Fuel & Advanced Combustion

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1

Fuel & Advanced Combustion

Lecture

Chemical Reaction

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2

Ideal Gas Model

The ideal gas equation of state is:

T R n M T

m R mRT

PV  



 



 



where is the Universal Gas Constant (8.314 kJ/kmol K), is the molecular weight and n is the number of moles.

R M

 c T dT T

h( ) _p( )

 c T dT T

u( ) _v( )

Specific enthalpy (units: kJ/kg)

Specific entropy (units: kJ/kg K) s(P,T)  s^o(T) Rln(P P_o) (P_o = 1 bar, s^o entropy at P_o) Specific internal energy (units: kJ/kg)

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3

Ideal Gas Model for Mixtures

 

 n i mi

m

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

i i i

n

i miui H mh m h

mu U

1 1

i

i h

u and

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

  

 n

i i

i

i x

m x m

1

and



 



 



    

n

i i i

n i

i n i

i i i

n i

i

i x h

m h m m

h H u

m x u m m

u U

1 1

The mixture specific internal energy u and enthalpy h is:

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4

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

 



n

i i i

n

i niui H n h

U

1 1

where are molar specific values (units: kJ/kmol) ui and hi

 

 n i ni

n

1

The total number of moles in the mixture is:

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

  

 n

i i

i

i y

n y n

1

and

 



n

i i i

n

i yiui h y h

u

1 1

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

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5

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

P y P P

P PV RT

RT V P n

y n ⁱ _i _i

i i

i    or 

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

  

_o



n

i i i

o i

i s R P P R P P

x

s ln( / ) ln

1

  

 

  ^

o

^

n

i i i

o i

i s R y R P P

y

s ln ln

1

  

 

The mixture molecular weight, M, is given by:

i n

i i

n i

i i n

i i

M n y

M n n

m n

M m     



  



1 1

1

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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 (y_O2=0.21, y_N2=0.79).

76 . 21 3

. 0

79 . 0

2 2

2

2     

O N O

tot tot

N O

N

y y 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:

1 0

) (

²

2 F   F 

 P T

P m

m

sat O H air

O

w H

kg/kmol 84

. 28 )

28 ( 79 . 0 ) 32 ( 21 . 0

2 2

2

1 2













 

  i O O N N

n

i i

air y M y M y M

Molecular weight of air is M

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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

O cH bCO

aO H

C₃ ₈  ₂  ₂  ₂

# of moles species

The overall chemical equation for the complete combustion of one mole of propane (C₃H₈) with oxygen is:

O H CO

O H

C₃ ₈ 5 ₂ 3 ₂ 4 ₂ Thus the stoichiometric reaction is:

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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.

2 2

2 3.76 4

) 2 76

. 3

4 (O N CO H O N

H

C 



 

 





 



 

 

      





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

 

₂ ₂ ₂ ₂ ₂

18

8H 12.5 (O 3.76N ) 8CO 9H O 47N

C     

2 2

2 3.76 )

(O N bCO cH O dN

a H

C_ _     

The complete reaction of a general hydrocarbon C_H_ with air is:

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9

Note above equation only applies to stoichiometric mixture The stoichiometric mass based air/fuel ratio for C_H_ fuel is:

   

 

_C _H

N O

i fuel i

i air i

fuel air

s M M

M M

M n

M n m

F m

A  

 





 

 

 



 

 

 

 

 ² 3.76 4 ²

/ 4

 

¹

12

) 28 76 . 3 32 4 (

1 )

/ ( / 1







 



 

 



  





s

s F A

F A

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) ≈ 14.6

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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

2 2

2

2 3.76 ) 2

4)(

( O N CO H O dN eO

H

C_ _ 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

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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.

2 2

2

2 3.76 ) 2

4)(

( O N CO H O dN eCO fH

H

C_ _ 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.

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The equivalence ratio, f, is commonly used to indicate if a mixture is stoichiometric, fuel lean, or fuel rich.

