BASIC ENGINEERING CHEMISTRY [3 ed.] 9899107446, 9911310888, 8121923549

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BASIC ENGINEERING CHEMISTRY For B.E. Ist YEAR (Common to all Branches) [According to the latest Syllabus of Rajiv Gandhi Proudyogiki Vishwavidyalaya, Bhopal]

Dr. S.S. DARA M.Sc., Ph.D.

Dr. A.K. SINGH

Former Registrar and Director Nagpur University, NAGPUR - 440 010

M.Sc., Ph.D.

Head of the Chemistry Department University Institute of Technology Rajiv Gandhi Proudyogiki Vishwa Vidyalaya BHOPAL

Dr. ABHILASHA ASTHANA M.Sc., (Chemistry) Gold Medallist, Ph.D. Associate Professor Deptt. of Engineering Chemistry Gwalior Engineering College, Gwalior M.P.          

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© 2004, Dr. S.S. Dara, Dr. A.K. Singh & Dr. A. Asthana

All rights reserved. No part of this publication may be reproduced or copied in any material form (including photo copying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction : All desputes with respect to this publication shall be subject to the jurisdiction of the Courts, tribunals and forums of New Delhi, India only.

First Edition 2004 Second Revised and Reprint Edition 2008 Third Revised Edition 2012

ISBN : 81-219-2354-9



Code : 10B 279

Dedicated to Our Parents

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PREFACE TO THE THIRD REVISED EDITION The specular respone received by the earlier editions, both from teacher and students has encouraged the authors to present this third revised editions. The latest revised syllabus of RGPV Bhopal is fully covered in this new edition. Every aspect written in this textbook is simple and presented in a lucid manner, so as to make students conversant with the requirements of the subject. To enchance the understanding of problem, diagrams figures and numebrical are included at required place. The authors believe that the book will largely meet the requirements of students with simple understanding. The authors received support and encouragement from many peoples. First, we wish to thank all readers of our previous editions of this book who appreciate & compliments. We sincerly acknowledge the encouragement moral support useful suggestions given by fellow teachers and students. Particularly Dr. Namrata Jain Professor and HOD RKDFIST Bhopal. We would like to extent our most sincere appreciation to all persons who have contributed his valuable time in best possible way for the completion of this book. Finally the author Dr. Abhilasha Asthana is particularly thankful to her parents Dr. K.K. Asthana, Mrs. Usha Asthana and brothers Mr. Rajeev and Mr. Reetesh Asthana for being supportive in times of stress and for assisting with the task of manuscript preparation. No major proffesional project can be understand without the cooperation of one’s family. Being a teacher, we appreciate the value of criticism. The quality of science improves through critical discussion. Therefore constructive suggestions comments and criticism on the subject matter of the book will be greatfully acknowledged as they will certainly help to improve the editions of the book. Dr. S.S. Dara Dr. A.K. Singh Dr. A. Asthana



Email: [email protected]

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SYLLABUS Unit I : WATER AND ITS INDUSTRIAL APPLICATIONS

Sources, Impurities, Hardness & its units, Industrial water characteristics, softening of water by various methods (Exrernal & Internal treatment), Boiler trouble causes, effect & remedies, Characteristics of municipal water & its treatment, Numerical problems based on softening methods. Unit II : FUELS & COMBUSTION



Fossil fuels & classification, Calorific value, Determination of calorific value by Bomb calorimeter Proximate and Ultimate analysis of coal and their significance, calorific value Computation based on ultimate analysis data, Carbonization, Manufacturing ofcoke & recovery of by products. Cracking of higher Hydrocarbons & mechanism of cracking, Knocking, relationship between' knocking & structure of hydrocarbon, improvement of anti knocking characteristics of IC engine fuels, Diesel engine fuels, Cetane number, combustion and it related numerical problems. Unit III : A. LUBRICANTS



Introduction, Mechanism of lubrication, Classification of lubricants, Properties and Testing of lubricating oils, Numerical problems based on testing methods. B. CEMENT & REFRACTORIES



Manufacture, IS-code, Setting and hardening of cement, Refractory : Introduction, classification and properties of refractories Unit IV : POLYMER



Introduction, types and classification of polymerization, Reaction Mechanism, Natural & Synthetic Rubber; Vulcanization of Rubber, Preparation, Properties & uses of the following- Polythene, PVC, PMA, PMMA, Teflon, Poly acrylonitrile, PVA, Nylon, Nylon 6:6, Terylene, Phenol formaldehyde, Urea - Formaldehyde Resin, Glyptal, Silicone Resin, Polyurethanes; Butyl Rubber, Neoprene, Buna N, Buna S. Flow sheet manufacturing diagram of Nylon 6:6 & Decoran. Unit V : A. INSTRUMENTAL TECHNIQUES IN CHEMICAL ANALYSIS



Introduction, Principle, Instrumentation and applications of IR, NMR,UV, Visible,Gas Chromatography, Lambert's and Beer's Law. B. WATER ANALYSIS TECHNIQUES



Alkalinity, hardness (Complexo-metric), Chloride, Free chlorine, DO, BOD and COD, Numerical problems based on above techniques. (vi)

CONTENTS



1. WATER AND ITS INDUSTRIAL APPLICATIONS



2. FUELS AND COMBUSTION

1 – 60 61 – 118



3A. LUBRICANTS

119 – 144



3B. CEMENT AND REFRACTORIES

145 – 173



4. POLYMERS

174 – 228



5A. INSTRUMENTL TECHNIQUES IN CHEMICAL ANALYSIS

229 – 250



5B. WATER ANALYSIS TECHNIQUES

251 – 271

QUESTION BANK

272 – 280

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UNIT

1

Water and Its Industrial Applications

1.1 INTRODUCTION Water is the basic necessitiy of life. It is necessary for the survival of all livings. Approximately 80% of the earth surface is covered by water, but only 1% water is available for the uses of different purposes such as domestic, agaricultural, municipal and industrial work. 1.2 SOURCES OF WATER The various sources of water are : (i) Surface waters : (a) Flowing water e.g. streams and rivers etc. (b) Still water e.g. ponds, lakes and reservoirs etc. (ii) Underground water (a) Water from shallow and deep springs and wells. (b) Water from lower measures of coal mines (iii) Rain water (iv) Esutarine and sea water Sources of water

(i) Surface water

Flowing water e.g. streams and rivers

Still water e.g. Ponds, lakes and reservoirs

(ii) Underground (iii) Rain water water

Water from shallow and deep springs and wells

1

Water from lower measure of coal mine

(iv) Esutarine and sea water

2

BASIC ENGINEERING CHEMISTRY

1.3 TYPES OF IMPURITIES FOUND IN WATER The impurities present in natural waters may be broadly classified as follows (1) Dissolved impurities (a) Inorganic salts e.g., (i) Cations : Ca2+, Mg2+, Na+, K+, Fe2+, Al3+ and sometimes traces of Zn2+ and Cu2+ (ii) Anions : Cl–, SO42–, NO3–, HCO3–, and sometimes F – and NO3– (b) Gases e.g., CO2, O2, N2, oxides of N2 and sometimes NH3, H2S (c) Organic salts (2) Suspended impurities (a) Inorganic e.g., clay and sand (b) Organic e.g., oil globules, vegetable and animal matter. (3) Dissolved impurities Finely divided clay and silica, aluminium hydroxide, ferric hydroxide, organic waste products, humic acids, colouring matter, complex protein, aminoacids, (which are generally classified as albunoid ammonia). (4) Bacterial impurities Bacteria, other micro-organisms and other forms of animal and vegetable life. Table– 1.1. Specifications of Purpose

ater Used for Different Industries and Other Purposes Specifications and Remark

(1) Paper Industry

(a) Free from alkalinity (alkaline water consumes more alum, thereby increasing the cost of production). (b) Free from hardness: (Calcium and magnesium salts increase the ash content of the paper produced). (c) Free from colour, turbidity and salts of Fe and Mn : (colour and brightness of the paper are affected by the above impurities). (d) Free from Silica : (Silica causes cracks in the paper).

(2) Textile industry

(a) (b) (c)

(3) Thermal Power Generation industry

(a) (b)

Free from turbidity : (turbidity causes uneven dyeing). Free from colour, and salts of Fe and Mn : (these impurities cause stains on the fabric). Free from hardness and organic matter : (Hard water reduces the solubility of acidic dyes and causes precipitation of basic dyes. They also render the dyeing non-uniform. Organic matter may cause foul smell of the product). Boiler feed Water : Free from hardness : (hard water causes scaleformation on boiler metal surface, thereby reducing heat transfer efficiency and causing shut-down or even accidents). Cooling water : The water should be non-scale forming, non-corrosive, and should not permit the growth of algae. Scale and algae reduce the heat transfer efficiency and interfere with free flow of wat .

(4) Dairy industry

– The water should be colourless, odourless, and tasteless. It should be free from pathogenic organisms.

(5) Beverage industry

–

The water should be pure. It should not be alkaline, because alkalinity inwater tends to neutralise the fruit acids and distorts the taste.

(6) Laundry

–

The water should be free from colour, hardness and salts of Fe and Mn : (Hardness of water increases the consumption of soaps and detergents. Fe and Mn salts impart undesirable colour to the fabric.

(7) Ice making, brewing, canning and distillery industry

–

Free from hardness and bacteria.

WATER AND ITS INDUSTRIAL APPLICATIONS

3

1.4 HARDNESS OF WATER The waters which do not produce lather or produces very little lather with soap are known as hard water. On the other hand soft waters readily produces a lot of lather when mixed with a little of soap. Therefore the study of hardness of water has great importance. Hardness. Hardness was originally defined as the soap consuming capacity of a water sample. Soaps generally consists of the sodium salts of long chain fatty acids such as oleic acid, palmetic acid and stearic acid. The soap consuming capacity of water is mainly due to the presence of calcium and magnesium ions. These ions react with the sodium salts of long chain fatty acids present in the soap to form insoluble scums of calcium and magnesium soaps, which do not possess any detergent value. 2 C17H35COONa + CaCl2   --→   (C17H35COO)2 Ca    + 2 NaCl Soap (soluble)               Calcium soap (insoluble)

Other metal ions like Fe2+, Mn2+ and Al3+ also react with the soap in the same fashion, thus contributing to hardness but generally these are present in natural waters only in traces. Further, acids such as carbonic acid can also cause free fatty acid to separate from soap solution and thus contribute to hardness. However, in practice, the hardness of a water sample is usually taken as a measure of its Ca2+ and Mg2+ content. 1.4.1 Types of Hardness It is of following two types: (i) Temporary Hardness: Temporary hardness is mainly due to the presence of bicarbonate of calcium and magnesium. It is also known as carbonate hardness or alkaline hardness. It can be removed easily be boiling the water. On boiling bicarbonates of calcium and magnesium get decomposed and formed insoluble carbonates or hydroxides which can be removed in the form of precipitates ∆  → CaCO3↓ + H2O + CO2↑ ∆ Mg(HCO3)2  → Mg(OH)2↓ + 2CO2↑ (ii) Permanent Hardness: It is due to the presence of chlorides and sulphates of calcium, magnesium, iron and other heavy metals such as Al2(SO4)3 etc. It is also known as non-carbonate or non-alkaline hardness. It cannot be removed by simply boiling the water. However it can be removed by using various chemical agents. Total Hardness = Temporary Hardness + Permanent Hardness



Ca(HCO3)2

1.4.2 Units of Hardness Various units used for expressing hardness of water are as under: 1. Parts per million (ppm) 2. Milligrams per liter (mg/L) 3. Degree french (°Fr) 4. Degree Clark (°Cl) 1. Parts per million (ppm) It is defined as number of parts of calcium carbonate equivalent hardness present per 106 parts of water. This is most common unit for expressing the hardness of water. 1 ppm = 1 part of CaCO3 equivalent hardness of 106 parts of H­2O 2. Milligrams per litre (mg/L) It is defined as the number of milligrams of CaCO3 equivalent hardness present per litre of water. 1 mg/L = 1 mg of CaCO3 equivalent per 106 mg of water

4

BASIC ENGINEERING CHEMISTRY

= 1 part of CaCO3 equivalent per 106 parts of water = 1 ppm Thus, mathematically both the units are euqal. 3. Degree French (°Fr) It is defined as the number of parts of CaCO3 equivalent present per 105 parts of water. 1°Fr = 1 part of CaCO3 equivalent hardness per 105 parts of water 4. Degree Clark (°Cl) It is the number of grains (1/7000 1b) of CaCO3 equivalent present per gallon (10 lbs or 70,000 grains) Always remember that, 1 lb = 7000 grain and one gallon = 70,000 grains It is defined as the number of parts of CaCO3 equivalent hardness present per 70,000 parts of water. 1°Cl = 1 parts of CaCO3 equivalent hardness per 70,000 parts of water Relationship among various units of hardness

1 ppm = 1mg/L = 0.1 °Fr = 0.07°Cl

1 mg/L = 1ppm = 0.01 °Fr = 0.07­°Cl 1°Cl = 1.433 °F = 14.3ppm = 14.3 mg/l 1°F = 10ppm = 10 mg/L = 0.7°Cl The above relation can be derived as follows: We know that, 1 ppm = 1 part per 10,00,000 parts of water 1°Cl = 1 part per 70,000 perts of water 1°Fr = 1 part per 1,00,000 parts of water \ 10,00,000 ppm = 70,000°Cl = 1,00,000°Fr or 100 ppm = 7°Cl = 10°Fr or 1ppm = 0.07°Cl = 0.1°Fr 1.4.3 Degree of Hardness Hardness of water is never present in the form of calcium carbonate, because it is insoluble in water. Hardness of water is expressed as equivalent of calcium carbonate CaCO3. Degree of hardness is defined as the part of calcium carbonate equivalent hardness per a definite number of parts of water depending upon the units in which hardness is expressed. 1.5 TREATMENT OF HARD WATER OR SOFTENING OF WATER BY VARIOUS METHOD

The hardness producing salts can be removed from water by following two method. (i) By External Treatment (ii) By Internal Treatment Treatment of Hard Water



External Treatment (i) By lime soda process (ii) By Zeolite (iii) By Ion Exchange process

Internal Treatment (i) By colloidal conditioning (ii) By phosphate conditioning (iii) By calogen conditioning

WATER AND ITS INDUSTRIAL APPLICATIONS

5

1.5.1 (i) External Treatment The external treatment of water is carried out before it is fed into the boiler. This treatment prevent boiler troubles. It can be done by lime soda process, zeolite process, ion exchange process etc. (A) Lime Soda Process: It is the most important method of chemical water softening. In lime soda process all the soluble hardness causing impurities are chemically converted into insoluble precipitates, which may be removed by settling and filtration In this method calculated quantity of lime [Ca(OH)2] and soda (Na2CO3) are added into water. The precipitated formed are finely divided, so they do not settle easily and cannot be filtered easily. Therefore we add a small amount of coagulants. (i) lime removes the temporary hardness: → 2CaCO ↓ + 2H O Ca(HCO ) + Ca(OH)  3 2

2

3

2

Mg(HCO3)2 + 2Ca(OH)2

 → Mg(OH) ↓ + 2CaCO + 2H O 2 3 2

MgSO4 + Ca(OH)2

 → Mg(OH) ↓ + CaSO 2 4



(ii) lime removes the permanent magnesium hardness: → Mg(OH) ↓ + CaCl MgCl2 + Ca(OH)2  2 2

(iii) lime removes dissolved iron and aluminium salts: → Fe(OH) ↓ + CaSO FeSO4 + Ca(OH)2  2 4 → 2Fe(OH) ↓ 2Fe(OH) + H O + O  2

2

3

Al2(SO4)3 + 3Ca(OH)2 (iv) lime removes free mineral acids:

2HCl + Ca(OH)2

 → 2Al(OH) ↓ + 3CaSO 3 4  → CaCl 2  → CaSO

H2SO4 + Ca(OH)2 4 (v) lime removes dissolved CO2 and H2S → CaCO ↓ Ca(OH) + CO  2

2

3

 → CaS↓

+ 2H2O + 2H2O + H2O

Ca(OH)2 + H2S + 2H2O (vi) Soda removes all the soluble calcium permanent hardness (i.e., that which is originally present as well as that which is introduced during the removal of Mg2+, Fe2+, Al3+, HCl, H2SO4 etc., by lime). → CaCO3↓ CaCl2 + Na2CO3  + 2NaCl → CaCO ↓ CaSO4 + Na2CO3  + Na2SO4 3 Since natural waters generally contain a large proportion of temporary hardness, it is often convenient and economical to remove temporary hardness by lime treatment. Lime is rather cheap and it removes temporary hardness efficiently without introducing any soluble salts in the water. Magnesium hydroxide produced in the above reactions precipitates as an insoluble sludge. The reaction of soda with the permanent calcium hardness produces insoluble CaCO3. Addition of a coagulant such as sodium aluminate or alum helps in accelerating the coagulation of the carbonate sludge, which is subsequently removed by filtration. Water softened by this process contains appreciable concentrations of soluble salts, such as sodium sulphate, and cannot be used in highpressure boiler installations.

6

BASIC ENGINEERING CHEMISTRY

Types of cold lime-soda softeners

There are four basic types of cold lime soda softeners: (1) The intermittent type (batch process) (2) The conventional type  (3) The catalyst or spiractor type  continous process  (4) The sludge blanket type 

(1) Intermittent or batch process The intermittent type of cold lime-soda softener consists of a set of two tanks which are used in turn for softening of water. Each tank is provided with inlets for raw water and chemicals, outlets for softened water and sludge, and a mechanical stirrer (Fig. 1.1). Raw water and calculated quantities of the chemicals are slowly sent into the tank simultaneously under agitation with the help of the stirrer. Fig.1.1. Intermittent cold lime-soda softener. Some sludge from a previous operation is also added which forms nucleus for fresh precipitation and thus accelerates the process. Thus by the time the tank is full, the reaction is more or less complete. Stirring is stopped and the sludge formed is allowed to settle. The clear softened water is collected through a float pipe and sent to the filtering unit. The sludge formed in the tank is removed through the sludge outlet. By employing a set of tanks planned for alternate cycles of reaction and settling, continuous supply of softened water may be ensured. (2) Conventional type In this process, the raw water and the chemicals in calculated quantities are continuously fed from the top into an inner chamber of vertical circular tank provided with a paddle stirrer (Fig. 1.2). The raw water and the chemicals flowing down the chamber come into close contact and the softening reactions take place. The sludge formed settles down to the bottom of the outer chamber from where it is periodically removed through the sludge outlet. The softened water rising up passes through the fibre filter where traces of sludge are removed and filtered soft water passes through the outlet provided. Water treated by the cold lime-soda process generally produces softened water containing about 50 – 60 ppm of residual hardness.

Raw water

Chemicals

Softened water

Fibre filter Paddle stirrer

Sludge Sludge outlet

Fig. 1.2. Conventional type of lime-soda softner.

WATER AND ITS INDUSTRIAL APPLICATIONS

(3) Catalyst or spiractor type

7 Air vent

The spiractor consists of a conical tank which is about two-thirds filled with finely divided granular catalyst (Fig. 1.3). The tank Softend used may be either open (for gravity operation) water or closed (for operation under pressure). In both the cases the raw water and the calculated Catalyst (crushed and graded quantities of chemicals enter the tank tangentially calcite or green sand) Raw near the bottom of the cone and spiral upwards water through the suspended catalyst bed. The catalyst Chemicals employed is a finely granuled (0.3 to 0.6 mm Draw-off valve diameter) insoluble mineral substance such for enlarged catalyst as graded calcite or sand or green sand. The Fig. 1.3. Catalyst or spiractor type cold limeretention time is about 8 to 12 minutes. The soda water softener. sludge formed during the softening reactions deposits on the catalyst grains in an adherent form and hence the granules grow in size. The softened water rises to the top from where it is drawn off. The catalyst or spiractor type of continuous water softener is of interest as it gives a granular sludge which drains and dries rapidly and can be handled easily. (4) The sludge blanket type The sludge blanket type of water treatment equipment is extensively used for coagulation and settling as well as water softening by cold lime soda process. These softeners differ from the conventional type in that the treated water is filtered upwardly through a suspended sludge blanket composed of previously formed precipitates. Thus in a single unit, all the three processes namely mixing, softening and clarification take place In the conventional type of equipment, some of the added lime suspension is carried down in the sludge formed by the precipitates, before it has time to dissolve and react with the hardness causing impurities of the raw water and thus some of the lime is wasted. In the sludge blanket type, this does not happen because the upward filtration through the suspended sludge blanket ensures complete utilization of the added lime. With the conventional type of equipment, it is generally observed that after precipitates or after deposits form on the granules or filter media employed, and in pipe lines or distribution systems carrying the filtered effluents. This usually necessitates recarbonation with CO2 to obviate the formation of such deposits. However, in the sludge blanket type of equipment, the intimate contact of the treated water with a large mass of solid phase mostly prevents super-saturation or the formation of after deposits. This results in the production of the effluent which is clear enough (turbidity usually less than 10 mg/l) for many industrial applications, so that subsequent filtration is often unnecessary. The retention period required with sludge blanket type equipment is one hour as against four hours with the conventional units. Further, silica is removed better in sludge blanket units. The sludge blanket type of water softening equipment, owing to its higher efficienc , shorter detention period and smaller space requirements, is rapidly displacing the conventional type.

8

BASIC ENGINEERING CHEMISTRY

Hot lime-soda process

143.0

Parts per million

The reactions during water softening take place in very dilute solutions (about 114.4 0.001 M) and hence proceed very slowly. 85.8 The rate of these precipitation reactions CaC o can be greatly accelerated by increasing 3 a t 10° 57.2 C C Mg(O aC the temperature, because, this not only H) O 2 at 10 3 °C at 96°C increases the rate of the ionic reactions M g(O 28.5 H) themselves, but also the rate at which 2 at 96°C particles of measurable size are formed. 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.5 The effect of temperature on the veloTime in hours city and completeness of precipitation reacFig. 1.4. Effect of temperature on rate of precipitation of tions involving the removal of scaleforming CaCO3 and Mg(OH)2 from CaSO4 and MgSO4 solutions respectively. constituents is shown in Fig. 1.4. It can be seen that at 96°C the precipitation is more complete in 10 minutes than after several hours at 10°C. Thus effect is more pronounced with the precipitation of magnesium compounds. Hot lime-soda plants carry out softening at 94° – 100°C which has several advantages. For efficient softening, cold lime-soda softening plants must be of considerable area and water-storage capacity, whereas hot lime-soda softeners are much more rapid in operation and therefore for a given through-put, much more compact. Elevated temperatures not only accelerate the actual chemical reactions but also reduce the viscosity of the water and increase the rate of aggregation of the particles. Thus, both the settling rates and filtration rates are increased. Thus the softening capacity of the hot lime-soda process will be several times higher than the cold process. Since the sludge formed settles down rapidly, there is no need of adding any coagulants. A smaller excess of chemicals is needed than with the cold process. Further, dissolved gases are driven out of the solution to some extent at the high temperature. The hot lime-soda process yields softened water having relatively lower residual hardness (about 17 to 34 ppm) as against the cold process (about 50 – 60 ppm). A typical hot limesoda water softening unit is shown in Fig. 1.5, Raw water which includes a reaction cum settling tank and a filte . If the water is alkaline, filtration through sand and gravel beds might contaminate the Exhaust or Chemicals water with dissolved silica, particularly if the Live quartz used is of inferior quality. Other filtering steam media used are anthracite coal, calcite and magnetite. If the precipitation is incomplete in the softening tank, “after-precipitation” occurs Fines Anthrafilt Coarse in pipes, storage tanks and even in boiler itself. If slight excess of chemicals are used over that Sludge Clarified theoretically required, more rapid and more water complete removal of hardness will result. But Sludge outlet if larger excess of chemicals are used, naturally they will appear in the softened water. LimeFig. 1.5. Hot lime-soda water softner. soda plants do not produce water of zero hardness. Tips for the solvening problems on water treatment by Lime-soda process. On the basis of the various reactions taking place in lime-soda process given earlier, the following deductions can be made:

WATER AND ITS INDUSTRIAL APPLICATIONS

9

(i) One equivalent of calcium temporary hardness requires one equivalent of lime → 2CaCO3↓ + 2H2O Ca(HCO3)2 + Ca(OH)2  (ii) One equivalent of magnesium temporary hardness requires two equivalents of lime Mg(HCO3)2 + 2Ca(OH)2  → Mg(OH)2↓ + 2CaCO3 + 2H2O (iii) One equivalent of calcium permanent hardness requires one equivalent of soda, → CaCO3 + Na2SO4 CaSO4 + Na2CO3  → CaCO3 + 2NaCl CaCl2 + Na2CO3  (iv) One equivalent of magnesium permanent hardness requires one equivalent of lime and one equivalent of soda. → MgSO4 + Ca(OH)2 + Na2CO3 

Mg(OH)2 + CaCO3 + Na2SO4

→ MgCl2 + Ca(OH)2 + Na2CO3 

Mg(OH)2 + CaCO3 + 2NaCl

(v) Lime reacts with HCl, H2SO4, CO2, H2S, salts of iron, aluminium, etc. Accordingly, their respective equivalents must be considered for calculating the lime requirement.

Ca(OH)2 + 2HCl Ca(OH)2 + H2SO4 Ca(OH)2 + CO2 Ca(OH)2 + H2S



 →

CaCl2 + 2H2O

 →

CaSO4 + 2H2O

 →

CaCO3 + H2O

 →

CaS + 2H2O

 →

Ca(OH)2 + FeSO4 CaSO4 + Fe(OH)2 1 → 2Fe(OH)3 2Fe(OH)2 + H2O + O2  2 → 2Al(OH)3 + 3CaSO4 2Ca(OH)2 + Al2(SO4)3  (vi) Lime, while reacting with HCl, H2SO4, MgSO4, MgCl2, Mg(NO3)2, salts of Fe, Al etc., generates the corresponding quantities of calcium permanent hardness. Accordingly, these constituents also should be considered while calculating the soda requirement. (vii) Two equivalents of HCO3– reacts with two equivalents of lime as follows:

2 HCO3– + (2 equivalents)

Ca(OH)2

 →

CaCO3 + 2H2O + CO32–

(2 equivalents)

(2 equivalents)

It is evident that in the above reaction, 2 equivalents of CO32– are generated. Thus for every one equivalent of HCO3– present, the corresponding reduction in the dose of soda has to be made in the calculations for soda requirement. For solving numerical problems on lime-soda requirements for softening of hard water, the following steps may be followed: 1. The units in which the impurities analysed are expressed i.e., ppm (or mg/l), grains per gallon (or degrees Clark), etc., are to be noted. 2. Substances which do not contribute towards hardness (e.g., KCl, NaCl, SiO2, Na2SO4, Fe2O3, K2SO4, etc.) should be ignored while calculating lime and soda requirements. This fact should be explicitly stated. 3. All the substances causing hardness should be converted into their respective CaCO3 equivalent, as a matter of convention and convenience. Wt. of the impurity × 50 CaCO3 equivalent of a hardness causing imputity = Chemical equivalent wt. of the imurity (since chemical equivalent weight of CaCO3 = 50).

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BASIC ENGINEERING CHEMISTRY

For instance, 136 parts by weight of CaSO4 would contain the same amount of Ca as that of 100 parts by weight of CaCO3. Hence, in order to convert the weight of CaSO4 as its CaCO3 equivalent, 100 50 the weight of CaSO4 should be multiplied by a factor of or . 136 68 Conversion factors for some of the impurities in water which commonly come across are given in Table 1.2. Table 1.2

Salt

(1)



Ca(HCO3)2 Mg(HCO3)2 CaSO4 CaCl2 MgSO4 MgCl2 Mg(NO3)2 Ca2+ Mg2+



HCO3–



HCl



H2SO4 CO2



Al2(SO4)2



FeSO4 . 7H2O CaCO3 MgCO3



NaAlO2

Multiplication factor to convert into its CaCO3 equivalent (2) 100/162 100/146 100/136 100/111 100/120 100/95 100/148 100/40 100/24 100 100 = 61 × 2 122 100 100 = 73 36.5 × 2 100/98 100/44 100 100 = 342/3 114 100/278 100/100 100/84 100 100 = 82 × 2 164

Notes :

(a) If the impurities are given as CaCO3 or MgCO3, these should be considered to be due to Ca(HCO3)2 or Mg(HCO3)2 respectively and they are only expressed in terms of CaCO3 and MgCO3. (b) The amount expressed as CaCO3 does not require any further convertion. However, the amount expressed as MgCO3 should be converted into its CaCO3 equivalent by multiplying with 100/84.

WATER AND ITS INDUSTRIAL APPLICATIONS

11

4. Calculate the lime and soda requirements as follows: Temporary calcium hardness + (2 × Temporary magnesium  74 Lime required hardness) + Perm. Mg hardness + CO 2 + HCl + H 2SO 4  (A) = ×  2+ 3+ for softening 100 + HCO3 + salts of Fe , Al etc., - Na AlO 2 ; all expressed  in terms of their CaCO3 equivalents  × Volume of water × Purity

}

Permanent Ca hardness + permanent Mg hardness + Salts  (B) Soda required 106   = × of Fe 2+ , Al3+ , etc., + HCl + H 2SO 4 - HCO3- - Na AlO 2 ;  for softening 100 all expressed in terms of their CaCO equivalents  3 

}

× Volume of water × Purity (C) If the analytical report shows the quantities of Ca2+ and Mg2+, then; 1 eq. of soda is required for Ca2+; whereas one eq. of lime and 1 eq. of soda is required for Mg2+. 5. If the lime and soda used are impure and if the % purity given, then the actual requirements of the chemicals should be calculated accordingly. Thus, if lime is 90% pure, then the value obtained 100 to get the actual requirement of lime. Similarly, if under 4-A above should be multiplied by 90 100 soda is 95% pure, then the value obtained under 4-B above should be multiplied by to get the 95 actual soda requirement. (B) By Zeolite Process The word zeolite is derived from two greek words (zein + lithos), which means “boiling stone”. The name was first used by cronstedt (a Swedis Geologist in 1736) to a certain group of natural minerals, which released their water of hydration (or combination) in the form of steam. They are also known as permutits. They are of two types: (a) Natural Zeolites: They are non-porous and derived from green sands by washing, heating and treatment with sodium-hydroxide (NaOH). They are more durable. Some important natural zeolites are : 1 (a) Thomsonite (Na2O, CaO).Al2O3.2SiO2.2 H2O 2 (b) Natrolite Na2O, Al2O3.3SiO2.2H2O (c) Laumontite CaO.Al2O3.4SiO2.4H2O (d) Harmotome (BaO. K2O).Al2O3.5SiO2.5H2O (e) Stilbite (Na2O.CaO). Al2O3.6SiO2, 6H2O (f) Brewsterite (BaO. SrO. CaO).Al2O3.6SiO2.5H2O (g) Pitlolite (CaO. K2O. Na2O).Al2O3.10SiO2.5H2O

(b) Synthetic Zeolites: They are porous and posses a get structure. They are prepared from solution of sodium silicate and aluminium hydroxide. They may also be prepared by heating together (a) china clay, felspar, and soda ash (or (b) solution of sodium silicate, Al2(SO4)3 and NaAlO3 or (c) solution of sodium silicate and Al2(SO4)3 or (d) solution of sodium silicate and NaAlO2) and granulating the resultant mass after cooling. Water softening by zeolites Sodium zeolites are used in water softening process. It may by represented as: Na2O.Al2O3.X(2-10).SiO2.Y(2-6).H2O For simplification they are represented as Na2Z. Where Z = insoluble zeolite radical frame work.

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BASIC ENGINEERING CHEMISTRY

Theory of zeolite softening On passing hard water through a bed of active granular of sodium zeolite, ion exchange occurs between water and zeolitle particles. The Ca2+ and Mg2+ ions are taken up by the zeolite and simultaneously relasing the equivalent sodium ions in exchange for them. Since zeolites are capable of exchanging basic radicals, they are generally known as base exchanger. The various reaction taking place are given below: Brine inlet → CaZ + 2NaHCO3 Ca(HCO3)2 + Na2Z  → MgZ + 2NaHCO3 Mg(HCO3)2 + Na2 Z  Hard → CaZ + Na2SO4 CaSO4 + Na2 Z  water → MgZ + Na2SO4 MgSO4 + Na2 Z  Hard water → CaZ + 2NaCl CaCl2 + Na2 Z  Sodium Zeolite → MgZ + 2NaCl MgCl2 + Na2 Z  Fine Gravel Relatively small quantities of iron and manganes, present as the Coarse Gravel divalent bicarbonate, may also get removed simultaneously.

 → FeZ + 2NaHCO3



Fe(HCO3)2 + Na2Z

 → MnZ + 2NaHCO3

Mn(HCO3)2 + Na2 Z

(Note : Soluble iron and manganes are always present in the divalent form in waters containing bicarbonate alkalinity).

Softend water Fig. 1.6. Zeolite softener.

Regeneration: When the zeolite bad is exhausted (i.e., saturated with Ca2+ and Mg2+), it can be regenerated by washing the zeolite bed with concentrated sodium chloride (brine). Sodium nitrate, sodium sulphate potassium chloride or potassium nitrate could also be used. → Na2Z + CaCl2 CaZ + 2NaCl  → Na2Z + MgCl2 MgZ + 2NaCl  Brine

Hard water

–

Ca 2+ Mg 2+

Cl SO42–

Na+ Ca Z and Mg Z

Na2 Z –

Na+

Cl SO42–

Softening progresses

–

Cl

Ca Z and Mg Z

Na2 Z

Exhausted

Regenerated

Softened Water

Ca2+ Mg2+ – Cl

(a) Softening Cycle

(b) Regeneration Cycle

Fig. 1.7. Schemes for softening of water by Zeolite process.

Advantages of Zeolite Process

1. Hardness is completely removed. 2. Equipment used is compact and occupies less place. 3. It automatically adjusts itself to waters of different hardness. 4. It can work under pressure. Hence the plant can be installed in the water supply line itself, avoiding double pumping.

WATER AND ITS INDUSTRIAL APPLICATIONS



13

5. In this process, the hardness causing ions are simply exchanged with sodium ions. As the process does not involve any precipitation, there is no problem of sludge formation and after precipitation in the softened water at later stages.

Limitation of Zeolites Process

(i) Water with turbidity should not be used as pores of Zeolite get dogged. (ii) pH of water should not be too high or two low as it effects Zeolite. (iii) Water containing Fe2+ and Mn2+ ions should be avoided because Fe2+ or Mn2+ will form their Zeolite which cannot be easily regenerated. (iv) Hot water can not be used as it dissolves Zeolites. (v) Acid radicles are not removed by this process. (vi) Water treated by Zeolite process contains about 25% more dissolved solid that softened by lime soda process. Table–1.3 Comparison between the Zeolite process and the lime-soda process

The salient features of Zeolite process and lime-soda process are summarized in the following table. Zeolite process 1. This process produces water of almost zero hardness 2. The cost of the plant and the Zeolite are higher. Hence the capital cost is higher.   3. The exhausted Zeolite bed can be regenerated with brine which is very cheap. Hence the operating cost is lesser   4. The plant is compact and occupies less space. The size of the plant depends on the hardness of water being treated.   5. Cannot be used for hot water, acidic waters, and waters having turbidity and suspended impurities.   6. This process can operate under pressure and can be designed for fully automatic operation.   7. This process does not involve a number of operations such as settling, coagulation, filtration. 8. Water treated with the Zeolite process contains larger amounts of sodium salts and greater percentage of dissolved salts than the raw water since Ca2+ (eq. wt. 20) and Mg2+ (eq. wt. 12) are replaced by Na+ (eq. wt. 23).

Lime-soda process



1. This process produces water having hardness of 15 to 60 ppm depending on whether it is hot process or cold process. 2. The capital cost is lower. 3. The chemicals needed viz., lime, soda, and coagulant are consumed in the process. Hence the operating cost is higher. 4. The plant occupies more space. The size of the plant depends on the amount of water being handled 5. The process is free from such limitations.

6. This process cannot be operated under pressure.

7. This process involves all the problems associated with settling, coagulation and filtration

8. Treated water contains lesser percentage of dissolved solids and lesser quantities of sodium salts.

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BASIC ENGINEERING CHEMISTRY

         





9. This process adjusts itself to waters of 9. Reagent doses must be adjusted for waters of different hardness. different hardness. 10. Salts causing temporary hardness are 10. Temporary hardness is completely reconverted to NaHCO3 which will be moved in the form present in the softened water. Such of insoluble CaCO3 and Mg(OH)2. a water creates problems when used as feed water in boilers. 11. There may be problems of after precipi11. No problems of after precipitation. tation in distribuion systems and even in boilers when used as boiler feed water.

(C) Ion Exchange Process (or Deionization or Deminerabization Process) The process of complete removal of all ions present in water is called demineralization. Demineralization of water is carried out by ion-exchange resins. Ion-exchange resins are cross-linked, long chain organic polymers with a microporous structure. The functional groups attached to the polymeric chain are responsible for ion exchange. On the basis of functional groups resins are classified into two groups (a) Cation Exchange Resin: Resin containing acidic groups like-COOH or –SO3H are known as cation exchange resin. These are usually styrene divinyl benzene copolymers which on sulphonation or carboxylation become capable of exchanging their H+ ions with the cations of the solution. They are also called H form cation exchangers.

CH — CH2 — CH — CH2— CH

SO3H H2C — CH— CH2

SO3H

CH— CH2—

—H2C — CH

–

SO3 H+

–

SO3 H+

Fig. 1.8. An Cation Exchanger Resin.

Resins having SO3H group are known as strongly acidic resins. While resins having COOH group are known as weakly acidic resins. (b) Anion Exchange Resins: Resing having basic functional groups like amine, substituted amine or quaternary ammonium groups as their hydroxide salts are known as anion exchange resins. They are styrene divinyl benzene complexes which are capable of exchanging their basic functional groups with the anions of the solution.

WATER AND ITS INDUSTRIAL APPLICATIONS

15

— CH — CH2 — CH — CH2— CH

+

–

CH2NR3 OH H2C — CH— CH2

+

–

CH2NR3 OH

CH— CH2—

H2C — CH

+

–

CH2NR3 OH

+

–

CH2NR3 OH

Fig. 1.9. An Anion Exchanger Resin.

For effective water treatment, ion exchangers should possess the following properties. 1. They should be non-toxic. 2. They should not decolourise the water being treated. 3. They should possess a high ion-exchange capacity. (It depends upon the total number of ion active groups per unit weight of the exchanger and is expressed as mill equivalents per gram of the exchanger). 4. They should be physically durable. 5. They should be resistant to chemical attack. 6. They should be cheap and commonly available. 7. They must be capable of being regenerated and back-washed easily and economically. 8. They should have a large surface area since ion-exchange is a surface phenomenon. At the same time, their resistance to flow must be compatible with hydraulic requirements Process: The ion exchange unit consists of two tanks. Cation exchange resins are placed in first tank and anion exchange resins are placed in second tank. At first hard water is passed through a cation exchange resin. All the Ca2+ and Mg2+ ions are exchange by H+ ions of resin.

2RH+ + CaSO4

2RH+ + MgSO4 2RH+

+ CaCl2

 → R2Ca2+ + H2SO4

 → R2Mg2+ + H2SO4  → R2Ca2+ + 2HCl

→ R2Mg2+ + 2HCl 2RH+ + MgCl2  The effluent (from this step) is then passed through an anion exchanger tank. Here the anions like SO42–, Cl– etc are present in water exchanged from OH– ions of the resin.

ROH– + Cl–



2ROH– + SO42–

 → RCl– + OH­–

 → R2SO42– + 2OH­–

→ R2CO32– + 2OH­– 2ROH– + CO32–  The water coming out of the anion exchanger is completely free from cations and anions responsible for hardness. It is known as deiomized or demineralized water. H+ released from cation exchanger combine with OH– released from anion exchanger and formed H2O

H+ + OH–

 → H2O

16

BASIC ENGINEERING CHEMISTRY Raw water To vacuum pump steam jacket

Anion exchanger

Cation exchanger

Alkali for regeneration Acid for regeneration

Deionised water

Fig. 1.10. Demineralization of water.

Regeneration. After some time the cation and the anion exchangers get exhausted and stop working. The anion exchanger may be regenerated by treating it with NaOH solution.

R2SO42– + 2NaOH

 → 2ROH– + Na2SO4

→ ROH– + NaCl R2Cl– + NaOH  The cation exchanger may be regenerated by passing a solution of HCl or H2SO4

or

R2Ca2+ + H2SO4

 → 2RH+ + CaSO4

R2Ca2+ + 2HCl

 → 2RH+ + CaCl2

R2Mg2+ + H2SO4

 → 2RH+ + MgSO4

→ 2RH+ + MgCl2 R2Mg2+ + 2HCl  Mixed bed deionizer: Demineralisation of water can be better achieved by using a “mixed bed” of cationic and anionic resins. This method produces an effluent which is far superior to that produced by the two bed operation. When water is passed through a “mixed bed” consisting of anion and cation exchange resins, each pair of contrasting resin particles functions as a stage in the treatment and thus the total effect is that of a multiple cycle deionisation. Thus the process is highly efficient When the resins are exhausted the bed is backwashed when the two resins are separated in different layers due to difference in their densities. Then the resins are separately regenerated, washed and mixed again by injecting air and reused for a fresh cycle.

Advantages

(i) Highly acidic or alkaline water can be softened. (ii) Water of very low hardness is produced (about 2 ppm) (iii) Anions as well as cations are removed thereby problems like caustic embrittlement and corrosion are reduced.

Disadvantage

(i) The process is costly. (ii) Turbid water decreases the efficiency of the process

1.5.2 (b) Internal Treatment Internal treatment consists of adding chemicals directly to the water in the boilers for removing dangerous scale forming salts which were not completely removed in the external treatment for water softening. This is mainly used as a corrective treatment to remove the slight residual hardness and also sometimes to remove the corrosive tendencies in water. This treatment is not usually

WATER AND ITS INDUSTRIAL APPLICATIONS

17

applied to raw waters, except for small boilers, but it is usually practised in larger power stations. In modern heavy-duty high pressure boilers, water of zero hardness is required, since even an egg-shell thickness of scale may be extremely detrimental. (A) Carbonate conditioning: In this, method we add sodium carbonate to boiler water, so that salt like CaSO4 etc are converted into calcium carbonate and can be removed

→ Na2SO4 + CaCO3 CaSO4 + Na2CO3  For a salt to be precipitate, sufficient amount of the ions forming the salt must be present, so that the product of their ionic concentration is high from their solubility product. Thus, for a salt like CaCO3 to be precipitated, the product of the concentration of Ca2+ and CO3– – must high from the solubility product of CaCO3. The above principles are used in the carbonate conditioning when sodium carbonate solution is added to boiler water, the concentration of CO32– ion increased and when it becomes greater than the solubility product of SO4– –(i.e. greater than K' × [SO4– –], only CaCO3 get precipitated and CaSO4 remains in solution. Thus the deposition of scale forming CaSO4 is prevented → CaCO3↓ + Na2SO4 Na2CO3 + CaSO4  Carbonate conditioning is used only for low pressure boiler. In high pressure boiler the excess Na2CO3 might be converted into NaOH due to hydrolysis as follows  2NaOH + H2CO3 Na2CO3 + 2H2O 

 H2O + CO2↑ H2CO3  NaOH causes caustic embrittlement in high pressure boilers. (B) Phosphate conditioning: Phosphate conditioning is applicable to high pressure boilers. In this method an excess of soluble phosphate is added to boiler water. It react with Ca and Mg salts and form soft sludge of Ca and Mg phosphate which can be removed by blow down process

→ M3(PO4)2↓ + 6NaCl 3MCl2 + 2Na3PO4  The three sodium orthophosphates viz., Na3PO4, Na2HPO4 and NaH2PO4 have been used for phosphate conditioning. Sodium pyrophosphate (Na4P2O7) and sodium metaphosphate (NaPO3) are also used for the same. The typical reactions of the various phosphates with the hardness represented as CaCO3, may be summarized as follows:



→ Ca3(PO4)2 + 3 Na2CO3 2 Na3 PO4 + 3 CaCO3 

→ Ca3(PO4)2 + 2 Na2CO3 + CO2 + H2O 2Na2HPO4 + 3 CaCO3 

→ Ca3(PO4)2 + Na2CO3 + 2 CO2 + 2 H2O 2 NaH2PO4 + 3 CaCO3 

→ Ca3(PO4)2 + Na2CO3 + 2 CO2 2 NaPO3 + 3 CaCO3  The quality of the feed water decide the choice of a particular phosphate to be used. For instance, if the feed water tends to produce an acidic condition in the boiler, the alkaline Na3PO4 should be chosen. This treatment could be supplemented with NaOH if the required alkalinity could not be maintained with Na3PO4 alone. If the feed water produces almost the right alkalinity desired in the boiler, it is preferable to use Na2HPO4 which is practically neutral. If the boiler water becomes too alkaline, the acidic NaH2PO4 would be selected. Both sodium pyrophosphate and metaphosphate are rapidly hydrolysed under boiler water temperatures to orthophosphate.



NaPO3 + H2O =

NaH2PO4

Na4P2O7 + H2O = 2 Na2HPO4 Thus, their behaviour within the boiler is identical with that of orthophosphates mentioned above. However, NaPO3 solutions are practically neutral, whereas NaH2PO4 solutions are acidic. Hence the former would be preferred if the use of NaH2PO4 causes feedline corrosion.

18

BASIC ENGINEERING CHEMISTRY

The use of internal treatment combined with suitable blow down to remove sludge has contributed largely to the operation of the modern high-pressure steam boilers without the formation of hard scales. However, precaution should be taken to inspect them at least once in six months and remove the scale and sludge accumulations. (C) Colloidal conditioning : Scale formation can also be minimised by adding some colloidal conditioning agents such as glue, agar agar, tannins, starches and sea-weed extract into the boiler feed water. These substances act as protective colloids. They function by surrounding the minute particles of CaCO3 and CaSO4 and prevent their coalescence and coagulation. Thus, the precipitated scale-forming salts are maintained in loose suspended form which can easily be removed by blowdown operation. Thus the scale formation is prevented. (D) Calgon conditioning : Another approach for preventing scale formation is to convert the scale forming salts into highly soluble complexes which are not easily precipitated under the boiler conditions. In order to achieve this, sodium hexameta phosphate (Na PO3)6 or Na2 [Na4P6O18] (its trade name is calgon) is generally employed. This substance interacts with the residual calcium ions forming highly soluble calcium hexametaphosphate and thus prevents the precipitation of scaleforming salts.

Na2 [Na4P6O18]



2 Ca2+ + [Na4P6O18]2–

 2 Na+ + [Na4P6O18]2–  → 4 Na+ + [Ca2P6O18]2–

1.6 BOILER PROBLEMS A proper quality of water for use in boilers is very important. If impure water is made use of as boiler feed water, then following boiler problems may occurs (i) Sludge and scale formation (ii) Corrosion of boiler metal (iii) Caustic embrittlement (iv) Carry over: priming and foaming (i) Sludge and scale formation In a boilder, water is continuously evaporated to form steam. This increases the concentration of dissolved salts. Finally a stage is reached when the ionic product of these salts exceeds their solubility product and hence they are thrown out as precipitates. If the precipitates formed are soft loose and slimy, these are known as sludges; while if the precipitate is hard and adhering on the inner walls, it is called as scale. Sludge : Sludge is a soft, loose and slimy precipitate formed within the boiler. Sludges are formed by substances which have greater solubilities in hot water than in cold water, e.g., MgCO3, MgCl2, CaCl2, MgSO4 etc. They are formed at comparatively colder portions of the boiler and get

Fig. 1.11. Sludge and scale formation in boilers

WATER AND ITS INDUSTRIAL APPLICATIONS

19

collected at place where the flow rate is slow, they can be easily removed (scrapped off) with a wire brush. If sludges are formed along with scales, then former gets entrapped in the latter and both get deposited as scales. Disadvantage of sludge formation

(i) Sludge are poor conductors of heat, they tend to waste a portion of heat generated and thus decrease boiler efficienc . (ii) Excessive sludge formation hampers working of the boiler. Sludge sittles in the regions of poor water circulation e.g. pipe connection, plug opening, gauge glass connection, there by causing choking of the pipes.

Prevention of sludge formation

(i) By using softened water (ii) By frequent blow down operation i.e. partial removal of concentrated water through a outlet at the bottom of boiler, when hardness of water in the boiler becomes high. Scale : Scales are hard deposits firmly sticking to the inner walls of the boiler. They are difficult to remove, even with the help of hammer and chisel, and are the main source of boiler troubles. CaSO4, Ca(OH)2, CaCO3, Ca3(PO4)2 etc are the main salts which causes scales. Scales may be formed inside the boiler due to : (i) Decomposition of calcium bicarbonate Ca(HCO3)2 → CaCO3 ↓ + H2O + CO2 ↑ scale However, scale formed by calcium carbonate is soft and is the main cause of scale formation in low-pressure boilers. But in high-pressure boilers, CaCO3 is soluble due to the formation of Ca(OH)2 CaCO3 + H2O → Ca(OH)2 + CO2 ↑ soluble (ii) Deposition of calcium sulphate The solubility of CaSO4 in water decreases with increase in temperature. CaSO4 is soluble in cold water, but almost completely insoluble in super-heated water. Consequently, CaSO4 gets precipitated as hard scale on the hotter parts, of the boiler. This type of scale causes troubles mainly in high-pressure boilers. Calcium sulphate scale is quite adherent and difficult to remove, even with the help of hammer and chisel (iii) Hydrolysis of magnesium salts Dissolved magnesium salts get hydrolysed (at prevailing high temperature inside the boiler) forming magnesium hydroxide precipitate, which forms a soft type of scale, e.g. MgCl2 + 2H2O → Mg(OH)2 ↓ + 2HCl ↑ scale (iv) Presence of silica Even if a small quantity of SiO2 is present, it may deposit as calcium silicate (CaSiO3) and/ or magnesium silicate (MgSiO3). These deposite adhere very firmly on the inner side of the boiler surface and are very difficult to remove. One important source of silica in water is the sand filt . Disadvantage of Scale Formation (i) Wastage of Fuel. Scales are the poor conductor of heat. They reduces the rate of heat transfer from boiler to inside water. Thus scale formation will result in wastage of fuel and reduction in boiler efficienc .

20

BASIC ENGINEERING CHEMISTRY

(ii) Lowering of Boiler Safety. Scale formation on the boiler tubes or other heated surfaces insulates the metal so well that it becomes overheated. The metal becomes soft and weak thus making the boiler unsafe particularly at high pressures. The overheating also causes burning out of the metal plates and tubes and breakdown of the expanded joints. (iii) Thinning of The Tube Wall. In addition to the loss of strength due to overheating, rapid reaction between water and iron occurs at high temperatures, causing additional thinning of the tube wall. 3Fe + 4H2O → Fe3O4(s) + 4H2(g) (iv) Danger of Explosion. Since the scale acts as heat insulator, the metal of the boiler is overheated. Under the high pressure of steam existing in the boiler, the metal expands until the scale on it cracks. Sudden entry of the water through these cracks to the very hot metal causes sudden cooling of the boiler metal with the simultaneous conversion of water into steam. The sudden increase in pressure due to this large quantity of steam thus formed may lead to explosion. (v) Decrease in Efficienc . Scales deposited in the valves and condensers of the boiler, block them partially, which lowers down the efficiency of the boile . Prevention of Scale Formation Scale formation can be prevented by the following methods. (a) External treatment. It involves removal of hardness causing impurities (such as calcium and magnesium salts) and silica from the water before entering the boiler. The various methods of external treatment of water are already discussed earlier. (b) Internal treatment. Internal treatment consists of adding chemicals directly to the water in the boilers for removing dangerous scale forming salts which were not completely removed in the external treatment for water softening. This is mainly used as a corrective treatment to remove the slight residual hardness and also sometimes to remove the corrosive tendencies in water. This treatment is not usually applied to raw waters, except for small boilers, but it is usually practised in larger power stations. In modern heavy-duty high pressure boilers, water of zero hardness is required, since even an egg-shell thickness of scale may be extremely detrimental. Table 1.4. Difference Between Sludge And Scales.

Sludge 1. They are soft, loose and slimy precipitate 2. They are non-adherent deposits and can be easily removed 3. Produced by salt like CaCl2, MgCl2, MgSO4, MgCO3 etc. 4. It formed at comparatively colder portion of the boiler 5. They are less dangerous

Scales They are hard deposits They stick very firmly to the inner surface of boiler and are very difficult to remov Produced by salt like CaSO4, Mg(OH)2 etc It formed generally at heated portion of the boiler They are more dangerous as compare to sludge.

(ii) Corrosion of Boiler Metal. Boiler corrosion is disintegration of boiler body material either due to chemical or electrochemical reaction with its environment. Boiler corrosion takes place due to following factors: (a) Presence of free acids in water. (b) Acids generated as a result of hydrolysis of some salts in water. (c) Acids formed by the hydrolysis of fatty lubricating oils. (d) Presence of dissolved gases such as O2, CO2, H2S etc., in water.

WATER AND ITS INDUSTRIAL APPLICATIONS

21

(e) Presence of salts like MgCl2 which directly attack the boiler metal: MgCl2 + Fe + 2H2O → Mg (OH)2 + FeCl2 + H2 (f) Presence of salts like MnS2 which may generate H2SO4 due to oxidation and hydrolysis. (g) Formation of galvanic cells. Some of these factors are discussed below: (a) Dissolved oxygen. Dissolved oxygen is the main source of corrosion in boilers. The concentration of oxygen in boiler waters should be below 0.05 ppm for low pressure boilers and less than 0.01 ppm for high pressure boilers (~ 500 psi). Oxygen enters the boilers through raw make up water and also through infiltration of air into the condensate system. When the water containing dissolved oxygen is heated in the boiler, the free gas is evolved which corrodes the metal parts under the conditions obtaining in the boiler. Dissolved oxygen can be removed by: Mechanical de-areation of water. The solubility of a gas is directly proportional to pressure and inversly proportional to temperature (Daltons law and Henry’s law). These two principles are made use of in the design of mechanical de-areratior. In mechanical dearators, dissolved oxygen is removed by injecting hot feed-water as a fine spray into a vacuum chamber heated externally by steam. Chemical Treatment For complete removal of dissolved oxygen, chemical methods of treatment are adopted. Sodium sulfite is suitable for boilers operating below 650 psi. The sulphite reacts with the dissolved oxygen to form sulphate. 1 Na2SO3 + O → Na2SO4 2 2

Fig. 1.12. Mechanical de-areration of water.

This method is very effective for removing oxygen in low pressure boilers but cannot be used in high pressure boilers because (i) increasing dissolved salts’ concentrations produce foaming and priming and (ii) sodium sulphate decomposes and liberates SO2 and/or H2S. Ferrous sulphate is also used sometimes. It reacts with dissolved oxygen giving a precipitate of Fe (OH)2 which is oxidized to Fe(OH)3. FeSO4 + 2NaOH → Fe(OH)2 + Na2 SO4 2Fe (OH)2 + H2O + O → 2Fe(OH)3 Hydrazine, N2H4 is now extensively used to remove dissolved oxygen in high pressure boilers. The pure compound is an explosive inflammable liquid (B.Pt. 113.5°C) but the 40% aqueous solution used for water treatment is quite safe to handle. However, rubber gloves must be worn to guard against the danger of dermatitis in some cases. One of the important advantage of hydrazine treatment is that combination with oxygen does not produce any salts. Nitrogen and water are the only reaction products obtained. N2H4 + O2 → N2 + 2H2O The molecular weight of hydrazine and oxygen are the same (viz. 32), so that complete elimination of 1 ppm of oxygen would require only 1 ppm of hydrazine (which is 1/8 of the necessary concentration of sodium sulphite). The residual hydrazine can be measured easily by a colorimeter.

22

BASIC ENGINEERING CHEMISTRY

The amount of hydrazine added must be closely controlled. Any excess reagent decomposes in the boiler, liberating ammonia: 3N2H4 = 4NH3 + N2 Dissolved ammonia can bring about corrosion of some alloys, e.g., copper alloy condenser tubes. (b) Dissolved mineral acids Most of the natural waters are alkaline excepting waters from mining areas or those polluted from acidic industrial wastes or those in which wet oxidation of sulphide minerals occur. Some inorganic salts like magnesium chloride and calcium chloride are also corrosive agents. MgCl2 hydrolyzes completely at 200°C producing hydrochloric acid as follows: MgCl2 + 2H2O → Mg (OH)2 + 2HCl CaCl2 also undergoes hydrolysis but to a lesser extent. At 600°C, its hydrolysis is about 25%. Silicic acid catalyzes the reaction so that in water containing silica, appreciable quantities of HCl may be formed at lower temperatures. The HCl thus produced reacts with iron as follows: Fe + 2HCl = FeCl2 + H2 FeCl2 + 2H2O = Fe(OH)2 + 2HCl Hence, even a small amount of MgCl2 can cause considerable corrosion of the metal. If the amount of HCl formed is small, it might get neutralized by the alkalinity present in the water, but otherwise it should be neutralized by the addition of alkali. (c) Dissolved carbon dioxide Natural waters contain CO2. Water containing bicarbonates release CO2 on heating. If CO2 is released inside the boiler, it will go along with the steam and as the steam condenses, the CO2 is dissolved in water forming carbonic acid. This produces intense local corrosion called pitting. CO2 along with oxygen in water, can be removed by mechanical deaeration. CO2 in water can be removed by lime treatment. Another method of removing CO2 from water is to filter it through lime stone CaCO3 + CO2 + H2O → Ca (HCO3)2 but this reaction produces temporary hardness in the water. CO2 can be converted into ammonium carbonate by the addition of ammonia. CO2 + 2NH3 + H2O → (NH4)2 CO3 However, the excess ammonia added to the boiler feed water may go along with steam to the condenser. If some O2 is also present in the condensate, the ammonia may attack the condenser tubes made of copper. Hence, a safe limit of 10 mg of NH3/litre of the condensate is prescribed. (d) Formation of galvanic cells Corrosion can also occur because of galvanic cell formation between iron and other metals present in the alloys used in boiler fittings. This may also lead to pitting corrosion. This can be prevented by suspending zinc plates which act as sacrificial anodes (iii) Caustic embrittlement : Caustic embrittlement is a form of corrosion caused by a high concentration of sodium hydroxide in the boiler water. It is characterised by the formation of irregular intergranular cracks on the boiler metal, particularly at places of high local stress, such as riveted seams, bends and joints. It is caused by the high concentration of NaOH which is capable of reacting with steels stressed beyond their yield point. It is most likely to occur in boilers operating at higher pressures, where NaOH is produced in the boiler by the hydrolysis of Na2CO3 as follows: Na2CO3 + H2O → 2NaOH + CO2 The extent of the hydrolysis increases with temperature and may reach even 90% of the carbonate present. The rate and extent, of corrosion by caustic embrittlement increases with the concentration of NaOH and temperature and hence with increasing operating pressure.

WATER AND ITS INDUSTRIAL APPLICATIONS

23

NaOH so formed along with the natural alkalinity of water make the boiler water caustic. The NaOH travels to the minute hair cracks, bends and joints. Water evaporates and the concentration of NaOH at these points keep on increasing with time due to poor circulation of water at these points. This caustic soda attracks the surrounding areas and corroding the boiler material or dissolving the iron of the boiler. This causes caustic embrittlement of the boiler parts. The caustic embrittlement can be explained by the formation of the concentration cell as Iron at Rivets

Conc. NaOH

Dil NaOH

Iron at plane surfaces

The iron surrounded by concentrated NaOH act as anode while iron surrounded by diltue NaOH act as cathode. The concentrated alkali dissolves the iron as sodium ferroate (Na2FeO2). Sodium ferroate decomposes (a short distance away from its place of formation) as 3Na2FeO2 + 4H2O → 6NaOH + Fe3O4 + H2 Due to regeneration of NaOH further dissolution of iron takes place. The caustic embrittlement occurs more rapily in following cases. (i) Due to high concentration of NaOH (ii) Presence of compounds such as sodium silicate etc. (iii) Material subjected to great stresses beyond then yield point. Prevention of Caustic Embrittlement

(a) By using sodium sulphate or sodium phosphate as softening agent in place of sodium carbonate: It is observed that boiler water containing sodium sulphate or sodium phosphate prevent the caustic embrittlement by blocking the capillaries, thereby preventing the infiltration of caustic soda solution to these. The concentration of Na2SO4 : NaOH are maintained at 1 : 1, 2 : 1 or 3 : 1 for operating pressure of 10, 20 and > 20 atmospheres respectively to check the caustic embrittlement. (b) By adding lignins or tannins which help in blocking the infiltration of NaOH through the hair cracks. (c) By adding NaNO3: It neutralize the excess amount of alkali and therefore help in preventing the caustic embrittlment. (iv) Carry over. As steam rises from the surface of the boiling water in the boiler, it may be associated with small droplets of water. Such steam, containing liquid water, is called wet steam. These droplets of water naturally carry with them some suspended and dissolved impurities present in the boiler water. This phenomenon of carrying of water by steam along with the impurities is called “carry over”. This is mainly due to priming and foaming. Steam used for power production is usually superheated for achieving greater efficiencies in the turbine or engine in which steam is used. Wet steam causes corrosion in the inlet ends of the superheaters. If the steam contains high percentage of moisture, the extent of superheating will decrease with the consequent reduction in efficiency of the turbine or engine. Further, the water carried over with the steam contains salts and sludges, these are carried into the superheater where they may deposit as the water evaporates. This will seriously restrict the flow of steam. Moreover, due to the insulating effect of these deposits, the superheater tubes also may burn out. A part of the dried salts may be carried along with the steam farther and deposit on the high-pressure turbine blades or in engine valves. Even a small amount of deposit on the turbine blades decreases its efficiency considerably. In order to eliminate the bad effects of moisture, mechanical steam purifiers are often installed in the steam drums of the boiler or between the boiler and the superheater. These devices force the steam to take curved paths whereby due to centrifugal action, the moisture is thrown out of the steam.

24

BASIC ENGINEERING CHEMISTRY

(a) Foaming. Foaming is the formation of small but presistent bubbles at the water surface. These bubbles are carried along with steam leading to excessive priming. According to Bancroft, foams are formed when there is a difference in concentration of solute or suspended matter between the surface film and the bulk of the liquid. Substances which increase the viscosity of the film favour production of foam. The bubbles may also be protected by finely divided solids forming a protective “shell” around each of them. Any material which lowers the surface tension of the water will collect at the interface and thus increase the foaming tendency of the liquid. Pure water has no tendency to form foam but almost all impurities causes foaming, specially the dissolved impurities like soap which reduces the surface tension of water. Foaming can be avoided by : (i) By the removal of the foaming and stabilizing agents from the water. (ii) By controlling the concentration of salts and sludge in the boiler by intermittent or continuous blow down process. (iii) By adding antifoaming agents e.g., castor oil etc. (iv) A foam may be destroyed by the addition of another good foaming agent. The foam is finally destroyed by mutual antagenistic effect of the difference in charge (Positive and negative) on the colloidal partical of the two foams. e.g., When foam obtained by a solution of an anionic detergent (e.g., Aerosol OT) mix with foam obtained by a solution of an cationic detergent (Ethyl cetab), they destroy each other. (b) Priming. When the steam is generated rapidly in the boilers, some droplet of the liquid water are carried along with steam. This process is known as wet steaming or priming. Causes of priming: The main causes of priming are given below: (i) It is due to the presence of suspended impurities and to some extent to dissolved impurities in the water. (ii) Due to alage and vegetable growth (iii) Feed water containg even a small quantity of scale forming salts may causes priming (iv) Due to sudden boiling (v) Due to very high water level in the boiler. (vi) Due to high velocity of steam which carries water droplets alongwith it in the steam pips. Minimization of primining. It can be minimised by: (i) By using a proper designed boiler (ii) By maintaing low water level (iii) By avoiding rapid changes in the steaming rate (iv) By minimizing foaming (v) By using only soft water 1.7 CHARACTERISTICS OF MUNICIPAL WATER AND ITS TREATMENT Municipal water is mainly used for drinking purposes and for cleaning, washing, etc. Water that is fit for drinking purposes is called potable water. Water used for drinking and other domestic used should be colourless, odourless, free from suspended impurities, free from germs, bacteria and other pathogenic organisms and should be free from harmful dissolved impurities. Therefore, the raw or impure water obtained by municipalities, from sources such as rivers, likes, wells, tube wells, etc. has to be properly treated before supplying for domestic purposes. The various steps involved in the treatment are as follows :

WATER AND ITS INDUSTRIAL APPLICATIONS

Water

25

Water

Sand

Sand

Gravel

Gravel

Porous drain tiles Fig. 1.13. Set of Water Filtration Units.

(i) Aeration : The raw water is first aerated by bubbling compressed air through it. This removes bad odours, CO2, etc. and also removes iron and manganese by precipitating them as their respective hydroxides. (ii) Settling : The water is then allowed to stand in large settling tanks. At this stage, some of the heavier impurities present in water settle down by gravity. Also, the bacteria present are partially eliminated due to the UV radiation from sunlight. (iii) Coagulation : The suspended impurities present are then removed by coagulation using lime, soda ash and aluminium sulphate (or ferric alum) as the case may be. The suspended impurities are trapped by the resulting precipitate of Al(OH)3 and settle down at the bottom, thereby bringing about partial clarification of the water. Also, the negatively charged colloidal impurities neutralized by the trivalent aluminium cations, followed by agglomeration and settling down by gravity. (iv) Filtration : The partially clarified water is then passed through sand gravity filters. These comprise of rectangular tanks which contain : (a) a top layer (about meter thick) of fine sand (b) a middle layer (0.3 – 0.5 meter thick) of coarse sand, and (c) a bottom layer (0.3 – 0.5 meter thick) of graded gravel. A series of porous drains are provided at the bottom of the gravel layer through which filtered water is collected. The slimy surface layer comprising of finely divided clay, algae, bacteria etc. formed on the filter bed acts as an effective filtering medium which filters the finely divided residue, suspended matter and bacteria. The filters are backwashed periodically to remove the precipitated matter from the surface, so as the ensure efficient filtration. Activated carbon may be used for filtration if the water contains undesirable odours (v) Chlorination : The filtered water is sterilized by chlorination (by adding chlorine or bleaching powder) to destroy the pathogenic micro-organisms. The water is now pumped to over-head tanks for domestic distribution. Use-based Classification of Waters Adopted By Water Pollution Control Boards in India Ans. 1. Fresh Waters



Classificatio

Best Use



A B C

— — —



D

—

Drinking water source after disinfection, without any other treatment Outdoor bathing Drinking water source with conventional treatment followed by disinfection. Propagation of wild-life and fisheries



E

—

Irrigation, industrial cooling and controlled waste disposal.

26

BASIC ENGINEERING CHEMISTRY

2. Sea Waters (including estuaries and tidal waters)



Classificatio A B C D E

— — — — —

Best Use Water sport (contact), shell fishing, salt pans Commericial fishing, non-contact recreatio Industrial cooling Harbour Navigation, controlled waste disposal.

1.8 METHOD OF TREATMENT OF WATER FOR DOMESTIC AND INDUSTRIAL PURPOSES Municipal water supply has been one of the most challenging problems of water technology. Water supplied by municipalities for domestic purposes must be free from pathogenic bacteria. It should be clear, colourless and pleasant to taste. It should be free from excessive dissolved salts, suspended impurities and harmful microorganisms. Water for domestic purposes should be obtained from such a source which is least contaminated by animal and vegetable matter as well as industrial effluents. Rivers, lakes and wells are the most common sources of water used by municipalities. Generally the treatment of these waters involves removal of suspended impurities and removal of colloidal impurities if any, followed by sterilization. If the water is very hard, certain amount of softening may be needed, which is very rare. Sedimentation, coagulation, filtration and sterilization are the treatment techniques usually employed depending on the requirements of the situation. 1.8.1. Sedimentation. Sedimentation is a process of removing relatively large particles (suspended solids) into large reservoirs of settlement tanks in which it is left for a few days or even weeks, where the suspended impurities partially sink to the bottom. The principle involved is to slow down the flow of water so that substances held up by the turbulence of fast moving water can fall gravitationally to the bottom of the tank when water flow is stilled. Periodically the accumulations of the debris are to be scraped away. In order to remove floating impurities, screens of various kinds (e.g., Bar screen, Band and drum screens and microstrainers) are employed. These screens also must be continuously cleaned. The rate of settling in still water at 10°C is known as the hydraulic settling value of a particle and is generally expressed in millimeters/second. During sedimentation, solid particles settle by gravity on the bottom of a settling tank in which the water being clarified is at rest or in slow horizontal or upward motion. The velocity with which a particle in water will fall under the action of gravity depends upon (i) the horizontal flow velocity of the water, (ii) the size of the particle, (iii) the specific gravity of the particles, (iv) the shape of the particle and (v) the temperature of the water. Accordingly, several formulae have been given to calculate the velocity of falling spherical particles in slowly moving water on the basis of which several types of sedimentation tanks have been designed. The sedimentation tanks commonly used are of horizontal flow rectangular type and circular shaped upward flow typ Sedimentation takes a long time, requires large-capacity settling tanks and cannot ensure complete removal of coarse-dispersed impurities from water. Plain sedimentation usually removes only 70 to 75% of the suspended matter. 1.8.2. Coagulation. Finely divided silica, clay and organic matter does not settle down easily and hence cannot be removed by simple sedimentation. Most of these are in colloidal form (e.g., sols, gels or emulsions) and are generally negatively charged and hence do not coalesce because

WATER AND ITS INDUSTRIAL APPLICATIONS

27

of mutual repulsion. Such impurities are generally removed by chemically assisted sedimentation, in which certain chemicals are added which produce ions of right electrical charge that neutralize the oppositely charged colloidal particles and brings about their coalescence. This process is called coagulation. This permits the particles to aggregate together until a denser particle is formed which falls through still water at a reasonable rate and is called flocculation Aluminium sulphate is the most common coagulating agent used for removing clay particles and is generally called filter alum. Other coagulants which also find application in water treatment include ferric sulphate, ferrous sulphate (copperas), chlorinated copperas, alum, ammonia alum or potash alum and sodium aluminate. Aluminium sulphate, when added to natural waters, hydrolyses to form colloidal aluminium hydroxide and an equivalent quantity of sulphuric acid as follows: Al2(SO4)3 + 6H2O → 2Al (OH)3 + 3H2SO4 The Al(OH)3 so formed acts as a floc or coagulant, which has an enormous surface area per unit volume and removes the finely divided and colloidal impurities by neutralizing the charge on them as well as by other mechanisms like adsorption and mechanical entrainment. Thus the smaller particles join together to form denser particles which settle down to the bottom. Some bacteria and colour associated with these particles also get removed simulataneously. In order to render the Al(OH)3 filterable and also in order to neutralize the H2SO4 liberated to permit the hydrolysis reaction to completion, some alkali will have to be added if the water is not sufficiently alkaline. The reaction taking place in alkaline waters may be represented as follows: Al2 (SO4)3 + 3Ca(HCO3)2 → 2Al(OH)3 + 3CaSO4 + 6CO2 With water having a little or no natural alkalinity (e.g., moorland waters) an alkali such as calcium hydroxide or sodium carbonate is added; the latter is more commonly used as it does not increase the hardness of the water. Al2(SO4)3 + 3Na2CO3 + 3H2O → 2Al (OH)3 + 3Na2SO4 + 3CO2 For treatment of acidic waters, sodium aluminate can be used as a source of Al(OH)3 and it is often used in conjunction with aluminium sulphate. The reactions taking place may be represented as under: Na AlO2 + 2H2O → NaOH + Al(OH)3 Al2(SO4)3 + 6Na AlO2 + 12H2O → 8Al(OH)3 + 3Na2SO4         Thus the alkalinity due to aluminate is neutralized by the acidity of the sulphate and Al(OH)3 is thus precipitated from both the reagents. In order to increase the efficiency of the coagulation process, coagulant aids such as lime, Fuller’s earth, bentonite clay and polyelectrolytes are also added. The coagulants are generally added in solution form to the water with the help of mechanical flocculators provided with slow moving rotating baffles or stationary baffles turning the flow of water; thus ensuring a gentle contact with the water and the reagents. After any sedimentation process, especially after that utilizing a chemical flocculent, there will be a substantial reduction in the bacterial count in the water. In addition, many coagulants release oxygen to the water as the chemical transformations take place. This oxygen helps in destroying some bacteria, in breaking up some organic compounds and also in partial removal of colour and taste producing organisms present in water. Thus the process of coagulation, flocculation and sedimentation has many benefic al results and is one of the most important processes of purification of water. Coagulation and settling equipment should be so designed that quick and thorough mixing of the coagulant and raw water should be achieved so that the coagulant dosage required is minimised.

28

BASIC ENGINEERING CHEMISTRY

For this purpose, mechanical type mixers and baffled mixing troughs are widely used. Modern types of coagulation and settling equipment include the floc-former type and the sludge-blanket type 1.8.3. Filtration. Filtration is a process of clarific tion of water by passing the water through a porous material, which is capable of retaining coarse impurities on its surface and in the pores. The porous material used is called the filtering medium and the equipment used for filtration is known as a filte . Filtration of water takes place due to the difference between the pressures at the of the bed of the filtering material and that underneath it. The difference in the two pressures (∆h) is called the pressure drop through the filtering medium. This pressure drop through the filtering bed depends on the rate of filtra ion, height of the filtering bed, diameter (size) of the grains of the filtering material and the extent of its contamination by the trapped impurities. Greater the numerical values of these factors (excepting grain size), greater will be the pressure drop. When contamination of a filter cannot be tolerated any longer, it is taken out of service for subsequent washing, in the course of which the impurities trapped in the filtering bed are washed out. Then the filter can be pressed into service again. The common materials used for the filtering medium are quartz sand (grain size 0.5 – 1.0 mm) containing not more than 96% SiO2, crushed anthracite (piece size 0.8 – 1.5 mm) and porous clay (piece size 0.8 – 1.5 mm). It should be remembered that when quartz sand filters are used with alkaline water, the filtrate (filtered water) gets enriched with silicic acid (due to the solubility of quartz sand in alkaline water). Slow sand filtratio Water for domestic use may be filtered through large area of finely graded sand beds at a slow rate (about 2 gal/sq. ft./hour). The rate of filtration slowly diminishes due to the accumulation of sediment in the capaillaries of the filter bed, and finally the rate becomes so slow that the bed must be cleaned. Cleaning is usually done by scarping the surface of the sand bed, or excavating, washing, and then relaying the entire filtering medium Slow gravity filters are not capable of removing colloidal impurities. These can be filter d out only when the water has been chemically treated (such as alum treatment) Fig. 1.14. Municipal water treatment with coagulation, and the slow sand filters cannot deal with settling and filtration tanks the gelatinous type of precipitates which are produced in such treatments. A typical equipment for Municipal water treatment with coagulation, settling and filtration tanks is shown in Fig. 1.14 Rapid-gravity filtratio Rapid gravity filters are capable of producing potable waters at flow-rates as high as 100 gal/sq. ft./hour. This is achieved by using carefully graded quartz sand and collecting the filtrate as evenly as possible over the entire bottom area of the sand bed which avoids undesirable channeling. The sand bed may be cleaned either by agitation with compressed air or in small units mechanically. This is followed by a flush back with clean water to wash away the accumulated impurities Rapid Gravity filtration has the following advantages (1) The filter bed and the quanlity of the filtrate can be easily inspected (2) The filter is unaffected by pressure variations on either the inflow or draw-off sides, and (3) Large reinforced concrete filters can be constructed at relatively low cost.

