COURSE NAME: ENGINEERING CHEMISTRY
Course code
lecture series no
Credits
Mode of delivery
Faculty
Email-id
: CY 1001
: 01 (one)
:
3
: OFLINE
: Dr. Suranjan De
Suranjan.de@Jaipur.manipal.edu
Syllabus
Unit-I Classification of Fuels, Gross Calorific value and Net Calorific
value. Solid, Liquid and Gaseous fuels.
Unit-II: Advanced materials and polymers: Liquid crystals, ceramics,
composites, bio-materials, nanomaterials, thin films and their
properties and applications.
Unit-III: Water treatment technology.
Unit- IV: Concept of corrosion and its importance, types of corrosion,
factors affecting corrosion, Corrosion control methods. Chemistry of
primary and secondary batteries. Working principles of fuels cells and
their applications.
Unit-V: Theory and application phase rule (up to two component
system).
Unit VI; General methods of chemical analysis, Instrumental methods.
Introduction to spectroscopic methods of analysis: Electromagnetic
radiation (EMR), Interaction of EMR with matter, Numerical Problems.
Concepts of rotational, vibrational and electronic spectra, Laws of
spectrophotometry.
Assignment
Quiz
Mid term examination –I/II
End Term Examination
Assessment Criteria’S
Criteria
Internal Assessment(Summative)
End Term Exam
(Summative)
Attendance(Formative)
Homework/ Home Assignment/
Activity Assignment
(Formative)
Description
Maximum Marks
Mid Term Examination I
Quizzes (03) and Assignment (03)
20
20
20
End Term Exam (Closed Book)
40
Total
100
Mid Term Examination II
A minimum of 75% Attendance is required to be maintained by a student to be
qualified for taking up the End Semester examination. The allowance of 25%includes all
types of leaves including medical leaves.
There are situations where a student may have to work at home/ hostel especially
before a quiz test or MTE-I/MTE-II. A student is expected to participate andperform
these assignments with full zeal since the activity.
Course Outcomes:
At the end of the course, students will be able to:
1. Understand and apply concepts of various types of fuel technology
2. Develop skill of synthesis and applications of polymer and some advanced
materials.
3. Applying knowledge about the water softening methods.
4. Understand and apply the concepts in electrochemistry and corrosion science in
protecting metallic objects.
5. Develop concept of phase rule
6. Gain skill in various modern analytical techniques.
“Recall chemical fuels
SESSION OUTCOME
and its types. Describe
types of calorific
values and their
classification”
Text Books
1. P.C. Jain and Monika Jain, Engg. Chemistry, Dhanpat Rai and Sons, Delhi, Revised, 2006.
2. J.C. Kuriacose, J. Raja ram, Chemistry in Engg. and Technology, Vol I/II Tata McGraw –
Hill(TMH), New Delhi 1988.
3. Engineering chemistry, Wiley India Pvt. Ltd., 2018.
Reference Books
1. Fuel Science &Technology Hand Book, James G Speight; Marcel Dekker, New York.
Lecture 1: Chemical fuels
Introduction
Fuel –defined as any substance used to produce heat or power by combustion. Any
chemical process accompanied by the evolution of light and heat is called
combustion.
Fuel + oxygen
Combustion products + Heat
Ex.
C (s) + O2 (g)
H2
+ ½O2
CO2 (g) + Δ H = - 94.1 Kcal/mol
H2O + Δ H = -67.5 Kcal/mol
During the process of combustion, the atoms of carbon, hydrogen etc. combine with
oxygen with the simultaneous liberation of heat at a rapid rate. Energy is liberated
due to rearrangement of valence electrons.
https://www.sciencephoto.com/media/736202/view/methane-combustionreaction-animation
Classification of Fuels
Classification of Fuels
Fuels are classified on the basis of their occurrence and physical state
Classification of fuels Based on occurrence
1) Based on their origin they are classified into
a) Primary fuels
b) Secondary fuels.
a) Primary Fuels: There are naturally occurring fuels which serves as
source of Energy without any chemical processing.
Ex: Wood, Coal, Crude oil, Natural gas, Peat, Lignite, Anthracite.
b) Secondary Fuels: - They are derived from primary fuels & serves as
source of Energy only after subjecting to chemical processing.
Ex: Charcoal, Coke, produser gas, Petrol, Diesel etc.,
Units of heat
Units of heat
The quantity of heat can be measured in the following units:
(i) Calorie: It is defined as the amount of heat required to raise the temperature of
1gm of water by 1oC
1 calorie = 4.2 Joules
(ii) Kilo Calorie: 1 k cal = 1000 cal
(iii) British thermal unit: (B. Th. U. or B.T.U) It is defined as the amount of heat
required to raise the temperature of 1 pound of water through 1oF.
1 B.T.U = 1,054.6 Joules
(IV) Centigrade heat unit (C.H.U): It is defined as the amount of heat required to raise
the temperature of 1 pound of water through 1oC.
Interconversion of the various units of heat:
1K cal = 1000 cals = 3.968 B.T.U = 2.2 C.H.U.
1 B.T.U. = 252 Cal = 0.252 k cal
Calorific value: It is defined as the total quantity of heat liberated when a unit mass of
a fuel is burnt completely
Units of Calorific value:
These units can be interconverted as follows:
1 cal/g = 1 K cal/Kg = 1.8 B.Th.U./Ib
1 K cal/m3 = 0.1077 B.Th. U./ft3
1 B.Th.U./ft3 = 9.3 Kcals/m3
Calorific Value
Gross Calorific Value: It is the total amount of heat generated when a unit
quantity of fuel is completely burnt in oxygen and the products of combustion
are cooled down to the room temperature.