 

   



^mixture



_s

mixture s

A F

A F F

A

F A

/ / /

/ 

f 

Off-Stoichiometric Mixtures

stoichiometric f = 1 fuel lean f < 1 fuel rich f > 1

Products )

76 . 3

4 ( ²  ² 



 

 

 O N

H

C_ _   Stoichiometric mixture:

Off-stoichiometric mixture:

Products )

76 . 3 4 (

1

2

2  



 

 

 O N

H

C  

 f

 g  f¹

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13

Off-stoichiometric Conditions

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

Stoichiometric : C₈H₁₈ 12.5(O₂ 3.76N₂) 8CO₂ 9H₂O47N₂

10% excess air is: C₈H₁₈ 1.1(12.5)(O₂ 3.76N₂) 8CO₂ 9H₂O aO₂ bN₂ 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)(

( ₂ ₂

8

3H g    O  N g 

C → mixture is fuel lean

 

1.1(12.5)(4.76)/1 ⁰^.⁹¹ 1

/ ) 76 . 4 ( 5 . 12 /

/  



mixture s

F A

F

f A ₁_.₁

91 .

1

1  

f

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14

First Law Analysis for Reacting System

Consider a constant pressure process in which n_f moles of fuel react with n_a moles of air to produce n_p moles of product:

P n A

n F

n_f  _a  _p

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

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15



 



















R i i R

P i i p

R P

T h n T

h n Q

H H

V P U

) ( )

(

) (

1 2

1 1 1

2 2 2

First Law Analysis for Reacting System

) (

)

( ₂ ₁ ₂ ₁

2

1 U U P V V

Q

W U

Q

















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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=P_o T₁=T₂=T_o

The heat released under this situation is referred to as the enthalpy of reaction, H_R,

fuel of

kmol or

kg per kJ : units

) ( )

(

) ( )

(















R i i o

P i i o

R i i R

P i i p

R

T h n T

h n

T h n T

h n H Q

T_o

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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

2 2

2 3.76 4

) 2 76

. 3

4 (O N CO H O N

H

C 



 

 





 



 

 

      





H_R = H_P – H_R < 0 (exothermic)

H_R = H_P – H_R > 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

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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.

T_p T

S h_g

h_f From steam tables:

h_fg = h_g – h_f> 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 (h_H20 = h_f) 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 (h_H20 = h_g)

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19 0

h(kJ/kg fuel)

T Products with H2O (g)

Reactants

Products with H2O (l)

h_fg,H2O per kg fuel

h_low

h_high

Heat of Combustion, graphical

298K

(20)

Source: Pulkrabek 20

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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

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22

Heat of Formation

Consider the following reactions taking place at atmospheric pressure and with T_P = T_R= 298K

2 2

2

2 2

/ 000 , 394

)

( )

(

/ 000 , 286

) ( )

( )

( 2 / 1

CO kmol kJ

Q g

CO g

O s

C

O H kmol kJ

Q l

O H g

H g

O

















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.

o

hf

kmol kJ

Q h

kmol kJ

Q h

o f, CO o

O f, 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:

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23

Heat of Formation for Different Fuels

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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

( atm K  h_f^o  i.e, h

Consider the following identity:

)]

298 , 1 ( )

, ( [ ) 298 , 1 ( )

,

(P T h atm K h P T h atm K

h   

)]

298 , 1

( )

, ( [ )

,

(P T h _, h P T h atm K

h_i  _f^o_i  _i  _i

sensible enthalpy

chemical enthalpy 298K  ₂₉₈^T _K c_p_,_idT Therefore, the enthalpy of the i’th component in a mixture is:

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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

i

hf

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26

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27

Adiabatic Flame Temperature

Consider the following adiabatic constant pressure process:

For a given reaction where the n_i’s are known for both the reactants and the products, T_a can be calculated explicitly.

Fuel Air

Products (T_a) Reactants

(T₁)

) ( )

(

0 ) ( )

(

 1

 

 



R i i P i i a

R i i R P i i p

T h n T

h n

T h n T

h n Q

For a constant pressure process, the final products temperature, T_a, is known as the adiabatic flame temperature (AFT).

