c
298
p
dT = 1000
(6)
Important Notes and Examples
Equation (6) is solved for T by using three different methods.
Ø (1) Use an iterative method utilizing the tabulated values of h (T)
h2 (T) = h1 (Tref ) + Q
Ø  Second approach is to express c p (T) as a polynomial
function of T, and
€
integrate Eq. (6), which yields a polynomial in terms of T.
Ø  Third simpler but less accurate approach is to assume c p (T) as constant
which€yields
€
c p (T − 298) = 1000
€
€
Standard Enthalpy of Formation
§ 
§ 
Standard enthalpy (or heat) of formation of a compound (or substance) is
defined as the change of enthalpy when one mole of this substance is formed
from its elements, with all substances in their standard states.
Standard or reference state is defined at T=298.15 K and p=1 atm
Consider the decomposition of one mole of O2
to form two moles of O atom at a constant
temperature of 298.
(1)
Q
(2)
Q = h2 − h1 = 2hO (298) − hO2 (298)
§  O2 molecule has a double bond between O atoms with a bond energy of about 498
kJ/mol. Thus the energy or heat required to decompose O2 and form two O atoms at
standard state is 498 kJ/mol.
§  Since we are only interested in change in enthalpy and not its absolute value, by
convention we assign a value of zero for the enthalpy of O2, and a value of 249 kJ/
mole for the enthalpy of O. Thus the standard enthalpy of formation of O is 249 kJ/
mole, and that O2 is zero. These values for many substances are available in the
combustion literature (thermodynamics and combustion books)
h f ,O2 = 0
h f ,O = 249 kJ / mol
h f ,CO2 = −393.546 kJ /mol
€
Enthalpy of Reaction
Enthalpy of a substance at a given temperature is the sum of the standard
enthalpy of formation and sensible enthalpy.
T
hi = h f ,i (Tref ) + Δh (T) = h f ,i +
∫c
pi
dT
298.15
Values of the standard enthalpy of formation, specific heat, and sensible enthalpy
for many common substances are provided in textbooks, see, for example, Turns
Combustion book.
Enthalpy of reaction is the heat release during a reaction occurring at constant
pressure and temperature. Generally p= 1atm, and T=298K. For example,
consider reaction C + O2 => CO2. The enthalpy of reaction is
Q = h2 − h1 = h f ,CO2 − h f ,C − h f ,O2 = −393.546kJ /mol
This is also the standard enthalpy of
formation for CO2 since it is formed from C
and O2 at their standard states. Note, a
negative value implies exothermic reaction.
(1)
Q
(2)
For combustion analysis, thermodynamic data such as standard enthalpy of
formation and specific heat of each species are required
Enthalpy of Reaction
Consider reaction between CO and O2 to produce CO2. Then the enthalpy of
reaction is
Q = h2 − h1 = h f ,CO2 − h f ,CO − 0.5h f ,O2
Q = −393.546 +110.541 = −283.005kJ /mol
However, this is not the standard enthalpy of formation for CO2 since it is not formed
from
€ C and O2 at their standard states.
a stoichiometric propane-air mixture reacting at standard conditions.
€ Consider
Then the enthalpy of reaction is
C3 H 8 + 5(O2 + 3.76N 2 ) ⇒ 3CO2 + 4H 2O +18.8N 2
Q = H p − H R = (3.h f ,CO2 + 4h f , H 2O +18.8h f ,N 2 ) − (h f ,C 3 H 8 + 5h f ,O2 +18.8h f ,N 2 )
Q = (−3.(393.5) − 4.(−241.85)) − (−103.85) = −2044kJ /mol
€
Here water is assumed to be in gaseous state (steam). For water in liquid state, the
standard enthalpy of formation also includes the enthalpy of vaporization (44kJ/
mol), and its value is -285.85kJ/mol.
€
Heat of combustion is the negative of the enthalpy (or heat) of reaction. It is also the
fuel heating value. What are LHV and HHV?
Adiabatic Flame Temperature
Two adiabatic flame temperatures are defined, at constant pressure and at
constant volume. The former is defined as as the temperature resulting from a
complete combustion process (involving the conversion of a fuel to CO2, H2O, and
other product species) occurring at constant pressure without any heat transfer,
work, or changes in kinetic or potential energy. Similarly, the constant-volume
adiabatic flame temperature is defined for a analogous process at constant volume.