WATER AND ITS INDUSTRIAL APPLICATIONS

29

Rapid pressure filtratio Pressure filters are much more widely used than gravity filters, particularly in industrial in-stallations. Rapid pressure filters are used preferentially where filtration is to be effected in a rising main without breaking the hydraulic head, or where water from an elevated source can be passed through the filter and delivered in storage. The operation and cleaning of these filters are more or less similar to gravity sand filters. Filtration rates are of the order of 80 to > 200 gallons/sq. ft./hour. Pressure filters are manufactured in vertical and horizontal types. These filters consist of cylindrical steel shells fitted with dashed heads, containing a layer of a granular filter medium (sand or anthrafilt) Fig. 1.15. Mechanical filters supported by graded gravel (or anthrafilt), and is equipped with the required accessories e.g., piping, underdrains, valves, etc., for carrying out the cycle of operations, viz., filtra ion, backwashing and filtering to waste. Filter installations may consist of one or more units depending upon the requirements. Several types and designs of mechanical filters are employed Clarification for domestic and industrial purposes is often achieved by a combination of two to four of the different processes as listed below: (a) Sedimentation (b) Filtration (c) Coagulation + filtratio (d) Coagulation + settling (e) Coagulation + settling + filtration (f) Sedimentation + coagulation + settling + filtration (g) Chlorination and (h) Special filters (e.g., activated carbon for removing taste and odour; manganese zeolite for removing Fe and Mn; and Neutralizing filters such as graded calcite for removing C 2). 1.8.4. Sterilization of water. Clarification of water by sedimentation, filtration and storage removes suspended solids and also reduces the number of bacteria in the water. Total elimination of bacteria can be achieved only by sterlization. The organisms which should be eliminated by disinfection are divergent in character. They include (a) the enteric bacteria belonging to Salmonella, Shigella and Vibrio groups (b) the intestinal protozoa such as Entamoeba histolytica (c) some types of worms such as Schistosomes (d) Viruses such as those of infectious hepatitis and (e) Coliform organisms which indicate water pollution, though not pathogentic. Coliforms and enteric bacteria can be easily destroyed while viruses and cysts of E. histolytica are resistant to disinfection. Chlorine is the most common sterilizing agent in water treatment. It is capable of removing B. Coli and substantially reducing other bacteria. Chlorine may be added in the form of bleaching powder, or directly as a gas or in the form of concentrated solution in water. Whatever may be the method employed, the treatment should give accurate dosage, good distribution and sufficient time of contact (~ 30 minutes) so as to ensure effective sterilization. CaOCl2 + H2O → Ca (OH)2 + Cl2 Cl2 + H2O → HOCl + HCl HO Cl → [O] + HCl Hypochlorous Nascent acid oxygen The nascent oxygen so liberated destroys the germs and bacteria by oxidation. The chlorine itself, the hypochlorous acid and other chlorine compounds are also believed to have powerful

30

BASIC ENGINEERING CHEMISTRY

germicidal properties. The OCl– ions are capable of rupturing the cell membranes of the disease producing microbes. Bleaching powder has the limitations of being unstable during storage and also it increases the calcium content of water rendering it more hard. Chlorine as a sterilizing agent has the advantages of being economical, efficient, limited space requirement and convenience. Further, it does not introduce any other impurities in the water. Both chlorine as well as bleaching power when used in excess produce disagreeable odour in the water. Too much excess may cause irritation to the mucous membranes. The unpleasant taste of this excess chlorine can be removed by treatment with ammonia, which reacts with chlorine to form the tasteless compound chloramine (NH2.Cl). Thus, the ammonia-chlorine treatment (chloramine process) is particularly useful where traces of impurities are present (e.g., phenols) which produce unpleasant tastes when chlorine alone is used. Further, chloramine provides a more lasting effect than that of chlorine. Cl2 + NH3 → NH2Cl + HCl                        Monochloramine



2Cl2 + NH3 → NHCl2

+ 2HCl

                       Dichloramine



NH2Cl + Cl2 → NHCl2 + HCl NH2Cl + H2O → HOCl + NH3 HOCl → HCl + [O]

                          

nascent oxygen

The chloramine process consists of adding ammonia to water in the form of gas together with chlorine resulting in the formation of dichloramine and monochloramine as shown above. Dichloramine is a relatively stable compound and is perhaps not a sterilizing agent by itself. However, it slowly decomposes with evolution of chlorine. Thus, the addition of ammonia stabilizes the chlorine to provide a prolonged effect. This is particularly useful when the water is passed into storage after treatment. Chlorine, ammonia-chlorine, or sodium hypochlorite are used widely for the sterilization of swimming-pool water. A disadvantage of chlorination is the potential formation of trihalomethanes (such as chloroform, bromodichloromethane, dibromochloromethane and bromoform) which are carcinogenic. Trihalomethanes (THMs) may be formed when chlorine combines with natural organic substances, such as decaying vegetation, etc., that may be present in water itself. One approach to tackle this problem is to remove the organics completely before subjecting it to water chlorination. In future, the actual removal of THMs from the treated water, perhaps by aeration or adsorption on activated carbon, may become necessary. Superchlorination In superchlorination, a large excess of chlorine is added to the water, thereby destroying not only the micro-organisms but also the other organic impurities present. This process ensures rapid and complete sterilization and successfully used for waters derived from wells and rivers. This process is usually followed by dechlorination by NH3 or SO2. Break point chlorination It is a more precisely controlled process in which just sufficient chlorine is added to oxidize all the organic matter, destroy bacteria and react with any ammonia leaving a slight excess of free chlorine. The amount of chlorine required for disinfecting water depends upon the inorganic and organic impurities present in water. A relationship between the amount of chlorine added to water and the free residual chlorine is shown in Fig. 1.16.

Residual chlorine

WATER AND ITS INDUSTRIAL APPLICATIONS

Oxidation of reducing agent by chlorine

Formation of chloro-organic compounds and chloroamines

31

Destruction of chloro-organic Free residual and chloroamines chlorine

d b

c a

Break-point

Amount of chlorine added Fig. 1.16. Break point chlorination.

If chlorine is added to a sample of water and after a few minutes, the residual chlorine available in the water is estimated, it will be found that the residual chlorine in water is less than the amount added initially. This is due to the fact that some of the chlorine added initially is consumed by oxidising bacteria and other organic matter. Now, if we take a few more aliquots of the same volume of the water sample and add increasing doses of chlorine to different samples and analyse the residual chlorine after the same interval of time (a few minutes), a curve of the type shown above is obtained (Fig. 1.16). When chlorine is added to water, initially it reacts with ammonia and there will be formation of chloramines. Thus chlorine is consumed in the formation of chloramines. Then the oxidation of chloramines and other impurities start and there is fall in combined chlorine content. Only when chlorine demand for all such impurities has been met, free chlorine content (Cl2 HOCl) increases and acts as a disinfectant. The point at which all other impurities have been oxidised away is called the ‘breakpoint’ chlorination. Hence, to achieve chlorine as adisinfectant, dosages are to be given more than ‘break point’ and then the presence of free chlorine is felt and effective chlorination takes place. The presence of free chlorine is tested by using the reagent O-toludine, which develops an yellow colour within 5 seconds. The combined chlorine, on the other hand develops this colour after a long time (after 5 minutes). Thus break point chlorination of water may be defined as the chlorination to such an extent that micro-organism as well as bad tastes and odours present in wates are also destroyed. Determination of free chlorine in a water sample The principle involved in the estimation of free chlorine in water is that when a measured quantity of water is treated with excess of potassium iodide, the free chlorine present in the water oxidizes the corresponding amount of potassium iodide to iodine. The liberated iodine is estimated by titrating against standard sodium thiosulphate solution using starch as indicator Cl2 + 2KI → 2KCl + I2 I2 + 2Na2S2O3 → Na2S4O6 + 2NaI Dechlorination The water treated by the process of break-point chlorination may be filtered through activated carbon, in order to remove the decomposition products formed and the excess chlorine remaining. Other methods of dechlorination include treatment with SO2 or Na2SO3. SO2 + Cl2 + 2H2O → H2SO4 + 2HCl Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl

32

BASIC ENGINEERING CHEMISTRY

Sterilization by Ozone Ozone is a powerful disinfectant and is readily absorbed by water. Ozone being unstable decomposes as follows giving nascent oxygen which is capable of destroying the bacteria. O3 → O2 + [O] However, this process is relatively expensive but has the advantage of removing bacteria, colour, odour and taste without leaving any harmful residual effects in the water being treated. Sterilization by ultraviolet radiation Ultraviolet radiation emanating from electric mercury vapour lamp is capable of sterilizing water. This process is particularly useful for sterilizing swimming pool waters. However, this process cannot be economical for water works. Irradiation of water by ultraviolet light is commonly used for disinfection in food industries. Water for domestic purposes on a smaller scale may be sterilized by boiling the filtered water for about 20 minutes. Chemicals like potassium permanganate and tincture iodine are also used occasionally but chlorine tablets and bleaching powder are more commonly employed. 1.9 NUMERICAL PROBLEMS (A) Numerical Based on Hardness Example 1. Calculate the temporary and permanent hardness of a water sample, having the following analysis: Mg (HCO3)2 —   73 mg/l Ca (HCO3)2 — 162 mg/l CaSO4 — 136 mg/l MgCl2 —   95 mg/l CaCl2 — 111 mg/l NaCl — 100 mg/l Solution. Salt CaCO3 equivalent 100 Mg (HCO3)2 — 73 × = 50 mg/l 146 100 Ca (HCO3)2 — 162 × = 50 mg/l 162 100 — 136 × = 100 mg/l CaSO4 136 100 — 95 × = 100 mg/l MgCl2 95 100 — 111 × = 100 mg/l CaCl2 111 NaCl — Does not contribute to hardness and hence ignored. Temporary hardness ≡ [Mg(HCO3)2] + [Ca (HCO3)2] = 50 mg/1 + 100 mg/1 = 150 mg/1 or 150 ppm

= 150 × 0.07° Clark = 10.5° Clark

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33

Permanent hardness ≡ [CaSO4] + [MgCl2] + [CaCl2] = 100 mg/l + 100 mg/l + 100 mg/l = 300 mg/1 or 300 ppm = .. . Total hardness = = =

300 × 0.07° Clark = 21° Clark 150 + 300 = 450 ppm 450 × 0.07° Clark 31.5° Clark

Alternative method of calculation Equivalent weights of the different salts involved are as follows:   Salt Equivalent weight 146 Mg (HCO3)2 — = 73 2 162 Ca (HCO3)2 — = 81 2 136 CaSO4 — = 68 2 95 MgCl2 — = 47.5 2 111 CaCl2 — = 55.5 2 100 CaCO3 — = 50 2 Also, Weight in ppm or mg/l Equivalents per million, e.p.m. = Equivalent weight Now, Salt Equivalents per million 73 Mg (HCO3)2 — = 1 epm 73 162 Ca (HCO3)2 — = 2 epm 81 136 CaSO4 — = 2 epm 68 95 MgCl2 — = 2 epm 47.5 111 CaCl2 — = 2 epm 55.5 NaCl — Does not contribute to hardness and hence ignored. Temporary hardness = 50 × {[Mg(HCO3)2] + [Ca (HCO3)2]} = 50 × {1 + 2} = 150 ppm as CaCO3 = 150 × 0.07° Clark = 10.5° Clark (since 1 epm of each salt 50 ppm of CaCO3)

34

BASIC ENGINEERING CHEMISTRY

Permanent hardness = 50 × {[CaSO4] + [Mg Cl2] + [CaCl2]} = 50 × {2 + 2 + 2} = 300 ppm as CaCO3 = 300 × 0.07° Clark = 21° Clark Total hardness = Temporary hardness + permanent hardness = 150 ppm + 300 ppm = 450 ppm = 450 × 0.07° Clark = 31.5° Clark. Example 2. An exhausted Zeolite softener was regenerated by passing 150 litres of NaCl solution, having a strength of 150 g/l of NaCl. How many litres of hard water sample, having hardness of 600 ppm can be softened, using this softener? Solution. 150 litres of NaCl solution contains 150 × 150 g = 22,500 g Na Cl 50 ≡ 22,500 × g of CaCO3 equivalent hardness. 58.5 Given that 1 litre of hard water contains 600 ppm hardness ≡ 600 mg of CaCO3 = 0.6 g of CaCO3 ... The amount of hard water that can be softened by this softener 22,500 × 50 = = 32,051 litres. 0.6 × 58.3 Example 3. A sample of water was found to contain the following species, on analysis: Ca2+ = 40 mg/l Mg2+ = 24 mg/l Na+ = 8.05 mg/l HCO3– = 18.3 mg/l 2– SO4 = 55.68 mg/l Cl– = 6.74 mg/l Express the results in terms of salts present as their CaCO3 equivalents. Solution. Impurity CaCO3 equivalents

Ca2+

→

40 ×



Mg2+

→

24 ×



Na+

→



HCO3– →



SO42–

→

100 = 100 ppm 40

100 = 24 100 8.05 × 23 × 2 100 183 × 61 × 2 100 55.68 × 96

100 ppm = 17.5 ppm = 150 ppm = 58 ppm

100 Cl– → 6.74 × = 9.5 ppm 35.45 × 2 Hence, Total alkalinity = 150 ppm

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35

Calcium alkalinity = 100 ppm Magnesium alkalinity =   50 ppm Total hardness = 200 ppm Calcium hardness = 100 ppm Magnesium hardness = 100 ppm Calcium temporary hardness = 100 ppm   (or calcium alkalinity) Magnesium temporary hardness =   50 ppm   (or magnesium alkalinity) Magnesium permament hardness =   50 ppm   (or magnesium non-carbonate hardness) Hence the salts presents in terms of their CaCO3 equivalent may be expressed as follows: Ca(HCO3)2 — 100 ppm Mg(HCO3)2 — 50 ppm MgSO4 — 50 ppm Na2SO4 — 8 ppm NaCl — 9.5 ppm Rules: 1. Ca-alkalinity = Ca-hardness or total alkalinity, which ever is small. 2. (a) Mg-alkalinity = Mg-hardness, if total alkalinity is equal to or greater than total hardness. (b) Mg-alkalinity = Total alkalinity — Ca-hardness, if total alkalinity is greater than Ca-hardness but less than total hardness. 3. Sodium alkalinity = Alkalinity — Total hardness. 4. Ca-non-carbonate hardness = Ca-harndess — Ca-alkalinity 5. Mg-non-carbonate hardness = Mg-hardness — Mg-alkalinity 6. Total non-carbonate hardness = Total hardness – Total alkalinity. If any of the above computation gives negative or zero result, none of that substance is present. Example 5. How many grams of FeSO4 dissolved per litre gives 210.5 ppm of hardness. Solution. FeSO4 ≡ CaCO3 56 + 16 + 64 100g = 136 g \ 100 ppm of hardness = 136 ppm of FeSO4 or 210.5 ppm of hardness 136 × 210.5 = 100 ≡ 286.3 ppm of FeSO4 = 286.3 mg/L or = 0.2863 g/L of FeSO4 Hence, 0.2863 g of FeSO4 dissolved per litre gives 210.5 ppm of hardness.

36

BASIC ENGINEERING CHEMISTRY

Example 6. A water sample contains 204 mg of CaSO4 per litre. Calculate the hardness in terms of CaCO3 equivalent. Solution. \

CaSO4 ≡ CaCO3 –1 136g mol 100g mol–1 136 mg/L of CaSO4 = 100 mg/L of CaCO3 eq 100 mg/L of CaCO3 eq × 204 or 204 mg/L of CaSO4 ≡ 136 = 150 mg/L of CaCO3 eq Hence, hardness = 150 mg/L or ppm. Example 7. Calculate the temporary and total hardness of a sample of water containing Mg (HCO3)2 = 73 mg/L, Ca(HCO3)2 = 162 mg/L, MgCl2 = 95 mg/L, CaSO4 = 136 gm/L. Solution. Constituent Multiplication factor CaCO3 equivalent Mg(HCO3)2 = 73 mg/L

100/146

Ca(HCO3)2 = 162 mg/L

100/162

MgCl2 = 95 mg/L

100/95

CaSO4 = 136 mg/L

100/136

100 = 50 mg/L 146 100 162 × = 100 mg/L 162 100 95 × = 100 mg/L 95 100 136 × = 100 mg/L 136 73 ×



Temporary hardness = Mg (HCO3)2 + Ca (HCO3)2 = 50 mg/L + 100 mg/L = 150 mg/L = 150 ppm Total hardness = Mg (HCO3)2 + Ca (HCO3)2 + MgCl2 + CaSO4 = (500 + 100 + 100 + 100) mg/L = 350 mg/L or 350 ppm. Example 8. A sample of water on analysis has been found to contain following in ppm Ca(HCO3)2 = 4.86, CaSO4 = 6.80, MgSO4 = 8.40 Calculate the temporary and permanent hardness of water. Solution. Constituent Multiplication factor CaCO3 equivalent Ca(HCO3)2 = 4.86 ppm

100/162

CaSO4 = 6.80 ppm

100/136

MgSO4 = 8.40 ppm

100/120

Hence

Temporary hardness = Ca(HCO3)2 = 3 ppm

100 = 3ppm 162 100 6.80 × = 5ppm 136 100 8.4 × = 7ppm 120 4.86 ×

WATER AND ITS INDUSTRIAL APPLICATIONS

37

And, permanent hardness = CaSO4 + MgSO4 = 5 + 7 = 12 ppm. Example 9. A sample of water on analysis was found to contain: Ca(HCO3)2 = 4 mg/L, Mg(HCO3)2 = 6 mg/L, CaSO4 = 8 mg/L, MgSO4 = 10 mg/L. Calculate the temporary, permanent and total hardness of water in ppm, °Fr and °Cl. Solution. Constituent Multiplication factor CaCO3 equivalent Ca(HCO3)2 = 4 mg/L

100/162

Mg(HCO3)2 = 6 mg/L

100/146

CaSO4 = 8 mg/L

100/136

MgSO4 = 10 mg/L

100/120

100 = 2.469 mg/L 162 100 8× = 4.110 mg/L 146 100 8× = 5.882 mg/L 136 100 10 × = 8.333 mg/L 120 4×

emporary hardness due to Ca(HCO3)2 and Mg(HCO3)2 T = (2.469 + 4.110) mg/L = 6.579 mg/L = 6.579 ppm = 6.579 × 0.1°Fr = 0.6579°Fr = 6.579 × 0.07°Cl = 0.4605°Cl Permanent hardness due to CaSO4 and MgSO4 = (5.882 + 8.333) mg/L = 14.215 mg/L = 14.215 ppm = 14.215 × 0.1°Fr = 1.4215°Fr = 14.215 × 0.07°Cl = 0.9950°Cl Total hardness = (6.579 + 14.215) mg/L = 20.794 mg/L = 20.794 ppm = 20.794 × 0.1°Fr = 2.0794°Fr = 20.794 × 0.07°Cl = 1.4556°Cl Example 10. 100 ml of water sample has a hardness equivalent to 12.5 mL of 0.08N MgSO4. What is its hardness in ppm. Solution. 100 mL of water sample = 12.5 mL of 0.08N MgSO4 = 12.5 × 0.08 mL of 1N MgSO4 = 1 mL of 1 N MgSO4 = 1 mL of 1 N CaCO3 eq = 0.001 L of 1 N CaCO3 eq = 0.001 × 50 g CaCO3 eq = 0.05 g CaCO3 eq (or 50 mg CaCO3) 500 mg CaCO3 eq ×1000 mL \ 1000 mL (or/L) of water sample = 100 mL = 500 mg CaCO3 eq Hence, the hardness of water sample is 500 mg CaCO3 eq. per litre or 500 ppm.

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Example 11. Hardness of standard hard water (A) in terms of CaCO3 is 1,000 mg/litre. 50.0 mL of water (A) require 50.0 mL of EDTA solution for titration. 50.0 mL of hard water (B) required 40.0 mL of EDTA solution for titration. In another titration, 50 mL of hard water (B) after boiling, cooling and filt ring required 20.0 mL of EDTA solution. Calculate the hardness of water (B) in different units. Solution. Hardness of SHW = 1,000 mg/L = 1 mg/mL Now 50 mL of EDTA soln = 500 mL SHW = 50 mg CaCO3 eq 1 mL of EDTA solution = 1 mg CaCO3 eq \ (i) Total hardness of water (A) Calculation : 50 mL hard water (HW) ≡ 40 mL EDTA solution = 40 mg CaCO3 eq 40 mg CaCO3 eq ×1000 mL 1 L (or 1,000 mL) of SHW ≡ \ 50 mL = 800 mg CaCO3 eq Hence, total hardness of water = 800 mg/L or ppm = 800 × 0.1 = 80°Fr = 80 × 0.7 = 56°Cl (ii) Permanent hardness water (B) calculation 50 mL of boiled water = 20 mL of EDTA solution = 20 mg CaCO3 eq = 20 mg/L or ppm = 20 × 0.1 = 2°Fr = 2 × 0.7 = 1.4°Cl 20 mg CaCO3 × 1000 mL \ Hardness of 1L or 1000 mL of water ≡ 50 mL = 400 mg CaCO3 eq Permanent hardness = 400 mg/L = 400 ppm Temporary hardness = (800 – 400) ppm = 400 ppm. Numerical based on Zeolite method Example 1. The hardness of 1,000,000 litres of a water samples was completely removed by a zeolite softner. The exhausted softener required 500 litres of sodium chloride solution, containing 110g/litre of sodium chloride for regeneration. Calculate hardness of water in different units. Solution. 500 L of NaCl solution contains = 500 L × 110 g/L = 55,000 g NaCl 50 g CaCO3 eq = 55,000 × 58.5 = 47,008 g CaCO3 eq Hardness of 1,00,000 H water = 47,008 g CaCO3 eq 47, 008 Hardness of 1L water = \ 1, 00, 000 = 0.4701 g CaCO3 eq = 470.1 mg CaCO3 eq

WATER AND ITS INDUSTRIAL APPLICATIONS

39

Hence, hardness of water = 470.1 mg/L or 470.1 ppm Or 47.01°Fr and 32.91°Cl (because 1 ppm = 1 mg/L = 0.1°Fr = 0.07°Cl) Example 2. A zeolite softener was comletely exhausted and then regenerated by passing 200 litres of NaCl solution containing 100 g per litre of NaCl. How many litres of a sample of water of hardness 500 ppm can be softened by the softner. Solution. 200 litres of NaCl = 100 L × 200 g/L = 20000 g of NaCl 100 g of CaCO3 eq = 20000 × 58.5 × 2 = 17094 g of CaCO3 eq. Now 500mg of hardness is present in = 1 litre of water 17094 × 103mg of CaCO3 eq 1 × 17094 × 103 L Hardness is present in = 500 = 34188 litres of hard water Hence, the softener can soften 34188 litres of hard waters. Example 3. The hardness of 10,000 litres of a water sample was completely removed by a zolite softener. The exhausted softner required 500 litres of NaCl solution containing 110 g/litre of NaCl for regeneration. Calculate hardness of water in different units. Solution. 1 L of NaCl solution contains 110 g of NaCl \ 500 L of NaCl solution will contain = 500 × 110 = 55000 g of NaCl 10,0000 L of hard water = 55000 g of NaCl 50 = 55000 × g of CaCO3 eq. 58.5 55000 50 × g of CaCO3 eq 1 L of hard water = 10,000 58.5 = 470.08 mg of CaCO3 eq = 47.08°Fr = 32.91°Cl and 9.40 M eq/L Example 4. 10,000 litres of hard water was softned by passing through a zeolite softener. The exhausted zeolite coloumn required 30 litres of 2 N sodium chloride solution for its regeneration. Calculate the hardness of water. Solution. 30 L of 2N NaCl solution ≡ 30 L × 2 × 58.5 g/L of NaCl 30 × 2 × 58.5 × 50 g of CaCO3 eq ≡ 58.5 = 3,000 g of CaCO3 eq = 3 × 106 mg of CaCO3 eq Hardness of 10,000 L water ≡ 3 × 106 mg of CaCO3 eq \ Or hardness of 1 L of water ≡ 3 × 106/10,000 mg of CaCO3 eq = 300 mg of CaCO3 eq Hence, hardness of water sample is 300 mg/L or 300 ppm. Example 5. A zeolite softener was 95% exhausted, when 10,000 L of hard water was passed through it. The softener required 150L of NaCl solution of strength 50g NaCl/L of solution to regenerate. What is the hardness of water.

40

BASIC ENGINEERING CHEMISTRY

Solution. 10,000 L of hard water = 150 L of NaCl solution = 150 L × 50 g/L of NaCl = 7500 g of NaCl 100 of CaCO3 eq = 7500 × 58.5 × 2 = 6410.26 g of CaCO3 eq 1 L of hard water = 6410.26 g of CaCO3 eq/10,000 = 0.6410 g of CaCO3 eq = 641 mg of CaCO3 eq Hence hardness of water = 641 mg/L Example 6. A zeolite softener was completely exhausted and then regenerated by passing 200 litres of NaCl solution containing 100 g per litre of NaCl. How many litres of a sample of water of hardness 500 ppm can be softened by the softener. Solution. 200 litres of NaCl = 100 L × 200 g/L = 20,000 g of NaCl 100 g of CaCO3 eq = 20,000 × 58.5 × 2 = 17094 g of CaCO3 eq Now 500 mg of hardness is present in = 1 litre of water 17094 × 103 mg of CaCO3 eq. 1 × 17094 × 103 L Hardness is present in = 500 = 34188 litres of hard water Hence, the softener can soften 34118 litres of hard water. Example 7. An exhausted zeolite softener was regenerated by passing 150 litres of NaCl, having strength of 150 g/litre of NaCl. How many litres of hard water of 600 ppm can be softened using this solution. Solution. 150mL of NaCl solution ≡ 150 L × 150 g/L = 22,500 g NaCl 22,500 × 50 = g CaCO3 eq. 58.5 = 19,230 g CaCO3 eq. Let x litres of water of 600 ppm hardness can be softened by this zeolite softener Hardness of water = xL × 600 mg/L \ = 600 x mg CaCO3 eq. = 0.6 x g CaCO3 eq. 0.6 x g CaCO3 eq. = 19,230 g CaCO3 eq. \ Hence, x = 19,230 L 0.6 = 32,051 L Example 8. After treating 104 L of water by ion exchanger, the cationic resin required 200 L of 0.1N HCl and anionic resin required 200 L of 0.1 N NaOH solution. Find the hardness of the above sample of water. Solution. In an ion exchanger, all hardness causing cations are removed by cation exchanger, while anion exchanger removes anions of the constituents present in water. Consequently, the amount of acid used for regeneration of cation resin refers hardness part

WATER AND ITS INDUSTRIAL APPLICATIONS

41

104

\ Hardness in L of water ≡ 200 L of 0.1N HCl = 200 L of 0.1 N CaCO3 eq. = 200 × 0.1L of N CaCO3 eq. = 20L of 1N CaCO3 = 20 × 50 g of CaCO3 eq = 1,000 g of CaCO3 eq. 1000 \ Hardness in 1L of water = g of CaCO3 eq. = 0.1 g of CaCO3 eq. 104 = 0.1 × 1000 mg of CaCO3 eq = 100 mg of CaCO3 eq. Hence, hardness of water = 100 mg/L or ppm. Numerical based on lime soda process Example 1. Calculate the amount of lime (84% pure) and soda (92% pure) required for treatment of 20,000 litres of water whose analysis is as follows: Ca(HCO3)2 — 40.5 ppm Mg(HCO3)2 — 36.5 ppm MgSO4 — 30.0 ppm CaSO4 — 34.0 ppm CaCl2 — 27.75 ppm NaCl — 10.0 ppm Also, calculate the temporary and permanent hardness of the water sample. Solution. Salt CaCO3 equivalent 100 Ca(HCO3)2 — 40.5 × = 25 ppm or mg/l 162 100 Mg(HCO3)2 — 36.5 × = 25 ppm or mg/l 146 100 MgSO4 — 30.0 × = 25 ppm or mg/l 120 100 CaSO4 — 34.0 × = 25 ppm or mg/l 136 100 CaCl2 — 27.75 × = 25 ppm or mg/l 111 NaCl — Ignored as it does not contribute to hardness. Temporary hardness = [Ca (HCO3)2 ] + [Mg (HCO3)2] = 25 + 25 = 50 ppm Permanent hardness = [MgSO4] + [CaSO4] + [CaCl2] = 25 + 25 + 25 = 75 ppm Lime is required for Ca(HCO3)2, Mg(HCO3)2 and MgSO4. Soda is required for CaSO4, CaCl2 and the CaSO4 generated as a result of reaction of lime with Mg SO4. Hence, 84% pure lime required for treating 20,000 litres of water 20, 000 100 74 [25  +  (2 × 25)  +  25] ×  × = 1, 000 84 100     Ca(HCO3)2  Mg(HCO3)2  MgSO4  Purity grams for factor  2,000 litres = 1761.905 g = 1.7619 kg.

42

BASIC ENGINEERING CHEMISTRY

92% pure soda required for softening 20,000 litres of the water sample 106 100 20, 000 = [25  +  25  +  25] ×    ×  100 92 1, 000     CaSO4  CaCl2  MgSO4  Purity grams for factor  2,000 litres = 1728.26 g    = 1.72826 kg Example 2. A water sample has the analytical report as under: MgCO3 — 84 mg/l CaCO3 — 40 mg/l CaCl2 — 55.5 mg/l Mg(NO3)2 — 37 mg/l KCl — 20 mg/l Calculate the amount of lime (86% pure) and soda (83% pure) needed for the treatment of 80,000 litres of water. Solution. CaCO3 and MgCO3 should be regarded as being present in the form of their bicarbonates and only their weights have been expressed in terms of CaCO3 and MgCO3. KCl does not react with lime or soda and also it does not contribute to hardness. On converting the weights of each constituent in terms of their CaCO3 equivalent, we get the following: Salt CaCO3 equivalent 100 MgCO3 — 84 × = 100 mg/l 84 100 CaCO3 — 40 × = 40 mg/l 100 100 CaCl2 — 55.5 × = 50 mg/l 111 100 Mg(NO3)2 — 37 × = 25 mg/l 148 KCl — Ignored 86% pure lime required for softening 80,000 litres of the water 74 100 80, 000 = [(2 × 100)  +  40  +  25] ×    ×  100 86 1, 000      MgCO3  CaCO3  Mg(NO3)2  Purity grams for factor  80,000 litres = 18241.86 g. = 18.24186 kg 83% pure soda required for softening 80,000 litres of the water 106 80, 000 100 [50  +  25] ×    ×  = 100 1, 000 83 CaCl2 Ca(NO3)2 Purity grams for generated from reaction factor 80,000 of lime with Mg(NO3)2 litres = 7662.65 g. = 7.6625 kg.

WATER AND ITS INDUSTRIAL APPLICATIONS

43

Example 3. A water sample, on analysis, gave the following data : MgCl2 —   95 ppm CaSO4 — 272 ppm MgSO4 — 120 ppm H2SO4 —   49 ppm SiO2 —    4 ppm Calculate the amount of lime (95% pure) and soda (97% pure) needed for treating 1 million litres of water. If the costs of lime and soda are Rs. 40 and Rs. 2000 per 100 kg each respectively, calculate the total cost of chemicals used for treating 1 million litres of the water. Solution. Impurity CaCO3 equivalent 100 MgCl2   95 × = 100 mg/l 95 100 CaSO4 272 × = 200 mg/l 136 100 MgSO4 120 × = 100 mg/l 120

H2SO4   49 ×

100 = 50 mg/l 98

SiO2 Ignored Amount of 95% pure lime required for softening 1 million litres of the water 74 106 100 [100  +  100  +  50] ×    ×  = 100 95 106 MgCl2 MgSO4 H2SO4 purity kg for factor 106 litres = 194.74 kg ...  Cost of the lime required at the rate of Rs. 40 per 100 kg 194.74 × 40 = = Rs. 77.90 100 Similarly, amount of 97% pure soda needed for 1 million litres of the water 100 106 [100 + 200 + 100 + 50] × × = 97 100 CaCl2 CaSO4 CaSO4 CaSO4 purity derived derived derived factor from from from MgCl2 MgSO4 H2SO4 during during during lime lime lime treatment treatment treatment = 491.75257 kg ...  Cost of the soda required at the rate of Rs. 2000 per 100 kg 491.75257 × 2000 = = Rs. 9835.05 100

106

106 kg for 106 litres

44

BASIC ENGINEERING CHEMISTRY

...  Total cost of the chemicals needed for softening 1 million litres of the water = Rs. (77.90 + Rs. 9835.05) = Rs. 9912.95 Example 4. A water sample, on analysis, gave the following constitutents in grains per gallon MgCl2 — 9.5 CaSO4 — 3.4 CaCO3 — 5.0 Mg(HCO3)2 — 7.3 MgSO4 — 6.0 SiO2 — 2.4 Calculate the cost of the chemicals required for softening 20,000 gallons of water if the purities of lime and soda are 95% and 90% respectively. The costs per 100 pounds each of lime and soda are Rs. 28 and Rs. 96 respectively. Solution. Salt CaCO3 equivalent 100 MgCl2 9.5 × = 10 gpg 95 100 CaSO4 3.4 × = 2.5 gpg 136 100 CaCO3 5.0 × = 5.0 gpg 100 100 Mg(HCO3)2 7.3 × = 5.0 gpg 146 100 MgSO4 6.0 × = 5.0 gpg 120 SiO2 Ignored Lime is required for [Temporary calcium hardness + (2 × Temporary magnesium hardness) + permanent magnesium hardness] = 5 + (2 × 5) + 10 + 5 = 30 gpg ...  Cost of 95% pure lime required for softening 20,000 gallons at the rate of Rs. 28 per 100 lbs 28 1 100 74 = × 30 × 20,000 × × × 100 7000 95 100 lime required gallons 1 lb = 7000 purity cost per per gallon treated grains factor 100 lb = Rs. 18.69 Similarly, soda is required for [Ca permanent hardness + Mg permanent hardness which generated equivalent quantity of Ca permanent hardness during lime treatment] = 2.5 + (10 + 5) = 17.5 gpg ...  Cost of 90% soda required for treating 20,000 gallons at the rate of Rs. 96 per 100 lbs 96 1 106 100 = × 17.5 × 20,000 × × × 100 7000 100 90 Soda required gallons 1 lb = 7000 purity cost per per gallon treated factor factor 100 lb = Rs. 56.53.

WATER AND ITS INDUSTRIAL APPLICATIONS

45

Example 5. A water sample, using FeSO4 . 7H2O as a coagulant at the rate of 139 ppm, gave the following results on analysis: Ca2+ — 160 ppm; Mg2+ —   72 ppm CO2 —   88 ppm; HCO3– — 488 ppm Calculate the lime and soda required to soften 1,00,000 litres of water. Solution. Impurity Equivalents of CaCO3 100 Ca2+ 160 × = 400 ppm 40 100 Mg2+   72 × = 300 ppm 24 100 CO2   88 × = 200 ppm 44 100 HCO3– 488 × = 400 ppm 61 × 2

FeSO4 . 7H2O



Lime required =

100 139 × = 50 ppm 278 74 × [Mg2+ + CO2 + HCO3– + FeSO4 . 7H2O] 100

74 × [300 + 200 + 400 + 50] 100 74 = × 950 = 703 ppm 100 = 703 mg/l Lime required for softening 1,00,000 litres of water =

= Similarly, soda required = = =

703 ×

105 105

= 70.3 kg

106 × [Ca2+ + Mg2+ + FeSO4 . 7H2O – HCO3– ] 100 106 × [400 + 300 + 50 – 400] 100 106 × 350 = 371 ppm 100

= 371 mg/l .. . Soda required for softening 105 litres of water 6 = 371 × 10 = 37.1 kg 106

Note 1 Note 2

Mg2+ + – 2HCO3 + (Two equivalents)

Ca(OH)2 → Ca(OH)2 → (Two equivalents)

Mg(OH)2 + Ca2+ CO3– – + CaCO3 + H2O (Two equivalents)

46

BASIC ENGINEERING CHEMISTRY

Example 6. Calculate the quantities of lime and soda required for cold softening of 2,00,000 litres of water using 16.4 ppm of sodium aluminate as a coagulant. The results of the analysis of raw water and softened water are as follows: Raw water Softened water 2+ 2– Ca — 160 ppm CO3 — 30 ppm Mg2+ —   72 ppm OH– — 17 ppm – HCO3 — 732 ppm Dissolved CO2 —   44 ppm Solution. Converting each of the constituents into their respective equivalents of CaCO3, we have Raw Water Ion or salt ppm CaCO3 equivalent Lime Soda   (ppm) required, required (ppm) (ppm) 100 Ca2+ 160 160 × = 400 — 400 40 100 Mg2+   72   72 × = 300 300 300 24 100 HCO3– 732 732 × = 600 600 – 600 61 × 2 100 CO2   44   44 × = 100 100 — 44 100 NaAlO2 16.4   16.4 × = 10 – 10 – 10 82 × 2 Softened Water 100 CO32–   30   30 × = 50 — + 50 60 100 OH–   17   17 × = 50 50 + 50 17 × 2 1040 190 74 Lime required = × 1040 mg/l 100 Lime required for 2,00,000 litres of water 74 2, 00, 000 kg × 1040 × = 100 106 = 153.93 kg Similarly, soda required 106 = × 190 mg/l 102 ... Soda required for 2,00,000 litres of water 106 2, 00, 000 kg × 190 × = 100 106 = 40.28 kg.

WATER AND ITS INDUSTRIAL APPLICATIONS

47

Notes: 1. 1 equivl. of Ca2+ requires 1 equiv. of soda. Ca2+ + Na2CO3 → CaCO3 + 2Na+ 2. 1 equiv. of Mg2+ requires 1 equiv. of lime and 1 equiv. of soda. Mg2+ + Ca(OH)2 → Mg(OH)2 + 2Ca2+ Ca2+ + Na2CO3 → CaCO3 + 2Na+ 3. 1 equiv. of HCO3– requires 1 equiv. of lime which simultaneously produces 1 equiv. of CO32–, which may be considered to be equivalent to 1 equiv. of soda. Ca (OH)2  +  2HCO3– → CaCO3  +  CO32 –  +  H2O



2 equiv.  

2 equiv.

2 equiv.



4. 1 equiv. of CO2 requires 1 equiv. of lime. 5. 1 equiv. of NaAlO2 requires neither lime nor soda. But, however, NaAlO2 produces 1 equiv. of OH–, which may be imagined to be 1 equiv. of lime Na AlO2 + 2H2O → NaOH + Al (OH)3 Hence the corresponding quantity of NaAlO2 in equivalents should be deducted both from lime as well as soda requirements. 6. Since the treated water is shown to contain OH– and CO32–, the required amount of OH– should have been supplied by its equivalent amount of Ca(OH)2. But, however, the corresponding amount of Ca2+ so incorporated should have been removed by adding equivalent amount of soda. The CO32– required to be present in the treated water must have been supplied by its equivalent amount of Na2CO3. Example 7. The analytical report of a raw water sample is as follows: Mg Cl2 — 47.5 mg/l Ca Cl2 — 55.5 mg/l Ca SO4 — 4.06 mg/l Turbidity — 120 mg/l 10 mg/l alum dose was found to be sufficient to remove the entire turbidity of the water sample. Calculate the total weight of the dry sediment in a lime-soda softening plant for 20,000 litres of water. Alum contains 7% Al. Solution. 1. MgCl2  +  Ca(OH)2 → Mg(OH)2↓  +  CaCl2



95

CaCl2  +  Na2CO3 111

58

111

→ CaCO3↓   +  2NaCl 100

95 mg of MgCl2 yields 58 mg of Mg(OH)2 58 × 47.5 ... 47.5 mg of MgCl2 yields mg 95   = 29 mg of Mg(OH)2 ...(1 a) Again, 95 mg of MgCl2 will produce 111 mg of CaCl2 which gives 100 mg of CaCO3 on treatment with soda, 100 mg of CaCO3 ...  29 mg of MgCl2 will produce 29 × 95 = 30.526 mg of CaCO3 ...(1 b)

48



BASIC ENGINEERING CHEMISTRY

2.



CaCl2  +  Na2CO2

→ CaCO3↓  +  2NaCl

111

100



111 mg of CaCl2 yields 100 mg CaCO3 . . . 55.5 mg of CaCl2 yields 100 × 55.5 mg of CaCO3 111 = 50 mg of CaCO3 3. CaSO4  +  Na2CO2 → CaCO3↓  +  Na2SO4

136

...(2)

100



136 mg of CaSO4 gives 100 mg of CaCO3 ... 4.08 mg of CaSO4 gives 4.08 × 100 408    = 3 mg of CaCO3 = 136 136

4.

Turbidity 120 mg/l

5.    

Al3+ 

   (From alum)

...(3) ...(4)

hydrolysed

 →   Al(OH)3↓ 78

27

100 mg of alum contains 7 mg of Al3+ 7 × 10 ... 10 mg of alum contains = 0.7 mg of Al3+ 100 (Since it is given that alum contains 7% Al and the alum dose used is 10 mg/l) Now, 27 mg of Al3+ (from alum) gives 78 mg of Al(OH)3 ...   0.7 mg of Al3+ gives 78 × 0.7 = = 2.222 mg of Al(OH)3 ...(5) 27 .. .  Dry sediment obtained from 1 litre of the water sample = (1a + 1b) + (2) + (3) + (4) + (5) = (29 + 30.526 + 50 + 3 + 120 + 2.222) mg = 234.748 mg .. . Total weight of dry sediment obtained from softening 20,000 litres of the water sample = (234.748 × 20,000) mg = 4.69496 kg. Example 8. A water sample gave the following analytical results: Total alkalinity = 290 ppm as CaCO3 Calcium hardness = 242 ppm as CaCO3 Magnesium hardness = 63 ppm as CaCO3 Calculate the lime and soda required to soften 1 million litres of such a water sample. Solution. Alkalinity present in water is first bound to C 2+, then to Mg2+, and only then to Na+. Further, in the present case, Ca-alkalinity = (Ca-hardness or Total alkalinity, whichever is small) = 242 ppm as CaCO3 ... Ca-Temporary hardness = 242 ppm as CaCO3 Mg-alkalinity = (Total alkalinity – Ca-hardness) since, the total alkalinity is greater than Ca hardness but lesser than total hardness

WATER AND ITS INDUSTRIAL APPLICATIONS

49

= (290 – 242) = 48 ppm as CaCO3 ... Mg-temporary hardness = 48 ppm as CaCO3 Mg-permanent hardness = (Mg hardness – Mg alkalinity) = (63 – 48) = 15 ppm as CaCO3 Ca-permanent hardness = (Ca-hardness–Ca-alkalinity) = (242 – 242) = 0 ... Lime required for softening 106 litres of water sample 74 = × [Ca Temp + (2 × Mg Temp) + Mg Perm] × 106 mg 100 74 = × [242 + (2 × 48) + 15] × 106 mg 100 74 = × 353 × 106 mg 100 74 = × 353 kg 100 = 261.22 kg. Soda reqd. for softening 106 litres of the water sample 106 = × [Ca perm + Mg perm] × 106 mg 100 106 = × [0 + 15] × 106 mg 100 106 1 = × 15 × 106 × 6 kg = 15.9 kg. 100 10 Example 9. Calculate the amount of lime and soda required for softening of 15000 litres of water which analysed as follows Temporary hardness = 20 ppm, Permanent hardness = 15 ppm, permanent Mg hardness = 10 ppm Calculate the lime and soda required to soften 1 million litres of such a water sample. Solution. Lime requirements 74 [Temporary hardness + Permanent hardness ] × Volume of water = 100 74 = 100 [(20 ppm + 10 ppm) ×15000 L] = 22.2 mg L–1 × 15000 L = 3.30000 mg = 330 g. Soda requirement 106 [permanent hardness] × volume of water = 100 106 = 100 [15ppm] × 15000 L = 15.0 mg L–1 × 15000 L = 238500 mg = 238.5 mg.

50

BASIC ENGINEERING CHEMISTRY

Example 10. Calculate the requirement of lime and soda for softening 25000 litres of water from the following analytical data. Dissolved CO2 = 30.8 ppm, HCO3– = 164.7 ppm, Ca2+ = 40 ppm, Mg2+ = 12 ppm Solution. Calculation of CaCO3 equivalents Salt CaCO3 equivalent 100 CO2 30.8 × = 70 ppm 44 100 = 135 ppm HCO3– 164.7 × 122 100 = 100 ppm Ca2+ 40 × 40 100 = 50 ppm Mg2+ 12 × 24 Lime requirement 74 [CO 2 + HCO3- + Mg 2 + ] × Vol of water = 100 74 [70 + 135 + 50] ppm × 250000 = 100 = 74 × 2.55 × 0.250 kg = 74 × 2.55 × 0.25 kg = 47.175 kg Soda requirement 106 [Ca 2 + + Mg 2 + - HCO3- ] × vol of water = 100 106 [100 + 50 - 135] × 250000 = 100 = 15.9 × 250000 mg = 15.9 × 0.25 kg = 3.975 kg. Example 11. A water sample contains the following impurities, Ca2+ = 20 ppm, Mg2+ = 18 ppm, HCO3– = 183 ppm and SO42– = 24 ppm. Calculate the amount of lime and soad needed for softening. Solution. Calculation of CaCO3 equivalents Salt CaCO3 equivalent 100 20 × = 50 ppm Ca2+ = 20 ppm 40 100 75 ppm = 24 100 HCO3– = 183 ppm 183 × 150 ppm = 122 Requirement of lime 74 [Mg 2 + + HCO3- ] = 100 74 [75 + 150] ppm = 100 = 166.5 ppm or mg/L.

Mg2+ = 18 ppm

18 ×

WATER AND ITS INDUSTRIAL APPLICATIONS

51

Requirement of soda 106 [Ca 2 + + Mg 2 + - HCO3- ] 100 106 [50 + 75 - 150] ppm = 100 = Nil. Example 12. Calculate the amount of lime and soda needed for softening 1,00,000 litre of water containing the following HCl = 7.3 mg/L, Al2(SO4)3 = 34.2 mg/L MgCl2 = 9.5 mg/L, NaCl = 29.25 mg/L Purity of lime is 90% and that of soda is 98%. 10% of chemicals are to be used in excess in order to comlete the reaction quickly. Solution. Calculation of CaCO3 equivalent Salt CaCO3 equivalent 100 7.3 × = 10 mg/L HCl = 7.3 mg/L 73 100 34.2 × 30 mg/L = Al2(SO4)3 = 34.2 mg/L 114 100 MgCl2 = 9.5 mg/L 95 × = 10 mg/L 95 Requirement of lime 74 [HCl + Al2 (SO 4 )3 + MgCl2 ] × Vol of water ×100/% purity = 100 74  100  [10 + 30 + 10] mg/L × 105 L ×  =  100  90  6 = 4.111 × 10 mg = 4.111 kg Lime required (using 10% excess) 110 = 4.111 kg × 100 = 4.522 kg Requirement of soda 106 = [HCl + Al2 (SO 4 )3 + MgCl2 ] × Vol of water ×100/% purity 100 = 106 [10 + 30 + 10] mg/L × 105 L × 100 100 98 =

= 5.488 × 106 mg = 5.408 kg \ Soda requirement using 10% excess 110 = 5.949 kg = 5.408 × 100 = 5.949 kg. Example 13. A water sample on analysis gave the following data Ca2+ = 30 mg/L, Mg2+ = 24 mg/L, CO2 = 24 mg/L, HCl = 50 mg/L, K+ = 10 mg/L Calculate the quantities of lime (90% pure) and soda (94% pure) required to soften one million litres of water sample.

52

BASIC ENGINEERING CHEMISTRY

Solution. Calculation of CaCO3 equivalent Salt CaCO3 equivalent 100 Ca2+ = 30 mg/L 30 × = 75.0 mg/L 40 100 24 × 100.0 mg/L = Mg2+ = 24 mg/L 24 100 24 × 54.5 mg/L = CO2 = 24 mg/L 44 100 HCl = 50 mg/L 50 × = 68.5 mg/L 73 74 [Mg 2 + + CO 2 + HCl] × Vol. of water × (100/% purity) Requirement of lime = 100 74 100 [100 + 54.5 + 68.5 mg/L] × 106 L × = 100 90 = 1.834 × 108 mg = 1.834 × 102 kg = 183.4 kg 106 Requirement of soda = [Ca2+ + Mg2+ + HCl] × Vol of water × 100% purity 100 106 100 [75 + 100 + 68.5] mg/L × 106 L × 100 94 106 100 [243.5 mg/L] × 106 L × = 100 94 = 2.946 × 108 mg = 2.946 × 102 kg = 294.6 kg Example 14. Calculate the amount of lime and soda required per litre for the chemical treatment of water containing: Ca2+ = 80 ppm, Mg2+ = 36 ppm, K+ = 39 ppm, HCO3– = 244 ppm, FeSO4.7H2O added as coagulant = 69.5 ppm. Solution. Calculation of CaCO3 equivalent Salt CaCO3 equivalent

=

100 = 200 mg/L or ppm 40 100 36 × = 150 mg/L or ppm Mg2+ = 36 ppm 24 100 224 × 200 mg/L or ppm = HCO3– = 224 ppm 122 100 FeSO4.7H2O = 69.5 ppm 69.5 × = 25 mg/L or ppm 278 74 Requirement of lime = [Mg 2 + + HCO3- + FeSO 4 .7H 2 O] × Vol. of water 100 74 [150 + 200 + 25 mgL-1 ] × 1L = 100 = 277.5 mg 106 [200 + 150 - 200 + 25 mgL-1 ] × 1L Requirement of soda = 100 = 185.5 mg.

Ca2+ = 80 ppm

80 ×

WATER AND ITS INDUSTRIAL APPLICATIONS

53

Example 15. Explain with equation and calculate the quantity of quick lime and soda ash required to soften 10000 litres of water containing: (i) 219 ppm of Mg(HCO3)2 and 234 ppm of NaCl (ii) 36 ppm of Mg2+ and 18.3 ppm of HCO3– (iii) 1.5 ppm of the free acids, 144 ppm of sulphate ions and 71 ppm of chloride ions. Solution. Calculation of CaCO3 equivalent 65 → 2CaCO3 + Mg(OH)2 + 2H2O (2L) Mg(HCO3)2 + 2Ca(OH)2  Then, Then,

→ Mg(OH)2 + Ca2+ Mg2+ + Ca(OH)2 

→ CaCO3 + 2Na+ (L + S) Ca2+ + Na2CO3  → CaCO3 + H2O + CO32– (L–s) 2HCO3– + Ca(OH)2  → Ca2+ + 2H2O 2H+ + Ca(OH)2 

→ CaCO3 + 2Na+ (L + S) Ca2+ + Na2CO3  74 10000 Lime = [2Mg (HCO3 ) 2 + Mg 2 + + HCO3- + H + ] × kg 100 106 74  100 100 100 100  1 = 100  2 × 219 × 146 + 36 × 24 + 18.3 × 122 + 1.15 × 2  × 100 = 3.996 kg 106  2 + 10000 Mg - HCO3- + H +  × kg Soda =  100 106 106  100 100 100  1 = 100 36 × 24 - 18.3 × 122 + 1.5 × 2  × 100   = 2.226 kg. Example 16. Calculate the amount of lime and soda required to soften 24000 litres of water per day for a year containing the following CaCO3 = 1.85 mg/litres, CaSO4 = 0.34 mg/litre, MgCO3 = 0.42 mg/litre, MgCl2 = 0.76 mg/ litre, MgSO4 = 0.90 mg/litre, NaCl = 2.34 mg/litre, SiO2 = 2.32 mg/litre. The purity of lime is 88.3% and that of soda is 99.2%. Solution. Calculat of CaCO3 equivalent Salt CaCO3 equivalent 100 1.85 × = 1.85 mg/litre CaCO3 100 100 CaSO4 0.34 × = 0.25 mg/litre 136 100 MgCO3 0.42 × = 0.50 mg/litre 84 100 0.76 × = 0.80 mg/litre MgCl2 95 100 MgSO4 0.90 × = 0.75 mg/litre 120 Note: NaCl, Fe2O3 and SiO2 do not require lime soda) Amount of water to be softened per year = 24000 litres × 365 = 8760000 litres

54

BASIC ENGINEERING CHEMISTRY

74 [Temp Ca + 2 × temp Mg + Per Mg] × Vol of water 100 74 [1.85 + 2 × 0.50 + 0.80] mg/litre × 8760000 = 100 = 2.701 × 8760000 mg = 2.701 × 8.76 kg

Requirement of lime =

= 23.66 kg. 106 [Temp + Per Mg] × Vol of water 100 106 = 100 [0.80 + 0.75] × 8760000 = 1.643 × 8760000 mg

Requirement of soda =

= 1.643 × 8.76 kg = 14.39 kg. Example 17. Calculate the amount of lime and soda ash required to soften a million litres of water having the following composition by lime soda proces: Free CO2 = 20 ppm, Ca(HCO3)2 as CaCO3 = 200 ppm, Mg(HCO3)2 as MgCO3 = 5 ppm, CaSO4 as Ca2+ ions = 30 ppm, MgSO4 as Mg2+ ions = 10 ppm. Solution. Calculat of CaCO3 equivalent



Salt

CaCO3 equivalent



CO2

= 20 ppm

20 ×



CaCO3

= 200 ppm



MgCO3

= 5 ppm



Ca2+

= 30 ppm



Mg2+

= 10 ppm



Requirement of lime =



=



100 45.45 ppm = 44 100 200 × = 200 ppm 100 100 = 5.95 ppm 84 100 30 × = 75 ppm 40 100 10 × 41.67 ppm = 24 5×

74 [CO 2 + CaCO3 + 2 × MgCO3 + Mg 2 + ] × Vol of water 100

74 [45.45 + 200 + 2 × 5.95 + 41.67] × 106 litre of water 100 = 221.27 × 106 mg

= 221.27 kg 106 [Ca 2 + + Mg 2 + ] × Volume of water 100 106 [75 + 41.67] × 106 mg = 100 = 123.67 kg.

Requirement of soda =

WATER AND ITS INDUSTRIAL APPLICATIONS

55

QUESTIONS

1. Match the following and write appropriate statements in each case:

Group A Group B (i) Dissolved oxygen (ii) Sequestration (iii) Calcium sulphate (iv) Sodium fluorid (v) Turbidity (vi) Tannin (vii) Milligrams per litre (viii) Colour (ix) Grains per gallon (x) M = 2P (xi) M < 2P (xii) M > 2P

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m)

Lime and soda Carbonate alkalinity only Carbonate and bicarbonate alkalinity Carbonate and hydroxide alkalinity Degree Clark Silica scale Colloidal conditioning Parts per million Silica removal Sodium hexametaphosphate Hydrazine Hazen units Boiler scales

2. Complete the following statements with appropriate words: (a) To control total dissolved solids in boiler water during boiler operation .................... practice is adopted. (b) Inter-crystalline cracking occurs in boiler due to presence of .................... . (c) Sodium bicarbonate gives .................... colour with phenolphthalein. (d) .................... is used as an indicator in the estimation of calcium hardness by EDTA method. (e) Principle involved in the zeolite process of water softening is .................... . (f) Pulsating steam demand is believed to be the cause of .................... in a boiler. (g) Sludges are formed by the substances which have .................... solubilities in hot water than in cold water. (h) Zeolite plant usually occupies .................... space than a lime soda plant softening same volume of water. (i) Exhausted cation exchanger bed is regenerated by passing a solution of .................... while exhausted anion exchanger is regenerated by treating it with .................... . (j) 1 ppm = .......................... degree clark. (k) Temporary hardness is due to ...................... . 3. Answer the following as directed: (a) “If hard water contains sufficient magnesium hardness, silica can be removed effectively in lime soda softening process”. State whether the statement is true or false. Justify your answer. (b) “Phosphate conditioning is suitable at all-operating pressures”. Give reasons to support this statement. (c) Mention any two points of contrast in lime soda and zeolite water softening processes. (d) Explain any two major drawbacks of scale formation. (e) A water sample contains Ca++ – 20 mg/l, Mg++ – 12 mg/l. What is the total hardness of the sample?

56

BASIC ENGINEERING CHEMISTRY

4. What are boiler troubles and what are their consequences? How can be boiler troubles be minimized? 5. (a) What do you mean by hardness of water? How is it classified (b) What process of water softening would you recommend for obtaining feed water for the modern high pressure boilers, and why? 6. Compare the hot lime soda process and the zeolite process of water softening with respect to the principles involved, advantages and limitations. 7. Compare carbonate conditioning and phosphate conditioning of boiler water with respect to the principles involved, advantages and limitations. 8. (a) Establish the relation between ppm and mg/l. (b) Calculate the lime and soda required to soften 10,000 litres of the water sample having the following analysis: Calcium hardness —  250 ppm as CaCO3 Magnesium hardness —  100 ppm as CaCO3 Total alkalinity present —  300 ppm as CaCO3 9. What are the problems expected by using untreated water in a boiler? How can they be minimized? 10. A zeolite softener was 75% exhausted by removing the hardness completely when 7000 litres of a water sample are passed through it. The zeolite bed required 180 litres of 2.5% NaCl solution for complete regeneration. Calculate the harndess of the water sample. 11. A water sample on analysis gave the following data: Ca2+ — 20 ppm, Mg2+ — 24 ppm – CO2 — 30 ppm, HCO3 — 150 ppm K+ — 10 ppm Calculate the lime (87% pure) and soda (91% pure) required to soften 1 million litres of the water sample. 12. Discuss the different types of cold lime soda softeners and their relative merits and demerits. 13. Write short notes on the following: (a) Sludge blanket type of lime soda softener (b) Cold lime-soda process (c) Sedimentation (d) Coagulation (e) Filtration of water (f) Sterilization. 14. Write short notes on the following: (a) Caustic embrittlement (RGPV Bhopal 2009) (b) Boiler corrosion (RGPV Bhopal 2006) (c) Internal treatment (d) Blow-down (e) Break-point chlorination (RGPV Bhopal 2006) (f) Deep well water (g) Impurities in water and their effects (h) Slow sand filtratio

WATER AND ITS INDUSTRIAL APPLICATIONS

57

(i) Rapid pressure filtratio (j) Scall and sludge formation (RGPV Bhopal 2009) 15. (a) Write the constituents responsible for the permanent hardness of water. Discuss one treatment method. (b) Why does hard water consume a lot of soap ? (c) Why does magnesium bicarbonate required double amount of lime for softening. (U.P. Technical University 2001) 16. (a) Describe the causes, harmful effects and control of scale and sludge formation in boilers. (RGPV Bhopal 2009) (b) A water sample using FeSO4.7H2O as coagulant at the rate of 278 ppm. gave the following results on analysis : Ca++ – 80 ppm, Mg++ – 48 ppm, CO2 – 88 ppm; and HCO3– – 244 ppm Calculate lime soda required for softening one million litres of the water sample. or (a) What are zeolites ? Discuss the chemistry involved in zeolite process of softening hard waer. Also mention the limitations, advantages and disadvantages of this process. (b) A zeolite softener was completely exhausted and was regenerated by passing 120 litres of sodium chloride solution containing 150 g/litre of NaCl. How many litres of a sample of water of hardness 500 mg/liter can be softened by the softener before regenerating it again ? (RGPV Bhopal 2001) 17. (a) The analytical results of raw water and treated water are as follows : Ca2+ = 300 ppm 2+ Mg = 150 ppm HCO3– = 244 ppm OH– = 65 ppm 2– CO3 = 40 ppm CO2 = 60 ppm Calculate, (i) The amount of lime (80% pure) and soda (90% pure) required to soften one million litres of water using NaAlO2 as a coagulant at the rate of 41 mg/litre. (ii) If 10,000 litres of the same water sample is softened through a zeolite softener, how much NaCl will be required for its regeneration ? (Nagpur University, S-2001) 18. (a) What is meant by hardness of water ? Name the substances that cause permanent hardness in water? (b) Explain the importance of “phosphate conditioning ? in the “internal treatment” of water. (c) What is “demineralisation process” ? Point out its advantages and limitations ? (Mumbai University, 1994) 19. (a) Explain the principle and process of lime-soda softening of water giving the different chemical reactions involved in the process. Point out the advantages of hot lime-soda process. (b) A zeolite softener was completely exhausted and was regenerated by passing 100 liters of sodium chloride solution containing 100 g/l of sodium chloride. How many litres of sample of water of hardness of 500 ppm can be softened by the softener ?

58

BASIC ENGINEERING CHEMISTRY

20. (a) How does the formation of sludge and scales affect boiler performance ? (b) What is meant by “Phosphate Conditioning” of water ? 21. (a) Explain the demineralisation of hard water, with a neat diagram and appropriate equations. (RGPV Bhopal 2009) (b) A sample of water was found to contain the following impurities. Mg(HCO3)2 = 156 mg/1 H2SO4 = 4.9 mg/l MgCl2 = 23.75 mg/1 NaCl = 5.6 mg/l CaCl2 = 111 mg/1 SiO2 = 16.2 mg/l Calculate the amount of lime (90% pure) and Soda (95% pure) required to soften 50,000 litres of the above water sample. 22. (a) Three water samples A, B and C were analyses for their salt contents. Sample A was found to contain 168 gms of magnesium carbonate per liter. Sample B was found to contain 82 gms of calcium nitrate and 2 mg of silica per litre, sample C was found to contain 20 gms of potassium nitrate and 20 gms of calcium carbonate per 500 ml. Determine the hardness in all the above three water samples. (b) How does dissolved oxygen affect the quality of water used in boilers ? What are the various methods employed in deaeration of water ? 23. (a) Describe lime-soda process of softening of water giving diagram and reaction involved. (RGPV Bhopal 2009) (b) When do you recommend only internal treatment for boiler water and completely avoiding external treatment ? (c) Write a brief note on “caustic embrittlement”. 24. (a) Why feed water conditioning is necessary in boilers ? What are the methods available for the same ? (b) Calculate the quantity of lime needed for softening of 5000 litres of water containing : CaSO4 = 13.6 mg/l MgCO3 = 8.4 Mg/l CaCO3 = 5 mg/l KNO3 = 20 mg/l (Mumbai University, 1998) 25. (a) Give only the equations involved in the lime-soda process. Describe the advantages of hot-lime soda process over cold lime-soda process. (RGPV Bhopal 2006) (b) The hardness of 10,000 litres of a sample of water (containing 341.9 ppm hardness) was completely removed by passing it through a zeolite softener. The zeolite softener was regenerated by passing sodium chloride solution containing 20 gms/l of sodium chloride. How many litres of sodium chloride solution will be required to regenerate zeolite softener ? (Mumbai University, 1998) 26. (a) What is the difference between scale and sludge ? (b) What principle is applied to remove hardness of water. Explain giving chemical reactions. (c) Describe with the help of a neat labelled diagram ion-exchange process for purification of water. (Mumbai University, 1999) 27. (a) Write notes on industrial uses of water.

WATER AND ITS INDUSTRIAL APPLICATIONS

59

(b) A zeolite softener was completely exhausted and was regenerated by passing 100 litres of NaCl solution containing 60 gms/liter of NaCl. How many litres of a sample of water of hardness 400 ppm can be softened by this softener ? (Mumbai University, 1999) 28. (a) Describe with the help of a neat labelled diagram, the hot lime-soda process of softening of water. (b) The hardness of 10,000 litres of a hard water sample was completely removed by passing it through a zeolite softener. The zeolite softener requires 5,000 litres of sodium chloride solution containing 1170 mg of NaCl per litre for regeneration. Determine the hardness of the water sample. (Mumbai University, 2000) 29. (a) Calculate the quantities of lime and soda required to soften 1 million litres of hard water containing the following impurities. CaCO3 – 10.0 ppm Mg(HCO3)2 – 36.5 ppm Al2(SO4)3 – 17.1 ppm CaSO4 – 20.4 ppm MgCl2 – 19.0 ppm and SiO2 – 24.0 ppm (b) State the limitations of the zeolite process of softening of water. (Mumbai University, 2000) 30. (a) Explain with a neat sketch and all chemical reactions taking place, the zeolite permutit process for softening water. What are the advantages and disadvantages of this process (any two). (b) Write a brief note on “Boiler Corrosion”. (c) Calculate the quantity of lime and soda needed to soften 20,000 litres of hardwater containing the following salts : MgCl2 = 9.5 mgs/lit CaCl2 = 22.2 mgs/lit Ca(CHO3)2 = 81 mgs/lit FeSO4 = 151.8 mgs/lit Mg(HCO3)2 = 73 mgs/lit MgSO4 = 120 mgs/lit (d) Write brief notes on internal conditioning of boiler water. (Mumbai University, 2001) (e) A totally exhausted zeolite softener required 50 litres of NaCl solution containing 351 g of NaCl/litre. How many litres of a hard water sample containing 70° clark hardners, can be softened by this process. 31. (a) Mention the advantages of hot lime-soda process of softening of water. (b) What are the different units for expressing hardness of water ? 32. (a) What are the natural sources of water ? (b) How the presence of hardness of water affect the consumption of soap ? (c) A Zeolite softener was 80% exhausted by removing the hardness completely when 800 litres of water are passed through it. The Zeolite bed required 200 litres of 3% NaCl solution for complete regeneration. Calculate the hardness of water. (d) Write short notes on: (i) Sterilization

60

BASIC ENGINEERING CHEMISTRY

(ii) Regeneration of exhausted ion-exchange resins (iii) Coagulation. (Nagpur University, 2002) 33. Calculate the quantities of lime and soda required to soften 1,25,000 litres of water having the following analysis and using 50 mg/l of sodium aluminate as coagulant : Ca2+ – 300 mg/l Mg2+ – 96 mg/l CO2 – 44 mg/l HCO3– – 100 mg/l (a) What do you understand by the terms priming and foaming ? How can they be controlled ? (b) What are the methods available for deaeration of boiler feed water ? (Nagpur University, 2002) 34. Calculate the cost of lime and soda required for softening 1 million litres of water containing : Mg(HCO3)2 = 73 mg/l MgSO4 = 120 mg/l CaSO4 = 68 mg/l CaCl2 = 111 mg/l The cost of lime of 80% purity is Rs. 200/MT and that of soda of 90% purity is Rs. 12000/M.T. 35. (a) What are the disadvantages of scale formation ? Explain briefly the various methods adopted for prevention of scale formation. (b) Describe a treatment method for municipal water supply. (Nagpur University, 1997) 36. (a) How is water softened by lime-soda process ? Describe the types and chemical reactions involved. (b) What are the external and internal treatment required for prevention of scale formation in the boiler. (c) Explain the functions of the following in water treatment : (i) Coagulant (ii) Bleaching powder (iii) Calgon 37. A water sample has the analytical report as under: MgCO3 = 84 ppm, CaCO3 = 80 ppm, MgSO4 = 30 ppm, CaSO4 = 34 ppm, NaCl = 10 ppm Fe2O3 = 56 ppm. Calculate the amount of lime and soda needed for the treatment of 80000 litres of water. (RGPV Bhopal 2006)

UNIT

2

Fuels and Combustion

2.1 DEFINATION OF FUEL Fuel can be defined as any combustible substance which during combustion gives large amount of heat which can be used economically for domestic and industrial purpose. 2.2. CLASSIFICATION OF FUEL Fuels may be divided into two types: (i) primary fuels which occur in nature as such, and (ii) secondary fuels which are derived from the primary fuels. Fuels may also be classified into three groups (a) Solid fuels, (b) Liquid fuels, and (c) Gaseous fuels. The examples of each of these main classes of fuels are summarised in Table 2.1. This classification is of practical significance because the equipment used for handling and burning of each class of fuels are usually different for these three types of fuels.

Table 2.1 Classification of Fuel

Solid Fuels Primary Wood, peat, lignite, brown coal, bituminous coal, anthracite, oil shales, etc. Secondary Semicoke, coke, charcoal briquettes, petroleum coke, pulverised coal and colloidal fuels. Solid rocket fuels such as thiokol, hydrazine, nitrocellulose, etc.

Liquid Fuels Primary   Crude oil or petroleum.

Gaseous Fuels Primary Natural gas.

Secondary Gasoline or motor spirit, diesel oil, kerosene, fuel, oils, coal tar and its fractions, alcohols and synthetic sprits.

Secondary Coal gas, coke oven gas, water gas, producer gas, carburetted water gas, oil gas, blast furance gas, refinery oil gas, synthesis gas, acetylene and liquid petroleum gas (LPG).

2.3 CALORIFIC VALUE Calorific value of a fuel is “the total quantity of heat liberated from the combustion of a unit mass (or volume) of the fuel in air or oxygen. 61

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2.3.1 Units of Heat or Calorific alue (i) Calorie : Calorie is the amount of heat required to increase the temperature of 1 gm of water through one degree centigrade. 1 calorie = 4.185 Joules = 4.185 × 107 erg. (ii) Kilocalorie or kilogram calorie or kilogram centigrade unit (Kcal or Kg cal or K.C.U.) The amount of heat required to raise the temperature of 1 kg of water through 1°C (more precisely from 15°C to 16°C) is known as kilocalorie 1 K cal = 1000 Cal (iii) British thermal unit (B.Th.U. or B.T.U.) The amount of heat required to rise the temperature of one pound (1 lb) of water through 1°F (more precisely from 60°F to 61°F) is known as british thermal unit. 1 B.Th.U. = 1,054.6 Joules = 1,054.6 × 107 ergs. (d) Centigrade heat unit (C.H.U.) The amount of heat required to raise the temperature of one pound (1 lb) of water through 1°C is known as centigrade heat unit.

Interconversion of the various units These units can be interconverted as follows : 1 cal/g = 1K cal/kg = 1.8 B.Th.u/1b. 1 K cal/m3 = 0.1077 B.Th.u/1b 1 B.Th. u./ft3 = 9.3 K cals/m3 2.3.2 Gross and Net Calorific alues The Gross Calorific Value or Higher Calorific Value is the total heat generated when a unit quantity of fuel is completely burnt and the products of combustion are cooled down to 60°F or room temperature i.e., ≈ 25°C. When a fuel containing hydrogen is burnt, the hydrogen present undergoes combustion and will be converted into steam. As the products of combustion are cooled to room temperature, the steam gets condensed into water and the latent heat is evolved. Thus the latent heat of condensation of steam so liberated is included in the gross calorific value The calorific value determination by Bomb calorimeter gives the Gross or Higher Calorific Value. The Net Calorific Value or Low Calorific Value is the net heat produced when a unit quantity of fuel is completely burnt and the products of combustion are allowed to escape. Thus, Net Calorific Value = Gross Calorific Value — Latent heat of Condensation of the water vapour produced. = Gross Calorific Value — (Mass of Hydrogen per unit weight of the fuel burnt × 9 × latent heat of vapourization of water). 1 part by weight of hydrogen gives 9 parts by weight of water as follows: H2 + O → H2O 2g  16g 18g 1g   8g 9g The latent heat of steam is 587 Cal/g (or Kcal/Kg) or 1060 B.Th.U./lb of water vapour produced.

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Net C.V. = Gross C.V. – 9 ×

H × 587 100

Net C.V. or L.C.V. = Gross C.V. – 0.09 × H × 587

where H = % of hydrogen in the fuel. In actual practical use of a fuel, it is rarely feasible to cool the combustion products to the room temperature to allow the condensation of water vapour formed and utilise that latent heat; hence the water vapour formed also is allowed to escape along with the hot combustion gases. 2.3.3 Calorific value at constant pressur The calorific value at constant pressure, C.P. can be calculated on the basis of the following: QC.P. = QC.V. – (∆n). R.T. where QC.V. = Calorific value at constant volume (as determined in a Bomb Calorimeter). ∆n = increase in number of gaseous molecules after reaction. R = gas constant T = absolute temperature. If there is a decrease in the number of gaseous molecules formed after the reaction, then ∆n will have a negative value and consequently, QC.P. will be higher than QC.V. 2.3.4 Theoretical calculation of calorific value of a fue Theoretically, the calorific value of a fuel can be calculated, if the percentages of the constituent elements are known. According to Dulong, “the calorific value of a fuel is the sum of the calorific values of its constituents”. Constituent H C S HCV (K cal/kg) 34500 8080 2240 If oxygen is present in the fuel it combines with hydrogen to form H2O. Thus the hydrogen in the combined form is not available for combustion and is called fixed hyd ogen. Amount of hydrogen available for combustion = Total mass of hydrogen-hydrogen combined with oxygen. Now, 8 parts of oxygen combine with one part of hydrogen to form water i.e., for every 8 parts of oxygen 1 part of hydrogen gets fixed \ Amount of hydrogen available for combustion = Total mass of hydrogen –1/8 mass of oxygen, in fuel. On the basis of above chemical composition of fuel Dulong’s gaves the following formula for calculating the calorific value of fue

HCV =

1 100

 0   8080C + 34,500  H − 8  + 22405 K cal/kg     

and LCV = [HCV – 0.09 H × 587] K cal/kg where C, H O and S are the precentages of carbon, hydrogen, oxygen and sulphur in the fuel. 2.3.5 Determination of calorific valu Bomb Calorimeter is an apparatus used for calculation of calorific value of solid and liquid fuels. A known mass of a fuel is burnt and the quantity of heat produced is absorbed in water and measured. Now the quantity of heat generated by burning a unit mass of the fuel is calculated.

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An steel vessel called the bomb. It is covered with a gas tight cover having three values. One for oxygen inlet, second for pressure gauge and third for release of pressure. Two electrical leads made up of platinum are fixed in the cover. A Silica crucible is kept inside the apparatus in which weighed quantity of fuel is taken. A fuse wire of platinum is dipped in the fuel taken in the crucible and its end are connected to the battery outside through a key. The bomb is placed in the copper calorimeter, containing known quantity of water. The copper calorimeter is surrounded by an air jacket, which is further inclosed in a vessel containing water. One electric stirrer for stirring the water is provided in the calorimeter. Calorimeter also contains one accurate thermometer for noting down the temperature.

Fig. 2.1. Bomb calorimeter.

Calculations Let, Weight of the fuel sample taken = m grams Higher or Gross calorific value of the fuel = θ cals/gram Weight of the water taken in the calorimeter = W grams Water equivalent of the calorimeter, bomb, thermometer, stirrer, etc. } = w grams Initial temperature = t1°C Final temperature = t2°C Heat liberated by the combustion of the fuel = mθ Heat absorbed by the water, calorimeter, etc. = (W + w) (t2 – t1) Heat liberated = Heat absorbed ... mθ = (W + w ) (t2 – t1) ...

