ACADEMIC YEAR 2014-2015
COMBUSTION PROCESSES
In order to achieve this outcome you will need to become familiar with the
chemistry of combustion of hydrocarbon fuels. There is a requirement to
understand and use some of the concepts in chemistry such as atomic and
molecular mass. The mathematical level is that of simple arithmetic.
This outcome will be presented to you in one section.
The section includes
6.
COMBUSTION CHEMISTRY
7.
ENERGY OF COMBUSTION
8.
PRODUCTS OF COMBUSTION
Recommended textbooks are
Bolton W
Mechanical Science
Blackwell Science
Hannah and Hillier
Mechanical Engineering Science
Addison Wesley Longman Ltd
Rayner Joel
1999
Basic Engineering Thermodynamics in SI units
Longman
0 582 42423 2
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1993
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6.
COMBUSTION CHEMISTRY
6.1
INTRODUCTION
Chemistry is an essential part of our daily lives. The air that we
breathe, the liquids that we drink and the food that we eat is
converted by a series of chemical reactions to provide us with the
energies and substances that we need to sustain life.
During this outcome we will look at the basic nature of matter and
explore how the elements and compounds that surround us have been
utilised to provide us with so much useful energy. As engineers we
can convert energy from one form to another, and in doing so do
some useful work.
6.2
BASIC NATURE OF MATTER
The most basic building block of any substance is the atom. A
quantity of any substance comprises an enormous numbers of atoms.
Frequently these atoms combine together to form larger units. When
they combine with atoms of their own substance they are known as
molecules. A molecule is the smallest amount of a substance that
displays its properties. When they combine with atoms of another
substance they are called compounds, although you can, of course
have a molecule of a compound.
As far as we are concerned there are only three principal components
to an atom. The NEUTRON, which is a relatively large lump, and is
found in the centre of the atom. The PROTON, which has the same
mass as a NEUTRON, and is also found in the centre. Finally, the
ELECTRON, which is about 0.0005 of the mass of the others, and
orbits around the others. A bit like a solar system.
Most of the mass of an atom is clearly in the centre.
Most of the activity of an atom is focussed on the electrons.
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The arrangement of the electrons determines many of the substances
properties, both mechanical and physical. It contributes to its degree
of reactivity, and to the type of compounds that it will form.
The simplest atom is that of hydrogen. It consists of one PROTON and
one orbiting ELECTRON. The PROTON has a single positive charge
associated with it, and the ELECTRON a single negative charge. On an
atomic scale therefore the hydrogen atom is neutral. The orbiting
ELECTRON is kept in its orbit by these opposite charges that are
analogous to the centripetal and centrifugal forces that we have met
before.
As atoms get bigger they have more PROTONS and ELECTRONS, but
also NEUTRONS. Generally there are as many protons as electrons,
and as many neutrons as protons. As the neutrons do not affect the
charge balance of the atom it is possible to have elements that have
a range of neutrons in their nucleus. When that occurs they are
known as ISOTOPES. The best-known isotopes are those of uranium,
which exist as uranium 237 and uranium 235. The numbers refer to
the atomic mass that is the total number of neutrons and protons in
the nucleus. In a simple manner we could conclude that an atom of
uranium 235 is 235 times heavier than an atom of hydrogen. This
does ignore the electron mass, but as we would need 1840 electrons
to match the mass of a proton it is a reasonable assumption.
To further complicate the issue we need to consider the orbits of the
electrons. The electrons of any one atom do not all orbit at the same
distance from the nucleus. Orbits get full. The number of electrons
needed to fill a particular orbit is given by 2n2, where n is the number
of the orbit. So, for example, an atom will have its first orbit filled
by 2(1)2 electrons, or 2. Similarly the second orbit needs 8 electrons
to fill it, and the third 18.
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This arrangement is not quite as simple as I have made it, as the
electrons in the larger orbits tend to exist at different energy levels.
This means that the third orbit that would be filled by 18 electrons
will have ‘sub orbits’ and the completion of a sub orbit can affect the
behaviour of the element. I am not going into this any deeper as it is
outside the scope of our course. Any ‘A’ level text book on Chemistry
will provide the more curious of you with a full explanation of the
s,p,d and f energy levels within the atom, and the implications for
behaviour.
