1.    Background

This Combustion File (CF) presents tools with which the user may calculate [GLOSS]comburent[/GLOSS] requirements for combustion of solid [GLOSS]fuel[/GLOSS]s in terms of pure oxygen and regular atmospheric air requirements, both in volumetric and mass terms, and with variation of “excess air”.  This tool is based on Combustion File (CF) 226

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2. Comburent requirements for combustion

Initially the oxygen requirement must be calculated.

The reactants

Solid fuels contain the combustible elements C, H, and S. The oxidation of these elements provides the vast majority of energy released in solid fuel combustion.

Solid fuels also contain the element O, which will contribute to the heat release as a “resident” oxidant.

Finally the element N occurs in solid fuels, along with water and mineral matter (ash), none of which react in significant amounts with oxygen.

The chemical reactions

The chemical reactions are:

            C + O2 -> CO2                                      (1)

            H2 + O2 -> H2O                                   (2)

            S + O2 -> SO2                                     (3)

In all subsequent calculations, gases are assumed to be perfect; therefore 1 kilo-[GLOSS]mole[/GLOSS] (kmole) of gas has a volume of 22.41 m3 at 1 atm and 273K ([GLOSS]STP[/GLOSS]).

The calculations to produce the volumetric requirements of oxygen and the resultant product of combustion are described in CF226. The basic results of these calculations are given in Table 1 below

Table 1:  Oxygen requirements and products of combustion for the main fuel elements in solid fuels

An example solid fuel

As an example of a solid fuel, a typical steam [GLOSS]coal[/GLOSS] is shown in Table 2 below, as sourced from CF225.

The composition of solid fuel is given in weight fractions (kg/kg fuel).

Table 2: Example solid fuel – a typical steam coal

Calculation of stoichiometric oxygen requirement

Using these reactions described above plus the solid fuel analysis, calculation of the [GLOSS]stoichiometric[/GLOSS]oxygen requirement for the complete combustion of the fuel, can be made. It should be recalled that fuels can also contain oxygen, which will reduce the amount of oxygen required.

In the following calculations:

O2 refers to the oxygen requirement

Subscript v refers to result expressed in m^3 at STP

Subscript m refers to result expressed in kg

Subscript s refers to the stoichiometric requirement

Other lower case symbols are defined above for Table 2.

O2vs     = (1.867)c +(5.603)h + (0.700)s – (0.700)o                                        (4)

            = (1.867)(0.76) + (5.603)(0.05) + (0.700)(0.02) – (0.700)(0.03)

            = 1.692 m^3 oxygen/kg example coal at 1 atm and 273 K (STP)

O2ms    = (2.667)c +(8.000)h + (1.000)s – (1.000)o                                        (5)

            = (2.667)(0.76) + (8.000)(0.05) + (1.000)(0.02) – (1.000)(0.03)

= 2.417 kg oxygen/kg example coal

Calculation of the stoichiometric air requirement

Although the use of relatively pure oxygen as a comburent is often practiced for gaseous fuels, the use of atmospheric air as a comburent for coal combustion is much more frequent. To calculate [GLOSS]stoichiometric air requirement[/GLOSS]s, dry air is assumed; i.e. it contains no water. Since dry air contains approximately 21% oxygen and approximately 78% nitrogen on a volumetric basis, 4.76 as much air as oxygen must be used to achieve complete combustion (1/0.21=4.76).

Similarly, since dry air contains approximately 23% oxygen and approximately 76% nitrogen on a mass basis, 4.35 as much air as oxygen must be used to achieve complete combustion (1/0.23=4.35).

In the following calculations:

A refers to the air requirement

Subscript v refers to result expressed in m^3 at STP

Subscript m refers to result expressed in kg

Subscript s refers to the stoichiometric requirement

Subscript a refers to the actual air used

Other lower case symbols are defined above for Table 2.

Avs       = 4.76 ((1.867)c +(5.603)h + (0.700)s – (0.700)o)                                           (6)

Avs       = (1.692)(4.76) m^3 dry air/kg example coal at 1 atm and 273 K (STP)

            = 8.054 m^3 dry air /kg example coal at 1 atm and 273 K (STP)

Ams      = 4.35 ((2.667)c +(8.000)h + (1.000)s – (1.000)o)                                           (7)

Ams      = (2.417)(4.35) kg dry air/kg example coal

            = 10.514 kg oxygen/kg example coal

Calculation of the excess air requirement

In industrial combustion it is necessary to use more oxygen (air) than required in order to achieve complete combustion. This is particularly true for heterogeneous combustion reactions, which occur with solid fuels. In the following paragraph, Subscript a refers to the actual amount of oxygen or air used in the combustion process

O2va/O2vs or Ava/Avs – the ratio of actually used oxygen/air to the stoichiometric requirement is called the [GLOSS]excess air ratio[/GLOSS] and has the symbol l (Lambda).

The excess air ratio is dimensionless and therefore can be applied similarly to volumetric basis calculations and mass basis calculations

For theoretical stoichiometric combustion l = 1.

To calculate the actual oxygen or air requirement for a chosen excess air ratio and for a given solid fuel, the stoichiometric requirement is simply multiplied by the value of chosen excess air ratio.

O2va    = O2vs * l                                                                                                         (8)

where l > 1 or l < 1

Similarly:

Ava     = Avs * l                                                                                                           (9)

Acknowledgements

The author would like to acknowledge the original authors of the CF from which the present CF is derived.

Sources

[1] Schramek, Taschenbuch für Heizung und Klimatechnik, Munich, 1999

[2] Perry, Perry’s Chemical Engineers’ Handbook ed., 1997