Combustion: Fundamentals and Applications 1.pdf · These are generally produced by the fractional distillation of crude oil ! Gasoline, diesel, aviation fuel, natural gas ! ... typical

Combustion: Fundamentals and Applications

Suresh K Aggarwal Department of Mechanical and Industrial Engineering

University of Illinois at Chicago, Illinois, USA [email protected]

Contents §  Introduction: why study 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

§  Droplet vaporization and combustion 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? §  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: §  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.

4

Sustainable Energy Future

n  Global energy demand is expected to increase 50% by 2030 with fossil fuels (oil, coal, gas) providing about 80% of the world’s energy.

n  Climate change (GHG and other pollutants) and diminishing supply of fossil fuels are the major drivers for combustion research.

n  Focus is on improving efficiencies, reducing emissions, and developing renewable fuels.

US EIA: World Energy Outlook 2013

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

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 hand-wheel 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

Biodiesels (Fatty acid esters): made from transesterification or by chemically reacting a lipid (vegetable oil or animal fat) with an alcohol

Bio-Fuels and Synthetic Fuels Biofuels-Alcohols

OHCH2

H3C

Ethanol (C2H5OH) N-butanol (C4H9OH) Iso-butanol (i-C4H9OH)

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

C

H

H

H H C C

H

H

H

H

H

H

C C

H

H H

H

C CH H

H3C

H2CCH2

CH3

Straight chain hydrocarbons: alkanes (CnH2n+2), alkenes (CnH2n), alkynes (CnH2n-2)

n-butane (n-C4H10) Methane-CH4

Ethene (ethylene)-C2H4

n-heptane (n-C7H16)

Ethyne (acetylene)-C2H2

1-heptene (C7H14)

Ethane (C2H6)

3-heptyne–C7H12

Alkenes (Olifins) with 1 double bond

Alkynes--1 triple bond

H3CCH

CH3

CH3

Iso-butane (i-C4H10)

Branched Chain Alkanes

Iso-octane (C8H18) 2,2,4-Trimethylpentane (Iso-octane)

§  N-heptane and iso-octane are considered reference fuels, and surrogates for diesel fuel and gasoline, respectively.

Cycloalkanes form closed ring structure with carbon atoms

Cyclopropane (C3H6) Cyclohexane (C6H12)

Biofuels-Alcohols

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

Benzene (C6H6) Toluene (C7H8) Naphthalene (C10H8)

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 O CH3

Dimethyl ether (C2H6O) R: alkyl

Aldehydes: Formed during hydrocarbon combustion, used in fragrances, contribute to smog, small amounts emitted from engines

H3C CH

O

Acetaldehyde (C2H5O) Formaldehyde (CH2O)

Phenyl: simplest aryl group

Carbonyl group: C=O

Acetone (C3H6O)

O

C CH3H3C

Methyl acetate (C3H6O2)

H3C C

O

O CH3

Ketones: Formed during hydrocarbon combustion, have many applications in industry, also used in solvents

Ketones contain a carbonyl group bonded to two other carbon atoms; R and R’ can be a variety of carbon containing species

Esters: Consist of a carbonyl group adjacent to an ether linkage

R and R’ are the hydrocarbon parts of the carboxylic acid and alcohol

Methyl decanoate (C9H19COOCH3)

Biodiesel Components

Biodiesels can also be produced from biomass and organic waste

methyl butanoate (butyrate) methyl butenoate

Review of Classical Thermodynamics §  Review of classical thermodynamics

§  Thermochemistry: thermodynamic analysis of a gaseous reacting mixture §  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 /Mp=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

Assumptions:

Other forms: pV = nkBT = NRuT =CRuTV=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

§  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

€

p = ρRT

€

pv = RT

€

R = Ru /M

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)

§  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)

h=u + p.v

Specific Heats at Constant Pressure and Volume

§  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

€

cp =∂h∂T#

$ %

&

' ( p

€

cv =∂u∂T#

$ %

&

' ( v

For an ideal gas:

