Stoichiometric Analysis of Fuel Combustion

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CACHE Modules on Energy in the Curriculum
Hydrogen as a Fuel
Module Title: Stoichiometric Analysis of Fuel Combustion
Module Author: Dr. Daniel A. Crowl
Author Affiliation: Michigan Technological University
Course: Stoichiometry, process design
Text Reference: Felder and Rousseau, Elementary Principles of Chemical Processes, 3rd ed.,
New York: John Wiley and Sons, 2005.
Concepts: Stoichiometric fuel ratios
Problem Motivation:
In order to obtain the best combustion in an internal combustion engine it is important to control
the concentrations of fuel and air in the engine cylinder. A small change in the concentrations
can result in poor engine performance and/or poor combustion emissions.
Basic Definitions:
 Stoichiometric or theoretical combustion
The minimum amount of oxygen required for the complete combustion of a fuel.
 Stoichiometric or theoretical air
The minimum amount of air required for the complete combustion of a fuel.
 Air fuel ratio
The ratio of the mass of air to the mass of fuel for a combustion process.
Problem Information
Example Problem Statement:
Consider the general combustion reaction:
Cm H x Oy  zO2  mCO2 
x
H 2O
2
(1)
a. Derive an equation for z in terms of m, x and y.
b. Show that for combustion of this fuel in pure oxygen the mole fraction of fuel required is
given by
1
yF 
(2)
1 z
c. Show that for combustion of this fuel in air the mole fraction of fuel required is given by
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D. A. Crowl
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yF 
1
z
1
0.21
d. Derive the following equation for the air fuel ratio in terms of the fuel mole fraction, yF:
1  yF  M A
m
Air fuel ratio  A 
mF
yF M F
Where:
(3)
(4)
m is mass
M is the molecular weight
y is the mole fraction
Subscripts A and F refer to air and fuel, respectively.
e. Gasoline is a mixture of hydrocarbons. However, for simple calculations it can be
represented as octane, C8H18. Calculate the stoichiometric concentration in pure oxygen and
in air.
f. For gasoline combustion the desired air fuel ratio is 14.7. Calculate the mole fraction of
gasoline and the volume percent gasoline for this air fuel ratio. Gasoline is composed of a
number of lighter hydrocarbons, so the molecular weight of gasoline is about 108.
Example Problem Solution:
a. From the combustion equation it is clear that the oxygen required is m moles from the carbon
x/2
dioxide,
for the water, minus y / 2 for the oxygen in the fuel. The net result is
2
x y
z  m 
4 2
b. From the definition of the mole fraction:
Mole of fuel
1
yF 

Moles of fuel + moles of oxygen 1  z
c. From the definition of the mole fraction:
Moles of fuel
1
1
1
yF 



Moles of fuel + Moles of air
 Moles air 
 1  Moles oxygen  1  z
1 
 1 


0.21
 Moles fuel 
 0.21  Moles fuel 
d. The molar density of an ideal gas is given by:
P

 mols/volume
RgT
Where P is the pressure, Rg is the ideal gas constant, and T is the absolute temperature.
The number of moles of fuel per unit volume is given by
nF  yF 
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The mass of fuel per unit volume is calculated by multiplying the number of moles per unit
volume by the molecular weight
mF  yF  M F
Likewise, the moles of air per unit volume is given by,
nA  y A 
But, by conservation of mass, yF  y A  1 and it follows that y A  1  yF and then
nA  1  yF  
The mass of air per unit volume is then
mA  1  yF   M A
Then
mA (1  yF )  M A (1  yF ) M A


mF
yF  M F
yF M F
e. For C8H18, m = 8, x = 18 and y = 0. Thus,
x y
18 0
z  m    8    12.5
4 2
4 2
For pure oxygen combustion,
1
1
yF 

 0.0741
1  z 1  12.5
The volume percent fuel is the same as the mole percent. Thus, the volume percent gasoline
in oxygen is 7.41%
For combustion in air,
1
1
yF 

 0.0165
z
12.5
1
1
0.21
0.21
The volume percent fuel in air is 1.65%
f. In this case,
Air fuel ratio 
mA 1  yF  M A
=14.7

mF
yF M F
The molecular weight of air is 29 and the molecular weight of the gasoline is 108.
Thus,
1  yF  M A  (1  yF )(29)
14.7 
yF M F
yF (108)
(14.7)(108) yF  (1  yF )(29)
54.74 yF  1  yF
55.74 yF  1
yF  0.0179
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The volume percent gasoline in the fuel –air mixture is 1.79%.
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Home Problem Statement:
Calculate the air fuel ratios for the following fuels:
a. hydrogen,
b. ethanol,
c. methane.
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D. A. Crowl
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June 25, 2010
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