THERMOCHEMISTRY OF FUEL‐AIR MIXTURES

THERMOCHEMISTRY OF FUEL‐AIR MIXTUREScourtesy (prof. S. Krishnan and K. Srinivasan – University of Alabama)

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Outline

• What is combustion and what is thermochemistry?• Types of flames – premixed vs nonpremixed, etc.• Composition of air and fuels• Combustion stoichiometry• First law of thermodynamics applied to combustion• Adiabatic combustion

– Adiabatic flame temperature• Combustion efficiency• Chemical equilibrium• Basics of chemical kinetics

2

Combustion and Thermochemistry

• Combustion is a fast, gas-phase exothermic oxidation reaction; oxygen is a required reactant

• Thermochemistry of combustion deals with the composition and thermodynamic properties of pre-and post-combustion gases (e.g., in IC engines)

• Flame is a combustion reaction that propagates subsonically through space (a “deflagration wave”)– Premixed vs. Non-premixed (aka diffusion) flames– Laminar vs. Turbulent flames– Steady vs. Unsteady flames

• Motion of flames relative to unburned gas is an important feature (e.g., laminar burning velocity)

3

Flames in IC Engines

• Homogeneous SI (gasoline) engines– Premixed, turbulent, unsteady flame propagation

• Direct Injection Spark Ignition engines– Partially premixed, turbulent, unsteady flame propagation

• Heterogeneous CI (diesel) engines– Lifted, turbulent, non-premixed (diffusion) unsteady flame

anchored away from the injector nozzle during injection• Homogeneous charge compression ignition (HCCI)

– HCCI engines have volumetric autoignition, no flamefront• End-gas knock in SI engines causes transition from a

propagating flame to a “detonation,” which is a supersonic wave

4

Composition of Air

• Typical composition of dry air– Nitrogen – 78.09% by vol. (or mole), 73% by mass– Oxygen – 20.95% by vol. (or mole), 27% by mass– Argon – 0.93% by vol. (or mole)– Trace amounts of CO2 (~0.03% by vol. or mole), neon,

helium, methane, etc.• Moist air can have 1-4% by mass of water vapor• Oxygen is the reactive component of air• Usually sufficient to regard air as consisting of 21% O2

and 79% N2 on a molar basis• So, for each mole of O2 in air, we have (79/21) or

3.76 moles of N2

5

Composition of Fuels

• Typical gasoline and diesel fuels are blends of (tens to hundreds) of different hydrocarbon compounds obtained from crude oil refinement

• Gasoline and diesel are predominantly composed of carbon (~86%) and hydrogen (~14%) by mass

• Diesels used to contain ~1% sulfur (now <15 ppm!)• Gasoline can contain up to 10% by volume of bio-

ethanol in many parts of the World• Other fuels of interest are alcohols (e.g., E85),

gaseous fuels (e.g., natural gas, LPG), and single component fuels (e.g., methane, propane, isooctane)

6

Classes of Organic Compounds

• Alkanes (paraffins) – CnH2n+2

• Alkenes (olefins) – CnH2n

• Alkynes (acetylenes) – CnH2n-2

• Cycloalkanes (cycloparaffins)

• Aromatics – CnH2n-6

• Alcohols – CnH2n+1OH

7

Classes of Organic Compounds

• Alkanes are saturated (no more hydrogen can be added); cycloalkanes, alkenes, and alkynes are unsaturated

• Isomers – compounds that contain the same number of carbon and hydrogen atoms but different structures (straight- vs. branched-chain)– E.g., C8H18 can be n-octane (which is a straight-chain C8

hydrocarbon) or several “isooctanes” (which are branched-chain; normally 2,2,4-trimethylpentane);

– Isooctane structure is much more stable than n-octane

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Classes of Organic Compounds (cont.d)

• The ring structure in aromatics (e.g., benzene) is very stable but can accommodate additional –CH2groups in side chains

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Combustion Stoichiometry

• For a general hydrocarbon fuel (CxHy)– For ɸ = 1, we can write the stoichiometry equation as:

CxHy + (x+y/4) (O2 + 3.76N2) xCO2 + (y/2)H2O + 3.76(x+y/4)N2

– For ɸ < 1, we can rewrite the stoichiometry equation as:CxHy + [(x+y/4)/ɸ] (O2 + 3.76N2) xCO2 + (y/2)H2O +

[(x+y/4)(1/ɸ - 1)]O2 + (3.76(x+y/4)/ɸ) N2– For ɸ > 1, there will be leftover fuel and products of

incomplete combustion (e.g., CO, H2) due to insufficient O2; so, element balance alone not sufficient to find product composition; need additional assumptions

• (A/F)molar = [(x+y/4)/ɸ]• (A/F)mass = {[(x+y/4)/ɸ]*(MWO2+3.76*MWN2)} / (x*MWC+y*MWH)• MWO2=32, MWN2=28, MWC=12, MWH=1; all in kg/kmol

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ɸ=(ma/mf) s/(ma/mf)

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Stoichiometric air/fuel ratio for air-hydrocarbon fuelmixtures as a function of fuel molar H/C ratio.

First Law of Thermo Applied toCombustion

• For closed reactive system of mass m, 1st law yields– QR-P – WR-P = UP – UR

• For const. vol. process WR-P =0, QR-P = (UP – UR)V, T’ = (ΔU)V, T’• For const. pressure process, QR-P – P(V’P-V’R) = UP – UR QR-P =

(UP + PV’P) – (UR + PV’R) QR-P = (HP – HR)P, T’ = (ΔH)P, T’

12Source: Heywood (1988), P. 73

ΔU or ΔH due to combustion

• (ΔU)V, T’ and (ΔH)P, T’are negative for exothermic reactions

• -(ΔU)V, T’ is heat of reaction at constant volume at T’

• -(ΔH)P, T’ is heat of reaction at constant pressure at T’

• (ΔH)P, T’ – (ΔU)V, T’ = P(VP-VR)


Effect of Product Phase on ΔU or ΔH

• With hydrocarbon fuels, the H2O in the products can be in gaseous or liquid phase

• For all vapor H2O (ΔU)V’, T’, H2O vap

• For all liquid H2O (ΔU)V’, T’, H2O liq

• |(ΔU)V’, T’, H2O liq| –|(ΔU)V’, T’, H2O vap| = mH2Ou’fgH2O

• |(ΔH)V’, T’, H2O liq| –|(ΔH)V’, T’, H2O vap| = mH2Oh’fgH2O


Effect of Reactant Phase on ΔU or ΔH

• For some fuels, the reactants may contain either liquid or vapor phase fuel

• |(ΔU)V’, T’, Fuel liq| –|(ΔU)V’, T’, Fuel vap| = mfu’fgf

• |(ΔH)V’, T’, Fuel liq| –|(ΔH)V’, T’, Fuel vap| = mfh’fgf

• Typically, the enthalpy of vaporization of the liquid fuel is provided by the intake or in-cylinder gases


Enthalpy of Formation (Δh0fi)

• Enthalpy of formation (Δh0fi) of species i is

defined as the enthalpy change associated with the reaction forming one mole of species i from its elements, with each substance being in its “thermodynamic standard state” at a given temperature

• Standard state is typically the state at 1 atm (or 1 bar) and a given temperature; it is referred to by a superscripted o

• By definition, (Δh0f) for elements in their

standard state (e.g., O2, N2) is zero

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Typical Enthalpies of Formation (Δh0fi)

• Given Δh0fi of each

reactant and product species i in a reaction, the enthalpy of reaction can be calculated

• Some fuels (e.g., C2H2) may have positive Δh0

fi• For each species i, the

total enthalpy is the sum of Δh0

fi and the sensible (temperature-dependent) component of enthalpy

17Source: Heywood 2nd Ed. 2018, P. 77

Species State* Δ˜h∘f, MJ/kmol

O2 Gas 0

N2 Gas 0

H2 Gas 0

C Gas 0

CO2 Gas −393.52

H2O Gas −241.83

H2O Liquid −285.84

CO Gas −110.54

CH4 Gas −74.87

C3H8 Gas −103.85

CH3OH Gas −201.17

CH3OH Liquid −238.58

C8H18† Gas −224.1

C8H18† Liquid −259.3

*At 298.15 K (25°C) and 1 atm.†Isooctane.