HCV or Gross CV (θ) =

(W + w) (t2 − t1 ) m

However, for more accurate results, the following corrections will have to be incorporated in the above equation : (i) Acids correction, tA, (ii) Fuse wire correction, tF, (iii) Cotton thread correction, tT and (iv) Cooling correction, tC. Accordingly, the above equation will have to be modified as [(W + w) (t2 − t1 + tc )] − [t A + t F + tT ] HCV or Gross CV (θ) = m Calculation of net calorific valu Net C.V. = Gross C.V. – Latent heat of water vapour, formed during the combustion of m grams of the fuel

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= Gross C.V. – 0.09 × H × 587 (where H is the percentage of hydrogen present in the fuel and latent heat of steam is 587 cals/g). Determination of water equivalent of the apparatus The water equivalent of the apparatus is best determined by burning a known weight (preferably about 1.2g) of pure and dry benzoic acid (in a pellet form) in the bomb under identical conditions as described above. The rise in temperature is noted. The standard calorific value of benzoic acid is taken as 6324 calories per gram. Since all the other values in the formula are known, the water equivalent of the apparatus can be calculated. Other substances suitable as standards are: ... Calorific alue Naphthalene ... 9622 cals/g Salicylic acid ... 5269 cals/g Camphor ... 9292 cals/g Corrrections: For calculation of more correct values in the results, following three correction must be made, while calculating the calorific value of a fuel (i) Acids correction (tA). The sulfur present in the coal is converted into H2SO4 in the bomb. S + O2 → SO2 2SO2 + O2 + 2H2O → 2H2SO4; ∆H = – 144,000 (4 × 49) calories

(Eq. Wt. of H2SO4 = 49)

Similarly the nitrogen present in the coal and part of that in the air in the bomb are converted into HNO3. 2N2 + 5O2 + 2H2O → 4HNO3; ∆H = – 57,160 (4 × 63) calories

(Eq. Wt. of



HNO3 = 63)

Since the above two reactions are exothermic and since the heat thus liberated is not obtainable in practical use of coal (because SO2 and NO2 pass off into the atmosphere) correction must be made for the heat liberated in the bomb by the formation of H2SO4 and HNO3, as follows: (a) 3.6 calories should be subtracted for each ml of N/10, H2SO4 formed. (b) 1.43 calories must be deducted for each ml of N/10 HNO3 formed       (as per the equations given above) (ii) Fuse wire correction. The heat liberaled as measured above, include the heat given out by ignition of the fuse wire used. Hence it has to be substracted from the total value. The amount to be substracted are given in the instruction provided by the supplier of the fuse wire. (iii) Cooling correction, (tC). If the time taken for the water in the calorimeter to cool from the maximum temperature attained to the room temperature is x minutes and the rate of cooling is dt°/minute, then the cooling correction = x × dt. This should be added to the observed raise in temperature. Hence, [(W + w) (T2 − T1 + cooling correction)] − (Acid + Fuse correction)] L= Mass of the fuel ( x)

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2.4. CRITERIA FOR SELECTING A FUEL The following characteristics are taken into consideration for the selection of a fuel for a particular purpose: 1. The fuel selected should be most suitable for the process. For instance, coke made out of bituminous coal is most suitable for blast furnace and also as a foundry fuel. 2. The fuel should posses a high calorific value 3. The fuel should be cheap and readily available. 4. It should possess a moderate ignition temperature. Too high ignition temperatures cause difficulty in kindling while too low ignition temperatures may create safety problems during storage, transport and use of the fuel. 5. The supply position of the fuel should be reliable. 6. The velocity of combustion should be moderate. 7. The fuel should have reasonable flexibility and control 8. The fuel should be such that a safe and clean operation is ensured. Too much smoke and obnoxious (harmul) odours are not desirable. 9. It should be safe, convenient and economical for storage and transport. 10. It should have low moisture content. 11. In case of a solid fuel, the ash content should be less and the size should be more or less uniform. 2.5. COAL 2.5.1. Origin of coal formation. Coal is regarded as a fossil fuel produced from large accumulations of vegetable debris due to partial decay and alteration by the action of heat and pressure over millions of years. The formation of coal is explained by the following two theories: (1) “In situ” theory states that coal seams are formed in the same area where vegetation grew and accumulated. The great purity of many coal seams holds testimony to this theory. (2) The “drift” or “transportation” theory contends that the coal seams are not formed where the vegetation grew and accumulated originally. These materials were drifted or transported by rivers to lakes or estuaries and got deposited there. The great thickness of coal seams support this theory. Thus evidences are available in support of both the above theories. The various agencies responsible for the conversion of plant tissues to coal include (i) Bacteria (under water), (ii) Time (millions of years), (iii) Temperature (> 300°C) and (iv) Pressure. The time required for the formation of young brown coals is of the order of 107 years while that for the most mature coals is 3 × 108 years. The vegetable matter fallen on the ground undergoes microbial degradation in presence of air and eventually gets converted to carbon dioxide and water without leaving any organic matter remaining. However, the course of decay when it is buried under water is different. The transformation of the vegetable debris to coal takes place in two stages; (i) the biochemical or peat stage and (ii) the metamorphic stage during which peat is transformed into coal. The effect of temperature and pressure caused by the depth of burial on the rank of a coal is brought out by Hilt’s law which states that in any vertical section the rank of the seams increases with depth. The formation of coal from decaying plant debris to bituminous stage is explained by two alternative theories.

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(i) Serial evolution. This is the commonly accepted theory according to which the evolution of coals occurs through geochemical metamorphism of peat to anthracite as follows:     (peat → lignite → bituminous coal → anthracite) (ii) Parallel evolution. This theory is based on the concept of entirely biochemical origin of coals of various ranks. According to this theory, lignites, bituminous coals and anthracites may not form a continuous series but may be the end-products resulting from the differences in the extent of the aerobic decomposition of peat, the subsequent composition of the overlying strata and the depth of the burial. This theory may be represented as follows:

Vegetable Matter Aerobic decay Acid medium H Peat stage prolonged 2 eliminated as CH4 continued elimination of H2 Peats



   Burial under sedimentary rocks

Peat (low in H2) Burial under sedimentary deposits Acid condition maintained Alkaline decay (sodium aluminosilicate (Calcium aluminosilicate roof) roof) anaerobic condition, loss of O2 as Consolidation, dewatering, Loss of O2 as H2O and CO2 H2O and CO2 Anthracite Bituminous Lignites  coals 2.5.2. Analysis of coal

The quality of a coal is determined by the following two types of analysis. (1) The proximate analysis, which includes the determination of moisture, volatile matter, ash and fixed carbon. This gives quick and valuable information regarding commercial classification and determination of suitability for a particular industrial use. (2) The ultimate analysis, which includes the estimation of ash, carbon, hydrogen, sulfur, nitrogen and oxygen. The ultimate analysis is essential for calculating heat balances in any process for which coal is employed as a fuel. Procedure for proximate analysis It involves the following determination 1. Moisture. Moisture is generally determined by heating a known quantity of air dried coal to 105°C to 110°C for one hour and calculating the loss in weight as percentage. Loss in weight ×100 Percentage of moisture = Weight of coal taken 2. Volatile Matter. Volatile matter is determined by heating 1 g of air dried coal exactly for 7 minutes in a translucent silica crucible of specified dimensions at a steady temperature of 925°C in a muffle furnace. The loss in weight calculated as percentage minus the % moisture gives the % volatile matter. Loss in weight due to removal of volatile matter ×100 % of V. M. = Weight of coal sample taken Here, Loss in weight due to removal of volatile matter = Weight of the coal (or residue) after removal of moisture – Weight of coal (or residue) after removal of volatile matter

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3. Ash. Ash is determined by heating at 400°C a known quantity of the powdered sample until most of the carbonaceous matter is burnt off and then heating for 1 hour at 750°C to complete the combustion. The weight of the residue remaining in the crucible corresponds to the ash content of the coal, which is reported on percentage basis. Weight of ash left ×100 % of Ash = Weight of coal taken 4. Fixed Carbon. The sum total of the percentages of volatile matter, moisture and ash subtracted from 100 gives the percentage of fixed carbon Percentage of fixed carbon = 100 – % of (moisture + volatile matter + ash Significance of proximate analysi Each constituent determined under proximate analysis, has its own importance in the assessment of the coal sample. (i) Moisture. Excessive surface moisture may cause difficulties in handling the coal. Moisture reduces the calorific value. A considerable amount of heat is wasted in evaporating the moisture during combustion. Hence high percentage of moisture is undesirable. (ii) Volatile matter. The volatile matter content of a coal is related to the length of the flame, smoke forming tendency and the ignition characteristics. High volatile matter coals give long flames, high smoke and relatively low heating values. Coal with low volatile content burns with a shorter flame. Thus, the higher the volatile matter content the larger is the combustion space required. Hence, the volatile matter content of a coal influences the furnace design. Further, the % of volatile matter in a coal denotes the proportion of the coal which will be converted into gas and tar products by heat. Hence, high volatile matter content is preferable in coal gas manufacture and in carbonization plants, particularly when the main objective is the byproduct recovery. For the manufacture of metallurgical coke, a coal with low volatile matter and high fixed carbon is preferred. The volatile matter content is more in bituminous coals than in anthracite coals. The volatile matter percentage gives some idea about coking property of the coal. (iii) Ash. It is non-combustile matter it reduces the calorific value of the coal. It causes hindrance to the flow of air and heat and decreases the efficiency of coal. Fused ash lumps are called clinkers. It causes obstruction to air circulation, thus burning of coal become irregular. Fixed carbon. It is reported as the difference between 100 and the sum of percentages of moisture, volatile matter and ash content of a coal. The fixed carbon content increases from low ranking coals such as lignite to high ranking coals such as anthracite. It is the Fixed Carbon which burns in the solid state. Hence, information regarding the percentage of fixed carbon helps in designing of the furnace and the fire box Ultimate Analysis: Methods of determination (B) Ultimate analysis involves in the following determinations : (1) Carbon and hydrogen : About 1-2 g of accurately weighed coal sample is burnt in a current of oxygen in a combustion apparatus. C and H of the coal are converted into CO2 and H2O respectively. The gaseous products of combustion are absorbed respectively in KOH and CaCl2 tubes of known weights. The increase in weights of these are then determined. C + O2 → CO2 : 12 44 1 H2 + O2 → H2O 2 2 18

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2KOH + CO2 → K2CO3 + H2O CaCl2 + 7 H2O → CaCl2.7 H2O ∴ Percentage of C =

Increase in weight of KOH tube × 12 × 100 Weight of coal sample taken × 44

and Percentage of H =

Increase in weight of CaCl2 tube × 2 × 100 Weight of coal sample taken × 18

(2) Nitrogen : About 1 g of accurately weighed powdered coal is heated with concentrated H2SO4 along with K2SO4(catalyst) in a long necked flash (called Kjeldahl’s flask). After the solution becomes clear, it is treated with excess of KOH and the liberated ammonia is distilled over and absorbed in a known volume of standard acid solution. The unused acid is then determined by back titration with standard NaOH solution. From the volume of acid used by ammonia liberated, the percentage of N in coal is calculated as follows : Volume of acid used × Normality × 1.4 Percentage of N = Weight of coal taken (3) Sulphur : Sulphur is determined from the washings obtained from the known mass of coal, used in a bomb calorimeter for determination of a calorific value. During this determination, S is converted into sulphate. The washings are treated with barium chloride solution, when barium sulphate is precipitated. This precipitate is filtered, washed and heated to constant weight Weight of BaSO 4 obtained × 32 × 100 Weight of coal sample taken in bomb × 233 (4) Ash : Ash determination is carried out as in proximate analysis. (5) Oxygen : It is obtained by difference. Percentage of O = 100 – Percentage of (C + H + S + N + ash)

Percentage of S =

Significance of ultimate analysi Significance Carbon and Hydrogen in coal directly contribute towards the calorific value of the coal. Higher the percentages of C and H, better is the quality of the coal and higher is its calorific value. Hydrogen is mostly associated with the volatile matter of the coal and thus influences the use of coal for the byproduct manufacture or otherwise. Nitrogen in the coal does not contribute any useful value to the coal and since it is generally present only in small quantities (~ 1%), its presence is not of much significance. Sulfur present in coal contributes towards the heating value of the coal but its combustion products (SO2 and SO3) have corrosive effects on the equipments, particularly in presence of moisture. Further, the oxides of sulfur are undesirable from the atmospheric pollution point of view. Sulfur containing coal is not suitable for the preparation or metallurgical coke as it adversely affects the properties of the metal. Oxygen content of coal is generally associated with moisture. The lower the oxygen content, the more is the maturity of the coal and greater is its calorific value. As the oxygen content increases, the capacity of the coal to hold moisture increases and the caking power decreases. Use of proximate and ultimate analysis in the theoretical determination of the calorific value of coal The calorific value is determined by burning 1 g of coal sample in an oxygen bomb calorimeter equipment and measuring the rise of temperature thus produced in the water content of the calorimeter. However, quite often, an engineer may have to estimate the thermal efficiency of a process when the calorific value of the fuel has not been determined. In such circumstances, formulae for the calculation of calorific value from ultimate and proximate analysis are very helpful

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Formulae Based on Ultimate Analysis 1. Dulong’s Formula 0  Calorific value in B.Th.U./lb = 14,544 C + 62,028  H −  + 4,050 S



8

where C, H, O and S represent the respective percentages of carbon, hydrogen, oxygen and sulfur. Several modifications to it have been proposed to this formula and one of them is as follows:

1  0   Gross C.V. = 8, 080 C + 34,500  H −  + 2240 S K cals/kg 100  8   2. Davies Formula O −S  Calorific value in  C = (6.543H + 403)  + H −   B.Th .U.per lb  8  3 where C, H, O and S are their respective percentages in the coal. 3. Seyler’s Formula Calorific value in  = 223.1 C + 698.6 H – 7684 + 0.45 O B.Th .U.per lb  where C, H and O are their respective percentages in the coal. Formula Based on Proximate Analysis 1. Gouthal’s Formula Calorific value in  = 147.6 C + aV B.Th .U.per lb  where C is the % of carbon, V is the % of volatile matter and ‘a’ is a constant depending on V. The relation between V and a is as follows: V 1– 4 10 15 20 25 30 35 40 a 270 261 210.6 196.2 185.4 176.4 171 144 2. Nakamura’s Formula Calorific value in  % Ash   = a V −  + 140.4 C, B.Th .U.per lb  10   where ‘a’ depends on the % volatiles and caking propensity as shown. The main use of all these formulae is that they provide a means for calculating the calorific values of coals approximately when their compositions are known but of which the samples are not available. 2.6. CARBONIZATION The process of converting coal into coke is called carbonization when the coking coal is heated in absence of air, the porous hard and strong residue left is called coke. It is white lustrous, porous and coherent mass. When coal is heated in absence of air as per its behaviour it can be classified into the following categories (i) Non-coking coal (ii) Coking coal Non-coking coals. The coal which do not fuse at all when heated is known as non coking coals. Anthracite, sub-bituminous, lignite coals are the example of non-coking coal. Coking coals and caking coals. There are some coals which have a tendency to soften and swell at higher temperature and form a solid coherent mass with porous structure, such coals are

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called caking coals. While the coals which give porous, hard and strong residue after heating in the absence of air, the residue is (used for metallurgical purpose) called coke. If the coke so produced is hard, porous and strong, then the coal from which this coke is derived is called coking coals. Obviously, all coking coals are caking coals but all caking coals are not coking coals. 2.6.1 Properties of good metallurgical coke The quality requirements of a good metallurgical coke are given below. 1. High purity. The best metallurgical coke should contain lowest possible percentage of moisture (< 4%), ash (< 6%), sulfur (< 0.5%) and phosphorous (< 0.1%). Mositure and ash reduce the calorific value. Sulfur and phosphorous in the coke may contaminate the metal and adversely affect its properties. They tend to make the metal brittle. 2. Porosity. The metallurgical coke should be porous to provide intimate contact between the carbon and oxygen and to ensure efficient combustion of the fuel in the furnace 3. Strength. The coke should be strong enough to withstand the abrasion and over burden of the ore, flux and the fuel itself in the furnace. If the coke breaks into fine particles during charging of the furnace, they may hinder the flow of gases and choke the air passages 4. Uniformity. The coke should be uniform and medium in size. If the lumps are too big, combustion is irregular. If they are too small, choking may result. 5. Calorific value The coke should possess a high calorific value 6. Cost. The coke should be cheaply available near the plant site. 7. Calorific intensity. The calorific intentisty of the fuel should be high enough to melt the metal. 8. Combustibility. The coke should burn easily but at the same time should not be very reactive. 9. Reactivity. Reactivity of coke refers to its ability to react with CO2, steam, air and O2. The reactivity of the coke should not be very high. Coke of low reactivity gives a higher fuel bed temperature than what is produced by a coke of high reactivity. Coal cannot be used as a metallurgical fuel (excepting in reverberatory furnaces) because it does not have the necessary purity, porosity and strength. During the process of carbonisation from coking coals, much of the volatile matter and sulfur compounds are removed and a strong and porous coke is produced. 2.6.2 Types of carbonization of coal Carbonization is of two types: (i) Low temperature carbonization. It is done at 500°–700°C and it produces mainly domestic fuel. The yield of coke 75–80% calorific value 6500 to 9500 Kcal/m3 and percentage of volatile matter is 5-15%. (ii) High temperature carbonization. It is done at 900–1200°C and is mostly carried out for the manufacture of metallurgical coke. The yield of coke is 65 to 75%, calorific value 5400 to 6000 °C Kcal/m3 and percentage of volatile matter 1 to 3%. 2.6.3 Manufacture of metallurgical coke There are two methods of manufacture metallurgical coke (i) Beehive oven method (ii) Otto-Hoffman oves or chamber ovens or by product oven method (i) Beehive oven. It is primitive method for the production of coke. A beehive oven is a fire brick chamber having a dome shaped structure. It is called beehive oven because it resembles in

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shap to that of a beehive. Its dimensions are 4 m wide and 2.5 m high. It has two doors, one charging door for coal at top and other at the side for taking cut coke. It also acts as an inlet for air as and when required. Coal chaigina door Refractory lining Zone of combustion 2.5 m Door for air supply or coke discharging

0.6 m

4m Fig. 2.2 Beehive coke oven.

The coal is introduced through the top hole in the dome and is spread evenly on the floor to produce a layer of about 60 to 90 cm deep. Initially some air is supplied to ignite the coal carbonization starts and volatile matters burns inside the partially closed side door. The complete carbonization takes place in about 3-4 days. The oven is then allowed to cool down and coke is quenched with water and taken cut through discharge door. Since heat is supplied by the burning of the volatile matter and hence no by products are recovered. The yield of the coke is about 75-80% of the coal charged. Demerits of beehive oven method

(i) Lower coke yield due to partial combustion. (ii) No recovery of by products. (iii) Lack of flexibility of operation (iv) Time consuming process HYDRAULIC COAL HOPPER Inspite of the above demerits MAIN it is still in use because of their low capital and running costs. OVEN (ii) Otto-Hoffman or N E FLU chamber ovens or byproduct OVEN N E oven method. FLU E OVEN FLU These ovens consist of END narrow rectangular chambers made of silica bricks having length, height and width as 12 m, 4 m and 0.5 m respectively (Fig. 2.2). They are tightly closed so that no air is admitted. The heat for GENERATIVE CHAMBER coking is furnished by burning CHECK WORK OF FIRE BRICK gas (e.g., coke oven gas, GAS AND producer gas or blast furnace AIR gas) in flues contained in the Fig. 2.3. By product coke oven.

GAS

TAR AND LIQUOR COKE CAR WASTE GAS

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side walls of the oven. The chambers are also fitted with charging doors having 3 to 4 openings at the top and discharging doors at the bases. Each oven is separated from the neighbouring one by a vertical flue in which the fuel gas burns. Thus the ovens get heated from both sides. These ovens work on the regenerative principle of heat economy. Generally, a battery of such ovens (consisting about 25 to 1000 ovens) are used which are placed over the regenerators having a chequer brickwork which helps in the utilization of the heat of the flue gases. The hot gases leaving the flues are allowed to pass through the chequer brick-work in the regenerator of the next oven while air and fuel gas pass through other chambers which have been already heated. After some time, the directions are reversed so that the waste hot flue gases preheat other chambers through which the air and fuel gas were passing.

Fig. 2.4. Chamber ovens with byproduct recovery system.

As the coal is heated in the coke oven, moisture is first expelled, then the decomposition of the coal substance takes place at 300 to 450°C. At about 500°C, the coal passes through a plastic state but at about 550° to 600°C, the plasticity ceases and semicoke is produced. This is black in colour and is low in strength. As the temperature raises further above 600°C, the semicoke decomposes with loss in volatile matter and gets transformed into steel grey hard coke. The process of carbonisation takes place layer by layer in the coal charge starting from the two side walls of the oven and moving towards the centre. Each oven holds about 20 tonnes of coal charge and the time taken for carbonisation is about 12 to 20 hours. The temperature goes around 1100°C and the yield of the coke is about 70% of the coal charged. Nearly 40% of the coke oven gas generated is sufficient to heat the ovens and the rest is available for other uses in the steel plant or is sold out. This gas has high calorific value and can be transported to distant places. After the carbonisation is complete, the discharging doors are lifted by a crane and the red hot coke is pushed out mechanically into a coke car. The car carries it to quenching station where the coke comes into contact with a spray of cooling water. The excess water on the coke is allowed to get evaporated and the coke is screened to different sizes. Then it is supplied for different uses in the plant e.g., foundry and blast furnace and also for domestic purposes. The gas coming out of the coke ovens contains ammonia, sulfur (as H2S) volatile hydrocarbons, tar, etc. Important byproducts such as high calorific value gas, light oil, tar, ammonium sulfate, finely divided sulfur and ammonium thiocyanate can be recovered from the byproduct-oven gases. (1) Recovery of tar. The gas from the coke ovens is passed through a tower where liquor ammonia trickles from the top. Tar and dust are removed into a tar tank. Tar and ammonia are recovered. Ammonia liquor is again sent to the top of the trickling tower. (2) Recovery of ammonia. The gases now enter another tower when water is sprayed. Ammonia goes into solution forming ammonium hydroxide. Sometimes, instead of water dilute

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H2SO4 is sprayed when ammonium sulfate is recovered. NH3 + H2O → NH4OH 2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O (3) Recovery of naphthalene. The gases then pass to a cooling tower where water at a low temperature is sprayed. Condensation of some gases takes place and naphthalene is recovered. (4) Recovery of benzene. The gases then pass through an oil (petroleum) scrubber where benzene and its homologues are recovered. (5) Recovery of H2S. The gases then enter a purifying chamber packed with moist Fe2O3 Fe2O3 + 3H2S → Fe2S3 + 3H2O After all the Fe2O3 exhausted, it is exposed to atmospheric air to recover the sulfur (as SO2), and regenerate the Fe2O3 Fe2S3 + 4O2 → 2FeO + 3SO2 4FeO + O2 → 2Fe2O3 (6) Recovery of gas. The gas after passing through the various scrubbers and condensers to remove the various byproducts mentioned above is finally collected in a gas holder. The gas has the calorific value of about 5000 Kcal/ 3. 2.7. CRACKING In cracking process, higher saturated hydrocarbon molecules are converted into simpler molecules such as paraffinic and olefinic hydrocarbons, as follow cracking C10 H 22  → C5 H12 + C5 H10

paraffin

olefin

These simpler molecules may be still further decomposed until at very high temperatures, the hydrocarbons may be cracked to completion, giving carbon and hydrogen only: CH4 → C + 2H2 However, in actual practice, the cracking of heavy hydrocarbons results in the formation of a complex mixture of saturated and unsaturated hydrocarbons in liquid and gaseous state. In addition, hydrogen and carbon may be produced; and aromatic hydrocarbons may also be formed by polymerisation. The higher the boiling range of petroleum fractions being cracked, (i.e., the larger the molecules and their chain lengths), the lower the cracking temperatures. Cracking is usually done by two methods: 1. Thermal cracking 2. Catalytic cracking 1. Thermal cracking This is the oldest method and is being replaced by other methods. In this method, the heavy oil is subjected to high temperature and pressure where the high molecular weight hydrocarbons are decomposed to lower hydrocarbons of paraffinic and olefinic series. Some of the molecules so formed may undergo polymerisation to yield larger molecules. The cracked products are then separated by fractional distillation. Generally, the yields is from 7 to 30% but sometimes, higher yields may be obtained. Some coke is also formed during this process along with other liquid and gaseous products. There are two types of thermal cracking which are usually distinguished. (a) Liquid-phase cracking. By this method, any type of oil (residues, fuel oil or gasolines) can be cracked. In this method, the charge is kept in the liquid form by applying high pressures of the range 15 to 100 kg/cm2. The temperature maintained is 420° to 550°C. The octane rating of the product formed is 65 to 70. (b) Vapour phase cracking. By this method, only those oils which can vaporize at low temperatures can be cracked. The time required for cracking in this method is lesser, the stability

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of the products poorer and the octane rating of the product higher, as compared to liquid-phase cracking process. The temperature maintained is 600 to 650°C, while the pressure is 10 to 20 kg/cm2. 2. Catalytic cracking In this method, cracking is brought about in presence of a catalyst at much lower temperatures and pressures (300° to 450° and 1 to 5 kg/cm2 pressure). There are two main types of catalytic cracking: (i) cracking carried out only in the presence of a catalyst (porous solid particles of definite composition and structure); (ii) cracking carried out in presence of a catalyst, but in a hydrogen atmosphere at a slightly reduced temperature but higher pressure. This type of cracking is known as “hydro-cracking.” The earliest cracking catalysts, were acid clays, but they were replaced by crystalline aluminosilicates (zeolites). In these, alumina and silica are the major constituents, but they also contain minor amounts of oxides of Ca, Mg, Na, Fe, Cr and rare earths. Catalyst cracking is done by following two method. (1) In fixed-bed cracking, the oil vapours, heated to cracking temperatures, are passed on to the fixed catalyst bed. When the catalyst gets carbonised, it is reactivated by burning off the carbon deposited. (2) In Fluid-bed cracking, the catalyst in the form of a fine powder, is circulated through the cracking reactor with the help of oil vapours or air. The catalyst accelerates and directs the cracking and also acts as a heat transfer medium. The catalyst is continuously regenerated. Fixed bed catalytic cracking The essential features of this process are represented in Fig. 2.11. Vapours Cooler Hot air for reactivation

Charge of heavy oil Preheater

Cracked vapours Gases

heaters Catalyst towers

Heavy oil Gasoline Fractionating Stabiliser Column condensed gasoline + dissolved gases

Fig. 2.5. Fixed-bed catalytic cracking.

The heavy oil charge is passed through a heater, where the oil is vaporized and heated to 400 to 500°C. The silica alumina gel (SiO2, Al2O3), or bauxite catalyst, is mixed with clay and zirconium oxide and packed in catalyst towers. The hot vapours are passed over fixed bed of catalyst in the catalyst towers maintained at 400 to 500°C and a pressure of 1 to 5 kg/cm2. Cracking of the oil takes place in the reactor. About 30-40% of the charge is converted into low molecular weight hydrocarbons, conforming to the composition of gasoline. About 4% of carbon is formed during the cracking process which gets deposited on the catalyst bed. The cracked vapours now enter the fractionating column where the gasoline vapours and other gaseous products are recovered from the top while the heavy gas oil fractions are condensed at the bottom of the column. The vapours are admitted into a cooler where the gasoline and some of the other gaseous products are condensed, while the uncondensed gases move on further. The condensate is now sent into a stabilizer where the dissolved gases are removed and gasoline is recovered. (The light gases produced in cracking are no longer allowed to escape into air or burnt. They form valuable raw material for the synthesis of new types of fuels and other materials such as antifreeze, plastics and synthetic rubber). When substantial amount of carbon is deposited on the catalyst bed, the catalyst ceases to function. It is reactivated by burning off the carbon deposited in a stream of hot air. This heat is utilized for heating the reactor. During the reactivation of catalyst, the vapours are diverted through a stand-by catalyst chamber. Thus, the reaction can proceed without interruption.

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Fluid-bed catalytic cracking The modern refineries employ fluid-bed of the catalyst in their huge and efficient “cat-crackers”. These consist of a reactor and a regenerator, generally placed side by side, as in Fig. 2.6.

Catalyst regenerator Fresh catalyst

cooler Lighter fractions

clean hot catalyst

Cracked vapours Smoke stock Reacter

Uncondensed gases Gases

Air

Heavy oil charge

Stabilizer

Spent catalyst

Heavy Gasoline fuel + dissolved gases Fractionating oil column

Gasoline

Fig. 2.6. Fluid-bed catalytic cracking.

In this process, the cracking stock (e.g., gas, oil and other fractions from the straight run still) is preheated and as it enters the reactor, hot catalyst, (in the form of a fine powder), is introduced from the regenerator. Cracking takes place on the surface of the turbulent catalyst bed as it circulates with the oil vapours in the reactor at a temperature of about 530°C and pressure of about 3 to 5 kg/cm2. The low-boiling lighter molecules move up to the top of the reactor and enter into the fractionating column. The cracked gases and gasoline are removed from the top of the fractionating column and sent to a cooler, where gasoline is condensed. It is then sent to a stabilizer to recover pure gasoline. The product contains a higher proportion of aromatics and iso-paraffins, and less of gum-forming diolefines than that obtained from thermal cracking process A part of the fluidised catalyst is continuously removed from a sump in the reactor and is forced into the regenerator with the help of air under high pressure. The carbon and the tarry residue deposited on the catalyst are burnt off in the regenerator and the temperature rises to about 590°C or more. The hot flue gases are allowed to pass through a waste heat boiler to generate steam. Then it passes through a cyclone and precipitator to remove any powdered catalyst. In the modern “catcrackers”, about 35 tons of catalyst are regenerated per minute. Any catalyst carried along with the cracked vapours into the fractionating column, collects at the bottom along with the heavy bottom fraction, which is recycled to the cracking stock subsequently (In some of the units, centrifugal separators or cyclones are attached near the top of the reactor and regenerator, to allow only the vapours and gases to pass on, but retaining the catalyst powder. Hydrogenation cracking. In this process, the heavy oil is cracked in the presence of hydrogen under high temperature (450 to 525°C) and a pressure of 25 kg/cm2. Although the process is expensive and intricate, complete cracking can be achieved by recycling. Advantage of catalytic cracking over thermal cracking

(i) The yield of petrol is higher, because catalyst are selective in action. (ii) The quality of petrol product is better because : (a) The product of cracking contain a higher amount of aeromatics and hence posses better antiknocking property. (b) Isomerization of branched chain compounds occur, there by better petrol is produced. (c) Decomposition of aeromatic removes only the side chain but not the ring itself. (iii) No external fuel is needed for cracking. The heat required is derived from coal present in the catalyst. (iv) A much lower pressure is needed in catalytic cracking (1-5 kg/cm2). (v) The process can be controlled to get the desired products.

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(vi) The percentage of gum forming compound is low. (vii) The production cost is low since high temperature and high pressure are required. Table. 2.2. Comparison of liquid and vapour phase thermal cracking

Characteristics Cracking temperature Pressure Yield of product Octane rating of petrol Criteria Time required for cracking

Liquid phase 475-530°C 100 kg/cm2 50-60% 60-70% Any heavy oil can be used More

Vapour phase 600-650°C 10-20 kg/cm2 – Better (> 70) Oil should be radily vapourised Less

2.8 REFORMING Reforming is a process of bringing about structural modifications in the components of straight run gasoling. Reforming mainly carried out for improving the anti-knocking characteristics of gasoline. Straight run petrol contains unbranched saturated molecules which contribute to low-octane rating. Aviation gasoline is manufactured, either from straight run-petrol, or from specially treated cracked gasolines blended with high-octane number components such as alkylate, iso-pentane and aromatics. Lead tetraethyl (TEL) is also added. A typical 100-octane number gasoline may be made by blending the following: Catalytically cracked gasoline — 40 to 60% Isopentane (octane number 90 to 95) — 10 to 20% Lead tetraethyl (TEL) — 4 ml/gallon High octane component (alkylate) — 30 to 40% The high octane components may be obtained by processes such as polymerisation, alkylation and isomerisation of the by-products obtained from the cracking unit. Thermal reforming comprises of heating the gasoline under pressure, which causes some of the molecules to crack and reform by alkylation. The resulting branched structure contributes higher octane rating to the product. Catalytic reforming consists of heating the gasoline under pressure of a catalyst. This process generates a product having superior fuel characteristics at lower pressures and with less cracking and coke formation as compared to thermal reforming. The process brings about several reactions such as isomerization, hydrogenation, dehydrogenation and aromatization, during which straight chain paraffins are converted into aromatics or benzene or other highly branched molecules. Several types of reforming reactions can be achieved, using different catalysts and different process parameters. Hydroforming is another reforming technique which gives a product high in aromatics, low in olefins and more responsive to tetraethyl lead. This gives a product with a much higher octane rating as compared to that obtained by thermal reforming. In this process, the stock is treated with a calculated quantity of hydrogen at about 500°C and 15 kg/cm2 pressure in presence of molybdenum as catalyst. Platforming is another type of reforming process, using platinum with a little fluorine on alumina. In this process, losses due to the formation of methane are small. Other types of reforming processes include houdriforming, catforming and ultraforming all of which use platinum on alumina catalysts at different process conditions; while thermoforming employs Cr2O3–Al2O3 beads as catalyst and hyperforming utilizes a cobalt-molybdenum oxide catalyst.

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50 40 30 20

ted ula d c l Ca erve obs

10

High

Power output

Efficiency %

2.9. KNOCKING Cylinder V1 In an internal combustion engine (spark ignition type), a piston mixture of air and petrol vapour is compressed and ignitied by an V2 electric spark and the essential chemical reaction is the oxidation piston of hydrocarbon molecules. It is essential that combustion of the fuel in the cylinder of an internal combustion engine should proceed in a regular way. After the reaction is initiated by a spark, a flame should spread Top-dead centre Bottom-dead centre rapidly and smoothly through the gas mixture and the expanding Compression ratio = V2/V1 gas drives the piston down the cylinder. In certain circumstances, Fig. 2.7. Compression ratio. however, the rate of oxidation is so great that the mixture produces a typical sound called engine “knock”. And this phenomena is known as knocking. The rate of oxidation of a hydrocarbon molecule depends on the number of carbon atoms in the molecules, on the structure and on the temperature. The temperature, in turn, depends on the “compression ratio” i.e., the ratio of the cylinder volume at the end of the suction stroke to that at the end of the compression stroke of the piston. In other words the “compression ratio” is the ratio of the volume of gas above the piston in its bottom-dead centre position to the gas volume above the top-dead centre position, as shown in Fig. 2.7. Theoretically, the power output and efficiency of an internal combustion engine should increase continuously with increase in the compression ratio (Fig. 2.8).

0 0 1 2 3 4 5 6 7 8 9 10 3 Compression ratio (i)

HUCR Medium HUCR Low 4

7 5 6 Compression ratio (ii)

Fig. 2.8. Variation of (i) thermal efficiency (ii) and power output with compression ratio.

However, H.R. Ricardo, with the help of a variable compression engine showed that in actual practice, the power increases to a maximum and then falls rapidly with further increase in the compression ratio. The compression ratio, corresponding to the maximum power output, is known as highest useful compression ratio (HUCR) at which a slight metallic “knock” or “pink” can be heard. This becomes more pronounced and heavy as the compression ratio is further increased above this optimum value and finally pre-ignition of the fuel will occur (i.e., the fuel ignites even before the regular spark occurs). It was found that for a given engine, the HUCR largely depends upon the type of the fuel used. For instance, aromatic fuels such as benzol can be used at higher compression ratios without knocking than normal straight-run paraffinic petrol Knocking is due to the spontaneous ignition of the last unburnt portion of the charge giving a detonating shock wave. (Knock occurs after the passage of the firing spark, while pre-ignition is the spontaneous combustion of the explosive mixture before the spark). The tendency to “knock” depends not only on the fuel but also on the engine design, shape of head, location of plug, ports, etc., and also upon the running conditions. The knocking tendency is increased by reducing the engine speed, advancing the ignition etc. 2.9.1. Consequence of knocking

(i) Decreassed power output (ii) Mechanical damage by overheating of cylinder parts (iii) Huge loss of energy

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2.9.2. Chemical structure and knocking The knocking tendency decreases with increase in the compactness of the molecules, doublebond and cyclic structure. For straight chain hydrocarbon the knocking tendency increase with increasing the molecular weight and boiling point. e.g. n-hexane n-pentane n-butane 90 60 29 Branched chain paraffins have lower knock propertie than their normal isomers The resistance to knock increase with the number of branches and their position. Thus 2-methyl hexane has an octane number of 55 while 2 : 2 dimethy pentane has an octane number of 80. Olefines have lower knocking properties than the corresponding paraffins. Further the knocking tendency decreases as the position of the double bond approaches the centre of chain. In general the knocking tendency is in the following order Straight chain paraffins > Branched chain Paraffins > Olefins > Cyclo paraffins (naphthenes) > Aromatic hydrocarbon. 2.10 OCTANE NUMBER The resistance offered by gasoline to knocking can not be defined in absolute terms. It is expressed on an arbitrary scale, known as octane rating or octane number. This scale is given by Graham Edgar in 1926. It was observed that n-heptane knocks very badly and hence it was assigned an antiknock value of zero. On the other hand, iso octane (2, 2, 4 trimethyl pentane) has a high resistnce to knocking and hence it was assigned an antiknocking value of 100. H3C—CH2—CH2—CH2—CH2—CH2—CH3 n-heptane (Antinknock value = 0) CH3 CH3 H3C—C—CH2—CH—CH3 CH3 Iso octane (2,2,4 trimethyl pentane) (Antinknock value = 100) Thus “the percentage of iso octane in the n-heptane-isooctane blend which has the same knocking characteristics as the gasoline sample, under the same set of conditions is called as octane number”. The octane rating of some common hydrocarbons are given below. Table 2.3

S. N. 1. 2. 3. 4. 5. 6.

Hydrocarbon Benzene Isopentane Cyclohexane 2-methyl pentane n-pentane n-hexane

Octane Number 106 90 77 71 62 26

2.11 CETANE NUMBER It is a measure the case with which a fuel will ignite under compression. The hydrocarbon centane (n-hexa decare) has a very short ignition delay as compared to any disel fuel and is assigned an arbitary value of 100. On the other hand a-methyl-naphthalene has a longer ignition delay as compared to any other diesel fuel and hence is assigned a value of 0.

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Diesel Fuel C16H34 n-hexadecane

Cetane No. 100 0

Remarks Very short ignition delay Longer ignition delay

CH3 a-methyl naphthalene

Thus, the centane number of a disel oil may be defined as the percentage of centane in a mixture of centane and a-methyl naphthalene which will have the same ignition characteristics as the fuel under lest, under same set of conditions. The centane number of high speed (1500-2000 rpm), medium speed (1500-500 rpm) and low speed (150 to 500 rpm) diesel engines should be atleast 45, 35 and 25 respectively. Diesel engine requires a fuel of cetane number greater than 45 Cetane no. of fuel primarily depends on the nature and composition of its hydrocarbons. For instance consider the following: n-alkanes > naphthalene (cycloalkanes) > alkenes > branched alkanes > aeromatics (cyclo alkanes)

(i) ignition delay increases from left to right,



(ii) ignition quality incrases from right to left,



(iii) cetane no. increases from right to left.

The straight chain hydrocarbons ignite easily (high ignition quality) but the aeromatics do not ignite easily on compression. Cetane number fuels eliminate diesel knock. The cetane number of diesel fuel may be raised by addition of pre ignition dopes. e.g., alkyl nitrites, ethyl nitrite or amyl nitrite etc. HMV (hepta methyl nomane) with cetane rating of 15 is now considered as the low quality diesel in view of its easy availability and purity.



CH3 CH3 CH3 CH3     | | | | CH3  C  CH 2  C  CH 2  C  CH 2  C  CH3    | | | |   CH3 CH3 CH3 CH3

On the revised scale (H.M.N.) the cetane number (C.N.) represents the % cetane in the blend 15 with H.M.N. plus of the % H.M.N. 100 Thus, a blend of 50% cetane and 50% H.M.N. has a cetane rating of 15 50 + × 50 = 57.5 100

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Table 2.5. Difference Between Octane Number And Cetane Number.

S. N. 1. 2. 3. 4.

Octane Number It is the % of iso-octane in a mixture of n-heptane and iso-octane With the help of octane number we can calculate the knocking character of petrol fuel Octane number can be increassed by adding tetraethyl lead

Cetane Number It is the % of hexadecane in a mixture of 1-methyl napthalene and hexa decane. With the help of cetane number we can calculate the knocking character of diesel fuel.

It can be increased by adding substances called “dopes” like ethyl nitrite isoamyl nitrates etc. Straight chain hydro-carbon molecules Straight chain hydrocarbon molecules are are worst fuel best fuel.

2.12 FLUE GAS ANALYSIS The mixtures of gases which comes out of the furnace (combustion chamber) is called flue gas. They are mostly the mixture of CO2, CO and O2 coming out of the combustion chamber. If flu gas contain large amount of CO, it indicates the incomplete combustion or the less supply of oxygen than, required for combustion. The flue gas analysis is generally carried out by Orsat’s apparatus (Fig. 2.23). Two types of orsat apparatus are commercially available. (1) A portable model, known as “short orsat” which is used where the number of components to be determined is relatively limited and where moderate accuracy is sufficient (2) A much larger apparatus known as a “long orsat” or a precision model orsat, which can be used for the analysis of quite complicate mixtures of gases and which gives very accurate results in competent hands. Principle. The principle involved in the gas analysis by orsat apparatus is that the gas under investigation is taken in the burette and is brought into intimate contact with the absorbent liquids in the absorption pipettes one after another following the specific order viz. KOH, alkaline pyragallic acid and ammoniacal cuprous chloride solutions respectively. In each case, the volume of the unabsorbed gases is measured separately at atmospheric pressure. The reduction in volume in each case corresponds respectively to the amounts of CO2, O2 and CO present in the gas under test. The result is usually expressed as the percentage composition of the gas by volume. Working. At first clean the apparatus and check leakage. If leakage occurs it is removed by applying greases etc. Now fill all the three chambers with ammonical CuSO4, alkaline pyarogallic acid and KOH solution respectively. Now the apparatus is connected to the gas supply and turn three way stop cock on. At first gas passes through CaCl2 tube, which absored all the moisture present in the flue gas, the burette is completely filled with the gas. The meniscus of the liquid in the burette is at zero mark. It means 100 cc dry gas has been taken in. Now three way stop-cock is closed and the stop-cock of the tube containing KOH solution is opened and the pressure water lifted. The gas moves in the KOH solution completely. Here KOH absorbed all the CO2 present in the flue gas. Now the water in the burette and the pressure bottle is brought to the same level and the volume noted. This is at atmospheric pressure. The difference between the two readings one before and other after the absorption of CO2 gives the volume of CO2 absorbed. By repeating the similar process with the remaining pipette, the volume of O2 and CO can be noted being absorned in pyrogallic acid and ammonical cuprous chloride respectively. By knowing the volumes of gases absorbed and value of original gas as 100 cc, percentage of CO2, CO and oxygen in the flue gas can be determine. If some gas still remains in the burette it is supposed to be nitrogen.

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2.13 TIPS FOR COMBUSTION CALCULATION The following points should be remembered in solving numerical problems based on combustion and flue gas analysis (1) Main Objectives

The problems are generally based on calculating: (a) The weight or volume of air theoretically required or used for the combustion of 1 kg of fuel. (b) The percentage of excess air used per kg. of fuel burnt, (c) The percentage composition of flue gas obtained (by weight or by volume) by burning a known quantity of fuel, and (d) The percentage composition of the fuel burnt.

(2) Concept of mole A mole of a substance is that quantity whose weight is numerically equal to its molecular weight. Number of moles,  n = Weight of the substance, W Molecular weight, M. The “mole” (or “mol”) is a general unit; when expressed in grams, it is called “gram mole”; when expressed in kilograms, it is called “kilogram mole” or “kilo mole”; and when expressed in pounds, it is called “pound mole”. Hence, the term “mole” should be interpreted as “gram mole”, “kilogram mole” or “pound mole” depending upon whether the basis of calculation is in grams, kilograms or pounds respectively. Also, “mol” is a volume unit and hence composition in terms of mols can be taken as the same as composition by volume (or composition in terms of m3) i.e., volume % = mole %. Further, 1 gram mole (g mol) of a gas at NTP occupties 22.4 litres (or dm3); 1 kilogram mole occupies 22.4 m3 and 1 pound mole occupies 359 ft3. (3) Composition and mean molecular weight of air The composition of air is taken as 21% of O2 and 79% of N2 (by volume); and 23% of O2 and 77% of N2 (by weight). The mean molecular weight of air is taken as 28.95. It may be noted that air consists of 21.00 mols of O2, 78.06 mols of N2 and 0.94 mols of Ar (argon). Since argon is inert, it is considered together with nitrogen for combustion calculations. That is why, the % N2 in air by volume is taken as 79%. However, the mean molecular weight of air is taken as 28.95 on the basis of the following:  (21× 32) + (78.06 × 28) + (0.94 × 39.34)    = 28.9522 100   1 × 100 = 4.76 m3 of air; and 1 kg. of O2 is supplied by Further, 1 m3 of O2 is supplied by 21 1×100 = 4.35 kg. of air. 23 (4) Density of air The density of air at N.T.P. is 1.290 kg. m–3 or 1.290 × 10–3 g cm–3. (5) Minimum Oxygen Required The minimum O2 required) = (Theoretical O2 required — O2 present in the fuel) The minimum O2 required should be calculated on the basis that complete combustion is taking place according to theoretical and stoichiometric combustion reactions.

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In case of partial combustion, the combustion products contain CO. In case of irregular combustion, the combustion products contain both CO and O2. In such a case, the excess O2 is calculated after subtracting the amount of O2 required to burn CO to CO2. (6) Combustion of Carbon The combustion of carbon in air may be represented as: C + (O2 + N2) → CO2↑ + N2 1 mol 1 mol 3.76 mols 1 mol 3.76 mols or or or or or 12 kg. 32 kg. 107.2 kg. 44 kg. 107.2 kg. Thus, combustion calculation can be done on mol basis or by using the stoichiometric weight relationships between the reactants and products. However, the mol method is considered to be more convenient and simpler. (7) Combustion of hydrogen The hydrogen in coal is present as (a) Combined hydrogen in the form of H2O 2H2 = O2 = 2H2O 4 kg. 32 kg. 2 × 18 = 36 kg. 1 kg. 8 kg.         9 kg. (b) Available hydrogen: Combined hydrogen in coal present as moisture does not undergo combustion. It is only the available hydrogen which is equivalent to (H—O/8) that takes part in combustion. (8) Weight of theoretical amount of air required For complete combustion of 1 kg. of solid or liquid fuel, the theoretical amount of air required 100  32 O   × C + 8  H −  + S kg.  23  12 8   where C, H, O and S are the respective weights of carbon, hydrogen oxygen and sulfur present in 1 kg. of the fuel.



=

(9) Calorific Value If the ultimate analysis of coal is available, its calorific value may be calculated by Dulong’s formula as follows: Calorific value (Kcals/kg)  O   = 8080 C + 34460  H − 8  + 2250S     where C, H, O and S represent the respective weights of carbon, hydrogen, oxygen and sulfur per kg. of coal. (10) % Excess Air (Actual air used − Theoretical air) × 100 Theoretical air (10) The mass of dry flue gases formed should be calculated by balancing the carbon in the fuel and the carbon present in the flue gases (11) The composition of a solid or liquid fuel is usually expressed on weight basis whereas the composition of a gaseous fuel is expressed on volume basis, unless otherwise stated. (12) The composition of flue gases is usually given on dry basis and on volume basis, unless otherwise stated.

% Excess Air =

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2.14 COMBUSTION CALCULATIONS In all combustion reactions, definite relationships exist between the masses of the fuel fired, the air used and the flue gases formed as well as the amounts of heat evolved or absorbed during the reactions, all of which are governed by certain well-known laws. These are: 1. The law of conservation of mass 2. The law of definite proportio 3. The gas laws (Boyle’s law and Charle’s law) 4. The law of conservation of energy. As the composition of air used is taken as uniform, the relationship existing between the above mentioned quantities can be accurately calculated. Thus, if the values of any two of the following are given the third can be found out: (1) the composition of the fuel, (2) the composition of the flue gas and (3) the nature of combustion. Nature of combustion indicates whether the combustion is complete or not and whether any excess air was used. The General Gas Equation: All perfect gases obey the equation of state: PV = nRT where P = absolute pressure V = Volume n = number of moles of gas R = The “Gas constant”, and T = absolute temperature. Gram mol and pound mol The composition of a solid or liquid fuel is usually expressed by weight whereas the composition of a gaseous fuel is given by volume. However, the composition of the flue gases is generally given by volume. The masses of gaseous substances can be calculated from their volumetric compositions and vice versa in accordance with the gas laws and Avogadro’s Hypothesis. From the corollary of the Avogadro’s law, it follows that gram molecular volume of all gases at N.T.P. (i.e., Normal temperature and pressure which are 0°C and 760 mm abs, or 32°F and 14.7 lbs/sq. in abs) is 22.4 litres. The corresponding volume in English system is 359 cu. ft., which is known as pound molecular volume. Thus, a gram mol of carbon is 12 grams; a kilogram mol of carbon is 12 kilograms, and a pound mol of carbon is 12 lbs. Similarly, a kilogram mol of CH4 is 16 kgs and a pound mol of CH4 is 16 lbs and a gram mol of CH4 is 16 g. Each gram mol, kilogram mol and pound mol occupy 22.4 litres, 22.4 m3 and 359 cu. ft. at N.T.P. respectively. The Avogadro’s hypothesis is true not only for gases but also for mixtures of gases. A mixture of gases behaves as if it were single with a molecular weight equal to the average molecular weight of 1 1 1 mol of N2 + mol of H2 + mol of O2 will have an average molecular its constituents. Thus, 3 3 3 1 1 62 1  = 20.67  . Hence, 20.67 grams of this gaseous weight of 20.67  × 28 + × 2 + × 32 = 3 3 3 3   mixture at 0°C, and 760 mm pressure occupies 22.4 litres. Similarly, 20.67 lbs of the gaseous mixture at 32°F and at 1 atmosphere pressure (i.e., 14.7 lbs/sq. in) occupies 359 cu. ft. Thus, we can easily calculate the weight per cubic foot (c. ft.) or cubic centimeter (c.c.) or cubic meter (m3) of any gas or any gaseous mixture of known composition.

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The molecular weights of some common gases are as follows: ... 2; Oxygen O2 ... Hydrogen, H2 Carbon monoxide, CO ... 28; Carbon dioxide, CO2 ... Methane, CH4 ... 16; Ethylene C2H4 ... Nitrogen, N2 ... 28; Water vapours, H2O ... Weight of the substance Number of moles = Molecular weight Composition of air

32 44 28 18

In all combustion calculations, the composition of air is taken as 21% O2 and 79% N2 (by volume); or as 23% O2 and 77% N2 (by weight). The average molecular weight of air is taken as 28.952.* Calculation of volume of a gas at a given temperature and pressure Boyle’s and Charle’s Laws can be used for reducing the volume of a gas at a given temperature and pressure to the corresponding volume at any other specified conditions of temperature and pressure with the help of the equation P1V1 PV = T T 1 where T and T1 are absolute temperatures of the gases and P and P1 and V and V1 denote the initial and final values of pressures and volumes expressed in identical units. If the temperature is i°C, then the absolute temperature, T = 273 + i°. Similarly, if the temperature is i°F, then the corresponding value in absolute scale is 460 + i°. Combustion reactions The reactions most commonly encountered in combustion calculations are given below: C + O2 → CO2 ...(1)

1 mol

1 mol

1 mol



2C + O2 2CO + O2 2H2 + O2 CH4 + 2O2

→ → → →

2CO 2CO2 2H2O CO2 + 2H2O



1 mol

2 mol

1 mol



...(2) ...(3) ...(4) ...(5)

2 mol

C2H4 + 3O2 → 2CO2 + 2H2O ...(6) 1 C2H2 + 2 O2 → 2CO2 + H2O ...(7) 2 1 C2H6 + 3 O2 → 2CO2 + 3H2O ...(8) 2 Note: In air, the volume of O2 and N2 are in the ratio of 21 : 79 = 1 : 3.76. Equation (1) shows that 1 mol of carbon (12 lbs) combines with 1 mol of oxygen (32 lbs or 359 cu. ft. at N.T.P.) to form 1 mol of CO2 (44 lbs or 359 cu. ft. at N.T.P.). * O2 = 21 mols = (21.0 × 22) = 672 N2 = 78.06 mols = (78.06 × 28) = 2135.68 Ar = 0.94 mols = 0.94 × 39.94 = 37.54

Average Mol. wt. =

2895.2 100

= 28.952.

86

BASIC ENGINEERING CHEMISTRY

Similarly, Equation (5) shows that 1 mol of CH4 reacts with 2 mols of O2 to give 1 mol of CO2 and 2 mols of H2O. From this equation, we can calculate the air required for the complete combustion of 1 mol of methane. Since 21 mols of oxygen come from 100 mols of air and also since 100 2 mols of oxygen are required to burn completely 1 mol of CH4, the air required = 2 × = 9.52 21 mols. This also means that 1 cft of CH4 required 9.52 cft of air (since volume % = mol %). Analyses Analyses of solids are always reported on weight basis. In order to convert them into mol basis, it is necessary to divide each constituent by its molecular weight. Thus a coal containing 72% C, 4% 4 6 72 = 6 mols of C; = 2 mols of H2; and = 0.187 mol of O2 H2 and 6% O2 would contain 2 32 12 per 100 kg of the coal sample. Analysis of gases is always reported on volume basis and hence directly gives the number of mols of each constituent present per 100 mols of the mixture. Thus, the analyses of gases obtained by the Orsat apparatus are always expressed as % by volume and also on dry basis. Hence, this gas analysis gives molar composition directly since the mol is a volume unit. (A pound mol corresponds to 359 cft at N.T.P.). For example, 100 mols of air (100 × 359 = 35900 cft) contains 21 mols of O2 (21 × 359 cft) and 79 mols of N2 (79 × 359 cft). Excess air Combustion seldom takes place efficiently with the theoretically minimum quantity of air. Invariably, an excess of air is used in the furnace. Excess air is the amount of air used over and above that required for complete combustion. Sample problem Let us consider the following data obtained with methane, CH4 as a fuel: Fuel gas CH4 = 100%

Flue gas

Constituent CO2 O2 N2

% 5.5 11.1 83.4

Mols of C 5.5 — —

Mols of O2 5.5 11.1 —

Total

100

5.5

16.6

(A) Ratio of Flue gas: Fuel gas If we take 100 mols of dry flue gas as the basis for calculation, there are 5.5 mols of C, 16.6 mols of O2 and 83.4 mols of N2. It is obvious that the entire amount of carbon present in the fuel gas must have been present in the flue gas. Since all the carbon came from the fuel gas, it is clear that 5.5 mols of CH4 (containing 5.5 mols of C) were used to form 100 mols of dry flue gas. Therefore, Dry flue gas 100 = = 18.2 Fuel gas 5.5 i.e., 18.2 mols (or cft) of dry flue gas per mol (or cft) of fuel gas (B) Dry flue gas: dry ai We know that N2 in air is equal to N2 in flue gas. In the above example, there are 83.4 mols of N2 per 100 mols of dry flue gas. In order to obtain 83.4 mols of N2, the air used must have been 100 83.4 × mols. 79

FUELS AND COMBUSTION

87

100 Dry flue gas = 100 Dry air 83.4 × 79 = 0.948 i.e., 0.948 mols of dry flue gas per mol of air or 0.948 cft. of dry flue gas per cft. of a . ...

(C) Air: Fuel gas ratio From (A) and (B) above, we have Air Flue gas Flue gas 18.2 = ÷ = = 19.2 Fuel gas Fuel gas Air 0.948 i.e., 19.2 mol or cft of air per mole or cft of fuel (CH4). (D) Excess air If means the % of air in excess of that is theoretically required for complete combustion. It can be determined by the following methods: CH4 + 2O2 → CO2 + 2H2O (i) 100 i.e., 1 mol of methane requires 2 mols of O2 which can be obtained by 2 × = 9.52 mols 21 of air. From (C) above, we calculated that the air actually used was 19.2 mols (or cft.) per mol (or cft.) of fuel gas. (19.2 − 9.52) % Excess air = × 100 = 101.7% ... 9.52 (ii) The N2 in the gas analysis indicates the total air used and the free oxygen indicates the excess air (provided that the fuel does not contain N2 and O2 respectively. In case they are present in fuel, the corresponding quantities will have to be accounted for in the calculation). Thus in the 83.4 present example, 83.4 mols of N2 are present which should have come from × 100 mols of air, 79 which contains 83.4 × 21 mols of O2 = 22.17 mols of O2. However, the free oxygen (O2) in the flue 79 gas is 11.10 mols. Hence the O2 required = (22.17 – 11.10) = 11.07 mols. (22.17 −11.07) ... Excess air =  × 100 = 100.3% 11.07 Notes: 1. The difference in the value of excess air in the above two methods of calculation might be due to the slight errors in the gas analysis. 2. When carbon monoxide is present in the flue gas, the amount of O2 necessary to burn it must be deducted from the free O2 in the gas before determining excess air: 1 2CO + O2 → 2CO2. Hence for every mole of CO, mol of O2 needed. 2 For example, if the gas from combustion of a nitrogen free fuel has 12% CO2, 5% O2, 2% CO 21 and 81% N2, then the total oxygen (O2) corresponding to N2 is 81 × = 21.5 mols and the excess 79 2  O2 would be  5 −  = (5 – 1) = 4 mols. 2  4 Hence, the % excess air = × 100 = 22.85%. (21.5 − 4.0)

88

BASIC ENGINEERING CHEMISTRY

(E) Net hydrogen The net hydrogen in fuel gas, or the hydrogen in excess of that which can combine with the O2 present in the fuel can also be calculated from the flue gas analysis. H2 burns with O2 in air to form H2O, which does not appear in the gas analysis. Hence it has to be calculated from the oxygen balance. For example, in the above sample problem on methane, while the O2 appearing in the flue 21 gases is only 16.6 mols, the O2 corresponding to 83.4 mols of N2 is 83.4 × = 22.17 mol. Hence, 79 (22.17 – 16.6) = 5.57 mols of O2 must have reacted with (2 × 5.57) = 11.14 mols of H2 forming Net hydrogen 11.14 11.14 mols of H2O from each 100 mols of flue gas. Further, = 2, i.e., 2 mols = Carbon 5.5 of H2 per mol of C. (F) Moisture The moisture in the flue gas is the sum of the water from free and combined mositure in the fuel, that from combustion of H2 and that from moisture in the air used. The last one is so small that it may be neglected. Thus if the fuel analysis is known, the moisture in the flue gas (moisture in the fuel + H2O from combustion of H2) can be calculated. Thus, if the fuel used is dry methane, the volume of water vapour would be 2 mols per mol of fuel gas (since CH4 + 2O2 → CO2 + 2H2O); 2 and × 100 = 11.1 mols per 100 mols of dry flue gas i.e., for every 100 cft of dry flue gas, there 18.2 would be 11.1 cft of wet flue gas 2.15 NUMERICAL PROBLEMS (A) Numerical based on calorific valu Example 1. The following data is obtained in a bomb calorimeter experiment: Weight of the crucible = 3.649 g Weight of the crucible + fuel = 4.678 g Water equivalent of the calorimeter = 570 g Water taken in the calorimeter = 2200 g Observed rise in temperature = 2.3°C Cooling correction = 0.047°C Acids correction = 62.6 calories Fuse wire correction = 3.8 calories Cotton thread correction = 1.6 calories. Calculate the gross calorific value of the fuel sample. If the fuel contains 6.5% hydrogen, determine the net calorific value Solution. (W + w) (t2 − t1 + tc ) − (TA + TF + Tt ) High C.V. = m (220 + 570) (2.3 + 0.047) – (62.6 + 3.8 + 1.6) = (4.678 – 3.649) (2770 × 2.347) − 68 1.029 6510 − 68 = 1.029

=

FUELS AND COMBUSTION

89

= 6261 cals/g Low C.V. = (H.C.V. – 0.09 × H × 587) cals/g = 6261 – 0.09 × 6.5 × 587 = 6261 – 343.4 = 5917.6 cals/g. Example 2. Determine the water equivalent of the bomb calorimeter apparatus which gave the following data in an experiment: Weight of benzoic acid taken = 1.346 g Weight of the calorimeter can = 1025 g Weight of the calorimeter can + water = 3025 g Initial temperature = 11.872°C Final temperature = 14.625°C Cooling correction = 0.015°C Heat from fuses = 22 cals. The washings of the bomb on analysis indicated the presence of 3.5 ml of N/10 HNO3 and there was no H2SO4 formed. Solution. Weight of benzoic acid taken = 1.346 g Weight of water taken in the calorimeter can = (3025 g – 1025 g) = 2000 g Corrected rise in temperature = (14.625 – 11.872) + 0.015 = 2.753 + 0.015 = 2.768°C Heat evolved due to benzoic acid = (Wt. of benzoic acid × calorific value = 1.364 × 6324 = 8626 cals ...(1) Heat from fuses = 22 cals ...(2) Heat from HNO3 = 3.5 × 1.43 = 5 cals ...(3) Total heat given to the calorimeter = (1) + (2) + (3) = 8626 + 22 + 5 cals = 8653 cals (Heat given to the calorimeter) Total water equivalent = (Corrected rise of temperature) 8653 = = 3126 g. 2.768 Water taken in the calorimeter = 2000 g ... Water equivalent of the calorimeter and auxiliaries = (3126 – 2000)g = 1126g. Example 3. A sample of coal contains: C = 93%, H = 6% and ash 1%. The following data were obtained when the above coal was tested in bomb calorimeter. Weight of coal burnt = 0.92 (i) (ii) Water equivalent of bomb and calorimeter = 2,200 g (iii) Weight of water taken = 550 g (iv) Rise in temperature = 10 Cal (v) Acid correction = 50.0 Cal

90

BASIC ENGINEERING CHEMISTRY

Calculate gross and net calorific value of the coal, assuming the latent heat of condensation of steam as 580 cal/g. Solution. We know that (W + w) (t2 − t1 ) − [Acid + fuse wire correction] GCV = x (550 + 2, 200) + 2.42 − [50 + 10] Cal = 0.92 g = 7,168.5 cal/g LCV = [GCV – 0.09H × latent heat of steam] = (7,168.5 – 0.09 × 6 × 580) = 6,855.3 cal/g. Example 4. In a bomb calorimeter experiment the following data was obtained. Weight of coal = 0.95 g Weight of water taken in the colorimeter = 3000 g Water equivalent of the calorimeter = 550 g Observed rise in temperature = 2.15°C Cooling correction = 0.055°C Fuse wire correction = 6.4 calories Acid correction = 46.8 calories Cotton thread correction = 1.7 calories Calculate GCV and NCV of coal if it contains 6.0% of hydrogen. Solution. (W + w) (t2 − t1 + c.c.) − ( AC + CTC + FWC ) HCV = x (300 + 550) (2.15 + 0.055) – (46.8 + 1.7 + 6.4) HCV = 0.95 (3550) (2.205) − (54.9) = 0.95 = 8181.94 Kcal/kg. Example 5. During the determination of calorific value of a gaseous fuel in the bomb calorimeter, the following result were obtained Weight of the coal burnt = 0.95 g Weight of water taken = 700 g Water equivalent of bomb and calorimeter = 2000 g Increase in temperature = 2.48°C Acid correction = 60.0 Cal Cooling correction = 0.02°C Fuse wire correction = 10.0 Cal Calculate the net and gross calorific value of coal, given latent heat of condenstion of steam = 587 cal/g. Solution. (W + w) (T2 − T1 + Cooling correction) − (Acid + Fuse wire correction) GCV = Mass of fuel

FUELS AND COMBUSTION

91

(2000 + 700) (2.48 + 0.02) − (60 + 10) 0.95 = 7031.6 cal/g LCV = GCV – 0.09H × 587 cal/g = 7031.6 – 0.09 × 5 × 587 = 6767.45 cal/g Example 6. Calculate the gross and net calorific value of a coal sample having the following composition. C = 82%, H2 = 8%, O2 = 5%, S = 2.5% N2 = 1.4% and ash 2.1%. Solution. 1  %O    8080 × % C + 34500  % H − GCV =  + 2240 × % S 100  8    1  5   = 100 8080 × 82 + 34500  8 − 8  + 2240 × 2.5     1 [662560 + 254437.5 + 5600] = 100 = 9225.9 Kcal/kg LCV = GCV – 0.09 H × 587 = 9225.9 – 0.09 × 8 × 587 = 8803.26 Kcal/kg. Example 7. A coal has the following analysis C = 84%, S = 1.5%, N2O = 6%, H2 = 5.5% and O2 = 8.4%. Find the gross and net calorific values with the help of Dulong s formula. Solution.

=



GCV =

=

1  O   8080 × C + 34500  H −  + 2240 S  100  8  

1  84    8080 × 84 + 34500  5.5 −  + 2240 × 15  100  8   

= 8356 Kcal/kg or cal/g.

9H × 587 Kcal/kg 100 9 × 5.5 × 587 = 8356 − 100 = 8356 – 290.57

LCV = GCV −

= 8065.43 Kcal/kg. Example 8. A coal has the following composition by weight. C = 92%, O = 2.0%, S = 0.5%, N = 0.5% and ash = 1.5%. Net calorific value of the coal was found to be 9,430 Kcal/kg. Calculate the percentage of hydrogen and higher calorific value of coal Solution. 1  2   8080 × 92 + 34500  H −  + 2, 240 × 0.5 Kcal/g HCV =  100  8   = [7433.60 + 345 H – 86.25 + 11.2] Kcal/kg = [7358.55 + 345H] Kcal/kg ...(i) Also, HCV = (LCV + 0.09H × 587) Kcal/kg

92

BASIC ENGINEERING CHEMISTRY

= 9430 + 0.09H × 587 = 9430 + 52.83H From eq (i) and (ii) [7358.55 + 345H] = [9430 + 52.83H] 292.17H = 2071.45

... (ii)

2071.45 292.17 = 7.09% \ HCV = (7358.55 + 345H) Kcal/kg = 7358.55 + (345 × 7.09) Kcal/kg. = 9804.6 Kcal/kg.

% of H =

(B) Numerical Based on Analysis of Coal Example 1. A sample of coal was analysed as follows: Exactly 2.500 of was weighted into silica crucible. After heating for one hour at 110°C, the residue weighed 2.415 g. The crucible next was converted with a vented lid and strongly heated for exactly seven minutes at 950 ± 20°C. The residue weighted 1.528 g. The crucible was then heated without the cover, until a constant weight was obtained. The last residue was found to weight 0.245 g. Calculate the percentage result of above analysis. Solution. Mass of moisture in coal sample = 2.500 – 2.415 = 0.085 g Mass of valatile matter = 2.415 – 1.528 = 0.887 g Mass of ash = 0.245 g 0.085 × 100 % of moisture = 2.500 = 3.40% 0.887 × 100 % of volatile matter = 2.500 = 35.48% 0.245 × 100 % of ash = = 9.80% 2.500 % of fixed carbo = 100 – (3.40 + 35.48 + 9.80) = 51.32%. Example 2. 0.5 gm of a sample of coal was used in a bomb calorimeter for the determination of calorific value. Calorific value of coal was found to be 8,600 cal/g. The ash formed in the bomb calorimeter was extracted with acid and the acid extracted was heated with barium nitrate solution and a precipitate of barium sulphate was obtained. The precipitate was filte ed, dried and weighed. The weight of precipitate was found to be 0.05 g. Calculate the percentage of sulphur in the coal sample. Solution. Percentage of sulphur in the coal sample

=



=



Weight of BaSO 4 ppt obtained × 32 × 100 233 × Weight of coal sample taken in bomb

0.05 × 32 × 100 233 × 0.5 = 1.3734%.

FUELS AND COMBUSTION

93

Example 3. A sample of coal was analyzed as follows: Exactly 2.000 g was weighed into a silica crucible. After heating for one hour at 110°C, the residue weighed 1.975 g. The crucible next was covered with a vented lid and strongly heated for exactly seven minutes at 950 ± 20°C. The residue weighed 1.328 g. The crucible was then heated without the cover, until a constant weight was obtained. The last residue was found to weigh 0.205 g. Calculate the percentage result of the above analysis. Solution. Mass of moisture in coal sample = 2.000 – 1.975 = 0.025 g Mass of volatile matter = 1.975 – 1.328 = 0.647 g

Mass of ash = 0.205 g

0.025 × 100 2 = 1.25% 0.647 × 100 Percent of volatile matter = 2 = 32.35% 0.205 × 100 Percent of ash = 2 = 10.25%

Percent of moisture =

Percent of fixed carbo = 100 – (1.25 + 32.35 + 10.25) = 56.15%. Example 4. 3.12 g of coal was Kjeldahlized and NH3 gas thus evolved was absorbed in 50 ml of 0.1 N H2SO4. After absorption, the excess (residual) acid required 12.5 ml of 0.1N NaOH for exact neutralization. Determine the percentage of nitrogen in the sample of coal. Solution. Amount of sulphuric acid used to neutralize the ammonia evolved = (0.1 × 50 – 0.1 × 12.5) milli equivlents 0.1 (50 − 12.5) × equivalents 1000 0.1 (50 − 12.5) × 14 Weight of nitrogen = 1000 Weight of nitrogen ×100 % of nitrogen = Weight of coal sample taken

= \

0.1 (50 − 12.5) × 14 1 × × 100 1000 3.12 = 1.683%.

=

Example 5. 1.0 g of a sample of coal was used for nitrogen estimation by kyeldahl method. The evolved ammonia was collected in 25 ml N/10 sulphuric acid. To neutralize excess acid, 15 ml of 0.1N NaOH were required. Determine the percentage of nitrogen in the given sample of coal. Solution. 15 ml of 0.1N NaOH solution = 15 ml of 0.1N H2SO4 \ Volume of H2SO4 used to neutralize the evolved ammonia = 25 ml of 0.1N

94

BASIC ENGINEERING CHEMISTRY

Volume of H 2SO 4 Used × Normality ×1.4 Weight of coal sample taken 10 × 0.1 × 1.4 = 1.4% = 1.0 Example 6. A coal has the following composition by weight: C = 90%, O = 3.0%, S = 0.5%, N = 0.5% and ash = 2.5%. Net calorific value of the coal was found to be 8,490.5 Kcal/kg. Calculate the percentage of hydrogen and higher calorific value of coal HCV = (LCV + 0.09 H × 587) Kcal/kg Solution. = (8,490.5 + 0.09H × 587) Kcal/kg = (8,490.5 + 52.8H) Kcal/kg ...(i)

Percentage of N =

\

Also

HCV =

1  3.0    8, 080 × 90 + 34,500  H −  + 2, 240 × 0.5 Kcal/kg  100  8   

= [7,272 + 345H – 129.4 + 11.2] Kcal/kg = [7,754.8 + 345H] Kcal/kg ...(ii) From equation (i) and (ii), we get 7,754.8 + 345H = 8,490.5 + 52.8H or 292.2H = 8,490.5 – 7,154 – 8 = 1,335.7 or percentage of H = 1335.7/292.2 = 4.575% ...(iii) \ HCV = (8,490.5 + 52.8 × 4.575) Kcal/kg (From equation (i) and (iii) = (8,490.5 + 241.3) Kcal/kg = 8,731.8 Kcal/kg. (C) Numerical Based on Combustion Example 1. A coal sample has the following percentage composition; C = 84.0%; H2 = 3.5%, O2 = 3.0%, S = 0.5%, Moisture = 3.5%, N2 = 0.5% and ash = 5.0%. Calculate (a) the theoretical weight of air required for the complete combustion of 1 kg of coal (b) its volume in m3 at NTP, and (c) percentage composition by weight and volume of the dry products of combustion. Solution The composition of solid fuels is expressed by weight. Method-1 (Mole Method) Let the basis of calculation be 100 kg. of coal. Constituent % By wt. Mol. wt. No. of Kmols

C

84.0



H2

3.5



O2

3.0



S

0.5



Moisture

3.5



N2

0.5



Ash

5.0

84 = 7.0 12 2 3.5/2 = 1.75 3.0 32 = 0.094 32 0.5 32 = 0.016 32 18    — 0.5 28 = 0.0178 28 —    — 12



Kmols of O2 reqd.

Kmols of dry products

7.0

CO2 – 7.0

0.875

—

– 0.094

—

0.016

SO2 — 0.016

—

—

—

N2 – 0.0178

—

—

7.797 k mols

FUELS AND COMBUSTION

95

)

K mols of O 2 required = 7.797 Kmols. per100 kg of coal (a) Hence, the theoretical quantity of O2 required for the combustion of 1 kg of coal = 7.797 × 10–2 Kmols = 7.797 × 10–2 × 32 kg.

Weight of air required for  100 the complete combustion  = 7.797 × 10–2 × 32 × kg. ...  23 of 1kg of coal  = 10.848 kg. (b) Theoretical quantity of O2 required per kg. of coal = 7.797 × 10–2 K mols. ... Theoretical quantity of air required per kg. of coal 100 = 7.797 × 10–2 × = 0.3713 K mols 21 ... Volume of air supplied at NTP for 1 kg of coal = 0.3713 × 22.4 m3 = 8.3171 m3 (c) Composition of dry products of combustion by volume The dry products formed by the combustion of 1 kg of coal are: CO2 = 0.07 K mols SO2 = 0.00016 K mols 79  N2 = 0.000178 +  0.3713 ×  = 0.2935 K mols 100   from fuel from air ... Total volume of the dry products of combustion = 0.3637 K mols ... Volumetric composition of the products of combustion is: 0.07 ×100 CO2 = = 19.25% 0.3637 0.00016 ×100 SO2 = = 0.04% 0.3637 0.2935 ×100 N2 = = 80.7% 0.3637 By weight The dry products of combustion formed from 1 kg of coal are; CO2 = 0.07 × 44 = 3.08 kg



SO2 = 0.00016 × 64 = 0.01024 kg N2 = 0.2935 × 28 = 8.218 kg Total weight = 11.3082 kg. .. . Gravimetric composition of the dry products of combustion is: 3.08 ×100 = 27.237% 11.3082 0.01024 ×100 SO2 = = 0.091% 11.3082 8.218 ×100 N2 = = 72.673% 11.3082

CO2 =

96



BASIC ENGINEERING CHEMISTRY

Method 2: (Stoichiometric method) C + O2 → 12 kg



H2



2 kg

32 kg

1 O 2 2

+

→

16 kg

S

+

32 kg

O2

→

32 kg

CO2

44 kg

H2O 18 kg

SO2

64 kg

Let 100 kg of the coal be the basis of calculation: Constituent

Weight in kg

C

84.0



H2

3.5



O2

3.0



S



Moisture N2



Ash

Weight of O2 required, kg.

Weight of dry products, kg

84 × 44 84 × 32/12 = 224 CO2 = = 308 12 16 3.5 × = 28.0      — 2 – 3.0      —

32 64 = 0.5 SO2 = 0.5 × = 1.0 32 32 3.5    —      — 0.5    — N2 = 0.5 0.5

0.5 ×

5.0    —      —

Weight of O 2 required  for the combustion of  = 249.5 kg. 100 kg coal 



(a) Hence, the theoretical quantity of O2 required for the combustion of 1 kg of coal = 2.495 kg. 100 ... The theoretical quantity of air required for the combustion of 1 kg. of coal = 2.495 × kg 23 = 10.848 kg. (b) 32 kg. of O2 occupies 22.4 m3 at N.T.P. ... 2.495 kg. of O2 occupies

2.495 × 22.4 = 1.7465 m3 32

Now, 21 m3 of O2 are present in 100 m3 of air at N.T.P.

1.7465 ×100 = 8.317 m3 of air. 21 ... Volume of the air at N.T.P. = 8.317 m3. (c) Composition of dry products of combustion. By weight The dry products of combustion formed from 1 kg of coal are: CO2 = 3.08 kg SO2 = 0.01 kg. N2 = [0.005 + (10.848 – 2.495)] kg. = 8.358 kg. ... 1.7465 m3 are present in

from fuel

from air

Total weight = 11.448 kg.

