Combustion Equation - Air to Fuel Ratio / Gaseous Fuels

      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, 

               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 (NO_X) 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-NO_X 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

                (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 %