Enthalpy of Reaction (ΔH or –QP)

• The enthalpy of reaction (ΔH = –QP) can be calculated as follows from the enthalpies of formation (Δh0

fi) of each reactant and product species i in a reaction:

• When all of the heat release is used to raise the temperature of the product gases, ΔH = –QP = 0. The product temperature (T2) in this case becomes the adiabatic flame temperature (discussed later)

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)(

00

010

000

020

0

evolvedPR

ireac iTTTfijprod jTTTfjRP

Q

hhhnhhhnHHH

Heating Values

• For fuels of unknown composition, the enthalpy of the reactants cannot be determined from the enthalpies of formation of the reactant species (because they are unknown!)

• So, the heating value (calorific value) of the fuel is directly measured

• Heating value, QHV, is the heat of reaction at constant pressure or constant volume at a standard temperature (T0=25°C) for the complete combustion of a unit mass of fuel

• QHVp = -(ΔH)P, T0• QHVv = -(ΔU)V, T0

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Higher and Lower Heating Values

• Higher heating value, QHHV, is used when H2O in products is in liquid phase

• Lower heating value, QLHV, is used when H2O in products is in vapor phase (as in IC engines)

• QHHVp = QLHVp + (mH2O/mf)*hfgH2O

• QHHVv = QLHVv + (mH2O/mf)*ufgH2O

• The heating value at constant pressure is more commonly used

• Heating values are measured in calorimeters (continuous flow for gaseous fuels, constant volume for solid/liquid fuels)

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Adiabatic Combustion Processes

• Adiabatic combustion is an idealization used to determine maximum possible product temperatures for a given reacting system

• Applying first law to an adiabatic const. volume combustion process, we get: QR-P = UP – UR = 0

• Similarly, for an adiabatic const. pressure combustion process, QR-P = HP – HR = 0

• Thus, for constant pressure adiabatic combustion:

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ireac iTTTfijprod jTTTfj

RPRP

hhhnhhhn

HHHH00

010

000

020

0

0

Adiabatic Flame Temperature

• While for reactants, U and H are functions only of T, for products, they are dependent on both P & T

• Typically, constant-volume Tad is higher than constant-pressure Tad

• Tad is dependent on fuel type, oxidizer type, initial P & T, and fuel-oxidizer ratio


Adiabatic Flame Temperature – CH4-Air

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Tad for Const. V and Const. P Combustion

• Tad shown at typical IC engine conditions

• Maximum Tad occurs slightly rich of stoichiometric at these conditions

• Tad is higher for constant volume adiabatic combustion because final pressure is higher and dissociation is less


Isooctane‐air mixture with Pin = 10 atm, Tin = 700 K, TP,V and TP,P are Tadat const. V & const. P; PP,V is const. V equil. pressure

Approximate Tad for various stoich.mixtures

25Source: Glassman, 4th Ed. (2008), P. 28

Tinitial = 298 K

Combustion Efficiency (ηc or ηcomb)

• Total chemical energy into engine is mfQHV

• ηc measures the fraction of fuel chemical energy released in combustion

• For SI ICEs, ηc~95-98% for ɸ<1, ηc for ɸ>1; engine instability ηc

• For CI ICEs, ηc>98%

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HVf

APARc

jprodjTfj

ireaciTfiAPAR

QmTHTH

hnhnmTHTHAA

,0

,0

Source: Heywood (1988), P. 81

m = mass of working fluid; n = # moles / mTA = ambient temperature, mf = fuel mass

Combustion Inefficiency (1 – ηc)