At this juncture it might be helpful to look at the periodic table of
the elements, which tells us their names, their number and their
mass. We will also look at some of the behaviours. The table is
overleaf.
Let us consider the first two elements, hydrogen and helium.
Hydrogen has one electron in its first orbit. For that orbit to be
complete there needs to be two electrons. What this means is that
hydrogen is likely to react with any elements that enable it to fill
that orbit. This will make hydrogen a reasonably reactive substance,
particularly if it can react with an element that is going to gain from
the relationship.
Helium, however, has two electrons in the first orbit and is therefore
satisfied. Helium is an inert gas, reacting with little because it
simply has no need to do so.
Look at two other elements, sodium (Na,11) and chlorine (Cl,17).
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Select an element from the periodic table.
Group
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Period
1
1
He
3
2
Li
3
Na
4
2
H
11
19
K
4
Be
Rb
6
Cs
7
Fr
55
87
B
12
13
Mg
Al
20
21
Ca
Sc
38
39
37
5
5
Sr
56
Ba
88
Ra
Y
*
*
*
71
40
Zr
72
Lu
Hf
103
104
Lr
57
*Lanthanoids
*
La
**Actinoids
*
*
Ac
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22
Ti
89
Rf
58
Ce
90
Th
23
V
41
Nb
73
Ta
105
Db
59
24
Cr
42
Mo
74
W
106
Sg
60
Pr
Nd
91
92
Pa
U
25
Mn
43
Tc
75
26
Fe
44
27
Co
45
Ru
Rh
28
Ni
46
Pd
Re
Os
76
77
Ir
Pt
107
108
109
110
Bh
61
Pm
93
Np
Hs
62
Sm
94
Pu
Mt
63
Eu
95
Am
78
Ds
64
Gd
96
Cm
29
Cu
47
Ag
79
Au
111
Rg
65
Tb
97
Bk
30
Zn
48
Cd
80
Hg
112
Uub
66
Dy
98
Cf
31
Ga
49
In
81
Tl
113
Uut
67
Ho
99
Es
6
7
8
C
N
O
14
15
16
Si
32
Ge
50
Sn
82
Pb
114
Uuq
68
Er
100
Fm
P
33
As
51
Sb
83
Bi
115
Uup
69
Tm
101
Md
S
34
9
F
17
Cl
35
Se
Br
52
53
Te
84
Po
116
Uuh
70
Yb
102
No
I
85
At
117
Uus
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn
118
Uuo
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The eleven electrons of sodium will occupy the first orbit (2), the
second orbit (8) and leave one electron in the third orbit. What
sodium would really like to do is either lose this electron to leave a
completely filled second orbit, or obtain some more so that the third
orbit contains a stable number of electrons.
Along comes chlorine. It has seventeen electrons. Two fill the first
orbit, eight fill the second, and there are seven left in the third orbit.
At this stage you may see that the seven electrons in chlorine’s third
orbit could be added to the one in sodium’s third orbit to make a
total of eight. This may then satisfy both of their needs in that there
is a sub-orbit of the third orbit that has eight electrons in it, and
consequently might be less reactive.
That is what happens. Two highly reactive elements, sodium and
chlorine, combine to form the relatively unreactive compound sodium
chloride, which is common salt. Taken individually these elements
will kill us. In combination they take a lot longer!
SELF TEST 6.2a
Examine the periodic table above. Try to identify five pairs of
elements that might combine to form an unreactive compound.
Do not use the same element twice. Name the compound.
Briefly explain its use. (maximum imagination required here!)
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6.3
VALENCY
This is a term used to describe the way in which atoms can join
together. It is connected with the number of electrons that are
available to be shared between atoms. It is a complicated issue with
the full and detailed explanation better left elsewhere. We can look
at a simple example to try to get an understanding of the type of
thing happening.
Consider the element hydrogen (H). We already know that it has one
electron in its outermost orbit. If it had two it would be fairly
unreactive, so hydrogen meets this condition by going around in pairs.
