Chapter 3pThermochemistry of Fuel Air Mixtures3-1 Thermochemistry3 2 Ideal Gas Model3-2 Ideal Gas Model3-3 Composition of Air and Fuels3 4 C b ti St i hi t3-4 Combustion Stoichiometry3-5 The1st Law of Thermodynamics and Combustion3-6 Thermal conversion efficiency3-7 Chemically Reacting Gas Mixtures

1

• The composition and thermodynamic properties of the pre- and postcombustion working fluids inthe pre and postcombustion working fluids in engines

The energy changes associated with the• The energy changes associated with the combustion processes

2

An “ideal” gas exhibits certain theoretical propertiesAn ideal gas exhibits certain theoretical properties. Specifically, an ideal gas …Does not condense into a liquid when cooledDoes not condense into a liquid when cooled.Shows perfectly straight lines when its V vs. T & P vs. T relationships are plotted on a graphT relationships are plotted on a graph.In reality, there are no gases that fit this definition perfectly We assume that gases are ideal to simplifyperfectly. We assume that gases are ideal to simplify our calculations.

If temperature is not too low or pressure not to high, the ideal gasmodel can be applied.

Gas species (working fluids) in engines (e.g., O2, N2, fuel vapor CO water vapor etc ) may be treated as idealvapor, CO2, water vapor, etc.) may be treated as ideal gases.

R~ TRnTMRmmRTpV ~===

where m = mass of gasR = gas constant (J/kg⋅K)n = number of moles~ = Universal gas constant

8.3143 J/mol⋅K or 1,543.3 ft⋅lbf/lb-mole⋅°RR~

4

M = Molecular weight

•Air typically contains about 1-3 percent by mass of vapor

irair

Nor

mal

Ai

norm

al a

ion

from

%

var

iati

%

5

TABLE 3.1 Principle constituents of dry air

Gas ppm by volume

Molecular weight

Mole fractions

Molar ratio

O2 209,500 31.998 0.2095 12 ,

N2 780,900 28.012 0.7905 3.773

A 9 300 38 948A 9,300 38.948

CO2 300 44.0092

Air 1,000,000 28.962 1.0000 4.773

6

Atmospheric Nitrogen

2222

~~aNaNOOair MXMXM +=

iX~ is a mole fraction of gas i with respect to total number of mole of the mixture gas.

where

22

2 ~

~OOair

aN X

MXMM

−=

2aNX

209501998.312095.0962.28 ×−

=

Molecular weightof air

2095.01−

16.28=2aNM

7

2

TABLE 3-2 Molecular weights

8

The fuels most commonly used in IC engines are blends of many

different hydrocarbon compounds obtained by refining petroleum or

crude oil. These fuels are predominantly carbon and hydrogen

though diesel fuels can contain up to about 1 percent sulfur. Other

f l f i t t l h l f l ( t l d li fi dfuels of interest are alcohols, gaseous fuels (natural gas and liquefied

petroleum gas), and single hydrogen compounds (e.g., methane,

propane, isooctane).

9

Substances contained in a crude oilSubstances contained in a crude oil have different boiling points, the substances in crude oil can be separated using fractional distillation. The crude oil is evaporated and its vapours are allowed to condense at different temperatures in the fractionatingtemperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms.

10

- Paraffins

11

12

13

- Olefins

14

- Alcohol What is the component of alcohol that differsWhat is the component of alcohol that differs from other fuels ?

15

Relationships between composition of reactants (fuel and air) and composition of products

For example, overall chemical equation for the complete combustion of one mole of propane C3H8 :

OHCOOHC 22283 435 +=+ OHCOOHC 22283 435 ++

16

What mass of O2 is required for the complete combustion of 8.58 g of C4H10?

2C4H10 + 13O2 --> 8CO2 + 10H2O

combustion of 8.58 g of C4H10?