As the products of combustion are cooled down to room temperature, the steam gets
condensed into water and latent heat is evolved. Thus in the determination of gross
calorific value, the latent heat also gets included in the measured heat. Therefore,
gross calorific value is also called the higher calorific value.
The calorific value which is determined by Bomb calorimeter gives the higher calorific
value (HCV)
Net Calorific Value: It is defined as the net heat produced when a unit quantity
of fuel is completely burnt and the products of combustion are allowed to
escape.
The water vapour do not condense and escape with hot combustion gases. Hence,
lesser amount than gross calorific value is available. It is also known as lower calorific
value (LCV).
LCV=HCV-Latent heat of condensation of the water vapours formed
= HCV – (weight of hydrogen × 9 × latent heat of steam)
LCV= HCV-Latent heat of condensation of the water vapors formed
= HCV – (weight of hydrogen × 9 × latent heat of steam)
1 part by weight of hydrogen gives 9 parts by weight of water as follows:
H2
2g
1g
+
O
8g
8g
H2O
18 g
9g
The latent heat of steam is 587 Cal/g (or Kcal/Kg) or 1060 B.Th.U./Ib of water vapor
produced.
Net C.V. or L.C.V = Gross C.V. – 9 × H/100 × 587
= Gross C.V. – 0.09 × H × 587
Where H = % of hydrogen in the fuel.
Calorific value: It is defined as the total
quantity of heat liberated when a unit
mass of a fuel is burnt completely
Determination of Calorific value
Bomb calorimeter
Boy’s calorimeter
Working of calorimeter, video:
https://www.youtube.com/watch?v=wwJG2JVg6qM
Bomb Calorimeter
Determination of Calorific value
1. Bomb calorimeter
Determination of calorific value of solid and nonvolatile liquid fuels
Principle: A known mass of the solid sample is burnt in excess oxygen. The surrounding water
and the calorimeter absorb the heat liberated. Thus, the heat liberated by the fuel is equal to
the heat absorbed by the water and the calorimeter.
Construction: The bomb calorimeter consists of a stainless-steel vessel with an airtight lid. This
vessel is called bomb. The bomb has an inlet valve for providing oxygen atmosphere inside the
bomb and an electrical ignition coil for starting of combustion of fuel. The bomb is placed in an
insulated copper calorimeter. The calorimeter has a mechanical stirrer for dissipation of heat
and a thermometer for reading the temperature.
Working: A known mass of the solid fuel is placed in a crucible. The crucible is placed inside the
bomb. The lid is closed tightly. The bomb is placed inside a copper calorimeter. A known mass of
water is taken in the calorimeter. The bomb is filled with oxygen at a pressure of 25-30 atm. The
temperature t1 in the thermometer is noted. On passing an electric current through the ignition
coil, the fuel gets ignited. The fuel burns liberating heat. The water is continuously stirred using
the stirrer. The maximum temperature attained by the water, t2 , is noted.
Calculations
Let weight of the fuel sample taken = x gm
Weight of water taken in the calorimeter = W gm
Water equivalent of the Calorimeter, stirrer, bomb, thermometer etc. = w gm
Initial temperature of water = t1oC
Final temperature of water = t2oC
Higher or gross calorific value of the fuel = C cal/gm
Heat gained by water = W x Dt x specific heat of water
= W (t2-t1) x 1 cal
Heat gained by Calorimeter = w (t2-t1) cal
Heat liberated by the fuel = x cal
Heat liberated by the fuel = Heat gained by water and calorimeter
x C = (W+w) (t2-t1) cal
𝑾+𝒘 (𝒕𝟐−𝒕𝟏)
C=
𝒙
cal/g
Net Calorific value:
Let percentage of hydrogen in the fuel = H
mass of water produced from 1 g of the fuel = 9H/100 g
Heat liberated during condensation of steam
= 0.09H  587 cal
Net (Lower calorific value) = GCV-Latent heat of water formed
= GCV-0.09H  587 cal/g
Corrections: For accurate results the following corrections will have to be
incorporated:
(a) Acid Correction (tA): During combustion, sulphur and nitrogen present in the fuel
are oxidized to their corresponding acids under high pressure and temperature.
S
+
O2
SO2
2SO2
+
O2 + 2H2O
2 H2SO4 + Heat; Δ H = - 144,000 cal
2N2
+
5O2 + 2H2O
4 HNO3
+ Heat; Δ H = - 57,160 cal
The amount of H2SO4 and HNO3 is analyzed by washings of bomb by titration while H2SO4 is
determined by precipitation as BaSO4.
For each ml of 0.1 N H2SO4 formed, 3.6 calories should be subtracted.
For each ml of 0.1 N HNO3 formed, 1.43 calories must be subtracted.
(B) Cooling correction (tc): As the temperature rises above the room temperature, the loss of
heat does occur due to radiation, and the highest temperature recorded will be slightly less than
that obtained. A temperature correction is therefore necessary to get the correct rise in
temperature.
If the time taken for the water in the calorimeter to cool down from the maximum temperature
attained, to the room temperature is x minutes and the rate of cooling is dto/min, then the cooling
correction = x  dt. This should be added to the observed rise in temperature.
