Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 36
Combustion Reactions
Combustion Processes
Why do mechanical engineers have to know about
combustion? Consider a combustion chamber in a gas
turbine cycle,
Our current model
What really happens
Fuel input
Q in
3
2
Air from
compressor
2
Qin  m a  h3  h 2 
Air to
turbine
2
Air from
compressor
3
Combustion
products to
the turbine
How is this analyzed??
Combustion
• Fuels
– Stored chemical energy
• Combustion Reaction
– Transforms the chemical energy stored in the
fuel to thermal energy (heat)
• Goals of this section of the course
– Understand combustion chemistry
– Use combustion chemistry to determine the
heat released during a combustion process
• Heat of reaction
3
Combustion
The combustion of a fuel requires oxygen,
fuel + oxidant  products
Fuel
In the most general sense, a fossil fuel makeup is,
C H  S O N
The Greek letters signify the atomic composition of the fuel.
For example ...
C8H18  octane
C2 H5OH  ethanol (ethyl alcohol)
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Combustion
fuel + oxidant  products
Oxidant
The oxidant must contain oxygen. The most abundant ‘free’
source is atmospheric air. By molar percent, atmospheric air
is considered to be ...
Nitrogen
79%
Oxygen
21%
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For every mole of
oxygen involved in a
combustion reaction,
there are 79/21 = 3.76
moles of nitrogen.
Combustion
fuel + oxidant  products
Products (for fuels with no sulfur content)
Complete Combustion: CO2, H2O, and N2
Incomplete Combustion: CO2, H2O, N2, CO, NOx
Combustion with Excess Oxygen: CO2, H2O, N2, and O2
NOTE: Fuels containing sulfur have the potential of
introducing sulfuric acid into the product stream.
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Combustion Terminology
• Theoretical or Stoichiometric Air
– The amount of air required for complete
combustion of the fuel
• Determined by balancing the combustion reaction
• Excess or Percent Theoretical Air
– The amount of air actually used in the
combustion process relative to the
stoichiometric value
• Can cause incomplete combustion or excess oxygen
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Combustion Terminology
In many combustion processes, one of the parameters we
are interested in is how much air (or oxygen) is required per
unit quantity (moles or mass) of fuel.
Air-Fuel and Fuel-Air Ratios
 A / F mol 
moles of air
moles of fuel
 A / F mass   A / F mol
M air
M fuel
 F / Amol 
 F / Amass 
1
moles of fuel