4 10 2 2 2

mole1476.0/ l12358g58.8

=Moles of C4H10 :g/mol123.584 10

2 moles of C4H10 require 13 moles of O2

Molecular weight of C4H10 2 moles of C4H10 require 13 moles of O2

0.1476 mol x (13/2) = 0.9595 mole O2

4 10

0.9595 mol x 31.998 g/mol = 30.7 g O2Ans

17

Stoichiometric Mixture of products contain CO2, H2O and N2

22222 )4

(773.32

)773.3)(4

( NbaOHbaCONObaHC ba +++=+++424

orAir

22222 41773.3

2)773.3)(

41( NyOHyCONOyCH y ⎥⎦

⎤⎢⎣⎡ +++=+++

or

aby =where

18

yyy ⎤⎡22222 4

1773.32

)773.3)(4

1( NyOHyCONOyCH y ⎥⎦⎤

⎢⎣⎡ +++=+++

The stoichiometric air/fuel or fuel/air ratios1−

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

AF

FA

Molecular weight of air⎠⎝⎠⎝ ss AF

y)4(5634 +y 9628)77331)(4/1( ×++A⎟⎞

⎜⎛

Molecular weight of air

yy

008.1011.12)4(56.34

++

=y

y008.1011.12

96.28)773.31)(4/1(+

×++=

sFA⎟⎠⎞

⎜⎝⎛

19

Fuel/air equivalence ratio φ

s

actual

AFAF

)/()/(

=φ

λ

FA )/(

Relative air/fuel ratio

s

actual

FAFA

)/()/(1 == −φλ

1,1 >< λφFuel-lean mixtures

Stoichiometric mixtures

Fuel rich mixtures 11 <> λφ

1== λφ

20

Fuel-rich mixtures 1,1 <> λφ

Examples of Stoichiometric combustion equation:

Methyl alcohol (methanol)

222223 66.52)773.3(5.1 NOHCONOOHCH ++=++Air molecular weight

47.6)116312(

96.28)773.31(5.1)F/A( s =+++×+

=

Air molecular weight

Hydrogen

)116312( +++

Fuely g

22222 887.1)773.3(5.0 NOHNOH +=++

3.34)/( =sFA

21

Hydrocarbon fuel of composition 84.1 percent by mass C and 15.9 percent by mass H has a molecular weight of 114.15. Determine the number of moles of air required for stoichiometric combustion and the number of moles of products produced per mole ofnumber of moles of products produced per mole of fuel. Calculate (A/F)S and the molecular weight of the reactants and products.

•Assuming fuel chemical formulation is of the form CaHb, we have the relation

b008.1a011.1215.114 += (1)

From mass fraction,

b252008.1/9.15

a25.2

011.12/1.84== (2)

From equations (1) and (2), we obtain,

a = 8 and b=18

Chemical balance equation

)773.3(5.12 22188 NOHC ++ ⇒ 222 16.4798 NOHCO ++

23

By mol:

)77331(5121 . . ++ 16.4798 ++⇒

66.591 + 16.64⇒By mass: Molecular weight of Air

96.2866.5915.1141 ×+× 16.2816.4702.18901.448 ×+×+×⇒

3.18428.172715.114 =+ (both sides are equal)

Based on 1 kg of fuel:Based on 1 kg of fuel:3.18428.172715.114

=+

14.1614.151 =+

15.11415.11415.114+

24

From the above equation, 1 mol of fuel requires 59.66 mol of air to get 65 16 mol of products

14.15114.15)/( ==sFA Ans

65.16 mol of products.

1

0661.01415

1)/( ==sAF Ans14.15

RM MMolecular weight of reactants ( ) and molecular weight of products ( ) areRM PMMolecular weight of reactants ( ) and molecular weight of products ( ) are

RM 1 1 (1 114.15 59.66 28.96)i in M= = × + ×∑Rtot

( )60.66i in ∑

36.30= Ans

PMtot

1 1 (8 44.01 9 18.12 47.16 28.16)64.16i in M

n= = × + × + ×∑

2571.28= Ans

3.5 THE FIRST LAW OF THERMODYNAMICS AND COMBUSTION

3.5.1 Energy and Enthalpy Balances

Reactants

ProductsCombustion

System changing from reactants to products

26

The first law of thermodynamics: relating changes in internal energy (enthalpy) to heat and work transferinternal energy (enthalpy) to heat and work transfer interaction

