1. Unit 42: Heat Transfer and Combustion
LO4.1: Combustion Chemistry

2. Aim LO4: Be able to analyse the combustion processes.
At the end of this Learning Outcome the students will be able to…
… derive combustion equations
… determine calorific value
… analyse products of combustion.

3. Types of Chemical Reactions
To put combustion reactions into context view the following video…

4. Combustion
A combustion reaction takes place when fuel is burned with oxygen
The most common fuels are hydrocarbons which are made up of hydrogen and carbon atoms
Combustion usually takes place in air (a mixture of nitrogen, oxygen and other gases).

5. Combustion
The hydrogen atom in the fuel bond and the carbon atoms react with oxygen to form carbon dioxide.
Every combustion reaction consists of reactants and products….
FUEL
CO2
Oxidizer
H2O
Reactants
Products

6. Combustion
The mass of the reactants is equal to the mass of the products because mass like energy cannot be created or destroyed but can change form.
Thus with consideration to mass, chemical reactions are written in molar quantities.
The molar mass i.e. 1-mole (mol) is the mass of 12-grams of Carbon 12 (12C) and it contains x 1023 molecules (Avogadro’s number)

7. Combustion
Sometimes however you need to convert molar quantities to mass quantities for thermodynamic analysis.
Thus if the number of moles of a substance N, is related to the mass of a substance m, using the molar mass (molecular weight) M, is related by…
M = m/N

8. Combustion
It is to be noted that air is composed of about 21% O2 and 79% N2.
Thus 79/21 = Thus for each mole of O2 there is 3.76 moles of N2.
One of the key questions in combustion is ‘how much air needs to be burned to completely burn all the fuel’.

9. Combustion
The amount of air that supplies just enough O2 to complete combustion of fuel is called theoretical air.
Combustion reactions completed with theoretical air are called stoichiometric reactions.
If the amount of air supplied for a combustion reaction is less than theoretical air, the reaction is incomplete.
When incomplete combustion occurs, carbon monoxide is formed along with carbon dioxide and soot (carbon particulates) may also appear in the exhaust.

10. Combustion
You can write the stoichiometric reaction for any simple hydrocarbon fuel reaction process using the general combustion reaction equation…
CxHy + (x + y/4)(O N2) -> xCO2 + (y/2)H2O (x + y/4)N2

11. Combustion
CxHy + (x + y/4)(O N2) -> xCO2 + (y/2)H2O (x + y/4)N2
The reactants are on the left side and the products are on the right.
This format is used for all hydrocarbon reactions
Note 1: the general reaction equation uses the molar ratio for oxygen and nitrogen in air.
Note 2: even though chemical reaction equations are written on a molar basis, the number of moles of reactants doesn’t necessarily equal the number of moles of products. However the mass of the reactants equals the mass of products.

12. Combustion
When writing a chemical reaction equation, keeping track of individual elements on both sides of the equation is helpful….
Element: Reactants:(♯moles)(♯atoms) = Products:(♯moles)(♯atoms)
CxHy + (x + y/4)(O N2) -> xCO2 + (y/2)H2O (x + y/4)N2
C: (1)(x) = (x)(1)
H: (1)(y) = (y/2)(2)
O: (x + y/4)(2) = (x)(2) + (y/2)(1)
N: (3.76)(x + y/4)(2) = (3.76)(x + y/4)(2)

13. Combustion
A simple hydrocarbon fuel has only carbon and hydrogen atoms like methane (CH4) and Octane (C8H18)
For more complex fuels such as ethanol (CH3OH) and Nitro-methane (CH3NO2), you cannot use the general combustion equation – you need to write out the balanced reaction equation by balancing the atoms in the reactants with those in the products.

14. Combustion Consider Octane (C8H18) in this case x = 8 and y = 18, then
CxHy + (x + y/4)(O N2) -> xCO2 + (y/2)H2O (x + y/4)N2
C8H (O N2) -> 8CO2 + 9H2O + 47N2
C8H O2 + 47N2 -> 8CO2 + 9H2O + 47N2

15. Combustion Thus with C8H18 + 12.5O2 + 47N2 -> 8CO2 + 9H2O + 47N2
then,
Element: Reactants:(♯moles)(♯atoms) = Products:(♯moles)(♯atoms)
C: (1)(8) = (8)(1)
H: (1)(18) = (18/2)(2)
O: (12.5)(2) = (8)(2) + (9)(1)
N: (47)(2) = (47)(2)

16. Combustion
Consider if we increased the theoretical air by 120% then for Octane (C8H18) where x = 8 and y = 18, then
CxHy + (1.20)(x + y/4)(O N2) -> xCO2 + (y/2)H2O (x + y/4)N2
C8H18 + (1.2)(12.5)(O N2) -> 8CO2 + 9H2O + (0.2)(12.5)O N2
C8H O N2 -> 8CO2 + 9H2O + 2.5O N2
Note: the coefficient for oxygen in the product is equal to the extra amount of oxygen being supplied. In this case 20-percent more than the theoretical amount making the coefficient 20% of 12.5 = 2.5 moles of oxygen per mole of octane