• Combustion inefficiency is calculated from measured exhaust concentrations of incomplete combustion products such as CO, H2, UHC, and particulates

• xi = mass fraction of species i (CO, UHC, etc.), QHV,CO= 10.1 MJ/kg, QHV,H2 = 120 MJ/kg, QHV,PM(solid C) = 32.8 MJ/kg, QHV,UHC is assumed to be equal to QHV,f, subscripts f and a refer to fuel and air

• Bottomline: The higher the UHC, CO, H2, and PM emissions, the lower the calculated value of ηc

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fHVfaf

i iHVi

LHVf

APARc Qmmm

Qx

QmTHTH

/11

Source: Heywood (1988), P. 154

Chemically Reacting Mixtures

• Engine working fluids are mixtures of gases• Depending on the portion of the engine cycle, the

working fluid composition may be taken to be:– “Frozen”– “Chemical Equilibrium”– “Kinetically Controlled”

• Typically, unburned gases are considered to be frozen• Burned gases (post-combustion) may be considered

to be in chemical equilibrium• Gas composition during combustion is kinetically

controlled

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Chemical Equilibrium

• A chemically reactive system is said to be in chemical equilibrium if the chemical reactions, by which individual species react with each other, produce and consume each species at equal rates– At chemical equilibrium, no net change in species

composition occurs– In ICEs, post-combustion burned gases can be assumed to

be in chemical equilibrium– In CxHy combustion, combustion products (CO2, H2O, etc.)

dissociate at high temperatures (> 2200 K)– Typical product mole fractions for ɸ = 1 : N2 ~ 0.7; CO2, H2O

~ 0.1; CO, OH, O2, NO, H2 ~ 0.01; H, O ~ 0.001

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Chemical Equilibrium (Continued)

• Second law defines criterion for chemical equil.• For constant P and T combustion:

– 1st law: δQ = dH– 2nd law: δQ ≤ TdS– Combining, we get, dH – TdS ≤ 0– For finite changes, we get ΔH – TΔS = ΔG ≤ 0– So, for constant P and T combustion, Gproducts ≤ Greactants– At equilibrium: (ΔG)P,T = 0 criterion for equilibrium

• Consider a reactive mixture of ideal gases (species Miwith stoichiometric coefficients i) :– aMa + bMb + …= lMl + mMm + …– We can write:

30

0i

iiM


• If δni of reactant species Mi react to produce δnj of product species Mj, we get the proportionality– δni = iδn, where δn is the amount of species that react

• For a reactive mixture of ideal gases, the change on Gibbs function is:

• Here, the chemical potential, μi is defined as:

•• μi is equal to the partial molal Gibbs function at a given

temperature and pressure• For an ideal gas, , where is the

standard specific Gibbs function of formation

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i

iiTP nG ~,

)(,,

~ijnTPi

ij

nG

0

ln~~~PPTRT io

ii oi~


• Substituting for μi in the (ΔG)P,T equation at equilibrium, we get:

• Dividing throughout by δn and rearranging:

, where KP is Equil. Const. @ Const. P

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0ln~~

0ln~~~

0

0,

ii

ioi

ii

ioi

iiiTP

nPPTRT

nPPTRTnG

i

iP

P

oi

ioi

i

i

i

i

PPK

KTR

GTRP

P

0

0ln~~

~ln


• From KPi given in tables for formation of each species i, the KP for a reaction can be obtained as:

• We can determine the effect of P on KP as follows:

– If Σi = 0, changes in pressure have no effect on equilibrium composition

– If Σi > 0 (dissociation), then as pressure increases, the mole fractions of the dissociation products decrease

– If Σi < 0 (recombination), then as pressure increases, the mole fractions of the dissociation products increase

33

i

iPireactionP KK 1010 loglog

ii

ii

i

iP

ii

iii

xPP

PPx

PPK

~~

000

ix~

ix~


• We can also define the equilibrium constant in terms of concentrations [Mi] :

