WORKSHOP CALCULATION AND SCIENCE

NATIONAL INSTRUCTIONAL

MEDIA INSTITUTE, CHENNAI

WORKSHOP CALCULATION

AND SCIENCE

COMMON FOR ALL ENGINEERING TRADES

Post Box No. 3142, CTI Campus, Guindy, Chennai - 600 032

Semester

DIRECTORATE GENERAL OF TRAINING

MINISTRY OF SKILL DEVELOPMENT & ENTREPRENEURSHIP

GOVERNMENT OF INDIA

NSQF

Workshop Calculation & Science (NSQF) - 1st Semester

Common for All Engineering Trades

First Edition

First Reprint : January 2019 Copies : 10,000

: December 2018 Copies : 10,000

Rs. 130/-

No part of this publication can be reproduced or transmitted in any form or by any means, electronic or mechanical,

including photocopy, recording or any information storage and retrieval system, without permission in writing from the

National Instructional Media Institute, Chennai.

Published by:

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Guindy, Chennai - 600 032.

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email : [email protected]

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Website: www.nimi.gov.in

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(iii)

FOREWORD

The Government of India has set an ambitious target of imparting skills to 30 crores people, one out of every

four Indians, by 2020 to help them secure jobs as part of the National Skills Development Policy. Industrial

Training Institutes (ITIs) play a vital role in this process especially in terms of providing skilled manpower.

Keeping this in mind, and for providing the current industry relevant skill training to Trainees, ITI syllabus

has been recently updated with the help of comprising various stakeholder's viz. Industries, Entrepreneurs,

Academicians and representatives from ITIs.

The National Instructional Media Institute (NIMI), Chennai, has now come up with instructional material to

suit the revised curriculum for Workshop Calculation & Science 1

Semester NSQF Common for all

engineering trades will help the trainees to get an international equivalency standard where their skill

proficiency and competency will be duly recognized across the globe and this will also increase the scope

of recognition of prior learning. NSQF trainees will also get the opportunities to promote life long learning

and skill development. I have no doubt that with NSQF the trainers and trainees of ITIs, and all stakeholders

will derive maximum benefits from these IMPs and that NIMI's effort will go a long way in improving the

quality of Vocational training in the country.

The Executive Director & Staff of NIMI and members of Media Development Committee deserve appreciation

for their contribution in bringing out this publication.

Jai Hind

RAJESH AGGARWAL

Director General/ Addl. Secretary

Ministry of Skill Development & Entrepreneurship,

Government of India.

New Delhi - 110 001

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PREFACE

The National Instructional Media Institute(NIMI) was set up at Chennai, by the Directorate General of Training,

Ministry of skill Development and Entrepreneurship, Government of India, with the technical assistance

from the Govt of the Federal Republic of Germany with the prime objective of developing and disseminating

instructional Material for various trades as per prescribed syllabus and Craftsman Training Programme(CTS)

under NSQF levels.

The Instructional materials are developed and produced in the form of Instructional Media Packages (IMPs),

consisting of Trade Theory, Trade Practical, Test and Assignment Book, Instructor Guide, Wall charts,

Transparencies and other supportive materials. The above material will enable to achieve overall improvement

in the standard of training in ITIs.

A national multi-skill programme called SKILL INDIA, was launched by the Government of India, through a

Gazette Notification from the Ministry of Finance (Dept of Economic Affairs), Govt of India, dated 27th

December 2013, with a view to create opportunities, space and scope for the development of talents of

Indian Youth, and to develop those sectors under Skill Development.

The emphasis is to skill the Youth in such a manner to enable them to get employment and also improve

Entreprenurship by providing training, support and guidance for all occupation that were of traditional types.

The training programme would be in the lines of International level, so that youths of our Country can get

employed within the Country or Overseas employment. The National Skill Qualification Framework

(NSQF), anchored at the National Skill Development Agency(NSDA), is a Nationally Integrated Education

and competency-based framework, to organize all qualifications according to a series of levels of Knowledge,

Skill and Aptitude. Under NSQF the learner can acquire the Certification for Competency needed at any

level through formal, non-formal or informal learning.

The Workshop Calculation & Science 1

Semester (common to all Engineering Trades) is one of the

book developed as per the NSQF syllabus.

