THE COMTHERM ON-LINE TECHNICAL MANUAL

COMBUSTION CHEMISTRY

Combustion is the chemical reaction of oxygen with a fuel which contains the two basic elements (carbon and hydrogen) to produce heat and usually a visible flame.

Most fuels contain varying amounts of carbon and hydrogen, and in the cases of liquid and solid fuels, some sulphur.
Some gaseous fuels, namely Towns gas (1st family gases) may contain some sulphur, but this is very small amounts and is closely controlled by gas purification.

Combustion reactions can be very complex; however the basic reactions can be considered to be :-

Hydrogen + Oxygen --------- to Water Vapor + heat.

1. 2(H₂) + (O₂) = 2(H₂O)

Carbon + Oxygen --------- to Carbon Dioxide + heat

2. (C) + (O₂) = (CO₂)

It can be seen from the above reactions that the carbon content of the gas fuel when combusted with oxygen produces carbon dioxide (CO₂) and the hydrogen content produces water vapour (H₂O).

Unlike oils and solid fuels gas can be burned with a minimum amount of excess air. Due to the difficulty of getting the air required for combustion in intimate contact with the fuel, oil and solid fuel burning requires significant quantities of excess air.

When fuel and air are mixed and burned having exactly the proper proportions, we achieve what is called perfect combustion and the reaction is considered going to completion without any excess fuel or air.

The gas-air mixture required for perfect combustion is referred to as the stoichimetric mixture.

If we supply an excess amount of air in the reaction, the mixture is termed "lean" and the flame is oxidizing, resulting in shorter and clearer flames with a significant reduction in flame temperature.

It is possible that too much fuel is supplied, in which case the fuel/air mixture is considered to be "rich" and the flame then is termed "reducing" resulting in a longer flame length which may be smoky in appearance.

Under normal circumstances the oxygen for combustion is obtained from air. Air contains 21% oxygen (O₂) and 79% nitrogen (N₂); the nitrogen is relatively inert and takes no part in the combustion reaction, however its presence must be taken into consideration as it affects the volumes of gases and temperatures produced by the combustion of the gas.

The calculation of the combustion air requirements is a vital part of the design of any fuel burning system.

A simple approximate 'rule of thumb’ sometimes used by practical fuel engineers is to allow 0.268 cubic metres of air for every Megajoule of gross heat input; in pure metric 1.124 cubic metres of air for every thousand Kilocalorie.
This equates to 1 cobic foot of air for every 100 Btu.

If accurate combustion air calculations are to be undertaken, then knowledge of the chemical reactions that take place with each individual component gas is required; the stoichimetric air gas ratios can then be calculated.

This chemical reaction knowledge can also be used to calculate the ultimate CO₂ percentage in the products of combustion, information which is required to ensure that the combustion is stoichimetric and that combustion efficiency is at its maximum.

It is useful to remember that the oxygen: nitrogen volume ratio of air is = (1 : 3.76)

If we consider the simple combustion of hydrogen (H₂) gas with oxygen we have the following reaction

2(H₂) + 1(O₂) = 2(H₂O)

(Note - 2 volumes of hydrogen require 1 volume of oxygen)

If air is considered then the complete volumetric reaction would therefore be

(2(H2) + 1(O2) + 3.76(N2) = (2(H2O) + 3.76(N2)

The AIR/GAS ratio for the combustion of hydrogen would therefore be 4.76 / 2 = 2.38
The volume of combustion products per volume of gas (wet) would be 5.76 / 2 = 2.88
The volume of combustion products per volume of gas (dry) would be 3.76 / 2 = 1.88
Note that these products of combustion contain no CO₂. (0%)

The most common combustion reaction is of course the combustion of methane (CH₄) - this can be represented as:-

1(CH₄) + 2(O₂) = 1(CO(sub>2,/sub>) + 2(H₂0)

or when air is considered :-

1(CH₄) + 2(O₂) + 7.52(N₂) = 1(CO₂) + 2(H₂O) + 7.52(N₂)

The AIR/GAS ratio for the combustion of methane would therefore be 9.52 / 1 = 9.52
The volume of combustion products per volume of gas (wet) would be 10.52 / 1 = 10.52
The volume of combustion products per volume of gas (dry) would be 8.52 /1 = 8.52
The CO₂ content (dry) would be 1 / 8.52 = 11.73%

Another common combustion reaction is the combustion of propane (C₃H₈) - this can be represented as :-

2(C3H8) + 10(O2) = 6(CO2) + 8(H₂O)

or when air is considered -

2(C₃H₈) + 10(O₂) + 37.6(N₂) = 6(CO₂) + 8(H₂O) + 37.6(N₂)

The AIR/GAS ratio for the combustion of propane would therefore be 47.6 / 2 = 23.8
The volume of combustion products per volume of gas (wet) would be 51.6 / 2 = 25.8
The volume of combustion products per volume of gas (dry) would be 43.6 / 2 = 21.81
The CO₂ content (dry) would be 6 / 43.6 = 13.76%.
The combustion reactions of all other fuel gases has be treated in this way and theoretical AIR/GAS ratios for perfect combustion (stoichimetric) calculated. See the following table:-

In general all commercially available fuel gases are mixtures of component gases and the stoichimetric AIR/GAS ratio has to be calculated using the data given in the above table, using the relationship given below.

AIR:GAS RATIO =

= 0.0238(%H₂) + 0.0238(%CO) + 0.0952(%CH₄) + 0.1666(%C₂H₆)
+ 0.238(%C₃H₈) + 0.3094(%C₄H₁₀) etc.

In the unusual situation where the fuel gas actually has an oxygen content then the result of the above AIR:GAS ratio calculation should be reduced by

0.0476(%O₂).

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