€

cp =dhdT

and dh = cp (T)dT

€

cv =dudT

and du = cv (T)dT

Using the ideal gas equation and h = u +R.T

€

cp = cv + R

€

γ = cp /cv

€

cv =R

γ −1 and cp =

γRγ −1

Specific Heat at Constant Pressure

cp / Ru = a1 + a2T + a3T2 + a4T

3 + a5T4

Turns: An Introduction to Combustion, 3rd Edition

Extensive data is available for the values of 5 constants for various species

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:

By definition:

€

Yii=1

Ns

∑ =1 and xii=1

Ns

∑ =1

€

Yi =mi

mii=1

Ns

∑ and xi =

ni

nii=1

Ns

∑

Some useful relations:

Partial pressure of species (i): €

mi = niMi and Yi =niMi

niMii=1

Ns

∑=

xinMi

niMii=1

Ns

∑=xiMi

M

M =niMi

i=1

Ns

∑

nii=1

Ns

∑ =

niMii=1

Ns

∑n

= xiMii=1

Ns

∑

Yi /Mii=1

Ns

∑ =1/M and

€

piV = niRuT

€

pV = RuT nii=1

Ns

∑

€

h = Yihii=1

Ns

∑ and h = xih ii=1

Ns

∑

€

cp = Yicpii=1

Ns

∑ and c p = xic pii=1

Ns

∑Mixture enthalpy and specific heat

Stoichiometric Equation

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.

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:

€

CH4 + 2(O2 + 3.76N2)⇒ CO2 + 2H2O+ 7.52N2

Stoichiometric fuel-air ratio:

€

F /A( )s =Mf

9.52Mair

General stoichiometric equation for any fuel

€

CxHy + a(O2 + 3.76N2)⇒ xCO2 + (y /2)H2O+ 3.76aN2

€

a = x + y /4

€

F /A( )s =Mf

4.76aMair

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 < 1: fuel lean φ > 1: fuel rich

€

φ =F /AF /A( )s

φ=1: stoichiometric mixture

Percentage excess air:

€

EA =A /F( ) − A /F( )s

A /F( )s.100 =

1−φφ.100

In gas turbine combustors, typically φ ≈0.1

€

CxHy + (a /φ)(O2 + 3.76N2)⇒ xCO2 + (y /2)H2O+ 3.76(a /φ)N2 + a(1/φ −1)O2

φ can be included in the stoichiometric equation:

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

Stoichiometric Equation: C7H16 +11(O2 +3.76N2 )⇒ 7CO2 +8H2O+ 41.36N2

An Example

State 1: Reactants at p=101325 n/m2, and T=298 K

ni =1, 11, 41.36 n=53.36 moles

pV = RuT nii=1

Ns

∑State is defined by p, T, and ni: V=1.304 m3

xi = 0.01874, 0.206, 0.775 mass =mr = nii=1

Ns

∑ Mi M=1.61 Kg

State 2: Products at p=101325 n/m2, and T=2086 K

ni = 7, 8, 41.36 n=56.36 moles

Total number of moles and volume change during this process, but total mass does not.

xi = 0.124, 0.142, 0.734 mp = nii=1

Ns

∑ Mi m=1.61 Kg V=9.648 m3

C7H16 +11(O2 +3.76N2 )⇒ 6CO2 + 7.2H2O+CO+ 0.8H2 + 41.36N2

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

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.

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.

state 1 state 2

€

dU = δQ−δW

€

U2 −U1 =Q1−2 −W1−2or

(1)

(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

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.

€

du = δQ (3) or

€

u2 − u1 =Q

€

dh = δQ+VdpUsing h=u+pv, Eq. 1 yields

For a constant pressure process, this becomes

(4)

€

h2 − h1 =Q

€

dh = δQ or

First Law for a Control Volume

Ø  Here is mass flow rate and h is enthalpy per unit mass Ø  is the rate of heat supplied to the control volume.

