lecture chemical reaction

1

Fuel & Advanced

Combustion

Lecture

Chemical Reaction

2

Ideal Gas Model

The ideal gas equation of state is:

TRnTM

RmmRTPV

where is the Universal Gas Constant (8.314 kJ/kmol K), is the molecular

weight and n is the number of moles.

R M

dTTcTh p )()(

dTTcTu v )()(

Specific enthalpy (units: kJ/kg)

Specific entropy (units: kJ/kg K) )ln()(),( o

o PPRTsTPs

(Po = 1 bar, so entropy at Po)

Specific internal energy (units: kJ/kg)

3

Ideal Gas Model for Mixtures

n

iimm

1

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

where are mass specific values (units: kJ/kg)

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

n

iii

n

iii hmmhHummuU

11

ii hu and

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

n

ii

ii x

m

mx

1

1 and

n

iii

n

i

iin

iii

n

i

ii hxm

hm

m

Hhux

m

um

m

Uu

1111

The mixture specific internal energy u and enthalpy h is:

4

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

n

iii

n

iii hnHunU

11

where are molar specific values (units: kJ/kmol) ii hu and

n

iinn

1

The total number of moles in the mixture is:


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

n

ii

ii y

n

ny

1

1 and

n

iii

n

iii hyhuyu

11

The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is:

5


The partial pressure of a component, Pi, in the mixture (units: kPa) is:

PyPP

P

RTPV

RTVP

n

ny ii

i

i

ii or

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

o

n

iii

o

ii PPRPPRsxs ln)/ln(1

o

n

iii

o

ii PPRyRsys lnln1

The mixture molecular weight, M, is given by:

i

n

ii

n

i

ii

n

ii

Myn

Mn

n

m

n

mM

11

1

6

Composition of Standard Dry Air

Air is a mixture of gases including oxygen (O2), nitrogen(N2), argon (Ar),

carbon dioxide (CO2), water vapour (H20)….

For combustion dry air is taken to be composed of 21% O2 and 79% N2

by volume (yO2=0.21, yN2=0.79).

76.321.0

79.0

2

2

2

2

2

2 O

N

O

tot

tot

N

O

N

y

y

n

n

n

n

n

n

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

The amount of water in moist air at temperature T is specified by the

specific humidity (w or the relative humidity (F) defined as follows:

10 )(

22 FFTP

P

m

m

sat

OH

air

OHw

kg/kmol 84.28)28(79.0)32(21.0

22221

NNOOi

n

iiair MyMyMyMMolecular weight of air is

7

Combustion Stoichiometry

If sufficient oxygen is available, a hydrocarbon fuel can be completely

oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen

is converted to water (H2O).

Elements cannot be created or destroyed, so

C balance: 3 = b → b= 3

H balance: 8 = 2c → c= 4

O balance: 2a = 2b + c → a= 5

OcHbCOaOHC 22283

# of moles species

The overall chemical equation for the complete combustion of one mole of

propane (C3H8) with oxygen is:

OHCOOHC 22283 435

Thus the stoichiometric reaction is:

8


Air contains molecular nitrogen N2, if products are at a “low” temperature

the nitrogen is not significantly affected by the reaction, it is considered inert.

222224

76.32

)76.3(4

NOHCONOHC

C balance: = b → b = H balance: = 2c → c = /2

O balance: 2a = 2b + c → a = b + c/2 → a = + /4

N balance: 2(3.76)a = 2d → d = 3.76a/2 → d = 3.76( + /4)

Example: The stoichiometric reaction of octane (C8H18) = 8 and = 18

22222188 4798)76.3(5.12 NOHCONOHC

22222 )76.3( dNOcHbCONOaHC

The complete reaction of a general hydrocarbon CH with air is:

9


Note above equation only applies to stoichiometric mixture

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

HC

NO

fuelii

airii

fuel

airs

MM

MM

Mn

Mn

m

mFA

22 476.3

4/

112

)2876.332(4

1

)/(

1/

s

sAF

FA

Substituting the respective molecular weights and dividing top and bottom

by one gets the following expression that only depends on the ratio of the

number of hydrogen atoms to hydrogen atoms (/) in the fuel.

Example: For octane (C8H18), / = 2.25 → (A/F)s = 15.1

(gasoline (A/F)s≈ 14.6

10

Fuel Lean Mixture

Fuel-air mixtures with more than stoichiometric air (excess air) can burn

With excess air you have fuel lean combustion

a fuel lean mixture has excess air, so g > 1

2222222

)76.3)(4

( eOdNOHCONOHC

g

At low combustion temperatures, the extra air appears in the products as

O2 and N2:

Above reaction equation has two unknowns (d, e) and we have two

atom balance equations (O, N) so can solve for the unknowns

11

Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.

With less than stoichiometric air you have fuel rich combustion, there is

insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.