FUELS AND COMBUSTION

...  Gravimetric composition of the dry products of combustion is 3.08 ×100 CO2 = = 26.90% 11.448

By Volume

97

0.01×100 = 0.087% 11.448 8.358 ×100 N2 = = 73.01% 11.448

SO2 =

3.08 = 0.07 Kmol 44 0.01 SO2 = = 0.0001563 Kmol 64 8.358 N2 = = 0.2985 Kmols 28 Total volume = 0.3687 Kmols ... Volumetric composition of the dry products of combustion is: 0.07 ×100 CO2 = = 18.986% 0.3687 0.001563 ×100 SO2 = = 0.0424% 0.3687 0.2985 ×100 N2 = = 80.96% 0.3687 CO2 =

Example 2. Find the volume of air required for complete combustion of 1 m3 of acetylene and the weight of air necessary for the combustion of 1 kg of fuel. Solution 2C2H2 + 5O2 = 4 CO2 + 2H2O 2 vols 5 vols 4 vols 2 vols O2 required per m3 of C2H2 = 2.5 m3 ... Air required per m3 of C2H2 = 2.5 ×

100 = 11.9 m3. 21

Further, 2 mols of C2H2 require 5 mols of O2 .. . (2 × 26) kg of C H requires (5 × 32) kg of O 2 2 2 (Since mol. wt. of C2H2 = 26 and Mol. wt. of O2 = 32) 160 kg O2 to be supplied per kg of fuel (C2H2) = 52 160 100 × ... Air to be supplied per kg of C2H2 = = 13.378 kg. 52 23 Example 3. The % analysis by volume of producer gas is H2 – 18.3%, CH4 – 3.4%, CO – 25.4%, CO2 – 5.1%, N2 – 47.8%. Calculate the volume of air required m3 of the gas.

98

BASIC ENGINEERING CHEMISTRY

Solution. 100 mols of the producer gas contains:

Constituents

% and Mols



H2

18.3

9.15



CH4

3.4

6.80



CO

25.4

12.70



CO2

5.1

—



N2

47.8

—



Mols of O2 required

Total O2 required ... 28.65

... Air required for 100 mols of fuel gas = 28.65 × ... Air required per m3 of gas =

100 = 136.43 mols 21

136.43 = 1.3643 m3. 100

Example 4. A gas has the following composition by volume: H – 22%, CH4 – 4%, CO – 20%, CO2 – 6%, O2 – 3% and N2 – 45%. If 25% excess air is used, find the weight of air actually supplied per m3 of this gas. Solution. 100 mols of the fuel gas contains

Constituents



H2

22 4

8

CO

20

10

CO2

6

—

3

–3

45

—



CH4

O2 N2

% and Mols



Mols of O2 required 11

Total O2 required ...

... Air required for 100 mols of fuel gas = 26 ×

26

100 = 123.8 mols 21

But air supplied actually (25% excess) for 100 mols of gas = 123.8 × ... Air supplied for 1 mol of gas =

154.75 = 1.5475 mols 100

... Air supplied for 1 m3 of gas = 1.5475 m3 =

125 = 154.75 mols. 100

1.5475 × 28.97 = 2 kg. 22.4

(Since 22.4 m 3 of air i.e. 1 kg mol at N.T.P. weighs 28.95 kg). Example 5. A sample of coal contained: C – 81%, H2 – 4%, O2 – 2% and N2 – 1%. Estimate the minimum quantity of air required for complete combustion of 1 kg of the sample. Find the composition of the dry flue gas by volume if 40% excess air is supplied

FUELS AND COMBUSTION

99

Solution. Let the basis of calculation be 100 kg of coal. Element Kg

K Mols



C

81

6.7500

6.7500

CO2 – 6.750



H

4

2.0000

1.0000

H2O – 2.000



O

2

0.06250

–0.0625

—



N

1

0.0360

Nil







K mols of O2 required

Total

Product

N2 – 0.036

7.6875

Oxygen required for 100 kg of coal = 7.6875 k mols. 100 K mols 21 = 36.6 k mols = 36.6 × 28.95 kg = 10.59.6 kg. 10596 ... Air required per kg of coal = = 10.596 kg. 100 Air actually used is 40% excess, and hence it is equal to 140 36.6 × = 51.24 k mols 100 ... Excess air used = (Actual air used – theoretical quantity of air required) = (51.24 – 36.60) k mols. = 14.64 k mols 79 Nitrogen in the total air used = 31.24 × = 40.48 k mols. 100 Oxygen in the excess air used which remains in free state in the flue ga 21 = 14.64 × = 3.0744 k mols 100 The dry flue gases thus contai CO2 = 6.75 k mols N2 = (40.48 + 0.036) = 40.51 k mols from from air fuel O2 = 3.0744 k mols ...

Air required for 100 kg of coal = 7.6875 ×



Total = 50.3344 k mols

Percentage composition of the dry flue gases would b 6.75 CO2 = × 100 = 13.4% 50.3344 40.51 N2 = × 100 = 80.5% 50.3344 3.0744 O2 = × 100 = 6.1% 50.3344

100

BASIC ENGINEERING CHEMISTRY

Notes 1. Air contains 21% by volume of O2 and 79% by volume of N2. 2. Average mol. wt. of air is 28.95. 3. H2O in flue gases is neglected for dry flue gas analysi Example 6. A gas fi ed engine uses a fuel gas having the following composition: CO2 – 9%, CO – 42%, H2 – 33%, N2 – 16% The combustion takes places with the theoretical quantity of air. Calculate the volume of air used per cubic meter of fuel gas burnt. Determine the dry flue gas analysis Solution. 100 mols of fuel gas contains

%

Mols

Mols of O2 required

Products and Mols

Nil 21 16.5 Nil

CO2 — 9  CO2 — 42 H2O — 33 N2 — 16

CO2   9   9 CO 42 42 H2 33 33 16 16 N2

Total

37.5 mols

Mols of O2 required per 100 mols of fuel gas = 37.5 ... Mols of air required per 100 mols of fuel gas 100 = 37.5 × = 178.5 21 178.5 ... Mols or cubic meters (m3) of air required per 1 mol or 1 cubic meter of the fuel gas = 100 = 1.785. The dry product of combustion:



CO2



N2

→ 9 + 42 = 51 mols 79   → 16 +  37.5 ×  = 157 mols 21   (from fuel) (from air)



Total

208 mols

... The dry exhaust gas contains 51×100 CO2 – = 24.52% 208 157 ×100 N2 – = 75.48% 208 Example 7. A furnance fi ed with a hydrocarbon fuel oil has a stack gas analysing as follows: CO2 – 10.2%, O2 – 8.3%, N2 – 81.5%. Calculate (a) the composition of the original fuel oil, (b) % excess of air used and (c) volume of air supplied per kg of oil. Solution. Basis for calculation: 100 k mols of dry stack gas

% and mols

K Mols of C

K mols of O2

K mols of N2

CO2 10.2 O2   8.3 N2 81.5

10.2 10.2 —   8.3 — —

— — 81.5



10.2

81.5

Total

18.5

FUELS AND COMBUSTION

101

(a) 100 k mols of dry flue gas contains 18.5 k mols of O2 and 81.5 k mols of N2 which must have come from the air supplied. 100 = 103.17 k mols; 79 21 = 21.69 k mols. and the O2 accompanying it = 103.17 × 100 The difference in quantity of O2 supplied and that used in the formation of water which is not appearing in the dry flue gas analysis will b = 21.69 – 18.5 = 3.19 k mols .. . Water vapour formed = 2 × 3.19 = 6.38 k mols Hence, the air supplied = 81.5 ×

1   → H2O   H 2 + O 2  2  

... Hydrogen present in the fuel oil = 6.38 k mols The same quantity of fuel oil contains = 10.2 k mols of C ... The fuel oil contains 10.2 k mols of C or (10.2 × 12) = 122.4 kg of C and (6.38 × 2) = 12.76 kg of H. That means, the oil weighing (122.4 + 12.76) = 135.16 kg was burnt and it should be composed of 122.4 kg of C and 12.76 kg of H. Hence the composition of the fuel oil is: 122.4 C = × 100 = 90.5% 135.16 12.76 H = × 100 = 9.5% 135.16 (b) Total O2 supplied = 21.69 k mols O2 needed for   C = 10.20 k mols O2 needed for  H = 3.19 k mols .. . O 2 actually needed for C and H = 13.39 k mols present in100 mols of fuel oil .. . Excess O2 = (21.69 – 13.39) = 8.30 k mols

}

8.30 × 100 = 61.99% 13.39 (c) Total oil burnt = 135.16 kg 100 Total air supplied = 21.69 × = 103.29 k mols 21 = 103.29 × 22.4 = 2313.7 m3 at NTP (Avogadro’s law) 2313.7 ... Air used per kg of fuel oil = = 17.12 m3. 135.16 Example 8. A coal containing 62.4% C, 4.1% H, 6.9% O, 1.2% N, 0.8% S and 15.1 moisture and 9.7% ash was burnt in such a way that the dry flue gases contained 12.9% CO2, 0.2% CO, 6.1% O and 80.8% N. Calculate (a) the weight of air theoretically required per kg of coal, (b) the weight of air actually used and (c) the weight of dry flue gas p oduced per kg of coal. ... Excess O2 or air used =

102

BASIC ENGINEERING CHEMISTRY

Solution. (a) Basis of calculation. 100 kg of coal, which contains

Analysis



C H O N S H2O Ash

Kg

K mols

62.4 4.1 6.9 1.2 0.8 15.1 9.7

5.2 2.05 0.217 0.043 0.025 0.840 —



K mols of O2 required 5.2 1.025 – 0.217 Nil 0.025 Nil Nil Total

K mols of product formed CO2 — 5.200 H2O — 2.050 Nil N2 — 0.043 SO2 — 0.025 H2O — 0.840 Nil

6.033 k mols

Thus, O2 theoretically required for complete combustion of 100 kg of coal = 6.033 mols. ... Air theoretically required for complete combustion 100 = 6.033 × = 28.73 mols 21 = (28.73 × 28.95) = 831.7 kg 831.7 Wt. of air required per kg of coal = = 8.317 kg 100 (b) Actual air used: Basis : 100 k mols of dry flue gas, which contain



Constituent CO2 CO O N

% 12.9 0.2 6.1 80.8

K Mols 12.9 0.2 6.1 80.8



K Mols C

K Mols of C

K Mols of N

12.9 0.2 — —

12.9 0.1 6.1 —

— — — 80.8

Total 13.1

19.1

80.8

13.1 13.1 k mols of C must have been supplied by × 100 kg of coal = 252 kg of coal containing 5.2 252 0.043 × = 0.108 k mols of N. 100 (Since 100 kg of coal contains 5.21 mols of C and 0.043 k mols of N). ... K Mols of N from air in 100 k mols of flue ga = (80.800 – 0.108) = 80.692 k mols ... K Mols of air actually supplied to obtain 100 k mols of dry flue gas or to burn 252 kg of coal would therefore be 100 = 80.692 × = 102.15 k mols 79 .. . Wt. of air supplied = (102.15 × 28.97) kg = 2959.3 kg ... Weight of air supplied per kg of coal = 2959.3 = 11.74 kg 252

FUELS AND COMBUSTION

103

(c) Wt. of dry flue gas obtained pe kg of coal: 100 kg mols of dry flue gas contains Constituent





% K Mols

Weight in kg

CO2 12.9 12.9 × 44 CO   0.2   0.2 × 28 O2   6.1   6.1 × 32 N2 80.8   80.8 × 28



= = = =

567.6    5.6 195.2 2262.4

Total = 3030.8 kg

This much flue gas is obtained by burning 252 kg of coal ... Wt. of dry flue gas obtained per kg of coal burn 3030.8 = = 12.03 kg. 252 Example 9. A hydrocarbon fuel, on burning, gave a flue gas having the following volumetric analysis: O2 = 3.56%, CO2 = 13.53%; N2 = 82.91%. What is the composition of the fuel by weight and the percentage of excess air used? What is the volume of air supplied per kg of the fuel oil? Solution. Basis of calculation: 100 k mols of dry flue gas Constituent





% and K Mols

Mols of C

CO2   13.53 O2    3.56 N2   82.91



100.00

K Mols of O2

K Mols of N2

13.53 13.53 —   3.56 — —

— — 82.91

13.53

82.91

17.09

100 k mols of dry flue gas contions 17.09 k mols of O2 and 82.91 k mols of N2 which must have 100 come out from the air supplied, i.e., 82.91 × = 104.95 k mols; and the O2 accompanying this air 79 21 = 104.95 × = 22.04 k mols. 100 Hence, the difference in quantity of O2 supplied and used in the formation of water vapour, which is not appearing in dry flue gas analysis, will be (22.04 – 17.09) = 4.95, k mols ... The water vapour formed = 2 × 4.95 = 9.90 k mols. ... Mols of H2 present in the oil should also be = 9.90 k mols. (2H2 + O2 → 2H2O) The same quantity of the fuel oil contains 13.53 k mols, of C, (C + O2 → CO2). ... The fuel oil contains 13.53 mols of C ≡ 13.53 × 12 kg = 162.36 kg of C and 9.9 × 2 = 19.8 kg of H2. ... The composition by weight of the hydrocarbon fuel is as follows: 162.36 ×100 C = 182.16 = 89.13% 19.8 ×100 H = = 10.87% 182.16 Now,

Total O2 supplied O2 needed for C O2 needed for H

= 22.04 k mols, = 13.53 k mols, =   4.95 k mols.

104

BASIC ENGINEERING CHEMISTRY

... Total O2 actually needed} = (13.53 + 4.95) = 18.48 k mols ... Excess O2 = (22.04 – 18.48) mols = 3.56 k mols. ... % of Excess O2 (or air) used 3.56 ×100 = = 19.264% 18.48 Further, Total fuel oil burnt = 182.16 kg 100 Total air supplied = 22.04 × kg mols 21 100 = 22.04 × × 22.4 m3 21 = 2319.4476 m3 .. . Air used per kg of fuel oil 2319.4476 = = 12.733 m3. 182.16 Example 10. A example gave the following analysis: C – 84%, H – 4%, O – 4%, ash – 8%. The composition of the dry flue gas obtained by using the above coal is as follows CO2 – 9%, CO – 1%, O2 – x% and N2 – (90 – x)%. What will be the value of x? Solution. The problem can be solved by using the principle of oxygen balance. 100 kg of the coal sample contains:

Constituent

Kg



C

84



H   4



O   4



k Mols 84 =7 12 4 =2 2 4 = 0.125 32

100 mols of the flue gas contain



Constituent

k Mols

k Mols of C

k Mols of O

k Mols of N



CO2 CO O2 N2

9 1 x (90 – x)

9 1 — —

9 0.5 x —

— — — (90 – x)



Total

100

10

(9.5 + x)

(90 – x)

The coal burnt to produce 100 k mols of the flue ga 100 × 10 = 142.8571 kg 7 k mols of H present in 142.857 kg of coal =

= *

2 × 142.8571 k mols = 2.8571 k mols 100

For other worked out examples on combustion calculations, the students may refer the “Text Book on Experiments and Calculation in Engineering Chemistry” By S.S. DARA.

FUELS AND COMBUSTION

105

As the fuel does not contain any N2, it may be assumed that the entire N2 present in the flue gas is obtained from air only. Mols of O2 supplied for the combustion of 142.8571 kg of coal 21 = (90 – x) × = 0.2658 (90 – x) k mols 79 Further, the k mols of O2 supplied = k mols of O2 present in dry flue gas + k mols of O2 that combined with H2 to form water vapour. i.e., ... ... ...

1  0.2658 (90 – x) = (9.5 + x) +  2.8571 ×  2  23.922 – 0.2658 x = 9.5 + x + 1.4285 1.2658 x = 23.922 – 9.5 + 1.4285 x = 10.265.

Example 11. A petroleum gas has the following composition. Ethane – 5%, Propane – 10%, Butane – 40%, Butene – 10%, Isobutane – 30%, Propene – 5%. Calculate the volume of air required for complete combustion of 100 M3 of the gas and the percentage composition of the dry flue gases if 35% excess air is supplied Solution. The combustion reactions of the constituents of the fuel gas are: C2H6 + 3.5 O2 = 2CO2 + 3H2O





1 mol

C3H8





3.5 mols

+

2 mols

3 mols

5O2 = 3 CO2 + 4H2O

Propane

C4H10 + 6.5 O2 = 4 CO2 + 5H2O



Butane or Isobutane



C4H8



Butene



C3H6



Propene

+ 6O2 = 4CO2 + 4H2O + 4.5O2 = 3CO2 + 3H2O

Let 100 mols of the petroleum gas be the basis for calculation:

Constituent



Mols O2 reqd.

Products and Mols

C2H6   5

5 × 3.5 = 17.5

10

10 × 5 = 50

CO2 = 10; H2O = 15

C4H10

40 (Butane)

40 × 6.5 = 260

C3H8

Mols



30 (Isobutane)

30 × 6.5 = 195



10

10 × 6 = 60



C4H8

C3H6   5



Total

5 × 4.5 = 22.5 605 mols

CO2 = 30; H2O = 40

CO2 = 160; H2O = 200 CO2 = 120; H2O = 150

CO2 = 40; H2O = 40

CO2 = 15; H2O = 15 375 mols

Now, O2 required for complete combustion of 100 mols of the petroleum gas = 605 mols.

106

BASIC ENGINEERING CHEMISTRY

100 ... Air required for the complete combustion of 100 mols of the petroleum gas = 605 × 21 mols = 2880.9524 mols. .. . Air required for the complete combustion of 100 m3 of the petroleum gas = 2880.9524 m3. Dry flue gas analysis with 35% excess ai

135 If 35% excess air is used, the actual air supplied would be = 2880.9524 × mols = 3889.2857 100 mols. The N2 contributed by the above quantity of air 79 = 3889.2857 × = 3072.5357 mols. 100 35 Excess O2 supplied during the combustion of 100 mols of the petroleum gas = 605 × = 100 211.75 mols. ... The flue gases produced by the combustion of 100 mols of the petroleum contain

CO2 = 375 mols N2 = 3072.5357 mols O2 = 211.75 mols.

Total = 3659.2857 mols .. . The dry flue gas analysis would b

CO2 =

375 × 100 = 10.25% 3659.2857



N2 =

3072.5375 ×100 = 83.96% 3659.2857



O2 =

211.75×100 = 5.79%. 3659.2857

Example 12. The following data for flue gas analysis has been recorded in a Lancashire boiler with an economiser.

Constituent



CO2 CO O2 N2

Percentage Analysis of Flue gas Entering the Economiser Leaving the Economiser 9.3 0.4 10.2 80.1

7.2 0.3 12.6 79.9

Calculate the air leakage into the economiser per kg of the fuel stoked if the carbon in the fuel was 0.735 kg/kg of the fuel. Solution. Let 100 ml of dry flue gas be entering the economiser. If x mols of air leaked into the economiser during a certain time, the volume of the gases leaving the economiser will be (100 + x) mols. 9.3 ×100 which should be equal Hence, the % CO2 in the flue gas leaving the economiser = (100 + x) to 7.2 mols as per the data. 9.3 ×100 = 7.2 ... (100 + x) ... x = 29.16 mols = (29.16 × 28.97) kg = 846 kg.

Hence 846 kg of air leaked into 100 mols of dry flue gas entering the economise .

FUELS AND COMBUSTION

107

Fuel burnt to produce 100 mols of flue gas entering the economiser Mols of Cin flue gas      = × 100 Moles of Cin100 kg of fuel (9.3 + 0.4) = × 100 = 158 kg 6.125 73.5 (Since 100 kg fuel contains 73.5 kg carbon ≡ mols = 6.125 mols of C) 21 ... Air leakage into the economiser per kg of fuel 846 = kg = 5.42 kg. 168 Example 13. A coal sample of the following analysis by weight was used for a boiler trial: C = 82%, H = 4.9%, O = 4.8% The flue gas analysis by Orsat s apparatus gave the following data: CO2 = 10%, O2 = 8%, CO = 1.5% and N2 = 80.5%. The temperature of ambient air in the boiler house is 18°C and that of the flue gases is 320°C. Calculate the heat carried away by the excess of air per kg of coal. The average specific heat of air is 0.238. Solution. Vol. of constituents per    m3 of flue ga

Proportional weight

CO2 = 10/100 = 0.1 m3 O2 = 8/100 = 0.08 m3 CO = 1.5/100 = 0.015 m3 N2 = 80.5/100 = 0.805 m3

0.1 × 44 = 4.40 kg 0.08 × 32 = 2.56 kg 0.015 × 28 = 0.42 kg 0.805 × 28 = 22.54 kg Total = 29.92 kg

4.4 = 0.147 kg 29.92 2.56 Weight of O2/kg of flue gas = = 0.085 kg 29.92 0.42 Weight of CO/kg of flue gas = = 0.014 kg 29.92 22.54 Weight of N2/kg of flue gas = = 0.747 kg. 29.92 Total weight of carbon/kg of flue gas = (Wt. of C in C 2 + Wt. of C in CO/kg of flue ga 12 × 0.147 12 × 0.014  =  +  44 28   Weight of CO2/kg of flue gas =

= (0.04 + 0.006) kg = 0.046 kg. (Since 12 kg of C is present in 44 kg of CO2 and in 28 kg of CO respectively). But, weight of C per kg of fuel = 82/100 = 0.82 kg. ... Wt. of dry flue gas/kg of fuel burnt = 0.82/0.046 = 17.82 k 16 × 0.014 Weight of O2 required for the combustion of CO per kg of flue gas = = 0.008 kg 28 But, wt of O2/kg of flue gas = 0.085 kg as shown above ... Excess O2/kg of flue gas = (0.085 – 0.008) = 0.077 k

108

BASIC ENGINEERING CHEMISTRY

... Excess O2/kg of fuel burnt = 0.077 × 17.82 = 1.37 kg 100 ... Excess air/kg of fuel burnt = 1.37 × = 6.0 kg 23 ... Heat carried away by the excess air = M × S × (t2 – t1) Kcals

Mass

Sp. heat

rise in temp.

= 6 × 0.238 × (320 – 18) Kcals = 431.256 Kcals. Example 14. A gaseous fuel has the following volumetric composition : CH3 = 40%; C2H2 = 6%; C2H6 = 24%; C2H8 = 16%; O2 – 3.0%; CO = 5% and N2 = 6%. If 40% excess air was used for the combustion, calculate the air; fuel ratio and the analysis of dry flue gas Solution. The combustion reactions of the constituent gases in the fuel are: CH4 + 2O2 → CO2 + 2H2O

1 mol



2 mole

1 mol

2 mols

C2H4 + 3O2 → 2 CO2 + 2H2O C2H6 + 3.5 O2 → 2 CO2 + 3 H2O C4H8 + 6O2 → 4CO2 + 4H2O 1 CO + O → CO2 2 2 Let 1 mol of the fuel gas be the basis for calculation: Constituent & mols

Mols of O2 reqd.

Mols of Products formed



0.4 × 2 = 0.80

CO2 — 0.4; H2O — 0.8

C2H4 — 0.06

0.06 × 3 = 0.18

CO2 — 0.12; H2O — 0.12

C2H6 — 0.24

0.24 × 3.5 = 0.84

CO2 — 0.48; H3O — 0.72

C4H8 — 0.16

0.16 × 6 = 0.96

CO2 — 0.64; H2O — 0.64



CH4 — 0.4

O2 — 0.03           



CO — 0.05



N2 — 0.06



Total

= – 0.03      —

0.05 × 0.5 = 0.025

CO2 — 0.05

—      — = 2.775 mols

... O2 required/mol of the fuel gas = 2.775 mols or O2 required/m3 of the fuel gas = 2.775 m3 (Since mol% = m3% = Vol%) But 40% excess air was actually used. ... O2 actually supplied/m3 of fuel gas 140 = 2.775 × = 3.885 m3 100 ... Air actually supplied/m3 of fuel gas 100 = 3.885 × = 18.5 m3 21 ... Air : fuel ratio = 18.5 m3. Analysis of dry products of combustion/m3 of fuel gas CO2 = 1.69 m3

CO2 — 1.69 mols.

FUELS AND COMBUSTION

109

40 21 3 × m = 1.554 m3 100 100 [from excess air)  79   3 3 N2 = 18.5 ×  + 0.06  m = 14.675 m 100    (from air + fuel) Total volume of combustion products = 17.919 m3 Whence the volumetric composition of the flue gas i 1.69 ×100 CO2 = = 9.43% 17.919 1.554 ×100 O2 = = 8.67% 17.919 14.675 ×100 N2 = = 81.9%. 17.919 Example 15. A sample of coal contains C = 60%, H = 4%, O2 = 6%, N2 = 2% and ash = 28%. Calculate the % age composition of the dry products of combustion, assuming that 40% excess air is used. Solution.

O2 = 18.5 ×

Constituents

Amount percent in 1 kg of fuel

Combustion Reaction

Amount of O2 (in kg)

Carbon 0.60 C + O2 → CO2 32/12 × 0.6 = 1.6 Hydrogen 0.40 H2 + 1/2O2 →H2O   16/2 × 0.4 = 0.32 Oxygen 0.60 — — Nitrogen 0.20 — —

Amount of dry pradud 44/12 × 0.60 = 22 — — —



Amount of oxygen required = 1.6 + 0.32 – 0.06 = 1.86 kg 100 Amount of air = 1.86 × = 8.087 kg 23 140 = 11.322 kg Amount of air actually supplied = 8.087 × 100 Example 16. The percentage composition of coal was found to be as C = 54%, H = 6.5%, O = 3.0%, N = 1.8%, moisture = 17.3% and remaining is ash. This coal on combustion with excess of air gave 21.5 kg of dry flue gases per kg of coal burnt. Calculate the percentage of excess air used for combustion. Solution. Constituents Amount/kg of Combustion Amount of Amount of the fuel Reaction O2 (in kg) dry products (kg) 32 12 = 1.44 16 Hydrogen 0.065 H2 + 1/2O2 →H2   0.065 × 12 = 0.52 Nitrogen 0.018 — —



Carbon

Oxygen

0.54

0.03

C + O2 → CO2

—

0.54 ×

—

44 12 = 1.98

0.54 ×

— — —

110

BASIC ENGINEERING CHEMISTRY

Amount of oxygen required = 1.44 + 0.52 – 0.3 = 1.93 kg Amount of air required = 1.93 × 100 23 = 8.39 kg Example 17. A gaseous fuel has the following composition by volume H2 = 20%, CH4 = 25%, C2H6 = 16%, C2H4 = 9.5%, butene = 2.5%, CO = 4.0%, CO2 = 8%, O2 = 4% and N2 = 12%. Find the air required for perfect combustion of 1m3 of this gas. If 50% of excess air is used, find the volume analysis of the dry products of combustion. Solution. Constituent

Amount in 1m3 of fuel

Combustion reaction

Vol of Oxygen (m3)

H2 0.20 H2 + 1/2O2 → H2O 0.20 × 0.50 = 0.1 CH4 0.25 CH4 + 2O2 →2CO2   0.25 × 2 + 2H2O = 0.50 C2H6 0.16 C2H6 + 3.5O2→ 0.16 × 35 2CO2 + 3H2O = 0.56 C2H4 0.095 C2H4 + 3O2→ 0.095 × 3 2CO2 + 2H2O = 0.285 C4H8 0.025 C4H8 + 6O2→ 0.025 × 6 4CO2 + 4H2O = 0.15 CO 0.08 CO + 0.5O2→ 0.04 × 0.5 CO2 = 0.2 O2 0.04 — — N2 0.12 — —

Vol of Dry products (m3)

CO2 = 0.25 × 1 = 0.25 CO2 = 16 × 2 = 0.32 CO2 = 0.95 × 2 = 0.19 CO2 = 0.25 × 4 = 0.1 CO2 = 0.4 × 1 = 0.04 — —

Example 18. A petrol sample contains 84% carbon and 16% hydrogen by weight. Its flue gas compositional data by volume is as under CO2 = 12.1%, CO = 1.1%, O2 = 1.3% and N2 = 85.5% Calculate (i) Minimum air for complete combustion of 10 kg of petrol (ii) Calculate actual air supplied per kg of petrol (iii) Calorific values of the pet ol sample Solution. (i) 1 kg of petrol contain C = 840 g, H = 160 g Constituent

Amount

Combustion Reaction

C

840

C + O2 → CO2

H

85.5

H + 0.5O2 → H2O

Weight of O2 needed (g)



32 = 2240 12 16 160 × = 1280 2 Total = 3520

840 ×

Minimum wt. of air required for complete combustion of 1 kg of petrol

100

= 3520 × 15304.35 g = 15.30 kg = 23 3 (ii) 1m of flue gas contain CO2 = 0.121 m3, CO = 0.011m3, O2 = 0.013m3, N2 = 0.855 m3

FUELS AND COMBUSTION

111

Conversion : Gas (V)

Molecular mass (m)

V × m

0.121/m3

CO2 = 44 5.324 CO = 0.011/m3 28 0.308 O2 = 0.013/m3 32 0.416 N2 = 0.855/m3 28 23.940 E = 29.99

Mass/kg of flue ga 5.324/29.99 = CO2 = 0.1776 kg 0.308/29.99 = CO = 0.01027 kg 0.416/29.99 = O2 = 0.0139 kg 23.94/29.99 = N2 = 0.7983kg Total = 1.0000

Mass of excess O2 per kg of flue ga = Mass of O2 (in flue gas – required to converte CO to C = [0.0139 – 0.01027 × (16/28)] kg = [0.0139 – 0.0059] kg = 0.008 kg

2)

  12   12   Mass of C/kg of flue ga = Mass of CO 2 ×   + CO ×     44   28    12  12  kg − 0.0103 ×   kg = 0.1776 × 44  20  = 0.0440 kg =

Weight of C/kg of petrol Weight of C/kg of flue gas



0.84 kg = 19.09 kg 0.044 kg

=

\ Excess O2/kg of petrol burnt

100 kg = 0.664 kg 23 Hence, actual air supplied per kg of petrol burnt = 0.153 ×

= Mass of air (For combustion + Excess in the gas) = 15.30 + 0.664 kg = 15.96 kg 1 [8080 × 8.4 + 34500 × 16] Kcal/kg (iii) HCV = 100 = 12307 Kcal/kg Since the fuel does not contain hydrogen, so its HCV = LCV. Example 19. The percentage composition of petrol by weight was found to be C = 84% and H = 16%. Calculate (i) Minimum air required for complete combustion of 1 kg of petrol (ii) % composition by weight of dry products of combustion corresponding to minimum air per 1 kg of petrol burnt (iii) Calorific value of pet ol

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Solution. (i) 1 kg of petrol contain C = 840 g, H = 160 g Constituent

Amount in (g)

Combustion Reaction

C

840

C + O2 → CO2

H

85.5

H2 + 0.5O2 → H2O



Weight of O2 required 32 2240g = 12 16 160 × = 1280g 12 840 ×

Total = 3520 g

Minimum weight of air required for complete combustion of 1 kg of petrol 100 = 3520 × = 15304.35 g = 15.30 kg 23 (ii) Weight of dry products of combustion 44 (i) CO2 = 840 3079.99 g 3.08 kg ×= = 12 77 × 15.30 = 11.78 kg (ii) N2 = 77% of weight of air = 100 Total weight of dry products of combustion 3.08 + 11.78 = 14.86 kg 3.08 × 100 % of CO2 = = 20.73 14.86 11.78 × 100 = 79.27% % of N2 = 111.86 1 [(8080 × 84) + (34500 × 16)] (iii) GCV = 100 1 [678720 + 552000] = 12307.2 Kcal = 100 NCV = GCV – 0.09H × 587 Kcal/kg = 12307.2 – 0.09 × 16 × 587 = 12307.2 – 845.28 = 11461.92 Kcal. Example 20. A gaseous fuel has the following composition by volume CH4 = 5%, H2 = 20%, CO2 = 6%, CO = 25% and rest nitrogen. If 20% excess air is used for combustion, then calculate volume of air per m3 of fuel and composition of dry fuel gas. Solution. 1 m3 of gaseous fuel contains CH4 = 0.05 m3, H2 = 0.20 m2, CO = 0.25 m3, CO2 = 0.06 m3 and N2 = 0.46 m2 by difference) Constituent Amount (m3) 0.05 CH4 H2 0.20 CO 0.25 CO2 0.06 0.46 N2

Combustion Reaction

Volume of O2 required

CH4 + 2O2 0.05 × 2 = 0.10 → CO2+ H2O H2 + 0.5O2 0.20 × 0.5 = 0.100 → H2O CO + 0.5O2 0.25 × 0.5 = 0.125 → CO2 — — — —

Total = 0.325

Vol of dry product (m3) CO2 = 0.005 × 1 = 0.5 — CO2 = 0.25 × 1 = 0.25 CO2 = 0.06 from N2 = 0.46 fuel CO2 = 0.36 N2 = 0.46

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Volume of air required per

m3

of gaseous fuel (using 20% excess air) 3  100   120  1.85% m3 = 0.325m ×  × =  21   100  Calculation of dry products of combustion CO2 = 0.36 m3, O2 = 20% of 0.325 m3  20  3 0.065 m3 =   × 0.325 m =  100  0.46 m3 (of gaseous fuel) + 1.857 m3 (77/100) of air 1.027 m3 (0.36 + 0.065 + 1.927) m3 2.352 m3 0.36 × 100 = 15.306% CO2 = 2.352 1.927 × 100 = 81.93% N2 = 2.352 0.065 × 100 = 2.764% O2 = 2.352

N2 Total volume of dry products

= = = =

QUESTIONS 1. How is calorific value of a solid fuel determined using a bomb calorimeter The following data is obtained in a bomb calorimeter experiment : Mass of the fuel pellet = 0.85 g Mass of water taken in the calorimeter = 2000 g Water equivalent of the calorimeter = 540 g Difference in final and initial temperature = 1.9° Cooling correction = 0.041°C Fuse wire correction = 3.8 calories Acids correction = 48.8 calories Calculate the net calorific value of the fuel if it contains 3.5% hydrogen and 1.2% oxygen 2. Distinguish clearly between the following : (a) Proximate and ultimate analysis (b) Coal and coke (c) Coking coals and caking coals (d) Octane number and cetane number (e) Thermal and catalytic cracking 3. How is the proximate analysis of a coal conducted and what is its significance in determining the utility of a coal for a particular purpose ? 4. Fill up the blanks : (a) 1 Calorie = ______________ Joules = ______________ ergs (b) 1 B. Th. U = ______________ Joules = ______________ergs (c) 1 B. Th. U = ______________ Cals = ______________ ergs (d) 1 K cal = ______________ B. Th. U. (e) 1 Kcal/kg = ______________ B. Th. U./Ib.

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(f) 0.1077 B. Th. U./ft3 = ______________ Kcal/m3. (g) Net calorific value = Gross calorific value mines _____________ (h) QC.V. = QC.P. + ______________. Heat of Combustion + ________ (i) Theoretical flame temperature = 5. Write short notes on the following : ________________ (a) Criteria for selecting a fuel. (b) Selection of coal. (RGPV Bhopal 2009) (c) Characteristics of coal. (d) Commercial types of coal. (e) Blending of coal. (f) Storage of coal. 6. What are the requisites of a metalurgical coke ? 7. What are the types of carbonization and for what purposes they are employed ? 8. Describe the Otto-Hoffman’s process for preparing coke ? What are its advantages over the earlier methods ? (RGPV Bhopal 2006) 9. Write informative notes on the following : (a) Refining of petroleu (b) Thermal and catalytic processes. (c) Reforming 10. What is cracking and what for it is used ? What are the types of cracking ? Describe the working of fixed bed catalytic cracking (RGPV Bhopal 2009) 11. What is meant by knocking in a petrol engine and what is it due to ? What is octane number of a petrol ? How is knocking related to chemical structure of the constituents of petrol and how can it be reduced? 12. How do you explain knocking in a diesel engine ? How can it be controlled ? What is cetane number ? 13. Write informative notes on the following : (a) Natural gas (b) Liquid petroleum gas (LPG) (c) Coal gas and coke oven gas (d) Power alcohol (e) Cracking (RGPV Bhopal 2006) 14. Complete the following sentences with appropriate answers : (a) Peat is not considered as an economic fuel because_________. (b) Lignite is susceptible for spontaneous combustion because_________. (c) _________are the most widely used coals in the world. (d) Anthracite is used for drying malt and hops because_________. (e) Coking coals are blended with non-caking coals because_________. (f) Low temperature carbonization yields coke which is suitable for_________. (g) High temperature carbonization yields coke which is_________. (h) Highest ranking coal is _________. 15. A producer gas has the following composition by volume : CH4 = 35%, CO = 25%, H2 = 10% CO2 = 10.8%, N2 = 50.7%. Calculate the theoretical quantity of air required per cubic meter of the gas. If 22% excess air is used, find the percentage composition of the dry products of combustion.

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16. A liquid hydrocarbon fuel contains 89.4%C and 10.6%H by weight. If 100 kg of the fuel is burnt, calculate (a) the theoretical amount of air required and (b) the volumetric composition of the products of combustion if 20% excess air was used. (Ans. : 48.1 K mol; CO2 = 12.34%, H2O = 8.78%, N2 = 75.53% and O2 = 3.35%) 17. A coal sample gas the following analysis by weight : C = 88%, H = 4%, S = 1%, 0 = 5%, ash and moisture 1% each. The coal sample on combustion in a boiler gave a flue gas having the following composition by volume CO2 = 11%, O2 = 7, CO = 1.5%, SO2 = 0.5% and N2 = 80%. Calculate (a) the weight of air required for complete combustion of 1 kg of the coal (b) weight of the wet flue gases per kg coal (c) percentage of excess air used and (d) percentage of carbon burnt to CO. (Ans. : 11.42 kg/kg of coal; 18.0 kg/kg of coal; 49.4% and 12% C burnt to CO) 18. How does knocking occur in I.C. engines. How can it be prevented ? 19. Establish the relationship between knocking in I.C. engines and the nature and molecular structure of the constituents in petrol and diesel fuel. 20. Describe a method for catalytic cracking of petroleum fractions and discuss the advantages of cat-cracking. Indicate the mechanism of catalytic cracking. 21. Define gross and net calorific values. Calculate the approximate calorific value (by Dulong’s formula) of a coal sample having the following ultimate analysis : C = 80%, H = 3.5%; S = 2.8%; O = 5.0%; N = 1.5% and ash = 7.2%. 22. A hydrocarbon fuel on combustion gave the flue gas having the following composition by volume : CO2 = 15%; O2 = 6.5%; and N2 = 80.5%. Calculate (a) the composition of fuel by weight and (b) percentage of excess air used. 23. A producer gas has the following composition by volume; CO = 30%; H2 = 12%; CO2 = 4%; CH4 = 2% and N2 = 52%. When 100 m3 of the gas is burnt with 50% excess air used, what will be the composition of the dry flue gases obtained 24. A coal sample has the following composition; C = 90%; H = 3.5%; O = 3%; S = 0.5% and N2 = 1%; the remaining being ash. Calculate the theoretical volume of air required at 27°C and 1 atm pressure when 100 kg of the coal is burnt. (Ans. : 1100 m3) 25. The composition by weight of a coal sample is : C = 81%; H–; O = 8; S = 1; N = 1.5%; Ash = 30%. (a) Calculate the amount of air required for the complete combustion of 1 kg of coal (b) Calculate the gross and net calorific values of the coal sample. Give that the calorific values of C, H and S are 8060 Kcals/kg; 34000 Kcals/kg and 2200 Kcal/kg respectively. (Ans. : 10.8 kg; 7.910 Kcal, 7.667 Kcal/kg) 26. A coal gas has the following volumetric composition; CO2 = 2.5%; C2H4 = 3.5%; CO = 8%; H2 = 50%; CH4 = 34% and N2 = 2%. Calculate the gross calorific value of the gas at N.T.P. Given that 1 kg of C while burning to CO develops 2440 Kcals; while burning to CO2 develops 8060 Kcals of heat; and 1 kg of H2 while burning to H2O develops 34,400 Kcals 1 kg mol of gas at N.T.P. occupies 22.4 m3. (Ans. : 5,848 Kcals/m3) 27. A coke oven gas of the following volumetric composition was used for firing gas retorts CO2 = 1.4%; CO = 5.1%; O2 = 0.5%; H2 = 57.4%; CH4 = 28.5%; C2H4 = 2.9% & N2 = 4.2%. The dry flue gases on analysis by Orsat s apparatus gave the following results : CO2 = 15%; CO = 0%; O2 = 2.5%; N2 = 82.5%. Calculate the excess air supplied for burning 100 m3 of the gas. (Ans. : 32.4 m3) 28. A coal containing 62.4% C; 4.1% H; 6.9% O; 1.2% N; 0.8%S; 15.1% moisture and 9.5% ash was burnt in such a way that the dry flue gases contained 12.9% CO2; 0.2% CO; 6.1% O2; and 80.8% N2. Calculate (a) the weight of air theoretically required per kg of coal (b)

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Percentage of excess air used and (c) the weight of the dry flue gas produced per kg of coal. (Ans. : 8.3246 kg; 41.0984%; 12.0306 kg) 29. A gasoline sample contains 86% C and 14% H2. If the air supplied is only 95% of the theoretically required value, calculate the dry flue gas analysis. Assume that all the H2 underwent complete combustion and no carbon is left free. [Hint : For every 0.5 mol of O2 supplied less, 1 mol of C is converted into CO] (Ans. : CO2 = 13.47%; CO = 2.36% and N2 = 84.17%) 30. Explain the following : (a) The knocking tendency of petrol or diesel oil can be predicted on the basis of the nature and molecular structure of its constituent compounds. (b) All coking coals are caking but all caking coals are not coking. 31. A furnace utilises a mixture of blast furnace gas and coke oven gas as a fuel. The gas mixture is fired along with air to obtain the required temperature. The analysis of the individual gases used and the flue gas evolved due to the combustion of the above gas mixture are given on dry basis and in volume percentage as below : Constituent

Coke oven gas

Blast furnace gas

Air

Flue gas



2.4 2.0 7.2 30.8 51.38 6.3

13.4 0.2 24.2 — 3.0 59.2

— 21.0 2.0 — — 79.0

13.5 6.9 0.4 — — 79.2

CO2 O2 CO CH4 H2 N2

Calculate (a) the ratio of blast furnace gas to coke oven gas used, and (b) the percentage excess of air used. 32. (a) Discuss briefly octane and cetane numbers. How is octane number related to chemical constitution of various hydrocarbons ? (b) A petrol sample contains 84% C, 16% H by weight. Its flue gas compositional data by volume is as under : CO2 – 12.1%; CO – 1.1% O2 – 1.3% and N2 – 85.5% Calculate minimum air for complete combustion of 1 kg of petrol. Also calculate actual air supplied per kg of petrol. Also, calculate actual air supplied per kg of petrol and the calorific value of the petrol sample. Or (a) What are the characteristics of metallurgical coke ? How is coke manufactured by Otto-Hoffman process ? (b) A sample of dry coal has the following composition by weight : C = 84%; H = 5%; O = 6%; N = 2% and the rest is ash. If 50% excess air is supplied, calculate the percentage composition by weight of the dry products of combustion and the calorific values of the coal sample (RGPV Bhopal 2001) 33. A boiler is fired with a coal having the following percentage composition : C = 75%; H = 9%; S = 2%; O = 4%; N = 3%; Ash = 7%. Calculate : (i) Gross and net calorific value of 1 kg of coal. (Latent heat of steam = 587 cals/gm (ii) Minimum theoretical air required for combustion of 1 kg of coal (by weight and by volume).

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117

(iii) Percentage composition of dry flue gas if 25% excess air is used (Nagpure University, 2001) 34. (a) What is the significance of ultimate analysis of coal ? How is this analysis carried out in the laboratory? (b) “All working coals are caking coals but all caking coals are not coking coal” Justify this statement. (Nagpur University, 2001) 35. (i) Define a chemical fuel (ii) What is meant by calorific value of a fuel ? What is the difference between gross calorific value and not calorific value (iii) Why are gaseous fuels more advantageous than solid fuels ? (U.P. Technical University, 2001) 36. (a) 0.5 gm of a sample of coal was used in a bomb calorimeter for the determination of calorific value. The calorific value of the coal was found to be 8600 cals/gm. The ash formed in the bomb calorimeter was extracted with acid and the solution. The barium sulfate precipitated formed was filtered, dried and weighed. The weight of the precipitate was found to be 0.05 gm. Calculate the percentage of sulfur in the coal sample. 37. Calculate the volume of air required to complete combustion of 1 cu.m. of gaseous fuel having the following composition : CO = 46%, CH4 = 10%; H2 = 40%; C2H4 = 2%; N2 = 1%; and remaining being CO2.

(Mumbai University, 1997)

38. What are antiknocking agents ? Give examples.   39. (a) A sample of coal contains 92% C, 5% H and 3% ash. On analysis using Bomb Calorimeter, the following data were obtained : Weight of coal burnt = 0.95 gms. Weight of water taken = 700 gms. Water equivalent of the calorimeter = 2000 gms Rise in temperature = 2.48°C. Cooling Correction = 0.02°C. Fuse wire correction = 10 cals Acid correction = 60 cals. Calculate the net and gross calorific value of the coal (b) The percentage composition of a coal sample by mass is as follows : C = 81%; H = 5%; O = 12%; D = 1%; N = 1%. Calculate the quantity of air required for complete combustion of 1 kg of the coal. (Mumbai University, 1999) 40. Explain the method of estimation of sulfur in coal. (Mumbai University, 1999) 41. (a) Calculate the gross and net calorific value of a fuel having the following composition : C = 78%; H = 7.5%; O = 12.5%, S = 2%. (b) The percentage composition by mass of a coal sample is as follows : C = 80%; H = 6%; O = 8%; S = 1,5%; N = 1%; Ash = rest. Calculate : (i) Amount of oxygen for complete combustion of 1 kg of fuel, and (ii) Gross and net calorific value of the fuel (Mumbai University, 1999) 42. (a) Describe the methods of estimating the following elements : (i) Percentage of sulfur in coal. (ii) Percentage of C and N in coal. (b) What is carbonisation ? What are coke even gases ? Describe the steps involved in the recovery of precious by products from the coke oven gases.

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(c) Explain the term fermentation. Discuss the manufacture of alcohol by fermentation of sugar. (Mumbai University, 1999) 43. What is cracking ? What are the advantages of catalytic cracking over thermal cracking ? (Mumbai University, 1999) 44. (a) Discuss with the help of a labelled diagram, the determination of calorific value of solid fuel by Bomb calorimeter. (RGPV Bhopal 2009) (b) The fuel oil used in a diesel engine has 85.8% C; 12.5% H2; 1.5% S and 0.2% O2. Analysis of the products of combustion by Orsat’s apparatus shows 8% CO2 and 10% O2 by volume. Calculate the weight of air supplied per kg of oil. (RGPV Bhopal 2003) 45. (a) What are the economic aspects of Otto-Hoffman’s coke oven method of manufacture of coke. (b) What is ultimate analysis of coal and its significance (c) Discuss in brief the proximate analysis of fuel. (RGPV Bhopal 2006) (d) A sample of coal was found to have the following percentage composition : C–81%; H2–5%; S–0.5%; N2–3%; Ash – 10.5%. Calculate the percentage composition by volume of the flue gases, when air, 30% in excess by weight is supplied for combustion. (RGPV Bhopal 2003) 46. (a) Describe the Otto-Hoffman’s process for the manufacture of coke. (b) What is cracking ? Describe the fluidized bed catalytic cracking process (Nagpur University, 2002) 47. (a) Describe flue gas analysis by Orsat s apparatus and state its significance (b) Calculate the volume of air in m3 of coal gas containing 16.5% CO, 51.5% H2, 18.5% CH4, 3.5% CO2, 0.5% O2 and 9.5% N2. Also calculate the volume of air actually used when the dry flue gas contains 8% C 2, 7.3% O2 and 84.7% N2 by volume. (Nagpur University, 2002) 48. A producer gas has the following composition by volume : CO – 81%; H2 – 5%; N2 – 3%; S = 0.5% and ash 10.5%. (RGPV Bhopal 2006) Calculate the percent age composition by volume of the flue gas, when air 30% in excess by weight is applied for combustion. 49. Significance of ultimate analysis of coal 50. Write informative notes on the following : (i) Flue gas analysis and its significanc (ii) Octane number and cetane number and their relationship with the chemical constitution of the fuel constituents. (Nagpur University, 2002) 51. Establish the relationship between knocking in I.C. engines and the nature and molecular structure of the constituents in petrol and Diesel fuel. 52. A sample of coal was analysed as follows: 0.9824 g of an air dried coal sample was weighed in a silica crucible. After heating for 1 hr. at 105-110°C, the dry coal residue weighed 0.9668 g. The residue was covered with a vented lid and then heated strongly for exactly 7 min. at 950°C ± 20°C. The residue weighted 0.7900 g. The crucible was then heated strongly in air until a constant weight was obtained. The last residue was found to weight 0.1200 g. Calculate the proximate analysis of the coal sample. (RGPV Bhopal 2009)

UNIT

3

PART–A

Lubricants

3.1 INTRODUCTION Friction and wear arise from the relative motion of two or more surface in contact. Friction is a force of resistance to the relative motion of two contacting surfaces. Wear results when this resistance is overcome by applied forces. The substances which are used to remove friction between two or more moving surfaces is known as Lubricant. The word ‘‘lubricant’’ derive from a latin word lubricus which means slippery. Hence lubricants may be define as a substance which introduced between moving part of machinery make the surface slippery and reduce friction, eliminates asperities and prevent cohesion. The phenomeon is known as lubrication. 3.2 SURFACE ROUGHNESS Even the best polished metal surface, when examined under a microscope giving high magnification (e.g., electron microscope), is seen to be more or less rough, having peaks and valleys of different heights and depths as shown in Fig. (3.1a). This is known as surface roughness. The highest peaks are called “asperities”. When two flat surfaces are placed over one another, the asperities of the upper surface rest on those of the lower surface. Thus, the surfaces make contact at these points only while over most of the area they are separated. Thus the real area of contact is very much smaller than the apparent area of contact (Fig. 3.1b).

Fig. 3.1. Surface roughness and intimacy of contact.

3.3 LUBRICATION Lubrication is a process in which we remove friction and wear between two or more moving surfaces by applying a substance between them. And the substance is known as lubricant. 3.4 FUNCTION OF LUBRICANTS The important function of Lubricants are given below: 1. It reduces friction and wear

2. It acts as a coolant by dissipating the frictional heat generated because of the rubbing 119

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120



3.



4. 5.



6.

surfaces. Thus the expansion of metal by local frictional heat and the resultant deformation and damage will be reduced. In internal combustion engines, the lubricant also acts as fuel gasket between the piston and cylinder wall at the compression rings and prevents the leakage of gases at high pressure in the combustion chamber, thus minimizing the power loss. It prevents the entry of moisture, dust and dirt between the moving parts. It acts as a cleaning agent and as a scavenger to wash off and transport solid particles produced in combustion or wear. In aircraft, the lubricating oil may be used as a hydraulic fluid to change the pitch of the propeller or to operate other mechanisms. The viscosity index of the oil is the important property in such cases.

3.5 MECHANISM OF LUBRICATION 1. Fluid or Hydrodynamic Lubrication: When the distance between two moving surfaces is sufficient, than in this case we apply fluid or hydrodynamic lubrication. In this type of lubrication we apply a thick film (~ 1,000 A°) of lubricant between two moving surfaces so that the surface to surface contact and welding rarely occur. Thus, the friction is reduced and prevents wear. A small friction (if any) is only due to the internal resistance between the particles of lubricant moving over each other.

Bearing lining

Lubricant

Shaft Fig. 3.2. Journal bearing hydrodynamic lubrication.

Coefficient of Friction f = F /W where

F = Force required to cause motion and



W = Applied load

Here the value of coefficient of friction is very lo . It ranges from 0.5 to 1.5. In hydrodynamic lubrication (Fig. 3.2), a film of the lubricating oil covers the shaft as well as the bearing surfaces. The oil film is sufficiently thick to cover the irregularities of the surfaces and the metal surfaces do not come into contact with each other. Thus practically there is no wear. The resistance to movement is only due to the resistance between the particles of the lubricant moving over each other. Thus, in hydrodynamic lubrication, the lubricant chosen must have enough viscosity so that the bearing force due to the rotation of the journal, will be sufficient to drag enough oil between the journal and the bearing and at the same time, it should not be too much viscous to offer resistance for the free motion of the lubricant particles over each other. The efficiency of lubrication by this mechanism depends on the design of the bearing, the loading, the rate of rotation of the shaft and on the viscosity of the lubricant. For a given load and rate of rotation, the greater the lubricant viscosity, the greater the hydrodynamic pressure developed. However, there is a practical limit to the viscosity which can be used, since a large amount of energy would be needed to circulate and maintain a very viscous lubricant film Hydrodynamic lubrication is maintained in case of delicate mechanical systems such as watches, sewing machines and scientific instruments The selection of a suitable fluid lubricant is complicated by changes of viscosity with temperature. The viscosity of a typical hydrocarbon oil decreases as the temperature rises, so that an oil which is satisfactory when an engine is cold may become too ‘thin’ to maintain an adequate lubricant film at normal running temperatures. In order to maintain suitable viscosity of the oil for

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121

adequate lubrication in all seasons of a year, ordinary hydrocarbon lubricants are usually blended with selected long chain polymers. Hydrocarbon oils are considered to be satisfactory lubricants. Their viscosity increases with increasing molecular weight. Suitable blends of appropriate fractions from petroleum refining plants can be selected for different applications. However, these fractions generally contain small quantities of unsaturated compounds which will oxidize under operating conditions, forming gums or lacquers. Antioxidants, such as aminophenols, must therefore be blended with these oils. Further, these oils may also undergo some decomposition in practice with the formation of solid carbon particles. In order to keep these carbon particles in suspension in the lubricating oil, organometallic “detergent” compounds are generally added. (ii) Thin Film or Boundary Lubrication When the thickness between two moving surfaces is very much smaller, than in that case we use boundary or thin film lubrication. Here, a thin layer of lubricant is absorbed on the surface of the metal which avoid metal to metal contanct. The contact between the metal surface is possible by taking out of lubricating oil film. When this happens the load will be taken on the high spots on of the journal and bearing, and the two surfaces tent to become welded by appreciable heat that is generated. This prevents the motion as the two surfaces adhere together. This is called seizure. If the motion takes place with the removal of some metal from one of the surfaces the result is called scuffing. In practice seizure and scuffing are delayed because metal tend to form film on their surface by chemical action leading to adsorption, which prevents the damage metal to metal contact temporarily. Boundary lubrication occurs whenever a continuous fluid film cannot be maintained which happens (a) when a shaft from rest comes into operation, (b) the speed is very low, (c) the load is very high and (d) the viscosity of the oil is very low. Under such conditions, a fluid film cannot be maintained between the surfaces, and if lubrication has to be maintained, it is essential that a layer of the lubricant be adsorbed on the rubbing surfaces by physical and/or chemical forces. Then the metal surfaces come very close to each other but yet are separated by a layer of the lubricant.

Fig. 3.3. Boundary Film Mechanism

(iii) Extreme Pressure Lubrication If the rubbing surfaces are subjected to very high pressure and speed, excessive frictional heat will be generated. The high local temperatures thus produced at the surfaces render the commonly used liquid lubricants ineffective due to decompositon or evaporation. In order to provide effective lubrication under these high local pressures and temperatures and onerous conditions, special additives called “extreme pressure additives” are used along with the lubricants. The active chemicals that are in general use are compounds of chlorine (e.g., chlroinated esters), sulfur (e.g., sulfurized fats and oils), and phosphorous (e.g., tricresol phosphate). Through chemical reactions with the metal surfaces, (at the prevailing temperatures) these additives form solid surface films

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of metallic chlorides, sulfides and probably phosphides. These boundary films have a relatively low shear strength (e.g., iron chloride, 0.2, iron sulfide 0.5) so that rubbing between the interacting surfaces occurs in the additive film and thus protects the underlying metal. The melting points of the extreme pressure layers are high (e.g., iron chloride 1,200°F; iron sulphide 2,150°F) so they will remain attached to the base metal even under extreme conditions of temperature. Further, these lubricants have an additional advantage that if the low shear strength films formed on the moving parts are broken by the rubbing action, they are immediately replenished. Extreme pressure additives are also needed in the “cutting fluids” used as lubricants in machining of tough metals, a continuous stream of the fluid, which may contain a hydrocarbon oil, a small amount of fatty acid as a boundary lubricant, and an organic chloride or sulfide additive, is fed to the cutting surface. In light cutting operations, a simple oil-water emulsion may be adequate. 3.6 CLASSIFICATION OF LUBRICANT The lubricants are classified according to their physical states as given below

L U B R I C A N T S

1

2 3 4

5

(a) Vegetable oils e.g. Castor oil, palm oil etc.

L I Q U I D S

(a) Animal oils e.g. Whale oil, Lard oil, Tallow oil etc. (c) Mineral oil e.g. Petroleum Products such as Paraffins, naphthalenes etc. (d) Blended oils e.g. Mineral oils + Vegetable or animal oils. Lubricating oils + additives

SOLIDS

e.g. Graphite, Talc, Chalk, Molyhbdenum disulphide etc.

SEMI-SOLIDS e.g. Greases, Vaselines, waxes etc. SYNTHETIC e.g. Silicones, chloro carbons, polypropylenes etc.

E M U L S I O N S

(a) Oil in water type (O/W) e.g. cutting emulsion

(b) Water in oil type (W/O) e.g. cooling liquids

3.6.1 Liquid Lubricant. They are of following types: (a) Animal fats and oils. Animal fats and oils are extracted from the crude fat by a process called “rendering” in which the enclosing tissue is broken by treatment with steam or with the combined action of steam and water. Important animal fats and oils are : Lard and lard oil Tallow and tallow oil Neatsfoot oil Fish oils

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Sperm oil Whale oil Seal oil Vegetables oils are obtained by crushing the seed. Some of the important vegetable oils used as lubricant are: (a) Olive oil (or sweet oil) (b) Castor oil (c) Rape oil or colza oil (d) Palm oil and Palm Kernel oil (e) Cotton seed oil Refining of fatty oils. The animal and vegetable oils require further treatment before use. The oil is cooled until stearine separates out. The oil is then filtered through animal carbon or treated with fuller’s earth for removing the colour and brightening the oil. Free fatty acids are neutralized by adding calculated quantity of caustic soda. Chemical refining consists of treatment with H2SO4 followed by removal of traces of acid by washing with water. H2SO4 removes suspended impurities by carbonizing and causing them to coagulate and settle out. The oil is finally filtere (b) Mineral Oils: They are lower molecular weight hydrocarbons with 12 to 50 carbon atoms. They are obtained during distillation of petroleum. Thus they are cheap, available in abundance and are stable under service conditions. Therefore they are widely used. They have less oilness. Thus the substance like oleic acid, stearic acid are add to increase their oilness. Purification: The impurities are generally removed by the following method. (i) Dewaxing:The petroleum oils contain a number of waxes in the dissolved form and these make it unfit for use, because waxes increase the pour point. For dewaxing mix the oil with proper volume of a suitable solvent (e.g. propane, mixture of benzene and acetone, trichloroethylene, a mixture of benzene and ethylene dichloride) and refrigerating the mixture to the required temperature when the wax in the oil precipitates out. The oil and wax are separated using continuous filters or centrifuge. The oil is separated from the solvent by distillation. (ii) Acid Refining of Lubricant: The dewaxed lubricating oil fractions still contain naphthenic, asphaltic and other undesirable constituents, which must be removed to produce a finished lubricating oil. This is done by thoroughly agitating the oil with the required amount of concentrated sulfuric acid, which acts as a solvent for some of the constituents and react chemically with others to form a tarry sludge. The oil is separated and neutralized with ammonia or caustic soda or by being brought in contact with finely divided fuller’s earth at 108 to 230°C. The latter treatment also helps to decolourize and stabilize the oil. The sludge can be used as a fuel. Percolation filtration through fuller’s earth is the final refining step in many case (iii) Solvent Refining of the lubricating oils: In this process, a suitable solvent is selected in which the solubility of the undesirable impurities such as naphthenic, asphaltic, and resinous constituents is more than that in the oil. The important solvents in use today include furfural, phenol, pp´- dichloro-diethyl ether (chlorex) and nitrobenzene. Mixtures of SO2 and benzene; propane (which dissolves the desired paraffinic fraction and decreases the solubility of undesirable constituents in this fraction) and a mixture of phenol and cresol (which is immiscible with the oil-propane solution) is a good solvent for the naphthenic, asphaltic and resinous constituents are also used. Mineral oils have largely replaced animal and vegetable oils because the former are cheaper, available in bulk quantities, stable under service conditions and can be re-refined after use

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Blended or compound oils: Desirable characteristics of lubricating oil can be improved by adding small quantities of various additive. The oil thus obtained are known as Blended or compound oils. They are of two types: (i) Chemically Active Additive: They are chemically interact with metals e.g. detergents, antiwear agent, dispersant oxidation inhibitors etc. (ii) Chemically Inert Additives: They improve the physical properties which are critical to the effective performance of the lubricant. e.g.: viscosity index improver, foam inhibitors, emulsifie , demulsifier etc Table 3.1 Additives and their functions

Sr. No. Name of the Chemical used additive (1) (2) (3) 1. Detergents and Normal or basic calcium deflocculent and barium salts of Phos- phonates and sulfonates, some salts of phenol, etc. 2. Dispersants Polymers such as nitrogen containing Polymethacry- lates, alkyl succinimides and high molecular weight amines and amides. 3. Anti-oxidants Phenols, amines, organic sulfides, o ganic phosphides, etc. 4. Corrosion inhibitors Organo metallic compounds, zinc dithiophosphates, sulfurized terpenes, phosphorized terpenes. 5. Rust inhibitors Amine phosphates, alkyl succinic acids, fatty acids, sodium and calcium petro leum sulfonates. 6. Anti-wear additives Zinc dialkyl dithiophos- phate, Tricresyl phosphate, alkyl earth phenolates. 7. Metal deactivators Triaryl phosphites, sulfur compounds, diamines. 8. Extreme pressure Sulfurized fats, chlorinated additives hydrocarbons, lead salts of organic acids, organic pho- sphorous compounds, metallic soaps such as lead naphthenates.

Functions (4) Reduce or prevent deposits in engines operated at high temperature. Prevent or retard sludge formation and deposition under low temperature operating conditions. Retard the oxdiation of oils, Minimize the formation of resins, varnish, acids, sludges and polymers. Protect bearings and other metal surfaces from corrosion. Protects ferrous metals from rusting. Reduce rapid wear in steelon-steel applications. Stop the catalytic effect of metals on oxidation and corrosion. Adsorbed on the metal surface or react chemically with the metal forming a surface layer of low shear strength which prevent tearing up, seizure and welding of the metals.

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Name of the additive (2)

125 Chemical used

Functions

(3)

(4)

9. Oiliness Fatty acids, fats and fatty Increase the strength of the amines, vegetable oils oil film and prevent the rupture of the oil film 10. Polymeric thickeners Long chain polymers such Reduce the rate of change and viscosity index as polyisobutylene, poly- of viscosity with improvers. styrene or alkyl styrene temperature. polymers, long chain alkyl acrylates, polyesters, “Plexol”, “Paratone” and “Exanol” are the trade names of such products commercially available. 11. Pour-point depressants Wax alkylated naphthalene, Lower the pour-point of the wax alkylated pheols, poly- oil. methacrylates. “Para flow” and “Santapour” are the patented commercial products available. 12. Antifoam additives Silicone polymers, oil in- Prevent formation of stable soluble liquids like glycols foams. and glycerols 13. Emulsifier Sodium salts of sulfonic Make mineral oil miscible acids, sodium salts of with water or help the for organic acis, fatty amine mation of emulsions of salts, nonionic emulsifiers lubricating oils with water. such as monoesters of polyhydric alcohols. 14. Thickness Soaps, polyisobutylene and Provide the oil with greater improvers polyacrylate polymers cohesion. 3.6.2 Solid Lubricant In boundary lubrication we use solid lubricant. They are used either in dry powder form or mixed with oil or water. The usual coefficient of friction of solid lubricant is in between 0.005 to 0.01. Solid lubricants are used in situations such as: 1. Heavy machinery working on a crude job at very high loads and slow speeds. 2. Where a liquid or a semi-solid lubricant film cannot be maintained or their presence is undesirable as in the case of commutator blades of electric motors and generators. 3. Where parts to be lubricated are not easily accessible, and 4. Where the operating temperatures and pressures are too high to use the easily combustible liquid lubricants. Many solid lubricants contain grains of particles which may damage delicate parts of the machinery. Hence, they are used only in special cases similar to those mentioned above. The commonly used solid lubricants include graphite, molybdenum disulfide, talc, mica, french chalk, boron nitride, etc. Amongst these, the most widely used solid lubricants are graphite (in colloidal form) and molybdenum disulfide (Mo 2), both having laminer structure.

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Structure of Graphite: It consists a number of hexagonal plates. In which each carbon atom present in sp2 hybridization state. Further, each carbon atom is linked by covalent bond to three other atoms but its distance from the fourth one is more than double. Hence, this fourth Valency bond is not fixed but moves about These plates are separated from each other by 3.4 Å and are held together by weak Vander Waal’s forces. So that even a small force is sufficient to slide the layers parallel to each other. Hence it is very soapy to touch, non-inflammable and not oxidized in air below 375°C. It can be used in the powdered form or as suspension in oil or water with tanning as emulsifying agent. Graphite dispersed in water is known as aquadag and dispersed in oil is known as oildag. Aquadag is useful where a lubricant free from oil is needed, while oildag is used in internal combustion engines, because it gives a tight-fit contact by forming a film between the piston rings and the cylinder. Generally Graphite is a used as lubricant in air-compressors, foodstuff industry, railway trackjoint, open gears, cast iron bearing, general machine shop works etc.

Fig. 3.4. Structure of Graphite

Structure of Molybdenum Disulphide Molybdenum disulphide has a sandwitch like structure in which a layer of molybdenum atoms lies between two layer of sulphur atoms which are 0.26 Å apart. It has low shear strength in a direction parallel to the layers due to poor interlaminar attraction. That’s why it has very low coefficient of friction. It is stable in air upto 400°C. A solid film lubricating surface useful for space vehicles is made from (70% MoS2 + 7% graphite) bonded with 23% silicates, which can withstand extreme temperature, low pressure and nuclear radiations.

Molybdenum atom Sulphur atom Fig. 3.5. Structures of graphite and molybdenum sulfide

3.6.3 Semi Solid Lubricant or Greases A lubricant obtained by combining lubricating oil with thickening agents are known as semi solid or grease. Greases are essentially thixotropic gels in which the fibrils or structural elements are metallic soap, and the liquid entrained is the lubricating oils. Some important aspect about grease are

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1. Greases show higher coefficients of friction than oils because of the greater amount of work that must be done in shearing the lubricating film. Therefore, wherever possible, it is better to use an oil instead of grease, barring the special situations listed above. 2. Greases cannot effectively dissipate heat from the bearing as a result of which the greaselubricated bearing works at relatively higher temperature as compared to the oil-lubricated bearing. 3. Greases on storage tend to separate into oil and soap. 4. On constant use, oil in the grease volatalizes off. 5. Greases do not require as much attention as oils and are thus more convenient in use. 6. Greases are capable of supporting greater load at lower speed due to their high shear resistance.

Types of greases 1. Lime or calcium soap base greases: Lime or calcium soap base greases are generally known as “cupgreases”. They are the cheapest and widely used greases. They have good resistance to displacements by water and are suitable for lubricating water pumps, tractors, caterpillar treads, etc. They can be prepared in a wide range of consistency, from soft paste to hard, smooth solid by varying the amount of lime soap from 10 to 30%. However, these greases cannot be used above 65°C as they deteriorate losing combined water. Cup-greases are prepared from fats, lubricating oil and slaked lime. 2. Sodium-soap greases: They have good high temperature properties as they can hold water more firmly due to their high melting point and fibrous structure and hence can be used upto 175°C. However, as the sodium soaps are soluble in water, these greases are not suitable for bearings exposed to wet conditions. 3. Rosin soap grease: These grease are prepared from rosin oil which contains several saponifiabl acids such as abietic acid. The rosin oil is dissolved in the lubricating oil and allowed to react at 58°C with a slurry of slaked lime, emulsified oil and water called “sett”. The resulting grease is known as “cold set grease”. It is mainly used as axle grease for farm wagons and low-speed machinery. This is the cheapest of the greases. In order to prevent the grease from being squeezed out under heavy loads, fillers like talc, mica, etc. are added to these “set” greases. Rosin greases are also used on any heavy, slow moving bearings as gear greases and for lubricating the curves in street railway tracks. Non-soap greases Non-soap greases are usually prepared from non-soap thickeners such as carbon black, silica gel, modified clays, organic dyes, etc. Many of them are suitable for high temperature applications. Lubricating greases are used in the following conditions: 1. When a machine is worked at slow speeds and high pressures. 2. In situations where spilling or spurting oil from the bearings is deterimental to the product being manufactured as in the case of textile mills, paper and food product manufacture, etc. Greases are ideal in such cases because they do not spill or splash as they are designed to “stay put”. 3. In situations where oil cannot be maintained in position due to bad seal or intermittent operation. 4. In situations where the bearing has to be sealed against entry of dirt, water, dust and grit. 3.6.4 Synthetic lubricating oils Synthetic lubricants are oily liquids which are not found naturally or not produced directly during the normal manufacturing and refining processes of the petroleum industry. Synthetic lubricants are designed for special jobs e.g.,

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(i) In metal forming process such as die casting (ii) Hot running bearing and hot rolling mills (iii) Air craft turbine In reactive environments. In order to meet the lubricating requirements under such peculiar operating conditions, viscous fluids have been prepared from various organic and inorganic substances, which are called synthetic lubricants. Many of these were previously known to industries other than the lubrication industry. The most important specific chemical classes of compounds which have been found to possess useful lubricating functions are: (a) Diabasic acid esters (b) Organo-phosphate esters (c) Polyalkylene glycols and their derivatives (d) Chlorinaled and flourinated hydrocarbo (e) Silicate ester (f) Silicones etc. 3.6.5 Lubricating Emulsion In various machining operations such as milling, threading, turning and boring, the tool employed gets heated to a very high temperature, particularly at the cutting edge. In a cutting process, the pressure at the knife-edge may sometimes reach as high as 100,000 psi and a lot of heat is generated. This may lead to oxidation and rusting of the metal under work. In order to prevent overheating in such cases and the consequent injury to the tool, efficient cooling and lubrication have to be provided. This is usually done by employing emulsions of oil droplets in water, which are called cutting oils or cutting fluids or cutting emulsions. Oil has a poor specific heat, but it has good lubricating properties. On the other hand, water is a poor lubricant but is an excellent cooling medium because of its high specific heat and a high heat of vaporization. Hence, the combination of the two in the form of an emulsion can provide both lubrication and cooling effects. The corrosive action of water on the tools, the machines and the work piece are objectionable and is therefore checked by the addition of soaps or other inhibitive alkaline substances. Even then, the use of water is generally limited to simple operations such as grinding and rough turning. The important criteria of a cutting emulsion include (a) to get itself drawn between the chip and the face of the tool and to provide efficient lubrication, (b) to conduct off the heat so as to prevent wear and damage of the metal, (c) to wash away the fragments of the metal, (d) to give a stable emulsion with water, (e) it should not cause rusting of the metal, (f) it should be antiseptic so that in case the worker gets injured, the wound should be rendered asceptic. A good cutting oil increases the accuracy of the cuts and reduces the cost of the work, (a) by making possible to achieve higher cutting speeds, (b) by prolonging the life of the cutting tool, and (c) by reducing the power demand and the number of rejects. Straight chain petroleum oils are very poor cutting lubricants. Fatty oils such as lard oil and sperm oil are very good for cutting oil although blended oils, pine oil, turpentine and rosin oil also find some use. In a good cutting oil, a sulfonated additive is present. For low speeds and light cuts, a chlorinated lubricant may be used. For machining brass, the emulsified oil would be a paraffin oil containing copper oleate or free fatty acid because a sulfurized oil may discolour the work piece. Two types of emulsions are used for lubricating jobs: 1. Oil-in-water type emulsions or cutting emulsions are prepared by mixing together an oil containing about 3 to 20% of a water soluble emulsifying agent (e.g., water soluble soap, alkyl or aryl sulfonate, alkyl sulfates, etc) and a suitable quantity of water. Chemicals like glycols, glycerols and triethanol amine are also added sometimes. Oil-in-water type emulsions are used as coolant cum lubricant for cutting tools and in diesel motor pistons and large internal combustion engines.

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2. Water-in-oil type emulsions or cooling liquids which are prepared by mixing together water and oil containing 1 to 10% of water insoluble emulsifiers (e.g., alkaline earth metal soaps) Emulsions containing 50% lube oil and water are used for the lubrication of steam cylinders, giving cooler walls and lesser oil consumption. Such emulsions have also been successfully used in lubricating compressors handling fuel gases. 3.7 PROPERTIES OF LUBRICANTS 1. Colour. The colours of lubricating oils vary from almost complete transparency to pitch black with all intermediate shades of yellow, red and brown. Some mineral oils exhibit green or blue fluorescence in reflected ligh The colour of an oil, to some extent, indicates its orgin. Paraffin base oils show ageen bloom while napthenic base oils have rather a bluish appearance. In general, the higher the boiling point of a petroleum fraction, the darker it will be. Sometimes, the degree of deterioration or contamination of a lubricating oil is reackoned by comparing with the colour of unused oil. 2. Specific gravity and A.P.I. gravity. The specific gravity of an oil virtually conveys no information regarding its lubricating properties, but since oil is sold by volume, this information on weight to volume ratio may be useful. Specific gravity is a dimensionless quantity which expresses the ratio of the density of the oil to the density of water at a specified temperature. In the petroleum industry, the specific gravity of oils is usually determined at 60°F (15.55°C). Most lubricating oils have specific gravity values between 0.85 to 0.9 at 60°F/60°F. In the U.S.A. specific gravity is sometimes replaced by A.P.I. (American Petroleum Institute) gravity, to provide a simple scale eliminating decimal points. In this, pure water has degrees A.P.I of 10 and for the zero a specific gravity of 1.076 has been adopted. Then, 141.5 A.P.I.° = – 131.5 Sp. Gr. 60°F/60°F 3. Specific heat. The specific heats of most lubricating oils lie in the range 0.44 to 0.49. Information on specific heat is required in heat transfer problems such as those pertaining to the design of plain bearings where the lubricating oil functions both as a lubricant and also as a coolant. 4. Neutralization number. The acidity or alkalinity of lubricating oil is determined in terms of neutralization number. Determination of acidity is more common and is expressed as acid value or total acid number (T.A.N). “It is defined as the number of milligrams of potassium hydroxide required to neutralize all the free acid present in 1 gram of the oil”. Even the most carefully refi ed oil may have a slight acidity. This is due to the presence of minute amounts of organic constituents not completely neutralized during the refining treatment or to traces of residues from the refining process. This small intrinsic acidity may not be harmful in itself but the degree to which it increases in used oil is usually taken as a measure of the deterioration of the oil due to oxidation or contamination. In fact, acid number greater than 0.1 is usually taken as an indication of oxidation of the oil. And the “Total number of milligrams of HCl needed to neutralize any base present in one gm of oil is known as total base number (T.B.N)”. Determination at T.A.N is more common and its procedure is given below: Reagents

(i) 0.1 N alcholic HCl solution (ii) 0.1 N alcholic KOH solution.

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(iii) Titration solvent: 500 ml toluene + 5 ml of distilled water + 495 ml isopropyl alcohol. (iv) Para-naphtholbenzene indicator solution (1 gm of dry indicator + 100 ml of isopropyl alcohol.