RPPRPR UUWQ −=− −− PRU −or

+ work transfer to the system, - work transfer from the system

C iCombustion process► Constant volume

initial and final temperature = T ′

∫ ==−

P

PR pdVW )0(27

∫R

TVRPPR UUUQ ′Δ=′−′= )( TVRPPRQ − ,)(

)(0 RPRP UUUU ′<′<′−′ )(0 RPRP UUUU <<

TVU ′Δ− )( T ′= heat of reaction at constant volume at temperature TV ,)( p

►► Constant pressure

W ∫ ′−′==P

RP VVppdV )(PRW − ∫

RRP VVppdV )(

PRPR WQ −− − RP UU −=PRPRQ −− RP

)( RPPR WWQ −−− RP UU −=

28

)( RPPR VVpQ ′−′−− RP UU ′−′=)( RPPR pQ − RP

PRQ − )( RPRP VVpUU ′−′+′−′=PRQ − )( RPRP VVpUU +

)()( RRPP VpUVpU ′+′−′+′=PRQ − )()( RRPP VpUVpU ++

HHH Δ=′−′= )(

PRQ

TpRP HHH ′Δ=−= ,)(

T ′= heat of reaction at constant pressure at TpH ′Δ− ,)(

29

ReactantsorProducts

or

or

or

or

Schematic plot of internal energy (U) or enthalpy (H) of reactants and products as a function of temperature

30

TpH ′Δ ,)( TVU ′Δ ,)(The difference between and

pvuh += <- ThermodynamicsFrom

TVTp UH ′′ Δ−Δ ,, )()( )()( ,, RPTRTP UUHH ′−′−−= ′′

TVRRPP UVpUVpU ′Δ−′+′−′+′= ,)()()(

TVRPRP UUUVVp ′Δ−′−′+′−′= ,)()()(

thereforecanceled out !

TVTp UH ′′ Δ−Δ ,, )()( )( RP VVp −=

31

If all the reactant and product species are ideal gasesp p g

TVTp UH ′′ Δ−Δ ,, )()( TnnR RP ′′−′= )(~

One of the products H2O, can be in the gaseous or liquid phase. The internal energy (or enthalpy) of the products will depend on the relative proportions of water in gaseous or liquid phase.

32

Internal energy difference between curvesgy

OfgHOHOvapHTVOliqHTV umUU )()( ′=Δ−Δ ′′ OfgHOHOvapHTVOliqHTV 2222 ,,,, )()(

OHm2

= mass of water in the products

u ′ i t l t T & f th d tOfgHu2 = internal energy at T & p of the products

The relationships applied for enthalpy

2 2 2 2, , , ,( ) ( )p T H Oliq p T H Ovap H O fgH OH H m h′ ′ ′Δ − Δ =

33

Products

Reactants ReactantsProducts

VaporVapor fuel

Liquid

Vapor Liquid Fuel

(a) (b)

Schematic plots of internal energy of reactants and products as a function of temperature. (a) Effect of water in products as either vapor or liquid.

(b) Effect of fuel in reactants as either vapor or liquid.

34

( ) p q

3.5.2 Enthalpies of Formation οfh~Δ

The enthalpy of formation of a chemical compoundf

is the enthalpy increase associated with the reaction ofis the enthalpy increase associated with the reaction of forming one mole of the given compound from its element in the thermodynamics standard [298.15 K (25°C), 1 atm]in the thermodynamics standard [298.15 K (25 C), 1 atm]

Products: ∑ Δ=~οο hnHProducts: ∑ Δ=

products,ifiP hnH

Reactants: ∑ Δ=reactants

,~ οο

ifiR hnH

35

οοRPT HHH −=Δ )( ⇒ ∑∑ Δ−Δ ifiifi hnhn οο ~~RPTp HHHΔ

0,)( ∑∑r

ifiP

ifi ,,

(enthalpy that increases)

from )()()( RPTVTP VVpUH −=Δ−Δ ′′ )()()( ,, RPTVTP VVpU

(already discussed previously) ( y p y)

)()()(00 RPTpTV VVpHU −−Δ=Δ )()()(00 ,, RPTpTV p

The case of ideal gasThe case of ideal gas

0)(~)()(00

TnnRHU RPTpTV −−Δ=Δ36

0,, )()()(00 RPTpTV

TABLE 3.2 Standard enthalpies of formation (25°C, 1 atm)Species State

MJ/kmol

O2 Gas 0

οfh~Δ

N2 Gas 0

H2 Gas 0

C Gas 0

CO2 Gas -393.52

H O Gas -241 83H2O 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

C H G 208 45

37

C8H18 Gas -208.45

C8H18 Liquid -249.35

3.5.3 Heating Values HVQ3.5.3 Heating Values

Heating value of a fuel is a magnitude of

HVQ

heat of reaction that is measured directly at a standard temperature (25°C or 77 °F) for complete combustion of

i f funit mass of fuel.