This hydrogen molecule is fairly stable because both atoms are
sharing the two electrons. Because the hydrogen molecule consists of
two atoms it is represented as H2. As there is only one electron
available to be shared by an atom of hydrogen we say that hydrogen
has a valency of one.
Oxygen (O), on the other hand, has eight electrons. Two will be in
the first orbit and six in the second. You will recall that the second
orbit is satisfied when it has eight electrons in it. So oxygen has a
need of two electrons, which it can get, amongst other ways, by
sharing the two electrons from an hydrogen molecule. Oxygen has a
valency of two, as it needs to share two electrons. The hydrogen
molecule and oxygen atom obtain satisfaction from the mutual
sharing of these two electrons, so the compound formed, H2O
(water), is unreactive.
Diagrammatically we can represent a water molecule as below.
H ----- 0 ----- H
Where the lines represent the bonds, or valency.
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Our studies of combustion will focus on hydrocarbon fuels. Because
carbon is such an important element we will spend a little time
looking at the peculiarities of this element.
Carbon is element number six, with six electrons and an atomic mass
of 12. The atomic mass refers to the sum of the protons and the
neutrons in the element. As the electrons are so small, they are an
insignificant part of the atoms mass. Where carbon is unusual is that
it can exhibit a valency of either one, two, three or four. This is
because carbon has only four electrons in its second orbit and
therefore requires sharing four more. It is happy to share with
hydrogen, or another atom of itself or with some other elements.
There are therefore numerous ways in which carbon can unite with
hydrogen. The study of carbon and its various compounds is a science
all of its own, and is known as organic chemistry.
Let us look at the simplest compound of carbon and hydrogen, called
methane.
H
H
C
H
H
Here one atom of carbon has bonded with four atoms of hydrogen to
form one molecule of methane. Each hydrogen atom is sharing it’s
electron with the outer orbit of carbon, giving it effectively eight,
and at the same time each hydrogen atom is sharing one of the four
electrons in carbon’s outer orbit giving it effectively two.
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Methane is a gas that is most commonly associated with the decaying
of vegetation or decomposition in swamps. It is the first member of a
family called the paraffins. The next member of that family is called
ethane and the structure uses two carbon atoms sharing one of each
atoms electrons. The diagram below shows this substance.
H
H
H
C
C
H
H
H
In shorthand notation methane is written as CH4, and ethane as C2H6.
SELF TEST 6.3a
Construct the diagrams similar to the ones above for methane and
ethane that shows the next three members of the paraffin family.
The periodic table shows us that carbon has an atomic weight of
12, and hydrogen an atomic weight of 1. When methane is formed
the molecule consists of one carbon atom and four hydrogen
atoms, which gives it a molecular weight of 16.
Determine the molecular weights of the three paraffins that you
have constructed. See if you can find their names and their boiling
points.
Draw a graph to see if there is a connection between the molecular
weights and their boiling points for all of the first five paraffins.
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As has been mentioned, carbon can have a valency from one to four.
You will realise that the previous family uses a valency of one. If we
look at the simplest arrangement where carbon exhibits a valency of
two, then we will get the following diagram.
H
H
C
C
H
H
Here we have the double bond arrangement that carbon can utilise.
Another way to look at it is shown below.
H
H
C
C
H
H
Where the blue dots represent the electrons belonging to hydrogen,
the red dots show the electrons belonging to one carbon atom and
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the green dots the electrons belonging to the other carbon atom. In
this way stability occurs in this molecule. This substance is ethylene,
and is the first member of the olefin family.
SELF-TEST 6.3b
Determine the structure, calculate the molecular weight
and find the boiling point of the first four members of
the olefin family. Add this data to the graph drawn for
SELF-TEST 6.3.a
6.4
BASIC CHEMICAL REACTIONS
We will now spend some time examining the basis for calculating the
quantities involved in some basic chemical reactions.
Using the information that oxygen has an atomic weight of 16, and
hydrogen has an atomic weight of 1, we can easily determine that a
molecule of water (H2O) will have a molecular weight of 16+1+1=18.
To write this in its accepted format we acknowledge the fact that
both hydrogen and oxygen exist as di-atomic molecules, that is they
cannot exist as single atoms.