Procedure

(i) Take 20 gm of sample of light coloured oil or 2 gm of dark coloured oil into a conical flask (ii) Now add 100 ml of titration solvent and 30 drops of indicator solution to the flask and carefully swril the mixture until the sample is completely dissolved. (iii) If the solution turns yellow-orange or deep oragne in colour, it means the oil sample is acidic (but if it is green or green-blue in colour it means the oil sample is basic). (iv) Slowly add alc. KOH (or alc. HCl in case the sample is basic) from the burette drop by drop until green or green blue end point is obtained (in case of basic oil sample the end point should by yellow orange or deep orange in colours) which persist for 15 seconds. (v) Note the final reading of burette

Calculation: Neutralization Number =

Total ml of titrating solution × 5.61 Wt. of sample used

Significance This test shows the relative changes in oil due to oxidation. Comparing T.A.N. or T.B.N with the values of a new oil, indicate the development of harmful products or effect of additive depletion. Lubricating oil should possess acid value less than 0.1. If the acid number is greater than one oil is usually taken as an indication of oxidation of the oil. This will lead to corrosion, besides gum and sludge formation. 5. Saponification number. The saponification value of an oil is defined as the number of milligrams of potassium hydroxide required to saponify one gram of the oil. Determination: Saponification number is determined by refluxing a known quantity of oil with a known excess of KOH solution and titrating the unused KOH against the standard acid. Signification: Animal and vegetable oils undergo saponification but mineral oils do not. Further, most of the animal and vegetable oils possess their own characteristic saponification values. Hence, the determination of saponification value helps to ascertain the presence of animal and vegetable oils (i.e., fixed oils) in a lubricant. Conversely since each of the fixed oil has got its own specific saponification number, any deviation from this value in a given sample indicates the probability and extent of adulteration. 6. Oxidation. Oxidation of straight mineral oils proceeds slowly even at room temperature but is greatly accelerated at higher temperatures (particularly above 200°C). Oxidation is also accelerated by the presence of moisture in the environment as well as by the presence of oxidation catalysts like Fe, Al and especially Cu, particularly when they are in finely divided state (e.g., as wear products). The resistance of various oils to oxidation depends largely on the nature of the crude oil and the method of refining. In most commercial oils, the rate of oxidation is retarded by adding oxidation inhibitors such as phenyl-b-naphthylamine. Oxidation products are undesirable because: (1) the insoluble oxidation products or sludge may clog oil holes, oil pipe lines, filters and other parts of the lubricating system, (2) The soluble oxidation products circulating with the oil have an acidic tendency and may corrode or pit bearing surfaces or may form harmful and tenacious varnish like deposits and gums. Several tests have been suggested for testing oxidation resistance of the oil but none is universally accepted. The only reliable test is the one in which nearly all the service conditions are simulated.

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7. Emulsification or steam Emulsification number. When pure oil is mixed with pure water, the liquids separate out into layers fairly quickly. But, if the oil is contaminated by finely divided dust, dirt, metal particles or acids, alkalis or soaps, the rate of separation is decreased and an emulsion of either oil in water or water in oil may be formed. Emulsions tend to collect impurities which may cause abrasion, or to form sludges which clog the oil lines, etc. Hence, in a large number of situations, it is essential that the lubricating oil should form such an emulsion with water which breaks off rapidly. This particular property of the lubricant is called “demulsification number” and is determined by noting the time required in seconds for a given volume of oil to separate out in distinct layer from an equal volume of condensed steam under standard conditions. It is also called “steam emulsion number”. The lower the steam emulsion number, the quicker the oil separates out from the emulsion formed and the better the lubricating oil for most purposes. Determination: Steam at 100° C is bubbled through the test tube containing 20 ml oil, till the temperature increases to 90°C. Now note the time in which the oil and water separate out in distinct layers, is called steam emulsion number (S.E.N) or demulsification number Significance Quicker the oil separates out from the emulsion, lower the steam emulsion number and the better the lubricating oil for most purposes. 8. Aniline point. Aniline point of an oil gives an indication of the possible tendency of deterioration of an oil when it comes into contact with packing, rubber sealing, etc. Generally, aromatic hydrocarbons have a tendency to dissolve natural and certain types of synthetic rubbers. Hence, the aromatic hydrocarbon content of the oil, has much significance from this point of view. This is usually determined on the basis of “Aniline point” of an oil which is defined as “the minimum equilibrium solution temperature for equal volumes of aniline and oil sample”. A higher aniline point means lower percentage of aromatic hydrocarbons. A higher aniline point is therefore desirable. Determination: Aniline point is determined by thoroughly mixing (mechanically) equal volumes of aniline and the oil sample in a tube and heating the mixture until a homogenous solution is obtained. Then it is allowed to cool at a specified rate until the two phases just separate out. The temperature corresponding to this particular observation is reported as the “Aniline Point”. 9. Corrosion. Lubricating oils are frequently employed in contact with systems containing Cu and brass. Hence the corrosive properties of lubricating oils are determind by copper corrosion test and steel corrosion test. Corrosive substances like sulfur, H2S and polysulfides are found in petroleum and are removed or converted into relatively harmless organic sulfides by the refining processes. Refinery chemists usually test for sulfur both during refining and in the final product The copper corrosion test is a valuable criterion for products like cutting oils used for machining of non-ferrous metals and for lubricants used in rolling contact bearings which have non-ferrous cages. The so-called “copper strip test” comprises of keeping a polished copper strip in the lubricating oil at a specified temperature for a specified time and then examining the strip after taking it out. Any tarnishing of the strip indicates the presence of corrosive substances in the oil. The steel corrosion test for oils is designed to determine the ability of the oil to prevent corrosion of ferrous parts in the presence of water. Corrosion inhibitors like zinc dithiophosphate and organometallic compounds are usually added to the lubricating oils. 10. Oiliness. Oiliness is the property of the lubricant by virtue of which a lubricating oil can stick on to the surface of the machine parts operating under high pressures. Oiliness of an oil is the most important property of a lubricant under boundary or thin film lubrication conditions. Mineral oils have very poor oiliness whereas animal and vegetable oils have good oiliness. Hence, oiliness of mineral oils is generally improved by adding small quantities of high molecular weight fatty acids

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132 Thermometer

Dry air

Furnace Coiled copper tube Hot air oil in Platinum tray

Gaseous fuel Fig. 3.6. Vaporimeter for determining volatility of a lubricating oil.

like oleic acid, stearic acid, chlorinated esters of these acids, etc. There is no perfect method for the determination of absolute oiliness of on oil. Only relative oiliness is considered while selecting a lubricating oil for a particular job. 11. Volatility. If the lubricating oil is exposed to high temperatures as in heavy machinery, some of it may volatalize off. Apart from loss of the volatalized lubricant, the residual oil left behind may have different properties (such as high viscosity and different viscosity index) than the original oil. A good lubricant naturally should have a low volatility. The volatility of a lubricating oil is usually determined by an apparatus called vaporimeter shown in Fig. 3.7. It consists of a furnace heated by a gaseous fuel in the middle of which a coiled copper tubing is placed. Air can be passed through the copper tubing. A known weight of oil sample is taken in a platinum crucible and it is introduced at the centre of the copper tube as shown in the diagram. Now, dry air at a rate of 2 litres per minute is passed through the copper tube for an hour. Then the tray is withdrawn and it is cooled and weighed. The loss in weight of the oil, calculated as percentage weight of the oil taken, is reported as volatility of the oil. 12. Ash Content. The ash content of a perfectly refined mineral oil is very low. In the case of used oils the ‘ash’ will include metal particles and attempts have been made to assess the rate of cylinder wear from the iron content of the ash from used cylinder oils. 13. Decomposition stability. The stability of the oil towards hydrolysis and pyrolysis reactions is also important because the products from these reactions are detrimental to the machine parts. 14. Precipitation number. This shows the percentage of asphalt present in an oil. This is determined by dissolving a known weight of the oil in petroleum ether and separating the asphalt precipitated by centrifugation. The asphalt is dried and weighed and reported as weight percentage of the oil taken. 15. Cloud point and pour point. Petroleum oils are complex mixtures of chemical compounds and do not show a fixed freezing point. When they are sufficiently cooled, they become plastic solids due to the formation of solid crystals or due to congealing of the hydrocarbons present. The cloudpoint is the temperature at which this crystallization of solids in the form of a cloud or haze first becomes noticeable, when the oil is cooled in a standard apparatus at a standard rate. The pour point is the temperature at which the oil just ceases to flow when cooled at a standard rate in a standard apparatus. Even if the jacket is put in horizontal position alteast for 5 second.

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The apparatused used is showin in fiq. It consists of a test jar, which is cylindrical with flat bottom, made of clear glass and is about 3 cm in diameter and 12 cm high. It is enclosed in a glass or metal jacket which is firmly fixed in a cooling bath The cooling baths used are as follows: Upto 10°C — Ice and water Oil thermometer Both Upto — 12°C — Crushed ice and salt thermometer Upto — 26°C — Ice and CaCl2 Cork Upto — 57°C — Solid CO2 and petrol Jacket The oil is poured into the test jar to a height of 2 Cooling bath to 2¼ inch. Thermometers are introduced in the oil and Cork ring the cooling bath. As the cooling takes place via the airjacket, the temperature of the oil falls. At every degree fall of temperature of the oil, the test jar is withdrawn for Cork disc examination and replaced immediately. The temperature at which cloudiness or hazyness is fist noticed represents the cloud-point. As the cooling is further continued, at a Fig. 3.7. Pour-point apparatus. particular temperature, the oil just ceases to flow or pour as observed from tilting the test jar. This particular temperature at which the oil does not flow in the test jar for 5 seconds on tilting it to horizontal position is reported as the pour-point. Significance: For lubricating oils, the pourpoint has a greater significance. It determines the suitability of a lubricant or a hydraulic oil for low temperature conditions. Important examples are refrigerator plants and air-craft engines, which may be required to start and operate at sub-zero temperatures. Cloud and pour point determination is also helpful in find out the minimum working range of a lubricating oil. 16. Carbon Residue Test. Lubricating oils contain high percentage of Carbon in combined form. On being subjected to high temperatures, they decompose and form a carbonaceous deposit. There are two method for measuring the amount of carbon residue. (i) The Conradson Method: It is conducted in the absence of air and is applicable for heavy residuals, crudes and non volatile stock.

Bridge (Flame height guide) vent

Hood Outer iron crucible Iron crucible Procelain crucible Heat insulator Triangular wire support Meker burner

Procedure: Take weighed amount of the sample in a silica crucible. Now put this silica Fig. 3.8. Conradson’s apparatus for the crucible into a skid more iron crucible having a determination of carbon residue. close-fitting cover with a small horizontal opening. The crucibles are then placed into a larger third crucible fitted with a cover. Now heat the crucible with a help of burner at certain prescribed rates till vapours of all volatile matter are burnt completely. After 30 minutes the silica crucible is removed, cooled in a desicator and weighed. % Caron Residue =

Wt. of residue in crucible × 100 Wt. of original oil sample

(ii) The Ramsbottom Method: A weighed sample is placed in a special glass bulb with a capilary opening. Now put this bulb in a electrically heated small furnace (known as Ramsbottom

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apparatus) maintained at approximately 550 °C. At this temperature all the volatile matter present in the sample escape through the bulp capillary. A residue left in the bottom of the bulb. This residue undergoes cracking and formed cake. After the heating, the bulb is taken out, cooled in a desiccator and weighed Weight of residue in bulb × 100 Weight of original oil sample Significance of carbon residue test: The Capillary dia. 15 mm carbon forming tendency of a lubricating oil on Capillary combustion is significant particularly for internal length 9.5 mm combustion engines. Oils which deposit minimum amount of carbon are naturally preferable. Carbon deposition in an internal combustion engine results both from incomplete combustion of the fuel as well as the carbonizing of the lubricating oil 1 mm carried up past the piston rings into the combustion chamber. Excessive build-up of carbon deposits in the combustion chamber results in decreased Hemispherical end volume of the charge at the end of the compression 24.9 mm stroke giving increased compression ratio which ± 0.4 mm eventually leads to detonation. Deposition of carbon residues by the lubricant may be objectionable in Fig. 3.9. Coking bulb used in Ramsbottom carbon residue test. other situations also. 57 mm

% Carbon residue =

17. Flash and fi e point: The flash-point of an oil is defined as the minimum temperature at which the oil gives off sufficient vapour to ignite momentarily when a flame of standard dimension is brought near the surface of the oil (at a prescribed rate in an apparatus of specified dimensions) is known as flash point. And the lowest temperature at which the vapours of the oil burn continuously for at least 5 seconds when the standard flame is brought near the surface of the oil (which is heated in a specified apparatus at a specified rate) is know as fire point. In a majority of the cases, the fire point of an oil is about 5 to 40°F higher than its flash point Determination: Flash point of an oil is determined by either open cup or closed cup apparatus. In the open cup apparatus, the oil is heated with its upper surface exposed to the atmosphere. The open cup apparatus commonly used is Cleveland’s apparatus. The closed cup apparatus in common use are Abel’s apparatus and Pensky-Martens apparatus. The closed cup apparatus gives more reproducible results. The flash-point obtained with an open cup apparatus is generally about 10 to 30°F higher than that obtained with a closed-cup apparatus. (A) Closed Cup Apparatus (i) Pensky-Marten’s Apparatus It is the most commonly used apparatus for determination of flash-points of oils having flash points between 50°C to 370°C. The essential features of the apparatus are shown in Fig. 3.9. It consists of a brass cup which is 5 cm in diameter and 5.5 cm in depth. The level upto which

Fig. 3.9. Pensky - Martin flash & Fire point Apparatus.

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oil is to be filled in the cup is marked at about 1 cm below the top of the cup. The cup is supported by its flange over a heating vessel in such a way that there is clearance between the cup and the heating vessel. The cover for the cup is provided with four openings of standard dimensions, which are meant for a special type of stirrer, a standard thermometer, an air inlet and a device for introducing the standard flame. The shutter provided at the top of the cup has a lever mechanism. When the shutter is turned, openings for the test flame and air are opened and the flame exposure device dips into the opening over the surface of the oil. The test flame gets extinguished when it is introduced into the opening for the test, but as soon as it returns to its original position on closing the shutter, the flame is automatically lighted again by the poilt burne . Procedure: The oil sample is filled up to the mark in the cup and covered with its cap or cover. The cover incorporating the stirring device and the thermometer. The test flame is lighted and adjusted until it is the size of a bead approximately 4 mm in diameter. The apparatus is heated so that the oil temperature increases by about to 6°C per minute while the stirrer is rotated at approximately 60 revolutions per minute. When the temperature rises to within upto 15°C of the anticipated flashpoint, the test flame is dipped into the oil vapour for about 2 seconds at every degree rise of temperature. This is done by twisting the knob which lowers the test flame and simultaneously opens the shutter. These spring back to their original positions as the knob is released. The flash point is taken as that minimum temperature at which, on introducing the test flame into the oil cup, a distinct flash is observed Oils containing minute quantities of volatile organic substances are liable to flash below the true flash-point of the oil. Although a small flash may be observed in such cases, it should not be confused with the true flash, since its intensity does not increase with increased temperature, as occurs when the true flash-point is reached (ii) Abel’s closed cup apparatus: It is best used for oil having flash point below 120° F. Since water bath is used hence the lubricating oil which have their flash point below 50°C can be tested by this apparatus (B) Open Cup Appratus (i) Cleveland open cup apparatus: It is generally used for the determination of flash point of fuel oil and other oils having flash point below 175° F. It consists a open brass cup (known as cleveland cup), a thermometer and a gas burners or electric heaters which act as a source of heat. Procedure: The oil sample is filled up to the mark in the cup. There should be no oil outside the cup. The thermometer is immersed in the sample. Now heat it with a rate of 9° to 11° F per minute. At every 5° F rise in temperature a small flame is passed over the oil surface. When a flash appears at any point on the surface of the oil, the temperature reading is noted as flash point. The heating of the oil is continued at the same rate. The test flame is applied again for every 5° F rise in temperature. The temperature at which oil ignites and continues to burn for atleast 5 second is known as fire point Limitation: The flash point of lubricating oil should be greater that 175 . Significance: A good lubricating oil should not volatalise under the working temperatures. Even if some volataliation takes place, the vapours formed should not form inflammable mixture with air under the conditions of lubrication. From this point of view, the flash point and fire point of a lubricating oil are of significance A lubricating oil selected for a job should have a flash-point which is reasonably above its working temperature. This ensures safety against fire hazards during the storage, transport and use

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of the lubricating oil. In addition, the flash point of an oil is often used as a means of identification and also for detection of contamination of the lubricating oils. 17. Viscosity Viscosity is one of the most important properties of a lubricating oil. Viscosity is a measure of the internal resistance to motion of a fluid and is mainly due to the forces of cohesion between the fluid molecules. Absolute viscosity may be defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel planes in the fluid unit distance apart. The unit of absolute viscosity h (eta) in C.G.S. system are poise and centipoise (1/100th of a poise). Poise is equal to one dyne per second per square centimeter. The viscosity of water at 20°C is about 1 centipoise. The ratio of absolute viscosity to density for any fluid is known as the absolute kinematic viscosity. It is denoted by h and in C.G.S. system, its units are stokes and centistokes (1/100th of a stoke). v =

where

h r



n = absolute kinematic viscosity



h = absolute dynamic viscosity



r = density of the fluid

The dimensions of dynamic viscosity are ML–1 T –1, and the dimensions of kinematic viscosity are L2T–1. For academic purposes, viscosity is usually expressed in centipoise or centistoke, but a more common practical measure of the viscosity of an oil is the time in seconds for a given quantity of the oil to flow through a standard orifice under the specified set of conditions. Thus, viscosities are usually determined with Redwood Viscometer in the Commonwealth countries, with Engler’s Viscometer in the Europe and with Seybolt’s viscometer in the U.S.A. Measurement: Many types of viscometer are used to measure the viscosity of a lubricating oil. Saybolt viscometer is used in United States of America and Redwood Viscometer is used in England. The Redwood Viscometer The Redwood Viscometer is made in two sizes. The Redwood–1 Viscometer is commonly used for determination of viscosities of lubricating oils and has an efflux time of 2,000 seconds or less. The Redwood 2 viscometer is similar to the 1 type but the jet for the outflow of the oil is of a larger diameter and hence gives an efflux time of approximately 1/10th of that obtained with 1 instrument under otherwise identical experimental conditions. Redwood 2 instrument is therefore used for the oils having higher viscosities, such as the fuel oils. Dimension of orifi Receiving flask ha Useful for

Redwood No. 1 Lenth 10 mm Diameter 1.62 mm smaller mouth low viscous

Redwood No.2 50 mm 3.80 mm larger mouth high viscous

Description: The Redwood Viscometer does not give a direct measure of viscosity in absolute units but it enables the viscosities of oils to be compared by measuring the time of efflux of 50 ml of oil through the standard orifice of the instrument under standard conditions. The results given by these two viscometers are reported as “Redwood 1 Viscosity” or “Redwood 2 Viscosity” followed by the efflux time in seconds at the experimental temperature

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Fig. 3.10. The Redwood Viscometer.

The Redwood viscometer no.1 consists of a standard cylindrical oil cup made up of brass and silvered from inside. It has 90 mm height and 46.5 mm in diameter. It is fitted with an agate jet in the base. The diameter of the orifice is 1.62 mm. The lid of the cup is provided with an arrangement to fix a thermometer to indicate the temperature of oil. The oil cup is surrounded by water bath for adjusting the temperature. A thermometer and stirrer are also provided in the water bath. A calibrated receiving flask (known as Kohlraush flask) is provided for receiving the oil from the orifice. When the sample reaches test temperature the time for 50 ml of the sample to flow through the orifice is measured. Result are reported in redwood second. Saybolt Viscometer It was the first viscometer, which is based upon the principle of letting a known volume of the liquid flow through a standard orifice. In the Saybolt Viscometer the fluid is contained in the container. The container is surrounded by a water jacket in which water can be heated and stirred electrically. The container is filled with the oil. After sometimes, the orifice at the bottom is opened and the time taken for 60 cc of the liquid to pass through the standard orifice in the flask is noted and the mean of the few reading recorded as viscosity. Saybolt viscometer gives the viscosity of any liquid as the number of second required for 60cc. of it to pass through the orifice under the given set of conditions For very viscous fuels, a viscometer with a larger jet known as the Saybolt Furol Viscometer is used. The Saybolt Universal viscometer can be used for oils having flow times of more than 32 seconds. There is no maximum limit, but in general, for liquids having flow times over 1000 seconds, Saybolt Furol Viscometer is better. Viscosity Index The viscosity of an oil decreases with increase of temperature as a result of decrease in intermolecular attraction due to expansion. Hence it is always necessary to state the temperature at which the viscosity was determined.

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Viscosity index is the numerical expression of the average slope of the viscosity temperature curve of a lubricating oil between 100°F to 210°F. The oil under examiantion is compared with two standard oils having the same viscosity at 210°F as the oil under test. Oils of the Pennsylvanian type crudes thin down the least with increase of temperature; whereas oils of Gulf coast origin thin down the most as the temperature is increased. Hence the viscosity index of Pennsylvanian oil is taken as 100 and that of the Gulf oil as zero. Then the viscosity of the oil under investigation is deduced as follows: VL – VX × 100 Viscosity index = VL − VH where VL = Viscosity at 100°F of Gulf oil standard which has the same viscosity at 210°F as that of the oil under test VX = Viscosity of the oil under test VH = Viscosity at 100°F of Pennsylvanian standard oil which has the same viscosity at 210°F as that of the oil under test. Thus, the higher the viscosity index the lower the rate at which its viscosity decreases with increase of temperature. Hence, oils of high viscosity index i.e., those having fl t viscosity temperature curves are demanded for air-cooled internal combustion engines and aircrafts engines. In general, oils of high specific gravity have steeper viscosity-temperature curves. However, all oils tend to attain the same viscosity above 300°C. By and large, light oils of low viscosity are used in plain bearings for high-speed equipment such as turbines, spindles and centrifuges whereas high viscosity oils are used with plain bearings of low speed equipment. Conversion of Redwood, Engler and Saybolt viscosities into absolute units Redwood, Engler and Saybolt viscosities can be converted to absolute units (centistokes). However, since these instruments are not the ideal methods of determining the absolute viscosities, the conversion values are only considered as good approximations and that too only when taken at the same temperature. For instance, Redwood viscosities at 30°C cannot be converted into absolute units at say 40°C, because, different fluids have di ferent viscosity temperature relationships. The conversion of the above relative viscosities to absolute viscosities is done with the help of the following equation: v = Ct – b/t where v = kinematic viscosity in centistokes, t = time of flow in seconds, and C and b are constants. The following values are taken for the constants: Instrument Value of C Value of b Redwood No. 1 0.25 172 Redwood No. 2 2.72 1120 Saybolt Universal 0.22 180 Engler 0.147 374 Notes: 1. t = (Degrees Engler) × 52 2. Redwood Seconds No. 1 = 0.88 Saybolt Seconds Universal 3. Degrees Engler = 0.0328 Redwood Seconds No. 1 4. For accurate results, conversion constants should be determined experimentally at the temperature under consideration. 19. Mechanical Tests. Several mechanical tests have been devised to test the performance of a lubricating oil under a given set of conditions of temperature, load etc; one among them is the “four ball extreme pressure lubricant test.” The working portion of such a machine consists three steel balls held in a ring and an upper fourth ball in contact with them and this ball is held at the end of a

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vertical shaft which is rotated at a fix d speed by an electric motor (Fig. 3.11). The three stationary balls are pressed upwards against the fourth ball by a lever carrying an adjustable load. The torque thus transmitted to the three fixed balls can be measured and if required, the coefficient of friction can be continuously recorded throughout a test. The points of the balls are lubricated by the test lubricant contained in a cup surrounding the ball assembly. In a simple lubricant test, the highest load that a ball can stand for 1 minute without squeezing can be taken as a measure of lubricant quality.

Ball chuck Rotating ball

R S

S

S

Stationery balls

Fig. 3.11. Four-ball extreme pressure testing machine.

3.8 SELECTION OF LUBRICANTS Industrial oils can be broadly classified into one of the following types 1. Machine and Engine oils. 2. Spindle oils 3. Refrigeration oils 4. Circulating oils 5. Gear oils 6. Steam cylinder oils. Obviously, the properties required for each of the above class of lubricating oils are different. The selection of industrial lubricants in any mechanised industry involves a consideration of the requirements of the equipments, available methods for handling and application of the lubricant itself and environmental conditions. In general, it is wiser to use the lubricants as recommended by the manufacturers of the equipment being used or by the standard oil companies in order to ensure maximum life of the operating equipment. In selecting a lubricant for a given application, it is essential to consider the various properties of the lubricant required in relation to the service conditions. The main machine elements that require the use of lubricants in any equipment are bearings, gears and cylinders. These simple elements work under a variety of operating conditions in different machines and equipment, and lubricants have to be specially designed to provide adequate protection to them. Some of the main factors to be taken into account are the effects of load, temperature, and speed at which these elements operate and also the contaminants which may affect the performance of the lubricants in use. The properties of lubricants required for different types of machinery are summarized in the following Table 3.2. Table 3.2 Properties of lubricants required for different types of machinery



Type of Machinery

Functions and properties of the lubricating oil required

1. Automotive Engine Oils (Internal Combustion Engines)

Automative engines are the most difficult piece of equipmen from the point of view of lubrication. The properties of the foil required are:



(a) Lubrication over a wide range of temperatures.



(b) Thermal stability and good heat transfer properties



(c) Wear protection of piston rings and cylinder liners subjected to high pressures.

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Type of Machinery

Functions and properties of the lubricating oil required



(d) Provide a seal between piston rings and cylinder walls against high pressure combustion gases.



(e) Detergency to prevent deposits and lacquer formation due to thermal decomposition of the oil at high temperatures.



(f) Prevention of corrosion and rusting of internal parts of the engine.



(g) Prevention of contaminants from precipitating to form sludge deposits (Dispersancy).

2. Spindle oils

For the lubrication of lightly loaded spindles at very high speeds, thin oils are essential. It is desirable to add oxidation and rust inhibitors. Oils with viscosities ranging from 30 to 105 SUS (Saybolt Universal, Seconds) at 100°F are suitable for many applications.

3. Refrigeration oils

Oils with low cloud and pour points are needed in refrigeration systems. Naphthenic base oils only have such characteristics. Many manufacturers also stipulate a minimum dielectric strength so as to ensure that the oil is dry and free from free or dissolved moisture. ISI specification IS 4578–1968 for, refrigeration oils cover 4 grades of viscosities of 85, 160, 200 and 325 SUS at 100°F. The pour-point requirements are – 40°F Max. for the lightest grade and –13°F for the heaviest grade.  Oils with viscosities around 150, 220 and 320 SUS at 100°F are commonly used for direct driven and geared turbines. Oils for marine applications have viscosities of 400 SUS at 100°F. Lubrication conditions in steam turbines are stringent. Very high oxidation stability and chemical stability are needed. Anti-oxidation, anti-rusting and anti-foaming additives are also required. The important characteristics required are proper viscosity, high viscosity index, good demulsibility characteristics, good oxdiation stability, rust preventive properties, anti-wear characteristics and sufficiently low pour-point. Viscosities of oils required in hydraulic systems are 150, 210, 310 and 400 SUS at 100°F for the highest grade, medium, medium heavy and heavy grades respectively. Various types of gears like spur, bevel, helical, herring, bone hypoid and worm are used in enclosed gearboxes for the transmission of power. The oils should have good oxidation resistance, proper viscosity, high viscosity index, extreme pressure characteristics, water separation properties and good foam resistance. Premium quality high viscosity index oils are required. Viscosities of 165, 220 and 300 SUS at 210°F are adequate to cover the requirements of most cylinder oils. Straight mineral oils are used for super-heated steam while compounded oils have to be used for wet or saturated steam. Compounded oils contain additives of fixed oils and emulsifiers which help to form an inverted emulsion with water.

4. Circulating oils (a) Turbines



(b) Hydraulic systems

5. Gear oils

6. Steam cylinder oils

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Type of Machinery 7. Cutting oils

8. Transformer oils

9. Lubricants used for machinery running at extreme pressures and low speeds e.g., tractor rollers, lathes, concrete mixers, and railway track joints where a film of lubricating oil or grease cannot be maintained. 10. Machines operating at high pressures and low speeds, as in wire ropes and rail axle boxes 11. Delicate equipment such as watches, clocks, scien- tific equipment and sewing machines.

Functions and properties of the lubricating oil required

Good lubricating property, low viscosity to enable it to fill the cracks formed on the work piece, high thermal conductivity, chemical stability, and anti-corrosive and antiseptic properties are required. Good lubrication and cooling are essential. The lubricating oils used in electrical transformers must possess high dielectric properties to insulate the windings. It should have low viscosity, optimum oxidation resistance, and good chemical stability under the operating conditions. Highly refined oils, without even traces of moisture and dirt, and possessing good dielectric properties, optimum oxidation resistance and good chemical stability are used. Solid lubricants such as graphite are used either as dry powder or in the form of emulsion e.g., aquadag or oil dag.

Thick blended oils or greases are used.

Fixed Oils (animal or vegetable oils) such as clarified sperm oil, neatsfoot oil, olive oil, palm oil, hazelnut oil are used. Apart from other conventional types of solid, liquid or semisolid lubricants, gases are also used as lubricants. The wear problems during starting or shut down of machinery in this case can be minimized by choosing wear resistant compounds or by coating machinery with molybdenum disulfid or Teflon

3.9 METHODS OF LUBRICATION Intermittent lubrication is used for low speeds whereas in other cases, continuous lubrication is employed. Various methods of lubrication such as gravity feed methods, force feed methods and mechanical feed methods are available to suit different purposes. 3.9.1 Degradation of lubricating oils and re-refinin Lubricating oils are degraded or deteriorated due to depletion of additives during use due to thermal stress and contamination from external and internal sources. The impurities present are sludge, saponifiable matter, aldehydes, carbon grit, dirt, metallic particles, moisture, diluents etc. Thermal stresses generate oxidation products and degradation products which promote further oxidation, particularly at high temperatures. The spent lubricating oil can be re-refined by the various processes commercially adopted in different parts of the world. These processes include (1) Acid-Clay process, (2) Propane clarification, (3) Distillation/Clay process, (4) Distillation/Hydrotreating process and (5) Caustic process. Standard specifications are available for re-refined oils als

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3.10 NUMERICAL PROBLEMS. Example. An oil sample under test has a Saybolt universal viscosity of 64 seconds at 210°F and 564 seconds at 100°F. The low viscosity standard (Gulf oil) possesses a Saybolt universal viscosity of 64 seconds at 210°F and 774 seconds at 100°F. The high viscosity standard (Pennsylvanian oil) gave the Saybolt Universal viscosity values of 64 seconds at 210°F and 414 seconds at 100°F. Calculate the viscosity index of the oil sample under test. Solution. VL − VX Viscosity index of the oil under test = × 100 VL − VH (774 − 564) 210 = = × 100 = 58.33. (774 − 414) 360 Example 2. An oil sample under-test has a saybolt universal viscosity same as that of standard Gulf oil (low viscosity standard) and Pennsylvanian oil (high Viscosity index standard at 210° F. The saybolt universal viscosities at 100° F are 61,758 and 420 s respectively. Calculate the viscosity index of the sample oil. Solution. Here, L = 758 s H = 420 s and U = 61 s So, viscosity index of the sample oil L–U V.I. = × 100 L–H 758 – 61 V.I. = × 100 = 206.2 758 – 420 Example 3. An oil of unknown viscosity index has a saybolt universal viscosity of 90 seconds at 210 °F. The high viscosity index standard (i.e. pennysylvanian) oil has saybolt viscosity of 90 seconds at 210 F and 450 seconds at 100 °F. The low viscosity index standard (i.e.Gulf) oil has a saybolt universal viscosity of 90 seconds at 210 °F and 785 seconds at 100 °F. Calculate the viscosity index of unknown oil. Solution. Here, Therefore

L = 785 s H = 450 s U = 850 s

L–U × 100 L–H 785 – 850 V.I. = × 100 = 19.4 785 – 450 V.I. =

QUESTIONS 1. 2. 4.

Explain the following properties of lubricants and discuss their significance (a) Viscosity and viscosity index (b) Flash Point (c) Aniline Point (d) Saponification Value Discuss the important properties of lubricating oils which are useful for their evaluation. Distinguish between fluid film and boundary lubricatio

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5. A lubricating oil has the same viscosity as standard naphthenic and paraffinic type oils at 210°F. Their viscosities at 100°F are 320 S.U.S., 430 S.U.S. and 260 S.U.S. respectively. Find the viscosity index of the oil. 6. What do you mean by viscosity index of a lubricating oil ? A lubricating oil has a S.U.S. os 64 seconds at 210°F and of 600 seconds at 100°F. The high viscosity index standard (i.e. Pennsylvanian) oil has S.U.V. of 64 seconds at 210°F and 400 seconds at 100°F. The low viscosity index standard (i.e. Gulf) oil has a S.U.V. of 64 seconds at 210°F and 700 seconds at 100°F. Calculate the viscosity index of the oil. (RGPV Bhopal 2006) 7. Write an essay on solid lubricants with emphasis on their classification, mechanism of action, examples and applications. 8. How are semi-solid lubricants prepared ? In what situations a semisolid lubricant is preferred? Mention some important tests for evaluating semisolid lubricants. 9. How are liquid lubricants classified ? Discuss the various methods available for refining mineral oils. 10. What are the various types of synthetic lubricants available ? Discuss their merits and demerits. 11. Discuss the use of lubricating emulsions. 12. Write informative notes on the following : (a) Graphite (b) Cup greases (c) Neutralization number (d) Extreme pressure lubrication (e) Oiliness (f) Carbon residue test (g) Pensky Martin’s apparatus (h) Redwood Viscometer. 13. Justify the following statements : (a) Flash point determination by the closed cup apparatus gives a lower value than that determined by an open cup apparatus. (b) Closed cup apparatus gives a more reliable value than the open cup apparatus for the determination of flash point (c) The relative viscosity determined by Saybolt viscometer or Redwood viscometer can be converted into absolute kinematic viscosity by calculations. 14. Define lubrication. Explain various types of lubrication and discuss any three important properties of lubricants. (RGPV Bhopal 2001, 2006) 15. (a) How does viscosity determine the operating characteristics of a lubricant ? (b) Suggest the suitable properties of the lubricating oil used for steam engines and transformers. (c) Describe with their significance the following (i) Aniline Point (ii) Cloud point and pour point (RGPV Bhopal 2009) (iii) Steam emulsification numbe (Nagpur University, 2001) (iv) Important function of lubricant. 16. (a) What are flash point and fire point of a lubricant. Point out their significan (b) Explain the following properties of lubricants : (i) Viscosity Index (ii) Aniline Point (iii) Neutralisation Number (c) Write short note on Redwood Viscometer 17. (a) Define (i) Acid Value (ii) Saponification iii) Pour Point (b) What are different mechanisms of lubrication ? Explain the boundary lubrication. (RGPV Bhopal 2009) (c) Write a note on determination of flash point by Pensky Martin Method. What is the significance of flash poin (RGPV Bhopal 2009) 18. (a) Define grease. Under what situations it is used as a lubricant (RGPV Bhopal 2006)

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(b) How are lubricants obtained and refined from crude petroleum (c) Write notes on boundary lubrication. (Mumbai University, 1998) 19. (a) What is the purpose of dewaxing a lubricant ? How is it carried out ? (b) Explain the following : (i) Flash point and fire poin (ii) Pour point and cloud point (c) Write notes on : (i) Solid lubricants (ii) Working of Redwood Viscometer. (Mumbai University, 1998) 20. (a) Define lubrication. Explain the mechanism of boundary lubrication (b) Explain the following properties of lubricants and give their significance (i) Emulsification ii) Aniline Point (iii) Cloud Point and Pour Point (c) Describe the conditions under which solid lubricants are used. (Mumbai University, 1999) 21. (a) How would you determine viscosity of a lubricating oil using Redwood Viscometer ? (b) Describe how do you determine neutralisation number and saponification value of a lubricant. (c) State essential properties of a lubricant. (Mumbai University, 2000) 22. (a) What are the essential properties of a lubricant ? (b) How would you determine flash and fire points of a lubricant (c) Explain the extraction and purification of mineral oil (Mumbai University, 2001) 23. Explain the following : (a) Synthetic lubricants (b) Boundary lubrication 24. Write informative notes on any two of the following : (a) Lubricating Emulsions (b) Solid Lubricants (c) Synthetic Lubricants 25. Write informative notes on any two of the following : (a) Extreme Pressure Lubricants. (b) Lubricating Emulsions (c) Blended Oils. 26. What do you mean by viscosity index ? Discuss the importance of oxidation stability and iodine value of a lubricant. 27. What are the factors to be considered while selecting a lubricant for a particular purpose. 28. How can we obtain lubricating oil from crude oil. What is the importance of dewaxing of oil fractions? 29. (a) “Closed cup apparatus gives a more reliable flash point than that given by the open cup apparatus”. Justify the statement. (b) Describe how relative viscosity of a lubricating oil is determined by Redwood viscometer. What is the difference between Redwood No.1. and Redwood No. 2 viscometers ? (c) What considerations should be made while selecting a lubricant for cutting oils. (Nagpur University, 2002) 30. (A) Explain the following properties of liquid lubricants emphasizing their significance (a) Neutralization number (b) Aniline point (c) steam emulsification numbe (B) Write short notes on graphite as a lubricant (C) Suggest suitable lubricants for the following with proper justification (i) Steam turbines (ii) I.C. engines (Nagpur University, 2002) 31. How viscosity index of a lubricating oil is determined by Redwood viscometer? Write the significance of the test (RGPV Bhopal 2009) 32. What are the characteristic features of lubricats? Write the criteria for selection of lubricants for specific purpose (RGPV Bhopal 2009)

PART–B

UNIT

3

Cement and Refractories CEMENT

3.11 Introduction Cement is a material which possesses adhesive and cohesive properties and capable of bonding materials like bricks, stones, building block etc. 3.12 Classification of cemen (i) Natural Cement: It is made by calcining a naturally occurring argillaceous lime stone at a high temperature and then Pulverizing the calcined product. Calcium silicate and aluminates are formed by the combination of silica and alumina with calcium oxide during calcination. Properties

(i) It possess hydrolic properties



(ii) It is a quite setting cement



(iii) It has low strength

Applications

(a) Combination of sand with natural cement (Known as Mortors) is used for laying bricks and setting stones.



(b) It is also used in large masses of concretes e.g. dams and foundation



(ii) Pozzolana cement: It is the oldest cement which was invented by Romans and was used for construction of dams etc. It is prepared by mixing and grinding pozzolana and slaked time. Natural pozzolana is deposit of volcanic ash produced by rapid cooling lava. Lava is a molten mixture of silicates of calcium, iron and alminium.

Properties. They posses hydrolic properties. Application Their mixture with Portland cement is used for different applications

(iii) Slag cement: It is prepared from blast furnace slag and hydrated lime. At first a mixture of calcium, aluminium silicates (blast furnace slag) is granulated by pouring it into the streams of cold water. Now dried the mixture and mix it with hydrated lime. Now pulvarized the mixture. Sometimes acceleratory like clay, salt or caustic soda are added to accelerate the hardening process. Properties. (i) They are low setting



(ii) They are poor in abrasion resistance.



(iii) They have low strength. 145

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BASIC ENGINEERING CHEMISTRY Applications



(a) It is used for making concrete in bulk construction. (iv) Portland Cement: It is made by calcining (at 1500 °C) an intimate mixture of clay and lime containing raw material in correct proportion. After calcinations, retarder like gypsum is added.

Other Types of Cements Some other types of cements are also used in special types of construction work, and these are not portland cements or their derivatives. (1) High alumina cement. It is manufactured by heating, until molten, a mixture of limestone and bauxite to produce essentially calcium aluminates (CA and C12A7) and quickly cooling the product. The finely ground product is quite dark in colour. It gains strength very quickly and is particularly useful under extremely low temperature conditions where ordinary Portland cement does not gain strength. This kind of cement is known as “Ciment Fondu” in western countries. Another remarkable feature about this cement is that it can be used as refractory material at 1000°C and is used for making castable refractories and for making ‘in situ’ castings. (2) Magnesium oxychloride cement. This is also called Sorel cement. It is prepared by the reaction of magnesia and a solution of MgCl2. The reaction is exothermic. The composition of the product formed is roughly 3 MgO.MgCl2.11 H2O. This kind of cement is generally prepared ‘in situ’ or in moulds. It can be used as flooring material with coloured pigments and other inert fillers producing very pleasing surfaces. However, it is rapidly eroded by water and hence is limited in its use. (3) Strontium and Barium Cements. These cements are extensively used in concrete shielding for atomic piles where resistance to penetration of radioactive emanations is essential. These are manufactured by employing strontium and barium salts instead of calcium salts during the cement manufacture. 3.13 PORTLAND CEMENT At first Joseph Aspidin in 1824 find out that the hard clinkers when ground and mixed with water, produced a highly superior hydraulic cement. Which are similar in colour with a natural stone Portland therefore it is known as Portland cement. Raw Materials The raw materials used in the manufacture of Portland cement are: (a) Calcareous materials (which supply lime) e.g., lime stone, cement rock (a soft argillaceous lime stone), chalk, marl or marine shells, and waste calcium carbonate from industrial processes. A lime stone high in magnesia cannot be used unless its magnesia content is reduced by some means, as by flotation, or dilution with low-magnesia rock so that the product will not contain more than 5% MgO. Similarly, chalk containing flint has to be freed from that impurity, and seams of gypsum or other materials such as pyrite may require selective handling before use. (b) Argillaceous materials (which supply silica, alumina and iron oxide) e.g., clay, shale, blastfurnace slag, ashes and cement rock. Clay or shale are most commonly used. Cement rock was sometimes used as such without any further addition, since it contains both limestone and clay minerals. The modern demands, however, have made necessary more precise control of composition and hence, such a simple procedure is rarely employed today. 3.13.1 Important Process Parameters for Manufacturing a Good Cement Clinker (1) The lime saturation factor CaO 2.8SiO 2 + 1.2 Al2 O3 + 0.65 Fe 2 O3

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should be in the range 0.66 to 1.02. This will ensure the formation of C3S, C2S and C3A, which are responsible for giving strength, in desired proportions. SiO 2 (2) Silica modulus, should be 2.2 to 3.5. Al2 O3 + Fe 2 O3 (3) Fine grinding of the raw materials which helps the kinetics of reaction. (4) Maintaining the MgO content below the specified limits which ensures that the cement is “sound”. (5) Maintaining alkali chlorides within the specified limits 3.13.2 Methods of Manufacturing Cement Wet process. This process was in predominant use in India and Europe until recently. In this process, the raw materials are finely ground and blended in the desired proportion and the mix is brought to the condition of a free flowing slurry containing 30 – 40% water. It involve the following steps. (i) Crushing and Grinding: The raw materials are crushed in large jaw crushers or coarse gyratory crushers. Then they are ground in the raw mills. (ii) Formation of slurry: The fine mixture of raw materials is mixed with water to form slurry. The slurry is thoroughly homogenized with the help of compressed air and stored in storage basins. (iii) Burning: The slurry from storage, is fed to the rotary kiln. Rotary kiln is a long horizontal steel cylinder lined with refractory bricks and rotating at a speed of 0.5 to 2 rotations per minutes. The slurry of the raw materials enters from the upper end of the rotary kiln while the burning fuel (pulverised coal, oil or natural gas) and air are introduce from the lower end of the kiln. The slurry gradually descends in the kiln into different zones of increasing temperature:

1. The upper part of the kiln is known as drying zone where the temperature is about 400°C. In this zone, most of the water is driven out of the slurry because of the hot gases. 2. The upper central part having a temperature of about 400–700°C is known as pre-heating zone. In this zone, clay and magnesium carbonate decompose. 3. The lower central part of the kiln is known as calcining or decarbonating zone where the temperature ranges from 700 to 1000°C. Here, lime stone is decomposed to give CaO and CO2. CaCO3 CaO + CO2 4. The material then enters the hottest zone (1350 to 1500°C) known as burning and clinkering zone, where lime and clay react with each other forming aluminates and silicates:

2 CaO + SiO2 → 2 CaO . SiO2 (C2S)



3 CaO + SiO2 → 3 CaO . SiO2

(C3S)



3 CaO + Al2O3 → 3 CaO . Al2O3

(C3A)





dicalcium silicate

tricalcium silicate tricalcium aluminate

4 CaO + Al2O3 + Fe2O3 → 4 CaO . Al2O3 . Fe2O3 (C4AF) tetracalcium alumino ferrite

The compounds then combine together to form small, hard, greyish pellets called cement clinkers.

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Fig. 3.12. Rotary Kiln for Cement manufacture.

C3 A + C4 AF , which is known as the C3 S burnability index. This is usually kept in the range 0.45 to 0.85. Too much of flux leads to balling and too little, to bad coating. The clinker formation is an exothermic reaction. 5. Cooling of hot clinker. The hot clinker emerging from the kiln is cooled by various systems such as rotary coolers, planetary coolers or air quench type coolers. In the coolers, the clinker is cooled with atmospheric air. The hot air so produced is used for drying the coal before pulverization. The composition of the clinker depends upon the ratio

The quality of cement produced also depends upon the rate of cooling. Cooling of the clinker should be controlled to produce a definite degree of crystallization of the molten clinke . Portland cement flow char

6. Grinding the clinker with gypsum. The cooled clinker is then finely pulverised together with 2 to 6% gypsum (which acts as a setting time retarder of cement water paste) in long tube mills. The finer the cement, the greater is the strength of the concrete made from it

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7. Storage and packing. The cement coming out of the grinding mills is stored in concrete storage silos. Moisture free compressed air is used to agitate the cement and to keep it free from compaction by its own weight. In India, cement is usually packed in jute bags each holding 50 kg nett of cement. Dry process. In this process, the calcareous and argillaceous materials are crushed in gyratory crushers to small pieces, dried and mixed in proper proportion, pulverised in tube mills and homogenised in a mixing mill with the help of compressed air. This “raw meal” is introduced into the upper end of the rotary kiln while a blast of burning coal dust is blown from the other end. The reactions taking place and the rest of the process is same as described under wet process. Semi dry process. In this process, the raw materials are initially ground dry, but instead of feeding as a powder the ‘raw meal’ is nodulised with 10–14% water in a pan or drum type noduliser. The nodules are fed on a travelling grate where they get dried and preheated before entering a short rotary kiln where they are burnt to form cement clinker. Dry Process Vs Wet Process

Dry Process 1. It is slow and costly process. 2. Cost of production of cement is less, as the fuel consumption is low. 3. The quality of cement produced is inferior.

Wet Process It is comparatively faster and cheaper process. Cost of production higher become of the higher fuel consumption. The quality of cement produced is somewhat superior. 4. This process is adopted when the raw This process is preferred when the raw materials materials are quite hard. are soft. 5 A shorter kiln is sufficien Longer kiln needed to drive off the excess water. 3.14 CHARACTERISTICS OF THE CONSTITUTIONAL COMPOUND IN CEMENT The properties of cement depend upon the relative proportions of the constitutional compounds present and each of them has different characteristic properties. These constitutional compounds are also called microscopic coostituents. (1) Tricalcium aluminate (C3A): The strength developed by different constitutional compounds in cement with time is represented in Fig. 3.2. Tricalcium aluminate undergoes hydration at a very fast rate. It is responsible for the initial set or flash set. Its early strength is good but the ultimate strength is quite low as shown in Fig. 5.2. Its heat of hydration is about 210 cals/ gram* (879 KJ/Kg), which is the highest amongst all the constitutional compounds of cement.** Its rate of hydration is 82.5%, as followed by X-ray diffraction studies. Fig. 3.13. Strength developed by different constitutional compounds in cement.

* KJ/Kg = Kilo joules per kilogram. This is the unit of heat in MKS system. 1 cal/g = 4.185 Joules/g = 4.185 KJ/Kg. ** The rate on hydration is the percentage of hydration over 7 days using Type I cement at water to cement ratio of 0.4. It is followed by X-ray diffraction studies.

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(2) Tricalcium silicate (C3S): It develops very high strength quite early and the ultimate strength is also the highest. Its rate of hydration is medium (73.5%). Its heat of hydration is about 120 cals/ gram (502 KJ/Kg). (3) Tetracalcium aluminoferrite (C4AF): It does not contribute much to the strength of cement because both its early strength and the ultimate strength are poor and the lowest among the constitutional compounds. Its rate of hydration is slow (57%) and hence it is slow setting. Its heat of hydration is about 100 cals/gram (418.4 KJ/Kg). (4) Dicalcium silicate (C2S): This hydrates very slowly. Its rate of hydration is 37.5%. Its heat of hydration is the lowest among all the constitutional compounds of cement and is about 60 cals/ gram (251 KJ/Kg). Its early strength is quite low but develops ultimate strength almost of the same order as C3S. 3.15 ADDITIVES FOR CEMENT Any material entering into concrete other than cement, water and aggregate is known as an admixture. Any material interground with the cement clinker (other than gypsum normally used in the manufacture of cement) is called an addition. Admixtures and/or additions are classified as under 1. Accelerators: These are added to increase the early strength development. Chemical accelerators commonly used include common salt, CaCl2, some organic compounds such as triethanol amine, some soluble carbonates, silicates and fluosilicates. CaCl2 is the most widely used accelerator. 2. Air-entrainment agents: These have assumed great importance primarily from the standpoint of pavement durability against alternating cycles of severe cold weather and the injurious action of salts used in snow removal. The action of air-entraining agents is similar to that of a foam or froth stabilizer. On account of favourable effects of air entrainment on workability and texture of the concrete, air entrainment is being frequently extended from pavement concrete to other constructions where durability is not a serious problem. Vinsol resin and Darex are the commonly used commercial air-entrainment agents which are introduced as “additions” during grinding of the clinker. 3. Retarders: These are used to offset the accelerating effect of temperature from hot weather concreting or hot water flows in grouting, to prevent the premature stiffening of some cements, or to actually delay the stiffening under difficult placing conditions. Admixtures of very small quantities of carbohydrate derivatives and calcium lignosulfonate are the more commonly used retarders. 4. Water repelling agents: These are used in 0.1 to 0.2% of the weight of the cement and are usually present in waterproofed Portland cements and many masonry cements. The commonly used water-repelling agents include soaps or other fatty acid compounds such as calcium–, ammonium–, aluminium, or sodium stearates or oleates and petroleum oils or waxes. 5. Workability agents: These are usually employed to offset deficiencies in grading that tend to produce harshness or segregation and jeopardize successful placement under inaccessible difficult conditions. Examples are bentonite clay and diatomaceous earth which are used upto 3 to 5% by weight of cement. Other examples are fly ash, clay, fi ely divided silica, fi e sand, hydrated lime, talc and pulverized stone, some of which are added even upto 20% by weight. Some of the commonly used air-entraining agents also increase workability. 6. Gas forming agents: Aluminium powder is the widely used gas forming agent. It reacts with the hydrating hydroxide in concrete to permeate the mass with minute hydrogen bubbles. Amounts added are of the order 0.005 to 0.02% by weight of the cement. However, larger quantities are used to produce the light weight, low strength, sound or heat insulation filler concrete known as Acrocrete.

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7. Pozzolanic materials: Finely divided siliceous and aluminous substances e.g., fly ash (fine flue dust, which is obtained as a byproduct of thermal power-plants), volcanic ash, heat treated diatomaceous earths, heat treated raw clays and shales which are not cementitious in themselves, combine with hydrated lime and water to form stable compounds of cementitious value. These are generally used upto 10 to 35% of the cement (as cement substitutes) in large hydraulic structures (mass concrete works) to lower the heat of hydration and to instill greater resistance to sea water, sulfate bearing soils, or natural acid waters. The specific gravity is lower than that of cement and substitution by equal weights increases the relative bulk of fine material thereby improving workability and reducing bleeding and segregation. The rate of gaining strength is slower but under favourable curing conditions, the later strengths are higher with most of the pozzolanic admixtures. 8. Natural cementing materials: These are the natural cementing materials such as hydraulic lime, water quenched blast furnace slag and lime. These are used upto 10 to 25% by weight of Portland cement. These may increase workability, decrease the bleeding and segregation, decrease the heat of hydration and usually decrease the strength when used in larger quantities. Some of them may contribute to the strength of the concrete through their own chemical activity. Natural cementing materials generally require a longer curing period for the development of their potential strength. 9. Miscellaneous admixtures. These include colouring pigments, integral floor hardeners, pore fillers and additives for resistance to wear and decrease of dusting 3.16 PROPERTIES OF CEMENT 3.16.1 Setting and Hardening of Cement When water is mixed with cement and allowed to stand, hydration of cement take place. The mixture eventually becomes stiff and hard. This process is called setting. After hydration, anhydrated compounds become hydrated. These anhydrated compounds have less solubility, hence they are precipitated as insoluble gets or crystals. These gels or crystals have the ability to surround sand crystal, stone and other inert materials and bind them very strongly. Hardening is the development of the strength because of crystallization. The first occurs within 24 hours and the subsequent hardening required 15-28 days. The chemical change that take place during setting of cement are not clearly understood. It is generally believed that the setting times of C2S, C3S, C3A and C4AF are 28 days, 7 days, 1 day and 1 day respectively. When cement is mixed with water, the paste becomes quite rigid within a short time which is known as initial set or flash set. This is due to C3A which hydrates rapidly as follows:

3 CaO . Al2O3 + 6 H2O → 3 CaO . Al2O3 . 6 H2O (crystals)

However, these crystals prevent the hydration reactions of other constitutional compounds forming barrier over them. In order to retard this flash set, gypsum or plaster of paris is added during the pulverisation of cement clinkers. Gypsum retards the dissolution of C3A by interacting with it forming insoluble complex sulfo aluminate which does not have quick hydrating property. 3 CaO . Al2O3 + x H2O + y CaSO4 . 2H2O → 3CaO . Al2O3 . y CaSO4 . z H2O

(1 – 3)

(10 – 33)

The tetracalcium aluminoferrite (C4AF) then reacts with water forming both gels and crystalline compounds as follow: 4 CaO . Al2O3 . Fe2O3 + 7 H2O → 3CaO . Al2O3 . 6H2O + CaO . Fe2O3 . H2O

gels

crystals

* The term “Pozzolanic” comes from “Pouzzoles”, a city near Naples where volcanic silico-aluminate calcium ash is found.

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These gels shrink with passage of time and leave some capillaries for the water to come in contact with C3S and C2S to undergo further hydration and hydrolysis reactions enabling the development of greater strength over a length of time. 3 CaO . SiO2 + x H2O → 2 CaO . SiO2 (x – 1) H2O + Ca (OH)2



gels

2 CaO . SiO2 + x H2O → 2 CaO . SiO2 . x H2O

crystals

gels

The setting and hardening of cement may be summarized diagrammatically as follows: Unhydrated cement Hydration

Metastable Gel

Crystalline hydration products

↓                      Stable Gel  →  Crystalline products

An abnormal type of set is sometimes encountered where the cement paste stiffens quickly, but without the evolution of considerable heat and may again be rendered fluid by remixing. This condition is called false set. This is due to (1) the dehydration of the gypsum during grinding process brought about by excessive temperature in the mills. The resulting cement then contains anhydrite, CaSO4 which quickly sets by hydrating to gypsum, CaSO4, 2H2O, and (2) presence of alkali carbonates in the cement, which may form during storage of alkali containing cements. The alkali carbonate could then react with Ca(OH)2 produced by the rapid hydrolysis of C3S, thus generating CaCO3 which brings about some rigidity of set. Cement + Water ↓ Plastic mass

Metastable gel (Colloidal) Stable get

Crystalline Products (hydrated) Crystalline Products (Coarser dimensions)

Fig. 3.14. Schematic Diagram of setting and Hardening of cement.

3.16.2 Heat of hydration When water is mixed with Portland cement, some amount of heat is liberated due to hydration and hydrolysis reactions leading to setting and hardening of cement. On the average, the quantity of heat evolved during complete hydration of cement is of the order of 500 KJ/Kg. As described earlier, the heats of hydration of the different constitutional compounds are in the following order: C3A > C3S > C4AF > C2S KJ/Kg 878 502 418 251

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Therefore, wherever large masses of concrete are poured into positions (such as construction of dams), it is necessary to dissipate the heat generated during hydration as quickly as possible to avoid the formation of shrinkage cracks on setting and hardening. 3. Soundness. If a cement on hydration produces only very small volume changes and that such volume changes are well within tolerance limits (laid down in the specifications) the cement is said to be “sound”. Presence of excessive quantities of crystalline magnesia contributes to delayed expansion or unsoundness. The soundness is determined by Le Chatlier’s test in which the expansion of a test piece in boiling water for 3 to 5 hours is measured. Recently this test is replaced by the “Autoclave test”. 3.16.3 Testing of Cement and ISI Specification In order to maintain the quality of cement, various tests are conducted from raw material stage right upto the cement in packing stage, at every half an hour to one hour intervals. The final product cement is tested for various physical and chemical characteristics. Different types of cements have to satisfy their relevant specifications. Some important specifications for ordinary portland cement, as per Indian Standard : 269–1967, are given below. Chemical requirements 1. Lime saturation factor. CaO − 0.7SO3 = 0.66 to 1.02 2.8SiO 2 + 1.2 Al2 O3 + 0.65 Fe 2 O3

2.



3. 4. 5. 6.

Al2 O3 Not lesser than 0.66 Fe 2 O3 Insoluble residue : Not more than 2% MgO : Not more than 6%. SO3 : Not more than 2.75% Loss on Ignition: Not more than 4%

Physical requirements 1. Setting time: Initial : Not less than 30 minutes Final: Not more than 600 minutes 2. Compressive strength: (1 : 3 cement mortar cubes cement and blended Ennore sand) 3 days  Not less than 1.6 Kgf/mm2 7 days  Not less than 2.2 Kgf/mm2 (Kgf = Kilogram force = 9.807 Newtons). 3. Soundness: By Autoclave method : Expansion not more than 0.8% By Le Chatiler method : Unaerated cement : max 10 mm Aerated cement : max 5 mm 4. Fineness: As specific surface b  Not less than  Blain permeability method  215 m2/Kg

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BASIC ENGINEERING CHEMISTRY Table 3.3 General Composition of Ordinary Portland Cement:

Average Chemical Composition CaO SiO2 Fe2O3 Al2O3 MgO Na2O & K2O SO3

: : : : : : :

60 to 66% 17 to 25% 0.5 to 6% 3 to 8% 0.1 to 5.5% 0.5 to 1.5% 1 to 3%

Average Composition with respect to Constitutional Compounds C3S C2S C3A C4AF Free CaO MgO CaSO4

– – – – – – –

48% 27% 10% 8% 0.9% 2.5% 2.8%

Reactions taking place during burning of the raw materials in the rotary kiln: < 800°C — Formation of CA, C2 F and C2S starts 800 – 900°C

— Formation of starts

900 – 1100°C

— Formation and decomposition of C2AS; Formation of C3A and C4AF starts; Full decomposition of CaCO3 takes place and formation of free CaO reaches a maximum.

1100 – 1200°C

— Maximum formation of the constitutional compounds viz., C3A, C4AF, and C2S

1200 – 1450°C

— Formation of C3S with gradual disappearance of free CaO takes place.

Table 3.4 Thermochemical changes taking places during cement formation:

Temperature range Reactions occurring

Nature of the heat-change

> 100°C ≥ 500°C ≥ 900°C ≥ 900°C 900 – 1200°C 1250 – 1280°C > 1280°C

Endothermic Endothermic Exothermic

Evaporation of free water Evolution of combined water from clay Amorphous dehydration products of clay start crystallizing Evolution of CO2 from CaCO3 Reaction between clay and lime Liquid formation starts Further liquid formation and completion of the formation of constitutional compounds

Endothermic Exothermic Exothermic The net heatchange may be endothermic

Table 3.5 Action of some chemicals on concrete:

Chemical

Reactions on concrete

Chlorine

Continuous exposure to water containing 5 to 10 ppm can cause surface etching of concrete. Acidic solutions of bleaching powder can attack concrete The acid types of ink, containing free organic acids and H2SO4, attack concrete. Solution of calcium bisulphite can attack concrete.

Bleaching Powder Ink Calcium

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bisulphite Sodium sulphide Sodium sulphite

High alumina cement is more resistant for attack. Solutions of moderate concentration can attake Portland cement, forming sulphides. High alumina cement is more resistant for attack. Solutions can attack concrete. High alumina cement is more resistant and bisulphite for attack. Sodium thiosulphate Leakage of photographic wastes containing hypo as well as pure hypo (hypo) solutions, can attack and disintegrate concretes, brickwork motors and renderings. Borax Slight attack may be there. Tri-sodium phosphate No appreciable effect upto 5% solution. Tan liquors Destructive action on concrete. Formaldehyde, Formic Aqueous solutions are destructive to concrete acid, or Acetic acid Detergents Acid detergents, such as those containing phosphoric acid, may attack Portland cement concrete slowly. High alumina cement is less resistant to detergents containing free alkali hydroxides and more resistant to acid detergents. Types of Portland Cement and its Derivatives The properties and uses of various types of Portland Cement and its derivatives are summarised in Table 3.6. Table 3.6 Types of portland cement and its derivatives

Sr. No.     (1)

Type (2)

True portland cements 1. Orindary cement (Type I) 2. Rapid hardening cement

Characteristics (3)

Uses and remarks (4)

As discussed earlier, C3S – 48%, C2S – 27%, C3A – 10%, C4AF – 8% CaO – 0.9%, MgO – 2.5%, CaSO4 – 2.8%, Loss on ignition – 0.8% Manufactured in the same manner as that of ordinary Portland cement excepting that the lime saturation factor is maintained relatively higher and the final product is ground to more fineness. Contains greater proportion of C3S than ordinary cement so that a more rapid gain of strength is achieved for the mortar or concrete. More expensive to produce composition: C3S – 54%, C2S – 18%, C3A – 12%, C4AF – 8%, CaO – 1.2%, MgO – 2.5%, CaSO4 – 3%

An all purpose construction material most widely used.

For emergency constructions for high early strength and for use in prestressed concrete constructions. Used for urgent constructions as in the constuction of border roads during emergency, where high early strength is desired. This is achieved by having greater amount of C3S (by introducing high percentage of lime in the raw material mix) and carrying out grinding to a greater fineness

156 (1) 3.

BASIC ENGINEERING CHEMISTRY (2) Low heat cement

(3)

Heat of hydration is low and within specified limits at specified ages. High percentages of C4AF and C2S and low percentages of C3S and C3A than those of ordinary type. The heat of hydration is only about half of that of the ordinary cement, so that the shrinkage cracks are reduced to a minimum. The raw materials are to be selected in such a way as to maintain C3A to minimum. The specficiation for low heat Portland cement is that, the heat of hydration measured in an adiabetic calorimeter should be less than 65 and 75 cals/g (272 and 314 KJ/kg) for 7 and 28 days respectively Composition : C3S – 20%, C2S – 53%, C3A – 5%, C4AF – 16%, CaO – 0.4%, MgO – 1.8%, CaSO4 – 3%. 4. Oil well cement Very expensive. Proportion of of C 3 A should be absolute minimum which is achieved by having a high iron content in the raw materials which helps in producing C 4 AF in preference to C 3 A. Special retarders like sugars, cellulose derivatives and organic acids are added in controlled quantities for lengthening the setting time. 5. Hydrophobic or water- This is nothing but ordinary port proof cements -land cement to which a waterrepellent agent (e.g., calcium stearate or rosin, oleic, lauric and stearic acids or pentachlorophenol which are hydrophobic in nature) is added during grinding. 6. White cement The raw materials used (lime- stone and clay) should be free from iron. 7. Sulfate-resisting cement Composition so adjusted to have higher C4AF and lower (Type V) C3A than ordinary cements.

(4) For mass concrete work where low liberation of heat is desirable such as in dams and other monolithic works. However, in recent times, blended cements are preferred to low heat cements for mass concrete works. For example, in Bhakra Dam, low heat cement was not used. Instead of that, both pozzolanic cement and chilling of concrete were employed in the mas concrete work.

For cementing steel castings of oil and gas wells which go to depths of the order 1000 m where temperature and pressure are very high.

For rendering concrete more impermeable to water and for imparting better storage properties under high humidity conditions.

For making decorative pastel shades for cement paints and for making coloured cements. Resistant to sulfate or chloride bearing waters.

B– CEMENT AND REFRACTORIES (1) 8.

(2) Expansive cements

9. Moderate heat cement (Type II) Blended Cements: (a) Portland Pozzolana cement

157 (3)

(4)

Ordinary portland cement has This is mostly used in France the drawback of certain and USA. Used where amount of shrinkage after shrinkage characteristics of setting and hardening, which the cements are undesirable. produces small cracks. In case of concrete pavements, some gaps are inevitably formed which have to be filled with special compounds. In order to overcome these drawbacks, expansive shrinkage compensating cements are prepared. The chief ingredient for this purpose is calcium aluminate (CA & C 5A 3) prepared by heating a mixture of high purity lime stone, bauxite and gypsum (25 : 25 : 50) to control the rate of formation of calcium sulfoaluminate during hardening. Contains lesser quantity of Used where the construction C3A which is more susceptible should resist moderate sulfor sulfate action. Composition : fate action and also where a C3S – 44%, C2S – 30%, lower heat of hydration is C3A – 7%, C4AF – 12%, desired. CaO – 0.7%, MgO – 2.9%, CaSO4 – 2.8% Pozzolana is a material which does not have any hydraulic property in itself but is activated into a hydraulic material. Portland pozzolana cement is obtained by grinding together Portland cement clinker and burnt clay or pulverised fly ash (from power plants) or bricks or burnt shale or any other pozzolanic material in the proportion of 3: 1 together with 6% gypsum. Such a product has lower heat of hydration, lower porosity and better sulfate resistance than ordinary Portland cement.

Particularly suited for mass concrete works such as construction of dams and piers. Because of lower porosity, this type of cement is useful for lining of canals. Offers resistance to sulfate attack. Improves workability and reduces liberation of heat. About 7 million tonnes of fly ash are currently produced annually. It has latent hydraulic properties. Fly ash having more than 6 – 7% unburnt carbon may cause difficulties in the final product

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(b) Portland blast furnace slag cement

(3)

Most important type of blended cements manufactured in India and in other countries also. As per the I.S.I. specifications, the granulated blast furnace slag should form 25 – 65% of the blend. In actual practice, the cement and granulated slag are in the ratio 1 : 1. When molten blast furnace slag at a temperature of about 1400°C is brought into contact with a jet of water under high pressure, a granulated slag is produced because of the sudden quenching and exfoliation. The product containing 7 to 22% moisture is dried and transported to cement factories. When this finely granulated slag powder is ground with portland cement or hydraulic lime, it is activated into a cementitious material so that the entire blend is as good as portland cement itself. According to the ISI specifications, this cement should have the same strength, soundness and setting time as that of ordinary portland cement. Further, it is claimed to have some special characteristics such as low heat of hydration and greater resistance to sea and sulfatic waters. (c) Super sulfated cement. Prepared by intergrinding a mixture containing 80–85% granulated slag, 10–12% anydride and 5 – 6% portland cement. This product is highly resistant to sulfatic and marine waters. When exposed to a temperature of ≥ 40°C over long periods, this type of cement deteriorates in strength. (d) Masonry cement

(4) This product meets the twin objectives of increased production of a good cementing material as well as utilizing a largely produced industrial waste material. It is an all purpose cement with low heat of hydration and volume stability concrete. Hence this is useful in mass concrete works and in grouting oil well casings. However, this type of blended cement has to be ground to greater fineness than the ordinary portland cement, to get identical strength. Thus the cost of grinding is higher.

Suitable in situations where the concrete is exposed to sea water and sulfate-bearing soils. Owing to its low heat liberation, it may be used for mass concrete jobs.

Produced by intergrinding portFor producing mortars having land cement with ground limestone better plasticity than that of or an inert filler along with air- ordinary portland cement. entraining and/or plasticiser additives. This improves plasticity and water retaining power and reduces shrinkage.

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3.17 GYPSUM Gypsum is CaSO4.2H2O. Pure gypsum is colourless to white but impure forms could be grey, pink brown or red. It is found in beds or bands in sedimentary rocks such as lime stone, sand stone or shales. On heating of gypsum a number of cementing materials related to CaSO4 are formed due to partial or full dehydration as follows: 1 heat heat CaSO 4 .2H 2 O  CaSO 4 H 2 O  150°C → 200°C → CaSO 4 2 Gypsum



Plaster of paris or Hemihydrite

  CaSO 4   CaO + SO3 800°C

anhydrite

Beyond 200°C, gypsum gets fully dehydrated and is converted to dead burnt gypsum which does not exhibit any setting properties.

REFRACTORIES 3.18 REFRACTORIES Refractories are such inorganic materials which can withstand very high temperatures without softening, melting or deformation. The essential function of a refractory is to serve as structural material and maintain its mechanical functions at high temperatures under the service conditions. 3.18.1 Requisites of a good refractory The requisites of a good refractory material are as follows : (1) They should possess good refractory properties i.e., their physical, chemical and mechanical properties should not undergo substantial changes at high temperatures. (2) They should be chemically stable under the service conditions in which they are employed i.e., they should not react with corrosive agents such as acidic or basic molten slags, hot gases etc. (3) They should possess good thermal strength i.e., they should be able to withstand thermal shock due to rapid and repeated temperature fluctuations (4) They should possess good resistance to abrasion by dusty gases and erosion by molten metals. (5) They should be able to withstand the charge load at the working temperature and other severe operating conditions. (6) They should possess low permeability. 3.18.2 Classification of Refractorie Classification of Refractorie

Acid Refractories They are resistant by acid but attacked by basic matrials e.g. silica, fire clay etc

Basic Refractories They are resistant by basic materials but attacked by acid material e.g. Magnesite, dolomite etc.

Neutral Refractories They are resistant by slightly acidic and slightly basic media e.g. carbon, graphite, etc.

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Physical forms Refractories are supplied to the trade in several physical forms which include bricks, finely ground cementing materials, plastics, castables, and granular materials in bulk. The principal refractory product is a brick or other preformed shape. However, there are numerous sizes and shapes to fit all types of construction: rectangular forms, wedges, arches, keys, skews, jambs, feather-edges, necks, bung arches and segmental shapes including circle brick, cupola and rotary kiln blocks. These are used in coke ovens, runners, tuyeres, burners, muffles, crucibles, saggers, glass pots, stoppers, nozzles, tubes, feeder parts for glass tanks, spark plug cores and highly specialized items of laboratory ware. Refractory materials are also available as mortars or cements, for laying-up, coating or patching brickwork. They are supplied either dry or wet in a ready mixed form for immediate application. They may be air setting at ordinary temperatures, or heat setting during furnace operations. Plastic refractories are essentially moist unformed brick mixes supplied for forming special shapes and solid jointless (monolithic) furnace sections at the installation point. They can be rammed into place with relatively low pressure and fired by the heat of the furnace in which they are installed Castables are refractory concretes and the aggregates now comprise of almost all common refractory materials. Highly porous refractory aggregates are used for insulating castables. Bulk products are prepared from refractories like grain magnesite, dolomite, chrome ore, fire clay, sand, and ganister and are supplied in different grain sizes for use in making bottoms, banks and fills of furnaces, as well as for other miscellaneous applications 3.18.3 Properties of Refractories (i) Refractoriness. It is the property of a material by virtue of which it can withstand high temperatures without appreciable softening or deformation under working conditions. Refractoriness is usually measured by the softening or fusion temperature of the material. Obviously, a refractory material should have softening temperatures higher than the operating furnace temperatures. Most of the commercial refractories soften gradually over a wide range of temperatures and do not exhibit sharp melting points because they are composed of several minerals, both crystalline and amorphous in nature. Fusion temperatures of some pure and commercial refractories are listed in Table 3.7. Table 3.7 Fusion Temperature of some Refractories

Refractory Material Silica (SiO2) Silica brick Fire clay brick Kaolinite (Al2O3.2SiO2.2H2O) Bauxite brick High alumina clay brick Alumina (Al2O3) Magnesia brick Magnesia Chromite (FeO.Cr2O3) Chromite brick Spinel (MgO.Al2O3) Silicon Carbide (SiC)

Fusion Temperature, °C 1710 1700 1600-1750 1785 1732-1850 1802-1880 2050 2200 2830 1770 1950-2200 2135 2700

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Forsterite (2MgO.SiO2) Sillimanite (Al2SiO5) Zirconia (ZrO2) Zirconia brick Boron nitride Boron Carbide (B4C) Zirconium boride (ZrB) Titanium boride Silicon nitride (Si3N4) Molybdenum disilicide (MoSiO2) Lime Carbon (C)

1890 1816 2710 2200-2700 2720 2450 3040 2940 1900 2100 2570 3500

The softening behaviour (fusion point or refractoriness) is commonly determined by means of the standard PCE (Pyrometric Cone Equivalent) test. The pyrometric cone were first developed by Dr. Herman August Segar to measure the refractoriness. Therefore these cones are also known as segar cones. The cones are small pyramid shaped pieces, prepared from carefully controlled mixtures of minerals. The cones are fixed on a base almost in the upright manner and heated in a definite manner. On heating, these cones are melt or fused. The temperature at which the cone finally touches the plaque is known as softening temperature or refractoriness. The softening behaviour of cones are compared with standard Pyrometric cones calibrating for testing the refractories. Standard data are available for each of the standard Pyrometric cones at various heating rates. Table. 3.8

Suger cone Number

Temperature in °C

1 5 10 15 20 30 35

1100 1180 1300 1435 1530 1670 1770

Fig. 3.15. Pyrometric Cone Equivalent Test

The PCE test is also used to check the uniformity of composition of refractory raw material and finished products, to classify fire clay refractories, and to determine the contamination of the refractories from the fluxes and other material encountered in serice conditions

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(ii) Refractoriness Under Load (Strength): The Commercial refractories which are used for lining high temperature furnace are expected to withstand varying load of the charge. Therefore, the refractory should have high strength and good temperature resistance. Refractoriness under load is an important consideration because usually a refractory fails at a lower temperature when subjected to load e.g. alumina brick and fire clay refractory they collapse for below their fusion temperature. This is supposed to be due to the fact that some crystals which have become fluidic act as a lubricant and deformation becomes easier when subjected to load. Some other refractory material like silica bricks when exert good bearing charactecteristics up to their fusion temperature. Thus for good results, refractoriness under load (R.U.L) test is performed by applying a load (3.5 or 1.75 Kg/cm2) to the refractory specimen (of size 5cm2 and 75 cm high). The sample is then kept in carbon resistance furnace and heating is started at the rate of 10 °C/ minute. The height of the specimen is plotted against temperature and R.U.L. is expressed as the temperature at which 10% deformation take place. (iii) Porosity: Porosity of a material is given by the ratio of its pores volume to that of its bulk volume. The higher the porosity of a refractory brick, the more easily it is penetrated by gases and molten fluxes. For a particular class of refractory brick, the one with the lowest porosity may be taken as the best because it will have the greatest strength, heat capacity, and thermal conductivity. Further, it will have greater resistance to abrasion and corrosion although the resistance to thermal spalling decreases with the decrease in porosity. Thus, porosity is an important property of a refractory material as it is related directly to many other physical properties. (iv) Thermal spalling: Rapid change in temperature, cause uneven expansion and contraction of refractory material. Due to which internal stresses and strains develops, which lead cracking, breaking or fracturing of refractory bricks under high temperature, this whole process is known as thermal spalling. Spalling may also occur due to penetration of molten slag into the pores of refractory brick which leads to cracking due to the differences in their coefficient of expansion and contraction. A good refractory must show a good resistance to thermal spalling. It can be minimized by

(a) Using a refractory bricks having low coefficient of expansion, good thermal conductivity and high porosity (b) By avoiding sudden change in temperature. (c) By over firing the refractory bricks during manufacture (d) By improving furnace design to minimize the stresses and strain.

