Chemical Fuels

 Introduction

Chemical fuels
Contents

 Classification
 Calorific value - GCV & NCV
 Determination of CV of solids - Bomb calorimeter
 Determination of CV of liquids/gases - Boys calorimeter
 Coal & its analysis - proximate & ultimate analysis
 Liquid & gaseous fuels
Chemical Fuels, Dept. of Chemistry

1

Learning objectives

 Define calorific value & classify fuels

 Perform simple calculations involving NCV & GCV
 Describe the procedures to obtain CV of S, L & G fuels
 Demonstrate an understanding of the working principle of Bomb &
Boys calorimeter
 Explain the procedure of proximate & ultimate analysis
 Know the merits & demerits of S, L & G fuels
Chemical Fuels, Dept. of Chemistry

2

Introduction
Chemical fuel: A combustible carbonaceous material which on proper
burning in the presence of oxygen liberates large amount of heat that can
be used for domestic & industrial purposes.
Eg. coal, petrol, water gas

Fuel+ oxygen → Oxidation products + heat

Organic matter + Oxygen  CO2 + H2O + heat
Chemical Fuels, Dept. of Chemistry

3

Combustion components

Ignition
Temperature
Time
Turbulence
Chemical Fuels, Dept. of Chemistry

4

Classification of fuels
Physical state

Primary fuel

Secondary fuel

Solid

Wood, Peat,
Coal, Lignite

Charcoal, Coke

Liquid

Crude petroleum

Petrol, Kerosene,
Diesel, Petrol

Gas

Natural gas

Producer gas,
Water gas, Biogas, LPG

Chemical Fuels, Dept. of Chemistry

5

Characteristics of a good fuel
•High calorific value.

•Products of combustion

•Moderate ignition temp.
• Low moisture content.
•Low content of non-

•Combustion should be easily
controllable
•Should be safe, convenient &

combustible matter
•Low ash content

economical

•Readily available in bulk at
low cost

should not be harmful

*Should not be more valuable
for other purpose than a fuel

Chemical Fuels, Dept. of Chemistry

6

Calorific value
The amount of heat liberated when unit mass/volume of fuel is
completely burnt in pure oxygen.
Units : S or L fuels : cal/g or kcal/kg or J/kg
G fuels : kcal/m3 or J/m3
Types : * Gross Calorific Value (GCV)
* Net Calorific Value (NCV)
GCV = NCV + Latent heat of condensation of steam
NCV = GCV - Latent heat of condensation of steam
Chemical Fuels, Dept. of Chemistry

7

CV measurement of S-fuels - Bomb calorimeter

Principle:
 Burn known mass of fuel completely in excess of oxygen
 Heat liberated by burning fuel is the heat absorbed by water &
calorimeter
Measure the increase in temp of calorimeter & water using sensitive
thermometer
 Calculate CV of fuel from measured data
Chemical Fuels, Dept. of Chemistry

8

Bomb calorimeter
6v battery

Beckmann
thermometer

Oxygen
valve

Stirrer

Electrodes
Cu calorimeter

Mg fuse
wires

Stainless steel bomb
Stainless steel
crucible
Air jacket

Fuel

Water jacket

Chemical Fuels, Dept. of Chemistry

9

Observations & calculations
Mass of fuel = m g
Mass of water in copper calorimeter = W g
Water equivalent of calorimeter = w g
Initial temp. of water = t1 0C
Final temp. of water = t2 0C
Heat liberated by burning of fuel = Heat absorbed by water
& calorimeter
GCV = (W + w) (t2 - t1) × 4.184 × 103 J/kg
m
NCV = GCV – 0.09H × 587 cal/g
Chemical Fuels, Dept. of Chemistry

10

Numerical Problems
1. Calculate the CV of coal with the following data:
Mass of coal= 0.6 g, water equivalent of
calorimeter 2200 g, specific heat of water = 4.187
kJ kg-1c-1, Increase in temp. = 6.52 0C.
2. Calculate the GCV and NCV of a fuel from the
following data: Mass of fuel = 0.75 g, water
equivalent of calorimeter = 350 g, mass of water =
1150 g, Increase in temp= 3.02 oC, % of hydrogen
in the fuel = 2.8 .

Chemical Fuels, Dept. of Chemistry

11

CV measurement of L/G fuels - Boys calorimeter

Principle
• Burn known volume of gaseous fuel sample
• Heat released is quantitatively absorbed by cooling water
circulated through copper coils.
* Note the mass of cooling water & its rise in temp.
• Calculate calorific value from the measured data

Chemical Fuels, Dept. of Chemistry

12

Boys calorimeter

Chemical Fuels, Dept. of Chemistry

13

Observations & calculations
Volume of fuel burnt at STP in time, t
= V m3
Mass of cooling water circulated in time, t = W kg
Steady temp. of incoming water
= t10 C
Steady temp. of outgoing water
= t20 C
Rise in temp.
= (t2 - t1) 0 C
Mass of water produced from steam condensation = m kg
Heat released by combustion of fuel = Heat absorbed by water
GCV = W (t2 - t1) × 4.184 kJ/m3
V
NCV = [W (t2 - t1) - m ×587 ] 4.184 kJ/m3
V
V
Chemical Fuels, Dept. of Chemistry