We have
2H2 + O2 = 2H2O
Using the molecular weights we get
2x2 + 32 = 2x18
If we consider the reaction between sodium (Na) and water (H2O) we
will find the following. The products will be sodium hydroxide and
hydrogen gas.
Na + H2O = NaOH + H
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However we need to ensure that H occurs as H2, so the equation
above needs to be doubled up to allow this to happen.
The equation becomes
2Na + 2H2O = 2NaOH + H2
If we now make use of the molecular masses we can calculate that 2
atoms of sodium (2x22units of weight) will react with 2 molecules of
water (2x18 units of weight) to form 2 molecules of sodium hydroxide
(2x39 units of weight) and 1 molecule of hydrogen (1x2 units of
weight).
A check will reveal that there are 80 units of weight before and after
the reaction.
Now that we have covered the fundamentals of reactions we will
move on to the combustion reactions.
6.5
COMBUSTION CHEMISTRY
We can carry out an investigation of the principles of combustion
chemistry without needing to examine the behaviour of the long
chained complex molecules that make up oil based fuels. As has been
stated before, the hydrocarbon fuels consist of the two elements
hydrogen and carbon. You have seen that carbon can exist with
several valencies, as we have noted with the paraffins and the
olefins.
We will now examine the well-known hydrocarbon acetylene. This is
widely used in welding as the combustion process generates a large
amount of heat, sufficient to melt metals such as steel.
This is perhaps the key point. The combustion of a hydrocarbon fuel
is an exothermic process. That means that a great deal of heat is
given out when a hydrocarbon fuel is burnt. The amount of heat
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given out is called the calorific value of the fuel. A typical value for
a hydcarbon fuel is 40 000kJ/kg.
Let us actually consider the implications of this value. If we burn 1kg
of a fuel with a calorific value of 40 000kJ/kg we will release
40 000kJ of heat energy. For a car travelling at a steady speed of 50
miles/hr (22m/s) the combined drag and friction would be about
400N. With no loss of efficiency the distance travelled would be
40 000kJ/400N = 100km.
Rough conversion gives us 225miles/gallon fuel consumption. Clearly
we cannot get 100% efficiency, but nevertheless the enormous
potential of harnessing the energy of combustion is clear.
SELF-TEST 6.5a
A small engine burns a hydrocarbon fuel with a calorific
value of 36MJ/kg. It is coupled to a small electrical
generator and the transmission efficiency is 80%. If it
burns the fuel at the rate of 2kg/hr, calculate the
generator output in kW. Assume the generator efficiency
is 100%.
(You will need to use knowledge gained in previous
outcomes here!)
Acetylene has the formula C2H2. The two carbon atoms are joined by
a triple bond that allows only one bond each for hydrogen.
The diagram is shown below.
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H
C
C
H
You may wish to represent this arrangement similarly to the lower
diagram on page 110.
When a hydrocarbon fuel is burnt completely in air, the products of
complete combustion are carbon dioxide and water vapour. The only
constituent of air that contributes to the combustion process is
oxygen. Let us look exactly at burning acetylene.
The combustion equations will show that
C2H2 + 5O = 2CO2 + H2O
As oxygen cannot exist as a single atom we need to double up as
before.
Correct is
2C2H2 + 5O2 = 4CO2 + 2H2O
Using numbers we find that 2 molecules of acetylene (2x26) will
combine with 5 molecules of oxygen (5x32) to give 4 molecules of
carbon dioxide (4x44) and two molecules of water (2x18).
52 + 160 = 176 + 36
The key information here is that contained on the left hand side of
the equation. It tells us that 52 units of acetylene need 160 units of
oxygen for complete combustion.
If we consider that air contains 23.3% oxygen by weight with the
remainder being nitrogen (not quite true, but near enough), then 52
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units of acetylene will need 160/0.233 units of air for complete
combustion.
So we have that 52 units of acetylene require 686.7 units of air for
complete combustion. This analysis being based upon weight.
If we now divide the units of air by the units of acetylene we will
obtain the air:fuel ratio by weight.
Air:fuel ratio for acetylene is 13.206:1 for complete combustion.
This ideal ratio is called the stoichiometric air:fuel ratio.
NOTE
Calculating the air:fuel ratio by volume is sometimes used.