(v) Thermal Conductivity: Thermal conductivity is an important property of a refractory because it determines the amount of heat that flows through a furnace wall under given service conditions. Insulating refractories with low thermal conductivity will not allow much loss of the furnace heat and hence they are used in blast furnace, open hearth furnace, etc. On the other hand, refractories with high thermal conductivity are used in the construction of muffle walls, retorts and recuperators where efficient heat transfer from the outer surface to the charge is needed. The densest and least porous bricks possess the highest thermal conductivity due to the absence of air in the voids. If the brick is porous, the air entrapped in the voids (or pores) provides an insulating effect. Porous bricks can be prepared by mixing copious amount of a carbonaceous material with the refractory mix before moulding. When the moulds are burnt, the carbonaceous material burns off leaving behind minute voids which provide the insulating effect.

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(vi) Heat Capacity: The heat capacity of a furnace depends upon the following three factors: (i) the thermal conductivity (ii) the specific heat and iii) the specific gravity of the refractor . Lightweight refractory bricks have low heat capacity and hence are suitable for intermittently operated furnaces because the working temperature of the furnace can be achieved in lesser time and with lesser consumption of fuel. Conversely, the dense and heavy fire clay bricks have higher heat capacity and as such are best suited for regenerative checkerwork systems used in coke ovens, stoves for blast furnaces, glass furnaces etc. (vii) Electrical Conductivity: Refractories are poor conductors of electricity with the exception of graphite. Refractories used for lining of electric furnaces should also have low electrical conductivity. (viii) Bulk Density: This influences many other important properties. A high bulk density will improve strength, volume stability, heat capacity and spalling resistance. However, for insulating refractories, a porous structure is needed which is provided by a low density of the refractory. (ix) Dimension Stability: It is resistance of a material to any change in volume when the refractory is exposed to high temperature, over a prolonged period of time. Types of Dimensional Changes (a) Permanent Contraction: When refractory is subjected to high temperature for long duration, either low fusible constituent melts away forming liquid which fills the pores of the refractory body causing shrinkage or vitrification or change of one crystalline form of material into another denser form can also take place e.g. magnesity bricks.

Amorphous MgO Sp. gr = 3.05

High temp → crystalline

sp gr = 3.54

With increase in density there is shrinkage in such bricks. (b) Permanent Expansion: When refractory is subjected to high temperature for a large time, transformation of one crystalline form of refractory into another form of low density take place. Due to which expansion of refractory take place. e.g. silica refractory when subjected to high temperature, crystalline transformation in silica bricks take place. Quarts

————→ Tridymite

Sp. gr = 2.65

S.p. gr = 2.26

————→ Cristobalite S.p. gr = 2.32

(x) Permeability: It measures the diffusion of molton solids, liquids and gases through the pores of refractory. Higher the porosity of a refractory bricks, more easily the gases penetrate. Permeability depends on the size and number of pores. The good refracting material should show permeability. (xi) Texture: Refractory material should be coarse or fine. Porosity of coarse or light textured bricks are higher than fine or dense textured bricks. Thus coarse textured refractory bricks have following properties. (a) good resistance to thermal spalling (b) low crushing strength and (c) low corrosion and abrasion resistance.

3.18.4. Types of Refractory Products 1. Acid Refractories (A) Fire Clay Refractories. Fire clays are the most widely used refractory materials and are well suited for a variety of applications. These contain 25 to 44% of Al2O3 and upto 70% SiO2. The

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chief raw materials used for the manufacture of this type of refractory are calcined fire clays, and since these clays can withstand high temperatures, these are called fire clay refractories. Calcined fire clay, known as the “Grog”, accounts for 50% or more of the batch mix. Grog is prepared by calcining fire clay at about 1450°C in rotary kilns, followed by crushing, sieving and grading. The graded fractions are suitably blended. Plastic clay is the usual bonding material used. For high grog bricks, additional bonding material like molasses, dextrin or sulphite may also be required. Addition of water may vary from 3 to 30% depending upon the method of forming proposed to be used subsequently. Moulding is done by hand or by machine. Hand moulding is done for non standard sizes and shapes. In this the mixture containing 15 to 20% water is rammed and beaten into mould taking care to see that air pockets, voids etc., are absent. Bricks of low porosity and high slag resistance are generally produced by dry pressing the mix containing 3 to 74 water in de-airing type mechanical or hydraulic toggle presses at a pressure of 75 to 150 MN/m2. Hollow bricks like sleeves, pipes and tuyers are prepared in specially designed vertical auger machines using semiplastic mixes having 10 to 12% water. Properties

1. The fire clay refractories are pale bu f to light brown in colour. 2. Their hardness depends upon the firing conditions. 3. Properly fired bricks are as hard as steel. 4. Their porosity varies from 8 to 30%. 5. At high temperatures, the fire clay refractories combine with soda, potash, lime, MgO, FeO, sulphates, chlorides and carbonates to form fusible salts. 6. Their crushing strength in cold is about 950 kg/cm2 which goes down with increasing temperature. 7. Their refractoriness depends upon the softening point of the fire clay used. Their safe working temperature is about 1545°C. 8. They have softening point of 1350°C under a load of 50 lbs sq.in. 9. Their spalling tendency can be decreased by using a coarse grog, by increasing the porosity, and decreasing the coefficient of expansion. 10. The specific heat of fire clay brick is as low as 0.25 which increases with temperature and reaches 0.264 at 1300°C. 11. Its thermal conductivity varies from 0.8 to 0.95 (in CGS units) from 300°C to 1100°C. 12. Though they are bad conductors of heat, their radiation power is high and hence fire chambers made of fire clay can be heated to very high temperature compared to silica bricks. 13. Their spalling resistance makes them particularly suitable for making checker work of regenerating furnaces which are susceptible to temperature fluctuations. 14. Fire clay refractories are classified as high duty (> 40% Al2O3), moderate heat duty ‘A’ (> 30% Al2O3 and < 65% SiO2) and moderate heat duty ‘B’ (> 25% Al2O3 and < 70% SiO2) types.

Uses

1. The steel industries are the largest consumers of fire clay refractories. 2. They are used for lining of blast furnaces, open hearths, stoves, ovens, flues, crucible furnaces etc. 3. Fire clay refractories are also widely used in foundries, lime kilns, regenerators, pottery kilns, continuous ceramic and metallurgical kilns, glass furnaces, cupolas, brass and copper furnaces, boiler settings, gas-generators, etc.

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(B) Silica Refractories. The raw material used for the manufacture of silica bricks are quartz, quartzites, sand stone, gannister etc. The rock is first crushed and ground with 2% lime and water (binding material). The thick paste is then made into bricks by machine pressing. Bricks are then dried slowly and than they are fired in the kiln for 24 hours. The temperature is slowly raised to 1500°C. This temperature is kept constant for twelve hours. During this period the quartzite is converted into crystobalite. Careful cooling is then taken in 1 to2 weeks. Cristobalite is slowly changed into tridymite and final brick is the mixture of the two form Properties

1. Silica bricks have homogeneous texture, free from air pockets and moulding defects. 2. They possess a low porosity (17 to 25%). 3. These are the properties desirable for resistance to slag penetration for a refractory used in furnaces. 4. The silica bricks can withstand a load of about 3.5 kg/cm2 upto about 1600°C. 5. They are resistant to thermal spalling below 800°C. 6. They possess low permeability to gases. 7. Their physical strength when heated is much higher than that of fire clay bricks. 8. Furnaces using silica bricks must be heated and cooled gradually to minimize spalling and cracking.

Uses

(i) The largest consumer of silica is the iron and steel industry. (ii) Due to their high Physical strength, they are used for arches in large furnace. (iii) They are used for side walls, port arches, and built-heads of open health furnaces, coproduct cake ovens, gas retorts, glass furnaces, electric furnace, roofs, copper stove, domes, acid converter linings etc. (C) High-alumina Refractories When the alumina content in fire clay brick reaches above 47.5%, then it is called high alumina refractory brick. These are made from clays rich in bauxite and diaspore. The refractoriness and the temperature of incipient vitrification increase with the alumina content. BASIC REFRACTORIES (A) Magnesite Refractories: It is made from dead burnt magnesite grain, which are properly crushed into powder form of proper size. Molasses or sulphite lye is used as a binder. 2 to 6% of alumina is added to impart thermal shock resistance of the refractory. Now add sufficient amount of water into it and ingredients are blended. The mix is aged for 1 to 10 days to ensure complete hydration of any free lime present. The mix is then moulded into bricks and temperature is slowly increased to 1500°C. The bricks are kept at this temperature for about 8 hrs and then slowly cooled. Fire bonded magnesite bricks are susceptible to spalling. This drawback is removed by chemical bonding or by metal casing. For preparing chemically bonded magnesite refractories, the mechanical strength is developed by adding a chemical bonding agent e.g., oxychloride or oxysulphate, instead of by firing. In this case, higher forming pressures are needed to minimize voids. Curing and drying are carried out in specially designed tunnel dryers. Properties



1. Magnesite bricks have low resistance to acid slag but high resistance to basic slag. 2. They have high thermal conductivity and low permeability.

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3. Their resistance to spalling for normal bricks is very poor. 4. However, chemically bonded and metal cased bricks have better spalling resistance. 5. Magnesite bricks can be used upto 2000°C without load and upto 1,500°C under load of 3.5 kg/cm2. Their abrasion resistance is poor. 6. They have a tendency to combine with water and CO2. Uses 1. Magnesite refractories are preferred where basic materials in molten state are to be heated at high temperatures. 2. These refractories are used in open hearth and electric furnace walls in the roofs of nonferrous reverberatory furnaces e.g., those used for Cu, Pb and Sb. 3. They are also used for lining of basic converters in steel industry, hot mixer linings, copper converters, refining furnaces for Ag, Au and Pt. etc. (B) Dolomite Refractories Dolomite is a mixed carbonate of calcium and magnesium CaMg (CO3)2.For preparing dolomite bricks, the mineral is washed, crushed and calcined. Due to which dolomite decomposed and form CaO and MgO. Now mix it with binder (e.g. silicate) and water. The mixture is allowed to stand for sometime and then moulded into bricks. The dried bricks are then fired at 1500°C in a kiln for about 24 hours. Properties:

(i) They posses less strength and low resistance to thermal shocks. (ii) They are more soft, more hygroscopic, more porous and possess less strength as compared to magnesia bricks. (iii) They have greater shrinkage and lesser resistance to thermal shock. The properties of dolomite bricks can be improved by mixing the dolomite with MgO SiO2 and calcining the mixture. Which brings about the formation of di and tri-calcium silicate. Now blend the mixture with silicate binder. Mould the mixture into bricks followed by fi ing at 1500°C for a day or two to give stabilized dolomite bricks. These bricks are more stable towards basic slags. Uses Dolomite refractories are quite cheap and used in granular form to pitch the bottom of openhearth furnace and for repairing works. Stabilized dolomite bricks are used in Bessemer convertors, ladle linings and for basic electric furnace lining. (C) Magnesite-chrome and Chrome-magnesite Refractories Magnesia bricks do not stand much load at elevated temperatures. In order to overcome this difficult , they are blended with chrome ores. If the blend contains more of magnesite, it is called magnesite-chrome and if chromite predominates, it is called chrome-magnesite. These refractories are made from dead burnt magnesite and chrome ore mixed in proper proportions. This may be fire bonded, chemically bonded or metal cased as in the case of magnesite bricks described earlier. A coarse grading provides better thermal shock resistance. Their resistance to acid slag is low and resistance to basic slag is high. They have high resistance to abrasion, erosion and spalling. Their load bearing capacity and volume stability at elevated temperatures is high. Chrome-magnesite refractories have a tendency to bursting in contact with iron oxide.

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(D) Forsterite Refractories Forsterite (2MgO.SiO2) is used both as a bonding material as well as a base for high-temperature refractories. When forsterite is used as a base, the refractories are usually made from olivine 2 (Mg, Fe) O.SiO2 which is characterized by its high refractoriness. In the manufacture of forsterite refractories, dead burned magnesite is usually added to convert some accessory

MgO.SiO2 Enstatite

+

MgO ¾ Magnesia

—→

2MgO.SiO2 Forsterite

Such materials are called super-refractories which possess high melting point, good volume stability and stability against transformations during heating. Their preparation does not require calcining. Uses Forsterite is used in glass tank super-structures and checkers because they possess high chemical resistance to the fluxes used and good strength at high temperatures allow increased tank output. These refractories are also used in open-hearth end walls and copper refining furnaces NEUTRAL REFRACTORIES (A) Carbon Refractories For preparing carbon refractories low ash content (0.15%) carbon (e.g. coke derived from coal) are used. At first crushed the coke and dried it in a rotary dryer. Now mix it with 15% tar or pitch in a dry pal mill. The mixing or blending carried out when the mixer is hot at 58 to 80°C. Moulding is carried out by dry pressing of the hot mixture at a pressure of 30 to 40 MN/m2. Hardening takes place within 1 to 6 days, depending upon the size. Firing is carried out in a reducing atmosphere at 800 to 1000°C packing the shapes in carbon. Properties

(i) Carbon refractories have high refractoriness, high resistance to acid slag, high resistance to thermal shock and spalling and high load bearing capacity at high temperatures. (ii) They have low resistance to basic slag, low resistance to oxidising atmosphere and possess low thermal expansion. (iii) Their thermal and electrical conductivity is wide ranging. (iv) They react with water above 600°C and CO2 above 700°C. (v) Carbon refractories are practically infusible, close textured and can withstand temperature fluctuations and chemical attack in neutral or reducing condition .

Uses Carbon refractories are used for making electrodes and for linings in chemically reactive equipments. (B) Graphite Refractories Graphite is a natural allotropic form of carbon having dark colour, bright metallic lustre. In addition to various other industrial applications, graphite is also used as a refractory material. Graphite refractories are produced by one of the following two methods : (1) by graphatising of pear shaped carbon products by prolonged heating (3 to 5 weeks) at about 2850°C in an electric furnace (2) by shaping of graphite flakes bonded with fire clay or carbon. This method is particularly used for preparing crucibles. The shapes are dried at 80 to 100°C for one day in controlled humidity

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driers and fired at 900 to 1200°C in down draught periodic kilns or continuous muffle kilns. In order to prevent oxidation, the crucible surface is coated with boric acid, borax, soda ash, sodium silicate, CaCl2 or MgCl2. The fired crucibles are slowly cooled. In India, clay or carbon bonded graphite crucibles upto 100 kg capacity are produced in small scale sector. The properties of graphite refractories are more or less similar to the carbon refractories excepting that they have higher oxidation resistance than the latter. Uses Graphite crucibles are widely used in industry. Graphite bricks are used for construction of electrodes, linings of chemically resistant equipments, atomic reactors, electric furnaces and in nonferrous metal smelting furnaces. (C) Carborundum or silicon carbide refractories It is a neutral refractories. Silica carbide or carborundum is made by heating a mixture of coke 40% and sand 60% together with some saw dust (it increase the porosity) and salt (it helps in the removal of iron in the form of volatile chlorides) in an electric furnace at 1500°C. The interpacked crystals of Sic are crushed, sized and graded suitably for having dense packing. These are then mixed with bonding agent like plastic fire clay, graphite or tar for imparting oxidation resistance. After mixing with bonding agents firing is done in reducing atmosphere at 1500°C. The silicon refractories generally contain 85-90% Sic. Self bonding type of silicon carbide refractory bricks can be prepared by mixing Sic particles with a temporary binding agent like glue. Then pressing and firing at 2000°C when inter crystalline bond develops Properties

1. Silicon carbide super-refractories are characterised by their chemical resistance, high refractoriness and ability to withstand sudden temperature fluctuations. 2. The bricks are extremely refractory and possess high resistance to spalling and abrasion. 3. They have a high thermal conductivity, low thermal expansion, high mechanical strength and can withstand loads in furnaces even at 1650°C and higher. 4. Their resistance to acid slag and to reducing atmosphere is high, whereas their resistance to oxidising atmosphere is medium and their resistance to basic slag is low. 5. Self-bonded silicon carbide refractory bricks have superior properties than silicon nitridebonded bricks which in turn are superior to the clay bonded bricks. Silicon carbide bricks have a tendency to oxidize to silicon in oxidizing atmosphere around 950°C. 6. This tendency can be counteracted by coating with a thin layer of zirconium.

Uses

1. Silicon carbide refractories on account of their good thermal conductivity, are mainly used in muffles. 2. Their ability to absorb and release heat rapidly and their resistance to spalling under repeated temperature fluctuations make them an ideal choice for recuperators. 3. Silicon carbide bricks are also used for partition walls of chamber kilns, coke ovens and furnace floors. 4. Owing to their high electrical conductivity, they are used as heating elements in furnaces in the form of rods and bars known as globars. 5. Silicon carbide bonded with tar are excellent for making high conductivity crucibles. 6. Clay bonded bricks are preferred for high conductivity bricks while lime-bonded bricks are used for muffle and electric furnace lining

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(D) Chromite Refractories Chromite is a mineral of spinel group which is chemically a double oxide of chromium and iron, FeO.Cr2O3. The general term spinel indicates any mineral given by the general composition M,O,M2O3 where M = Ca, Fe and Mg and M' = Cr and Al respectively.

Preparation: At first chromite (Cr2O3. FeO) is blended with clay and then firing is done at temperature of 1500°C to 1700°C. Due to firing process oxidation of Fe ii) to Fe (iii) takes place.

Fe2 Cr2 O4 + O2 —→ Fe2O3 + Cr2O3

Properties

(i) They are neutral refractory having good resistance for slag (ii) Their spalling resistance is average. (iii) They have high density and have moderate thermal conductivity (iv) They are used at 1800°C (v) They have refractoriness under load and have excellent crushing strength

Uses

These bricks are used in (i) Furnace lining (ii) Sodium carbonate recovery furnace (iii) Bottom of soaking pits.

SPECIAL REFRACTORIES (A) Single or Pure Oxide Refractories The refractory industry has been constantly facing demands for products having very high refractoriness and which can withstand very severe operating conditions. Pure oxide refractories are developed to meet these demands. Refractory oxides of interest under this category in the increasing cost per unit volume are: alumina, magnesia, zirconia, beryllia and thoria. All these have been developed commercially for light refractory products. The pure zirconia refractories are used in kilns needed for firing barium titanate resistors (i) Pure Alumina Refractories These are made from calcined or fused alumina having a purity greater than 97%. The usual additions are 2% of Kaolin (which reduces the sintering temperatures and improves pressing characteristics) and 2% of talc or magnesite (which controls the grain size and improves pressing characteristics). The binding agent used depends upon the method adapted for forming. For dry pressing, organic binders e.g., cellulose or polyvinyl alcohol are used. Firing temperatures are in the range of 1700-1800°C depending upon the purity of alumina. Forming can also be done by hot pressing in graphite moulds. Pure alumina refractories belong to neutral type having good resistance to oxidising and reducing atmospheres even upto 1900°C. High temperature kilns furnish alumina bricks which approach pure corundum in properties which have high slag resistances. Such refractories are used where severe slagging occurs. (ii) Magnesia Refractories These are prepared from fused magnesia grains. Cellulose is normally used as binder. Firing temperatures are in the range 1900-1950°C. Care must be taken in their setting to ensure that they do not react with the kiln material at this high temperature. Forming by hot pressing is similar as in the case of pure alumina refractories. This is a basic type of refractory having refractoriness of the order of 2300°C.

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(B) Mixed Oxide Refractories (i) Mullite (3Al2O3.2SiO2) Refractories These are prepared from high alumina minerals and fire clay in proper proportions. The grog is firs prepared by fusing or sintering the raw material mix at 1820°C or by calcining it at 1600°C. After blending the fine powder with a binder such as plastic clay or sulphite lye, the shapes are formed by the usual methods followed by firing at about 1750°C. Addition of a small quantity of aluminium fluoride during blending might reduce the mullitisation temperature Mullite refractories belong to neutral type of refractories. They possess high refractoriness and high resistance to acid slag, basic slag, oxidising atmosphere, reducing atmosphere and to thermal and structural spalling. (ii) Zircon (ZrO2.SiO2) Refractories These are prepared from micronised zircon mixed with finely ground clay or alumina. The grog is prepared by the partial sintering of zircon. Firing temperature is 1650-1700°C. These refractories belong to acid type and possess high resistance to acid slag. Their resistance to thermal shock or spalling is high. Their thermal expansion is low and resistance to molten metals is high. Their resistance to oxidising atmosphere is low. (C) Non-oxide Refractories These include silicon nitride, boron carbide, molybdenum disilicide, zirconium boride and titanium boride. They can be formed into dense shapes by hot-pressing or sintering in controlled conditions. Molybdenum disilicide exhibits oxidation resistance even at 1800°C, while others tend to oxidize from 1000°C. Non-oxide refractories are expensive, and are mostly used at present in nuclear and space research programmes. (D) Insulating Refractories These are characterised by low thermal conductivity and low bulk density. They are of two types : (1) Backing up insulations which are made from highly porous inorganic refractory materials like asbestos, mica, vermiculite and diatomite. (2) Hot face insulations which are similar in composition to dense refractories. The raw materials used for this type of insulating refractories are chromite, magnesite, fire clay, silica and alumina. The insulating value of this type of refractories is due to the method of their preparation. For example, waste cork is ground and sized and is mixed with fire clay. This is then moulded and burnt in a kiln. The cork burns out leaving a highly porous and light weight refractory brick. (E) Monolithic Refractories These are made from a volume stable refractory (e.g., calicined fire clay, SiC, silica, magnesia, alumina, calcined diatomite, vermiculite or expanded mica depending on the refractory type being made) and a bonding agent (e.g., high alumina, cement, phosphate, sulphate, sodium silicate, ethyl, silicate, polyvinyl alcohol or phenolic resins depending upon the-application). Monolithic refractories are in the form of unfired loose mixture which can be made into castables, mortars, gunning compounds or plastic and ramming mixes before application. IS specific tions are available for all these products. Industrial outlets for the Refractories The various types of refractories described above are widely used in iron and steel, non-ferrous metals, cement, gas, power, ceramics, glass and chemical industries as well as nuclear and space research programmes.

B– CEMENT AND REFRACTORIES

171 QUESTIONS

1. Match the following and write appropriate statements in each case:

Group A

Group B

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

Flash set High ultimate strength Lowest heat of hydrogen Retarder Lowest ultimate strength Chalk Shale Hydration and hydrolysis reactions Cement + Sand + Fine aggregates Cement + Sand + Coarse aggregates Superior chemical resistance to sea water Concrete shield for atomic piles

Tricalcium silicate Dicalcium silicate Tricalcium aluminate Tetracalcium aluminaferrite Gypsum Argillaceous material Setting of cement Calcareous material Mortar Concrete High alumina cement Barium and strontium cement

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)

Answers : 1b, 2c, 3a, 4c, 5d, 6g, 7h, 8f, 9i, 10j, 11k, 12l. 2. Draw a labelled diagram of a rotary kiln used for the manufacture of Portland cement by wet process and discuss the various reactions taking place in the furnace. 3. What are the microscopic constituents (or constitutional compounds) present in Portland cement? How do they contribute towards the properties of the cement? 4. What do you mean by setting and hardening of cement? Discuss the various reactions involved with the help of equations. 5. “The properties of Portland cement depend upon the relative proportions of its constitutional compounds”. Justify this statement. 6. What are the different methods of manufacturing cement? Discuss their relative merits and demerits. 7. Write informative notes on the following: (a) Important process parameters for the manufacture of good cement clinker (b) Sequence of operations in the manufacture of Portland cement (c) Reactions taking place in the rotary kiln (d) Constitutional compounds in cement and their properties (e) Additives for cement (f) Important properties of cement 8. Discuss the different types of Portland cement and its derivatives. 9. Write short notes on the following: (a) Setting and hardening of cements (b) Soundness of cements (c) Cement additives (d) Mortar and concrete 10. Write short notes on the following : (a) Curing (b) High alumina cement (c) Magnesium oxychloride cement (d) Strontium and barium cements 11. Discuss the significance of the following with respect to the manufacture and properties of Portland Cement: (a) Lime saturation factor (b) Silica modulus (c) Cooling of clinker (d) Heat of hydration (e) Soundness

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12. (a) Explain the role of gypsum in setting and hardening of cement : (b) Which type of cement will you recommend for the follwing and why : (i) construction of a dam. (ii) Lining of fire bricks in a blast furnace (iii) Oil and gas wells. (d) State the various reactions taking place in the rotary kiln during the manufacture of Portland Cement. 13. (a) What is the percentage of composition of the constitutional compounds of portland cement ? (b) Explain the reactions involved in setting and hardening of cement. (RGPV Bhopal 2001) 14. Explain briefly (a) Hardening and setting of cement (b) Constitutional compounds of portland cement and their role. (RGPV Bhopal 2003) 15. What are the requisites of a good refractory material ? 16. What are refractories ? How are they classified ? State some important industrial applications of refractories. 17. Discuss the important properties of refractories which have a direct bearing on their industrial use. 18. Write informative notes on (a) Fire clay refractories (b) Silica refractories (c) Magnesite refractories (d) Silicon carbide refractories (RGPV Bhopal 2009) 19. (a) Define the term “refractory”. How are refractories classified ? Give two examples of each type. (b) What are the causes for failure of a refractory? 20. Write short notes on the following : (a) Pyrometric Cone Equivalent (b) Thermal spalling (c) Graphite refractories. 21. (a) Discuss the essential characteristics of a good refractory. (b) What type of refractories are used for each of the following and why? (i) Bessemer converter (ii) Regenerative furnace (iii) Blast furnace (iv) Open-hearth furnace. 22. Give an important of the preparation, properties and uses of the following: (a) High alumina refractories (b) Dolomite refractories (c) Fire clay refractories 23. Write informative notes on: (a) Single oxide refractories (b) Mixed oxide refractories (c) Non-oxide refractories. 24. (a) What are the raw materials for refractories ? (b) What are the different steps in the manufacture of refractories ? (c) What do you mean by super-refractories ? 25. Discuss the significance of the following properties in the evaluation of a refractory brick (a) Refractoriness (b) Dimensional stability (c) Spalling (d) Thermal conductivity (RGPV Bhopal 2009) 26. Discuss the various physical and chemical factors which affect the industrial uses of refractories. 27. Write briefly on the preparation, properties and uses of the following (a) Carborundum bricks (b) Zirconia bricks (c) Chromite bricks

B– CEMENT AND REFRACTORIES

173

28. Write short notes on : (a) Insulating refractories (b) Monolithic refractories (c) Silica refractories 29. Fill up the blanks in the following: (a) Refractoriness of a material is def nes as .................. (b) Acid refractories should not be used in contact with .................. whereas basic refractories should not be used in contact with .................. (c) Porosity of a refractory material is given by the ratio of .................. (d) Thermal spalling means .................. (e) The heat capacity of a furnace depends upon the following three factors namely .................. (f) Dimensional stability of a refractory indicates .................. 30. Def ne and classify refractories and discuss any two important properties of refractories. (RGPV Bhopal 2001, 2006) 31. (a) Mentional different raw materials used in the manufacture of silica refractories. Discuss its manufacture giving the various steps involved in the process. Give their important properties and used. (b) Explain the following properties of refractory materials : (i) Refractoriness (ii) Thermal conductivity (iii) Spalling 32. (a) Def ne : (i) Refractoriness (ii) Refractoriness under food (iii) Porosity (b) With the neat f ow diagram, explain the manufacture of silica bricks. State their properties and uses. 33. (a) What are the essential constituents of a good refractory ? (b) Give the manufacture, properties and uses of silica refractories. 34. (a) Discuss with a neat f ow diagram the various steps involved in the manufacture of chrome magnestite refractory bricks. Describe their properties and uses. (b) Mention different causes that lead to the failure of refractories. 35. (a) Explain the term “refractoriness” of the refractories. (b) What are the refractories ? Give the classif cation of refractoriness with examples. 36. (a) What are the essential qualities of good refractory ? (b) Discuss, with the help of a f ow diagram, the manufacture of chrome magnesite bricks. What are their uses ? (c) State the causes that lead to the failure of refractories. 36. (a) Describe the manufacture of chrome magnesite refractories. Give f ow sheet. Give any two properties and industrial uses. (b) Write a brief note on failure of refractories. 37. Write informative notes on the following : (a) Types of lime and their properties (b) Setting and hardening of lime mortors (c) Gypsum plasters (d) Weathering and deterioration of cement concrete. 38. Explain brief y – (a) Silica refractories (b) Thermal spalling and thermal conductivity (RGPV Bhopal 2003)

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UNIT

4

Polymers

4·1 INTRODUCTION The word “polymer” is derived from two Greek words, Poly (= many) and meros (= parts or units). A polymer is a large molecule which is formed by repeated linking of small molecules called “monomers”. For example, the polymer polyethylene is formed by repeated linkages of simple ethylene molecules. n CH= CH 2 → ( −CH 2 − CH 2 − ) n 2 ethylene

polyethylene

And the process of formation of polymer is known as polymerization. The number of repeating units in the chain so formed is called the degree of polymerization. Polymers with a high degree of polymerisation are called “high polymers” and those with low degree of polymerisation are called “oligopolymers”. High polymers have very high molecular weights and hence are called “macromolecules.” The number of reactive site present in a molecule is known as its functionality. For a substance to act as a monomer it must have two reactive sites. e.g. Ethylene is a bifunctional molecule, because when double bond is broken, two single bonds become available for combination. CH2 = CH2 → 4.2 CLASSIFICATION OF POLYMERS (1) On the Basis of chemical composition:

(a) Organic polymer (b) Hetero Organic Polymer (c) Inorganic polymer

(a) Organic polymer: These include compounds containing, apart from carbon atoms, hydrogen, oxygen, nitrogen, sulfur and halogen atoms, even if the oxygen, nitrogen, or sulfur is in the back bone (or main) chain. Organic polymers also include polymeric substances containing other elements in their molecules provided the atoms of these elements are not in the main chain and are not connected directly to carbon atoms. Examples :— Polyethylene

...—CH2—CH2—...

Polyvinyl alcohol

...—CH2—CH—...

PVC

OH ...—CH2—CHCl– 174

POLYMERS

175 OH

Epoxy polymers

… O  R  O  CH 2  CH  CH 2 …

Polyurethane

… C  NH  ( CH 2 ) x  NH  C  O  ( CH 2 ) y  O …

Polysulfides

O O … R  (S) z  R '  (S) z …

starch, cellulose, etc. (b)  Elemento organic or hetero organic polymers: These include : (a) Compounds whose chains are composed of carbon atoms and hetero atoms (excepting N, S and O). (b) Compounds with organic chains if they contain side groups with C atoms connected directly to the chain. (c) Compounds whose main chains consist of carbon atoms and whose side groups contain hetero atoms (excepting N. S, O and halogen atoms) connected directly to the C atoms in the chain. Examples. :—Polysiloxanes Polytitoxanes R R R R | | | | ... — Ti — O — Ti — O — ... ... — Si — O — Si — O — ... | | | | R R R R (c) Inorganic Polymers: These are polymers containing no carbon atoms. The chains of these polymers are composed of different atoms joined by chemical bonds, while weaker intermolecular forces act between the chains. Examples :—

Fig. 4.1. Inorganic polymers.

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BASIC ENGINEERING CHEMISTRY

(ii) On the basis of chemical structure (a) Homopolymer: If the polymer consist of identical monomer it is known as homopolymer ~A~A~A~A (b) Copolymer: If the polymer is a mixture of more than one type of monomer, it is known as copolymer. ~A~B~A~B~A~B~ (c) Random polymer: If in a polymeric chain, the arrangement of monomer molecule is irregular fashion, it is known as random polymer. ~A~B~B ~A~B~A~ (d) Block polymer: Linear polymer in which the identical monomer units form a block (fairly long continuous sequeous) known as block polymer. —A—A—A—B—B—B—C—C—C—C—A—A—A (e) Graft Polymer: In this type of polymer the main chain contain one kind of monomer and side chain contain another type of monomer units.

(iii) On the Basic of their structure or shap (a) Linear: In which momeric units are joined in the form of long straight chairs. ~A~A~A~A~ (b) Branched: They are mainly linear in nature, but also posses some brached along the main chain.



(c) Cross-linked polymer: The contain monomers molecules connected to each other by only covalent bond. These monomers units are connected to each other by cross links.

Due to presence of cross links, they are hard, rigid, brittle and do not melt, but burn on heating. (iv) On the Basic of Physical State

(i) Amorphous: The polymer formed are amorphous in nature. (ii) Crystalline: The polymer formed are crystalline in nature.

POLYMERS

177

(v) On the Basic of synthesis

(a) Addition polymer: In addition polymerization, the polymer is formed from the monomer, without the loss of any material, and the product is an exact multiple of the original monomeric molecule.

e.g.



n

(b) Condensation Polymerization: In condensation polymerization, the chain growth is accompanied by elimination of small molecules, such as H2O, SO2, etc.

Examples :— (1) nHOOC (CH2)4 COOH + nH2N (CH2)6 NH2 Adipic Acid

−H O

2 → 

Hexamethylene diamene

Polyhexamethylene adipamide (Nylon 6-6).



(c) Copolymer: In copolymer, the polymer is formed from the addition of two different kind of monomer units.



(d) Co-ordination Polymer: Polymerization reaction taking place in the presence of organe metallic compounds as catalyst, are known as co-ordination polymerization and the polymer formed is know as co-ordinate polymer.

Karl Triegler in Germany developed a steriochemical process to polymerize ethylene at much lower temperature and pressure than what were required for free-radical polymerization by using Zeigler-Natta catalyst. The product thus formed had fewer branches and possess a higher degree of crystallinity than that of LDPE and hence is called high density polyethylene (HDPE). Zeigler and Natta were awarded the Nobel prize in 1963 for this work. ZieglerNatta catalysts are a novel type of transition metal catalysts, which exhibit a co-ordination or insertion-type of mechanism during the polymerization. A Ziegler-Natta catalyst, in general, comprises of a metal-organic complex of a metal cation from I to III Group of the periodic table [e.g. triethyl aluminium, Al(C2H5)3] and a transition metal compound from IV to VIII Group [e.g., titanium tetrachloride, TiCl4].

178

BASIC ENGINEERING CHEMISTRY For example, HDPE can be produced by bubbling ethylene into a suspension of Al(C2H5)3 and TiCl4, in hexane at ambient temperature. Similarly, polypropylene of about 90% isotacticity can be prepared by polymerising propylene in presence of TiCl3 and diethylaluminium chloride, Al (C2H5)2 Cl, at 50°C. It is postulated that the growing polymer-chain is bound to the metal atom of the catalyst and that monomer insertion involves a coordination of the monomer with the atom. Thus, it is this coordination of the monomer which is responsible for steriospecificity of the polymerization. Coordination polymerisation processes can be terminated by the introduction of hydrogen, water, metals, such as Zinc and aromatic alcohols.

(vi) On the Basis of Tacticity or Configuratio The orientation of monomer unit in a macromolecule can take an order or disorder fashion w.r.t. the chain on the basis of this polymer are of following three types.

(a) Isotactic: If all the side group lie on the same side of the chain (i.e. cis arrangement) it is called and isotactic polymer.



(b) Syndiotic: If the arrangement of side group is in alternating fashion (i.e. trans arrangement) it is called an syndiotic arrangement.



(c) Atactic: If the monomers have entered the chain in a random fashion it is called an atatic polymer.

POLYMERS

179

(v) On the Basis of Intermolecular Force of Attraction



(a) Elastomers: The polymers which are capable of being stretched rapidly atleast 150% of their original length without breaking and return to their original shape on release of stress are known as elastomers. (b) Fibres: They are thin, long threadlike materials which have great tensile strength in the direction of fibres. they posses strong intermolecular forces (c) Thermoplastic: Thermoplastic resins soften on heating and become plastic so that it can be converted to any shape by moulding. On cooling, they become hard and rigid. On re-heating, they soften again and the material can be remoulded to any desired shape. Thus, their softness and hardness are temporary phenomena which are attainable by heating or cooling. The resins that are formed by addition polymerization are thermoplastic and have linear long chain polymeric structure without any cross linkings. In thermoplastic resins, the chemical structure or the molecular weight are not changed during the heating or moulding operations. Only the secondary bonds between the individual molecular chains are broken on heating which results in their softening and flow properties. On cooling, these secondary bonds are re-established as a result of which they become hard again. Obviously, thermoplastic resins are weaker, softer and less brittle as compared to thermosetting resins. The molecular weights of thermoplastic resins are smaller than those of thermosetting resins. Also their inter-molecular forces are weaker than in thermosetting plastics. That is why thermoplastics swell or dissolve in some solvents. (d) Thermosetting resins. Thermosetting resins are those which set upon heating and cannot be reformed when once they are set. In general, those resins which are formed by condensation are thermosetting. The thermosetting resins have three dimensional network structure and have very high molecular weights. These resins have predominant covalent cross links between the long chain molecules which are responsible for the three dimensional network structure. When these materials are moulded, additional cross linkings are formed between the long chains leading to further increase in molecular weight. When cross linkings are formed, the thermosetting resins acquire some of their characteristic properties such as hardness, toughness, non-swelling and non-softening properties, brittleness, etc. The strength of these bonds are retained even on heating and hence they cannot be softened or remoulded or reclaimed when once they are cured as can be done in the case of thermoplastic resins.

(vi) On the Basis of Polarity

(i) Ionic (ii) Non-ionic

(vii) On the Basis of Miscellaneous (i) Steriospecific polymerization. Steriospecific polymerization is one which results in the production of sterioregular polymers. Sterioregular polymers are those in which all units and substituents are arranged in space in some definite order. It may proceed by an ionic or a radical mechanism. A polymer lacking a regular order of its units and substituents is called sterioirregular. These differences in spatial arrangement of units or substituents give rise to configurational isomerism (e.g., cis-trans and D—L isomerism) of polymers. The term coordination polymerisation is sometimes used to the processes that yield polymers with ordered structure (steriopolymers) because the coordination complex between the monomer and the organometallic compound regulates the polymer structure through coordination.

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BASIC ENGINEERING CHEMISTRY

(ii) Step polymerization. Step polymerization involves the combination of several molecules

Fig. 4.2. Step polymerization.

to one another as result of the migration of some mobile atom (generally a hydrogen atom) from one molecule to another. An example of step polymerization is provided by the polymerization of di-isocyanates and dihydric alcohols into linear polyurethanes. 4.3 PLASTICS AND RESINS Plastics are high molecular weight organic materials, which can be moulded or formed into stable shapes by the application of heat and pressure. Resing are the basic binding materials which form the major part of the plastics and which determines the type of treatment needed in the moulding operation. From engineering point of view there are two types of resins. (i) Thermoplastic resins (ii) Thermosetting resins On the basis of the type of the resin used in its preparation, the plastic itself is called thermoplastic or thermosetting plastic. (i) Thermoplastic resins. Thermoplastic resins soften on heating and become plastic so that it can be converted to any shape by moulding. On cooling, they become hard and rigid. On reheating, they soften again and the material can be remoulded to any desired shape. Thus, their softness and hardness are temporary phenomena which are attainable by heating or cooling. The resins that are formed by addition polymerization are thermoplastic and have linear long chain polymeric structure without any cross linkings. In thermoplastic resins, the chemical structure or the molecular weight are not changed during the heating or moulding operations. Only the secondary bonds between the individual molecular chains are broken on heating which results in their softening and flow properties. On cooling, these secondary bonds are reestablished as a result of which they become hard again. Obviously, thermoplastic resins are weaker, softer and less brittle as compared to thermosetting resins. The molecular weights of thermoplastic resins are smaller than those of thermosetting resins. Also their intermolecular forces are weaker than in thermosetting plastics. That is why thermoplastics swell or dissolve in some solvents. Examples : Polyethylene, polystyrene, polyvinyls, (such as nylons), acrylics and cellulose derivatives. (ii) Thermosetting resins. Thermosetting resins are those which set upon heating and cannot be reformed when once they are set. In general, those resins which are formed by condensation

POLYMERS

181

are thermosetting. The thermosetting resins have three-dimensional network structure and have very high molecular weights. These resins have predominant covalent cross links between the long chain molecules which are responsible for the three-dimensional network structure. When these materials are moulded, additional cross linkings are formed between the long chains leading to further increase in molecular weight. When cross linkings are formed, the thermosetting resins acquire some of their characteristic properties such as hardness, toughness, non-swelling and non-softening properties, brittleness, etc. The strength of these bonds are retained even on heating and hence they cannot be softened or remoulded or reclaimed when once they are cured as can be done in the case of thermoplastic resins. Examples. Phenol formaldehyde resins (Bakelite), urea and melamine formaldehydes resins, alkyds, polyesters resins, silicones, etc, 4.4 MECHANISM OF POLYMERIZATION 4.4.1 (i) Addition Polymerization: Addition polymerization is a typical chain reaction. The three type of mechanism for addition polymerization are given

(a) Free Radical Addition Polymerization (b) Cationic Addition Polymerization (c) Anionic Addition Polymerization All the three mechanism completes in following three steps



(i) Initiation (ii) Propagation (iii) Termination

Free Radical Mechanism The free radical mechanism involves the formation of free radical in the inition step. (i) Initiation: In free radical Polymerization, at first a free radical generated with the help of initiators. The generation of free radical takes place as a result of homolytic dissociation of an initiator, which may take place in the presence of heat energy, light energy or catalyst. In second step of initiation, the free radical so formed gets attached to the monomer molecule and form the chain initiating species. R—R → 2R* Initiator Free radicals



R* + CH2 = CH2 → R — CH2 — CH2 Free radical Monomer (ii) Propagation: In this step, additional monomer units are attached to the initiated monomer species and form a long chain 





R — CH2 — CH2 + CH2 = CH2 —→ R — CH2 — CH2 — CH2 — CH2



R — CH2 — CH2 — CH2 — CH2 + CH2 = CH2 —→



— R — CH2 — CH2 — CH2 — CH2 — CH2 — CH2 — The propagation step involves a continuing attack on fresh monomer molecules which in turn, keep successively adding to the growing chain one after another.

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BASIC ENGINEERING CHEMISTRY

Termination In this step the propagating polymer chain is terminated by any of the following reaction. (a) By Combination: The two growing chains may come close and colloide with each other and get deactivated. —→



(b) By disproportionation: In some cases one hydrogen atom from one growing chain is absorbed by the other growing chain. Due to which one saturate and one unsaturate polymer molecules are formed.



—→ Dead polymer

(c) Thus the product so formed do not contain any reactive sites known as dead polymer, because these cannot grow further. In addition to the above two method of chain termination, chain transfer by hydrogen abstraction from solvent molecule, may also terminate the growing chain. R—[ CH2 — CH2— ]n CH2 — CH2 + SH —→ R—[ CH2 — CH2— ]n CH2 — CH2 + S



Solvent

Ionic polymerization In ionic polymerization, the active centres initiating the chain reaction are “ions”. Ionic polymerization proceeds due to the presence of catalysts and hence it is called catalytic polymerization. Depending upon the charge of the ion formed, the polymerization may be cationic or anionic. (a) Cationic polymerization: Cationic polymerization (or carbonium polymerization) takes place with the formation of a carbonium ion, which is a polar compound with a tricovalent carbon atom having a positive charge : + R — CH | R' The catalysts used in carbonium polymerization (or cationic polymerization) are compounds with pronounced electron acceptor properties e.g., AlCl3, SnCl2, TiCl4 boron fluoride etc. The polymerizing monomer is an electron donor such as styrene in the presence of SnCl4. A carbonium ion interacts with a monomeric molecule, and the reaction of the chain growth is accompanied with the communication of a positive charge along the chain. Thus, the growing chain itself is a cation and the molecular mass increases in the course of polymerization. Cationic polymerization completed in following three steps. (i) Initiation: This step involves the formation of a carbonium ion by the attack of a proton on the monomer. Any strong lewis acid like BF3, AlCl3 etc can be used as a catalyst. In this case a cocatalyst like water is required to provide the proton source.

POLYMERS Here

183 H+

BOH]–

act as a initiator and [F3

anion faces the growing chain end +

CH2 = CH2 + [F3BOH] H+ —→ CH2 — CH2 [BF3OH] – (ii) Propagation: The Propagation reaction involves the addition of monomer units. More and more units are added up and the chain keeps on growing. The electron pair of the adding monomer units are pulled in a direction opposite to the growth of the chain and the positive charge keeps on moving in the direction of the chain growth. The counter ion [BF3OH]– moves along the positive charge all the time. +

+

–

CH2 – CH2 [BF3OH] + CH2 = CH2 —→ – CH2 – CH2 – CH2 – CH2 [F3BOH] – (iii) Termination: In this step the growing polymer chain having C + collides with an anion, which may be added deliberately or with the counter ion. The termination takes place– (a) By the donation of a proton to the counter ion by the growing polymer chain +

CH2 – CH2 – CH2 – CH2 [F3BOH] –

–

—→ CH2 – CH2 – CH = CH2 + [F3BOH] — H+

(b) By the formation of covalent bond between the carbonium ion and the counterion (coupling)

+

CH2 – CH2 – CH2 – CH2 [F3BOH] – —→ CH2 – CH2 – CH2 – CH2 – OH + BF3

In this case also the regeneration of the initiator takes place. (b) Anionic Polymerization: Anionic or carbanion polymerization involves the formation of a carbanion, a compound with a trivalent carbon atom carrying a negative charge. Naturally, anionic polymerization occurs in the presence of catalysts which readily yield electrons. e.g. electron donors such as sodium or potassium amide, triphenyl methyl sodium, alkali metals and alkyl alkalis. Carbanion polymerization takes place in case of monomers such as acrylonitrile and methyl methacrylate which contain electronegative substituents at one of the carbon atoms connected by a double bond. Anionic polymerization involves the following steps :

(i) Initiation: In anionic polymerization initiators are strong base such as organo alkali compounds (alky) or aryl derivatives of alkali metals) like n-butyl lithium, ethyl, sodium, alkali metal amides, hydroxide etc.

The electron pair in the case is pushed to the end of the molecule forming the carbanion Li+ is the counter ion. Propagation: The carbanion formed propagates the chain growth by attacking the second monomer unit, pushing its p electron pair further away to the end and forming a sigma bond with the new monomer unit.



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BASIC ENGINEERING CHEMISTRY

In the case of anionic polymerization the movement of the p electrons is towards the direction of the chain growth. Termination: The termination in anionic polymerization is not a spontaneous process. If the starting reagents are pure and no impurities are present, propagation can proceed indefinitely or until all the monomer is consumed. Thus anionic polymerization is also known as “living polymerization”, because the carbanions at the chain ends remain potentially active. If a fresh quantity of monomer is added, polymerization again goes on until the added monomer is consumed. Termination is generally carried out by the transfer of negative charge to a species which is not directly involved in the reaction. E.g. CO2, methanol,water etc.

It may be noted that like free radical polymerization, the initating specie (BU) has been in corporated as the end group of the terminated polymer. Thus Butyl-lithium is an initiator rather than a catalyst. Table 4.1. Difference between Addition and Condenstion Polymer

Addition Polymer/Polymerization

Condensation Polymer/Polymerization

1.

In addition polymerization, the In condensation polymerization, the chain growth is polymer is formed by the addition of accompained by elimination of small molecules, such as H2O, monomer unit, without the loss of any SO2 etc. material.

2

It requires the presence of double It requires two reactive functional group to be present at both bond in the monomer. end of the monomer. There should be atleast two different monomers having functional groups with affinity for each other.

3.

No byproduct is formed

4.

They are Homochain polymer, They are Heterochain polymer either thermoplastic or thermoset generally a thermoplastic is obtained. can be obtained.

5.

The growth of chain is at one active The growth of chain occurs at minimum of two active centre. centre

6.

Example: Polystrene PVC etc.

Generally a byproduct is formed.

Example: Nylon 6 : 6, Bakelite etc.

IMPORTANT THERMOPLASTIC RESINS 4.5 NATURAL RESINS Naturally occurring materials such as shellac, rosin or calophony, cobal, amber, some asphaltic and bituminous materials can act as resins and can be used as binders in plastics. Natural resins are generally hard and possess low thermal conductivity and low dielectric constant. They can be easily moulded. These resins find use as binders for phonographic disc, grading wheels, as electrical insulators and for several other applications. 4.6 CELLULOSE DERIVATIVES Cellulose is a naturally occurring polymeric material containing thousands of glucose like rings each of which contain three alcoholic OH-groups. Its general formula is represented as (C6H10O5)n. The OH-groups present in cellulose can be esterified or etherified. The most important cellulose derivatives are the esters e.g., cellulose nitrate, cellulose acetate and aellulose acetobutyrate, and the ethers, methyl and ethyl cellulose.

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185

(i) Cellulose Nitrate This is also known as nitro cellulose. This is prepared by reacting cellulose with nitric acid in presence of sulphuric acid which acts as a dehydrating agent. The partially nitrated cellulose mixed with camphor gives the so called celluloid, which can be easily softened and moulded. The function of camphor is to act as a plasticizer to enhance the moulding properties.

H 2 SO4 n [ C6 H 7 O 2 .(OH)3 ] + 2n.HNO3  → n [ C6 H 7 O 2 (OH) (NO3 ) 2 ] + 2n H 2 O



H 2 SO4 n [ C6 H 7 O 2 .(OH)3 ] + 3n.HNO3  → n [ C6 H 7 O 2 (NO3 )3 ] + 3nH 2 O

cellulose dinitrate ⋅ nitrogen) (contains 111%

cellulose trinitrate (contains 14⋅1% nitrogen)

Properties

(i) Cellucose nitrate plastics are transparent, flexible, strong and touch. (ii) The specific gravity is 1.35 to 1.40. Its dimensional stability is poor as it slowly shrinks due to slow loss of plasticizer and hence is not used for precision parts. (iii) It is inflammable. (iv) It is resistant to water but is attacked by strong acids and alkalis. (v) It tends to discolour with age. (vi) It becomes brittle at low temperatures.

Uses

(i) Nitrocellulose containing 10.5 to 11.5% N is used for moulded extruded articles; that containing 11.5 to 12.1% N is used in lacquers, and that with 12.1 to 13.8% N is used for explosive gun cotton. (ii) Cellulose nitrate plastics are used for tool handles, spectacle frames, drawing instruments, pens, toothbrush handles, table tennis balls, radio dials, picture films and many fancy article . (iii) The first ‘‘artificial leather’’ was made by coating fabric with nitrocellulose suitably covered and embossed, which is still used in coverings for books, luggage and upholstery.

(ii) Cellulose Acetate It is obtained by treating cellulose with concentrated acetic acid or acetic anhydride in presence of a catalyst such as H2SO4 H 2SO4 n [ C6 H 7 O 2 .(OH)3 ] + 3n.CH3COOH  → n [ C6 H 7 O 2 .(OCOCH3 )3 ] + 3n.H 2 O cellulose

cellulose triacetate

The resulting cellulose triacetate is partially hydrolysed to render it soluble in organic solvents such as acetone. Properties

(i) It has high dielectric strength (ii) It has good clarity, stability, toughness, impact strength and resistance to U.V. radiation (iii) It also has high tensile strength (iv) It shows resistance to mineral acid (v) It has good film and plastic strength

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Uses

(i) It is used for preparing fibres for textiles on one hand and for preparing transparent sheets on the other. (ii) Thick sheets are made from blocks and polishing. (iii) Thinner films can be cast from solution on to polished metal surfaces. (iv) The films are also widely used in photography, for wrapping, and for making small envelopes, bags and boxes for packaging. (v) It is also used for combs, goggles, handles, radio-appliances, steering wheels, etc. (vi) Since cellulose acetate is much less inflammable than cellulose nitrate it has replaced the latter in the manufacture of photographic films

(iii) Cellulose acetate butyrate It is a copolymer having lower water absorption that the cellulose acetate alone. It is prepared by treating cellulose with a mixture of acetic acid and butyric acid in presence of H2SO4 (catalyst). Its thermal stability and other properties are similar to that of cellulose acetate plastics but has better chemical and moisture stability, better dimensional stability and impact strength. It is used for toothbrush handles, combs, buttons, pens automobile hardware, etc. (iv) Ethyl cellulose It is prepared by treating cellulose first with caustic soda to produce the caustic cellulose and then allowing it to react with ethyl chloride to give an average of 2.5 ethoxyl groups per 1/2 ethyl cellulose monomer. ETHYL CELLULOSE CH2ONa

CH2OH O

C C

H

O C

OH H C

C

H

OH

O

H

C

+ NaOH H

O

C

O

H

H

C

C

H C

ONa ONa

1 Cellulose unit 2

1 Soda Cellulose unit 2

CH2OC2H5 O C H

O

C

H

H OC2H5 H C

C

H

OC2H5

C O

Triethyl Cellulose Fig. 4.3. Triethyl cellulose.

+ C2H5Cl Pressure

O

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Properties

(i) It is the toughest cellulose derivative among all the cellulose plastics (ii) Soluble in many cheap solvents (iii) It shows good chemical resistance (iv) Stable to heat and light (v) It shows good electrical properties

Uses

(i) In the formation of chisel handles mallet heads etc. (ii) As electrical insulators (iii) In the preparation of raincoats, surgical tapes, laminated glass sheet and straw etc.

4.7 POLYETHYLENE Polyethylene or polythene is a versatile plastic made from the polymerization of ethylene molecule. n CH2 = CH2



Ethylene

—→

(—CH2 —CH2)n Polyethylene

Polyethylene was first produced in the laboratories of Imperial chemical Industries Ltd. (England) in a fortuitous experiment in which ethylene was subjected to 1400 atmospheres of pressure at 70°C. Traces of oxygen caused the polymerization to take place. There are two variety of polyethylene (i) High Density Polyethylene (H.D.P) (ii) Low Density Polyethylene (L.D.P) (i) High Density Polyethylene (H.D.P.E): It is prepared by polymerizing ethylene at low pressure (1200-2000 atm) and at a temperature of 180-200°C in the presence of metal oxide catalyst. It is also prepared by heating ethylene molecule to 60-200°C under a pressure of 6-7 atmosphere in the presence of Ziegler-Natta catalyst. The polyethylene so produced consists of linear chains of polymer molecules. These molecules are pack close together hence the polyethylene so formed are high density polyethylene. = nCH 2 CH 2

60 − 200°C  → 6 − 7 atm

Ziegler - Natta Catalyst

(  CH 2  CH 2 ) n HDPF

Properties

(i) It shows higher melting point (145-150°C), higher density (0.941 to 9.65) and higher tensile strength (ii) It is much stiffer than LDPE and has high hardness (iii) Chemically resistant (iv) It shows lower gas permeability

Uses

(i) It is used in the manufacture of toys, housewares (ii) For wrapping material for food products

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(iii) Due to low gas permeability it is used in the formation of domestic gas pipes. (iv) It is used in the manufacture of overhead tanks for water storage. (v) It is also used for manufacturing bottles for milk, household chemicals and drug packing.

(ii) Low Density Polyethylene L.D.P.E It is prepared by polymerizing ethylene at high pressure (1000 to 3000 atm) at a temperature of 250°C in the presence of oxygen as a free radical initiator.

The polymer so formed posses branched structure and low density. Properties:

(i) It is a white, waxy translucent material (ii) It has low density (0.91 g/cm3) (iii) Chemically inert, slightly flexible and poor electrical conducto (iv) It shows moderate tensile strength and high toughness (v) Moisture resistant (vi) Its elasticity is preserved even at lower temperature

Uses

(i) It is used as films and sheet for packaging applications, including bags and wrappings, drapes, table cloth etc. (ii) It is also used for insulation for electric wires and cables. (iii) In the manufacture of ‘‘squeeze bottles’’ (iv) It is used for production of industrial containers, pipes for water supply etc.

4.8 TEFLON OR POLYTETRAFLUORO ETHYLENE Polytetrafluoro ethylene (Teflon) is prepared by polymerization of tetrafluoroethylene (CF2 = CF2) at elevated pressures in presence of water using initiators (catalysts) such as benzoyl peroxide, persulfates, H2O2, etc. F F  F F | | |  |   C  C  = nC C  →  | | | |  F F  n R F Tetrafluroethylene

PTFE (Teflon)

Polymerization can lead to reactions of explosive violence. Since the heat of polymerization is high, precautions must be taken to prevent local overheating leading to explosive disproportionation the monomer into carbon and carbon tetrafluoride It is a highly crystalline and orientable polymer. It has many remarkable properties.



(1) It is extremely resistant to attack by corrosive reagents and solvents. Of several hundreds of reagents tested upto their boiling points, only alkali metals either molten or dissolved in liquid ammonia attack the polymer, probably by removing fluorine from the chain. Fluorine itself can degrade the polymer under pressure on prolonged contact. (2) It is totally unaffected by water.

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189

(3) It has an excellent thermal stability even at 250°C. Its mechanical and electrical properties do not change for months together. It decomposes at still higher temperatures. (4) Its melting point is so high that techniques similar to those employed in powder metallurgy have to be used in forming sheets and moulding articles. (5) Moulded polytetrafluoroethylene articles have high impact strength but are easily strained beyond the point of elastic recovery. (6) Its density is unusually high (2.1 to 2.3). (7) Its refractive index is unusually low (1.375). (8) It is not hard, but is slippery and waxy to the touch. (9) It has a very low coefficient of friction (10) It has extremely good electrical properties. Its dielectric constant is low (2.0), and its loss factor for all frequencies tested, including radar and T.V., is one of the lowest known for solids. (11) Its mechanical properties including resistance to wear and deformation under load, stiffness and compressive strength can enhanced by the use of additives or fillers such as asbestos, glass fibre, graphite, bronze powde , zirconium oxide etc. (12) It can be machined, punched and drilled.

Uses Teflon is widely used because of its excellent toughness, chemical and heat resistance, electrical properties and low frictional coefficient

(1) Its electrical applications include wire and cable insulation, insulation for motors, generators, coils transformers and capacitors and high frequency electronic uses. (2) Its low friction and antistick applications include its use a solid lubricant film on ammunition, gun mechanisms and as a bearing surface for light loads. (3) It is also used in non-lubricated bearings, linings for trays in bakeries, etc. (4) It is used in mould release devices, package machines, etc. (5) It does not stick to other materials and is hence called an abhesive. (6) It is accordingly used for making non-sticking stopcocks for burettes and as a thin tape for severing pipe threads. (7) Low molecular-weight PTFE can be dispersed as aerosols for effective dry lubrication. (8) It is used in chemical equipment e.g., gaskets, pumps, valve packing, and pump and valve parts. (9) Its use as fibre include gasketing, belting, pump and valve picking, filter cloths and for various industrial functions where absolute chemical and thermal resistance upto 250°C is required.

4.9 POLYSTYRENE It is prepared by the polymerization of styrene monomer in presence of a peroxide (e.g.benzoyl peroxide). Here peroxide act as a initiator.

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Properties

(i) Polystyrene is a linear polymer which is relatively inert chemically. (ii) It is quite resistant to alkalis, halide acids and oxidizing an reducing agents. (iii) It can be nitrated by fuming HNO3 and sulfonated by con. H2SO4 at 100°C to a watersoluble resin. (iv) Polystyrene is transparent and its high refractive index (1.60) renders it useful for plastic optical components. (v) It is a good electrical insulator and has a low dielectric loss factor at moderate freequencies. Its tensile strength reaches about 8000 psi. (vi) On the other hand, polystyrene is readily attacked by a number of solvents including drycleaning agents. (vii) It has a poor stability for outdoor weathering. (viii) It turns yellow and crazes on exposure. (ix) Its two major mechanical defects are its brittleness and low heat distortion temperature (softening temperature) which is only about 82—100°C. Hence polystyrene articles cannot be sterilized.

Uses

(i) Polystyrene is used for injection moulding of articles such a combs, buttons, toys, buckles, high frequency electric insulators, lenses, radio, T.V. and refrigerator parts, indoor lighting panels, etc. (ii) High impact strength polystyrene made by compounding polystyrene with styrene butadiene synthetic rubber is used as an injection moulding material for producing all the above mentioned articles.

4.10 POLYMETHYL METHACRYLATE It is prepared by the polymerization of methyl methacrylate in presence of acetyl oxide.

Properties

(i) Polymethyl metacrylate is a clear, colourless transparent plastic with a higher softening point, better weatherability and better impact strength than polystyrene. (ii) It transmits 98% of the sunlight including the ultraviolet. (iii) Owing to internal reflection, light entering one end of a spiral or bent rod of this plastic will emerge from the opposite end with almost the same intensity. (iv) Hence it is used to pipe light behind opaque objects through bundle of fibers and appropriate lenses. Uses

(i) It is mostly used for automotive applications such as in tail and signal light lenses, dials, medallions, etc. (ii) It is also used for brush backs, jewellery, lenses and small signs. Its sheets are used for signs, glazing skylights and decorative purposes. (iii) It is also used in aircraft light fixtures, bone splints, paints, adhesives, etc

POLYMERS

191

4.11 POLYVINYL ACETATE Polyvinyl acetate. It is the most widely used polymer. It is used as a plastic and also for the preparation of polyvinyl alcohol as well as polyvinyl acetals, both of which cannot be prepared by direct polymerization. The monomer, vinyl acetate is prepared (1) by bubbling acetylene through hot glacial acetic acid in presence of a catalyst (such as mercuric salts plus H2SO4), or (2) by passing the mixed gases of acetylene and acetic acid at about 200°C over a catalyst containing Zn or Cd salts on charcoal (vapour phase synthesis).

CH ≡ CH + CH 3 COOH  → CH 3 COOCH = CH 2 acetylene

acetic acid

Vinyl acetate

Polyvinyl acetate is prepared from vinyl acetate by heating in presence of a small quantity of benzoyl peroxide or acetyl chloride as catalyst. Emulsion or suspension polymerization is most commonly used although bulk or solution polymerization can also be used.

Properties



(i) This polymer is atactic and hence amorphous. (ii) Its glass transition temperature is only 28°C as a result of which the polymer although tough and stable at room temperature (below 28°C), becomes sticky and undergoes severe cold flow at slightly higher temperatures. Hence the articles formed from polyvinylacetate are distorted even at room temperature under the influence of compressive and tensile forces. (iii) Lower molecular weight polymers are brittle but become gum like when masticated, and hence are used in chewing gums. (iv) Polyvinyl acetate is a clear, colourless and transparent material. (v) It has got water and heat resistance. (vi) It is fairly soluble in organic solvents.

Uses

(i) It is used for the manufacture of polyvinyl alcohol. (ii) Its major use is in the production of water based emulsion paints, lacquers and adhesives (iii) It is used for making chewing gums, surgical dressings, for coating on wrapping paper and cardboard for packing purposes, for the finishing of textile and other fabrics and for bonding papers, textiles, leather, metals, etc.

4.12 POLYVINYL CHLORIDE (PVC) It is one of the cheapest and most widely used plastic globally. Polyvinyl chloride (PVC) is prepared by the polymerisation of vinyl chloride by heating its water emulsion in presence of a small quantity of benzoyl peroxide or H2O in an autoclave under pressure.

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It is prepared from the monomer, vinyl chloride. The vinyl chloride is a gas, boiling at —14°C and is commercially produced by the catalytic addition of dry hydrogen chloride to acetylene. The two gases are passed together at 100 to 250°C over charcoal catalysts containing mercuric or other heavy metal salts. An alternative source is vapour phase cracking. PVC is available in two grade:

(i) Rigid PVC (ii) Plasticized PVC: For the preparation of plasticized PVC plasticizer are added .e.g. tricresly phosphate.

Properties

(i) Commercial PVC is a hard and stiff amorphous plastic (ii) Soluble in cyclohexanone and tetrahydrofuran (iii) Its glass transition temperature is 81 °C. (iv) Its shows flame resistanc

Uses Plasticized PVC (i) For wire coating (ii) As electrical insulator (iii) As tool handles, toys etc. Rigid or unplasticized PVC

(i) For tanks, linings, display and light fitting (ii) In safety helmets (iii) In refrigerator components (iv) Mudguards of cycle and motor cycle.

4.13 POLYAMIDE Polyamides are a group of polymers which contain the amide (—CONH—) linkage in the main polymer chain. Polymers of this type are the synthetic linear aliphatic polyamides which are capable of fibre formatio Nylon is a generic term for synthetic polyamides capables of forming fibres.Aminoacid polymers are designated by a single number, as Nylon : 6 for poly (w-amino-caproic acid) or polycaprolactum. Nylons from diamines and dibasic acids are designated by two numbers, the first representing the diamine, and second representing acid as Nylon 6 : 6 for the polymer of hexamethylene diamine and adipic acid and Nylon 6 : 10 for that of hexamethylene diamine and sebacic acid. (i) Nylon 6:6 It is prepared by the condensation of adipic acid and hexamethylene diamine in the absence of air.

Nylon 6:6

POLYMERS

193

It is known as Nylon 6:6, because both the monomer (i.e. hexamethylene diamine and adipic acid) contain six, six carbon atom. Properties

(i) It has high strength, elasticity toughness and abrasion resistance. (ii) Its solvent resistance is good, only formic acid, phenols and cresols dissolve the polymer at room temperature. Strong acids degrade it somewhat. (iii) It shows good mechanical properties upto 125 °C. (iv) Its m.p. is 264 °C and specific gravity is 1.14 (v) Resistant to moisture. (vi) Hydro carbons resistant.

Uses

(i) It is used as a plastic as well as fibres. Its most important fibre applications include automobile tire cords, ropes, threads, cords having high tenacity and good elasticity. (ii) It is also used to make textile fibres for use in dresses, unde garments, socks etc. (iii) Being a tough plastic it is used as a substitute for metals in gears and bearing etc. (iv) It is used in making rollers, slides and door latches. (v) It is also used in making thread guide in textile machinery (vi) It is used for jacketing electrical ware to provide a tough, abrasion resistant outer cover to protect the primary electrical ware to provide a tough, abrasion resistant outer cover to protect the primary electrical insulation.

(ii) Nylon 6 It is prepared by the self condensation of a-amino carporic acid





The formation of Nylon 6 take place in following four step: (i) Conversion of cyclohexane into cyclohexanone by oxidation

(ii) Conversion of cyclohexanone into cyclohexanone oxime by treatment with hydroxylamine NH2OH

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(iii) Conversion of cyclohexanone-oxime into caprolactum by Backmann rearrangement in the presence of H2SO4



(iv) Ring opening polymerization of caprolactum to give amino carproic acid, which on self polymerization formed Nylon 6

Selt nNH 2  (CH 2 )5  COOH  [ NH 2  (CH 2 )5  CO  ]n polymerization → 

Nylon 6

Properties

Its properties are similar to those of nylon 6:6 but its m.p. is 225 °C and it is less stiff and softer than Nylon 6:6. Uses It is mainly used for making tyre cords. (iii) Nylon 6:10 Nylon 6:10 (Poly hexamethylene sebacamide) is prepared by the condensation between hexamethylene diamine and sebacic acid.

Nylon 6:10 is generally not used as a fibre, but it is suitable for mono filaments, which are used for bristles, brushes etc. (iv) Nylon 11 Nylon 11 or poly w-aminoundecanoic acid is prepared by self condensation of w-amino undecanaic acid w-aminoundecanoic Acid

Nylon 6

Properties and uses It is less water sensitive than the other nylons because of its greater hydrocarbon character. It is used as textile fibre. It is also used for making flexible tubings for conveying petro

POLYMERS

195

4.14 PHENOLIC RESINS or BAKELITE These are the condensation products of phenol or phenolic derivatives (e.g., resorcinol) and aldehydes (e.g., formaldehyde and furfural). These resins are also called phenoplasts. Bakelite was the earliest thermosetting resin named after the Belgian American chemist, Bakeland, who patented it in 1909. It was prepared by condensation of phenol and formaldehyde. The reaction is catalyzed by alkalies and acids. The nature of the product depends mainly on the nature of the catalyst and the proportion of the reaction (i.e. phenol formaldehyde ratio). The formation of phenol formaldehyde resins involves the following steps: (a) Methylolation The first step in the reaction between phenol and formaldehyde is the formation of addition compounds known as methylol derivatives, the reactions taking place in ortho and para-positions. These products may be considered as the monomers for subsequent polymerization. These are formed most satisfactorily under neutral or alkaline conditions.

(b) Novolac In the presence of acid catalyst when the P/F ratio is greater than one, the methylol derivative condenses with phenol and form a liner polymer with little methylo groups. The product is thermosetting in nature and known as novolac.

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Resole Formation In the presence of alkaline catalyst and when the P/F ratio is less than 1, the methylol phenols condenses and formed a linear structure called resole. The resoles are soluble in the reaction mixture and have excess of methylol group, which are capable of further reaction during continued heating.

Formation of Phenolic Resins On heating further novolac in the presence of hexamethylenet etramine (it produces HCHO) as curing agent, it produces a three-dimensional, cross-linked networked polymer. Which knowns as Bakelite.

Cross linking of resoles can be done by heating in either neutral or just acidic conditions. Cross linked product from resole is called Resite.

POLYMERS

197

Properties (i) They have excellent heat resistance (ii) They have good dielectric property (iii) They have good chemical resistance (iv) They have good abrasion resistance (v) They are hard, rigid and scratch resistant (vi) They have remarkable adhesive properties and bonding strength (vii) They are hard rigid and strong materials (viii) They are resistance to cold flo Uses

(i) Because of their good dielectric properties, they are widely used in electrical, automotive, radio and TV parts. (ii) Due to their excellent adhesive properties and bonding strength, they are used for producing brake linings, abrasive wheels and sand paper, and sand filled foundry moulds (iii) Phenolic resins are widely used in varnishes, electrical insulation and protective coating (iv) They are also widely used in the production of ion exchange resins with a variety of functional groups such as amine, sulfonic acid, hydroxyl, phosphoric acid etc.

4.15 AMINO RESINS Amino resins or amino plasts are prepared by the reactions of formaldehyde with nitrogen compounds such as urea, melamine and other amino compounds. Because of their attractive light colours, these resins have great commercial importance. (i) Urea Formaldehyde Resins It is a important amino resins. It is prepared by the reaction of urea and formaldehyde in neutral or acidic conditions. The first products formed during the formation of resin are monomethylol and dimethyol ureas.

Now dimethylol urea undergoes condensation polymerization and formed cross linked polymer.

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Cross-linked urea Formaldyde

When monomethylol urea undergoes condensation polymerization, it formed linear polymer

Thermosetting of the resin takes place during the moulding process. For moulding, the methylol derivative are compound with fillers plasticizer, pigments etc and then are subjected to heat and pressure. The resulting product is hard and infusible substance. Properties

(i) They are clear and colourless. (ii) They have better hardness and tensile strength than phenol formaldehyde resins. (iii) They have good electrical insulating properties. (iv) They have good adhesive property. (v) They are resistant to oxidation to oil, grease and weak acids. (vi) Their heat and moisture resistance are lower than those of phenolic resins.

Uses (i) They are used for the production of light or pastel coloured object. (ii) They are widely used for moulding articles. (iii) They are also used to impregnate wood to prevent cracking. (iv) They are used for the manufacture of coated abrasive paper and binder for foundry cores. (v) They are used for the manufacture of cation exchange resins. (vi) They are used for making buttons, vaccum flask cups and jugs, colored toilet seats etc (vii) They are also used for the finishing of cotton textiles (for crease resistance, shrinkage control, water repellency and fire-retardation) (viii) They are used as an adhesive for plywood and furniture. (ix) They are widely used for making cosmetic container closures. 4.16 EPOXY RESINS Epoxy resins are basically polyethers, but they are given this name due to the fact that their starting material is epichlorohydrin and the polymer before cross linking contains epoxide groups.

POLYMERS

199

The epoxy resin most commonly used is prepared by the condensation of epichlorohydrin with bis-phenol A (diphenylol propane). An excess of epichlorohydrin is generally used to leave epoxy groups on each end of the low-molecular weight (900 to 3000) polymer. Depending on molecular weight, the polymer is a viscous liquid or a brittle solid having a high melting point, (about 150°C). Other hydroxyl containing compounds such as resorcinol, glycols, glycerol and hydroquinone may be used instead of bisphenol-A. However, epoxides other than epichlorohydrin are expensive. The epoxy resins are cured by many types of materials including polyamines, polyamides, polysulfides, phenol and urea formaldehyde and acids or acid anhydrides through coupling or condensation to give three dimensional crosslinked structures. Epoxy resins show outstanding toughness, chemical inertness, flexibility and adhesion in the finished products in this three dimensional cross linked state only. Epoxy resins are mainly used in surface-coating materials, which give outstanding toughness, flexibilit , adhesion and chemical resistance. 4.17 POLYESTERS (i) Decron or Terylene; It is most important polyester It is prepared by the action of terephthalic acid with ethyleneglycol. Terephthalic acid is first react with methyl alcohol and formed dimethyl terephthalate. Now this dimethyl terephthalate react with ethylene glycol and formed diethylene glycol terephthalate. Which on polymerization formed ethylene terephthalate polymer, commonly known as decron or terylene.

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Properties

(i) Its glass transition temperature is 80°C (ii) M.P. 265°C (iii) Chemical and solvent resistance (iv) Its retains its good mechanical properties up to 150-175°C (v) Crease resistance

Uses:

(i) The polyester fibres are generally blended with cotton or wool to make summer and medium weight suiting and other goods. (ii) In making magnetie tapes. (iii) In bypass surgery, decron tubes are good substitutes for human blood vessel (iv) In the preparation of films and for packing frozen food

4.18 SILICONES They are the very high molecular weight polymers in which the siloxane unit –si-o-si, is the continuing framework. Each silicon atom is attached to one or two organic groups. They are prepared by following 3 methods. (i) Direct Method: In this method methyl chloride is passed through a mixer of powdered silicon and powdered copper (act as catalyst). Heating is carried out about 230-290°C 2 CH3Cl + Si ——→ (CH3)2 SiCl2 Methyl Chlorosilane The exist gases are condensed and methyl chlorosilace can be separated by fractional distillation.

Note: Silver catalyst is used in the preparation of phenyl chlorosilane. (ii) Grignard Method: In the Grignard method, a solution of silicon tetrachloride in dry ether is allowed to react with the Grignard reagent which is pumped slowly into the reactor : 2CH3MgCl + SiC4 ——→ (CH3)2 SiCl2 + 2MgCl2 Depending on the amount of Grignard reagent added, 1 to 4 organic groups may be added to silicon. After filtering the MgCl2, the mixture of products formed is separated by fractional distillation. (iii) Olefin Hydrocarbon Method: In this method an olefin hydrocarbon is added to a chlorasilane in a bomb uder pressure and at a temperature of 400 °C without a catalysl or at a temperature of 45 °C using a peroxide catalyst. CH2=CH2 + HSiCl3 ——→ CH3CH2SiCl3

POLYMERS

201

Fig. 4.4. Preparation of silicones.

Properties (i) They shows remarkable stability over wide range of temperature (from-70° to 250 °C) (ii) They shows chemically and physiological inertness (iii) They are good water repellent (iv) They are resistance to the effect of weathering All these properties of silioxanes are due to (a) their molecular size(b) the number and type or organic groups attached to silicon (c) configuration of the molecules Classification of silico On the basis of their physical state and appearance at room temperature silicon are classified into following classes.

