MODULE DEN326 – Combustion in Automotive Engines

COMBUSTION – BASICS

1. Introduction A major part in thermodynamics is the conversion of heat into work. The heat energy often comes from a combustion process. Combustion accounts for approximately 85 percent of the world’s energy usage and is vital to our current way of life. Ground transportation, spacecraft and aircraft propulsion, electric power production, home heating and industrial processes all mainly rely on combustion to convert chemical energy to thermal energy or propulsive force. Combustion is a complex interaction of physical (fluid dynamics, heat and mass transfer) and chemical (thermodynamics and chemical kinetics) processes accompanied with the release of heat. It is a rapid reaction by which fuel is combined (burned) with a source of oxygen to release thermal energy.

Fuel + Oxidant Combustion Products + Heat (Q) In many cases a fossil fuel is burned in air either externally to the power cycle (e.g. steam plant) or internally as in a gas turbine or a reciprocating engine. For a cycle to operate correctly and on a maximum efficiency, the amounts of fuel and air need to be determined accurately to provide the required amount of heat energy. 2. Fuels These are mainly from fossil origin and can be classified by their physical state: solid (e.g. coal), liquid (e.g. petrol) or gaseous (e.g. natural gas). The last two categories are subject to refining processes before use which controls to some extent their composition and properties. Solid fuels are normally burned as mined apart from mechanical separation of incombustibles. Most fuels (whether solid, liquid or gas) consist mainly of carbon and hydrogen with trace amounts of other substances like sulphur. They may also contain small quantities of incombustibles (e.g. water vapour, nitrogen or ash). For solid and liquid fuels the chemical analysis is given by mass (percentage by mass of each chemical element in the fuel) whereas for gaseous fuels, the analysis is given by volume (percentage by volume of each gas or type of hydrocarbon in the fuel). Typical fuels and their composition Solid: Coal (by mass)

Carbon Hydrogen Oxygen Nitrogen/Sulphur Ash

74 % 5.98 % 13 % 2.27 % 4.75 %

Liquid: Oils (by mass)

Fuel Carbon Oxygen Sulphur

Petrol 85.5 % 14.4 % 0.1 % Kerosene 86.3 % 13.6 % 0.1 %

Gaseous: Natural gas (by volume)

Methane Ethane Propane Butane Nitrogen Carbon dioxide

93.6 % 3.6 % 0.8 % 0.3 % 2.6 % 0.1 %

3. Combustion equations Combustion processes obey the principle of conservation of mass: the total mass of products equals the total mass of reactants (fuel + oxidant). Each element has an atomic mass measured relative to carbon which has been given the atomic mass of 12. Values used by engineers are rounded to the nearest whole number. The following table gives the atomic masses of some elements:

Element Carbon Hydrogen Oxygen Nitrogen Sulphur

Atomic symbol

C H O N S

Mass 12 1 16 14 32

The following table gives the molecular masses of some compound substances (groups of atoms called molecules):

Substance Hydrogen Oxygen Water Carbon

monoxide Sulphur dioxide

Molecular grouping

H2 O2 H2O CO SO2

Mass 2 32 18 28 64

3.1. Chemical reactions The main constituents of fuels are carbon and hydrogen. When combined with oxygen during a combustion process, the reactions are defined by chemical equations. In a chemical reaction molecules are not conserved whereas atoms are.

• Global reaction (approximation): describes overall reaction, i.e. the ‘averaged’ or ‘net’ result of many elementary reactions.

OH2OH2 222

• Local reaction (exact): describe real reactions happening in steps and making up

the global reaction.