Reaction at constant pressure

)( HQ Δ−=0,)( TpHV HQ

PΔ=

Reaction at constant volumeReaction at constant volume

0,( )VH V V TQ U= − Δ

38

Higher heating value

OH formed in products is condensed to liquid phase

(HHV)

OH 2 formed in products is condensed to liquid phase

L h ti l (LHV)Lower heating value

OH 2 formed in products is in vapor phase

(LHV)

OH 2 formed in products is in vapor phase

39

HHV and LHV are related by

(constant pressure)

HHV and LHV are related by

OHhm

mQQ fg

OHLHVHHV pp 2)( 2+=

(constant pressure)

m f

(constant volume)

OHum

mQQ fg

f

OHLHVHHV vv 2)( 2+=

f

OH

mm

2 : ratio of mass of H2O produced to mass of fuel burnedfm

40

Heating values of fuels are measure in calorimeter

For gaseous fuels, using continuous-flow atmosphere pressure

g

calorimeter.

An entering fuel is saturated with water vapor and mixedAn entering fuel is saturated with water vapor and mixed with sufficient saturated air for complete combustion at reference temperature. p

For liquid and solid fuels, fuel is burned with oxygen under q ygpressure at constant volume in a bomb calorimeter.

41

Used for solid/liquid f lfuelsBurning under pressure at constant volumeObtain HHV

42

Example 3.2Liquid kerosene fuel of the heating value (d t i d i b b l i t ) f 43 2 MJ/k d(determined in a bomb calorimeter) of 43.2 MJ/kg and average molar CH / ratio of 2 is mixed with the,

t i hi t i i i t t 298 15 K C l l t thstoichiometric air requirement at 298.15 K. Calculate the enthalpy of the reactant mixture relative to the datum of zero enthalpy for C O N Hd t 298 15 Kzero enthalpy for C 2O 2N 2H, and at 298.15 K.,

In bomb, const volume {V 0

298.15

HHV V, TQ ( U)= − Δ = 43.2 MJ/kg

The combustion equation per mole of C can be written

222222 66.5).77.3(23 NOHCONOCH ++=++

43

6605)773(3 NOHCONOCH

kmolkmolkmol 66716071 ⎫⎫⎫

222222 660.5).77.3(2

NOHCONOCH ++=++

productskgkmolairkg

kmolfuelkgkmol

4.22166.7

4.207160.7

141

⎭⎬⎫

→⎭⎬⎫

+⎭⎬⎫

οVU )(Δ−The heating value given is at constant volume, = 43.2 MJ/kg

can be obtained, noting that the fuel is in the liquid phase:

οH )(Δ οTnnRU )(~)( +Δ=

οpH )(Δ

pH )(Δ ο TnnRU RpTV)()(

,−+Δ=

15.298)1607667(1031438243 3 ××+= −

14)160.766.7(103143.82.43 ×−×+−=

fuelkgMJ /1.4309.02.43 −=+−=

44

fg

The enthalpy of the products per kilogram of mixture is found from the enthalpies of formation (with H2O vapor):the enthalpies of formation (with H2O vapor):

οPH ∑ Δ=

productsifi hn ο

,~

products

4221)83.241(1)52.393(1 −+−

=οPH

4.221

2.87 /MJ kg= −

( )PH οΔ οοRPTp HHH −=Δ

0,)(οοο HHH )(Δ= pPR HHH )(Δ−=

⎟⎟⎞

⎜⎜⎛−−−=

fuelkgMJH R14.1.4387.2ο

AnsEnthalpy of the reactants per kilogram of mixture

⎟⎟⎠

⎜⎜⎝ productkgfuelkgR 4.221

..7.