(i) Silicone Fluid (ii) Silicon Resins (iii) Silicon Greases (iv) Silicon Rubber

Silicon Resins They contain ring structure. They have a much higher cross link density than silicon elastomer. Resins. The preparations of silicone resins is as much an art as a science. The resins are prepared by hydrolyzing and condensing (or polymerizing) mixtures of bifunctional or trifunctional alkyl chlorosilanes under carefully controlled conditions. The properties of the resins can be varied by controlling the relative amounts of bifunctional and trifunctional units, the type of organic units attached to silicon, the polymer size or configuration, and the addition of other organic resins. It is

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usually necessary to cure the silicone resin at a temperature somewhat higher than that at which it is likely to be used. The silicone resins can be broadly classified into six categories, although there is overlapping because some can be applied in more than one way. Coating resins. These have got distinctive properties of heat resistance, water repellency and resistance to most of the aqueous chemicals and corrosive gases. The pure silicone (unmodified) films are useful on hot stocks, exhausts, electric motors and turbines, stoves, boilers, steam pipes and other places where conditions are too severe for the usual type of protective coatings. Silicone coating resin formulations with aluminium flakes or zinc dust and with or without the addition of organic resin can withstand temperatures of the order of 500°C and can be applied on high temperature processing equipments. Silicone modified alkyd resins have got the additional advantages of giving finishes with greater flexibility and hardness, stability at elevated temperatures, better colour retention and toughness. They are recommended for application to refrigerators, washing machines, stoves and similar household equipments. Laminating resins. They possess high temperature stability and good electrical properties and can be prepared in the form of sheets or tubes. Resin laminated structures are widely used in electrical industry in the form of insulated board, coil former, rods, tubes and slotwedges in motors operating at high temperature. They can be used in aircraft instruments such as radomes. Release resins. The release characteristics of the silicone fluids and compounds are incorporated into the resins. The release resins are widely applied to bread pans, waffle irons, cooky sheets candy pans as well as to ‘‘smoke sticks’’ used in smoking of meats. Water repellent resins. These resins applied as dilute solution coat the pores (but do not fill them) in masonry and concrete. They do not allow water to enter the pores because of their water repellent character and as a result keep the masonry dry. Electrical resins. Silicone electrical insulations possess greater heat resistance, water repellency and resistance to combustion. They can be used with advantage where conditions are too severe for conventional insulators. Impregnating silicone resin compositions are also used in the treatment of capacitor insulations, transformer coils, motor windings and similar equipment demanding excellent chemical, mechanical and dielectric stability. Foamed resins. Silicone resins may be foamed in situ or may be foamed and then moulded to the desired shape by the use of blowing agents. They can be advantageously used for thermal and electrical insulation, firewall structures, buoyancy applications and high temperature vibration dampers. Uses: Silicon resins are used: (i) as electrical insulators (ii) for coating on metal and for coating cooking pans (iii) for lamination purpose. 4.19 ALKYD RESINS Alkyd or Glyptal resins are synthesized from polybasic acids or anhydrides with polyhydric alcohols (such as glycerol) by condensation polymerisation at elevated temperatures in presence of a catalyst.

POLYMERS

203

Since the above polymer is synthesized from glycerol and phthalic anhydride, the alkyd resin thus prepared is called “glyptal resin’’. O

n Phthalic anhydride

Alkyd or Glyptal Resin.

Properties

(i) Alkyd resins are hard (ii) Dimensionally stable (iii) It is infusible, insoluble and resistant to chemical and corrosion

Uses Alkyd resins are used: (i) In paints, lacquers, varnishes, and enamels (ii) In switches and gears (iii) As binder in asbestor (iv) In cement and automobile parts (v) In circuit breaker insulation 4.20 ELASTOMERS An elastomer is a linear polymer which exhibits elasticity and other rubber like properties. The main source of natural rubber is the species of tree known as “Hevea Brasiliensis”, although it is found in several other plants. Today, more than 95% of the rubber is obtained from the rubber tree. ‘‘Hevea Brasiliensis’’, grown on plantation mostly in Ceylon and the Malay peninsula. Small quantities of the rubber are produced in Brazil from the uncultivated wild rubber trees and from the ‘‘guayule’’ shrub in Mexico and southwestern United States.

204

BASIC ENGINEERING CHEMISTRY Natural rubber is a polymerized form of isoprene (2 methyl-1, 3-butadiene) :

CH 2 = C − CH = CH 2 | CH3 The polyhydrocarbon chain consists of 2000 to 3000 monomer links. The polymerization occurs by a biochemical reaction in which a particular type of enzyme acts as a catalytic agent.

For obtaining rubber from rubber plant a cut is made halfway through the bark, expending about 2/3rds around the tree. The liquid is collected in the containers. This liquid is known as latex. Latex is milky collidal emulsion which contain 25 to 45% of rubber. The flow of latex from the cut diminishes with time, necessitating the removal of another thin layer of bark. This process, called tapping, is continued at intervals throughout the life of the tree. A plantation grown tree continues to yield for as long as 40 years and gives latex to the extent that 3 to 6 lb of rubber can be obtained every year. Latex is treated in two ways to obtain rubber goods : (1) the crude rubber is coagulated from it by acids or heat, and then processed (2) the latex itself is mixed with appropriate compounding materials and then precipitated directly from solution in the shape to be used, e.g., rubber gloves. This is a new and important technique. Generally, a small amount of ammonia is added as a preservative to the latex collected. The latex is then coagulated by the addition of 5% solution of acetic acid or formic acid of 90% strength. Ammonium or potassium alum are also used as coagulants. The coagulum is washed and dried. Then it is subjected to any of the following processes : 1. Crepe rubber is prepared by adding a small amount of sodium bisulfite to bleach the rubber and the coagulum is then rolled out into sheets of about 1 mm thickness and dried in air at about 50°C. 2. Smoked rubber is made by eliminating the bleaching with sodium sulfite and rolling the coagulum into somewhat thicker sheets. These are then dried in smokehouses at about 50°C in the smoke from burning wood or coconut shells. 4.20.1 Vulcanization The process of vulcanization is discovered by Charles Goodyear in 1839. He observed that when rubber is heated with sulfur, its tensile strength, elasticity and resistance to swelling are increased tremendously. This process is named as vulcanization. The sulfur combines chemically at the double bonds in the rubber molecule bringing about excellent changes in its properties e.g., resistance to changes in temperature, increased elasticity and tensile strength, durability and chemical resistance. Vulcanization brings about a stiffening of the rubber by anchoring and Fig. 4.5. Unvulcanized and vulcanized rubber. restricting the intermolecular movement of the rubber springs. This is due to the chemical combination of the sulfur at the double bonds of different rubber and providing cross linking between the chains.

POLYMERS

205

The vulcanization (or the curing as it is some times called) can be carried out in several ways : 1. The articles to be vulcanized are heated with steam under pressure. 2. The article is immersed in hot water under pressure. 3. By heating the article in air or in carbon dioxide. 4. By passing steam directly into the article such as fire hose 5. By vulcanizing the article in the mould in which it is shaped. The temperature used in 110 to 140°C. The curing time may vary from a few minutes to 3 hours; overcured stock decreases stretch and tensile strength whereas undercured stock is too soft with excessive stretch but lower tensile strength. The amount of sulfur used for ordinary soft vulcanized rubber is 1 to 5% whereas for hard rubber, it is 40 to 45% of the rubber. Vulcanization of very thin sheets of rubber can be accomplished by either dipping the articles in S2Cl2 or exposing them to vapours of S2Cl2. Other vulcanizing agents used include Se, Te, ZnO, benzoyl chloride, trinitrobenzene, alkyl phenol sulfides, 2S, MgO, benzoyl peroxide etc. Advantage of Vulcanization

(i) Due to vulcanization the tensile strength of vulcanized rubber is 10 time more as compared to raw rubber. (ii) Vulcanized rubber shows resistant to organic solvent like CCl, benzene etc but it swell in them. (iii) It has only slight tackiness. (iv) It has excellent resilience (After the removal of deforming force, the articles made from vulcanized rubber regains their original shape. (v) It has useful temperature range -40 to 100°C. (vi) It has better resistant to moisture, oxidation and abrasion.

4.20.2 Mastication In 1824, Hancock discovered that rubber becomes a soft and gummy mass when subjected to severe mechanical working. This process is called mastication or plasticization. In 1824, Hancock discovered that rubber becomes a soft and gummy mass when subjected to severe mechanical working. This process is called mastication. This process greatly facilitated the addition of compounding agents to rubber, which is usually carried out on roll mills or in internal mixers or Banbury mixer or plasticators which are similar to extruders. Mastication is accompanied by a marked decrease in the molecular weight of the rubber. Oxidative degradation is an important factor in mastication, since the decrease in viscosity and the other changes in properties do not take place if the rubber is masticated in the absence of oxygen. After mastication is complete, compounding ingredients are added and the rubber mix is prepared for vulcanization process. 4.20.3 Compounding of Rubber The masticated rubber is mixed with other substances by the rolls to incorporate the desired properties. All the ingredients may be worked into the rubber thoroughly in a mill or Banbury mixer.



1. Vulcanizers: These are vulcanizing agent mainly sulphur or sulphur mono chloride, hydrogen sulphide, benzoyl chloride etc. The process of addition of vulcanizers into natural rubber is known as vulcanization. Due to vulcanization excellent changes in the properties of raw rubber like increased tensile strength, durability and chemical resistance take place. 2. Accelerators: These are meant for catalyzing the vulcanization process thus reducing the time required for vulcanization. The inorganic accelerators include lime, magnesia, litharge

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and white lead, whereas the organic accelerators are complex organic compounds such as aldehyde amines, thiocarbamates, Guanidines, zinc alkyl xanthate and 2-mercaptol benzothiozole, which are more useful and more commonly used. Sometimes, accelerator activators like ZnO are also added. Generally 0.5 to 1% of the accelerator is used. 3. Antioxidants. These substances when used in small quantities (about 1%) retard the deterioration of rubber by light and air. These are complex organic amines like phenyl naphthyl amine, phenolic substances and phosphites. 4. Reinforcing agents. These are usually added to give strength, rigidity and toughness to the rubber and may form as much as 35% of the rubber compound. The commonly used reinforcing agents include carbon black (for automobile tyres, etc.), ZnO, MgCO3, BaSO4, CaCO3 and some clays. 5. Inert fillers. The main function of the inert fillers are to alter the physical properties of the mix to achieve simplification of the subsequent manufacturing operations, or to lower the cost of the product. 6. Plasticizers or softeners. These are added to impart greater tenacity and adhesion to the rubber. The most commonly used plasticizers are vegetable oils, waxes, stearic acid, rosin, etc. 7. Colouring agents. These are added to impart to the rubber the desired colour, as follows: TiO2, zinc sulfide or lithophon … White Lead chromate … Yellow Ferric oxide … Red Antimony sulfid … Crimson Ultramarine … Blue Chromium trioxide … Green 8. Miscellaneous agents. These include baking soda for sponge rubber, abrasives (e.g., silica and pumice), stiffening agents to stiffen the stock until vulcanization, etc. 9. Calendering. The rubber compound is passed through a calendering machine to convert it into sheets. In this, the material is passed between rolls which press it into thin sheets (0.003 to 0.1 inch thickness). If thicker sheets are required, several thin sheets are rolled together which eliminates air pockets. 4.20.4 Properties of Rubber It have following properties: 1. Concentrated acids react with rubber and destroy it. 2. Rubber is attacked by oxygen in air and by sunlight and the rate of such attack is reduced by anti oxidants. 3. Rubber is soluble in various organic solvents such as petrol, petroleum ether, benzene, turpentine, CS2, CCl4 etc. 4. Solutions of rubber in benzene are used as adhesives. 5. When rubber is treated with a solvent, the solvent is first absorbed by the rubber, which swells and forms a jelly like mass and when enough of the solvent has been absorbed, the mass assumes the liquid state. 6. Thus, in a sense the solvent may be said to have dissolved in the rubber. 7. The specific heat of raw rubber at room temperature is 0.502 and its specific gravity at 0°C is 0.950 and that at 20°C is 0.934.

POLYMERS

207 –6.



8. Its coefficient of cubical thermal expansion is 670 × 1 9. When rubber is extended, heat is evolved and this is called Joule’s effect. When it is stretched to 82%. 680 calories of heat/g are liberated. 10. When rubber is cooled to low temperatures, it becomes stiff. 11. When it is frozen it attains a fibrous structure 12. When rubber is heated with about 1% organic sulfonyl chloride or an organic sulfonic acid at about 130°C, it is converted into a tough, thermoplastic resin which resembles Guttapercha. 13. Such products are known as thermoprenes.



14. Rubber reacts with chlorine giving chlorinated ruber.



15. With HCl, rubber forms rubber hydrochloride.



16. Rubber is oxidized by oxidizing agents such as HNO3, peroxybenzoic acid, peroxide and KMnO4.





17. The oxidation reactions are catalyzed by Cu and Mn.

18. During the normal oxidation of rubber, an unstable peroxide of rubber is first formed followed by transformation into a stable oxide.

4.20.5 Uses of Rubber

1. Rubber is used for the manufacture of gaskets used for sealing refrigerator cabinet doors, etc. 2. Rubber is mainly used for the manufacture of tyres. 3. It is used for preparing V-belts for the power transmission and conveyor belts for conveying several types of materials. These products are compact, non-slipping, clean and shockabsorbing. 4. Rubber lined tanks (steel, Al, etc.) are widely used in chemical plants where protection from corrosive chemicals is required. 5. Rubber mountings are prepared from sandwiching the rubber between two metal plates. They reduce machine vibrations and prolong the life of the machines besides reducing noise. 6. Rubber is used for manufacturing hoses. 7. Rubber threads and sponge rubber have good shock absorbing and thermal insulation properties. Rubber threads are used in shock absorber cords, heat bands for goggles and helmets, golf balls, etc. 8. Rubber is used for various related products like chlorinated rubber, oxidized rubber, rubber hydrochloride, cyclized rubber and ebonites. All these substances have many industrial uses. 9. Foam rubber is used in the manufacture of cushions, mattresses, paddings, etc. 10. Rubber is also used for manufacturing toys, sports items, etc. 11. Rubbers when blended with plastics give improved strength, hardness, flexibility and chemical and thermal resistance.

4.20.6 Manufacture of rubber articles directly from latex Manufacture of rubber goods directly from latex is a recent technique and has several advantages :

(1) There is no need of expensive machinery. (2) Mastication of rubber is not required in this process and hence higher tensile strength can be obtained.

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(3) Time required for vulcanization is much less. In this process, the compounding ingredients are first emulsified in water and then added to the latex. Then, the finished articles are made in any of the following ways (1) Insulated wires may be made by passing the wire through the latex compound. (2) Cords and fabrics can be impregnated by dipping. (3) Sponge rubbers can be prepared by forcing air into the latex compound mechanically, followed by adding a coagulant. (4) Rubber thread can be prepared by extruding the latex into a coagulating bath. Several other techniques are also available and in all these cases, the forming process should invariably be followed by vulcanization. 4.20.7 Polymers Related to Natural Rubber 1. Chlorinated rubber. Chlorinated rubber was traditionally prepared by allowing chlorine gas to react with a solution of masticated rubber in a chlorinated solvent. The newer methods of preparing chlorinated rubber involve the direct chlorination of the latex or passing of Cl2 over thin sheets of rubber swollen with a solvent like CCl4. The mechanism of chlorination involves substitution, addition as well as cyclization. Chlorinated rubber is mainly used for preparing thermal and chemical resistant paints, varnishes and lacquers. Films, impregnating solutions and adhesives also can be prepared from chlorinated rubber. 2. Oxidized rubber: This is prepared by controlled oxidation of rubber by mastication in air in presence of a catalyst. This is used for impregnating paper and cardboard and for protective coatings. Varnishes prepared with oxidized rubber have outstanding electrical insulating properties. 3. Rubber hydrochloride: When HCl gas is passed into a solution of previously milled rubber in benzene, rubber hydrochloride is produced, due to the addition of HCl to the double bonds of rubber as follows : CH 3 | — CH 2 — C — CH 2 — CH 2 — | Cl The presence of chlorine atom on the tertiary carbon has been confirmed by X-ray studies and is also in accordance with Markownikoff’s rule. Stretched and plasticized films of rubber hydrochloride have good mechanical properties, including high tear resistance. Rubber hydrochloride is highly resistant to chemical attack but it is susceptible to thermal and photochemical decomposition, stabilizers are helpful in retarding this decomposition. Rubber hydrochloride is extensively used for wrapping precision machines, machine parts, food materials, etc. 4. Cyclized rubber: Cyclized rubbers are commercially prepared by treating rubber with either H2SO4 or various sulfonic acids or sulfonyl chlorides, or chlorostannic acid. The products are nonelastic and are primarily used as compounding ingredients in shoe soles and heels and for rubber to metal bonding adhesives. 5. Hard rubber (ebonite): Ebonite was discovered in 1840 and the commercial production started since about 1860. If rubber reacts with excess of sulfur, the final product is a hard, inextensable solid, called ebonite. It contains about 32% combined sulfur.

POLYMERS

209

Fillers are usually used in the production of ebonite to reduce the difficulties in handling the rubber sulfur mix before vulcanization. Since reinforcement is not needed, only inert fillers and carbon black are added. Accelerators reduce the curing time from several hours to minutes. The commonly used ebonite dust is prepared by grinding the scrap ebonite. Ebonite stocks for vulcanization are prepared by calendering or extrusion. Ebonite can be machined well and is usually produced in bar, tube or stock for this purpose. Its main uses are dependent upon its chemical inertness, corrosion resistance, thermal and electrical insulating properties. However, it is not suitable for high temperature applications. 6. Guayule Rubber: It is obtained from Guayule shrub and is a source of natural rubber in North America. Rubber latex from Hevea tree exists in a canal system but the rubber Guayule is enclosed in the cells. Rubber obtained from the Guayule bush is recovered by cutting the shrub (after removing the leaves) into small pieces and then milling them in pebble mills in presence of water. The material is then sent to flotation tanks. The rubber floats to the top which is collected. This rubber material contains 70% rubber hydrocarbon, 20% resin and 10% cellulose, lignin and other insolubles. Hevea rubber and Guayule rubber are chemically identical and both exist in the form of cispolymer of isoprene. However, the molecular weight of Guayule rubber is lower than that of Hevea rubber.

Fig. 4.17. Hevea rubber and Guayule rubber (cis polymer of isoprene)

7. Gutta Percha and Balata: This is obtained from the mature leaves of the trees known as Dichopsis gutta and palaqium gutta belonging to Sapotaceae family, which are found mostly in Sumatra, Borneo and Malaya. The mature leaves are carefully ground in mills and treated with water at 70°C for about half an hour and then dropped into cold water. Gutta-percha floating on the surface is collected. Very pure Gutta percha can be recovered by solvent extraction, so that insoluble gums and resins are separated. Gutta-percha is tough and horny at room temperature but turns soft and tacky at about 100°C. It is soluble in chlorinated and aromatic hydrocarbons but not in aliphatic hydrocarbons. Gutta percha is used in the manufacture of submarine cables, golf ball covers, tissue for adhesive and surgical purposes. Balata is obtained from wild trees in Central and South America and the processing and uses are to Gutta percha. Both are the trans-polymers of isoprene.

Fig. 4.18. Gutta percha and Balata (transpolymers of isoprene)

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8. Reclaimed Rubber: Rubber can be reclaimed in a usable form from wornout rubber articles and rubber waste from the factories. This can be done by various methods but the most widely used one is the ‘‘alkali process’’. In this process, the used rubber is separated from metal (by electromagnetic separation) and fiber and heated at about 200°C under pressure for about 8 to 15 hours with an aqueous solution of an alkali (e.g., NaOH) in a closed iron vessel (or an autoclave). This treatment removes the remaining fiber and converts the free sulfur present into alkaline sulfide. The fibre gets hydrolysed and the rubber becomes devulcanized. The resulting material is carefully washed and dried. This may now be mixed with small quantities of reinforcing and processing agents such as clay, carbon black, softeners, etc. Reclaimed, rubber, known as ‘‘rubber shoddy’’, generally has much lower elasticity, tensile strength and wearing quality than fresh rubber. Nevertheless, it may be superior to some of the poorer grades of crude rubber. The properties of the reclaimed rubber mostly depend on the degree to which the plasticity is regenerated by the reclaiming process. Reclaimed rubber is cheaper than fresh rubber. Reclaimed rubber is quicker and easier to process, faster in curing and has better aging properties as compared with fresh rubber. Reclaimed rubber is used for the manufacture of tyres, tubes, belting, hoses, automobile floor mats, hard rubber containers for batteries, soles, heels, steering wheels, mountings, couplings, etc. 9. Sponge rubber: It can be produced by milling the elastomer to the desired extent and mixing with all the necessary ingredients (as in the compounded rubbers) in a mill or in Banbury along with chemical blowing agents (such as Na2CO3) and the modifiers (such as fatty acids) which react with the blowing agents to produce gas under vulcanizing conditions. The mixed stock is then moulded and cured. Chemically blown sponge may have an open (interconnecting) or closed cell structure depending on the process conditions. Closed cell sponge can be produced by adding nitrogen producing organic materials (e.g., diazoamino-benzene and benzenesulphonyl hydrazide) in place of CO2 producing chemicals (e.g., Na2CO3 with a fatty acid). 10. Foam rubber: This can be prepared by bubbling a gas into a compounded liquid latex. The product is gelled in a mould with the help of gelling agents like sodium or potassium silicofluoride to give it the desired shape, cured, washed and dried. Latex foam has an open cell structure. Foam rubber is generally produced by the following two processes.



(1) In the Talalay process, the foam is blown in the latex by adding chemical blowing agents which release oxygen (such as H2O2) and a modifier (such as yeast). The foam is quickly frozen in a mould and coagulated with carbon dioxide. (2) In the Dunlop process, the foam is formed by whipping air into the latex. Gelling agent like sodium silicofluoride is added. The latex foam is vulcanized or cured before it is dried.

4.20.8 Polymers Related to Synthetic Rubber 1. S.B.R (GR-S or Buna-S or Ameripol or Styrene-Butadiene Rubber) SBR or Buna-S is a copolymer of butadiene (75%) and styrene (25%). The two component are allowed to react in a mixing vessel containing an aqueous solution of an emulsifiying agent. Initiators like cumen hydroperoxide and P-methane hydroxide were used in presence of antifreeze components to produce “cold” SBR or “cold rubber’’.

POLYMERS

211

Properties It is resembles to natural rubber in processing characteristics, however, it gets readily oxidized especially in oils and solvents. It requires less amount of sulphur for vulcanization. Uses It is used for lighter duty tires, hoses, molded goods. Unvulcanized sheet, are used for floaring, rubber shoe soles and for electrical insulation. 2. Thiocol or Polysulphide Rubber It is prepared by the condensation polymerization of sodium polysulphide (Na2Sx) and ethylene dichloride. In these elastomers, sulphur forms a phart of the polymer chain.

Properties



1. Thiokols have outstanding resistance to swelling and disintegration by organic solvents. 2. Fuel oils, lubricating oils, gasoline and kerosene have no effect on Thiokols. However, benezene and its derivatives cause some swelling. 3. Thiokol films have low permeability to gases. However, it has some limitations such as (1). It tends to flow or lose shape under continuous pressure (2). Its tensile strength is lesser than that of natural rubber. Recent developments of Thiokols have made up for these deficiencies to some extent. 4. It has lower tensile strength and modulus than natural rubber. Under continuous pressure, it tends to lose shape 5. It has offensive mercaptan like odour, which restricts it use.

Uses

(i) Fabrics coated with Thiokol are used for barrage balloons, life rafts and jackets which are inflated by C 2. (ii) Thiokols are also used for lining hoses for conveying gasoline and oil, in paints, for gaskets, diaphragms and seals in contact with solvents and for printing rolls. (iii) It is also used in the making of containers for transporting solvents and solid propellant fuels for rockets etc.

3. Polyurethane They are prepared by the rearrangement polymerization of di (or poly) isocyanate with di (or poly) hydric alcohol. They contain the characteristic urethane linkages.

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e.g. Reaction between ethylene di-isocyanate and polyalcohols (e.g. ethylene glycol) gives polyurethane rubber.

Properties

(i) Because of their saturated character, they are highly resistant to oxidation. (ii) They have good resistance to many organic solvents, but are attacked by concentrated and hot acids and alkalis. (iii) In comparison with Nylons, polyurethanes are less stable at elevated temperature.

Uses

(i) For the coating of gymnasium and dance floors (ii) Resilient polyurethane fibres (spandex) are use for making swim suits and foundation garments. (iii) Elastomeric polyurethanes are used for enhancing the life of tyre treads. They are also used for making small industrial wheels and shoe soles. (iv) Flexible formed polyurethanes are used for making car and furniture cushions. They are also very good leather substitute and well known in market as car foams, used for making pillow, mattresses etc.

4. Nitrile Rubbers It is a polymers of butadiene and acrylonitrile.

Properties

(i) These rubber have low swelling and low solubility. (ii) They are abrasion resistance even after immersion in gasoline or oils (iii) The rubbers have good heat resistance (iv) They are inherently less resilient than naturel rubber (v) They are good heat resistance

POLYMERS

213

Uses

(i) Nitrile rubbers are extensively used for fuel tanks, gasoline hoses, creamery equipment etc. (ii) In adhesive (iii) In the form of latex, for impregnating paper, leather and textile.

5. Polyisobutylene and Butyl Rubber Butyl rubbers are copolymers of isobutylene with a small amount of isoprene added in order to render them vulcanizable. It is manufactured by mixing isobutylene with 1.5 to 4.5% isoprene and methyl chloride as solvent. The mixture is fed to stirred reactors cooled to —95°C by liquid ethylene. Catalyst solution made by dissolving anhydrous AlCl3 in methyl chloride is added. The polymer forms at once as fi ely divided product suspended in the reaction mixture.

Properties

(i) Under normal conditions these rubber are amorphous but crystallize on stretching. (ii) Unstabilized polyisobutylene are degraded by light or heat. (iii) It is resistance to heat, abrasion, ageing, chemicals and ozone. (iv) It has good electrical insulating properties.

Uses

(i) Due to its superior impermeability to gases, butyl rubber is used as inner tubes. (ii) It is used for wire and cable insulation. (iii) It is also used in the production of tyres.

6. Neoprene Neoprene is copolymers of chloroprene (2-Chloro-1, 3-butadiene).

Properties (i) Its oil resistance is inferior to nitrile rubber but superior to natural rubber. (ii) It have good weathering resistance and ozone resistance properties. (iii) It possess high tensile strength. (iv) It has excellent resistance to petroleum oils and gas oline. Uses

(i) It is excellent rubber for tyres, but more expensive than other rubbers (ii) It is mainly used in wire and cable coatings, industrial hoses and belts, shoe heels and solid tyres. (iii) Gloves and coated fabrics can be prepared from neoprene latex.

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4.21 POLYACRYLATES Polyacrylates are the esters of acrylic acid or its anhydride, halide, amide methyl or nitril derivatives. Among them, esters and nitriles are commercially important. (i) Polyacrylic Acid It is prepared by the polymerization of acrylic acid.

Properties It shows high viscosity in solution. Uses It is used as a thickener in adhesive. (ii) Polymethyl Acrylate (PMA) It is obtained by the polymerization of methyl acrylate.

Properties

(i) It is stable to light and moderate heat. (ii) It has good adhesive property.

Application It is used in making emulsion, paints, adhesive and other substances. (iii) PMMA or Plexiglass or Lucite It is prepared by the polymerization of methyl methacrylate in the presence of acetyle peroxide or hydrogen peroxide as catalyst.

Properties

(i) It is amorphous, colourless, transparent thermoplastic with high optical transparency. (ii) Due to presence of intermolecular dipole forces, it has high softening point (about 130-140° C). (iii) It shows excellent weather ability. (iv) Scratches resistened.

POLYMERS

215

Uses

(i) In making display signs. (ii) It used for making dome shaped covers of solar collectors (i.e. solar heater) and optical fibers (iii) It is also used in the making of motor cycle windscreen. (iv) It is used for light fittings for street lamp housing, ceiling lighting for school rooms, railway stations, factories and offices and automobile rear lamp housing

(iv) Polyacrylonitrile or Orlan or Acrilon It is prepared by the polymerization of acrylonitrile. Produced by vapour phase reaction between propylene, ammonia and excess of air in presence of a fluidised catalyst (BiO3 + MoO3) or Biophosphomolybdate on SiO2. 500°C → CH2 =CH – CN + H2O + heat CH3CH = CH2 + NH3 + O2  Catalyst

Polymerization

n CH 2 = CH  CH →  [ CH 2  CH  CN] n

Acrylonitrile

Acrilon or orlan

The heat produced in the reaction is removed by circulating cold water. A hot gaseous mixture of acrylonitrile and HCN are produced. The hot gases are passed through a wastewater boiler to cool the gaseous mixture to about 80°C. Then it is passed through an absorption chamber where H2SO4 is sprayed from the top. Here, acrylonitrile, acetonitrile and HCN are absorbed in water. These components are separated in a distillation chamber. The acetonitrile is dehydrated by azeotropic distillation. Acrylonitrile can also be prepared by the reaction between acetylene and ethylene oxide in presence of a catalyst, CuCl at 80°C.



CH 2 = CH | Catalyst CN CH ≡ CH + HCN → acrylonitrile



+ HCN → HOCH 2 − CH 2 CN → CH 2 = CH |

CN



ethylene oxide

cyanohydrin

acrylonitrile

Properties

(i) It is hard and has a high melting point (ii) It is chemical resistant (iii) It is also resistant to weathering (iv) It has good mechanical properties (v) Resistance to heat upto about 220° C

Uses

(i) In the form of latex, it is used for impregnating leather, paper and textiles. (ii) In the production of fibers (iii) The copolymer of acrylonitrile with butadiene (Known as nitrile rubber) has many industrial uses such as in fuel tanks, gasoline hoses, adhesives etc.

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4.22 CHART FOR THE MANUFACTURE OF SOME IMPORTANT POLYMER

1. H2

n(CH2 = CH2) ––→ n(CH2 – CH2)



Molecular rearrangement

CH ≡ CH +

2. + HCl



↓Polymerization (–CH2 – CH2–)n



n(CH2 = CH)

Polythene Polymerization

1 - 1.5 atm | small amount of –CH2 – CH – 60 - 80oC Cl H2O2 in an autoclave | Metal chloride vinyl chloride under pressure Cl n as catalyst PVC

3. + HCNn

CH2 = CH Polymerization –CH2 – CH – | | CN CN n Vinyl cyanide or Acrylonitrile Polyacrylonitrile

4. CH ≡ CH+ Hot glacial CH3 COOH CH2 = CH Polymerization CH2 – CH – Hg SO4 (catalyst) | | OOC CH3 OOC CH3 n Vinyl acetate PVA 5. H H H | | | CH3 – C = 0 + HCN ––→ CH3 – C – OH H·OH CH3 – C – OH | | Acetaldehyde CN COOH cyanohydrin hydroxy acid

– HOH

On heating

–CH2 – CH – Polymerization n CH2 = CH + CH3OH CH2 = CH | | | COOCH3 COO·CH3 COOH n methyl acrylate Acrylic acid PMA

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217

6. CH3 CH3 CH3 | | | CH3 – C = 0 + HCN ––→ CH3 – C – OH H·OH CH3 – C – OH | | Acetone CN COOH cyanohydrin – HOH Heating CH3 CH3 CH3 | | | –CH2 – C – Polymerization n CH2 = C + CH3OH CH2 = C | | | COO·CH3 n COOH

COO·CH3 COOH methyl methacrylate Methyl Acrylic acid

PMM A Plexiglass

4.23 FLOW SHEETS FOR DIFFERENT POLYMERS (i) Flow sheet for polyethylene

Fig. 4.6. Flow sheet for polyethylene



(ii) Flow sheet for Decron or Terylene

Fig. 4.7. Flow sheet for Decron

218

BASIC ENGINEERING CHEMISTRY (iii) Flow-Sheet diagrams for Cellulose Acetate

Fig. 4.8. Flow-sheet for producing cellulose acetate.

POLYMERS

219 (iv) Flow sheet diagrams for Teflo

4.9. Flow-sheet for producing polytetrafluroethylene ( eflon

220

BASIC ENGINEERING CHEMISTRY (v) Flow sheet diagram for Vinayl polymer

4.10. Flow-sheet for polymerization process for Vinyl Polymer.

POLYMERS

221 (vi) Flow sheet diagram for Vinyl chloride

4.11. Flow-sheet for producing vinyl chloride from acetylene.

(vii) Flow sheet diagram for Nylon 6 : 6

4.12. Flow sheet for producing Nylon 6 : 6

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BASIC ENGINEERING CHEMISTRY (viii) Flow sheet diagram for Acrylonitrile

4.13. Flow-sheet for producing Nylon 6.

(ix) Flow sheet diagram for Acrylonitrile

4.14. Flow-sheet for producing acrylonitrile.

POLYMERS

223 (x) Flow sheet Diagram for Bakalite Resin

4.15. Flow-sheet for producing phenol formaldehyde resin products.

224

BASIC ENGINEERING CHEMISTRY (xi) Flow sheet diagram for Silicones

4.16. Flow-sheet for producing silicones via direct monomer process.

4.17. Flow-sheet for producing silicones by Grignard’s method.

POLYMERS

225 (xii) Flow sheet for Polyethylene

4.18. Flow-sheet for producing Polylefins by low pressure Ziegler process

(xiii) Flow sheet diagram for SBR Rubber

4.19. Flow-sheet for the manufacture of Butadiene styrene (SBR) rubber process.

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1. Distinguish clearly between the following :

2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

(a) Thermoplastic and thermosetting resins (b) Addition and Condensation polymerization (c) Natural and synthetic rubbers. Write informative notes on the following : (a) Classification of polymer (b) Mechanism of radical polymerization (c) Anionic and cationic polymerization (d) Thermodynamics of a polymerization process. What are the common constituent of plastics and what are their functions ? Write informative notes on the following : (a) Fabrication of plastics (b) Vulcanization of rubber (RGPV Bhopal 2006) (c) Polyvinyl chloride (d) Compounding of rubber (e) Reclaimed rubber (f) Silicone rubbers Write an essay on the preparation, properties and uses of the following : (a) Polyvinyl acetate (b) Cellulose acetate (c) Phenol formaldehyde resins (d) Urea and melamine formaldehyde resins (e) Teflo (f) Polyethylene (g) Polystyrene (h) Polymethyl methacrylate. Discuss the various polymers related to natural rubber with emphasis on their preparation, properties and uses. Write an essay on the various types of synthetic rubber with brief description of the preparation, properties and uses of any three of them. Write an essay on fibre-reinforced plastics What are “High Polymers”? Give preparation, properties and uses of any four of the following : (a) Bakelite (b) Nylon 6 : 6 (c) PVC (d) Natural rubber from latex (e) Decaron (f) Vulcanisation of Rubber (g) Buna -S (h) Buna - N (i) Polymethyl methacrylate. (RGPV Bhopal 2001) (a) Write explanatory note on Silicones (b) Discuss the changes in structure due to vulcanisation. Write short notes on the following : (a) Synthetic rubbers. (b) Condensation Polymerisation.

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12. Answer any four of the following : (a) Define a Polyme . (b) What is meant by degree of Polymerisation ? (c) Why do all simple organic molecules not produce polymers ? (d) Why are plastics indispensible in everyday life ? (e) Why is Teflon highly chemical resistant 13. (a) Differentiate between thermoplastic and thermosetting resins. (b) Describe preparation and properties of (i) Teflo (ii) Plexiglass (iii) Bakelite (iv) Perlon-v (RGPV Bhopal 2009) (c) Write short note on vulcanization of rubber. (Nagpur University, Winter, 2000) 14. (a) What is fibre glass reinforced plastic ? Write its properties and uses : (b) Write short notes on (any two) : (i) Epoxy resin (ii) Neoprene (iii) Polystyrene (c) What is vulcanization? How it brings about the changes in properties of rubber. (RGPV Bhopal 2009) 15. (a) Distinguish between thermosetting resins and thermoplastic resins with suitable examples. (b) Write short notes on any three of the following : (i) Teflon ii) Silicones (iii) Fibre glass (iv) PVC 16. (a) Explain with examples the terms addition polymerization, copolymerisation and condensation and condensation polymerisation. (b) Discuss preparation, properties and uses of the following (any three) : (i) Silicones (ii) Teflon iii) Polyethylene (iv) Neoprene (v) PVC (RGPV Bhopal 2009) (v) Polyacrylonitrile (Nagpur University, 1999) 17. (a) Describe the preparation, properties and uses of polyvinyl chloride plastics. (b) What are natural rubbers ? What are their drawbacks ? How they are overcome ? (c) Explain the term fabrication of plastics. (Bombay University, 2000) 18. (a) What is compounding of rubber ? Name the ingredients used in compounding rubber. What are their functions. (b) How would you prepare polyethylene commercially ? Distinguish between LDPE and HDPE. What are their uses ? (Bombay University, 2000) 19. Distinguish clearly : (a) Addition and condensation polymerization (b) Mechanism of ionic and radical polymerization (c) Natural and synthetic rubber. (d) Thermoplastic and thermosetting resins. (RGPV Bhopal 2003) (e) Homopolymer and copolymer 20. (a) Discuss the properties of thermosetting and thermoplastic resins. (b) Discuss the properties and applications of silicone oils. (c) Describe the preparation, properties and uses of styrene Butadiene Rubber (SBR) (Nagpur University 2002)

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21. (A) Differentiate between the following : (a) thermoplastic and thermosetting materials (b) Natural rubber and synthetic rubber (B) Give the preparation, properties and uses of any two of the following : (a) Teflo       (b) Bakelite     (c) Polystyrene (Nagpur University, ’95) 22. “The chemical nature of the monomeric units and their molecular arrangement determine the properties of the polymer”. Justify this statement with suitable examples. (Nagpur University, ’96) 23. (A) Explain the following :      (a) Polyvinyl chloride is stronger and tougher than polythylene.      b)( Natural rubber needs vulcanisation. (B) How are the following polymers prepared ? Give their important industrial applications. (i) HDPE and LDPE (ii) Neoprene (iii) Polystyrene (Nagpur University, ’96) 24. (A) What kind of structural features are important for a polymer to behave like an elastomer ? (b) Distinguish between thermoplastic and thermosetting resins. (c) Give preparation, properties and industrial uses of any two of the following : (i) Buna-S rubber (ii) phenolic resins (iii) HDPE (iv) polytetrafluorethylen (Nagpur University ‘98) 25. Describe the preparation and flow sheet diagram of Nylon 66 (RGPV Bhopal 2006) 26. Write preparation and uses of the following: (i) Polyethene (ii) PVC (iii) Polyvinyl acetate (iv) Buna S (RGPV Bhopal 2009) 27. What is polymerisation? Discuss mechanism of polymerisation reaction. (RGPV Bhopal 2009)

UNIT

5

PART–A

Instrumental Techniques in Chemical Analysis

COLORIMETRY AND VISIBLE SPECTROSCOPY 5.1 INTRODUCTION The variation of the colour of a system with change in concentration is the basis of colorimetry. In colorimetry, the concentration of a substance is determined by measurement of the relative absorption of light with respect to a known concentration of the substance. In visual colorimetry, natural or artificial white light is generally used as a source of light, and determinations are made with the help of a simple instrument called a calorimeter or colour comparator. If the eye is replaced by a photoelectric cell, the instrument is called a photoelectric colorimeter. Colorimeter analysis is specially useful for systems in which substances or their solutions are colored. When a substance is colourless, then a suitable complexing agent is added to the solution so that a coloured complex is obtained. The later than absorbs light in the visible region. For instance, for the estimation of cuprous ions, complexing agent, ammonium hydroxide is used to get blue coloured solution. 5.1.1 Theoretical principles Lambert’s Law: If a monochromatic light passes through a transparent medium, the rate of decrease in intensity with the thickness of medium is proportional to the intensity of the incident light. In other words, the intensity of the emitted light decreases exponentially as the thickness of the absorbing medium increases arithmetically. The law may be expressed in the form of a differential equation; −d I = K I d1 where I is the intensity of the incident light of wavelength l, l is the thickness of the medium, and K is proportionality factor. On integration, and equating I with I0 if l = 0, we get −d ln 0 = K l …(1) dt or It = I0.e–Kl …(2) where I­t is the intensity of transmitted light, I0 is the intensity of the incident light, l is the thickness of the absorbing medium and K is the constant for the wavelength and the absorbing medium used. By changing from natural to common logarithms, we get It = I0 . 10–2.303 kl = I0 . 10–Kl …(3) where K = k / 2.3026 , which is known as the absorption coefficient The ratio It / I0 is called transmittance, T. log I0/It = A, the absorbance of the medium or optical density, D or the extinction, E. 229

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Beer’s Law: If a monochromatic light passes through a transparent medium, the rate of decrease in intensity is directly proportional to the concentration of the media or the intensity of a beam of monochromatic light decreases exponentially as the concentration of the absorbing substance increases arithmetically. This may be expressed in the form It = I0 . e–K'c = I0 . 10–2.303K'c = I0 . 10–k'c …(4) where c is the concentration, and K' and k' are constants. Combining equations 3 and 4, we get It = I0 . 10–a c l …(5) or

logI0/I = a c l               

…(6)

This expression is known as Beer-Lambert’s Law, which is the fundamental equation for colorimetry and sprectrophotometry. If c is expressed in mole L–1 and l in cm, ‘a’ is given a symbol ‘e’, which is known as the molar absorption coefficient or molar absorptivit . Since I0/It is called absorbance, A, we arrive at the following inter-relationship: A = e c l = log I0 /It = log l / T = –log T If l is constant (as in the case of matched cells used in a colorimeter or spectrophotometer), the Beer-Lambert’s may be written as c ∝ log I0 / It c ∝ log l / T or c ∝ A          Hence, by plotting A or log l/T as ordinate, against concentration as abscissa, straight line will be obtained and this will pass through a point (c = 0, A = 0 i.e., T = 100%). This calibration line can be used for determining unknown concentrations of the same material by measuring their respective absorbances, and obtaining their corresponding concentrations from the calibration line. Deviations from or exceptions to the Beer-Lambert’s Law The law does not hold when coloured solute ionises, associates, dissociates, or undergoes complexation, because the nature of the coloured species will vary with concentration. Additivity of absorbances According to Beer’s law, the absorbance at any particular wavelength is directly proportional to the number of absorbing molecules. If a solution contains more than one type of absorbing species, the total absorbance will be the sum of the absorbances of all the species, provided they do not interact chemically. 5.1.2. Instrumentation 1. Radiation Source: Tungsten filament lamp is most widely used for producing visible light in the wavelength range 400–750 nm. If colorimetric analysis is carried out in the UV range, (i.e., down to 200 nm) hydrogen discharge lamp or deuterium is used as radiation source, whereas for work in IR region, Nernst Glower is used. 2. Dispersing Device: The selection of a narrow band of wavelength, which is required for colorimetry and spectrophotometry, is accomplished with the help of a monochromator. Monochromators used in various spectral regions are: (a) filters b) prisms, and (c) gratings

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231

Plates of coloured glass can be used as filters in the visible region. From the stability point of view, gelatin filters are better. i.e., Glass prisms are used for work in the visual range, quartz prisms are used for UV and alkali halide prisms are generally used for IR range. Gratings are of two types: (a) Transmission gratings (b) Reflection gratings. A transmission grating comprises of a series of closely spaced parallel grooves ruled on a piece of glass or other transparent material. The greater the number of lines per square inch, shorter will be the wavelength dispersed by the grating and greater will be the dispersing power. A grating suitable for visible and UV work should have 15,000–30,000 lines per square inch. 3. Slits: A slit system is used for selecting the desired wavelengths from the light dispersed by a monochromator. An entrance slit and an exit slit are placed each side of the prism or grating. The slit jaws are made of metal in the shape of knife-edges, and they can be moved with respect to each other to control the width. The entrance slit chooses a small parallel beam of incident light, while preventing stray radiations entering the optical path. The light, after being dispersed (by the prism or grating) goes through the exit slit and travels through the sample or reference cell and finally reaches the detector system. 4. Sample holders: To hold the sample solution to be analysed and the reference solution, optically matched colour corrected fused glass cells are used in the wavelength range 300–2500 nm. Corex glass or quartz cells are used in the wavelength range 210–300 nm, and fused silica cells are used for measurements at somewhat lower wavelengths. 5. Detector: Generally we use following type of detectors. (i) Photovoltaic or barrier layer cells (ii) Photoemissive cells or phototube (iii) Photo multiplier tube (iv) Silicon diode or photodiode 5.1.3 Colorimetric method The fundamental principal for colorimetric analysis is to compare, under well designed conditions, the colour produced by the substance in unknown amount with the same colour produced by a known amount of material being determined. The following methods are available for the quantitative comparison of colours of known and unknown solutions: (i) Methods using visual comparators (a) Standard series method: In this method, Nessler tubes are used, which are colourless glass tubes of uniform cross section and flat bottoms. The solution of the substance being determined is made up to a known volume, and the colour is compared with that of a series of standards in the same way starting from known amounts of the substance being determined. 100 ml of the solutions of the unknown and each of the standard solutions are placed in Nessler tubes, and the solutions are viewed vertically through the length of the liquid columns. The concentration of the unknown solution is equal to that of the standard solution having the same colour. LOVIBOND–2000 comparator works on this principle. (b) Duplication method: In this method, a known volume (say x ml) of the test solution is treated in Nessler cylinder with a measured volume (say y ml) of appropriate reagent so as to develop a colour. Now, x ml of distilled water is placed in a second Nessler cylinder together with y ml of the reagent. A concentrated standard solution of the substance under test is now added to the second Nessler cylinder from a micro burette until the colour thus developed exactly matches the colour in the first Nessler cylinder already developed with unknown solution (c) Dilution method: In this method, a colour developed from the sample and standard solution are taken in two identical glass tubes of same diameter and are observed horizontally through the

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tubes. The more concentrated solution is diluted until the colours in the tubes are of identical intensity when observed horizontally through the same thickness of solutions. The relative concentrations of the two original solutions (known and unknown) are then proportional to the heights of the matched solutions in the tubes. (d) Balancing method: In this method, comparison is made in two tubes, and the height of the liquid in one tube is so adjusted that when both the tubes are observed vertically, the colour intensities in the tubes are equal. If the concentration of solution in one of the tubes is known then the concentration in the other can be calculated from the relation

c1 l1 = c2 l2

where c1 and c2 are the known and unknown concentrations and l1 and l2 are their respective lengths of the two liquid columns. Hence,

c2 = c1 l1 / l2

Duboscq colorimeter works on this principle. (e) Photoelectric photometer method: In this method, the human eye is replaced by a photoelectric cell. Instruments which incorporate photoelectric cells measure the light absorption and not the colour of the substance, hence the term ‘photoelectric colorimeters’ is a misnomer. They may be better called as photoelectric photometers or photoelectric comparators or absorptiometers. These instruments essentially consist of a light source, a suitable light filter to provide nearly monochromatic light, a glass cell for the sample/ standard solution, a photoelectric cell to receive the radiation transmitted by the solution, and a measuring device to determine the response of the photoelectric cell. The various types of each of these components used in visible spectrophotometry and colorimetry have been already discussed in earlier sections. The photometer is first calibrated by measuring the absorbance of a series of solutions of known concentrations (at the wavelength at which the coloured species has maximum absorbance). A calibration curve is plotted connecting the concentration with the readings of the measuring device employed (optical density). The concentration of the unknown solution is then determined by noting the cell response and referring it to the calibration curve. are

The photoelectric photometers available in different type

Source

Collimating lens

Filter

Movable holder Solution

Solvent

Barrier-type photocell

Fig. 5.1. Optical diagram of a simple photoelectric photometer.

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incorporating one or two photocells. In the one cell type of instruments, the absorption of light by the solution is usually measured directly by determining the current output of the photoelectric cell in relation to the value obtained with the pure solvent. In instruments of the one cell type, it is essential to ensure that the light source is of constant intensity and the photocell used doesn’t exhibit fatigue effect. The two cell type of photoelectric photometer is considered to be more trustworthy because any fluctuation of the intensity of light source will affect both cells alike as they are matched for their spectral response. In this type of instrument, the two photocells, which were illuminated by the same source of light, are balanced against each other through a galvanometer. The test solution is placed before one cell and the pure solvent before the other, and the difference in current output is measured. (f) Spectrophotometric method: This is the most accurate method for determining the concentration of a substance. However, it is more expensive. A spectrophotometer may be considered as a refined photoelectric photometer which provides a continuously variable and more nearly monochromatic light. The essential parts of a spectrophotometer are a source of radient light, a monochromator to produce narrow bands of radient energy, glass or silica cells for the solvent and for the test solution, a photocell or a photomultiplier, an amplifier and a recorder. Single-beam and double beam spectrophotometers are commercially available. The essential features of a Singlebeam spectrophotometer are illustrated in Fig. 5.2. B A

F

M E

D

C

G

H

Fig. 5.2. Essential features of a single-beam spectrophotometer. A - Light source, B - dondensing mirror, C - diagonal mirror, D - entrance slit, F - quartz prism, G - absorption cell, H - photocell, M - meter

An image of the light source, A is focussed by the condensing mirror, B and the diagonal mirror, C on the entrance slit, D. The light beam falls on the collimating mirror, E where it is rendered parallel and reflected to the quartz prism, F. The back surface of the prism is aluminised, so that the light refracted at the first surface is reflected back through the prism. This light beam undergoes further refraction as it emerges from the prism. The collimating mirror focuses the spectrum in the plane of the slit system, D. The light of the wavelength for which the prism is set then passes out of the monochromator through the exit slit, through the absorption cell, G and finally to the photocell, H. The response of the photocell is amplified and registered on the meter. In modern versions of the instrument, the prism is replaced by a diffraction grating. Double beam spectrometers are the most modern general purpose UV-visible instruments covering the spectral region from 200-800 mm by a continuous automatic scanning process and producing the absorption spectrum as a pen trace on calibrated chart paper. Salient features of colorimetric analysis:

(a) It gives accurate results at low concentrations. (b) It enables the analysis of such substances for which gravimetric and titrimetric procedures are not available (e.g. In case of some biological substances). (c) It is ideal for quick and routine analysis of components of a number of similar samples. When once the calibration curve is plotted, a large number of samples of the same component in different concentrations can be rapidly analysed.

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Criteria of a satisfactory colorimetric analysis:

(a) The colour reaction should be specific or atleast selective for the particular substance being analysed under the chosen experimental conditions. (b) Beer’s law should be obeyed in the desired concentration range. In other words, the colour should be proportional to concentration. (c) The colour should be sufficiently stable to enable accurate analysis (d) The colorimetric procedure should give reproducible results under the specific experimental conditions. (e) The colour reaction should be highly sensitive and the reaction product should strongly absorb in the visible region. (f) The coloured solution should be clear and free from turbidity.

5.2. SPECTROSCOPY Spectroscopy deals with the transitions that a molecule undergoes between its energy levels upon absorption of suitable radiation determined by quantum mechanical selection rules. Spectrophotometer is an instrument used for absorption measurements. It can be made to operate in the ultraviolet, visible and infrared regions, using suitable source of radiant energy. 5.2.1. Ultraviolet spectroscopy It involves the transition of electrons within a molecule or ion from a lower electronic energy level to a higher electronic energy level by the absorption or emission of radiations falling in the UV-visible range of electromagnetic spectrum i.e., When a molecule absorbs U.V. radiation of frequency V sec–1, the electron in that molecule undergoes transition from a lower to a higher energy level. The difference in energy in given by E = hv erg The actual amount of energy required depends on the difference in energy between the ground state (E0) and the excited state (E1) of the electrons. E1 – E0 = hv The total energy of a molecule is the sum of electronic energy (Eelec), vibrational energy (Evib) and rotational energy (Erot), Also Eelc > Evib > Erot When UV energy is quantised, the absorption spectrum arising from a single electronic transition is expected to consist of a single discrete line. But this does not happen because electronic absorption is superimposed upon vibrational and rotational subenergy levels. Due to which formation of bands in the electronic spectra of simple molecules in the gaseous phase take place. Classification of electronic transitio Electronic transition in molecules can be broadly classified into (i) s-s* transition. Such transitions occur in case of saturated hydrocarbons, which do not contain lone pairs of electrons. The energy required for this type of transitions is very large and the absorption band occurs in the far ultraviolet region (126 to 135 nm). For example, methane has λmax at 121.9 nm and ethane at 135 nm correspond to this transition. These transitions cannot be observed in commercial spectrophotometers which generally do not operate at wavelengths below 180 nm. (ii) p-p* transition. These type of transitions are related to the promotion of an electron from a bonding p orbital to an antibonding p* orbital.

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(iii) n-s* transition. The energy required for n → transition is generally less than that required for σ → σ* transition and their corresponding absorption bands appear at longer wavelengths in the near untraviolet region (180 to 200 nm). Saturated compounds with lone pair (non-bonding) electrons undergo n → π* transitions apart from σ − σ* transitions. (iv) n-π* transition. These transitions are shown by unsaturated molecules which contain atoms like N, O and S. These transtions show a weak band in their absorption spectrum. In aldehydes and ketones (having no C º C or C = C bonds), the band due to n → π* transition generally occurs in the range 270-300 nm. On the other hand, carbonyl compounds having double bonds separated by 2 or more single bonds exhibit the bands in the range 300 to 350 nm due to n → π* transitions. Instrumentation for UV spectroscopy The following are the important component of a UV spectrometer. (i) Source of radiation: The following are the most common radiation sources used in UV spectrometers: (a) Hydrogen discharge lamps (b) Deuterium lamps (c) Xenon discharge lamps (d) Mercury arcs. (ii) Monochromators: Monochromators are used to disperse the radiation according to the wavelength. The essential components of a monochromator are an entrance slit, a dispersing element (eg., a prism or a grating) and an exit slit. The prisms are generally of quartz or fused silica. The dispersing element disperses the heterochromatic radiation into its component wavelengths, whereas the exit slit allows the nominal wavelength to pass through. (iii) Detectors: The following three type of detectors are commonly used (a) Photovoltaic or Barrier-layer cell (b) Photocell or photoemissive cell (c) Photomultiplier cell (iv) Recorders: The signal from the detector is received by the recording system provided with a recorder pen. (v) Sample and reference cells: Matched pair of cells made of quartz or fused silica are used. Single-beam and double-beam UV-spectrophotometers are available commercially. Working of double-beam UV-Spectrophotometer The lay-out of a double-beam UV spectrophotometer is shown in Fig. 5.3.

Attenuators Sample Prism Photomultiplier M1 Reference Lamp Source

Rotating sector Beam spliter Monochromator

Sample chamber

Fig. 5.3. Double beam-ultraviolet spectrophotometer..

Detector

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The UV radiation emanating from the source is allowed to pass through a monochromator unit via a mirror system. The radiation of narrow range of wavelengths coming out of the monochromater (through the exit slit) is received by the rotator system which divides the beam into two identical beams, one passing through the sample cell and the other through the reference cell. The two beams, one emerging from the sample cell and the other from the reference cell are focussed on the detector. The output from the detector is connected to a phase sensitive amplifie . The signals transmitted by the amplifier are transmitted to a recorder which is connected to a recorder. The chart drive is coupled to the rotation of the prism. Thus the absorbance or transmittance of the sample is recorded as a function of wavelength. Applications of Ultraviolet Absorption Spectroscopy 1. Qualitative Analysis: Ultraviolet absorption spectroscopy is used for characterizing aromatic compounds and conjugated olefins. Identification is done by comparing the UV absorption spectrum of the sample with the UV spectra of known compounds available in reference books. 2. Detection of impurities: UV absorption spectroscopy is one of the best methods for detecting impurities in organic compounds. Examples: (a) Benzene, which is the most common impurity in cyclohexane, can be easily detected by the absorption band of benzene at 255 nm. (b) Nylon is manufactured from pure adiponitrile and hexamethylene diamine. If these raw material are not pure, the nylon produced will be of a poor quality. The aromatic and unsaturated impurities present in these starting materials can be detected from their UV absorption spectra. (c) Purification of organic compounds can be continued until the absorption bands characteristic of the impurities disappear in the spectrum. 3. Quantitative analysis: UV absorption spectroscopy can be used for quantitative analysis of such compound which are able to absorb ultraviolet radiation. The determination can be done on the basis of Beer-Lambert’s law. We know that absorbance (A) is given by A = ec x For a solution of same compound in the same sample cell e and x are constant. Absorbance of standard solution (As) and unknown solution (Au) is measured with the help of spectrophotometer. Knowing the value of concentration of standard solution (Cs) concentration of unknown (Cu) can be determined. As Au Since = Cs Cu Au × Cs or Cu = As 4. Determination of dissociation constants of weak acids or bases: UV spectroscopy can be used to determine the dissociation constants of acids or bases. For instance, the dissociation constant (pka) of an acid HA can be determined by determining the ratio [HA]/[A–] spectrophotometrically from the graph plotted between absorbance and wavelenghts at different pH values. This value can be substituted in the equation. pka = pH + log [HA]/[A–] 6. Determination of molecular weight: UV spectroscopy can be used for determining the molecular weight of a compound if it can be converted into a suitable derivative which shows an absorption band in its spectrum. For instance the molecular weight of an amine can be determined

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by converting it into picrate. One litre of the solution of the amine picrate is prepared by dissolving a known weight of the amine picrate in a suitable solvent and determining its optical density at 380 nm. The concentration of the amine picrate can be determined using the formula. log (I 0 / I t ) C = e max × 1 7. Study of tautomeric equilibria: UV spectroscopy can be used to determine the % of keto and enol forms present in compounds such as ethyl acetyl acetate by measuring the strength of the respective absorption bands. 8. Determination of Calcium in blood cerum: Calcium in the blood can be indirectly determined by converting the Ca present in 1 ml of the serum as its oxalate, redissolving it in sulphuric acid, and treating it with dilute ceric sulphate solution. The absorption of the excess ceric ion is measured at 315 nm. The amount of Ca in the blood serum can thus be indirectly calculated. 9. Determination of ozone in the environment: The ozone concentration present in smokefog (smog) in the environment can be determined by measuring its absorption at 260 nm. 10. Miscellaneous applications: (a) UV spectroscopy has been used for confirmation of the structure of chloral. (b) This technique has also been used to study charge-transfer transitions, such as the one observed with iodine and benzene in heptane. 5.2.2. Infrared spectrophotometry Introduction Infrared spectrophotometry is a powerful analytical technique useful for chemical identification. When coupled with intensity measurements, this technique can be used for quantitative analysis. Infrared spectra originate from the absorption of energy by a molecule in the infrared region and the transitions occur between two vibration levels. By measuring molecular vibrational frequencies, useful information regarding molecular structure can be obtained. The vibrational spectra, therefore are considered as molecular finge prints. The infrared region of the electromagnetic spectrum which extends from the red end of the visible spectrum to the microwave region, may be divided into the following three regions: 1. Near-infrared (overtone region)

0.8 to 2.5 µm (12,500 to 4000 cm–1)

2. Middle-infrared (vibration-rotation region)

2.5 to 50 µm (4000 to 200 cm–1)

3. Far-infrared (rotation region)

50 to 1000 µm (200 to 10 cm–1)

The important region of interest from analytical point of view is 2.5 to 25µm (4000 to 400 cm–1). Infrared spectra originate from the different modes of vibration and rotation of molecule. In the case of simple diatomic molecules, we can calculate the vibrational frequencies by treating the molecule as a harmonic oscillator. The frequency of vibration is given by the following relation: 1 f /µ S−1 n = 2 p Where n = the frequency (vibrations per second),

f = force constant (the streching or restoring force between two atoms) in newtons per meter and m is the reduced mass per molecule (in kilograms) defined by the following relationship

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238

µ=

m1 × m 2 = + m2 )

( m1

( A r1

A r1 × A r2 kg + A r2 ) × L × 1000

Where m1 and m2 are masses of the individual atoms, and Ar1 and Ar2 are the relative atomic masses, L is Avogadro’s constant. The absorption bands are usually quoted in units of wave numbers ( n ) which are expressed in reciprocal centimeters, cm–1. In some cases, wavelengths (l) measured in micrometers (nm) are also used. The inter-relationship between these units is given by. 1 V n= = l C 1 1/ 2 = n × ( f / µ ) cm −1 Therefore, 2pC Usually, reasonably good agreement is found between calculated and experimental values for wave numbers. However, this simple calculation has not taken into consideration any possible effects arising from other atoms in the molecule. More sophisticated methods of calculation have been developed to take these interactions into account. Vibrational Spectra of Polyatomic molecules A diatomic molecule has only one vibrational mode and hence it yields a rather simple system. But for a polyatomic molecule, several vibrational modes are possible and, therefore, it gives a complicated IR spectrum. The vibration of atoms in a polyatmoic molecule may be visualised from a mechanical model of the system. The atoms in a molecule can be seen as resembling a system of balls by varying masses and arranged in accordance with actual space geometry of the molecule. These balls are connected with the mechanical springs whose forces are proportional to the bending forces of the chemical bonds. These forces keep the balls in position of balance. From such a model, it can be visualised that a molecule has two types of fundamental vibrations; (i) in one type of vibrations, the distance between two atoms increases or decreases but the atoms remain in the same bond axis. This is known as stretching vibration. When the stretching and compressing occurs in a symmetric fashion, it is called symmetric stretching. On the other hand, when one bond is compressing while the other is stretching, then it is called asymmetric stretching (ii) The other type of vibration is known as bending or deformation in which the position of the atom changes relative to the original bond axis. This bending involves oscillation of the atoms perpendicular to its bond axis. Four types of deformations may be distinguished: (a) Scissoring: When the two atoms joined to a central atom deformation produced is known as scissoring. (b) Rocking: When the two atoms joining a central atom move back and forth in the plane of the molecule, the resulting deformation is called rocking. (c) Wagging: In this type of deformation, the structural unit moves back and forth, out of the plane of the molecule. (d) Twisting: In this type of deformation, the structural unit rotates about the bond which joins the rest of the molecule. Types of Vibration Stretching vibration (a) symmetric

(b) asymmetric

Bending vibration (a) scissoring

(b) rocking (c) wagging (d) twisting

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Group Mode C—H C—H C—H C—H

Wavelength (microns)

Wave number (cm–1)

Bending (in-phase) Streching Bending (out-of-phase) Rocking

6.8 – 7.7 3.0 – 3.7 12.0 – 12.5 11.1 – 16.7

O—H

Bending

6.9 – 8.3

1200 – 1500 (m,w)

O—H N—H N—H

Streching Bending Rocking

2.7 – 3.3 6.1 – 6.7 11.1 – 14.0

3000 – 3700 (m) 1500 – 1700 (s,m) 600 – 900 (s,m)

C—C Stretching C = C Stretching C ≡ C Stretching C = N Stretching C ≡ N Stretching C—O Stretching C = O Streching C—Br C—Cl C—F

8.3 – 12.5 5.9 – 6.3 4.2 – 4.8 5.9 – 6.3 4.2 – 4.8 7.7 – 11.1 17.0 – 21.0 15.0 – 20.0 13.0 – 14.0 7.4 – 10.0

800 – 200 (m,w) 1600 – 1700 (m) 2100 – 2400 (m,w) 1600 – 1700 (m,s) 2100 – 2400 (m) 900 – 1300 (m,s) 480 – 600 (s) 500 – 670 (s) 710 – 770 (s) 1000 – 1350 (s)

s – Sharp

m – Medium

1300 – 1500 (m,s) 2700 – 3300 (m,s) 800 – 830 (w) 600 – 900 (w)

w - Weak

Instrumentation An infrared spectrometer contains the following three major components: (1) Source of radiation: The main sources of mid-infrared radiation are (1) Nichrome wire wound on a ceramic support (2) Nernst glower, which is a filament containing oxides of zirconium, thorium, yittrium and cerium held together with a binder (3) Globar, which is a bonded Silicon carbide rod. When heated electrically at 1200 to 2000°C, they glow and produce mid-IR radiation. (2) Monochromator: Prisms and gratings are commonly used for this purpose. The prism materials used commonly are as follows:        Prisms and materials

Sodium chloride (rock salt) Potassium bromide Caesium iodide Calcium fluorid

4000 to 650 cm–1 1000 to 400 cm–1 1000 to 260 cm–1 50,000 to 1100 cm–1

Lithium fluorid

16,666 to 1670 cm–1

The most commonly used prism materials, NaCl or KBr are hygroscopic. (3) Detectors: The IR detectors generally convert thermal radiant energy into electrical energy. Two types of IR detectors are in use; (a) selective and (b) non-selective. In selective detectors, the response depends upon the wavelength of the incident radiation. Example: Photo cells, Photoconductive cells, Photographic plates, and IR phosphors.

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240 Sample preparation

(a) Solid Sampling: The following four techniques are generally used for preparing solid samples: (i) Solids run in solution: If a suitable solvent which is transparent in the IR region can be found to dissolve the solid, then the solution can be run in one of the cells for liquids. (ii) Solid films: If the solid sample is amorphous, it can be deposited on the surface of a KBr or NaCl cell by evaporation of a solution of the solid sample. This technique is suitable for quick quantitative analysis but not for quantitative work. (iii) Nujol mull technique: This is the most convenient and routine method. The finely powdered sample is mixed with a small quantity of a heavy paraffin oil, (usually a medicinal grade nujol) and mulled to form a thick paste. The paste is allowed to spread between IR transmitting windows. This is then mounted in the path of the IR beam and the spectrum is run. Nujol is transparent in the IR region, but it has absorption bands near 2857 cm–1, 1449 cm–1 and 1389 cm–1, where C—H streching and C—H bending absorption bands occur. In order to have the complete spectrum, a second mull has to be prepared using perfluoroker sene, fluorolube or halocarbon oil, and a separate spectrum is run. The mull technique is useful for quantitative analysis. (iv) Pressed pellet technique: In this method, a very small quantity of finely ground solid sample is intimately mixed with about 100 times its weight of pure and desiccated KBr (or less commonly with KI or CsBr), and then pressured in an evacuated die under high pressure. The resultant transparent disc is inserted into the sample holder of the spectrophotometer, while a blank KBr pellet of identical thickness is kept in the path of the reference beam. Advantages of pallet techniques (a) This method is free from problem of bands due to the mulling agent, appearing in the IR spectrum. (b) KBr pellets can be stored for long periods. (c) The resolution of the spectrum is superior (d) Since the concentration of sample taken in the pellet can be suitably adjusted, this technique is useful for quantitative analysis. Limitations (i) Owing to the traces of moisture present in the sample, a band at 3450 cm–1 always occurs due to OH- group from the moisture. Care should, therefore, be taken while investigations are carried out in this region. (ii) The high pressure involved in pressing of the disc may bring about polymorphic changes in the crystallinity in the samples (particularly inorganic complexes) which may cause complications in the IR spectrum. (iii) This method is not suitable for some polymeric substrances which cannot be powdered easily. In such cases, grinding is done at liquid nitrogen temperatures, at which the material becomes brittle. (b) Liquid sampling: Liquid sample are directly taken into rectangular cells of NaCl, KBr or for work in double beam spectrophotometers, matched cells of identical thickness are used. Absorption cells for liquids or solutions are available commercially. (c) Sampling of gases: Gas samples are directly introduced into 10 cm long special cells provided with IR transparent windows. For analysing dilute gases, long path cells (20 cm or 30 cm long) are employed.

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Single beam spectrophotometers The layout of a single-beam infrared spectrophotometer is shown in Fig. 5.4. Littrow mirror

Prism

Detector Exit slit

Source Sample Chopper

Entrance slit

Fig. 5.4. Lay-out of a single beam infrared spectrophotometer.

In this system, the radiation emitted from the source passes through the sample entrance slit, collimating mirror and then through a fixed prism and a littrow mirror. The prism and the littrow mirror select the desired wavelength and allows it to pass on to the detector, with the help of the collimating mirror. The detector measures the intensity of radiation. On comparing this with the original intensity of radiation, one can measure the fraction of radiation that has been absorbed by the sample. The absorption spectrum can be obtained by measuring the degree of absorption of radiation at different wavelengths in the desired range. The single beam instruments have the following disadvantages. (i) The intensity of the emission of radiation source may change with wavelength from time to time during the analysis, which results in sloping of the base-line and deformation of the spectra (ii) When the sample is analysed in solution, the bands of the solvent appear in the spectrum. This may lead to problems in interpreting the bands. These difficulties can be overcome by double beam spectrophotometers Double beam spectrophotometers These are constructed in such a way that the radiation emitted by the source is split into two identical beams having equal intensity, one of the beams passes through the sample, whereas the other passes through the reference (air or pure solvent) for compensation. The two half beams are recombined on to a common axis, and are alternately focussed on to the entrance slit of the monochromator. When there is no sample in the sample cell, the half-beam travelling along the sample cell is equal to that travelling through the reference cell, when these two identical half beams recombine, a steady signal reaches the detector. However, when the sample is present in the sample cell, the half-beam travelling through it becomes less intense (depending on the nature of the sample). When the two half beams (one coming from the reference and the other from the sample) recombine, an oscillating signal is produced, which is measured by the detector. The signal from the detector then passes through a servomotor to the recorder. Applications of Infrared spectrophotometry The important applications of infrared spectrophotometry are: (i) It is used for the identification of an unknown compound. This is generally done by finger print technique which involves matching the spectrum (IR) of unknown compound with that of known compound.

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(ii) Identification of functional groups in organic molecules: The different functional groups in organic compounds will absorb at characteristic frequencies in the IR spectrum. Thus, from the IR spectrum of a compound, the presence and absence of certain functional groups in the compound is know. (iii) Elucidation of structure: Structure elucidation is possible by IR spectroscopy because it gives valuable information regarding molecular symmetry, dipole moments, bond lengths, bond strength etc. (iv) Identification of an unknown compound: It is done by finger print technique which involves matching the spectrum (IR) of unknown compound with that of a unknown compound with that of a known compound. (v) To distinguish between intra and inter molecular hydrogen bonding: It is based on the fact that as the concentration is increased, the absorption band, due to intermolecular hydrogen bonding increase, while that due to intra-molecular Hydrogen bonding remains unchanged. (vi) IR spectroscopy is used in conformational studies of some compounds by studying the C-X stretching frequency present in equitorial and axial positions. It is also used in determining conformational equilibrium constant, k which is given by k = Ce / Ca Where Ce and Ca are the integrated intensities of the C – X stretching peaks in the equitorial and axial positions. (vii) IR spectroscopy can be used for studying the progress, of a chemical reaction. For instance during the oxidation of a secondary alcohol to a ketone, by examining the IR spectrum of aliquots withdrawn from the reaction mixture from time to time. As the reaction proceeds the O-H stretching band (at 3570 cm–1 ) of secondary alcohol slowly disappears and the C = O stretching band (at 1725 cm–1) due to the formation of ketone appears. (viii) IR spectroscopy is useful to study the progress of chromatographic separations. (ix) Determination of aromaticity: The difference in the wavelengths of overtones of C–H bands in different environments can be used to determine the relative proportions of unsaturated and saturated rings present in hydrocarbons and also to determine the percentage of aromatics or olefines present in the mixture. (x) In determining the shape or symmetry of a molecule such as NO2, which shows 3 peaks as per (3n-6) formula whereby it is confirmed that it is not a linear molecule (xi) In calculation of force constants of molecules. The force constant is a measure of the force (in dynes/cm) required to deform a bond. (xii) In studying tautomeric equilibria such as Keto-enol, lactum-lactum and mercaptothioamide tautomerism by examining the characteristic frequencies of groups such as C = O; O–H, H–H or C = S in the respective IR spectra. (xiii) Industrial Applications: (a) IR spectroscopy is used to determine bulk structure and incidental structure of industrial polymers. Bulk structure results from the normal polymerisation of the monomers, whereas the incidental structures arise from impurities in the monomers, side reactions, etc. IR studies helped in assigning bulk structure of ploymers such as butadiene polymers and incidental structures of polythylene, etc. (b) The degree of crystallinity of nylon-66 has been studied by IR spectra. Absorption band at 934 cm–1 is a measure of crystallinity while the band at 1238 cm–1 is used as a measure of the amorphous content. (c) IR spectroscopy has been used to determine molecular weight of polymers by measuring end group concentrations.

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5.3 NUCLEAR MAGNETIC RESONANCE (NMR) SPECTROSCOPY It is a branch of spectroscopy in which radio frequency waves induce transitions between magnetic energy levels of nuclei of a molecule, which are created by keeping the nuclei in a magnetic field For a nucleus to be magnetic, it must possess spin angular moment, whose magnitude is (h / 2p) I ( I + 1), where I is the spin quantum number of the particular nucleus and “h” is planck’s

constant. Nuclei with I = 0 are non-magnetic and hence is not important from NMR point of view. Nuclei with I = 1/2 give the best resolved NMR spectra. Important examples are 1H, 13C, 19F, and 31P nuclei. Nuclei with I > ½, which are of interest from NMR point of view, include 2H (I = 1), 14N(I = 1) and 11B (I = 3/2). The magnetic axis of the nucleus can assume (2I + 1) orientations with respect to the external magnetic field. Each orientation corresponds to a discrete energy level, given by the following relationship: mµ = × βH o E I Where E is the energy of transition, m is the magnetic number, µ is the magnetic moment of nucleus expressed in nuclear magnetons, I is the spin quantum number, and Ho is the external magnetic field strength in gauss. The spectrum of allowed values, in terms of spin quantum number, is I, (I – 1),…., - (I – 1), –I. Each value corresponds to a discrete orientation (or energy level). Therefore, a nucleus with spin 1 has three orientations, and so on. When a nucleus is placed in a system where it absorbs energy, it gets excited. It then loses energy and reverts to the unexcited state. It absorbs energy again and enters the excited state. Such a nucleus, which becomes excited and unexcited alternately, is said to be in a state of resonance. For determining the resonance frequency, the energy absorbed by the nuclei is measured while the magnetic field is varied. As the magnetic field is increased, the processional frequency of the nucleus increases. When this frequency becomes equal to the frequency of the oscillation field, transitions occur between the nuclear energy states. The energy absorbed in this process produces a signal at the detector, which after amplification is recorded as a band in the spectrum. An NMR spectrum is then plotted between absorption signal at the detector and the strength of the magnetic field

Fig. 5.5. Schematic diagram of NMR spectrometer.