14

Numerical Problems
• 1. The following results were recorded during the
determination of CV of a gaseous fuel. Calculate the GCV
& NCV of the fuel. Vol. of fuel burnt = 0.093 m3, Wt. of
water used = 30.5 Kg, Wt. of steam condensed = 0.031 Kg,
Inlet water temp = 26.1 oC, Outlet water temp. = 36.5 oC.
• 2. Calculate the GCV & NCV of a gaseous fuel from the
following data: Vol. of gas burnt = 0.05 m3, Temp. of gas =
20 oC, Wt. of water = 4.5 kg, Rise in temp.= 25 oC, Specific
heat of water = 4.2 kJ/kg-K, Wt. of water produced = 0.75
kg, Latent heat of steam = 2.5 kJ/m3.

Chemical Fuels, Dept. of Chemistry

15

Solid fuels
• Advantages

• Disadvantages

 Easy transportation

 High ash content

 Convenient storage without any
risk of spontaneous explosion

 Large amount of heat waste

 Combustion operation control
difficult

 Low production cost
 Moderate ignition temperature.

 High cost of handling

Chemical Fuels, Dept. of Chemistry

16

Analysis of coal – Procedure for proximate analysis
1. Moisture: Weigh coal sample into dry silica crucible. Heat for 1h at
110 0C. Cool & weigh. Calculate the loss in wt. as percentage.
% moisture = Loss in weight × 100
Wt. of coal taken
•Less moisture content  better quality coal
2. Volatile matter: Cover coal left in step 1. Place in a muffle furnace at
950 0C for 7 minutes. Cool & weigh. Calculate the loss in wt. as %age.
% Volatile matter = Loss in weight × 100
Wt. of coal taken
• Less volatile matter  better the rank of coal
Chemical Fuels, Dept. of Chemistry

17

3. Ash: Heat coal left in step 2 without lid in a muffle furnace at 750 0C
to complete the combustion & constant weight of residue is obtained.
% Ash = Wt. of ash left

× 100

Wt. of coal taken
•Ash-forming constituents  undesirable

4. Fixed carbon: Sum total of the % ages of volatile matter, moisture &
ash subtracted from 100.
i.e. 100 – (% moisture + % volatile matter + % ash)
• High percentage of fixed carbon  greater CV & better the quality
Chemical Fuels, Dept. of Chemistry

18

Analysis of coal - Ultimate analysis
1. C & H: Weigh coal & burn in pure O2. C & H convert into CO2 &
H2O on combustion. Absorb in weighed KOH & CaCl2 tubes. Increase in
the weights of tubes  amount of CO2 & H2O formed
% Carbon = Increase in wt .of KOH tube × 12 × 100
Wt. of coal taken × 44
% Hydrogen = Increase in wt. of CaCl2 tube × 2 ×100
Wt. of coal taken ×18
•High percentage of C & H  better coal quality

Chemical Fuels, Dept. of Chemistry

19

2. Nitrogen: Digest known quantity of coal in a Kjeldhal flask with
conc. H2SO4 & HgSO4 with K2SO4 catalyst. Treat clear soln. with excess
NaOH. Distill liberated NH3 into known volume of std. acid solution.
Determine volume of residual acid by back titration with std. NaOH
soln.
% Nitrogen = Vol. of acid used × Nacid ×14
Wt. of coal taken
* Less nitrogen content  Good coal quality

Chemical Fuels, Dept. of Chemistry

20

3. Sulfur: Weigh coal sample & burn in bomb calorimeter. Sulfur
converts to sulfates. Treat washings with BaCl2 soln. to obtain BaSO4 ppt
(at. wt. = 233). Filter, wash, ignite & weigh. Calculate S present in coal.
% Sulfur = Wt. of BaSO4 obtained × 32 × 100
Wt. of coal taken × 233
* Low sulfur-coal is useful
4. Ash: Determine as described under proximate analysis.
5. Oxygen: Determine by difference as follows:
% Oxygen = 100 – % (C + H + N + S + Ash)
Lower oxygen content  greater CV
Chemical Fuels, Dept. of Chemistry

21

Numerical Problems
• 1. About 1.5 g of coal was used for nitrogen estimation by
Kjeldahl method. The evolved ammonia was collected in 25
mL of 0.1 N H2SO4. the excess acid was neutralized by 15
mL of 0.1 N NaOH. Calculate the % nitrogen in the sample.

• 2. The residue obtained when 2.5 g of coal was heated in a silica
crucible for 1 h at 110 oC weighed 2.42 g. This was heated strongly
for 7 minutes at 950 oC covered with a vented lid. The residue
weighed 1.55 g & it was further heated until a const. wt. of 0.25 g was
obtained. Calculate % moisture, volatile matter, ash & fixed carbon.

