THE COMTHERM ON-LINE TECHNICAL MANUAL

THE BASICS OF COMBUSTION CHEMISTRY

Combustion is the reaction of oxygen with the two basic elements that make up the common fossil fuels (carbon and hydrogen) to produce heat.

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

1. 2(H₂) + (O₂) = 2(H₂O)

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)

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.

When calculating combustion air volumwa it is therefore useful to remember that the oxygen : nitrogen volume ratio of air is = (1 : 3.76)

If we consider the simple combustion of hydrogen 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 volume ratio for perfect (stoichimetric) combustion of hydrogen would therefore be = (5.76 ÷ 2) = 2.38

The volume of products of combustion (per unit volume of gas) would be 5.76 ÷ 2 = 2.88

The most common combustion reaction is of course the combustion of methane (CH₄)

- this can be represented as :-

1(CH₄) + 2(O₂) = 1(CO₂) + 2(H₂0)

or when air is considered :-

1(CH₄) + 2(O₂) + 7.52(N₂) = 1(CO₂) + 2(H₂) + 7.52(N₂)

The AIR/GAS volume ratio for the perfect combustion of methane would therefore be (2+7.52) ÷1 = 9.52

The total volume (per unit volume of gas) of products of combustion would be (1+2+7.52) = 10.52 - 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 volume ratio for the perfect combustion of propane would therefore be (47.6 ÷ 2) = 23.8

The total volume of products of combustion would be 51.62 ÷ 2 = 25.81 - the CO₂ content (dry) would be 6 ÷ 43.6 = 13.76%.

The combustion reactions of all other fuel gases can be treated in this way and theoretical AIR/GAS ratios for perfect combustion can be calculated.

The calculated AIR/GAS ratio is required for perfect combustion ( referred to as the stoichimetric mixture).
However gas and air mixtures within a certain range (Limits of flammability) of the stoichimetric mixture are combustible.

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