145.0−= MJ/kg product

45

Anskilogram of mixture

6605)773(3 NOHCONOCH

kmolkmolkmol 66716071 ⎫⎫⎫

222222 660.5).77.3(2

NOHCONOCH ++=++

productskgkmolairkg

kmolfuelkgkmol

4.22166.7

4.207160.7

141

⎭⎬⎫

→⎭⎬⎫

+⎭⎬⎫

οVU )(Δ−The heating value given is at constant volume, = 43.2 MJ/kg

can be obtained, noting that the fuel is in the liquid phase:

οH )(Δ οTnnRU )(~)( +Δ=

οpH )(Δ

pH )(Δ ο TnnRU RpTV)()(

,−+Δ=

314 298.1543 2 8 3143 10 (7 66 7 160)−× + × ×43.2 8.3143 10 (7.66 7.160)221.4 221.4

= − × + × − ×

d tkMJ /724200570732 +46

productkgMJ /724.20057.073.2 −=+−=

The enthalpy of the products per kilogram of mixture is found from the enthalpies of formation (with H2O vapor):the enthalpies of formation (with H2O vapor):

οPH ∑ Δ=

productsifi hn ο

,~

products

4221)83.241(1)52.393(1 −+−

=οPH

4.221

2.87 /MJ kg= −

( )PH οΔ οοRPTp HHH −=Δ

0,)(οοο HHH )(Δ= pPR HHH )(Δ−=

724.287.2 +−=οRH

Ans

Enthalpy of the reactants per kilogram of mixture

R

146.0−= MJ/kg product

47

Ans

3.5.4 Adiabatic Combustion Process

-> adiabatic constant volume processdi b ti t tadiabatic constant pressure process

-> Constant volume -> UWQ Δ=Δ−Δ

∫P

∫−ΔR

pdVQ UΔ=adiabatic

{0 0

( )P RQ p V VΔ − −14243 UΔ=

adiabatic

UΔ 0=

048RP UU − 0=

Reactants Productsor Products

or

Adiabatic combustion process: constant-volume on U-Tdiagram and constant-pressure on H-T diagram

49

3 5 5 combustion efficiency3.5.5 combustion efficiency

Control volume surrounding engine

50

Consider a mass m which passes through the controlConsider a mass m which passes through the control volume, the net chemical energy release due to combustion iscombustion is

[ ] ⎟⎟⎞

⎜⎜⎛

Δ−Δ=− ∑∑ ~~)()( οοifiifiAPAR hnhnmTHTH[ ] ⎟⎠

⎜⎝

∑∑productsi,

,reactantsi,

,)()( ifiifiAPAR

where οifh ,

~Δ enthalpy of formation of speciesif , py p

51

The combustion efficiency cη c

The fraction of fuel energy supplied to the control volume around the engine which is released by combustion

THTH )()(

HVf

APARc Qm

THTH )()( −=η

energy releasedHVf Q

where HVf Qm is the amount of energy supplied

due to combustion

where HVf Qm is the amount of energy suppliedto the control volume around the engine which can be released by combustionreleased by combustion

52

Variation of engine combustion efficiency with fuel/air equivalence ratio

53

equivalence ratio

3.6 Thermal conversion efficiency tη

Efficiency which relates the actual work per cycle to amo nt of f el chemical energ released in comb stion process

cycleperworkActual

amount of fuel chemical energy released in combustion process.

releaseenergychemicalFuelyp

=tη

ccct Qm

WHW

THTHW

ηΔη =−=

−=

)()()( HVfcTAPAR QmHTHTHA

ηΔ )()()(

h fTherefore tcf ηηη =Fuel conversion efficiency

( b f !!)54

(seen before!!)

Example 3.3 The brake fuel conversion efficiency is 0.3. Themechanical efficiency is 0 8 The combustion efficiency is 0 95 Themechanical efficiency is 0.8. The combustion efficiency is 0.95. Theheat losses to coolant and oil are 60 kW. The fuel chemical energyentering the engine per unit time, is 190 kW. What percentage of thisg g p , p genergy become ?

a) brake work b) friction work c) heat lossesa) brake work b) friction work c) heat losses

d) Exhaust chemical energy e) Exhaust sensible energy

Analysis:

Fuel chemical energy per unit time

⎫⎧ h i l⎭⎬⎫

⎩⎨⎧

+++)5()4(

)3()2()1(sensiblechemicalenergyExhaustlossheatfrictionb PP

55

a) From bf ,η b

QmP

&=)

HVf Qm

bP kW571903.0 =×=

%30= AnsbP

mηig

b

PP

=b) From

P

fb

b

PPP+

=

578.0

fP+=

5757

fP kW25.14=

251456%5.7100

19025.14

=×= Ans

c) Heat loss = 60 kW

%63110060 Ans%6.31100190

=×= Ans

d) Exhaust chemical energy Fuel chemical energy

d) Exhaust chemical energy

HVfc Qm&)1( η−=

kW4.11190)94.01( =×−=due to combustion inefficiency

%6100190

4.11=×=

e) Exhaust sensible energy = 190 - (a + b + c + d)