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.

€

˙ m (h2 − h1) = ˙ Q

For a reacting flow, a control volume is generally used, and the equivalent form of the above equation at steady state is

(2) (1)

€

˙ m

(5)

€

˙ Q

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.,

€

h 2 = h 1 + Q or

€

c pdT298

T

∫ =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

€

c p (T − 298) =1000

€

h (T)

Ø  Second approach is to express 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 as constant which yields

€

c p (T)

€

h 2(T) = h 1(Tref ) + Q

€

c p (T)

Standard Enthalpy of Formation

(2) (1) Consider the decomposition of one mole of O2 to form two moles of O atom at a constant temperature of 298.

€

Q = h 2 − h 1 = 2h O (298) − h O2(298)

§  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 Q

§ 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 hf ,O = 249 kJ /mol

€

h f ,CO2= −393.546 kJ /mol

Enthalpy of Reaction

(2) (1)

€

Q = h 2 − h 1 = h f ,CO2− h f ,C − h f ,O2

= −393.546kJ /molQ

Enthalpy of a substance at a given temperature is the sum of the standard enthalpy of formation and sensible enthalpy.

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

€

h i = h f ,i(Tref ) + Δh (T) = h f ,i + c pi298.15

T

∫ dT

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.

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.

For combustion analysis, thermodynamic data such as standard enthalpy of formation and specific heat of each species are required

Enthalpy of Reaction

€

Q = h 2 − h 1 = h f ,CO2− h f ,CO − 0.5h f ,O2

Consider reaction between CO and O2 to produce CO2. Then the enthalpy of reaction is

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.

However, this is not the standard enthalpy of formation for CO2 since it is not formed from C and O2 at their standard states.

€

Q = −393.546 +110.541 = −283.005kJ /mol

Consider a stoichiometric propane-air mixture reacting at standard conditions. Then the enthalpy of reaction is

€

C3H8 + 5(O2 + 3.76N2)⇒ 3CO2 + 4H2O+18.8N2

€

Q = H p −H R = (3.h f ,CO2+ 4h f ,H2O

+18.8h f ,N2) − (h f ,C3H8

+ 5h f ,O2+18.8h f ,N2

)

€

Q = (−3.(393.5) − 4.(−241.85)) − (−103.85) = −2044kJ /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 = Hp(p,Tf )−HR (p,TR ) = 0   Hp(p,Tf ) = HR (p,TR )

Tf can be calculated using the values of the standard enthalpy formation and sensible enthalpy of various species involved.

  ni (hfio

i=1

N

∑ + cpi298

Tf

∫ dT )#

$%%

&

'((P

= ni (hfio

i=1

N

∑ + cpi298

TR

∫ dT )#

$%

&

'(R

Generally TR=298 K, then   (nii=1

N

∑ cpi298

Tf

∫ dT )#

$%%

&

'((P

= ni (hfio

i=1

N

∑ ))

*+

,

-.R

− ni (hfio

i=1

N

∑ ))

*+

,

-.P

= −HR

This provides an equation for Tf assuming that the product mixture composition (ni) is known.

Adiabatic Flame Temperature

Example: Consider a stoichiometric hydrogen-air mixture reacting at standard conditions. Then the enthalpy of reaction is

H2 + 0.5(O2 +3.76N2 )⇒ H2O+1.88N2

Q = 0 = Hp(Tf )−HR (T0 ) = (hf ,H2O +1.88hf ,N2 )Tf − (hf ,H2 + 0.5hf ,O2 +1.88hf ,N2 )T0

(cp,H2O +Tref

Tf

∫ 1.88cp,N2 )dT = 241.85kJ

For constant volume process, the equation for adiabatic flame temperature can be obtained as :

UP (Tf ) =UR (T0 ) ⇒ HP (Tf ) = HR (T0 )− (NRT0 − NPTf )Ru

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

  nii=1

N

∑"