OOHOH 2

HOHOH2

HOHOHH 22

OOHOHOH 2

There are 2 types of general reactions:

• Exothermic: the overall reaction releases energy

• Endothermic: the overall reaction absorbs energy Example:

Combustion of carbon 22 COOC

1 atom + 1 molecule 1 molecule

Atomic mass: 12 + 32 = 44

In volumes: 0 volume + 1 volume = 1 volume

Combustion of hydrogen OHOH 2221

2

1 molecule + 1/2 molecule 1 molecule

Atomic mass: 2 + 16 = 18

In volumes: 1 volume + ½ volume = 1 volume (if vapour) Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. In other words, a quantity of a gas with a mass equal to its molecular weight and is at a given temperature and pressure then it will

occupy a standard volume. When the temperature is 0 C and the pressure is the atmospheric pressure, the volume is 22.4 litre (dm3). The amount of substance contained in this volume is known as 1 mole. In combustion analysis the unit kilomole (kmol) is often used and corresponds to a volume of 22.4 m3. The actual mass of a substance contained in a kmol (molar mass) is numerically equal to its atomic or molecular mass (e.g. 1 kmol of carbon has a mass of 12 kg and 1 kmol of O2 has a mass of 32 kg). Avogadro’s constant = 6.0236·1023. One mole of a substance corresponds to 6.0236·1023 particles (atoms or molecules). The unit kmol can also be used for solids and liquids. However, their volumes will be negligible in comparison with those of gaseous substances. Thus the combustion equation for carbon may be interpreted as:

22 COOC

1 kmol of carbon reacts with 1 kmol of oxygen to form 1 kmol of carbon dioxide.

Similar global equations may be written for other combinations of reactants:

COOC 221

2221 COOCO

22 SOOS

3.2. Mixture of fuels Combustion can also happen in mixtures of fuels. In this case we need to define a few more parameters. Consider a mixture of N different species: ni is the number of moles of species i and mi = Mi ni is the mass of species i with Mi the molar mass (g/mol or kg/kmol) of the species.

Mole number n is the total number of moles in the mixture:

N

1i

inn

Mole fraction Xi of the ith species is defined as n

nX i

i with 1XN

1i

i

Total mass m of the mixture is i

N

1i

i

N

1i

i nMmm

Mass fraction Yi of the ith species is defined as m

mY i

i with 1YN

1i

i

Mean molar mass is defined as 1

N

1i

iii

N

1i

i M/YXMn

mM

Relationship between mole and mass fractions: ii

i XM

MY

Mass concentration or partial density is defined as ii Y (g/m3 or kg/ m3)

Molar concentration is defined as M

X

M

Y

V

nXC i

i

iiii

(mol/m3 or kmol/ m3)

3.3. Stoichiometry A combustion is stoichiometric if the premixed reactants contain the right amount of oxidizer to consume (burn) the fuel completely (complete combustion).

If there is an excess of fuel: fuel-rich system

If there is an excess of oxygen: fuel-lean system

Stoichiometric coefficients: The stoichiometric coefficient vi (or stoichiometric number) of any given species i is the number of moles that participate in the reaction. For reactant i on the left hand side of the chemical equation: vi

’

For product i on the right hand side of the chemical equation: vi

’’

Si is species i

iN

1i

''

ii

N

1i

'

i SvSv

With the net stoichiometric coefficient vi = vi’’ - vi

’ we have 0Sv i

N

1i

i

.

Example:

COOC 221

Species C: i = 1 species O2: i = 2 species CO: i = 3

v1’ = 1 v1

’’ = 0 v1 = -1 v2

’ = 0.5 v2’’ = 0 v2 = -0.5

v3’ = 0 v3

’’ = 1 v3 = 1 4. Combustion in air The following analysis of air can be used: By volume: 21 % Oxygen (O2) 79 % Nitrogen (N2) By mass: 23.3 % Oxygen (O2) 76.7 % Nitrogen (N2) All other inert gases present in the air are included in the proportion of nitrogen. Note that the ratio of the percentage of volumes 79/21 = 3.76: with 1 kmol of oxygen there are 3.76 kmol of nitrogen making a total of 4.76 kmol. The stoichiometric air to fuel (A/F)stoich ratio by mass is mass of air/mass of fuel for a stoichiometric combustion. The actual air to fuel (A/F)actual ratio by mass is mass of air/mass of fuel for lean or rich fuel combustion.