 A / F mol moles of air
1
mass of fuel

 A / F mass mass of air
Equivalence Ratio
 F / Aactual

 F / Astoichiometric
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Equivalence Ratio and Products
• Stoichiometric (=1)
– CO2, H2O, N2
• Lean ( < 1 with T < 1800 R)
ME 322
– CO2, H2O, N2, O2
• Rich ( > 1 with T < 1800 R)
– CO2, H2O, N2, O2, CO, H2
• Rich ( > 1 with T > 1800 R)
– CO2, H2O, N2, O2, CO, H2, H,
O, OH, N, C(s), NO2, CH4
9
Advanced
courses
ME 422 & 433
Stoichiometric (Complete) Combustion
Stoichiometric Combustion of a General Fuel in Air
C H S O N +  0 (O2 + 3.76 N2 )  1CO2 +  2H2O +  3SO2 +  4N2
Atomic Balances
C:
H:
S:
O:
N:
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  1
  2 2
 3
  2 0  2 1  2  2 3
  2(3.76) 0  2 4
5 equations
5 unknowns
(0 through 4)
Lean Combustion
Lean Combustion of a General Fuel in Excess Air
C H S O N + 0 (O2 + 3.76 N2 )  1CO2 + 2H2O + 3SO2 + 4O2 + 5N2
 0   0 (PTA )
C:
11
  1
PTA = Percent Theoretical Air
expressed as a decimal
H:
  2 2
S:
  3
O:
  2 0  21   2  23  2 4
N:
  2(3.76) 0  25
6 equations
6 unknowns
(0 through 5)
Requires a stoichiometric
balance first (to get 0)
Example – Octane Combustion
Given: Gasoline (modeled as octane - C8H18) burns
completely in 150% theoretical air (or 50% excess air).
Find:
(a) the A/F ratios (mass and molar)
(b) the equivalence ratio
(c) the dew point of the products of combustion at assuming
that the products are at 1 atm
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Example – Octane Combustion
In order to calculate the air-fuel ratios and the equivalence
ratio, we need to know how much air is used in the
combustion reaction. This is determined by balancing the
combustion reaction.
In order to determine the dew point of the products, we need
to know the molar composition of the products. This is also
determined by balancing the combustion reaction.
Everything depends on the correct balance
of the combustion reaction!
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Example – Octane Combustion
Solution strategy ...
1. Balance the stoichiometric reaction to get 0
C8H18 +  0 (O2 + 3.76 N2 )  1CO2 +  2H2O +  4N2
2. Balance the reaction with 150% theoretical air
C8H18 + 1.5 0 (O2 + 3.76 N2 )  1CO2 + 2H2O 4O2 + 5N2
3. Calculate the required (A/F) ratios, the equivalence ratio,
and the dew point temperature of the products
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Example – Octane Combustion
Stoichiometric Reaction
C8H18 +  0 (O2 + 3.76 N2 )  1CO2 +  2H2O +  4N2
C:
H:
8  1
18  2 2
S:
0  3
O:
2 0  21  2  2 3
N:
2(3.76) 0  2 4
 2  9
  0  8  9 / 2  12.5
  4   3.7612.5  47
C8H18 + 12.5(O2 + 3.76 N2 )  8CO2 + 9H2O + 47N2
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Example – Octane Combustion
Combustion in 150% theoretical air
C8H18 + 1.5 0 (O2 + 3.76 N2 )  1CO2 + 2H2O 4O2 + 5N2
0   0 (PTA)  0  12.51.5  18.75
C:
8  1
H:
18  2 2
S:
0  3
O:
2 0  21   2  23  2 4
N:
2(3.76)0  25
 2  9
  4  18.75  8  9 / 2  6.25
 5   3.76 18.75  70.5
C8H18 + 18.75(O2 + 3.76 N2 )  8CO2 + 9H2O + 6.25O2 + 70.5N2
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Example – Octane Combustion
C8H18 + 12.5(O2 + 3.76 N2 )  8CO2 + 9H2O + 47N2
C8H18 + 18.75(O2 + 3.76 N2 )  8CO2 + 9H2O + 6.25O2 + 70.5N2
The molar (A/F) ratio can now be found ...
 A / F mol =
moles of air moles of O2 + moles of N2
=
moles of fuel
moles of fuel
 A / F mol ,stoich 
 A / F mol ,act 
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 0 1  3.76 
1
0 1  3.76 
1
 4.76 0  4.76 12.5  59.5
 4.760  4.76 18.75  89.25
Example – Octane Combustion
The mass-based (A/F) ratio can be found knowing the
molecular masses of the air and the fuel,
 A / F mass   A / F mol
The molecular mass of the air is,
M air
M fuel
M air  28.97
lbm
lbmol
Table C.13a
The molecular mass of the fuel is,
MWfuel   M C   M H   M S   M O   M N
MW fuel  8 12.01115 lbm/lbmol  18 1.00797 lbm/lbmol   114.23 lbm/lbmol
Table C.20
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Table C.20
Example – Octane Combustion
Now, the mass-based (A/F) can be found ...
 A / F mass ,stoich   59.5
 A / F mass ,act  89.25
28.97
 15.09
114.23
28.97
 22.63
114.23
Once the (A/F) ratios are determined, the equivalence ratio
can be found,

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 F / Aactual
 F / Astoich

1/  A / F act
1/  A / F stoich

1/ 89.25
 0.667 (molar based)
1/ 59.5

1/ 22.63
 0.667 (mass based)
1/15.09
Example – Octane Combustion
The dew point of the products is the temperature where the
water vapor condenses,
Tdp = Tsat at Pw (partial pressure of the water vapor)
yw 
Pw
P
 Pw  yw P
C8H18 + 18.75(O2 + 3.76 N2 )  8CO2 + 9H2O + 6.25O2 + 70.5N2
2
9
yv 

 0.096
1  2  3  4  5 8  9  0  6.25  70.5
Pw  yw P   0.0961 atm   0.096 atm  1.411 psia
 Tdp  Tsat  Pw   113.4F
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Example – Problem 15.42
Given: Combustion exhaust with 9.1% CO2, 8.9% CO,
82% N2, and no O2
Find:
a) fuel model CnHm
b) mass percent of carbon and hydrogen in fuel
c) molar air/fuel ratio and percent theoretical air (PTA)
d) dew point temperature at .106 MPa
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Example – Problem 15.42
STEP 1: Write balance equation using ORSAT data
CnHm + a(O2 + 3.76 N2)
 9.1 CO2 + 8.9 CO + bH2O + 82 N2
STEP 2: Solve for unknowns and write fuel model CnHm
n=?
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m=?
a=?
b=?
Example – Problem 15.42
STEP 3: Compute molar mass of fuel & mass composition
Mfuel = 18lbmolC/lbmolfuel*(12lbm/lbmolC)
+ 33lbmolH/lbmolfuel*(1lbm/lbmolH)
= 249 lbm/lbmolfuel
C: 18*(12)/249  87%
H: 33*(1)/249  13%
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Example – Problem 15.42
STEP 4: Calculate molar air/fuel ratio
21.8 * (1 + 3.76) /1 = 103.8 moles air / mole fuel
STEP 5: Write equation for stoichiometric combustion
C18H33 + 26.25(O2 + 3.76 N2)
 18 CO2 + 16.5 H2O + 98.7 N2
STEP 6: Find theoretical air
%TA = (21.8 / 26.75) * 100 = 83%
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Example – Problem 15.42
STEP 7: Find dew point of combustion products
# moles of H2O (in original equation) = 16.5
# moles of other combustion products = 100
# moles of all products (in original equation) = 116.5
Pw = .106 * (16.5/116.5) = .015 Mpa
Tsat (.015 MPa) = 54 C
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