#$

%

&'P

cp(Tf − 298) = −HR

Second Law and Entropy

€

dS ≥ δQ /T

€

dS = (δQ)rev /T

€

δQrev = dU + pdV TdS ≥ dU + pdV = dH −Vdp

€

ds =dhT− v dp

T

€

s(T2, p2) − s(T1, p1) = cpT1

T2

∫ dTT− R ln p2

p1

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

€

s0(T) = cp0

T

∫ dTT

€

s(T2, p2) − s(T1, p1) = s0(T2) − s0(T1) − R ln

p2p1

€

pV = nRuT⇒ pnMv = nRuT⇒ pv = RTUsing

Second Law and Entropy

Ø  Entropy of a species can be calculated from the tabulated values of given in Appendix A.

Ø  Examples: CO2 at T1=300K, p1=1atm: Ø CO2 at T1=1000K, p1=3atm: Ø O2 in air at T1=1000K, p1=3atm, xO2=0.21:

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s(T2, p2) − s(T1, p1) = s0(T2) − s0(T1) − R ln

p2p1

Considering 1 kg contains 1/M kmol of a species, this equation can be written on molar basis as

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s (T2, p2) − s (T1, p1) = s 0(T2) − s 0(T1) − Ru lnp2p1

s 0 (T )

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s (T) = s 0(T) = 213.97 kJ /(kmol.K)

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s (T) = 269.3 − 8.314 ln 3 = 260.1kJ /(kmol.K)

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s (T, p) = s 0(T) − Ru lnp

pref

Thus the molar entropy at any T and p:

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s i(T, p) = s i0(T) − Ru ln

pi

pref

Molar entropy of any species in a mixture:

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s O2(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

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.

CO2 ⇔CO+ 0.5O2

Conditions for chemical equilibrium can be obtained using the First and Second laws (assuming only work is pdV work):

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dS ≥ δQ /T TdS − dU − pdV ≥ 0For an isolated system under constant volume, this yields

dS( )U,V ≥ 0

Example: Equilibrium Composition

Combustion book: Turns

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) = nisi (Tf , pi ) = (1−α)sCO2 +∑ αsCO + 0.5αsO2

  (nii=1

N

∑ cpi298

Tf

∫ dT )#

$%%

&

'((P

= ni (hfio

i=1

N

∑ ))

*+

,

-.R

− ni (hfio

i=1

N

∑ ))

*+

,

-.P

For a given value of α, compute Tf and then compute entropy of mixture

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

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

gi0 (T )

Thus chemical equilibrium corresponds to the minimum value of G, and can be expressed as

dG( )p,T = gi (dni )i∑ = dni gi

0 (T )+ RuT lnpipref

"

#$$

%

&''i

∑ = 0

aA+ bB⇔ cC + dDThe 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 ΔG0 (T ) = −RuT lnKp

ΔG0 (T ) = cgC0 + dgD

0 − agA0 − bg 0B = G0 (T )( )P − G0 (T )( )R

KP =(pC / pref )

c (pD / pref )d

(pA / pref )a (pB / pref )

b = exp −ΔG0 (T )RuT

#

$%

&

'(

ΔG0 (T ) = (gCO0 + 0.5gO2

0 − gCO20 )T KP =

xCO (xO2 )1/2

xCO2(p / pref )

1/2

Examples CO2 ⇔CO+ 0.5O2(1)

CO+H2O⇔CO2 +H2(2) Water gas shift reaction:

ΔG0 (T ) = (gCO20 + gH2

0 − gCO0 − gH2O

0 )T KP =xCO2 xH2xCOxH2O

This along with the conservation of atomic species provide equations equal to the number of unknowns in the stoichiometry equation.

Provides an equation between equilibrium constant and change in G(T)

Equilibrium Temperature and Composition

Combustion: Turns

Propane-air mixture

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