Q = H p ( p,T f ) − H R ( p,TR ) = 0 H p ( p,Tf ) = H R ( p,TR )
Tf can be calculated using the values of the standard enthalpy formation and
sensible enthalpy of various species involved.
Tf
TR
#N
& #N
&
o
o
%∑ ni (h fi + ∫ c pi dT )( = %∑ ni (h fi + ∫ c pi dT )(
%$ i=1
('P $ i=1
'R
298
298
Generally TR=298 K, then
Tf
N
#N
& )N
,
)
,
o
o
%∑ (ni ∫ c pi dT )( = + ∑ ni (h fi ). − + ∑ ni (h fi ). = −H R
%$ i=1 298
('P * i=1
-R * i=1
-P
This provides an equation for Tf assuming that the product mixture composition (ni) is
known.
Adiabatic Flame Temperature
Since specific heats (cpi ) are function of temperature, an iterative procedure is
required to calculate Tf . In simplified analysis, all cpi may be assumed constant
and the same. Then
"N %
$∑ ni ' c p (Tf − 298) = −H R
# i=1 &P
Example: Consider a stoichiometric hydrogen-air mixture reacting at standard
conditions. Then the enthalpy of reaction is
H 2 + 0.5(O2 + 3.76N 2 ) ⇒ H 2O +1.88N 2
Q = 0 = H p (T f ) − H R (T0 ) = (h f ,H 2O +1.88h f ,N2 )T f − (h f ,H 2 + 0.5h f ,O2 +1.88h f ,N2 )T0
Tf
∫ (c
p,H 2O
+1.88c p,N2 )dT = 241.85kJ
Tref
For constant volume process, the equation for adiabatic flame temperature can be
obtained as :
U P (Tf ) = U R (T0 ) ⇒ H P (Tf ) = H R (T0 ) − (N RT0 − N PTf )Ru
Second Law and Entropy
dS ≥ δQ /T
dS = (δQ) rev /T
δQrev = dU + pdV
€ds =
dh
dp
−v
T
T
TdS ≥ dU + pdV = dH −Vdp
Using
€
pV = nRuT ⇒ pnMv = nRuT ⇒ pv = RT
T2
dT
p2
s(T2 , p2 ) − s(T1, p1 ) = ∫ c p
− Rln
T
p1
T1
€
In order to calculate entropy (state variable) change, a reference state is
defined at Tref=0o K and pref=1 atm, where s=0. Then the specific entropy at a
given temperature (T) and pressure of 1 atm can be written as
€
T
0
s (T) =
∫ cp
0
dT
T
p2
s(T2 , p2 ) − s(T1, p1 ) = s (T2 ) − s (T1 ) − Rln
p1
0
0
Second Law and Entropy
s(T2 , p2 ) − s(T1, p1 ) = s0 (T2 ) − s0 (T1 ) − Rln
p2
p1
Considering 1 kg contains 1/M kmol of a species, this equation can be
written on molar basis as
€
s (T2 , p2 ) − s (T1, p1 ) = s 0 (T2 ) − s 0 (T1 ) − Ru ln
Thus the molar entropy at any T and p:
p2
p1
s (T, p) = s 0 (T) − Ru ln
p
pref
pi
si (T, p) = si (T) − Ru ln
€ Molar entropy of any species in a mixture:
pref
0
s
(T )
Ø  Entropy of a species can be calculated
from
the
tabulated
values
of
€
given in Appendix A.
0
Ø  Examples: CO2 at T1=300K, p1=1atm: €s (T) = s (T) = 213.97 kJ /(kmol.K)
0
Ø  CO2 at T1=1000K, p1=3atm:
s (T) = 269.3 − 8.314 ln 3 = 260.1kJ /(kmol.K)
Ø  O2 in air at T1=1000K, p1=3atm, xO2=0.21: sO 2 (T) = 244 − 8.314 ln(0.63) = 247.8
€
€
Thermodynamic Equilibrium
For an isolated system, equilibrium implies thermal equilibrium (uniform
temperature),, mechanical equilibrium (uniform pressure), and chemical
equilibrium (uniform composition and not changing with time). Consider, for
example, a reacting mixture associated with the dissociation of CO2
CO2 ⇔ CO + 0.5O2
Equilibrium implies that the concentrations of these three species do not
change with time, while this reaction is occurring in both forward and
reverse directions. Thus at given time, the mixture contains (1-α) moles of
CO2, α moles of CO, and 0.5α moles of O2.