(C) Fuse wire correction(tf): As metal wire is used for ignition, the heat generated by burning of
metal wire is also included in the gross calorific value. Hence this amount of heat has to be
subtracted from the total value.
(D) Cotton thread correction (tt): The correction for the cotton thread used for firing the charge
is calculated from the weight of the dry cotton thread actually used. Hence this amount of heat
has to be subtracted from the total value.
Therefore, Gross calorific value
GCV =
(𝑾+𝒘) 𝒕𝟐−𝒕𝟏+𝒕𝒄 −(𝒕𝒂+𝒕𝒇+𝒕𝒕)
cal/g
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒇𝒖𝒆𝒍
Boy’s calorimeter
Determination of calorific value of gases and volatile liquid fuels
Principle: principle involved in this method is to burn the gas at a known constant
rate in a vessel under such conditions that the entire amount of heat produced is
absorbed by water which is also flowing at a constant rate. Thus the heat liberated by
the fuel is equal to the heat absorbed by the water and the calorimeter.
Construction: Boy’s calorimeter consists of a combustion chamber surrounded by
water tube made of copper with two thermometers T1 and T2 attached. There is a
burner in the chamber, which is connected to a gas tube through which a known
volume of gaseous fuel is burnt at a known pressure at a rate of 3-4 l/min.
Working: water at a constant rate is allowed to pass through the water tube. The
temperatures of incoming and outgoing water are recorded by thermometers t1 and t2
respectively. When the gaseous fuel is burnt, water passing through the tubes takes
all the heat produced. The water is collected in a measuring jar. The whole set up is
enclosed in an insulated container.
Boy’s Calorimeter
https://www.youtube.com/watch?v=neCpRT7XpqE
Calculations:
Let
W = wt of cooling water passed in time t
V = volume of gas burnt at S.T.P. in time t
t1 = temperature of the incoming water
t2 = temperature of the outgoing water
m = weight of steam condensed in time t in a graduated cylinder
θ = HCV of fuel
Heat absorbed by circulating water = W (t2-t1)
Heat produced by combustion of fuel = V θ
Assuming no heat loss
V θ = W (t2-t1)
HCV = θ =
𝑾(𝒕𝟐−𝒕𝟏)
𝑽
Kcal/m3
Mass of H2O condensed per m3 of gas =
Latent heat of steam per m3 of gas =
Net calorific value = [HCV –
𝒎
𝑽
𝒎
𝑽
Kg
× 587 Kcal
𝒎 𝑿 𝟓𝟖𝟕
] Kcal/m3
𝑽
In an experiment in a Bomb calorimeter, a solid fuel of 0.90
g is burnt. It is observed that increase of temperature is 3.8
oC of 4000 g of water. The fuel contains 1% of H. calculate
the HCV and LCV value (equivalent weight of water = 385 g
and latent heat of steam = 587 cal/g)
18514.5 cal/g
18461.6 cal/g
On burning 0.72 g of a solid fuel in a Bomb
calorimeter, the temperature of 250 g of water is
increased from 27.3 oC to 29.1 oC. If the water
equivalent is 150 g, calculate the HCV of the fuel.
1000 cal/g
HM
A coal sample with 93% carbon, 5% of Hydrogen and 2% Ash is
subjected to combustion in a bomb calorimeter. Calculate GCV and
NCV Given that. Mass of the coal sample = 0.95g. Mass of water in
copper calorimeter = 2000g. Water equivalent wt of calorimeter =
700g. Rise in temp = 2.80 C Latent heat of = 587 cal/g.
When 0.84 g of coal was burnt completely in Bomb
calorimeter the increase in temp of 2655 grams of
water was 1.850 C if the water equivalent calorimeter
is 156g. Calculate GCV.
The following data are obtained in a Bomb Calorimeter experiment.
Weight of coal burnt = 0.95 g
Weight of water taken = 700 g
Water equivalent of calorimeter = 2000g
Increase in temperature = 2.48 0C
Acid correction = 60.0 cal
Cooling correction = 0.02 oC
Fuse wire correction = 10.0 cal
Latent heat of condensation = 587 cal/g
Calculate the GCV and NCV of the fuel if the fuel contains 92% of C. 5%
of H and 3% of
ash.
Dulong’s formula for calculating the calorific value
is given as:
Gross calorific Value (HCV)
1
O
=
[8080C + 34,500( H − ) + 2,240S ]kcal / kg
100
8
Net Calorific value (LCV)
9H
= [ HCV −
 587]kcal / kg
100
= [ HCV − 0.09 H  587]kcal / kg
9650, 9227
Coal: coal is regarded as a fossil fuel produced from the vegetable debris under
conditions of high temperature and pressure over million of years.
The transformation of the vegetable debris to coal takes place in two stages:
(a) Biochemical or peat stage: During this stage, the plant materials were
attacked by various micro organisms.
(b) Chemical stage or metamorphism: In this stage, the peat deposit buried
under sedimentary deposits lose moisture and volatile components under the
effect of high temperature and pressure.
Classification of Coal: Coals are mainly classified on the basis of their degree of
coalification from the parent material, wood. When wood is converted into coal,
there is gradual increase in the concentration of carbon and decrease in the
percentage of oxygen and nitrogen.
Coal is given a ranking depending upon the carbon content of the coal from wood to
anthracite.
The main solid fuels are wood, peat, lignite, coal and charcoal.
Coal: - Coal is a fossil fuel which occurs in layers in the earths crust.
