55
CHAPTER 3
COMBUSTION
3.1 Introduction
The term ‘combustion’ refers to the fairly rapid reaction
accompanied by the evolution of heat (and maybe a flame) which
occurs between a fuel and an oxygen carrier such as air. For a given
amount of fuel, there is a definite amount of oxygen, and therefore
air, which is required for the complete combustion of the fuel. To
ensure complete combustion it is usual to supply air in excess of the
amount required for chemically correct combustion. The oxygen not
consumed in the reaction passes into the exhaust with the products of
combustion.
3.2 Combustion equations
By the conservation of mass, the mass flow remains constant (i.e.
total mass of the products equals total mass of the reactants), but the
reactants are chemically different from the products, and the products
leave at a higher temperature. The total number of atoms of each
element concerned in the combustion remains constant, but the atoms
are rearranged into groups having different chemical properties. This
information is expressed in the chemical equation which shows the
reactants and products of combustion and the relative quantities of
the reactants and products.
, 48
3.3 Fuel analysis
In combustion calculations it is necessary to know the
composition of a fuel and there are two ways of analysing fuel
composition:
- An accurate chemical analysis by mass of the important elements
in the fuel is called the ultimate analysis; the elements usually
being carbon, hydrogen, nitrogen, and sulphur.
- A proximate analysis of a fuel gives the percentages of moisture,
volatile matter, combustible solid (called fixed carbon), and ash
3.4 Stoichiometric air fuel ratio
- A stoichiometric mixture of air and fuel is one that contains just
sufficient oxygen for the complete combustion of the fuel. A
mixture which has an excess of air is termed a weak mixture, and
one which has a deficiency of air is termed a rich mixture.
stoichiometric A/F ratio
- Mixture strength =
actual A/F ratio
- The percentage of excess air
actual A/F ratio - stoichiometric A/F ratio
=
stoichiometric A/F ratio
(where A denotes ‘air’ and F denotes ‘fuel’).
- For gaseous fuels the ratios are expressed by volume; for solid
and liquid fuels the ratios are expressed by mass.
- Where a fuel contains some oxygen (e.g. ethanol C 2H6O) this
oxygen is available for the combustion process and so the fuel
requires a smaller supply of air.
, 49
3.5 Exhaust and flue gas analysis
The products of combustion are mainly gaseous. When a sample is
taken for analysis it is usually cooled down to a temperature which is
below the saturation temperature of the steam present. The steam
content is therefore not included in the analysis, which is then quoted
as the analysis of the dry products. Since the products are gaseous, it
is usual to quote the analysis by volume. An analysis which includes
the steam in the exhaust is called a wet analysis. The most common
method of analysis the exhaust gas is by successive absorption of the
CO2, O2 and CO as in the Orsat apparatus.
Example 3.1
Determine the stoichiometric air/fuel ratio for a petrol approximating
to hexane C6H14. Hence deduce the chemical equation if the petrol is
burned in 20% excess air, and the wet volumetric analysis of the
products (a) if all the water vapour is present, and (b) if the products
are cooled to an atmospheric pressure and temperature of 1 bar and
15oC. Determine also the dry volumetric analysis. Finally, estimate
the chemical equation if only 80% of the air required for
stoichiometric combustion is provided.
Solution
Since the combustion is theoretically complete with a stoichiometric
mixture, the products will consist only of CO2, H2O and the N2
brought in with the air. The amount of oxygen required can be found
by first balancing the carbon and hydrogen atoms on each side of the
, 50
equation and then summing the number of atoms of oxygen in the
products:
C6 H14 9 12 O2 6CO2 7 H 2 O
86 kg C6H14 + 304 kg O2 264 kg CO2 + 126 kg H2O
Since air contains 23.3% by mass oxygen, then:
304 100
stoichiometric air/fuel ratio 15.17
86 23.3
(A) Combustion with excess air
If there is 20% excess air, the equation, including the nitrogen
becomes:
79
C6 H14 1.29 12 O2 9 12 N 2
21
79
6CO2 7 H 2O 0.2 9 12 O2 1.2 9 12 N 2
21
(i) Wet volumetric analysis
(a) The amount-of-substance in the products is:
6 (CO2) + 7 (H2O) + 1.9 (O2) + 42.89 (N2) = 57.79 kmol
Now the mole fraction of a constituent equals the volume
fraction. The wet volumetric analysis if all the water vapour is
present is therefore:
6 100 7 100
CO2: 10.38% ; H2O: 12.11% ;
57.79 57.79
1.9 100 42.89 100
O2: 3.29% ; N2: 74.22%
57.79 57.79
CHAPTER 3
COMBUSTION
3.1 Introduction
The term ‘combustion’ refers to the fairly rapid reaction
accompanied by the evolution of heat (and maybe a flame) which
occurs between a fuel and an oxygen carrier such as air. For a given
amount of fuel, there is a definite amount of oxygen, and therefore
air, which is required for the complete combustion of the fuel. To
ensure complete combustion it is usual to supply air in excess of the
amount required for chemically correct combustion. The oxygen not
consumed in the reaction passes into the exhaust with the products of
combustion.
