Thermochemistry of fuel air mixtures
Dr. Primal Fernando
primal@eng.fsu.edu
Ph: (081) 2393608
1
Combustion process
• Thermodynamic aspects of particular type process involving
chemical reactions, is called combustion
• Usually occurs between fuel and an oxygen carrier (air)
• Energy stored in the bonds between constituent atoms of fuel
and air (form of internal energy) and in the combustion process
it will transformed to new molecules of lower energy level
combustion products plus release heat (exothermic reaction)
2
Combustion process
•
•
•
•
Controls the engine power
Efficiency
Controls the emissions
Different for SI and CI engines
3
Flames
• A flame is a combustion reaction propagate subsonically through
space; motion of gas relative to unburn gas is important.
• The existence of flame motion implies that the reaction is confined to a
zone which is small in thickness compared to the engine combustion
chamber.
• The reaction zone is usually called the flame front
• Flames can be categorized as premixed and diffusion flame (mixed
together at same place where the reaction takes place)
• Flames also categorized as laminar (mixing and transport done by
molecular process) and turbulent (enhanced by eddies and lumps)
• Flames also categorized by whether the flow is steady or unsteady
4
Fuel - fossil fuels mainly consists of H and C
By mass
Dry Coal
Anthracite
Bituminous
Lignite
C
90.27
74.00
56.52
H
3.00
5.98
5.72
O
2.32
13.01
31.89
N+S
1.44
2.26
1.62
Ash
2.97
4.75
4.25
By mass
C
H
S
Motor
Petrol
85.5
14.4
0.1
Vaporizing
oil
86.8
12.9
0.3
Kerosene
86.3
13.6
0.1
Diesel oil
(gas oil)
86.3
12.8
0.9
Light Fuel
oil
86.2
12.4
1.4
Heavy
Fuel oil
86.1
11.8
2.1
By volume
Coal gas
Producer gas
Blast furnace
North Sea gas
H2
49.4
12.0
2.0
-
CO
18.0
29.0
27.0
-
CH4
20.0
2.6
93.0
CnHm
C4H8 2.0
C2H4 0.4
C2H6 4.8
O2
0.4
-
N2
6.2
52.0
60.0
2.0
CO2
4.0
4.0
11.0
0.2
5
Chemical equation and conservation of mass
Consider a simple equation
C  O2  CO2
1 kmol C  1 kmolO2  1 kmolCO2
12 kg C  32 kg O2  44 kg CO2
0 Vol C  1 Vol O2  1 Vol CO2
Note: all gasses occupy equal volume for kmol when they are at same
pressure and temperature (exactly true for perfect gases, but for other
gasses substantially true). Volume occupied by liquid and solids are
negligibly small compared to gasses.
6
Chemical equation and conservation of mass
C  O2  CO2
1 kmol C  1 kmolO2  1 kmolCO2
12 kg C  32 kg O2  44 kg CO2
If insufficient O2 presents
1
C  O2  CO
2
1
CO  O2  CO2
2
0 Vol C  1 Vol O2  1 Vol CO2
1
O2  H 2 O
2
1
1 kmol H 2  kmol O2  1 kmolH 2 O
2
H2 
2 kg H 2  16kg O2  18 kg H 2O
1
1 Vol H 2  Vol O2  1 Vol H 2 O (Vapour)
2
 0 Vol H 2 O (liquid)
7
Necessary Oxygen is mainly obtained by
mixing fuel with are
Volumetric analysis %
Gravimetric analysis %
O2
N2
21
79
(20.95) (78.09)
23.3
76.7
Note: Molar mass of N2 is 28 kg/kmol (28.16),
and that for air 29 kg/kmol (28.962)
8
Example
Determine the stoichiometric air/fuel ratio for a petrol approximating to
hexane C6H14. Hence deduce the chemical equation if the petrol is burnt
in 20 percent excess air, and the wet volumetric analysis of the products
• If all the water vapor is present
•If products are cooled to an atmospheric pressure and temperature of 1
bar and 15 °C.
•Determine also the dry volumetric analysis.
•Estimate the chemical equation if only 80% of the air required for
stoichiometric combustion is provided
The partial pressure of saturated water vapor at 15 °C is 0.01704 bar
9
Solution
Products
C6 H14  O2  CO2  H 2O
C 6 H 14 
19
O2  6CO2  7 H 2 O
2
C6 H14  9.5 O2  6CO2  7H 2 O
86kg C6 H14  304kg O2  264kg CO2  126kg H 2O
23.3 kg of O2 containin  100kg of air
320 kg of O2 contain in 
100
 304 kg of air
23.3
100
 304kg
st ochiometric air/fuel ratio  23.3
 15.17
86kg
10
Solution
Wet volumetric analysis of the Products including N2
C 6 H 14  9.5 O2 
79
 9.5 N 2  6CO2  7 H 2 O         
21
79