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28

Adiabatic Flame Temperature



 

R i i P nihi(Ta) n h (T )

₁



 

  

  

P R

o i f i o

i f i R

T

i p i

P

T

i p

i c dT n c dT n h n h

n ₂₉₈^a _, ₂₉₈ⁱ _, _, _, OR

 

 

^

   

     

R i i

o i f

P i a i i

o i f

i h h T h K n h h T h K

n _, ( ) (298 ) _, ( ₁) (298 )

o

HR



Sensible heat of reactants (equal to 0 if T₁ = 298K) Sensible heat of products



^



^ ^



^



^_^^ ^^ _^



P R

o i f i o

i f

R i i i i

P ni hi(Ta) hi(298K) n h (T₁) h (298K) n h _, n h _,

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29

     



f^oCH



o O H f o

CO f

N a

N O

H a

O H CO

a CO

h h

h

K h

T h K

h T

h K

h T

h

1 8 8 2

2

2 2

, ,

, 9

8

) 298 ( )

( 47

) 298 ( )

( 9

) 298 ( )

( 8



















2 2

2 18

8H (l) 12.5(O 3.76N ) 8CO 9H O 47N

C     

Consider constant pressure complete combustion of stoichiometric liquid butane-air initially at 298K and 1 atm

Look up enthalpy of formation values and iterate T_a → 2410K Adiabatic Flame Temperature, example

 

_  _^



 



 

P R

o i f i o

i f i

R i i i

i

P ni{hi(Ta) h (298K)} n h (T₁) h (298K) n h _, n h _,

2 0

2 ,

,  _f^o_N 

o O

f h

Note: h

(30)

30 T_a,

Adiabatic Flame Temperature with products at equilibrium

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31

Now consider octane air with 10% excess air

Look up values and iteration gives T_a = 2262K

Adiabatic Flame Temperature, example

2 2

18

8H 1.1(12.5)(O 3.76N ) 8CO 9H O 1.25O 51.7N

C      

     

  

f^oC H



o O H f o

CO f N

a N

O a

O O

H a

O H CO

a CO

h h

h K

h T

h

K h

T h

K h

T h

K h

T h

18 8 2

2 2

2

2 2

, ,

, 9

8 )

298 ( )

( 7

. 51

) 298 ( )

( .25

1 ) 298 ( )

( 9

) 298 ( )

( 8























H_R 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.

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32

Adiabatic Flame Temperature with products at equilibrium

nitromethane

octane ethanol hydrogen

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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

 

 



R i i

P i i a

R i i R

P i i p

T u n T

u n

T u n T

u n Q

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

T) R ) (T h ( n T

R T

h n

R i i

P i i a    

 ( ( ) ) ₁

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34

   

 









 



 

 





p i a R i

P R

o i f i o

i f

R i i i i

P i i a i

T R n T

R n

h n h

n K

h T

h n K

h T

h n

1

, ,

1

) 298 ( )

( )

298 ( )

(

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

 

 

 















   

R i i i

o i f

P i a i i i

o i f

i h h T h K RT n h h T h K RT

n _, ( ) (298 ) _, ( ₁) (298 )

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

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35



 







 





 



 







 







i a R

p i

CV R

p R

p

p p p R

R R

P R

T T n

n P

P T

T n

n P

P

P T R n P

T R n

V V

Constant Volume Combustion Pressure

Assuming ideal gas behaviour:

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

2 2

2 18

8H (l) 12.5(O 3.76N ) 8CO 9H O 47N

C     

06 . 5 1 . 60

64 )

76 . 4 ( 5 . 12 1

47 9

8  



 

R p

n n

(36)

36

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

47 9

8 )

76 . 3 (

5 .

12 ₂ ₂ ₂ ₂ ₂

18

8H O N CO H O N

C     

1 . 8 604 . 7 06 . 298 1

2266 5

. 60

64    



 







 



 



 







 



 

i a R

p i

CV

T T n

n P

P

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

88 . 1 1

) 76 . 3 (

5 .

0 ₂ ₂ ₂ ₂

2 O N H O N

H    

8 . 6 0 . 8 85 . 298 0

2383 38

. 3

88 .

2    



 







 



 



 







 



 

i a R

p i

CV

T T n

n P

P

Engine Fuel Comparison

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

2 2

2 6

2H O 3.5(O 3.76N ) 2CO 3H O 13.2N

C     

5 . 7 37 . 7 02 . 298 1

2197 7

. 17

2 .

18    



 







 



 



 







 



 

i a R

p i

CV

T T n

n P

P