The Workshop Calculation & Science (common to all Engineering Trades as per NSQF) 1

Semester is

the outcome of the collective efforts of experts from Field Institutes of DGT, Champion ITI’s for each of the

Sectors, and also Media Development Committee (MDC) members and Staff of NIMI. NIMI wishes that the

above material will fulfill to satisfy the long needs of the trainees and instructors and shall help the trainees

for their Employability in Vocational Training.

NIMI would like to take this opportunity to convey sincere thanks to all the Members and Media Development

Committee (MDC) members.

R. P. DHINGRA

Chennai - 600 032 EXECUTIVE DIRECTOR

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ACKNOWLEDGEMENT

The National Instructional Media Institute (NIMI) sincerely acknowledge with thanks the co-operation and

contribution of the following Media Developers to bring this IMP for the course Workshop Calculation & Science

Semester as per NSQF.

MEDIA DEVELOPMENT COMMITTEE MEMBERS

Shri. M. Sangara pandian - Training Officer (Retd.)

CTI, Guindy, Chennai.

Shri. G. Sathiamoorthy - Jr.Training Officer (Retd.)

Govt I.T.I, DET - Tamilnadu.

Shri. K. Lakshminarayanan - Training Officer (Retd.)

Govt I.T.I, DET - Tamilnadu.

NIMI CO-ORDINATORS

Shri. K. Srinivasa Rao - Joint Director,

NIMI, Chennai - 32.

Shri. G. Michael Johny - Assistant Manager,

NIMI, Chennai - 32.

NIMI records its appreciation of the Data Entry, CAD, DTP Operators for their excellent and devoted services in

the process of development of this IMP.

NIMI also acknowledges with thanks, the efforts rendered by all other staff who have contributed for the develop-

ment of this book.

INTRODUCTION

The material has been divided into independent learning units, each consisting of a summary of the topic and an

assignment part. The summary explains in a clear and easily understandable fashion the essence of the mathematical

and scientific principles. This must not be treated as a replacment for the instructor’s explanatory information to be

imparted to the trainees in the classroom, which certainly will be more elaborate. The book should enable the

trainees in grasping the essentials from the elaboration made by the instructor and will help them to solve independently

the assignments of the respective chapters. It will also help them to solve the various problems, they may come

across on the shop floor while doing their practical exercises.

The assignments are presented through ‘Graphics’ to ensure communications amongst the trainees. It also assists

the trainees to determine the right approach to sove the problems. The required revelent data to solve the problems

are provided adjacent to the graphics either by means of symbols or by means of words. The description of the

symbols indicated in the problems has its reference in the relevant summaries.

At the end of the exercise wherever necessary assignments, problems are included for further practice.

Time allotment:

Duration of 1

Semester (26 weeks) : 52 Hrs

Effective weeks avaliable (22 weeks) : 44 Hrs

Revision and Test (4 weeks) : 8 Hrs

Total time allotment : 52 Hrs

Time allotment for each module has given below. Common to all Engineering Trades, Mechanic Refrigeration and

Air Conditioner and Sheet Metal Worker.

S.No Module Exercise No. Time allotment (Hrs)

1 System of units, factors and fraction 1.1.01 - 1.2.09 10 Hrs

2 Square root, Ratio and proportion, percentage 1.3.10 - 1.5.16 12 Hrs

3 Algebra (only for SMW in 1

Semester) 1.6.17 - 1.6.18 4 Hrs

4 Material science, Mass weight and density 1.7.19 - 1.8.26 14 Hrs

5 Speed and veloctiy, work, power and energy 1.9.27 - 1.10.30 8 Hrs

6 Heat and temperature (only for MR&AC and SMW

in 1

Semester) 1.11.31 - 1.12.37 6 Hrs

7 Basic electricity (only for MR&AC in 1

Semester) 1.13.38 - 1.13.41 9 Hrs

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Exercise No. Title of the Exercise Page No.