Just as a simple example, if we used the information above we would
find that 13.206N of air is needed to give complete combustion to 1N
of acetylene.
By dividing both sides by 9.81 we can refer to masses, which makes
our understanding a bit easier.
Therefore 13.206kg of air is needed to give complete combustion to
1kg of acetylene.
At sea level, 150C and normal atmospheric pressure (101325N/m2) the
density of air is 1.2kg/m3.
Therefore 13.206kg of air occupy 13.206/1.2 m3 = 11.005m3
(11005 litres)
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As acetylene is a gas at this temperature, we can make use of the
fact presented to us by Avogadro, that the gm molecular weight of
any gas occupies 22.4 litres at standard conditions. (As above)
I found this confusing! Which gas?
So 52g of acetylene occupy 22.4 litres.
52kg will occupy 22 400 litres.
1kg will occupy 22 400/52 litres, which equals 430.77 litres
Air:fuel ratio by volume = 11005/430.77 = 25.55:1
QUITE A DIFFERENCE!
SELF-TEST 6.5b
Ensure that you know whether the air:fuel ratio is needed by
weight or
volume.
A fuel
has the formula C4H4. Produce a diagram to show all
the possible arrangement of bonding of the carbon atoms
that will meet this formula.
Determine the stoichiometric air:fuel ratios by weight and
In the previous examples the fuel has been a pure hydrocarbon, that
is only the elements carbon and hydrogen have been present in the
compound. Some fuels contain oxygen within their structure, and
this is available for combustion as well as the oxygen in the air. The
presence of this oxygen will clearly affect the air:fuel ratios.
Example 6.5a
Determine the stoichiometric air:fuel ratios by weight and volume for
the fuel shown in the diagram below.
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H
O
O
H
C
C
C
H
H
C
O
Solution
This fuel has the composition C4H4O3. Note how each carbon atom
uses four bonds, each oxygen two bonds and the hydrogen one bond.
For complete combustion the products will be CO2 and H2O.
The four atoms of carbon need eight atoms of oxygen. The four
atoms of hydrogen need two atoms of oxygen. The total requirement
is for ten atoms of oxygen. However there are already three present
in the fuel, so the net requirement for additional oxygen is seven
atoms.
We know that oxygen exists as a di-atomic molecule so we need to
double up on these quantities.
The combustion equation becomes
2 C4H4O3 + 7O2 = 8CO2 +4H2O
By mass
2(48+4+48) + 7x32 = 8(12+32) + 4x18
200
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+ 224 =
352
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so 200kg of fuel need 224kg of oxygen, which equates to 224/.233kg
of air.
Air:fuel ratio by weight 961.37:200, or 4.8:1
By volume
Using the molecular weights in kg, and assuming that the fuel is
gaseous at these temperatures
2x22400 litres of fuel need 7x22400/(0.233x1.2) litres of air.
Air:fuel ratio by volume is 12.518:1
SELF-TEST 6.5.c
A lot of fuels contain Sulphur. Sulphur burns completely in air
to produce a gas called sulphur dioxide, SO2. Page 105 will
show that sulphur has atomic number 18 (2,8,8) with an
atomic weight (=molecular weight) of 36.
Why is the atomic weight the same as the molecular weight?
If Benzine sulphonic acid, C6H5SO3H was to burn completely in
air, what would be the air:fuel ratio by mass?
6.6
EXCESS/RESTRICTED AIR
So far we have looked at the exact amount of air required for
complete combustion. In practice the amount of air might be less
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than that, and incomplete combustion will occur. When that happens
the carbon may form carbon monoxide as well as carbon dioxide.
CARBON DIOXIDE
O
C
O
CARBON MONOXIDE
O C C O
Our combustion equations would change, for example, from
CH4 + 2O2
to
= CO2
2CH4 + 3O2
=
+ 2H2O
2 CO + 4H2O
SELF-TEST 6.6a
The hydrocarbon fuel shown below is burnt to produce
a.
CO2 only as well as water vapour.
b.
CO only as well as water vapour.
c.
50% CO2 and 50% CO as well as water
vapour.
Determine the air:fuel ratios by weight in each case.