• We can then relate KP and KC as follows for P0 = 1 atm:

• For Σi = 0, KP = KC• Several reactions can be simultaneously in

equilibrium during combustion

34

i

iCiMK

i

iii

TRKP

TRMPPK C

i

i

i

iP

~~][00

Mole fractions of equilibrium combustion productsof isooctane-air mixtures as a function of fuel/airequivalence ratio at 30 atmospheres

35John B. Heywood: Internal Combustion Engine Fundamentals, Second Edition

Introduction to Chemical Kinetics

• While chemical equilibrium is dictated by the thermochemistry of the fuel-air mixture, ICE combustion processes are not always in chemical equilibrium (e.g., flame zone in SI combustion, pollutant formation processes, etc.)

• During combustion, the burn rate of the fuel is normally dictated by chemical kinetics rather than chemical equilibrium

• Chemical kinetics describes the rates at which chemical reactions proceed based on the species concentrations, temperature, and pressure

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Chemical Reaction Rates

• We will consider binary (two-species) reactions of the form: Ma + Mb = Mc + Md. For example:– CO + OH = CO2 + H Very important combustion

reaction, where most of energy is released!– O + N2 = NO + N one NO formation reaction

• Law of Mass Action – rate at which products are produced and reactants are consumed is proportional to the concentrations of species raised to the power of their respective stoichiometric coefficients

37

ba

baffcfa

f

ddccbbaa

MMkdtMd

dtMd

RR

MMMMreactionFor

:


• We can also determine the backward reaction rate (RRb) as follows:

– kf and kb are the specific reaction rate constants for the forward and backward reactions, respectively

• Thus, we can write the net reaction rate (RR) as:

• Now, consider the following general reaction:

38

dcdcb

babcb MMk

dtMd

dtMd

RR

dcbadcbbafbf MMkMMkRRRRRR

prodreac n

jjj

n

iii MM

11


• Then the net reaction rate (RR) becomes

• The net rate of removal of reactant species i is:

• The net rate of production of product species jis:

39

prod

jreac

i

n

jjb

n

iifbf MkMkRRRRRR

11

bfii RRRR

dtMd

bfjj RRRR

dtMd

Specific Rate Constants & ActivationEnergy

• Specific reaction rate constants are usually expressed in the so-called “Arrhenius form”:

– A: preexponential factor, EA: activation energy– exp (-EA/RT) is the Boltzmann factor that encapsulates

the molecular collisions that have an energy greater than EA (i.e., sufficient energy for reactions to occur)

– The functional dependence of k on T is usually determined experimentally for many reactions

40

RTEAk Aexp

Concept of Activation Energy

• Activation energy of the forward reaction is EA = Ef• Activation energy of the backward reaction is EA = Eb• Generally, the more exothermic a reaction, the smaller the

activation energy• Low EA reactions proceed faster than high EA reactions at low

temperatures and are less T-sensitive and vice versa

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Relationship between kf,kb, and Kc

• At equilibrium, we know that the forward and backward reaction rates for a reaction are equal

• Thus, the forward and backward specific reaction rate constants can be related to each other by means of the equilibrium constant based on concentrations (KC), and in turn, to KP

42

Ci

iba

dc

b

f KMMMMM

kk

i

ba

dc

dcbadcbbaf

bf

MMkMMkSo

RRRR

,

Chemical Reaction Mechanisms• A simple stoichiometric reaction such as

CxHy + (x+y/4) (O2 + 3.76N2) xCO2 + (y/2)H2O + 3.76(x+y/4)N2

is not how hydrocarbon fuel combustion actually happens in combustion systems• Actual combustion processes are governed by

chemical kinetics and may involve tens of species and hundreds of reaction steps

• E.g.,, GRI-Mech, a popular reaction mechanism for methane (the simplest hydrocarbon) involves 53 species and 325 reactions! – See, for example:

http://combustion.berkeley.edu/gri-mech/version30/text30.html)

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THERMOCHEMISTRY OF FUEL‐AIR MIXTURES

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