Module 1

1.1.01 Classification of system

of units 1

1.1.02 Fundamental and Derived units F.P.S, C.G.S, M.K.S and SI units 2

1.1.03 Measurement units and conversion 4

1.1.04 Conversions of length, mass, force,work, power and energy 12

1.2.05 Factors, HCF, LCM

and problems 14

1.2.06 Fractions 15

1.2.07 Decimal fractions 18

1.2.08 Pocket calculator and its applications 21

1.2.09 Solving problems by using calculator 25

Module 2

1.3.10 Square and square root 28

1.3.11 Simple problems using calculator 29

1.3.12 Applications of pythagoras theorem and related problems 30

1.4.13 Ratio and proportion 31

1.4.14 Direct and indirect proportions 33

1.5.15 Percentage 37

1.5.16 Changing percentage to decimal and fraction 40

Module 3 (only for SMW in 1

Semester)

1.6.17 Algebraic symbols and fundamentals 41

1.6.18 Addition,subtraction,multiplication and division of algebra 44

Module 4

1.7.19 Physical and mechanical properties of metals 47

1.7.20 Introduction of iron and cast iron 50

1.7.21 Types of ferrous and non ferrous 53

1.7.22 Difference between iron & steel, alloy steel & carbon steel 55

1.7.23 Properties and uses of rubber, timber and insulating materials 57

1.8.24 Mass, units of mass, density and weight 60

1.8.25 Difference between mass & weight, density & specific gravity 61

1.8.26 Related problems with assignment of mass, weight & density 62

Module 5

1.9.27 Rest, motion, speed, velocity, difference between speed & velocity,

acceleration & retardation 68

1.9.28 Related problems with assignment of speed & velocity 72

CONTENTS

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Exercise No. Title of the Exercise Page No.

LEARNING / ASSESSABLE OUTCOME

On completion of this book you shall be able to

• Understand, explain different mathematical calculation & science

in the field of study including basic electrical and apply in day to

day work.[Different mathematical calculation & science -Units,

factors and fractions, square root, ratio and proportion,

percentage, material science, mass, weight, density, speed and

velocity, work, power & energy, algebra, heat & temperature, basic

electricity, pressure].

• SMW - Algebra and Heat & temperature.

• MR&AC - Heat & temperature and Basic electricity.

1.10.29 Units of work, power and energy, horse power of engines and mechanical

efficiency 79

1.10.30 Potential energy, kinetic energy and related problems with assignment 78

Module 6 (only for MR&AC and SMW in 1

Semester)

1.11.31 Concept of heat and temperature, effects of heat, thermometer scale, celcius,

fahrenheit, reaumer, kelvin and difference between heat and temperature 81

1.11.32 Conversion between centigrade, fahrenheit, reaumer, kelvin scale of temperature 83

1.11.33 Temperature measuring instruments, types of thermometer, pyrometer and

transmission of heat 84

1.11.34 Co-efficient of linear expansion and related problems with assignment 86

1.11.35 Problems of heat loss and heat gain with assignment 88

1.11.36 Concept of pressure and its units in different system 92

1.12.37 Thermal conductivity and insulations 93

Module 7 (only for MR&AC in 1

Semester)

1.13.38 Introduction, use of electricity, molecule, atom, electricity is produced,

electric current, voltage, resistance and their units 95

1.13.39 Ohm’s law, relation between V.I.R & problems, series & parallel

circuits problem 97

1.13.40 Electrical power, energy and their units, calculation with assignment 100

1.13.41 Magnetic induction, self and mutual inductance and EMF generation 104

SYLLABUS

First Semester Common for All Engineering Trades Duration: Six Months

(Except MR&AC and SMW)

S.no. Title

Unit

Fractions

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Square Root

Square and Square Root, method of finding out square roots, Simple problem using calculator.

Ratio & Proportion

Simple calculation on related problems.

Introduction, Simple calculation. Changing percentage to decimal and fraction and vice-versa.

Percentage

Fractions, Decimal fraction, L.C.M., H.C.F., Multiplication and Division of Fractions and Decimals,

conversion of Fraction to Decimal and vice versa. Simple problems using Scientific Calculator.

Systems of unit- FPS, CGS, MKS/SI unit, unit of length, Mass and time, Conversion of units.

Material Science

Properties - Physical & Mechanical, Types - Ferrous & Non-Ferrous, difference between Ferrous and

Non-Ferrous metals, introduction of Iron, Cast Iron, Wrought Iron, Steel, difference between Iron and

Steel, Alloy steel, carbon steel, stainless steel, Non-Ferrous metals, Non-Ferrous Alloys.

Mass, Weight and Density

Mass, Unit of Mass, Weight, difference between mass and weight, Density, unit of density, specific

gravity of metals.

Speed and Velocity

Rest and motion, speed, velocity, difference between speed and velocity, acceleration, retardation,

equations of motions, simple related problems.

