Applications of the Combustion Equation
(1) Stoichiometric
proportions for finding the correct air supply rate for a fuel
(2) Composition of the combustion products is useful during the design,
(2) Composition of the combustion products is useful during the design,
commissioning and routine maintenance of a boiler installation
On site measurements of flue gas composition and temperature
are used as a basis for calculating the efficiency of the boiler at routine
maintenance intervals.
Combustion Air Requirements: Gaseous
Fuels
Calculating the air required for gaseous fuels combustion is most
convenient to work on a volumetric basis.The stoichiometric combustion reaction of methane is:
CH4 + 2O2 → CO2 + 2H2O
which show that each volume (normally 1 m3) of methane requires 2 volumes of oxygen to complete its combustion.
•
If we
ignore the components which are present in the parts per million range, air
consists of about 0.9% by volume argon, 78.1% nitrogen and 20.9% oxygen
(ignoring water vapor). Carbon dioxide is present at 0.038%.
•
For the
purposes of combustion calculations the composition of air is approximated as a
simple mixture of oxygen and nitrogen: oxygen 21% , nitrogen79%
•
The
complete relationship for stoichiometric combustion:
CH4 + 2O2 + 7.52N2 → CO2 + 2H2O
+7.52N2
as the volume of nitrogen will be 2×79÷21=7.52.
•
A very
small amount of nitrogen is oxidized but the resulting oxides of nitrogen (NOX)
are not formed in sufficient quantities to concern us here. However, they are
highly significant in terms of air pollution.
•
It can
be seen that the complete combustion of one volume of methane will require
(2+7.52=9.52) volumes of air, so the stoichiometric air-to-fuel (A/F) ratio for
methane is 9.52.
•
In
practice it is impossible to obtain complete combustion under stoichiometric
conditions. Incomplete combustion is a waste of energy and it leads to the
formation of carbon monoxide, an extremely toxic gas, in the products.
Excess air is expressed as a percentage increase over the stoichiometric requirement and is defined by:
• Excess
air will always reduce the efficiency of a combustion system.
It is sometimes convenient to use term excess air ratio, defined as:
Where sub-stoichiometric (fuel-rich) air-to-fuel ratios may be encountered, for instance, in the primary combustion zone of a low-NOX burner, the equivalence ratio is often quoted. This is given by:
Flue Gas Composition-Gaseous
Fuels
•
The
composition of the stoichiometric combustion products of methane is:
1 volume CO2
7.52 volumes N2
2 volumes H2O
•
Given a
total product volume, per volume of fuel burned, of 10.52 if water is in the
vapor phase, or 8.52 if the water is condensed to a liquid.
The two cases are usually abbreviated to “wet” and “dry”.
•
The
proportion of carbon dioxide in this mixture is therefore
• The instruments used to measure the composition of flue gases remove water vapor from the mixture and hence give a dry reading, so the dry flue gas composition is usually of greater usefulness.
Considering the combustion of methane with 20% excess air, the excess air (0.2×9.52) of 1.9 volumes will appear in the flue gases as (0.21×1.9)=0.4 volumes of oxygen and (1.9-0.4)=1.5 volumes of nitrogen
•
The
complete composition will be:
constituent vol/vol methane
CO2 1
O2 0.4
N2 9.02
H2O 2
giving a total product volume of 12.42 (wet) or 10.42 (dry).
•
The
resulting composition of the flue gases, expressed as percentage by volume, is:
Constituent % vol (dry) % vol (wet)
CO2 9.6 8.1
O2 3.8 3.2
N2 86.6 72.6
H2O – 16.1
•
Example :
A gas consists of 70% propane (C3H8) and 30% butane (C4H10) by volume. Find:
(a) The stoichiometric air-to-fuel ratio and
(b) The percentage excess air present if a dry analysis of the combustion products shows 9% CO2 (assume complete combustion).
Solution:
The combustion reactions for propane and butane are
(a) Stoichiometric Air Requirement
On the basis of 1 volume of the fuel gas, the propane content requires 0.7 × (5 + 18.8) = 16.7 vols air and the butane requires0.3 × (6.5 + 24.5) = 6.3 vols air. Hence the stoichiometric air-to-fuel ratio is 23:1.
(b) Excess Air
The combustion products (dry) will contain
The combustion products (dry) will contain
(0.7 × 3) + (0.3 × 4) = 3.3 vols CO2
(0.7 × 18.8) + (0.3 × 24.5) = 20.5 vols N2
plus υ volumes excess air, giving a total volume of products of (23.8 + υ ).
Given that the
measured CO2 in the products is 9%, we can write:
•
hence υ = 12.87 vols
The stoichiometric air requirement is 23 vols so the percentage excess air is: 55.9 %