It is formed by the partial decay of plant materials accumulated
millions of years of ago and further altered by action of heat and
pressure.
The process of conversion of wood into coal can be represented as
Wood >Peat > Lignite> Bituminous Coal > Anthracite
Type of coal
Percentage
%
moist
ure
C
H
O
N
VM
peat
45-60
3.5-6.5
20-45
0.75-3
45-75 70-90
lignite
25-30
3.5-6.5
20-45
0.75-2
45-75 60-80
Sub-bituminous
35-45
4.5-5.5
17-35 0.75-2
45-60 30-50
Bituminous
45-86
4.0-5.5
20-30
0.75-2
11-50 6-10
Anthracite
86-97
3-4
2-3
0.5-2
3.8-8
1.5-3.5
Analysis of Coal
Coal is analysed in two ways:
1. Proximate analysis
2. Ultimate analysis
Proximate Analysis
The data varies with the procedure adopted and hence it is called proximate analysis.
Proximate analysis of coal determines the moisture, ash, volatile matter and fixed
carbon of coal.
1.
Moisture Content: moisture is determined by heating a known amount of coal to
105-110 oC in an electric hot air oven for about one hour. After one hour, it is taken
out from the oven and cooled in a desiccator and weighed.
Percentage of moisture
=
Loss in weight × 100
Weight of coal taken
2. Volatile Matter: consists of a complex mixture of gaseous and liquid
products resulting from the thermal decomposition of the coal. It is determined by
heating a known weight of moisture free coal sample in a covered platinum crucible at
950  20oC for 7 minutes.
Percentage of volatile matter = Loss of weight due to volatile matter × 100
Weight of coal sample taken
Ash: Coal contains inorganic mineral substances which are converted into ash by
chemical reactions during the combustion of coal. Ash usually consists of silica,
alumina, iron oxide and small quantities of lime, magnesia etc.
Ash content is determined by heating the residue left after the removal of volatile
matter at 700  50oC for ½ an hour without covering
Weight of the residue left × 100
Percentage of ash =
Weight of the coal
Fixed Carbon: Fixed carbon content increases from lignite to anthracite. Higher the
percentage of fixed carbon greater is its calorific value and better is the quality of coal.
The percentage of fixed carbon is given by:
Percentage of fixed carbon = 100 - % of [moisture + volatile matter + ash]
Significance of proximate analysis:
(i) Moisture:
•Excess of moisture is undesirable in coal. Increases transport costs.
•Moisture lowers the heating value of coal and takes away appreciable amount of the
liberated heat in the form of latent heat of vapourisation.
•Presence of excessive moisture quenches fire in the furnace.
(ii) Volatile mater:
▪A high percent of volatile matter indicates that a large proportion of fuel is burnt as a
gas.
▪The high volatile content gives long flames, high smoke and relatively low heating
values.
▪For efficient use of fuel, the outgoing combustible gases has to be burnt by supplying
secondary air.
▪High volatile matter content is desirable in coal gas manufacture because volatile
matter in a coal denotes the proportion of the coal which will be converted into gas
and tar products by heat.
(iii) Ash:
➢The high percentage of ash is undesirable. It reduces the calorific value
of coal.
➢In furnace grate, the ash may restrict the passage of air and lower the
rate of combustion.
➢Apart from loss of efficiency of coal, clinker formation also leads to loss
of fuel because some coal particles also get embedded in the clinkers.
(iv) Fixed carbon:
Higher the percentage of fixed carbon, greater its calorific value
Ultimate Analysis
It is carried out to ascertain the composition of coal.
Ultimate analysis includes the estimation of carbon, hydrogen, sulphur, nitrogen
and oxygen.
1.
Carbon and Hydrogen: A known amount of coal is taken in a combustion
tube and is burnt in excess of pure oxygen.
C + O2
H2
+
CO2
½O2
H2O
Fig. Estimation of carbon and hydrogen
2 KOH + CO2
CaCl2
+
K2CO3
7 H2O
+ H2O
CaCl2.7H2O
The % of C and H are calculated from the increase in weight of
the respective absorption tube
y 100
Percentage of carbon = 12 
44 weight of coal taken
z100
Percentage of hydrogen = 2 
18 weight of coal taken
Y = Increase in weight of KOH tube
Z = Increase in weight of CaCl2tube
2. Nitrogen: Nitrogen present in coal sample can be estimated
by Kjeldahl’s method
Nitrogen + H SO ⎯Heat
⎯⎯⎯⎯
⎯→( NH ) SO
2 4
42 4
( NH ) SO ⎯2⎯NaOH
⎯ ⎯ ⎯⎯→ Na SO + 2 NH + 2H O
42 4
2 4
3
2
NH + H SO → ( NH ) SO
3
2 4
42 4
The contents are then transferred to a round bottomed flask and solution is heated
with excess of NaOH.
The ammonia gas thus liberated is absorbed in a known volume of a standard
solution of acid used.
The unused acid is then determined by titrating with NaOH. From the volume of acid
used by NH3 liberated, the percentage of nitrogen can be calculated.
Vol. of acid used × normality of acid × 1.4
% of nitrogen =
wt. of coal taken
3 Sulfur: Sulfur is determined by conveniently from the bomb washing obtained
from the combustion of a known mass of coal in the bomb calorimeter experiment for
the determination of calorific value. The washings contain sulfur in the form of sulfate
from which it is precipitated as BaSO4. The precipitate is filtered, ignited and
weighed. From the weigh of BaSO4 obtained the sulfur present in the coal is
calculated.
wt. of BaSO4 obtained × 32 × 100
% of sulfur =
wt. of coal taken in bomb × 233
5. Oxygen: determined by difference
% of O = 100 - % of (C + H + S + N + ash)
Liquid Fuels
Liquid Fuels: The importance of liquid fuels is the fact that
almost all combustion engines run on them.