Conditions for chemical equilibrium can be obtained using the First and
Second laws (assuming only work is pdV work):
dS ≥ δQ /T
TdS − dU − pdV ≥ 0
For an isolated system under constant volume, this yields
( dS )U,V ≥ 0
€
Example: Equilibrium Composition
Consider a reactant stream containing CO and O2 at T1=298K p1=1atm, and product
stream containing (1-α)CO2 + αCO + (0.5α)O2 at temperature Tf, p=1atm. For
different values of α, one can compute T and S of the product stream as shown in the
plot. The equilibrium value of α corresponds to the maximun S.
smix (Tf , p) = ∑ ni si (Tf , pi ) = (1− α )sCO2 +α sCO + 0.5α sO2
Tf
#N
& )N
, )N
,
o
o
%∑ (ni ∫ c pi dT )( = + ∑ ni (h fi ). − + ∑ ni (h fi ).
%$ i=1 298
('P * i=1
-R * i=1
-P
For a given value of α, compute Tf and
then compute entropy of mixture
Combustion book: Turns
Gibbs Function and Chemical Equilibrium
For chemical reaction under constant pressure and temperature, it is
convenient to use the Gibbs free energy ot Gibbs function: G=H-TS, which
yields (after using the 2nd Law in terms of TdS)
dG −Vdp + SdT ≤ 0 ⇒ ( dG ) p,T ≤ 0
Thus chemical equilibrium corresponds to the minimum value of G, and can
be expressed as
( dG ) p,T
"
pi %
0
= ∑ gi (dni ) = ∑ dni $gi (T ) + RuT ln
'= 0
pref '&
$#
i
i
gi0 (T ) is defined as the Gibbs function of formation of species i at
temperature T and p= pref=1 atm. Its values can be found from
thermodynamic tables. For a general reaction
aA + bB ⇔ cC + dD
The above condition can be written in terms of the change in total Gibbs
function of formation and the equilibrium constant KP
Equation for Chemical Equilibrium
ΔG 0 (T ) = −RuT ln K p
ΔG 0 (T ) = cgC0 + dgD0 − agA0 − bg B0 = (G 0 (T )) − (G 0 (T ))
P
R
# ΔG 0 (T ) &
( pC / pref )c ( pD / pref )d
KP =
= exp % −
(
a
b
( pA / pref ) ( pB / pref )
RuT '
$
Provides an equation between equilibrium constant and change in G(T)
CO2 ⇔ CO + 0.5O2
1/2
x
(x
)
1/2
CO
O2
0
0
K
=
(
p
/
p
)
ΔG 0 (T ) = (gCO
+ 0.5gO02 − gCO
)
P
ref
2 T
xCO2
(1)
Examples
(2)
0
Water gas shift reaction:
0
CO2
ΔG (T ) = (g
0
H2
0
CO
+g −g
−g
0
H 2O T
)
CO + H 2O ⇔ CO2 + H 2
KP =
xCO2 xH 2
xCO xH 2O
This along with the conservation of atomic species provide equations equal
to the number of unknowns in the stoichiometry equation.
Equilibrium Temperature and Composition
Propane-air mixture
Combustion: Turns
Equilibrium Software-HPFLAME
Instructions for using the software
-Download (copy) all the files in Software folder in order to use
HPFLAME
- From the downloaded files in PC,
1. The file named ''Access to TPEQUIL, HPFLAME and UVFLAME
Software.exe'' will execute the software.
2. Clicking on this file will open a new window.
3. Now click on the ''File'' option in the new window.
4. For constant enthalpy and pressure (enthalpy) case,
select ''HPFLAME'' from the file menu.
5. This opens another new window, where we can change the
parameters required for your study.
6. After changing the required parameters, click ''Save and Run.''
7. This will open another new window with the results.
Equilibrium Software-HPFLAME
N-heptane-air mixture