2222222

)76.3)(4

( fHeCOdNOHCONOHC

g

Fuel Rich Mixture

Above reaction equation has three unknowns (d, e, f) and we only have

two atom balance equations (O, N) so cannot solve for the unknowns

unless additional information about the products is given.

a fuel rich mixture has insufficient air → g < 1

Get incomplete combustion where carbon monoxide (CO) and molecular

hydrogen (H2) also appear in the products.

12

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

stoichiometric, fuel lean, or fuel rich.

s

mixture

mixture

s

AF

AF

FA

FA

/

/

/

/f

Off-Stoichiometric Mixtures

stoichiometric f = 1

fuel lean f < 1

fuel rich f > 1

Products)76.3(4

22

NOHC

Stoichiometric mixture:

Off-stoichiometric mixture:

Products)76.3(4

122

NOHC

f

fg

1

13

Off-stoichiometric Conditions

Example: Consider a reaction of octane with 10% excess air, what is f?

Stoichiometric : 22222188 4798)76.3(5.12 NOHCONOHC

10% excess air is: 222222188 98)76.3)(5.12(1.1 bNaOOHCONOHC

O balance: 1.1(12.5)(2) = 16 + 9 + 2a → a = 1.25,

N balance: 1.1(12.5)(3.76)(2) = 2b → b = 51.7

Other terminology used to describe how much air is used in combustion:

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

1.1 )76.3)(4

( 2283 g

g NOHC → mixture is fuel lean

91.01/)76.4)(5.12(1.1

1/)76.4(5.12

/

/

mixture

s

FA

FAf 1.1

91.

11

f

14

First Law Analysis for Reacting System

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

na moles of air to produce np moles of product:

PnAnFn paf

Applying First Law with state 1 being the reactants at P1, T1 and state 2

being products at P2, T2:

Reactants Products

Reactants Products

Reaction

State 1 State 2

Q

W

15

RRii

Ppii

RP

ThnThnQ

HH

HH

VPUVPU

)()(

)()(

12

111222

First Law Analysis for Reacting System

)()( 121221 VVPUUQ

WUQ

16

Enthalpy of Reaction

Consider the case where the final temperature of the products is the same as

the initial temperature of the reactants (e.g., calorimeter is used to measure Q).

Reaction Q

W

P1=P2=Po

T1=T2=To

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

reaction, HR ,

fuel of kmolor kgper kJ :units )()(

)()(

Roii

Poii

RRii

Ppii

R

ThnThn

ThnThn

HQ

To

17

The maximum amount of energy is released from a fuel when reacted with a

stoichiometric amount of air and all the hydrogen and carbon contained in the

fuel is converted to CO2 and H2O

Heat of Combustion

222224

76.32

)76.3(4

NOHCONOHC

HR = HP – HR < 0 (exothermic)

HR = HP – HR > 0 (endothermic)

This maximum energy is commonly referred to as the heat of combustion, or

the heating value, given per unit mass of fuel or air

18

Heat of Combustion

There are two possible values for the heat of combustion depending on

whether the water in the products is taken to be saturated liquid or vapour.

Tp

T

S

hg hf

From steam tables:

hfg = hg – hf > 0

The term higher heat of combustion and heating value (HHV) is used

when the water in the products is taken to be in the liquid state (hH20 = hf )

The term lower heat of combustion and heating value (LHV) is used

when the water in the products is taken to be in the vapour state (hH20 = hg )

19

0

h(kJ/kg fuel)

T

Products with H2O (g)

Reactants

Products with H2O (l)

hfg,H2O per kg fuel

hlow

hhigh

Heat of Combustion, graphical

298K

20 Source: Pulkrabek

21

Fuel A/F Energy

density

(MJ/L fuel)

Specific

energy

(MJ/kg air)

Gasoline*

14.6 32 2.9

Butanol 11.2 29.2 3.2

Ethanol 9.0 19.6 3.0

Methanol

Hydrogen

6.5

34.3

16

**

3.1

4.1

* Diesel about 10% more energy per volume than gasoline

** liquefied hydrogen has roughly ¼ the energy density of gasoline

22

Heat of Formation

Consider the following reactions taking place at atmospheric pressure and

with TP = TR = 298K

222

2222

/ 000,394 )()()(

/ 000,286 )()()(2/1

COkmolkJQgCOgOsC

OHkmolkJQlOHgHgO

Values for standard heat of formation for different species are tabulated

In these reactions H2O and CO2 are formed from their elements in their

natural state at standard temperature and pressure (STP) 1 atm and 298K.

ofh

kmolkJQh

kmolkJQh

o

f, CO

o

Of, H

/ 000,394

/ 000,286

2

2

Reactions of this type are called formation reactions and the corresponding

measured heat release Q is referred to as the standard heat of formation

( ) so:

23

Heat of Formation for Different Fuels

24

Enthalpy Scale for a Reacting System

By international convention, the enthalpy of every element in its natural state

(e.g., O2(g), N2(g), H2(g), C(s)) at STP has been set to zero

0)298,1( o

fhKatmhi.e,

Consider the following identity:

)]298,1(),([)298,1(),( KatmhTPhKatmhTPh

)]298,1(),([ ),( , KatmhTPhhTPh ii

o

ifi

sensible enthalpy chemical enthalpy 298K T

K ip dTc298 ,

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

25

-14000

-5000

-9000

298 2800

H2O

CO2

Temperature K

Enthalpy (kJ/kg)

The data is also found in the JANNAF tables provided at course web site

,

o

ifh

27

Adiabatic Flame Temperature

Consider the following adiabatic constant pressure process:

For a given reaction where the ni’s are known for both the reactants and the

products, Ta can be calculated explicitly.

Fuel

Air Products

(Ta )

Reactants

(T1)

)()(

0)()(

1

Rii

Paii

RRii

Ppii

ThnThn

ThnThnQ

For a constant pressure process, the final products temperature, Ta, is known

as the adiabatic flame temperature (AFT).

28


R

iiP

aii ThnThn )()( 1

P R

o

ifi

o

ifiR

T

ipiP

T

ipi hnhndTcndTcn ia

,,298 ,298 ,OR

R

ii

o

ifiP

iai

o

ifi KhThhnKhThhn )298()()298()( 1,,

o

RHSensible heat of reactants

(equal to 0 if T1 = 298K)

Sensible heat of products

P R

o

ifi

o

ifiR

iiiP

iaii hnhnKhThnKhThn ,,1 )298()()298()(

29

o

HCf

o

OHf

o

COf

NaNOHaOHCOaCO

hhh

KhThKhThKhTh

18822

222222

,,, 98

)298()(47)298()(9)298()(8

22222188 4798)76.3(5.12)( NOHCONOlHC

Consider constant pressure complete combustion of stoichiometric liquid

butane-air initially at 298K and 1 atm

Look up enthalpy of formation values and iterate Ta → 2410K

Adiabatic Flame Temperature, example

P R

o

ifi

o

ifiR

iiiiP

aii hnhnKhThnKhThn ,,1 )298()()}298()({

022 ,, o

Nf

o

Of hhNote:

30

Ta,


with products at equilibrium

31

Now consider octane air with 10% excess air

Look up values and iteration gives Ta = 2262K

Adiabatic Flame Temperature, example

222222188 7.5125.198)76.3)(5.12(1.1 NOOHCONOHC

o

HCf

o

OHf

o

COfNaN

OaOOHaOHCOaCO

hhhKhTh

KhThKhThKhTh

1882222

222222

,,, 98 )298()(7.51

)298()(.251 )298()(9)298()(8

HR same as for f =1

Excess air adds 6 moles of diatomic molecules (O2 and N2) into the

products that does not contribute to heat release just soaks it up.

32


with products at equilibrium

nitromethane

octane

ethanol

hydrogen

33

Constant Volume AFT

Reaction Q

W

Consider the case where the piston is fixed and the cylinder is perfectly

insulated so the process is adiabatic (Q = 0)

)()(

0)()(

1

Rii

Paii

RRii

Ppii

TunTun

TunTunQ

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

T) R)(Th(nTRThnR

iiP

aii 1))((

34

p Riai

P R

o

ifi

o

ifiR

iiiP

iaii

TRnTRn

hnhnKhThnKhThn

1

,,1

)298()()298()(

Extra term compared to constant pressure AFT ( term > 0)

Constant Volume AFT

R

iii

o

ifiP

iiai

o

ifi TRKhThhnTRKhThhn )298()()298()( 1,,

The AFT for a constant volume process is larger than for a constant

pressure process.

The AFT is lower for constant pressure process since there is Pdv work

done

35

i

a

R

p

i

CV

R

p

R

p

R

p

p

pp

R

RR

PR

T

T

n

n

P

P

T

T

n

n

P

P

P

TRn

P

TRn

VV

Constant Volume Combustion Pressure

Assuming ideal gas behaviour:

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

22222188 4798)76.3(5.12)( NOHCONOlHC

06.15.60

64

)76.4(5.121

4798

R

p

n

n

36

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

4798)76.3(5.12 22222188 NOHCONOHC

1.8604.706.1298

2266

5.60

64

i

a

R

p

i

CV

T

T

n

n

P

P

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

88.11)76.3(5.0 22222 NOHNOH

8.60.885.0298

2383

38.3

88.2

i

a

R

p

i

CV

T

T

n

n

P

P

Engine Fuel Comparison

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

2222262 2.1332)76.3(5.3 NOHCONOOHC

5.737.702.1298

2197

7.17

2.18

i

a

R

p

i

CV

T

T

n

n

P

P

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