The largest source of liquid fuels is petroleum.
Petroleum: The term petroleum means rock oil (Latin words;
petra = rock and oleum = oil). It is also called mineral oil.
Petroleum is a dark colored viscous liquid found deep in the
earth’s crust and is a complex mixture of paraffinic, olefinic
and aromatic hydrocarbons with small quantities of organic
compounds containing oxygen, nitrogen and sulphur and
traces of metallic constituents.
Composition:
Element
Carbon Hydrogen Sulphur Oxygen Nitrogen
Percentage 80-87
11.1-15
0.1-3.5
0.1-0.9
0.4-0.9
Petroleum is found deep below the earth crust. The oil
is found floating over salt water or brine. Generally,
accumulation of natural gas occurs above the oil.
Mining of Crude Petroleum:
Fig. Mining of oil
Refining of Petroleum
Refining of Petroleum
Steps involved in refining of petroleum:
(i) Demulsification: The crude oil coming out from the well, is in the form of
stable emulsion of oil and salt water, which is yellow to dark brown in colour.
The demulsification is achieved by Cottrell’s process, in which the water is
removed from the oil by electrical process. The crude oil is subjected to an
electrical field, when droplets of colloidal water coalesce to form large drops
which separate out from the oil.
(ii) Removal of harmful impurities: Excessive salt content such as NaCl
and MgCl2 can corrode the refining equipment. These are removed by
washing with water.
The objectionable sulphur compound are removed by treating the oil with
copper oxide. The copper sulphide so formed is separated by filtration.
(iii) Fractional Distillation: It is done in tall fractionating tower
or column made up of steel.
• In continuous process, the crude oil is preheated to 400 oC in
specially designed tubular furnace known as pipe still.
Fig.: Fractional distillation of crude petroleum
•
The hot vapours from the crude are passed through a tall
fractionating column, called bubble tower.
•
Bubble tower consists of horizontal trays provided with a no
of small chimneys, through which vapours rise.
•
These chimneys are covered with loose caps, known as
bubble caps. These bubble caps help to provide an intimate
contact between the escaping vapours and down coming
liquid.
•
The temperature in the fractionating tower decreases
gradually on moving upwards.
•
As the vapours of the crude oil go up, they become
gradually cooler and fractional condensation takes place at
different heights of column.
The residue from the bottom of the fractionating tower
is vacuum distilled to recover various fractions
Fig.: Vacuum distillation of residual oil
A brief description of the various fractions
Uncondensed gases: composition range is C1 to C5. Boiling range 0 to 30 oC. It is
used as a gaseous fuel.
1. Petroleum ether: Composition range is C5 to C6 and boiling range is 30-70 oC.
2. Gasoline or petrol: Composition range is C5 to C8 and boiling range is 70-120 o C.
Its approximate composition is: C = 84%, H = 15%, N + S + O = 1% and calorific
value is about 11250 kcal/kg. It is highly volatile, inflammable and used as a fuel
for internal combustion engines of automobiles and aeroplanes.
3. Naphtha: Composition range is C8 to C10 and boiling range is 120-150 oC. Used
as solvent.
4. Kerosene oil: Composition range is C10 to C15 and boiling range is 150-250 oC. Its
approximate composition is: C = 84%, H = 16%, with less than 0.1% S and
calorific value is about 11100 kcal/kg. It consists of paraffines, naphthenes and
aromatic compounds. It is used as domestic fuel in stoves, as jet engine fuel and
for making oil gas.
5. Diesel oil or gas oil: Composition range is C12 to C18 and boiling range is 250-320
oC. It is used as a fuel for diesel engines, heating oil, for cracking to obtain petrol.
6. Heavy oil: Composition range is C17 to C40 and boiling range is 320- 400 oC.
https://www.youtube.com/watch?v=PYMWUz7TC3A
Therefore, heavier fractions are converted into more useful fraction, gasoline.
Catalytic cracking: Cracking is brought about in the
presence of a catalyst at much lower temperatures
and pressures. The catalyst used is mainly a mixture
of silica and alumina. Most recent catalyst used is
zeolite. The quality and yield of gasoline is greatly
improved by this method.
Types of catalytic cracking
(a) Fixed bed catalytic cracking: In this case, the
catalyst is in the form of granules or pellets. These
granules are used in the form of fixed beds in catalyst
towers.
1.
2.
3.
4.
5.
6.
7.
The heavy oil charge is converted into vapor form by passing through a
preheater, maintained at 425-450 oC.
The vapors are then passed through a catalytic chamber (containing silica
alumina gel mixed with zirconium oxide) maintained at 425-450oC and 1.5 kg/
cm2 pressure.
During their passage through the tower, about 40% of the charge is converted
into gasoline and about 2-4% carbon is formed. The carbon gets deposited on
the catalysts bed.
The cracked vapors now enter the fractionating column where the gasoline
vapors and other gaseous products are recovered from the top while the
heavy gas oil fractions are condensed at the bottom of the column.
The vapors are then admitted through a cooler, where some of the gases are
condensed along with gasoline and uncondensed gases move on.