4.116025.1457190 −−−−= 4.116025.1457190

kW35.47=

57%9.24100190

35.47=×= Ans

3.7 CHEMICALLY REACTING GAS MIXTURES

Depending on problem and portion of engine cycle, chemical reaction may:chemical reaction may:

1) to be very slow -> negligible effect on mixture i i (f )composition (frozen)

2) to be rapid > the composition remains in chemical2) to be rapid -> the composition remains in chemical equilibrium

3) rate-controlling -> determine how composition changes with timechanges with time

58

3.7.1 Chemical EquilibriumThe chemical reaction by which individual species in the burned gases react together, produce and remove species g g , p pat equal rates.

221 COOCO =+ 222

COOCO +

Consider a system undergoing a constant-pressure, constant temperature process.In the absence of work transfer, the first law gives

HQ Δ=Δ1st Law 0≤Δ−Δ STH2nd Law STQ Δ≤Δ Δ

43421G

0)( ≤ΔGor

59

0)( , ≤Δ TpG

Thus reactions can occur if Gibbs free energy GG <Thus reactions can occur if Gibbs free energy RP GG <

0)( , =Δ TpGAt equilibrium, ,pq ,

or RP GG =

Consider general reaction whose stoichiometry is given by

...... ++=++ mmllbbaa MMMM υυυυ

The mixture composition in equilibrium can be determined b ilib i t t Kby equilibrium constant, PK

GKlnοΔ

−=60TR

K P ~ln −=

++=++ MMMM υυυυ

The equilibrium constant for a specific reaction is obtained via

...... ++=++ mmllbbaa MMMM υυυυ

q pthe relation

∑=i

iPireactionP KK )(log)(log 1010 ν

reactionPK )( = equilibrium constant of reaction

iPK )( = equilibrium constant of formation of species iν = stoichiometric coefficientsiν = stoichiometric coefficients

Positive (+) for product species

61Negative (-) for reactant species

Ex 2221 COOCO =+ reaction r1 at 2,500 K2

r1reaction10 )(log pK ∑=i

iPi K )(log10ν

22)(log

21)(log1)(log1 101010 OPCOPCOP KKK −−=

PK can also be obtained from partial pressure

i

i

iP p

pK

ν

∏ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

0

...321 ννν

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛pp

pp

pp iii->

i p ⎠⎝ 0 000 ⎠⎝⎠⎝⎠⎝ ppp

Π = multiplicationΠ multiplication

ip = partial pressure of species i

62

( ) ( ) ii i ip pK ν ν νχ χ∑= =∏ ∏% %

0 0

( ) ( ) ii i ip i i

i i

Kp p

χ χ= =∏ ∏where iχ

~= mole fractionw e e iχ mole fraction

p0 = standard state pressure

If ∏∑ ==i

iPi

iiK νχν ~,0

gmolecK ( )gmole

3cmAn equilibrium constant, , based on concentrations

[ ]∏= iciMK ν[ ]∏

iic

Relation between CP KK & CP KK &∑= i iTRKK CPν)~( R~; -> universal gas const

63Hence if Cpi i KK ==∑ ,0ν

Example 3.4 In fuel-rich combustion product mixtures, equilibrium between the species CO H O CO and H isequilibrium between the species CO2, H2O, CO and H2 is often assumed to determine the burned gas composition. For φ 1 2 f C H i b ti d t d t iFor φ = 1.2, for C8H18-air combustion products, determinethe mole fractions of product species at 1,700 K

Analysis:

Th i l i h i f ll d “ hThe reaction relating these species often called “the water gas reaction”.

OHCOHCO +=+ OHCOHCO 222 +=+

64

Question

What is the chemical equation for a stoichiometric combustion of C H ?

22222188 N16.47OH9CO8)N773.3O(5.12HC ++→++

combustion of C8H18 ?

Remember that these gases are the products for stoichiometric combustion !!

φ

for stoichiometric combustion !!