Instrumentation Two types of NMR spectrometers are in use : (1) Wide line NMR spectrometers. (2) High resolution NMR spectrometers. The essential components of an NMR spectrometer are:

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(a) A strong magnet to provide the principal part of the magnetic field, Ho (b) A sweep circuit consisting of a set of Helmholtz coils superimposing the main magnetic field, to provide the additional field required to bring the total field to the resonance condition. (c) A transmitter to supply the desired radio frequency (r.f.) energy. (d) A detector amplifier circuit to pick up and amplify the resonance signal (e) A probe which serves to hold the sample between the pole pieces. (f) A device to receive and record the signal. For recording NMR spectra, the sample is placed between the poles of the strong magnet. The sample is then irradiated with radiowaves. At certain value of the magnetic field, absorption of r.f. energy occurs. A sensitive detector monitors the absorption energy which is recorded as a peak on the graph. The spectra can be plotted at low resolution or high resolution as desired. The peak areas are measured automatically by the modern NMR spectrometers. NMR spectra can be described in terms of chemical shifts and coupling constants. Chemical shifts: Chemical shift is an important feature of high resolution NMR spectra. In different chemical environments, the same type of nucleus will be shielded slightly from the applied field surrounding electrons. For a fixed external field, Ho, different screening factors cause slightly different frequencies. The magnitude of the effective field experienced by each group of nuclei can be expressed a Heff = Ho (1 – s) Where s is a non-dimensional shielding constant, which may be a positive or negative number. The value of the shielding constant depends on factors such as hybridization and electronegativity of the groups attached to the atom containing the nucleus. Thus, the shielding constant for protons in a methyl group is larger than that for protons in methylene, and it is zero for an isolated hydrogen molecule. For ethanol, the field applied must be always greater than the field for the resonance of an isolated protein, in order that the various protons may resonate. Since the value of shielding constant for protons in different functional groups is different, the required applied field would also be different for different groups. Thus, least shielded (i.e., low shield constant) proton of hydroxyl group resonates at the lowest field and of methyl group at the highest field. The areas under the peaks are in direct ratio to the numbers of protons, e.g., 1:2:3 on hydroxyl, methylene and methyl groups. Experimental Calibration: Since commercial NMR spectrometers employ different field strengths, it is desirable to express the position of resonance, in field independent units and with respect to the resonance of a reference compound. For proton spectra in non aqueous media, tetramethyl silica, (CH3­)4 Si (abbreviated at TMS) is used as reference material. Its position is assigned as 0.0 on the δ scale. TMS contains 12 protons but these are all chemically equivalent and hence give rise to a sharp signal. The magnitude of the chemical shift is expressed in ppm as follows: Hsample − H TMS δ= ppm v1

Where Hsample and HTMS are the positions of the absorption lines for the sample and reference respectively, expressed in frequency units (hertz) and ν is the operating frequency of the spectrometer. A positive δ value represents a greater degree of shielding in the sample than in the frequency. Chemical shifts are also expressed tau (τ ) units, with t = 10 – δ. Values for the chemical shifts (δ) of protons (hydrogen atoms) in some chemical groups are given in Table 5.2 given below.

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Table 5.2. Chemical shifts for protons in ppm relative to TMS = 0

Type of proton R—C—H || O R — OH R — OCH3 R — CH2 — R R — CH3

δ , ppm 9.7 5 3.8 1.3 0.9

Applications of NMR spectroscopy 1. It is used for identification of atomic configurations in molecule 2. It is used for quantitative analysis of materials for particular isotope content from the integrated area under the NMR absorption band. 3. It is a rapid, non-destructive method for analyzing proton content of fats and oils. 4. It is used for the determination of fluorine content in plastics and other chemical compounds. 5. It is used for the determination of water (H2O) in liquid N2O4 and in heavy water (D2O). 6. It is used to assay pharmaceutical formulations such as aspirin, phenacitin and caffeine. 7. It is used for the quantitative determination of water in food products, agricultural materials, paper and pulp, etc. 8. It is used in structural diagnosis and in study of Keto-enol tautomerism. 9. It is used in conformational analysis of molecules. 10. It is used in determination of activation energy. 11. It is used for the studies on inorganic complexes and their structural determinations. 12. It is used in structural studies of polyethylene. 13. It is used in investigating intra-molecular convertions. 5.3. CHROMATOGRAPHY Chromatography is a techniques used for the separation of a mixture of solutes brought about by the dynamic partition or distribution of dissolved or dispersed materials between two immiscible phases, one of which is moving past the other. Chromatographic processes may be conveniently classified broadly as follows (i) Partition chromatography; (ii) Adsorption chromatography. Chromatographic process may also be classified under Liquid chromatography and Gas chromatography.

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5.3.1 Gas Chromatography (G.C.) This is a process by which a mixture is separated into its constituents by a moving gaseous phase passing over a stationary sorbent. Gas chromatography may be of the following two types : (i) Gas Liquid Chromatography (GLC), in which the separation takes place by partitioning a sample between a mobile gas phase and a thin layer of a non-volatile liquid coated on an inert support. This is more important technique than GSC. (ii) Gas Solid Chromatograophy (GSC), in which a suitable solid material with a large surface area such as granular silica, alumina or carbon is used as a stationary phase, while employing a mobile gas phase. Gas chromatography technique was originally developed in 1941 by A.J.P. Martin and R.L.M. Synge for which they were awarded Nobel prize in 1952. Today this technique is the most important and extensively used analytical tool for the determination of number of components in a mixture, the presence of impurities in a substance, and identification of a compound. This technique is also becoming important for process control in chemical industries and refineries Although gas chromatography is limited to volatile materials, the applicability of this technique has been further extended because of (i) availability of column temperatures upto 450°C (ii) availability of pyrolytic techniques, and (iii) the possibility of converting non-volatile materials into volatile derivatives. Theoretical Principles Gas chromatography process is controlled by three physical transport phenomena namely, flo , diffusion and more importantly, the partition of the solutes between the stationary phase and the mobile phase. As the solute is introduced into the column, the molecules are distributed (or partitioned) between the stationary phase and the mobile phase (the carrier gas) and a dynamic equilibrium is soon established. The process of distribution of the solutes between the two phases continues further as the fresh mobile phase (carrier gas) passes over the column, thereby establishing a fresh dynamic equilibrium. The process goes on and on until a final equilibrium is established and at this stage, the concentration of molecules of each solute in the two phases is constant. Partition ratio, Concentration of solute molecules in the stationary phase K = Concentration of the solute molecules in the mobile phase Partition ratio depends upon (a) the nature of the solute (b) the nature of the solvent (i.e., the stationary liquid phase) (c) the concentration of the liquid phase, and (d) temperature. The solute components of the sample mixture travel down the column at their own rate depending upon their respective partition ratios and the extent of their band spreading, thereby allowing a clear and clean separation of the components and their subsequent detection and determination. Sequence of Gas Chromatographic process steps: A sample mixture containing different solutes is injected into a heating block where it is immediately vaporized and swept by the mobile phase (carrier gas) stream into the column inlet. The solutes are absorbed at the head of the column by the stationary phase and then desorbed by fresh carrier gas. This sorption-desorption process occurs repeatedly as the sample is moved by the carrier gas down towards the column outlet. Each

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solute band will travel at its own rate through the column. Their bands will separate to a degree that is determined by partition ratios of the individual solutes present in the sample, and the extent of their band spreading. The solutes are then eluted sequentially in the increasing order of their partition ratios and enter a detector attached to the column exit. If a recorder is used, the signals appear on the chart as a plot of time versus the composition of the carrier gas stream. The time of emergence of a peak is characteristic of each component. The peak area is proportional to the concentration of the component in the sample mixture. Equipment: A schematic representation of a typical laboratory gas chromatographic apparatus is shown in Fig. 5.6. Pressure control

Flow regulator

Flow meter

Drying tube Detector

Pure carrier gas (H2, He, N2 Ar, CO, etc.)

Sample injection

Sample collector Bridge Amplifier

Cylinder containing carrier gas

Thermostated column

Recorder

Fig. 5.6. Schematic diagram of a gas chromatograph.

The essential parts of a typical gas chromatographic assembly are as follows : (i) High pressure gas cylinder to supply pure carrier gas (mobile phase): The operating efficiency of a gas chromatographic process depends on the maintenance of a constant flow of pure carrier gas. The carrier gases commonly used are pure helium, nitrogen, hydrogen and argon. The choice of a carrier gas depends upon (a) the nature of the sample (b) the type of detector system employed and (c) the column efficienc . Hydrogen and helium are most suited for use with thermal conductivity detectors because of their high thermal conductivity (as compared to the vapours of most organic compounds) and low density. Hydrogen is seldom used as carrier gas because of fire hazard and its reactivity towards unsaturated or reducible sample components. The high pressure supply of carrier gas is associated with pressure regulators and flow meters to monitor and control the flow rate of the carrier gas (ii) Sample Injection System: The quantity of sample required for gas chromatographic analysis depends upon the nature and concentration of the solutes being analyzed, the size of the column and sensitivity of the detector. Generally, 0.1 to 50 microlitres are used for gases and liquid samples, whereas a fraction of a milligram is used for solid samples. The sample is introduced into the carrier gas using (i) a specially prepared micro-syringe having a hypodermic needle (ii) a glass ampoule (used for viscous liquids or solids) or (iii) a gas sampling valve. Analysis of non-voltile organic compounds can be done by first converting them into a volatile derivative before introducing into the carrier gas. High molecular weight compounds may be subjected to pyrolysis to break the large molecules into smaller and more volatile fragments and then analysed by gas liquid chromatography. Such a technique is called pyrolysis gas chromatography. (iii) The column: Actual separation of sample components takes place in the column. Important factors for obtaining the desired separation are the nature of the solid support, type and amount of

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the stationary phase, method of packing, length of the column and temperature. The column is enclosed in a thermostatically controlled oven to ensure reproducible conditions. The basic types of columns generally used in GC are (a) packed columns and (b) open tubular or capillary columns. Packed columns are made of stainless steels, copper, cupronickel, glass or even plastic (nylon or polythene) which may be coiled in U- or W-shape. Metal columns are preferred, though they are expensive. The lengths of the column may be anywhere from 120 cm to 5 m. Adsorption columns are used in gas solid chromatography (GSC) with suitable adsorbing materials such as: Activated carbon (for H2, N2, O2, NO, CO, CH4) Silica gel (for CH4, C2H4, C2H6, CO2) Alumina (for CH4, C2H4, C2H6) Linde Sieve 5A (for H2, N2, O2, CH4, C2H6, CO) Partition columns are used in gas liquid chromatography (GLC). They are packed with an inert support carrying a non-volatile liquid phase. The support material consists of a finely divided celite, ground fire brick or glass beads, which can hold adequate amount of the liquid phase Gas chromatograms are generally obtained with the column maintained at a constant temperature. Vapour jackets containing benzene (B.pt. 80.1°C) toluene (B.pt. 110°C), cyclohexane (B.pt. 161°C), tetralin (B.pt. 207.5°C) and bromonaphthalene (B.pt. 281°C) or liquid baths may be used to maintain the column temperature. The columns are seldom operated at room temperature except in case of extremely volatile sample mixtures. (v) Detectors: The detector is situated at the exit end of the separation column. The main functions of the detector are to sense and measure the small quantities of the separated components present in the carrier gas stream leaving the column. The output of the detector is fed to a recorder which produces a pen-trace, called chromatogram. The choice of the detector depends upon the nature and concentration level of the components being measured. (vi) Recorder: Almost all the detectors give small and weak electrical signal. Hence it is necessary to pass these signals through an amplifier before going to the recorder. The recorder comprises of a mobile recording pen activated by the signal and a recording chart strip moving at a pre-selected speed. The amplifi d signals drive the pen on the moving strip of paper and trace out a chromatogram containing a series of Gaussian (i.e., bell shaped) peaks. The time required for the sample components to travel through the column is compared with that for known compounds, which serves as a means of identific tion. The area of the peak is a direct measure of the concentration of the corresponding compound present in the sample. Applications of Gas Chromatography

(i) Identification and separation of compounds: petroleum products, waxes, nitrogen and sulphur compounds, saturated and unsaturated hydrocarbons have been separated by GC and identified by IR, U , NMR and mass spectrometric techniques. (ii) GC was used to analyse fatty acids, steroids, herbicides, pesticides, esters, esterogens, perfumes, plastics, cosmetics, beverages, fertilizers, rubber products and detergents. (iii) It can be used for the preparation of pure substances or narrow fractions as standards for further investigation. (iv) On an industrial scale, it can be utilized for process monitoring. (v) It can be used to determine specific surface area in absorption studies (vi) It is also used for the elemental analysis of organic components, etc. (vii) In environmental studies :Asthma, lung cancer, emphysema, bronchitis and other chronic

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respiratory diseases could result from air pollution. Air sample can be very complex mixtures. Gas chromatography generally used for the separation and analysis of such mixtures. (viii) It is also used for the searation and determination of many components in petroleum products. (ix) In clinical chemistry: Blood, urine and other biological fluids can be analyzed for proteins, carbohydrates, amino acid, fatty acid etc directly or preparation of appropriate volatile derivatives. (x) Preparative gas chromatography: Though originally developed as an analytical tool, GC has recently been proved to be a preparative tool too. (xi) Determination of carbon monoxide: Trace levels of CO (less than 10 ppm) can be determined by gas chromatography. The sample is subjected to catalytic reduction by hydrogen over a nickel catalyst at 360 °C and the resulting CH4 is measured by gas chromatography using flame ionization detecto . CO + 3H2

CH4 + H2O

Avdantages of gas chromatography

(i) It is easily used for the analysis of volatile solids, high boiling liquid on samples of permanent gases. (ii) It requires less time. (iii) It gives high relative precision of the order of 2-5%. (iv) The data obtain from gas chromatogram is easily recorded.

Disadvantages of gas chromatography

(i) It requires volatile and thermally stable (below 400 °C) sample. (ii) Published retention data are not always reliable for qualitative analysis. QUESTIONS



1. State Beer-lambert’s law and discuss its applications and limitations. (RGPV Bhopal 2006, 2009) 2. What are the essential parts of a spectrophotometer? What are its advantages over a colorimeter? 3. Discuss the origin of ultraviolet spectra and explain the working of UV-Spectrometer. 4. Discuss the essential components of a UV-spectrometer. Give four important applications of UV-Spectroscopy. 5. How do IR-spectra originate? Describe the working of an IR spectrometer. 6. What is the instrumentation required for IR spectroscopy? Give five important applications of IR spectroscopy. 7. Describe the essential components of HPLC system. State some important advantages of HPLC. 8. Discuss the principles involved in Gas Chromatographic. Describe the working of a gas chromatograph. 9. What are the essential parts of a Gas Chromatographic assembly? Give five important applications of gas chromatography.

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10. Write short notes on any two of the following : (a) Spectrophotometric estimation of cations (b) Various techniques of chromatography. (c) IR-spectroscopy and its significanc (d) UV-Spectroscopy and its significance (RGPV Bhopal 2001) 11. State the principles involved, instrumentation and applications of UV spectroscopy. 12. State the principles involved in visible spectrophotometry. What are the essential features of a photoelectric photometer. State some important applications of visible spectrophotometry. 13. What are the essential components of an IR Spectrometer and what are their functions ? State some important applications of IR spectroscopy. (RGPV Bhopal 2009) 14. Illustrate the essential features of an NMR Spectrometer. Give some important applications of NMR Spectroscopy. 15. (a) What is chromatography ? (b) What are the important types of chromatographic processes ? (c) What are the criteria for selecting a particular type of chromatographic process for a given separation problem ? 16. Write informative notes on (a) HPLC (b) Gas Chromatography (c) MNR spectroscopy

251

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PART–B

UNIT

5

Water Analysis Techniques

5.4 ALKALINITY Alkalinity. By alkalinity of water we mean the total content of those substances in water that cause an increased concentration of OH– ions upon dissociation or due to hydrolysis. The alkalinity of natural waters is generally due to the presence in them of HCO3–, SiO32 –, HSiO3 – , and sometimes CO32 – ions and also due to the presence of salts of some weak organic acids, known as humates, that bind H+ ions as a result of hydrolysis, thereby increasing the concentration of OH– ions. In addition to the above, the alkalinity of boiler water is also conditioned by the presence of PO43– and OH– ions. Also, the presence of salts of weak acids such as silicates and borates induces buffer capacity in water and resists the lowering of pH. Surface waters containing algae and also water treated by lime-soda process may contain considerable quantities of alkalinity due to CO32 – and OH–.

Depending on the anion that is present in water (HCO3 –, CO32 – or OH–), alkalinity is classified respectively as bicarbonate alkalinity, carbonate alkalinity or hydroxide alkalinity. Highly alkaline waters may lead to caustic embrittlement and also may cause deposition of precipitates and sludges in boiler tubes and pipes. With respect to the constituents causing alkalinity in natural waters, the following situations may arise : 1. Hydroxides only 2. Carbonates only 3. Bicarbonates only 4. Hydroxides and carbonates 5. Carbonates and bicarbonates

(Notes. The possibility of hydroxides and bicarbonates existing together is ruled out because of the fact that they combine with each other as follows forming the carbonates: → CO3– – + H2O OH– + HCO3–  The types and extent of alkalinity present in a water sample may be conveniently determined by titrating an aliquot of the sample with a standard acid to phenolphthalein end-point, P, and continuing the titration to methyl orange end point M. The reactions taking place may be represented by the following equations:

→ H O OH– + H+  2 + – –  → HCO – P CO3 + H 3 → H CO  → H2O + CO2 HCO3– + H+  2 3 251

M

...(1) ...(2) ...(3)

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252

The volume of acid run down upto phenolphthalein end point P corresponds to the completion of equations (1) and (2) given above, while the volume of acid run down after P corresponds to the completion of equation (3). The total amount of acid used from beginning of the experiment, i.e., the methyl orange end point M, corresponds to the total alkalinity present which represents the completion of reactions (1) to (3). Table 5.1.

S. No.

Result of



Titration

Hydroxide (OH–)

P = O P = M 1 P= M 2 (or V1 = V2) 1 P> M 2

Nil P or m

Nil Nil

M Nil

Nil

2P

Nil

(2P – M)

2 (M – P)

Nil

Nil

2P

M – 2P

1. 2. 3. 4.

(or V1 < V2)

5.

P< 1M 2 (or V1 < V2)



Alkalinity due Carbonate (CO32–) Bicarbonate (HCO3–)

Procedure. The alkalinity of a water sample may be determined volumetrically by titrating it with a standard acid using indicator. Solution Required : 1. Standard acid solution

2. Water sample

3. Phenolphthalein Indicator

4. Methyl orange Indicator

Necessary Equipment : Burette, pipette, funnel, conical flask, burette stand & beaka . Procedure : 1. Wash the Burette with water, rinse it with given acid solutions and then fill with N/50 HC 2. Wash the Pipette with water, pipette out 10 ml. of water sample & transfer it in conical flask. Add 1 drop of phenolphthalein indicator into solution turns pink. 3. Add acid from the burette to the conical flask till the disappearance of pink colour. This value is noted as Phenolphthalein end point[P]. 4. To the same conical flash add a drop of methyl orange indicator, & then continue the titration with same acid till the appearance of light pink colour. Methyl orange reading includes total titre value from begining to end. 5. Repeat the same procedure till we get concordant reading. Alkalinity is generally expressed as parts per million (ppm) in terms of CaCO3. 1000 ml of N/50 acid solution ≡ 1000 mg of CaCO3.

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253

Hence Alkalinity =

Vol.of N/ 50 acid × 1000 mg/l or ppm Vol.of sample taken for titration

On the basis of the analysis of water with respect of alkalinity and total hardness, the amounts of carbonate hardness (temporary hardness or alkaline hardness) and non-carbonate hardness (nonalkaline hardness or permanent hardness) present in the water can be determined as follows: 1. If the methyl orange alkalinity of the water equals or exceeds the total hardness, all the hardness is present as carbonate hardness. 2. If the methyl organe alkalinity of water is less than the total hardness, the carbonate hardness equals the alkalinity. 3. The non carbonate hardness, under conditions in (2) above, is equal to the total hardness minus the methyl organe alkalinity. Further, on the basis of hardness and alkalinity, natural waters can be subdivided into two groups : Non alkaline and alkaline. If the hardness is greater than alkalinity, the water is called nonalkaline. If the hardness of the water is lesser than alkalinity, the water is characterised as alkaline. Non alkaline waters are more frequently encountered in nature, which are characterized by different kinds of hardness as follows: Ht

= (Hc + Hnc) = (HCa + HMg)

where Ht Hc Hnc HCa HMg

= = = = =

total hardness, carbonate hardness, non-carbonate hardness, calcium hardness and magnesium hardness.

5.5 HARDNESS DETERMINATION BY COMPLEXOMETRIC METHOD Hardness in water is generally due to dissolved salts of Ca and Mg. It is easily estimated by titrating the water sample with standard EDTA solution using Eriochrome Black-T as an indicator. Ethylene diaminetetra acetic acid (EDTA) EDTA (ethylene diaminetetra acetic acid): For simplicity it is represented by symbol H4y. Because of its limited solubility, it is not used directly. It is usually used as its disodium salt dihydrate Na2H2y.2H2O, because it can be obtained in high state of purity and is a primary standard. The solution of EDTA are very valuable titrant because the reagent combines with metal ions in a 1 : 1 ratio. – NaOOC—CH2 CH2COO HN—CH2—CH2—NH – NaOOC—CH2 CH2COO Structure of Na2 EDTA

Indicator Eriochrome Black T (EBT) is used as a indicator: It is a typical metal ion indicator chemically it is sodium 1-(1-hydroxy-2-naphthylazo)-6-nitro-2-naphthol-4-sulphonate. The indicator Eriochrome Black T (a dye stuff) is effective between the pH (8 - II). It is therefore, essential while performing the titration of hard water with EDTA solution, the pH of the solution must be made (pH = 10) by adding a suitable buffer solution.

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254 OH NaO3S

OH N

N

NO2

Eriochrome Black-T sodium 1 (1 hydroxy-2 naphthyl azo)-6 nitro-2 naphthol-4 Sulphonate Buffer Solution : The optimum pH for the experiment is 10.0 ± 0.1 and is adjusted by NH4OH and NH4CI Buffer solution (Alkaline buffer). Take 17.5 gms of NH4CI and add 14.2 ml. of conc. NH3 solution to it. Dilute the solution to 250ml. with distilled water. Necessary Requirement : Glasswares, burette, pipette, conical flask EDTA solution Standard water sample (A) Buffer solution Eriochrome Black T Indicator Unknown water sample (B). Procedure :

1. Wash the burette with water, rinse it with given EDTA solution and then fill with EDTA solution. 2. Wash the pipette with water, pipette out 10ml. of water sample & transfer it in the conical flask 3. Add 5 ml of buffer solution and 2 drops of Eriochrome black-T indicator. 4. Titrate the solution with EDTA solution from the burette until the colour changes from wine red to clear sky blue at the end point. 5. Repeat the same procedure till we get concordant reading.



OH



(i)

NaO3S

OH N

N

+ Ca++/Mg++

Buffer solution pH = 10

Ca/Mg NO2 Eriochrome Black T

O NaO3S

O N

N

NO2 Weak complex wine red in colour

+ 2H+

B– WATER ANALYSIS TECHNIQUES

255

(ii)

At last thus releases Eriochrome Black T in free state. At this stage wine red colour of the solution changes into sky blue colour (End point) Permanent hardness can be determined by precipitating the temporary hardness by prolonged boiling for about 30 minutes followed by titration with the EDTA solution as above. The difference in the titre values corresponds to the temporary hardness of the water sample. Calculation Let volume of the water taken for each titration = 50 ml Volume of EDTA used when titrated against standard hard water = V1 ml Volume of EDTA used when titrated against sample hard water = V2 ml Volume of EDTA used when titrated against water having permanent hardness = V3ml (a) Strength of EDTA solution 1 ml of standard hard water (SHW) contains = 1 mg of CaCO3 50 ml of SHW contains = 50 mg of CaCO3 Volume of EDTA consumed for 50 ml of SHW = V1ml \ V1ml of EDTA is used for = 50 mg of CaCO3 50 Or strength of EDTA solution = mg/ml of EDTA. V1 (b) Total hardness Volume of EDTA used for 50 ml of sample hard water = V2 ml 50 mg of CaCO3 Since 1 ml of EDTA consume = V1 V2 ml of EDTA will consume =

50 × V2 mg of CaCO3 V1

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256

1000 ml of sample water =

50 × V2 1000 × mg of CaCO3 V1 50

V2 × 1000 mg V1 V × 1000 ppm. \ Total hardness = 2 V1 (c) Permanent hardness =

Volume of EDTA used for 50 ml of water, containing permanent hardness = V3 ml 50 × V3 mg of CaCO3 V2 ml of EDTA will consume = V1 Hence, 50 ml of water contains permanent hardness 50 V3 × × 1000 mg of CaCO3 = V1 50 V × 1000 mg = 3 V1 V3 × 1000 ppm Permanent hardness = V1 (d) Temporary hardness Temporary hardness = total hardness – permanent hardness. 5.6 DISSOLVED OXYGEN (DO) Determination of DO is important for boiler feed water and also in sanitary engineering practice. DO is needed for living organisms to maintain biological processes. DO test is the basis of BOD test which is an important parameter in evaluating the pollution potential of domestic and industrial wastes. Theory It is usually determined by Winkler’s method. The method is based on the fact that in presence of MnSO4 (act as a oxygen carrier to bring about the reaction between KI and oxygen) dissolved oxyge, oxidises potassium iodide (KI) to iodine. And the liberated iodine is titrated against standard hypo solution using starch as indicator. → Mn(OH)2 + K2SO4 MnSO4 + 2KOH   → 2MnO(OH)2 2Mn(OH)2 + O2 Basic Manganic Oxide → MnSO4 + 2H2O + O MnO(OH)2 + H2SO4 

→ K2SO4 + H2O + I2 2KI + H2SO4 + O 

→ Na2S4O6 + 2 NaI I2 + Na2S2O3  hypo Iron and nitrite interfere with this procedure. The interference of nitrite can be eliminated by adding sodium azide (NaN3). If ferric iron is present, phosphoric acid is used instead of H2SO4. Procedure Take water sample in a 300 ml BOD bottle. Now add 2 ml of MnSO4 and 2 ml of alkaline iodide-azide reagent into stopper the bottle and shake them thoroughly. After 2 minutes add 2 ml of concentrate H2SO4 into it and shake the bottle until the ppt has completely dissolved. Allow the yellow solution to stand for 5 minutes. Now take 100 ml of this solution and titrate against N/100 hypo solution using starch as indicator. At the end point blue colour of indicator disappear. The titre value, V is noted.

257

B– WATER ANALYSIS TECHNIQUES Calculation The D.O. content of the water sample can be calculated as follows: 1 × 8 × 1000 mg/l or ppm DO = V × N × 200 Where V = titre value, N = Normality of hypo solution 5.7 BIOCHEMICAL OXYGEN DEMAND (B.O.D.)

B.O.D. is defined as the quantity of dissolved oxygen required by aerobic bacteria for the oxidation of organic matter under aerobic conditions BOD is considered as the major characteristic used in stream pollution control. It gives valuable information regarding the self purification capacity of the streams and serves as a guideline for the Regulatory Authorities to check the quality of the effluents discha ged into such water bodies. The demand for oxygen is proportional to the amount of organic waste to degrade aerobically. When the BOD is high, the D.O. become low. Hence greater the BOD, greater is the polluties. Table 5.2.

S. No. 1. 2. 3. 4.

Source of effluen Domestic sewage Cowshed sewage Paper mill Tannery efflun

BOD (ppm) 320 3010 8190 12360

Procedure The BOD test essentially consists of measurement of DO content of the water sample, before and after incubation at 20 °C for 5 days. Take the diluted sample in two stoppered bottles. The D.O. content of one of the bottles is immediately determined. The another bottle incubated at 20 °C for 5 days. Then its D.O. content is determined. The depletion in D.O. caused is used as a measure of BOD. Calculation where

(DO0 − DO5 − B) × 100 % of the sample used DO0 = Initial DO content in mg/l DO5 = Do content after incubation at 20 °C for 5 days

B.O.D. (mg/l) =

B = Dilution factor i.e.,

ml of sample after dilution ml of sample before dilution

Note : While carrying out the BOD test on some types of industrial wastes, addition of small measured volume of sewage containing a good bacterial population (called seed) may be required. 5.8 CHEMICAL OXYGEN DEMAND (COD) According to the American Society of Testing and materials (ASTM), COD is defined as the amount of oxygen (expressed in mg/l) consumed under specified conditions in the oxidation of organic and oxidisable inorganic matter, corrected for the influence of chlorides The COD is a measure of the Oxygen equivalent to that portion of organic matter present in the wastewater sample that is susceptible to oxidation by potassium dichromate. This is an important and quickly measurable parameter for streams, sewage and industrial waste samples to determine their pollutional strength.

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258

The principle involved in the determination of COD is that when a wastewater sample is refluxed with a known excess of K2Cr2O7 in a 50% H2SO4 solution in presence of AgSO4 (as catalyst) and HgSO4 to eliminate interference due to chloride, the organic matter present in the sample is oxidised to CO2, H2O and NH3. The excess dichromate remaining unreacted in the solution is titrated against a standard solution of ferrous ammonium sulfate. The COD of the wastewater sample is calculated as follows : (V − V ) × N × 8 × 1000 COD in mg/l = 1 2 x where V1 and V2 are the volumes of ferrous ammonium sulfate of normality N, run down in the blank and test experiments respectively, and x is the volume of the sample taken for the test. Since in the COD test, both biologically oxidisable and the biologically inert matter are oxidised, the COD value for a sample is always higher than the BOD value. 5.9 FREE CHLORIDE OR RESIDUAL CHLORINE Chlorine is widely used for disinfection of potable or municipal water supplies to remove bacteria, fungus and other pathogenic micro-organisms. The sterilizing action of chlorine is due to its reaction with water, producing hypochlorous acid and nascent oxygen both of which have powerful germicidal properties. CaOCl2 + H2O —→ Ca(OH)2 + Cl2 Cl2 + H2O —→ HOCl + HCl HOCl —→ HCl + [O] hypochlorous acid nascent oxygen However, excess of free chlorine in drinking water is undesirable not only because of unpleasant taste but also because it is injurious to human metabolism. Hence, free chlorine present in municipal water is generally estimated prior to the domestic supply for adjusting the chlorine dose rate. Procedure When the water sample containing free chlorine is treated with KI, free chlorine oxidizes KI and liberates I2 in equivalent amount. The liberated I2 is titrated against hypo solution using starch as indicator. At the end point blue colour disappeared. Cl2 + 2KI —→ 2KCl + I2 I2 + 2Na2S2O3 —→ Na2S4O6 + 2 NaI sodium sodium

ulfate (hypo)



tetrathionate

The free chlorine present in the water sample can be calculated as follows : V × N1 Free chlorine, ppm = 1 × 35.45 × 1000 V2 where V1 = Volume of standard hypo solution run down at the end-point. N1 = Normality of the hypo solution (usually N/20), and V2 = Volume of water sample used for the test (usually, 100 ml) 5.10 DISSOLVED CHLORIDE Chlorides are present in water mostly as NaCl, MgCl2 and CaCl2. Chlorides when present in concentrations over 250 ppm impart aesthetically unacceptable tastes for drinking purposes. Presence of unusually high concentration of chloride in water indicates pollution from domestic or Industrial wastes. Salts like MgCl2 are unacceptable in boiler feed water. Chloride in water can be determined by titration with standard (about N/50) AgNO3 solution using potassium chromate as indicator (Mohr’s method).

B– WATER ANALYSIS TECHNIQUES

259

Theory The determination of dissolved chloride based on precipitation titrations. When AgNO3 solution is added to the water sample in presence of K2CrO4, the chlorides present in it are precipitated as AgCl. As soon as all the chlorides are precipitated out, even a drop of AgNO3 added in excess gives a red precipitate of silver chromate. This indicate the end point AgNO3 + NaCl —→ AgCl↓ + NaNO3 AgNO3 + K2CrO4 —→ Ag2CrO4↓ + 2KNO3 Red or pink Procedure Take 100 ml of given water sample in a titration flask. Now add 2-3 drops of K2CrO4 solution into it and titrated the solution against N/50 AgNO3 solution. At the end point brick-red colour appeared. Let the volume of AgNO3 used be V ml. Calculation N1V1 = N2V2 Nchloride × V1 = N2 × V2 N 2 × V2 = V1 N × V2 × 35.5 g/L Strength of chloride ions = 2 V1 N × V2 × 35.5 × 1000 mg/L or = 2 V1 where V1 = Volume of water sample taken for test V2 = Volume of AgNO3 solution used N1 = Normality of chloride ion N2 = Normality of AgNO3 solution. (i) Total dissolved solids (TDS) Water samples may contain both dissolved as well as suspended solids. (a) Suspended solids content in a water sample is determined by filtering an aliquout of the sample through a previously weighed sintered crucible or a tared Gooch Crucible and drying the crucible at 103°C to 105°C to constant weight. The different in weight gives the suspended solids content of the water sample, which is calculated and reported in terms of mg/l (or ppm). (b) The total solids content of the water sample is determined by evaporating known volume of the water sample taken in a dish, and drying the residue for 24 hrs. at 103°C to 105°C followed by weighing. The increase in weight of the evaporating dish gives the total solid content of the water sample. The total solids value is calculated and reported in terms of mg/l (or ppm). (c) The difference in the values of total solids and suspended solids gives the total dissolved solids (TDS) present in the water sample. Alternatively, the TDS can be determined by evaporating the filtrate obtained from part (a) above in a clean platinum dish in an overn at 103°C – 105° C for an hour and weighing the residue. w TDS, ppm = × 106 where w is the weight in grams of the dry residue and V is the volume of V the water sample taken for the test.

BASIC ENGINEERING CHEMISTRY

260 5.11 NUMERICAL PROBLEMS

(A) NUMERICAL PROBLEMS BASED ON ALKALINITY Example 1. 100 ml of a raw water sample on titration with N/50 H2SO4 required 12.4 ml of the acid to phenolphthalein end point and 15.2 ml of the acid to methyl orange end point. Determine the type and extent of alkalinity present in the water sample. Solution. P = 12.4 ml of N/50 H2SO4 M = 15.2 ml of N/50 H2SO4 1 Since P > M, the water sample must contain only OH– and CO3– – alkalinities and there cannot 2 be any HCO3– alkalinity (vide Table 2). Further, 1. The volume of N/50 H2SO4 equivalent to OH– present in 100 ml of the water sample



= [2 P – M] = [(2 × 12.4) ml – 15.2 ml] = 24.8 ml – 15.2 ml = 9.6 ml. and 2. The volume of N/50 H2SO4 equivalent to CO3– – present in 100 ml of the water sample = 2 [M – P] = 2[15.2 – 12.4] ml = 2 × 2.8 ml = 5.6 ml 3. Since the equivalent weight of CaCO3 = 50 1 ml of 1 NH2SO4 ≡ 50 mg of CaCO3 Alkalinity due to OH– Since 1 ml of 1N H2SO4 ≡ 50 mg CaCO3 1 9.6 ml of N/50 H2SO4 ≡ 50 × 9.6 × mg CaCO3 50 = 9.6 mg CaCO3 This corresponds to 100 ml of the water sample. ...  Amount of OH– present in 1 litre of the water sample 1000 = 9.6 × = 96 mg/l as CaCO3 100 ...  Alkalinity of the water sample due to OH– = 96 ppm Alkalinity due to CO3– – Similarly, 1 mg CaCO3 50 = 5.6 mg CaCO3 This corresponds to 100 ml of the water sample. ...  Amount of CO3– – present in 1 litre of the water sample 1000 = 5.6 × = 56 mg/l as CaCO3 100 ...  Alkalinity of the water sample due to CO3– – = 56 ppm 5.6 ml of N/50 H2SO4 ≡ 50 × 5.6 ×

B– WATER ANALYSIS TECHNIQUES

261

Report The given water sample contains: OH– alkalinity = 96 ppm CO3– – alkalinity = 56 ppm Total alkalinity = 152 ppm Example 2. A water sample is not alkaline to phenolphthalein. However, 100 ml of the sample, on titration with N/50 HCl, required 16.9 ml to obtain the end-point, using methyl orange as indicator. What are the types and amount of alkalinity present in the sample ? Solution. P = 0;  M = 16.9 ml. Hence, the alkalinity of the water sample is only due to HCO3– ions. ... Volume of N/50 HCl equivalent to the HCO3– present in 100 ml of the water sample = 16.9 ml Since 1 ml of 1N HCl ≡ 50 mg CaCO3 1 16.9 ml of N/50 HCl ≡ 50 × 16.9 × mg 50 = 16.9 mg CaCO3 This corresponds to 100 ml of the water sample ... Amount of HCO3– present in 1 litre of the water sample 1000 ≡ 16.9 × = 169 mg as CaCO3 100 .. . Alkalinity ≡ 169 ppm Report The given water sample consists of only HCO3– alkalinity and it is 169 p.p.m. Example 3. A water sample is alkaline to both phenolphthalein as well as methyl orange. 100 ml of the water sample on titration with N/50 HCl required 4.7 ml of the acid to phenolphthalein end-point. When a few drops of methyl orange are added to the same solution and the litration further continued, the yellow colour of the solution just turned red after addition of another 10.5 ml of the acid solution. Elucidate on the type an extent of alkalinity present in the water sample. Solution.   P = 4.7 ml of N/50 HCl M = (4.7 + 10.5) = 15.2 ml of N/50 HCl 1 Since P < M, the water sample must contain CO3– – alkalinities and HCO3– alkalinity only 2 and not OH– alkalinity. Further, (i) The volume of N/50 HCl equivalent to CO3– – present in 100 ml of the water sample ≡ 2 [P] = 2 × 4.7 = 9.4 ml, and (ii) The volume of N/50 HCl equivalent to HCO3– present in 100 ml of the water sample ≡ M – 2P = (15.2 – 9.4) ml = 5.8 ml Alkalinity due to CO3– – Since 1 ml of 1N HCl ≡ 50 mg of CaCO3 1 9.4 ml of N/50 HCl ≡ 50 × 9.4 × = 9.4 mg CaCO3 50 This corresponds to 100 ml of the water sample.

262

BASIC ENGINEERING CHEMISTRY .. .  Amount of CO – – present in 1 litre of the water sample 3 1000 = 9.4 × = 94 mg/l = 94 ppm as CaCO3 .. .  Alkalinity due to CO – – = 94 ppm 100 3

Alkalinity due to HCO3–

1 = 5.8 mg of CaCO3 50 This corresponds to 100 ml of the water sample. ...  Amount of HCO3– present in 1 litre of the water sample 1000 = 5.8 × = 58 mg/l as CaCO3 .. .  Alkalinity due to HCO – = 58 ppm100 3

5.6 ml of N/50 HCl ≡ 50 × 5.8 ×

Report The given water sample contains: CO3– – alkalinity = 94 ppm HCO3– alkalinity = 58 ppm Total alkalinity = 152 ppm Example 4. 100 ml of a water sample, on titration with N/50 H2SO4, gave a titre value of 5.8 ml to phenolphthalein end-point and 11.6 ml to methyl orange end point. Calculate the alkalinity of the water sample in terms of CaCO3 and comment on the type of alkalinity present. Solution. P = 5.8 ml;  M = 11.6 ml. 1 Since P = M, it means that all the alkalinity present in the water sample is due to CO3– – only: 2 while OH– and HCO3– are absent. Further, the volume of N/50 H2SO4 equivalent to CO3– – present in 100 ml of the water sample = 2P = 2 × 5.8 = 11.6 ml Since 1 ml of 1N HCl ≡ 50 mg of CaCO3 11.6 1 × 11.6 ml of N/50 HCl ≡ 50 × = 11.6 mg of CaCO3 1 50 This is the CO3– – present in 100 ml of the sample ... Strength of CO3– – in terms of CaCO3 1000 = 11.6 × = 116 mg/l 100 = 116 ppm Report The alkalinity of the water sample is 116 ppm, which is only due to CO3– –. Example 5. 100 ml of a water sample, on titration with N/50 H2SO4, using phenolphthalein as indicator, gave the end point when 5.0 ml of acid were run down. Another aliquot of 100 ml of the sample also required 5.0 ml of the acid to obtain methyl orange end-point. What is the type of alkalinity present in the sample and what is its magnitude? Solution. P = 5.0 ml;  M = 5.0 ml. Since P = M, it is obvious that the water sample contains only hydroxide alkalinity and it is not a natural water sample. Further, since 1 ml of 1N H2SO4 ≡ 50 mg of CaCO3

B– WATER ANALYSIS TECHNIQUES 5 1 × 1 50 = 5 mg of CaCO3 This corresponds to 100 ml of the sample only. ... The amount of OH– present in 1 litre of the water sample 1000 = 5× = 50 mg of CaCO3 100 ... Alkalinity of the water sample = 50 mg/l or 50 ppm Example 6. A sample of water was found to contain the following species, on analysis: Ca2+ = 40 mg/l Mg2+ = 24 mg/l Na+ =   8.05 mg/l HCO3– = 18.3 mg/l SO42– = 55.68 mg/l Cl– =   6.74 mg/l Express the results in terms of salts present as their CaCO3 equivalents. Solution. Impurity CaCO3 equivalents 100 Ca2+ → 40 × = 100 ppm 40 100 Mg2+ → 24 × = 100 ppm 24 100 Na+ → 8.05 × = 17.5 ppm 23 × 2 100 HCO3– → 183 × 61× 2 = 150 ppm 100 SO42– → 55.68 × = 58 ppm 96 100 Cl– → 6.74 × = 9.5 ppm 35.45 × 2

5 ml of N/50 H2SO4 ≡ 50 ×

Hence, Total alkalinity = 150 ppm Calcium alkalinity = 100 ppm Magnesium alkalinity =   50 ppm Total hardness = 200 ppm Calcium hardness = 100 ppm Magnesium hardness = 100 ppm Calcium temporary hardness = 100 ppm   (or calcium alkalinity) Magnesium temporary hardness =   50 ppm   (or magnesium alkalinity) Magnesium permament hardness =   50 ppm   (or magnesium non-carbonate hardness) Hence the salts presents in terms of their CaCO3 equivalent may be expressed as follows: Ca(HCO3)2 — 100 ppm Mg(HCO3)2 — 50 ppm MgSO4 — 50 ppm

263

BASIC ENGINEERING CHEMISTRY

264

Na2SO4 — 8 ppm NaCl — 9.5 ppm. Example 7. 50 ml of a sample of water required 5 ml of N/50 H2SO4 using methyl orange as indicator but did not give any colouration with phenolphthalein. What type of alkalinity is present? Express the same in ppm. Solution. As the water sample does not give any colouration with phenolphthalein (P = 0) hence only HCO3– ions are present. Now, 50 ml of water sample upto methyl orange end point = 5 ml of N/50 H2SO4 \ 50 mL × NM = 5 mL × N/50 Normality, N 1 1 × = N NM = 5mL × 50 50mL 500 Now, strength of alkalinity upto methyl orange end point (in terms of CaCO3 equivalents) = (NM × 50) g/L 1 × 50g/L × 1000 mg/L M = 500 M = 100 mg/L = 100 ppm Hence, alkalinity due to HCO3– = M = 100 ppm. Example 7. 500 ml of a water sample, on titration with N/50 H2SO4 gave a titre value of 20 ml to phenolphthalein end point and another 500 mL sample on titration with same acid gave a titre value of 58 ml to methyl orange end point. Calculate the alkalinity of water sample in terms of CaCO3 and comment on the type of alkalinity present. Solution. 500 ml of water upto phenolphthalein end point N H 2SO 4 = 29 mL of 50 N

500 mL × NP = 29 mL of 50 29 mL 1 × N or Normality, NP = 500 mL 50 Now, strength of alkalinity upto phenolphthalein end point in terms of CaCO3 equivalent = NP × 50 × 1000 ppm = P = 58 ppm ...(i) Give, 500 ml of water upto methyl orange end point N H 2SO 4 = 58 mL of 50 \

\ Normality,

N 50 58 mL 1 × N NM = 500 mL 50

50 mL × NM = 58 mL ×

Now, strength (in terms of CaCO3 equivalents) = M = NM × 50 × 1000 ppm = M 116 ppm From equation (i) and (ii) 1 P = M 2

...(ii)

265

B– WATER ANALYSIS TECHNIQUES CO32–

CO32–

Hence only ions are present and alkalinity of water sample is due to = 2P = M = 116 ppm. Example 9. Acid solution Elucidate on the type and extent of alkalinity present in the water. Solution. 200 ml of water upto phenolphthalein end point = 9.4 ml of N/50 HCl \ 200 ml × NP = 9.4 ml × N 50 9.4 ml 1 × N or Normality, NP = 200 ml 50 Now, strength of alkalinity upto phenolphthalein end point in terms CaCO3 equivalent = P = NP × 50 × 1000 ppm P = 47 ppm As, 200 ml of water upto methyl orange end point N = 9.4 + 21 = 30.4 ml of HCl 50 30.4 ml 1 \ 200 ml × NM = × N 200 ml 50 30.4 ml 1 × N or N M = 200 ml 50 Hence, strength of alkalinity upto methyl orange end-point in of CaCO3 equivalent hardness M = NM × 50 × 1000 M = 152 ppm 1 Since P < M , Hence CO32– and HCO3– ion re present in terms the given sample of water. 2 The alkalinity due to CO32– = 2P = 2 × 47 = 94 ppm and the alkalinity due to HCO3– = M – 2P = 152 –94 = 58 ppm Hence, the given water sample contains CO32– alkalinity = 94 ppm and HCO3– alkalinity = 58 ppm. (B) NUMERICAL BASED ON HARDNESS Example 1. 500 ml of a sample of water on EDTA titration with EBT as indicator consumed 18 ml of 0.045 M EDTA till end point is reached. Calculate the hardness of water. Solution. 1 ml of 0.01M EDTA = 1 mg of CaCO3 1 × 18 × 0.45 18 ml of 0.05M EDTA = .01 = 81 mg of CaCO3 500 ml of water contains 81 mg of CaCO3, so hardness present in 1 litre 81 × 1000 mg/L = 500 = 162 mg/L = 162 ppm.

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Example 2. One gram of CaCO3 was dissolved in dil HCl and the solution diluted to one litre 100 ml of this solution required 90 ml of EDTA solution, while 100 ml of the sample water required 40 ml of EDTA. On the other hand, 100 ml of the boiled sample water when titrated against EDTA consumed 20 ml of solution. Calculate each type of hardness in ppm. Solution. Strength of standard hard water = 1 mg/ml = 1g/L Strength of EDTA 90 ml of EDTA is required for = 100 ml of standard hard water = 100 mg of CaCO3 100 mg of CaCO3 \ 1 ml of EDTA is required for = 90 Total hardness 100 ml of sample required = 40 ml of EDTA 10 mg of CaCO3 Since 1 ml of EDTA is required for = 9 10 × 40mg of CaCO3 \ 40 ml of EDTA is required for = 9 Hence 100 ml of sample water contains 400/9 mg of CaCO3 400 1000 mg of CaCO3 = 444.4 mg/L × 1000 ml of sample water contains = 9 100 Total hardness = 444.4 ppm. Temporary hardness After boiling 100 ml of sample, water requires = 20 ml of EDTA 10 100 ml of sample water contains = 20 × 9 200 mg of CaCO3 = 9 200 1000 × mg of CaCO3 1000 ml of sample water contains = 9 100 = 222.2 Permanent hardness = 222.2 mg/L Temporary hardness = 444.4 – 222.2 = 222.2 ppm. Example 3. 100 ml of a sample of water required 20 ml of 0.01M EDTA for titration using eriochrome black T as indicator. After boiling 100 ml of same sample, required 10 ml of 0.01 M EDTA. Calculate (i) the total hardness (ii) permanent hardness (iii) temporary hardness Solution. Since EDTA forms 1 : 1 complex with the metal ions hence 1 meter solution of ETDA reacts with 1 mole of CaCO3 100 1 ml of 1M EDTA reacts with = × 1000 mg of CaCO3 1000 100 = 1 mg of CaCO3 1 ml of 0.1M EDTA reacts with = 100

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Strength of the given EDTA solution = 1mg/mL Total hardness Since 1ml of EDTA is required for = 1mg of CaCO3 20ml of EDTA is required for = 20mg of CaCO3 Hence 100ml of sample hard water contains = 20mg of CaCO3 20 × 1000 1000 ml of sample hard water contains = 100 = 200mg of CaCO3 Total hardness = 200mg/L = 200 ppm. Permanent hardness 10 ml of EDTA is required for = 1 × 10mg of CaCO3 Hence 100ml of boiled sample hard water contains = 10 mg of CaCO3 1000ml of boiled sample hard water contains = 100mg of CaCO3 \ Permanent hardness = 100mg/L = 100 ppm. Temporary hardness Total hardness – permanent hardness = 200 – 100 = 100 ppm. Example 4. A standard hard water contains 1000mg of CaCO3 per litre 50ml of this required 50ml of EDTA solution, 50ml of sample water required, 40 ml of EDTA solution. The sample after boiling required 20 ml EDTA solution. Calculate the temporary and permanent hardness of the given sample of water, in different units. Solution. Standardization of EDTA solution Given 1L of standard hard water contains 1000 mg CaCO3 \ 1ml of standard hard water contains 1 mg CaCO3 Now 50 ml of EDTA = 50 ml of standard water = 50 × 1 = 50 mg of CaCO3 50 1 ml of EDTA = = 1 mg of CaCO3 equivalent hardness 50 Determination of total hardness of water 50 ml of sample water = 40 ml of EDTA = 40 × 1 = 40 mg of CaCO3 equivalent hardness 40 × 1000 1L of sample water = \ 50 = 800 mg of CaCO3 equivalent hardness Hence, total hardness of water = 800 ppm. Determination of permanent hardness 50 ml of boiled water = 20 ml of EDTA = 20 × 1 = 20 mg of CaCO3 equivalent hardness 20 × 1000 \ 1L of boiled water = 50 = 400 mg of CaCO3 equivalent hardness Hence, permanent hardness of water = 400 ppm

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Determination of temporary hardness Temporary hardness = Total hardness – permanent hardness = 800 – 400 = 400 ppm. Temporary hardness 400ppm = 400 × 0.1 = 40 °Fr = 400 × 0.07 = 28 °Cl. Example 5. A standard hard water contains 15g of CaCO3 per litre. 20 ml of this required 25ml of EDTA solution, 100 ml of sample water required 18ml of EDTA solution. The sample after boiling required 12ml EDTA solution. Calculate the temporary hardness of the given sample of water, in terms of ppm. Solution. Standardization of EDTA solution Given 1L of standard hard water contains 15 gm CaCO3 \ 1ml of standard hard water contains 15 mg CaCO3 Now 25ml of EDTA = 20 ml of standard hard water = 20 × 15 = 300 mg of CaCO3 1ml of EDTA = 300/25 = 12mg of CaCO3 equivalent hardness Determination of total hardness of water 100ml of sample water = 18ml of EDTA = 18 × 12 = 216 mg of CaCO3 equivalent hardness \ 1L of sample water = 2160 mg of CaCO3 equivalent hardness Hence, total hardness of water = 2160 ppm Determination of permanent hardness 100 ml of boiled water = 12 ml of EDTA 12 × 12 = 144 mg of CaCO3 equivalent hardness \ 1L of boiled water = 1440 mg of CaCO3 equivalent hardness Hence, permanent hardness of water = 1440ppm Determination of temporary hardness Temporary hardness = Total hardness – permanent hardness = 2160 – 1440 = 720ppm \ Temporary hardness = 720ppm. Example 6. 50ml of a standard hard water containing 1.5mg of pure CaCO3 per ml consumed 44 ml of EDTA. 40ml of a water sample consumed 20ml of same EDTA solution using EBT indicator. Calculate the total hardness of water sample in ppm, °Cl and °Fr. 44ml of EDTA = 50ml × 1.5 mg/ml of CaCO3 eq. Solution. 1.5 mg of CaCO3 eq. 1ml of EDTA = 50 × 44 = 75/44 mg CaCO3 eq. Now, 40ml of hard water ≡ 20ml EDTA solution 75   ≡  20 × 44  mg CaCO3 eq.   20 × 75 × 1000 mg CaCO3 eq. \ 1L (or 1000ml) of hard water = 44 × 40 = 833.3 mg CaCO3 eq.

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Hence, total hardness of water = 833.3 ppm = 833.3 × 0.1 = 83.33 °Fr = 83.33 × 0.7 = 58.33 °Cl. Numerical based on COD, BOD etc. Example 1. 100 ml of waste water containing 920 ppm dissolved oxygen was diluted with 100 ml of distilled water and kept in a bottle at 20 °C for 5 days. The oxygen content of the resulting water was then found to be 260 ppm. Calculate the BOD of the sample. BOD = [Original – Final] dissolved O2 content × Dilution factor Solution. = (920 – 260) ppm × (200 ml/100 ml) = 1,320 ppm. Example 2. In a COD experiment, 30cm3 of an effluent sample required 9.8 cm3 of 0.001M K2Cr2O7 solution for oxidation. Calculate the COD of the sample. Solution. 1,000 cm3 (or 1L) of 0.001M K2Cr2O7 = 6 × 8 × 0.001 g of oxygen 6 × 8 × 0.001 × 9.8 \ 9.8 cm3 (or ml) of 0.001M K2Cr2O7 = g of oxygen 1000 or 30 cm2 of effluent contain = 4.704 × 10–4 = 0.4704 of oxygen 0.4704 mg × 1000 or 1,000 cm3 (or 1L) of effluent contain = 30 = 15.69 mg of oxygen Hence, COD of effluent sampl = 15.69 mg/L or ppm. Example 3. 25ml of sample for COD analysis was reacted with 15ml of 0.25N K2Cr2O7 solution and after the reaction, the unreacted K2Cr2O7 required 18ml of 0.1N FAS for reaction, under identical conditions 15ml of dichromate solution mixed with 25ml of distilled water required 30ml of 0.1N FAS. What is the COD of the sample. Solution. Difference in volume of FAS required in the blank (distilled water) and the sample = (30 – 18)ml = 12ml Now, 1000 ml of 1N FAS contains = 8g oxygen 8 × 12 × 0.1 g oxygen or 12ml of 0.1N FAS contains = 1000 8 × 12 × 0.1 g oxygen \ 25 mL of water sample contains = 1000 8 × 12 × 0.1 × 1000 g oxygen or 1,000 ml of water sample contains = 1000 × 25 = 0.384 g or 384 mg oxygen Hence, COD of sample = 384mg/L or 384 ppm. Example 4. A 100ml of sample contains 920 ppm of dissolved oxygen. After 5 days the dissolved oxygen value becomes 300 ppm after the sample has been diluted to 100 ml. Calculate the BOD of the sample. BOD = (DOb – DOi) × dilution factor Solution. ml of sample after dilution = (DOb – DOi) × ml of sample before dilution 100 = 620 ppm. = (920 − 300) × 100

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Example 5. A 50ml of sewage water sample was refluxed with 10ml of 0.25N K2Cr2O7 solution in presence of dil H2SO4, Ag2SO4 and HgSO4. The unreacted dichromate required 7.5ml of 0.1N ferrous ammonium sulphate, 10ml of the same K2Cr2O7 solution and 25ml of distilled water under the same conditions as the sample required 29ml of 0.1N FAS calculate the COD of the sewage water sample. (V − Vt ) × M × 8 × 1000 mg/L COD = b Solution. Ve where Vb = 29 ml Vt = 7.5 ml M = 0.1 Ve = 50 ml (29 − 7.5) × 0.1 × 8 × 1000 COD = 50 21.5 × 0.1 × 8 × 1000 = 50 17200 = 50 = 344ppm.

QUESTIONS

1. How does dissolved oxygen affect the quality of water used in boilers ? What are the vari-

ous methods employed in deaeration of water ? (RGPV Bhopal 2006) 2. Give the principle involved in the estimation of hardness by EDTA method. 3. Calculate the hardness of a water sample, whose 20 ml required 30 ml of EDTA. 10 ml of calcium chloride solution, whose strength is equivalent to 300 mg of calcium carbonate per 200 ml, required 20 ml of EDTA solution. (Mumbai University, 1998) 4. Describe briefly the methods used for the analysis of the following parameters in water or wartewater and discuss their significance (a) Alkalinity (b) Hardness (c) DO (d) BOD (e) COD 5. How are the following determined in a water sample ? (a) Chlorides (b) Sulfates (c) Dissolved CO2 (d) Residual Chlorine (e) TDS 6. ������������������������������������������������������������������������������������������� Evluate the type and extent of alkalinity due to various anions in the presence of phenolpthalein and methyl orange indicator in a titration (RGPV Bhopal 2009) 7. 0.5 gm of CaCO3 was dissolved in dilute HCl and diluted to 500 ml. 50 ml of this solution required 48 ml of EDTA solution. For titration 50 ml of given hard water sample require 15 ml of EDTA solution for titration. Calculate total hardness of water sample in ppm. (RGPV Bhopal 2006)

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–.

8. A sample of water contains 32 mg/l. of CO3 and 56 mg./L of HCO3 If the pH of the water sample is 10.0, what is the alkalinity of the water sample in mg/L as CaCO3. (Ans : 104.2 mg/L as CaCO3) 9. (a) How is the harness of water determined by the complexometric method using EDTA ? (b) 50 ml of a water sample when directly titrated with 0.01 m EDTA solution using the standard procedure gave 21 ml titre value at the end-point. Calculate the total hardness of the water sample in PPm as CaCO3. (Ans : 420 PPm) 10. (a) How is the alkalinity of a water sample determined ? (b) 100 ml of a water sample on titration with N/50 H Cl required 8.0 ml of the acid to phenolphthalein end-point and 9.0 ml of the acid to methyl orange end-point. Calculate the type and extent of alkalinity present in the water sample. (Ans. OH– alkalinity = 70 PPm as CaCO3 CO32–- alkalinity = 20 PPm as CaCO3). 11. Discuss briefly the principles involved in the determination of DO, BOD and COD in waste water and their significance from the point of view of environmental pollution. 6. How are the following parameters determined in a water sample ? (a) Chloride, Sulphate and TDS. (b) Briefly discuss their significance

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QUESTION BANK Water And Its Industrial Application 1. (a) Explain the significance of the following types of chlorination : (i) Break Point chlorination (ii) super chlorination (b) Distinguish between disinfection and sterilization of water. 2. (a) How can you determine the residual chlorine in a water sample ? (b) Discuss the importance of the following processes with reference to the treatment of water for domestic and industrial purposes : (a) sedimentation (b) coagulation (c) filtration (d) sterilization 3. Compare the lime-soda process and zeolite process of water softening with respect to the principles involved, merits and demerits. 4. (a) Discuss the ion-exchange process of water softening. (b) Why does the mixed bed ion exchange resin column provide very efficient demineralisation of water ? 5. State some important specifications of the water used for the following industries. (a) Thermal power generation    (b) Paper    (c) Textile    (d) Dairy 6. (a) State the reasons as to why water used in boilers should be softened. (b) Discuss the ion-exchange method of softening water. 7. (a) Explain the following : (i) Priming (ii) Foaming (iii) Blow down (iv) Units of hardness (b) What are the factors responsible for caustic embrittlement in boilers ? How can this be prevented ? 8. (a) What are boiler troubles ? (b) Why are they caused ? (c) How can they be prevented ? 9. (a) Why are scales formed in boilers ? What are they made up of ? Why are they undesirable ? (b) How can the scale formation in boilers be minimised ? 10. Discuss the various methods used for internal treatment of boiler feed water. 11. (a) Explain the zeolite process of water softening ? (b) An exhausted zeolite softener was regenerated by passing 200 liters of NaCl solution containing 0.2 g per liter of NaCl. How many liters of hard water sample having hardness of 500 ppm can be softened using this softener ? (Ans : 68,376 lit) 12. A water sample gave the following analytical results : Ca (HCO3)2 – 162 ppm, Mg (HCO3)2 – 7.3 ppm, CaCl2 – 111 ppm, MgCl2 - 9.5 ppm, Mg SO4 - 60 ppm, NaCl - 60 ppm, CO2 - 44 ppm, HCl - 36.5 ppm. Calculate the quantities of lime (80% pure) and soda (85 % pure required for softening 1012 litres of such a water sample. (Ans: 684.5 kgs lime ; 648..5 kg soda) 13. Write informative notes on any two of the following : (a) Phosphate conditioning (b) Boilder corrosion (c) Internal treatment (d) Sludge blanket type of lime - soda water softener (e) Temporary and permanent hardness (f) Colloidal conditioning. 272

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14. What are the requirements of feed water quality for modern high pressure boilers ? How can they be achieved ?

15. Write short notes on the following: (a) Caustic embrittlement (RGPV Bhopal 2009) (b) Boiler corrosion (RGPV Bhopal 2006) (c) Internal treatment (d) Blow-down (e) Break-point chlorination (RGPV Bhopal 2006) (f) Impurities in water and their effects (g) Slow sand filtratio (h) Rapid pressure filtratio (i) Scale and sludge formation (RGPV Bhopal 2009) (j) Priming and foaming. 16. (a) Write the constituents responsible for the permanent hardness of water. Discuss one treatment method. (b) Why does hard water consume a lot of soap ? (c) Why does magnesium bicarbonate required double amount of lime for softening. (U.P. Technical University 2001) 17. (a) Describe the causes, harmful effects and control of scale and sludge formation in boilers. (RGPV Bhopal 2009) (b) A water sample using FeSO4.7H2O as coagulant at the rate of 278 ppm. gave the following results on analysis : Ca++ – 80 ppm, Mg++ – 48 ppm, CO2 – 88 ppm; and HCO3– – 244 ppm Calculate lime soda required for softening one million litres of the water sample. or (a) What are zeolites ? Discuss the chemistry involved in zeolite process of softening hard waer. Also mention the limitations, advantages and disadvantages of this process. (b) A zeolite softener was completely exhausted and was regenerated by passing 120 litres of sodium chloride solution containing 150 g/litre of NaCl. How many litres of a sample of water of hardness 500 mg/liter can be softened by the softener before regenerating it again ? (RGPV Bhopal 2001) 18. (a) The analytical results of raw water and treated water are as follows : Ca2+ = 300 ppm Mg2+ = 150 ppm HCO3– = 244 ppm OH– = 65 ppm CO32– = 40 ppm CO2 = 60 ppm Calculate, (i) The amount of lime (80% pure) and soda (90% pure) required to soften one million litres of water using NaAlO2 as a coagulant at the rate of 41 mg/litre. (ii) If 10,000 litres of the same water sample is softened through a zeolite softener, how much NaCl will be required for its regeneration ? (Nagpur University, S-2001)

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19. (a) What is meant by hardness of water ? Name the substances that cause permanent hardness in water? (b) Explain the importance of “phosphate conditioning ? in the “internal treatment” of water. (c) What is “demineralisation process” ? Point out its advantages and limitations ? (Mumbai University, 1994) Fuels and Combustion 1. What do you mean by “proximate analysis” and “Ultimate analysis”? What are the parameters determined under them and what is their significance in evaluation of the coal quality ? 2. (a) What is meant by gross calorific value and net calorific value ? (b) How is the gross calorific value of a solid fuel determined by Bomb calorimeter ? 3. (a) Calculate the approximate calorific value of a coal sample using Dulong’s formula, given the following ultimate analysis of the sample : C-80%, H-3.5%, S-2.8%, N-1.10%, O - 5.0%, ash - 7.7%. (b) The following data is obtained in a Bomb calorimeter experiment for determination of calorific value of a coal sample : Weight of water taken in the calorimeter = 2500 g. Observed rise in temperature = 2.42°C Weight of coal sample taken for the experiment = 0.9 g. Water equivalent of the calorimeter = 500 g Fuse wire correction = 10 calories Cooling correction = 0.052 calories. Calculate the gross calorific value of the fuel. Also calculate the net calorific value if the fuel contains 6.5 % H and given that the latent heat of steam is 587 cals/g. (Ans. 8228 cals/g; 7884.6 cals/g) 4. (a) What is mant by carbonisation of coal and how is it performed ? (b) Describe the Otto - Hoffman’s process for manufacturing coke ? What are its advantages over the earlier processes used for this purpose ? 5. (a) What is meant by knocking and why does it take place in internal combustion engines ? (b) What is meant by octane number and cetane number and what is their significance ? 6. (a) What are the reasons for knocking in petrol engines and diesel engines? How can knocking be avoided ? (b) How knocking in petrol and diesel engines be related to the chemical structure of the respective fuel constituents ? 7. (a) How do you explain knocking in diesel engines ? (b) How can it be controlled ? (c) What is the significance of cetane number ? (d) How is knocking related to the chemical structure of the diesel fuel constituents ? 8. The analysis of coal is boiler trial was C-81%, H - 4.5%, O-8%, uncombustible matter - 6.5 % The Orsat analysis of the dry flue gas was CO2 - 8.3%, Co- 1.4%, O - 10% and N (by difference) - 80.3%. Calculate : (a) the weight of air supplied per kg of coal, (b) the percentage of excess air used. (Ans : 20.15 kg; 89.8 %)

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9. A producer gas has the following volumetric composition : CO - 30%, N2 - 52%, CH4 – 2%, H2 - 12% and CO2 - 4%. If 100 m3 of this gas was burnt using 58% excess air, what will be the percentage composition of the dry flue gases ? 10. A coal sample gave the following composition by weight : C - 74%, H - 5%, N-1.5%, O-6.5%, S-4%, ash - 4%, moisture - 5%. When this coal sample was burnt as a fuel in a boiler, the flue gas obtained has the following volumetric composition on dry basis : CO - 2%, CO2 - 9%, SO2 - 4%, O2-7%, and N2 - 78%. Calculate : (a) Wt. of the flue gases (including water vapour) formed by burning 1 kg of the coal (b) The % of excess air used, and (c) The % of carbon converted into CO. 11. A coal sample gave the following analysis : C - 66.2%, H-4.2%, O-6.1%, N-1.4%, S-2.9%, Moistrue - 9.7% and Ash - 9.5% Determine the quantities of products of combusion per kg of the coal is 25% excess air was used for combastion. 12. A hydrocarbon fuel on burning gave a flue gas having the following volumetric analysis : O2 - 3.56%, CO2 - 13.53% and N2 - 82.91 %, Determine : (a) The composition of the fuel by weight (b) The % of excess air used (c) Volume of air supplied per kg of fuel (Ans: (a) 89.13% C and 10.87% H; (b) 19.3%; (c) 12.7 m3) 13. Describe the Otto-Hoffman’s process for preparing coke ? What are its advantages over the earlier methods ? (RGPV Bhopal 2006) 14. What is cracking and what for it is used ? What are the types of cracking ? Describe the working of fixe bed catalytic cracking. (RGPV Bhopal 2009) 15. Complete the following sentences with appropriate answers : (a) Peat is not considered as an economic fuel because_________. (b) Lignite is susceptible for spontaneous combustion because_________. (c) _________are the most widely used coals in the world. (d) Anthracite is used for drying malt and hops because_________. (e) Coking coals are blended with non-caking coals because_________. (f) Low temperature carbonization yields coke which is suitable for_________. (g) High temperature carbonization yields coke which is_________. (h) Highest ranking coal is _________. 16. Establish the relationship between knocking in I.C. engines and the nature and molecular structure of the constituents in petrol and diesel fuel. 17. Write short notes: (a) Dulong’s formula (b) Coking coal and caking coal. Lubricants 1. (a) What are the functions of a lubricant ? (b) What are the important properties that an efficient lubricant should possess ? 2. How are lubricants classified ? Give examples of each type. 3. (a) Discuss the various mechanisms of lubrication ? (b) What type of lubricants are used in situations wher such mechanisms are in operation ?

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4. (a) What are greases ? Under what service conditions they are used ? (b) What are cup greases ? How are they manufactured ? (c) What are the important tests for evaluating greases and what is their significance ? 5. Write informative notes on any two of the following : (a) Extreme pressure lubrication. (b) Blended or compounded oils (c) Lubricating emulsions (d) Synthetic lubricants 6. (a) What are the characteristics of a good solid lubricant ? (b) How do the following act as lubricants ? (i) Graphite (ii) Molybdenum disulphide 7. Discuss the various additives used for lubricants and their functions. 8. What are the various factors that determine the choice of a lubricant for a given application ? 9. (a) What do you understand by viscosity index and why it is an important property for lubricant ? (b) A lubricating oil has the same viscosity as standard naphthenic and paraffinic type of oils at 210° F. Their viscosities at 100° F are 320 S U respectively. Find the viscosity index of the lubricating oil. 10. Write the significance of the following in evgluating a libricating oil : (a) Flash point (b) Pour point (c) Aniline pint (d) Oiliness 11. Write short note: (a) saponification value (b) Acid value (c) Iodine value (d) Carbon residue test 12. Describe with their significanc the following : (i) Aniline Point (ii) Cloud point and pour point (RGPV Bhopal 2009) (iii) Steam emulsificatio number (Nagpur University, 2001) (iv) Important function of lubricant. 13. (a) What are flas point and fir point of a lubricant. Point out their significance (b) Explain the following properties of lubricants : (i) Viscosity Index (ii) Aniline Point (iii) Neutralisation Number (c) Write short note on Redwood Viscometer 14. (a) Defin : (i) Acid Value (ii) Saponificatio (iii) Pour Point (b) What are different mechanisms of lubrication ? Explain the boundary lubrication. (RGPV Bhopal 2009) (c) Write a note on determination of flas point by Pensky Martin Method. What is the significanc of flas point. (RGPV Bhopal 2009) Cement and Refractories 1. Describe with the help of a neat diagram the various steps involved in the manufacture of portland cement.

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2. (a) What are the various reactions taking place in the rotary kiln used for the manufacture of portland cement. (b) Discuss the properties of the major constitutional compounds present in the portland cement. 3. Explain the following : (a) Reactions taking place during the setting and hardening of portland cement paste. (b) Microscopic constituents in portland cement and their significance. 4. What are the various additives and admixtures used in portland cement and what properties are conferred by them on the cement. 5. Discuss the different types of cement and its derivatives 6. Write informative notes on any two of the following : (a) High-alumina cement (b) Sorel cement (c) Strontium and Barium cements (d) Pozzolana cement 7. Explain the following : (a) Soundness of cement (b) Rapid hardening cement (c) Curing 8. Write short notes on any three of the following : (a) Composition of cement (b) Significance of ISI specifications for parland cement (c) Setting and hardening of lime mortar (d) Heat of hydration of cement 9. (a) What are refractories ? (b) How are they classified ? (c) What are the requisites of a good refractory material ? (d) Give some important industrial uses of refractories. 10. Discuss the industrial preparation, properties and uses of any two of the following : (a) Fire-clay refractories (b) Silica refractories (c) Magnesia refractories (d) Chromite bricks (e) Dolomite refractories 11. (a) What are the raw materials used for manufacture of refractories ? (b) What are the different steps involved in the manufacture of refractories ? (c) What are super refractories ? 12. Discuss the significance of the following properties of refractory bricks : (a) Refractoriness or sugar cones (b) Dimensional stability (c) Spalling (d) Thermal conductivity (e) R.U.L. Polymers 1. What are polymers? How are they classified ? Give examples of each type. 2. Mention any three important high polymers and their important uses. 3. How are polymerization reactions classified on the basis of their mechanism ? Discuss with examples. 4. What are addition and condensation polymerization processes ? Give examples.

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5. (a) Discuss the mechanism of chain polymerization. (b) Degree of polymerization (c) Homopolymer (d) Copolymer. 6. Describe the major manufacturing processes employed for polymerization. Give examples. 7. What are the basic differences between thermoplastics and thermosetting plastics ? Give examples. 8. What is Bakelite ? How is it manufactured ? State some of its important applications. 9. What are nylons ? How is nylon : 6 : 6 manufactured ? Give flow sheet, state some important uses. 10. How is terylene prepared ? State some important uses. 11. What are silicones ? How are they manufactured ? Give flow sheet diagram ? State some important applications of some important silicone products. 12. Discuss the preparation, properties and uses of any three of the following : (a) PVC (b) Teflon (c) High density polyethylene (d) Polyacrylates (e) Glyptals (f) Urea formaldehyde (g) Terylene (h) Polyurethanes (i) Thiokols (j) Neoprene 13. Discuss the various stages of processing of rubber ? Give some important uses of rubbers. 14. What are elastomers ? What type of structural features are necessary for a polymer to behave like an elastomer ? Give some important applications of elastomers. 15. Mention 3 important types of synthetic rubbers and discuss their synthesis, properties and applications. How do synthetic rubbers compare with natural rubbers with respect to their properties and service conditions ? 16. Write informative notes on any two of the following : (a) Vulcanization of rubber (b) Compounding of rubber (c) Reclamation of rubber (d) Fabrication of plastics 17. Describe the preparation, properties and uses of any three of the following : (a) Cellulose acetate (b) Phenol-formaldehyde resins (c) Polyethylene (d) Polystyrene (e) Melamine formaldehyde resins (f) Decaron (g) Silicone resins (h) Butyl rubber (i) Buna-s (j) Nitrile rubbers (k) Orlan (l) PMMA or lucite 18. Discuss the preparation properties and uses of various polymers related to synthetic rubber. 19. Draw the flow sheet for Nylon 6:6 and decron. Instrumental Techniques in Chemical Analysis 1. State Beer - Lamberts law and discuss its applications and limitations. 2. Describe the essential parts of a photoelectric photometer and explain its working. 3. What are the essential parts of UV - visible spectrophotometer ? What are its advantages over a simple colorimeter ?

QUESTION BANK

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4. Discuss the origin of UV Spectra. Give some important applications of UV - spectroscopy in analytical chemistry. 5. How do IR-spectra originate ? What is the instrumentation used for IR-spectroscopy ? State some important applications. 6. What are the essential components of a gas chromatographic assembly? State 5 important applications of gas chromatography. 7. Discuss the important principles, essential instrumentation and applications of NMR spectroscopy. 8. Write informative notes on the following: (a) Colorimetry (b) NMR Spectroscopy 9. Explain briefly : (a) Beer - Lamberts law (b) Applications of IR spectroscopy (c) Various techniques of chromatography (d) Applications of UV spectroscopy. (RGPV Bhopal 2003) Water Analysis Techniques 1. A sample of water contains 32 mg/l. of CO32– and 56 mg./L of HCO3–. If the pH of the water sample is 10.0, what is the alkalinity of the water sample in mg/L as CaCO3. (Ans : 104.2 mg/L as CaCO3) – + 2– – [Hint : Alkalinity = [CO3 ] + [HCO3 ] + [OH ] – [H ], all expressed in mg/1 as CaCO3) Alkalinity = [53.3 + 45.9 + 5.0 – (5.0 × 10–6)] = 104.2 mg/L as CaCO3 2. (a) If the total hardness in a water sample is greater than total alkalinity, what type of hardness does it contain ? (b) A water sample having a PH of 7.2 gives the following analysis : Impurity mg/l as CaCO3 Ca2+ – 80 Cl – 100 Na+ – 72 K+ – 6 – HCO3 – 165 SO42– – 201 Mg2+ – 30 Calculate the carbonate hardness, non-carbonate hardness, alkalinity (in mg/1 as CaCO3) and total dissolved solids (in mg/1) (Ans: CH - 135.2 mg/L, NCH - 187.8 mg/L, Alkalinity - 135.2 mg/L, all as CaCO3. TDS - 654 mg/L) [Hint : For nearly neutral water (PH around 6 to 8), the concentrations of H+ and OH– are insignificant and in such cases, alkalinity is determined entirely by [CO32–] and HCO3–]. 3. (a) How is the hardness of water determined by the complexometric method using EDTA ? (b) 50 ml of a water sample when directly titrated with 0.01 m EDTA solution using the standard procedure gave 21 ml titre value at the end-point. Calculate the total hardness of the water sample in PPm as CaCO3. (Ans : 420 PPm) 4. (a) How is the alkalinity of a water sample determined ?

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(b) 100 ml of a water sample on titration with N/50 H Cl required 8.0 ml of the acid to phenolphthalein end-point and 9.0 ml of the acid to methyl orange end-point. Calculate the type and extent of alkalinity present in the water sample. (Ans : OH– alkalinity = 70 PPm as CaCO3 CO32– alkalinity = 20 PPm as CaCO3). 5. Discuss briefly the principles involved in the determination of DO, BOD and COD in waste water and their significance from the point of view of environmental pollution. 6. How are the following parameters determined in a water sample ? (a) Chloride, Sulphate and TDS. (b) Briefly discuss their significance 7. Write short notes: (i) BOD (ii) COD