The gasoline containing some dissolved gases is then sent to a stabilizer where
dissolved gases are removed and pure gasoline is obtained.
The catalyst, after some time stop functioning, due to the deposition of black
layer of carbon, formed during cracking. This is reactivated by burning off the
deposited carbon.
(b) Moving bed catalytic cracking: This is also known
as fluid bed catalytic cracking. In this type of cracking
the catalyst is in the form of fine powder which flows
down to the cracking chamber in gas stream.
1. The vapors of cracking stock (gas, oil and heavy oil, etc) mixed with fluidized
catalyst is forced up into a large reactor bed in which cracking of the heavier into
lighter molecules occurs.
2. Cracking takes place on the surface of the turbulent catalyst bed as it circulates
with the oil vapors in the reactor at a temperature of ~ 530 oC and pressure of
about 3-5 kg/cm2.
3. Near the top of the reactor, there is a centrifugal separator, cyclone, which allows
only the cracked oil vapors to pass on to the fractionating column, but retains all
the catalyst powder in the reactor itself.
4. The catalyst powder gradually becomes heavier, due to coating with carbon, and
settles to the bottom from where it is forced by an air blast to regenerator,
maintained at 600oC.
5. In regenerator, carbon is burnt and the regenerated catalyst then flows through a
stand-pipe for mixing with fresh batch of incoming cracking oil. At the top of the
regenerator, there is a separator, which permits only gases (CO2 etc.) to pass out,
but holds back catalyst particles.
https://www.youtube.com/watch?v=Xsqlv4rWnEg
https://www.youtube.com/watch?v=GYRwWyG3Qqw
Knocking and Anti-knocking
In an internal combustion engine (spark-ignition type), a phenomenon that
occurs when unburned fuel-air mixture explodes in the combustion chamber
before being ignited by the spark. The resulting shock waves produce a metallic
knocking sound.
Relationship of knocking with the chemical constituents:
It has been found that the knocking tendency is higher, when straight-chain
hydrocarbon are used in engines, whereas the branched chain hydrocarbons,
cycloalkanes and aromatics have lesser tendency to knock
The knocking tendency follows the following order:
Aromatics < cycloalkanes < olefins < branched chain alkanes < straight chain alkanes
An antiknock agent is a gasoline additive used to reduce engine knocking and
increase the fuel's octane rating.
The typical antiknock agents in use are:
Tetra-ethyl lead (toxic)
MTBE – methyl tertiary butyl ether
Methyl cyclo pentadienyl manganese tricarbonyl (MMT)
Ferrocene, Iron pentacarbonyl, Toluene, Isooctane
Octane number: is defined as the percentage of iso-octane present in a
mixture of iso-octane and n-heptane, which has the same knocking
characteristics as that of fuel under examination, under same set of conditions.
Thus a gasoline with an octane no of 80, would give the same knocking as a
mixture of iso-octane and n-heptane containing 80% of iso-octane by volume.
Greater the octane number, greater is the antiknock property of the fuel.
Cetane number: Fuels required for diesel engine are in contrast to petrol
engine fuels, hence a separate scale is used to grade the diesel oils as they
cannot be graded on octane number scale.
The cetane number of a diesel oil is defined as the percentage of cetane in a
mixture of cetane and a-methyl naphthalene which will have the same ignition
characteristics as the fuel under test, under same set of conditions.
Cetane is n-hexadecane
Reforming: Change the structure of molecules
it is a process of bringing about structural modifications in the components of
straight-run-gasoline with the primary object of improving its anti-knock
characteristics.
It is accomplished by an increase in volatility (reduction of molecular size) or by the
conversion of n-paraffins to isoparrafins, olefins and aromatics and of naphthenes to
aromatics
Cracking converts heavier oils into gasoline constituents while reforming converts these
gasoline constituents into higher-octane molecule.
It is carried out either thermally or in the presence of a catalyst
1. Thermal reforming:
It is carried out in a reactor at 500-600 oC and a pressure of 85 atm. To avoid
the
formation of gas, the conditions are controlled carefully by rapid
cooling of the
products. The reformed are then fractionated to remove
residual gases. During
thermal reforming, some cracking also occurs to
yield alkanes and alkenes and
these might undergo dehydrogenation
followed by dehydrocyclization to yield
naphthelenes (aromatization) .
Conversion of n-alkanes to branched chain
alkanes also takes place during
thermal reforming.
2. Catalytic reforming:
To get better grade and yield of gasolines (thermal- octane number: 65-80
and
catalytic – octane number: 90-95), catalytic reforming (Pt (0.75%)
supported on
alumina) is carried out by using either a fixed-bed or fluidized-bed at 460-530 oC and
35-50 atm pressure. The main reactions
during catalytic reforming process are:
i)
-3H2
Dehydrogenation:
ii) Dehydrocyclization:
n-C7H16
-H2
iii) Hydrocracking
CH3-(CH2)8-CH3 + H2
i)
Isomerization:
2 CH3-(CH2)3-CH3
-3H2
Producer gas
Producer gas is essentially a mixture of combustible gases, CO and H2 associated with
large percent of non-combustible gases, N2, CO2 etc.
A typical producer gas obtained from coke contains 27% carbon monoxide (CO), 12%
hydrogen (H2), 0.5% methane, 5% carbon dioxide and 55% nitrogen, by volume. It has
a heating value of about 5,000 kJ/m3 [1].