Chemical Fuels, Dept. of Chemistry

22

Liquid fuels
 They are combustible molecules that can be harnessed to create
energy
 Formed by anaerobic decay of marine plant & animal life under the
influence of high T & P
 Complex mixture of hydrocarbons
 Refining- separation of crude oil into different useful fractions on the
basis of their boiling points
 Petroleum- An important primary liquid fuel

Chemical Fuels, Dept. of Chemistry

23

Petroleum cracking
Decomposition of higher mw into lower mw hydrocarbons.

Cracking

Thermal Cracking
[high T & P, Absence of catalyst]

Catalytic Cracking
Low T & P, Presence of catalyst e.g.
(Al2O3 + SiO2)

Chemical Fuels, Dept. of Chemistry

24

Catalytic cracking
CC is the breaking down of long-chain hydrocarbons into shorter chain
hydrocarbons
Advantages
•High octane number & yield of gasoline
•Better controlled process
•Low production costs - high T & P are not required
•External fuel not required
•Selective catalysts permit cracking of only high boiling hydrocarbons.

Chemical Fuels, Dept. of Chemistry

25

Fluidized bed catalytic cracking

Chemical Fuels, Dept. of Chemistry

26

Fluidized bed catalytic cracking
• Fluidize catalyst bed (Al2O3+SiO2) by upward passage of feed stock
vapors in reactor maintained at 550 0C & the pressure of 2 atm.
• Allow cracked oil vapors from centrifugal separator to pass onto the
fractionating column
• Allow catalyst to settle to the bottom, forced by an air blast to the
regenerator maintained at 600 0C
• Strip off adsorbed oil by spent catalyst -Pass steam & decarbonized
by a hot air blast in regenerator
• Heat liberated during regeneration used to raise steam & preheat the
catalyst.
Chemical Fuels, Dept. of Chemistry

27

Petroleum reforming
• Upgrading gasoline by increasing its octane number in presence of a
catalyst.
• Isomerization
H3C

(CH2)4

CH3

H3C CH CH2

CH2 CH3

CH3
n-hexane

2-methyl petane

• Dehydrogenation:

Chemical Fuels, Dept. of Chemistry

28

• Cyclisation & dehydrogenation:
CH3

(CH2)4

+

CH3

n-Hexane

H2

Cyclohexane

+
Cyclohexane

3H2

Benzene

• Hydro cracking:

CH3 (CH2)8
n-Decane

CH3

+

H2

Cat

2 CH3 (CH2)3

CH3

n-Pentane
Chemical Fuels, Dept. of Chemistry

29

Knocking
in ICEs
Sharp sounds caused
by premature combustion of part of the
compressed air-fuel mixture in the cylinder of IC engine (See next slide
for image)
Adverse effects:
* Increases fuel consumption
* Decreased in efficiency
* Causes mechanical damage to the engine parts & spark plug
* Unpleasant driving
* Undesirable rattling sound during travel
Minimization methods:
* Suitable change in engine design
* Using high octane number fuel
* Use of anti-knocking agents
Chemical Fuels, Dept. of Chemistry

30

Gaseous fuel
• Advantages

• Disadvantages

 Easier to handle than solid fuels

o Requirement of large storage
tanks

 Easily transportable through
pipelines
 No residue after burning

o Highly inflammable
o More expensive than S/L fuels

 Higher CVs than solids
 Produce little or no smoke
 Relatively low ignition temp.
Chemical Fuels, Dept. of Chemistry

31

Water gas
• Mixture of CO & H2
Manufacture :
• Produced by passing alternatively steam & air through a bed of red
hot coke maintained at 1000 0C.
• C + H2O (steam) → CO + H2
•

∆H = 131.4 KJ (endothermic)

Drop in temperature must be prevented by temporarily cutting
off steam supply & air is blown to raise the temp. to 1000 oC.
2C + O2 → 2CO

∆H = - 221.9 KJ (exothermic)

Chemical Fuels, Dept. of Chemistry

32

Schematic diagram of water gas plant

Chemical Fuels, Dept. of Chemistry

33

Summary

Types of fuels based on their origin & physical state
Methods to determine GCV & NCV of fuels using bomb & boys
calorimeters
Procedure of proximate & ultimate analysis of solid fuels like coal
Liquid fuels- Refining, Cracking, Reforming, Knocking
Gaseous fuels – manufacturing of water gas

Chemical Fuels, Dept. of Chemistry

34

Review questions
1. What are the differences between ultimate analysis & proximate
analyses of coal?
2. Describe the principle, procedure & calculation steps used to
determine the CV of a gaseous fuel.
3. Explain fluidized bed catalytic cracking technique.
4. How do you carry out catalytic reforming of gasoline?
5. Describe the manufacture & uses of water gas.

Chemical Fuels, Dept. of Chemistry

35