Work, Power and Energy

Work, unit of work, power, unit of power, Horse power of engines, mechanical efficiency, energy, use

of energy, potential and kinetic energy, examples of potential energy and kinetic energy.

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Mechanic Refrigeration and Air Conditioner

General simplifications

Fractions, Types of fractions, common fractions, Decimal fractions with examples Addition, subtrac-

tion, multiplication and division of fraction . conversion of Fraction to Decimal and vice versa.

S.no. Title

Square & Square root

Square root of perfect square, Square of whole number and decimal. Applications of Pythagoras theorem

and related Problems.

Unit & Measurements

Definition, classification of System of units, Fundamental & derived units. C.G.S, M.K.S,. F.P.S, & S.I

System of units. Metric system of weight and measurement unit and conversion factors, problems.

Percentage

Introduction, Simple calculation. Changing percentage to fraction and decimal & vice-versa.

Introduction, use of Electricity, Molecule, Atom, and How Electricity is Produced, Electric current,

voltage, Resistance and their units. Ohm’s law. Relation between V.I.R & Problems. Series & Parallel

circuits & Problems. Electrical Power and energy & their units & calculation.

Magnetic Induction, Self & Mutual Inductance, EMF generation.

Material Science

Properties of metals - Physical & Mechanical, Meaning of tenacity, elasticity, malleability brittleness,

hardness, ductility Types – Ferrous & Non-Ferrous, difference between Ferrous and Non-Ferrous metals,

introduction of Iron, Cast Iron, Wrought Iron, Steel, difference between Iron and Steel, Alloy steel,

carbon steel, stainless steel, Non-Ferrous Alloys. Effect of Alloying elements.

Properties and uses of copper, zinc, lead tin, aluminum etc., Properties and uses of Brass, Bronze

as bearing material.

Heat and Temperature

Measurement of Temperature, Boiling and melting points. Interchange of heat, (Principle of

calorimetry) Co-efficient of linear expansion, Related problems.

Vapours and gases

Saturated and superheated vapours, Critical pressures and temperatures. Heat transfer conduction,

Convection, Radiation. Thermal conductivity and Insulations.

Syllabus for Mechanic Refrigeration & Air Conditioner

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Sheet Metal Worker

Introduction and Importance of Science and Calculation to the Trade skill.

S.no. Title

- System of Units: British, Metric and S. I. Units for Length, Mass, Area, Volume, Capacity and

time.

- Conversions between British and Metric Systems.

3 - Density & Specific gravity.

- Mass, weight. Definition and units.

4 - Metals: Properties and uses of cast iron, wrought iron, plain carbon steels and alloy steels.

- Difference between metals, non-metals and alloys.

5 Properties and uses of Copper, Zinc, Lead, Tin and Aluminum.

6 Properties and uses of Brass, Bronze, Rubber ,Timber and insulating materials.

7 Concept of heat and temperature. Difference between heat and temperature. Effects of Heat,

Thermometric Scales such as a Celsius, Fahrenheit and Kelvin, Temperature measuring Instruments

- types of thermometers and pyrometers.

- Conversions between the above Scales of Temperature.

- Units of Heat-Calorie, B.Th.U & C.H.U., joule.

- Concept of Specific Heat, Latent Heat, problems on Heat Loss and Heat Gain.

- Definition of Force: Units of Force in M.K.S.& S.I. Systems.

- Concept of Pressure and its Units in different systems.

- General simplifications: BODMAS rule.

- Fraction: Addition, Subtraction, multiplication and Division-Problems.

- Decimal: Addition, Subtraction, Multiplication, and Division-Problems.

Conversion of Fraction to Decimal and vice-versa.

Square roots: The Square and Square root of a Whole Number and Decimal.

Percentage: Changing Percent to Decimal and Fraction and vice versa, applied problem.

Concept on Ratio and Proportion-Direct and Inverse Proportion, simple applied problems.

Algebraic Symbols and Fundamentals Addition, Subtraction, Multiplication and Division-Problems.

Syllabus for Sheet Metal Worker

Title of the Contents Common for all MR&AC SMW

Engineering Trades

1 Unit 

2 Fraction 

3 Square root 

4 Ratio & Proportion  ⎯ 

5 Percentage 

6 Algebra ⎯⎯

7 Material Science 

8 Mass, Weight & Density  ⎯ 

9 Speed & Velocity  ⎯⎯

10 Work, Power & Energy  ⎯⎯

11 Heat & Temperature ⎯ 

12 Basic Electricity ⎯  ⎯

Summary of Contents

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Classification of system of units Exercise 1.1.01

Necessity

All physical quantities are to be measured in terms of

standard quantities.