When coal is used as fuel the producer gas contains about 3% methane and 0.5%
higher hydrocarbons.
Applications:
It is cheap, clean and easily preparable gas and is used
(i) primarily as an industrial fuel for iron and steel manufacturing, such as firing coke
ovens and blast furnaces,
(ii) cement and ceramic kilns
(iii) for mechanical power through gas engines.
(iv) As a reducing agent in metallurgical operations
1. Himus, G. W. (1972) The Elements of Fuel Technology, Leonard Hill, London. Gas Making and Natural
Gas, BP Trading Ltd.
Water gas (blue gas)
Water gas is a poisonous flammable gaseous mixture that consists
chiefly of carbon monoxide (CO) and hydrogen (H2) with small
amounts of methane, carbon dioxide, and nitrogen.
Applications:
(i) Used as a source of hydrogen gas.
(ii)Used as fuel gas
(iii)Used as an illuminating gas
Synthetic petrol:
Direct conversion of coal to synthetic fuel was originally developed in Germany. The
Bergius process was developed by Friedrich Bergius, yielding a patent in 1913.
Indirect coal conversion (where coal is gasified and then converted to synthetic fuels)
was also developed in Germany by Franz Fischer and Hans Tropsch in 1923. During
World War II, Germany used synthetic oil manufacturing to produce substitute oil
products by using the Bergius process (from coal), the Fischer Tropsch process (water
gas) and other methods
Gasoline can be produced from coal by the following two methods:
1.
Fischer-Tropsch process:
2.
Bergius process
1.
Fischer-Tropsch method:
i)
Water gas (CO + H2) produced by passing steam over heated coke, is mixed with
hydrogen.
C +
i)
H2O
1200 𝑜𝐶
CO
+ H2
The gas is purified by passing through Fe2O3 (to remove H2S) and then into a
mixture of Fe2O3.Na2CO3 (to remove organic sulphur compounds).
ii) The purified gas is compressed to 5-25 atm and then led through a convertor
(containing a catalyst, consisting of a mixture of 100 parts cobalt, 5 parts thoria,
8 parts magnesia, and 200 parts keiselguhar earth), maintained at about 200300 oC.
iii) A mixture of saturated and unsaturated hydrocarbons results:
n CO +
n CO +
i)
2n H2
(2n + 1) H2
CnH2n + n H2O
CnH2n+2 + n H2O
The reaction is exothermic, so outcoming hot gaseous mixture is led to a cooler,
where a liquid resembling crude oil is obtained. The crude oil thus obtained is
then fractionated to yield: (i) gasoline (ii) high boiling heavy oil
2. Bergius process:
1. The low ash coal is finally powdered and made into a paste with heavy oil
and then a catalyst (composed of nickel and tin oleate) is incorporated.
2. The whole is heated with hydrogen at 450oC and under a pressure 200-250
atm for about 1.5 hrs, during which hydrogen combines with coal to form
saturated hydrocarbons, which decompose at prevailing high temperature
and pressure to yield low-boiling liquid hydrocarbons.
3. The gases from the reaction vessel are led to condenser, where a liquid
resembling crude oil is obtained, which is then fractionated to get: (i)
gasoline (ii) middle oil (iii) heavy oil.
4. The latter is used again for making paste with fresh coal dust. The middle
oil is hydrogenated in vapor-phase in presence of a solid catalyst to yields
more gasoline. The yield is about 60% of the coal dust used.
𝐶𝑜𝑎𝑙 𝑑𝑢𝑠𝑡 + 𝐻2
𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 450 𝑜𝐶,200 𝑎𝑡𝑚
Mixture of hydrocarbons
𝑖 𝐻2, 𝑖𝑖 𝑐𝑟𝑎𝑐𝑘𝑖𝑛𝑔
𝐶𝑟𝑢𝑑𝑒 𝑜𝑖𝑙
COMBUSTION
Combustion is an exothermic chemical reaction, which is accompanied by
evolution of heat and light and the temperature rises considerably.
Eg.
C(s)
+
O2 (g)
CO2(g)
+
97 kcal
Ignition temperature: the minimum temperature at which the substance
ignites and burns without further addition of heat from outside.
Note: a gaseous fuel undergoes combustion on ignition only if its volume
concentration in the fuel-air mixture is in between the lower and upper
combustion limits. This range is known as ignition range.
Eg.
Hydrogen = 6 to 71
Methane = 6 to 13
Petrol vapours = 2 to 4.5
1. Amount or quantity of oxygen required for the
combustion of fuel.
2) Amount/weight/quantity of air will be required for the
combustion of fuel
Air contains 21% oxygen by volume and 23% oxygen by
mass. From the amount of oxygen required by the fuel, the
amount of air can be calculated.
Volume of O2 required = (Quantity of Oxygen/32)*22.4 lit
Weight of Air required = (weight of Oxygen required/23)*100 Kg
Volume of Air required = (Volume of Oxygen/21)*22.4 lit
Volume of Air required = (Wt of air required/28.94)*22.4 lit
2295
9.60
A sample of coal was found to have the following percentage
composition. C= 75%, H= 5.2%,
O= 12.1 %, N= 3.2% and ash =4.5%. Calculate the minimum air
required for complete combustion
of 1 kg of coal
Calculation of Air Quantities
To determine the amount of oxygen and hence the amount of air required for
combustion for a unit quantity of fuel, the following chemical principles are applied.
(1) Substances always combine in definite proportions given by molecular mass.