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H
O O
O C C C C H
H C O
H
Practically excess air needs to be supplied to ensure that complete
combustion occurs. There is a time delay involved in the burning of a
fuel and in many practical situations the combustion process may be
inhibited by the other constituents of air. In a gas turbine excess air
is provided to ensure complete mixing of the air and fuel, so that a
uniform temperature distribution at entry to the turbine is obtained.
The excess air is also used to keep the temperature in the combustion
chambers down so that dissociation of CO2 into CO is minimised.
Excess air does mean that useful heat is lost as the temperature of
elements not involved in the combustion process is also increased.
Example 6.6a
An analysis of a sample of coal reveals that it contains by percentage
of weight
carbon: 84, hydrogen: 5, sulphur: 3,oxygen: 6 and ash: 2.
Air is supplied to this fuel at the ratio of 20:1 by weight. What is the
percentage excess air supplied?
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Solution
Consider 1 kg of the fuel.
There will be 0.84kg of carbon.
As C + O2 = CO2, then using molecular weights (masses) we deduce
that 12kg of carbon will need 32kg of oxygen for complete
combustion.
Therefore 0.84kg of carbon will need 2.24kg of oxygen for complete
combustion.
There will be 0.05kg of hydrogen.
As 2H2 + O2 = 2H2O, then our molecular weights (masses) show that
4kg of hydrogen needs 32kg of oxygen for complete combustion.
Therefore 0.05kg of hydrogen will need 0.4kg of oxygen for complete
combustion.
There will be 0.03kg of sulphur.
As S + O2 = SO2, then our molecular weights (masses) indicate that
32kg of sulphur need 32kg of oxygen for complete combustion.
Therefore 0.03kg of sulphur will need 0.03kg of oxygen for complete
combustion.
There will be 0.06kg of oxygen.
This is present in the fuel, and will reduce the amount necessary
from the air.
Ash is inert. It plays no part in the process of combustion.
So 1kg of this coal needs 2.61kg of oxygen.
For 2.61kg of oxygen we need 2.61/0.233kg of air = 11.202kg.
Air supplied is 20kg. Excess air is 20 – 11.202 = 8.796kg
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Percentage excess = 8.796/11.202 = 78.52%
SELF-TEST 6.6b
An analysis of a fuel reveals that it contained 76% carbon, 7%
hydrogen, 6% oxygen, 7% sulphur, 3%nitrogen by weight, the
remainder being ash.
Determine the air supply rate needed (kg air: kg fuel) to give
100% excess air by weight.
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7.
ENERGY OF COMBUSTION
7.1
CALORIFIC VALUES
We have met the term calorific value previously. Hydrocarbon fuels
release energy in the form of heat when burnt. Clearly the amount of
heat released by a fuel will determine how useful it is. The obvious
conclusion that the higher the calorific value the better the fuel may
not be the case. Too large a heat release may produce temperatures
that cause damage to the arrangement that is containing the
combustion.
Some typical calorific values of fuels are shown below.
COAL
35Mj/kg
WOOD
20Mj/kg
PETROL
45Mj/kg
COAL GAS
20Mj/kg
NATURAL GAS
38Mj/kg
It is important at this stage to note the fact that there are two
calorific values that can be defined.
a) Higher Calorific Value (HCV)also known as the Gross Calorific
Value
b) Lower Calorific Value (LCV) also known as the Net Calorific Value.
The HCV is the energy released when the products of combustion are
cooled to the original fuel temperature. As fuels contain hydrogen
one of the principal products of combustion is water. It is also
possible that there may be a small amount of moisture present in the
fuel anyway. On combustion the water will appear as water vapour in
the exhaust or flue. Only if this exhaust is cooled to the original fuel
temperature will the water vapour condense and then release its
enthalpy of evaporation.
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It is more usual for the water to remain as vapour and consequently
this heat is lost to the plant. The adjustment for the loss of this heat
indicates that the fuel calorific value is less than in the case where
the water vapour condenses. In most cases it is more appropriate to
use this lower value.
7.2.
DETERMINATION OF CALORIFIC VALUES.
The calorific value of the fuel is determined in a container called a
calorimeter. These are especially designed to deal with the differing
rates of burning, so for solid fuels and some liquid fuels the bomb
calorimeter is used. For other liquid fuels and gaseous fuels the gas
calorimeter is used.