Equivalence ratio stoich

actual

F/A

F/A

1 : stoichiometric mixture

1 : rich mixture

1 : lean mixture

In the same way we define:

The stoichiometric fuel to air (F/A)stoich ratio by mass is mass of fuel/mass of air for a stoichiometric combustion. The actual fuel to air (F/A)actual ratio by mass is mass of fuel/mass of air for lean or rich fuel combustion.

Equivalence ratio stoich

actual

A/F

A/F

1 : stoichiometric mixture

1 : lean mixture

1 : rich mixture

Note 1 .

Example Consider the combustion of hydrogen:

OHOH 2221

2

or OH2OH2 222

Molecular mass 4 + 32 = 36 In kilograms 4 kg H2 + 32 kg O2 = 36 kg H2O or 1 kg H2 + 8 kg O2 = 9 kg H2O 8 kg of oxygen are necessary to completely burn 1 kg of hydrogen. Oxygen is contained in air with a proportion of 23.3% by mass, hence the mass of air required to burn 1 kg of H2 is 8/0.233 = 34.5 kg. Of this 34.5 kg of air, there are 8 kg of oxygen and 34.5 – 8 = 26.5 kg of nitrogen. The equation can then be rewritten as: 1 kg H2 + 34.5 kg air = 9 kg H2O + 26.5 kg N2 The stoichiometric air-fuel (A/F) ratio by mass is 34.5 kg of air/1 kg H2 =34.5 The stoichiometric fuel-air (F/A) ratio by mass is 1 kg H2/34.5 kg of air =0.02899 The mix of combustion products varies with the equivalence ratio. Combustion may be

complete under stoichiometric and fuel-lean conditions ( ≤ 1) with some oxygen

remaining unreacted in the combustion products ( < 1). The composition of the products of fuel-lean combustion is, to a good approximation, determined by atom balances alone. The problem of specifying the products of combustion is more complicated for fuel-rich

combustion ( > 1). Since there is insufficient oxygen for complete combustion under fuel-rich conditions, some carbon monoxide, hydrogen, and possibly, unburned hydrocarbons remain in the combustion products. Thus there are at least five products present (CO, CO2, H2, H20, N2), but only four elemental balances are possible. An auxiliary condition based on thermodynamics or combustion kinetics is needed to determine the exhaust composition. The same applies to ‘real’ combustion where fuel does not burn fully.

Exercises 1. A sample of dry anthracite has the following composition by mass: C 90%, H 3%, O

2.5%, N 1%, S 0.5% and ash 3%. Calculate the stoichiometric A/F ratio and determine the dry and wet analysis of the combustion products by mass and by volume when 20% excess air is supplied. [11.24]

2. Ethyl alcohol has the following formula C2H6O. Calculate the stoichiometric air-fuel ratio

and the corresponding wet volumetric analysis of the products of combustion. Determine also the wet volumetric analysis when 10 % excess air is supplied with the fuel. [8.953]

3. The percentage composition by mass of a certain fuel is given as C 90%, H 3.5%, O

3% and remainder is incombustible. The fuel is burnt with air and the resulting analysis gave the following result by volume: CO2 12.7%, O2 7%, N2 remainder. Find the mass of air supplied per kilogram of fuel, and the percentage excess air. [17.16, 47%]

4. A certain fuel consists of 52% carbon, 13% hydrogen and 35% oxygen by mass. If the

fuel is burnt in a combustion chamber of a gas turbine plant with 120% excess dry air, find the stoichiometric air/fuel ratio by mass and determine the wet volumetric analysis (as percentages) of the products of combustion. [8.91]

5. A petrol has the following analysis by mass: C 83.7%, H2 16.3%. If the dry product

analysis by volume is CO2 11.8%, O2 3.7% and N2 84.5 %, determine the A/F ratio by mass. [18.17]

6. In an engine test the dry volumetric analysis of the combustion products was CO2

5.27%, O2 13.38% and N2 81.35%. Assuming that the fuel is a pure hydrocarbon and that it is completely burnt, estimate the ratio of carbon to hydrogen by mass. [5.29]