C + O2 → CO2
12 32 44
12 g of carbon requires 32 g of oxygen and 44 g of CO2 is formed.
22.4 L of a gas at 0°C and 760 mm pressure has a mass equal to 1 mol.
That is, 22.4 L of oxygen has a mass of 32 g.
(3) Air contains 21% oxygen by volume and 23% oxygen by mass. From the amount of
oxygen required by the fuel, the amount of air can be calculated.
1 kg oxygen is supplied by 1 x 100/23 = 4.35 kg of air
1 m3 of oxygen is supplied by 1x100/21= 4.76 m3 of air
(4) The molar mass of air is 28.94 g mol
(5) Minimum oxygen required for combustion is equal to the theoretical oxygen
required minus the oxygen present in the fuel.
(6) Mass of any gas can be converted to volume at certain temperature and pressure
by assuming that the gas behaves ideally.
(PV = nRT)
(7) The total amount of oxygen consumed is given by the sum of the amount of
oxygen required by individual combustible constituents present in the fuel.
Procedure for combustion calculations:
Reaction
Weight of oxygen
required (g)
Volume of oxygen
required (m3)
C + O2 → CO2
A gm or m3
A × 32/12
A×1
H2 + 1/2 O2 → H2O
B gm or m3
B × 16/2
B × 1/2
CO + 1/2 O2 → CO2
C gm or m3
C × 16/28
C × 1/2
S + O2 → SO2
D gm or m3
D × 1 × 32/32
D×1
CH4 + 2O2 → CO2 + 2H2O
E gm or m3
E × 2 × 32/16
E×2
C2H6 + 3.5O2 → 2CO2 +
3H2O
F gm or m3
F × 3.5 × 32/30
F × 3.5
C2H4+3O2 → 2CO2+2H2O
G gm or m3
G × 3 × 32/28
G×3
C4H10+6.5O2 → 4CO2+5H2O H × 6.5 × 32/58
H gm or m3
H × 6.5
Total
Less O2 in fuel
Y
= - w m3
X
= - w gm
Let oxygen required = X – w (g) or Y –w (m3)
Since air has 23% oxygen by weight and 21% oxygen by volume
Weight of air required = Net oxygen × 100/23 kg
Volume of air required = Net oxygen × 100/21 m3
Composition of Fuel Combustion Reaction Volume of 02
gas/m3
required
H2 = 0.5 m3
H2+ 1/2 O2 = H2O
C2H6 = 0.06 m3
C2H6 + 3.502 = 2C02 + 0.06 x 3.5 = 0.21 m3
3H20
0.30 x 2 = 0.6 m3
CH4 + 2O2 = C02 +
2H20
CO + 1/2 O2 = CO2
008 x 0.5 = 0.04 m3
CH4 = 0.30 m3
CO = 0.08 m3
Total
0.50 x 0.5 = 0.25 m3
1.1 m3
Solution:
Volume of air supplied = 1.1 × 100/21 × 120/100 = 6.6 m3 = 6600
L
Weight of air supplied = 28.94 × 6600/22.4 = 8.5Kg
Paraffins
Aromatics
Cracking is usually done by two methods:
1. Thermal Cracking: When it takes place simply by the application of heat
and pressure, the process is called thermal cracking. The heavy oils
are subjected to high temperature and pressure, when the bigger
hydrocarbons break down to give smaller molecules of paraffins,
olefins etc.
(a) Liquid Phase thermal cracking: The charge is kept in the liquid form by
applying high pressures of the range 30-100 kg/cm2 at a suitable
temperature of 475-530 oC. The cracked products are separated in a
fractionating column. The important fractions are: Cracked gasoline
(30-35%), Cracking gases (10-45%); Cracked fuel oil (50-55%). The
yield is 50-60 % and octane rating of the petrol produced is 65-70.
(b) Vapour phase thermal cracking: By this method, only those oils which
vaporize at low temperatures can be cracked. The cracking oil is
first vaporized and then cracked at about 600-650oC and under a low
pressure of 10-20 kg/cm2. petrol obtained from vapor phase
cracking has better anti-knock properties, but poor stability than
petrol from liquid-phase cracking.
Cracking: Gasoline is the most imp fraction of crude petroleum. The yield of
this fraction is only 20% of the crude oil. The yield of heavier petroleum fraction
is quite high.
This is achieved by a technique called cracking.
Simply defined as the decomposition of high boiling bigger hydrocarbon
molecules into simple low boiling hydrocarbons of lower molecular
weight.
e.g:
Cracking
C H
⎯⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯
⎯→ C
H
+ C H
5 12
5 10
10 22
Decane
n - pentane pentene
B.Pt =174 οC
B.Pt = 36 οC
Advantages of catalytic cracking over thermal cracking:
•High temp and pressure are not required in the presence of a catalyst.
•The use of catalyst not only accelerates the cracking reactions but also
introduces new reactions which considerably modify the yield and the nature of
the products.
•The yield of the gasoline is higher and quality is better
•No external fuel is required for cracking.
•The process can be better controlled so desired products can be obtained.
•The product contains a very little amount of undesirable sulphur because a
major portion of it escapes out as H2S gas, during cracking.
•It yields less coke, less gas and more liquid products.
•The evolution of by-product gas can be further minimized, thereby increasing
the yield of desired product.
•Catalysts are selective in action and hence cracking of only high boiling
fractions takes place.
•Coke forming materials are absorbed by the catalysts as soon as they are
formed.