A bomb calorimeter.
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A bomb calorimeter is used to measure the energy that is given out
when a substance is completely burned in oxygen. A known amount of
the substance is placed in the crucible, oxygen is added at 20
atmospheres pressure, and then the substance is ignited using the
heated wire. As the calorimeter heats, it heats the water. The rise in
water temperature is measured and from this the heat generated by
the burning substance is calculated.
The assumption is that the mass of fuel multiplied by its calorific
value will generate the heat required to raise the temperature of the
bomb (the steel? container), the water and any peripherals through
the recorded temperature rise.
Clearly accurate temperature measurement is required as is
confidence that equilibrium has been reached.
It is usual to calibrate the bomb calorimeter before evaluation of a
new fuel. This enables the thermal equivalent (mass x specific heat)
of the equipment to be confirmed by using a fuel of known calorific
value.
A file of a powerpoint slideshow of this process is included. It is the
work of Dr Roger Latham, Faculty of Applied Sciences, De Montfort
University.
calibration of a bomb calorimeter
The usual calorimeter used to evaluate the calorific value of a
gaseous fuel is Boys calorimeter. The Boys Gas Calorimeter has been
developed, from apparatus designed by the late Sir Charles Boys
F.R.S., to provide a simple but accurate method of ascertaining the
calorific values of a wide range of gaseous fuels currently in use.
The principle of operation is that gas is supplied at a regulated
pressure at a constant rate, which is metered. This gas is then
ignited at burners located at the base of the calorimeter, with the
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hot gases passing upwards over a set of cooling tubes through which a
controlled flow of water is passing.
The water vapour in the exhaust can be condensed and collected to
enable the LCV to be determined as well.
The simple relationship that the heat gained by the cooling water is
equal to the heat given up by the fuel enables the determination of
the calorific value.
7.3
EFFICIENCY OF POWER PLANT AND COMBUSTION PROCESS
The efficiency is usually simply defined as the ratio of what was got
out compared with what was put in. When dealing with power plant
as a whole it is necessary to consider the efficiency of the method of
producing heat, which involves whether or not the fuel is being
completely burnt, and whether the conditions are at an optimum.
An overall thermal efficiency therefore relates to the work output
compared to the energy stored within the fuel.
If the power plant is an open system with steady flow, then fuel and
air enter, exhaust products leave and there may be both work and
heat transfer
te
tf
W
fuel
air
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Q
exhaust
product s
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If the fuel and air enter at atmospheric temperature the maximum
value of the energy transferred will occur when all the fuel is burnt
completely, when inlet and outlet temperatures are equal and when
water products in the exhaust have condensed.
Under these conditions it is accepted that the efficiency can be
represented by  =
W
H.C.V
In practice it is not possible to reduce the exhaust product’s
temperature to inlet level. The condensing water would absorb the
sulphur dioxide that is a product of most solid fuels, to form sulphuric
acid. As the combustion process is denied the enthalpy of
evaporation of the water, it may be more appropriate to define
efficiency as  =
W
L.C.V
In the case of a boiler, for example, the heat supplied to the water is
determined by the difference in enthalpies between the feed water
and the steam at exit.
The boiler efficiency can therefore be assessed by
 = Δh
L.C.V
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8
PRODUCTS OF COMBUSTION
8.1
VOLUMETRIC ANALYSIS
If a fuel is a pure hydrocarbon the products of combustion when,
burnt completely in air, would be carbon dioxide and water vapour.
However under conditions when there is excess air present there will
be some excess oxygen present in the exhaust products. In a similar
way, if the amount of air is restricted, then not all the carbon will
form carbon dioxide and carbon monoxide will be present.
An analysis of the products of combustion can lead to a determination
of the effectiveness of the combustion process. By taking samples at
stages through a reaction it is possible to find variations of air-fuel
distribution. Analysis of the products of combustion also allows an
air:fuel ratio to be computed.
The most common way to analyse the exhaust gases is by means of
the Orsat apparatus.