3.2 Combustion equations
By the conservation of mass, the mass flow remains constant (i.e.
total mass of the products equals total mass of the reactants), but the
reactants are chemically different from the products, and the products
leave at a higher temperature. The total number of atoms of each
element concerned in the combustion remains constant, but the atoms
are rearranged into groups having different chemical properties. This
information is expressed in the chemical equation which shows the
reactants and products of combustion and the relative quantities of
the reactants and products.
, 48
3.3 Fuel analysis
In combustion calculations it is necessary to know the
composition of a fuel and there are two ways of analysing fuel
composition:
- An accurate chemical analysis by mass of the important elements
in the fuel is called the ultimate analysis; the elements usually
being carbon, hydrogen, nitrogen, and sulphur.
- A proximate analysis of a fuel gives the percentages of moisture,
volatile matter, combustible solid (called fixed carbon), and ash
3.4 Stoichiometric air fuel ratio
- A stoichiometric mixture of air and fuel is one that contains just
sufficient oxygen for the complete combustion of the fuel. A
mixture which has an excess of air is termed a weak mixture, and
one which has a deficiency of air is termed a rich mixture.
stoichiometric A/F ratio
- Mixture strength =
actual A/F ratio
- The percentage of excess air
actual A/F ratio - stoichiometric A/F ratio
=
stoichiometric A/F ratio
(where A denotes ‘air’ and F denotes ‘fuel’).
- For gaseous fuels the ratios are expressed by volume; for solid
and liquid fuels the ratios are expressed by mass.
- Where a fuel contains some oxygen (e.g. ethanol C 2H6O) this
oxygen is available for the combustion process and so the fuel
requires a smaller supply of air.
, 49
3.5 Exhaust and flue gas analysis
The products of combustion are mainly gaseous. When a sample is
taken for analysis it is usually cooled down to a temperature which is
below the saturation temperature of the steam present. The steam
content is therefore not included in the analysis, which is then quoted
as the analysis of the dry products. Since the products are gaseous, it
is usual to quote the analysis by volume. An analysis which includes
the steam in the exhaust is called a wet analysis. The most common
method of analysis the exhaust gas is by successive absorption of the
CO2, O2 and CO as in the Orsat apparatus.
Example 3.1
Determine the stoichiometric air/fuel ratio for a petrol approximating
to hexane C6H14. Hence deduce the chemical equation if the petrol is
burned in 20% excess air, and the wet volumetric analysis of the
products (a) if all the water vapour is present, and (b) if the products
are cooled to an atmospheric pressure and temperature of 1 bar and
15oC. Determine also the dry volumetric analysis. Finally, estimate
the chemical equation if only 80% of the air required for
stoichiometric combustion is provided.
Solution
Since the combustion is theoretically complete with a stoichiometric
mixture, the products will consist only of CO2, H2O and the N2
brought in with the air. The amount of oxygen required can be found
by first balancing the carbon and hydrogen atoms on each side of the
, 50
equation and then summing the number of atoms of oxygen in the
products:
C6 H14 9 12 O2 6CO2 7 H 2 O
86 kg C6H14 + 304 kg O2 264 kg CO2 + 126 kg H2O
Since air contains 23.3% by mass oxygen, then:
304 100
stoichiometric air/fuel ratio 15.17
86 23.3
(A) Combustion with excess air
If there is 20% excess air, the equation, including the nitrogen
becomes:
79
C6 H14 1.29 12 O2 9 12 N 2
21
79
6CO2 7 H 2O 0.2 9 12 O2 1.2 9 12 N 2
21
(i) Wet volumetric analysis
(a) The amount-of-substance in the products is:
6 (CO2) + 7 (H2O) + 1.9 (O2) + 42.89 (N2) = 57.79 kmol
Now the mole fraction of a constituent equals the volume
fraction. The wet volumetric analysis if all the water vapour is
present is therefore:
6 100 7 100
CO2: 10.38% ; H2O: 12.11% ;
57.79 57.79
1.9 100 42.89 100
O2: 3.29% ; N2: 74.22%
57.79 57.79