The combustion of C8H18 with air for φ = 1.2 can be written as

3039)7733(5.12 NdHCOObHCONOHC ++++→++ 222222188 30.39)773.3(2.1

NdHcCOObHaCONOHC ++++→++

65

A carbon balance: 8=+ ca (2)

A hydrogen balance: 1822 =+ db (3)

A oxygen balance: 83202 ++ cba (4)

( )

A oxygen balance: 83.202 =++ cba (4)

Combustion products are in chemical equilibrium.Combustion products are in chemical equilibrium.

OHCOHCO 222 +=+

5100log =K 2363=K

This chemical reaction is called “water gas reaction”. At 1,700 K,

The equilibrium relation then gives

510.0log10 =PK 236.3=PK

(5)0 0

( ) ( ) ii i ip i i

i i

p pKp p

ν ν νχ χ∑= =∏ ∏% %

66

0 0i ip p

What is Kp for this reaction?

OHCOHCO 222 +=+

p

?=∑i

iνIn this case,i

01111 =++−−=∑i

iν

0p ∑ ∏ ∏

i

0

( ) i i ip i

i

pKp

ν νχ∑= ∏ % ∏=i

iPiK νχ~Thus

67

OHCOHCO 222 +=+ 222

1111222

~~~~~OHCOHCOiP

iK χχχχχν −−==∏

3039)7733(5.12 NdHcCOObHaCONOHC ++++→++

222i∏

222222188 30.39)773.3(2.1

NdHcCOObHaCONOHC ++++→++

a b c d

PCO n

a=

2

~χP

OH nb

=2

~χP

CO nc

=χ~P

H nd

=2

~χ

⎟⎞

⎜⎛⎟⎞

⎜⎛ bc

30.39++++= dcbanp

Therefore⎟⎞

⎜⎛⎟⎞

⎜⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

= PPp da

nb

nc

K

68

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

PP nd

na

⎟⎞

⎜⎛⎟⎞

⎜⎛ bc

⎞⎛⎞⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

= PPp d

nb

nc

Kd

cb=236.3 (6)

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

PP

p

nd

na ad

From the equations (2), (3), (4) and (6) we can solve for1 b 69 2 86 d 1 31a = 5.14, b = 7.69, c = 2.86, d = 1.31

3039++++ db

30.3931.186.269.714.5 ++++=

30.39++++= dcbanp

30.56=

69

Mole fractions of products become

091.030.56

14.5~2

===P

CO naχ Ans30.56Pn

137.0305669.7~

2===OH n

bχ Ans30.56Pn

051.0305686.2~ ===CO

cχ Ans30.56PCO n

023.031.1~ ===Hdχ Ans023.0

30.562P

H nχ

3039~

s

698.030.5630.39~

2==Nχ Ans

70

For a chemical reaction that is not in equilibrium, the processes are controlled by rates at which the actual reactions occurby rates at which the actual reactions occur.

For a chemical reaction

ddccbbaa MMMM νννν +=+

The reaction rate in forward (+) direction:The reaction rate in forward (+) direction:

[ ] [ ] [ ] [ ] baba

ca MMk

dtMd

MdtdR υυ+

+++ ==−=

dtdt

The reaction rate in backward ( - ) direction:

[ ] [ ] [ ] [ ] dcdc

ac MMk

dtMd

MdtdR νν−

−−− ==−=

71

k+ and k- are the rate constants in forward and reverse directions

The net rate of production of products or removal of reactants:The net rate of production of products or removal of reactants:

[ ] [ ] [ ] [ ] dcbadcba MMkMMkRR νννν −+−+ −=−

⎟⎠⎞

⎜⎝⎛−=

RTE

Ak Aexp A: pre-exponential factor

E ti ti⎠⎝ EA: activation energy

At equilibrium, 0=− −+ RR

[ ][ ] [ ][ ]dcba MMkMMk −+ =

[ ] [ ][ ] [ ] c

dc KMMMM

kk

ba

dc

==−

+

νν

νν equilibrium constant, based on concentration

(learned before !!)

72

[ ] [ ]ba MMk (learned before !!)

NNOON fk +→+ 1 NNOON +⎯→⎯+2

ONOON fk +⎯→⎯+ 22

Reaction rates of NO and N

[ ]NOdtd [ ][ ] [ ][ ]2221 ONkONk ff +=dt

[ ]Nd [ ][ ] [ ][ ]k N O k N O= −[ ]Ndt

[ ][ ] [ ][ ]1 2 2 2

consumedf fk N O k N O=

14243

73