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measuring
buret t e
gas
levelling
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reagent bot t le
t ubes t o increase
wet t ed surface
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The Orsat apparatus is used to measure volumes of carbon dioxide,
oxygen and carbon monoxide within a fixed volume of a sample gas
(100cc). However it is not particularly accurate at detecting very low
concentrations.
The Orsat apparatus works using a very simple method. To find the
volume of a particular gas within the sample a fixed volume of the
sampled gas is passed through a specific solution which absorbs only
the required gas. The remaining volume of the gas can then be remeasured and compared with the original volume to find the
proportion of a specific gas within the sample.
The sampled gas is passed through a sequence of solutions each
removing one of the gases and the remaining volume calculated
afterwards. A solution of caustic potash is used to absorb the carbon
dioxide, a mixture of pyrogallic acid, caustic potash and water is used
to absorb the oxygen, and to remove the carbon monoxide a solution
of cuprous chloride is used.
8.1
PRODUCT PROPORTIONS AND AIR:FUEL RATIO
As has been mentioned before the air:fuel ratio can be an indicator
of the effectiveness of mixing of air and fuel, as well as a measure of
the efficiency of the combustion process. The analysis of the
products of combustion by the Orsat method, or any other, can give
valuable information about the combustion process.
For any hydrocarbon fuel the hydrogen will appear in the exhaust as
water vapour. The carbon will be either carbon dioxide or carbon
monoxide, or both.
If too little air is supplied the initial air:fuel ratio is low indicating a
rich mixture. Generally the hydrogen will be burnt off first leaving
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too little air for complete combustion of the carbon. Nitrogen from
the air will also appear in the exhaust products
If too much air is provided the air:fuel ratio will be high indicating a
weak mixture. Again the hydrogen will burn off first as water vapour,
the carbon will burn to carbon dioxide and there will be some oxygen
left in the exhaust along with the nitrogen.
This then explains the reasoning behind the Orsat analysis. The
water vapour is nearly always burnt off and consequently the degree
to which the products carbon dioxide, carbon monoxide and oxygen
appear will be the indicators of combustion efficiency.
Nitrogen in the air tends to get in the way of direct contact between
oxygen and the fuel. Improved combustion efficiency is obtained by
the introduction of turbulence in the combustion zone. This can be
effected by many methods – flare plates in the Olympus 301 engine,
secondary and tertiary air holes in annular combustion chambers,
piston head designs to encourage rotation of the mixture, etc, etc.
Similarly, any attempt to reduce the fuel droplet size will aid burning
in that the surface area will become larger in comparison to the
volume. Therefore fuel spraying or injection will generally improve
efficiency.
It will seem fairly safe to say that the efficiency of the combustion
process can be indicated by the level of CO. If the air content is
increased then the CO content will be reduced, until the point is
reached where excess oxygen is detected and the excess air condition
is reached.
The following diagram shows the percentage variation of CO 2, CO and
O2 against air:fuel ratio.
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C 02
O2
CO
FUEL : AIR RATIO
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ANSWERS TO SELF-TEST QUESTIONS.
6.2a
There is no set answer here. Your choices should be discussed with
your tutor to determine the likelihood of their validity. It is probable
that checks will be made to confirm suggestions.
6.3a
Name
Molecular weight
Boiling point 0C
Methane
16
-164
Ethane
Propane
Butane
30
44
58
-89
-44.5
-0.5
Pentane
72
36
BOILING TEMP verses MOLECULAR WEIGHT
50
0
0
10
-50
-100
-150
-200
6.3b
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20
30
40
50
60
70
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Name
Ethylene
Propylene
Butylene
Pentylene
Molecular Weight
26
42
56
70
Boiling Point 0C
-105
-40
-6.1
30.2
MOLECULAR WEIGHT verses BOILING POINT
40
20
0
0
10
20
30
40
50
60
70
80
-20
-40
-60
-80
-100
-120
combined
50
0
0
10
20
30
40
50
60
70
-50
-100
-150
-200
A linear correlation would be appropriate here!
6.5a 16kW
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6.5b 13.206 by weight
25.547 by volume
6.5c 6.085
6.6a a 4.292
b 1.609
c If 50% by volume then 2.95
If 50% by mass then 2.673
6.6b 11.44
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