C6 H 14  1.29.5 O2   9.5 N 2   6CO2  7 H 2 O         
21


79
79


C6 H 14  1.29.5 O2   9.5 N 2   6CO2  7 H 2 O  0.2  9.5 O2  1.2   9.5 N 2
21
21


1Vol C 6 H 14  1.2  9.5Vol O2  1.2 
79
79
 9.5 Vol N 2  6Vol CO2  7 Vol H 2 O  0.2  9.5Vol O2  1.2   9.5 Vol N 2
21
21
Amount-of-substance in the product
 6CO2  7 H 2 O  0.2  9.5 O2  1.2 
 6  7  0.2  9.5  1.2 
79
 9.5 N 2
21
79
 9.5  57.79 kmol or  57.79 Vol
21
11
Solution
1Vol C 6 H 14  1.2  9.5Vol O2  1.2 
79
79
 9.5 Vol N 2  6Vol CO2  7 Vol H 2 O  0.2  9.5Vol O2  1.2   9.5 Vol N 2
21
21
Amount-of-substance in the product
 6CO2  7 H 2 O  0.2  9.5 O2  1.2 
 6  7  0.2  9.5  1.2 
79
 9.5 N 2
21
79
 9.5  57.79 kmol or  57.79 Vol
21
The wet volumetric analysis
6
CO2 
 100  10.38%
57.5
N2 
79
 9.5
21
 100  74.22%
57.79
1.2 
7
H 2O 
 100  12.11%
57.79
O2 
0.2  9.5
 100  3.29%
57.79
12
Solution
If products are cooled to an atmospheric pressure and temperature of 1
bar and 15 °C.
Part of water will be condensed, if new amount of water is y,
total substance is = 50.79 + y
Volume fraction=Mole fraction=partial pressure/total pressure
molefraction
0.01704
y

1
50.79  y
y  0.88km ol
totalvolume 50.79  y  50.79  0.88 51.67
Above volumetric analysis repeats base on the total volume of 51.67
CO2 11.61%; H 2O 1.7%;O2 3.68%; N2 83.01%
13
Solution
Determine also the dry volumetric analysis
Analysis is done by assuming no water present, then the amount of
substance becomes = 50.79
Above volumetric analysis repeat base on total volume of 50.79
CO2 11.81%; O2 3.74%; N2 84.45%
14
Solution
Chemical reaction with insufficient air (80%)
79
79


C6 H 14  1.29.5 O2   9.5 N 2   6CO2  7 H 2 O  0.2  9.5 O2  1.2   9.5 N 2
21
21


When insufficient O2 is given there will be unburned C and H2. H2, However, has a
greater affinity for O2. If mixture is not too rich in fuel, it is reasonable to assume
that all the H2 will be burnt. Some of C will be burnt to CO and other to CO2.
79
79


C6 H14  0.89.5 O2   9.5 N 2   aCO2  bCO  7 H 2O  0.8   9.5 N 2
21
21


6  ab
0.8  9.5  2  a  2  b  7
a  2.2 and b  3.8
79
79


C6 H14  0.89.5 O2   9.5 N 2   2.2CO2  3.8CO  7 H 2O  0.8   9.5 N 2
21
21


15
General combustion stoichiometry
79
79


C6 H14  9.5 O2   9.5 N 2   6CO2  7 H 2O   9.5 N 2
21
21


79
79


C6 H 14  9.5 O2 
N 2   6CO2  7 H 2 O   9.5 N 2
21 
21

79.05
79.05


C a H b  .......... O2 
N 2   ............CO2  .........H 2 O 
 ............ N 2
20.95 
20.95

Ca Hb  ..........O2  3.773 N2   ............CO2  .........H 2O  3.773 ............ N2
b
b
b




Ca H b   a   O2  3.773 N 2  aCO2  H 2O  3.773 a   N 2
4
2
4


16
General combustion stoichiometry
b
b
b


Ca H b   a  O2  3.773 N 2   aCO2  H 2O  3.773 a   N 2
4
2
4


b

 a   O2  3.773 N 2 
4
 A

  
Ca H b
 F s
b

 a   32  3.773 28.16
4
 A 

 
12.011 a  1.008 b
 F s
Take, y=b/a = ratio of H2 to C
34.56(4  y)
 A
  
 F  s 12.011 1.008y
17
Energy balance
QR-P
Initial state reactants
TR, PR, VR, UR
WR-P
Combustion process; heat and
work transfer interaction
Final state products
TP, PP, VP, UP
Systems changes from reactants to products (since mass constant, can
apply first law for a close system)
P
QRP  WRP  U P  U R
WR  P   PdV  P(VP  VR )
R
QRP  PVP  PVR  U P  U R  PVP  U P  PVR  U R  H P  H R
 H P,T  heat of reactionat constantpressure
18