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Combustion

Introduction..... Composition of Air..... Combustion Notes..... Combustion Equations.....

Introduction

The notes below relate to the combustion process.   Combustion is a rapid reaction between a fuel and oxygen that produces heat (the chemical energy content of a fuel is converted to heat energy).  


Composition of Dry Air

Oxygen is involved in the majority of combustion reactions and this is present in the air.
The composition of dry air as mole fractions is 0,7809 N2, 0,2095 O2, 0,0093Ar, and 0,003 CO2.
For combustion equations it is convenient and practical to treat Ar and CO2 as Nitrogen.   The composition of molar air can then be taken as (approx.) 0,79kmol N2 and 0,21 kmol O2 per kmol of Air.

Note: I am using kmols (The molar volume is 22,414 m/kmol at 0 C and 101.325 kPa absolute pressure )

The equivalent mass fractions of air are 0,768N2 and 0,232 O2 per unit of air.   These values are conveniently represented as

1 kmol O2 + 3,76 kmol N2 = 4,76 kmol Air
1 kg O2 + 3,31 kg N2 = 4.31 kg Air


Combustion Notes

When considering the combustion of fuels with air equations are used to determine the proportions of the various chemicals involved.  The fuels generally are composed of carbon, hydrogen and sulphur with other substances including oxygen and ash.   The carbon, hydrogen and sulphur combine with oxygen in the air and the nitrogen and other gases in air are assumed to take no part in the combustion process.....

The proportion of air for complete combustion is called the stoichiometric air/fuel ratio.  Normally an excess of air is available and the mixture is weak or lean.  When insufficient air is available for complete combustion the mixture is rich...

Some of the terms used in combustion are identified below.....

  • Air/Fuel Ratio  R = (Amount on air) /(Amount of fuel)
  • Stoichiometric air/fuel Ratio for complete combustion   = R s
  • Percentage excess Air   E = [(R - R s) /R s]100%
  • >
  • Mixture Strength   M s = ( R s /R )100 %

Weak mixture M s less than 100%
Rich mixture M s greater than 100%

It is important to note that the total combustion of a fuel requires ideal conditions.    The fuel must be intimately mixed with the oxygen, the temperature must be appropriate, the ignition cannot start without a source of activation energy ( a spark, or flame, or local high temperature).   Once the ignition has commence the combustion will generally spread spontaneously. For an car engine it may be desireable to have a rich mixture to allow for maximum power or at start up.  A boiler requires an excess or air to ensure complete combustion of the fuel for efficient operation....





Combustion Equations

The following typical equations are used to determine the combustion process

  • Carbon ...C + O 2 -> CO 2
  • Carbon ...2C + O 2 -> 2CO...Incomplete Combustion
  • Hydrogen ...2H 2C + O 2 -> 2H 2O
  • Sulphur ...S + O 2 -> SO 2

Reaction equations are generally based on volumes for gases and masses for liquids and solids.   An reaction equation can include the enthalpy change Δ H at the end of the equation eg

C H4 + 2 O 2 -> CO 2 + 2 H2 - Δ H

The enthalpy term is negative if the reaction is exothermic and positive if the reaction is endothermic.


Consider the combustion of Propane gas (C3 H8 )as an example...
Writing down the basic reaction equation without identifying the quantities

C 3 H 8 +x O 2   ->  y C O 2 + z H 2 O..x, y z being unknown

There are three carbon atoms on the LHS and therefore y = 3.  There are 8 Hydrogen atoms on the left hand side ( H 8) and therefore z = 4 (4.H2) . The equation resulting is therefore

C 3 H 8 +x O 2   ->   3 C O 2 + 4 H 2 O..x being unknown

To determine x it is easily calculated that there are now 10 Oxygen atoms on the RHS (3 O2 + 4 O) x is therefore 5.   The reaction equation is therefore .

C 3 H 8 +5 O 2   ->   3 C O 2 + 4 H 2 O..

In terms of kmoles 1 kmol C 3 H 8 + 5 kmol O 2 =    ->   3 kmol C O 2 + 4 kmol H 2 O

Transfer into mass units ( 1 - kmol has a mass in kg = Molecular Weight)

[1 kmol of C 3 H 8 = 44kg + 5 kmol of O 2 = 160 kg ]  ->   [3 kmol of C O 2 = 132 kg + 4 kmol of H2 O = 72 kg ]

Now considering the equations with respect to Air.  For each volume of O2 there are 4,76 volumes of Air.   Therefore the stoichiometric ratio of for combustion of propane is 1 to 5.4,76 = 1:23.8 based on volume..
Table showing various substances involved in combustion with their Molecular Weights


Substance Formula Approx.
molecular
Weight
Benzene C 6H 6 78
Butane C 4H 10 -
Carbon C 12
Carbon Monoxide CO 28
Carbon Dioxide CO 2 44
Ethane C 2H 6 30
Ethanol C 2H 5OH 46
Ethene C 2H 4 28
Hydrogen H 2 2
Methane CH 4 16
Nitrogen N 2 28
Octane C 8H 18 114
Oxygen O 2 32
Pentane C 5H 12 72
Propane C 3H 8 44
Propene C 3H 6 42
Sulphur S 32
Sulphur Monoxide SO 48
Sulphur Dioxide SO 2 64
Water H 2O 18

For combustion Notes Combustion Notes ....More notes to follow....



Links To Boilers
  1. Combustion Notes..pdf Download - a short paper
  2. Combustion Reaction ..Set of informative clear notes
  3. Calculating Heat of reactions..Set of useful notes and diagrams
  4. Combustion..pdf download - Informative paper
  5. Calorific Value of fuels..Table of calorific value of fuels

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Last Updated 27/01/2013