Unit

A unit is defined as a standard or fixed quantity of one kind

used to measure other quantities of the same kind.

Classification

Fundamental units and derived units are the two classifica-

tions.

Fundamental units

Units of basic quantities of length, mass and time.

Derived units

Units which are derived from basic units and bear a

constant relationship with the fundamental units.E.g. area,

volume, pressure, force etc.

Systems of units

– F.P.S system is the British system in which the basic

units of length, mass and time are foot, pound and

second respectively.

– C.G.S system is the metric system in which the basic

units of length, mass and time are centimeter, gram and

seconds respectively.

– M.K.S system is another metric system in which the

basic units of length, mass and time are metre, kilo-

gram and second respectively.

– S.I. units are referred to as Systems International units

which is again of metric and the basic units, their

names and symbols are as follows.

Fundamental units and derived units are the two classifica-

tions of units.

Length, mass and time are the fundamental units in all the

systems (i.e) F.P.S, C.G.S, M.K.S and S.I. systems.

Example

Length: What is the length of copper wire in the roll , if the

roll of copper wire weighs 8kg, the dia of wire is 0.9cm and

the density is 8.9 gm/cm

Solution

mass of copper wire in the roll = 8kg (or)8000grams

Dia of copper wire in the roll = 0.9cm

Density of copper wire = 8.9 gm/cm

Area of cross section of copper wire

636.0

)9.0(

cm







Volume of copper wire

cm88.898

cmgm/ 8.9

8000grams

wirecopper ofDensity

wirecopper of Mass



Length of copper wire

0.636cm

898.88cm

wirecopper of section cross of Area

wirecopper of Volume



= 1413.33 cm

Length of copper wire =1413cm.

Time: The S.I. unit of time, the second, is another of the

base units of S.I it is defined as the time interval occupied

by a number of cycles of radiation from the calcium atom.

The second is the same quantity in the S.I. in the British

and in the U.S. systems of units.

Fundamental units of F.P.S, C.G.S, M.K.S and S.I

S.No. Basic quantity British units Metric units International units

F.P.S Symbol C.G.S Symbol M.K.S Symbol S.I Units Symbol

1 Length Foot ft Centimetre cm Metre m Metre m

2 Mass Pound lb Gram g Kilogram kg Kilogram Kg

3 Time Second s Second s Second s Second s

4 Current Ampere A Ampere A Ampere A Ampere A

5 Temperature Fahrenheit °F Centigrade °C Centigrade °C Kelvin K

6 Light intensity Candela Cd Candela Cd Candela Cd Candela Cd

Derived units of F.P.S, C.G.S, M.K.S and SI system

S.No Physical quantity British units Metric units International units

FPS Symbol CGS Symbol MKS Symbol SI Units Symbol

1 Area Square foot ft

Square centimetre cm

Square metre m

2 Volume Cubic foot ft

Cubic centimetre cm

Cubic metre m

3 Density Pound per cubic lb/ft

Gram per cubic g/cm

Kilogram per cubic kg/m

Kilogram per cubic Kg/m

foot centimetre metre metre

4 Speed Foot per second ft/s Centimetre per second cm/sec Metre per second m/sec Metre per second m/sec

5 Velocity (linear) Foot per second ft/s Centimetre per second cm/sec Metre per second m/sec Metre per second m/sec

6 Acceleration Foot per square ft/s

Centimetre per cm/sec

Metre per square m/sec

second square second second second

7 Retardation Foot per square ft/s

Centimetre per cm/sec

Metre per square m/sec

Metre square second m/sec

Second square second second

8 Angular velocity Degree per second Deg/sec Radian per second rad/sec Radian per second rad/sec Radian per second rad/sec

9 Mass Pound (slug) lb Gram g Kilogram kg Kilogram kg

10 Weight Pound lb Gram g Kilogram weight kg Newton N

11 Force Pounds lbf dyne dyn Kilogram force kgf Newton N(kgm/sec

)

12 Power Foot pound per ft.lb/sec Gram.centimetre/sec g.cm/ kilogram metre per kg.m/ - -

second sec second sec

Horse power hp Erg per second watt W watt W(J/sec)

13 Pressure,Stress Pound per square inch lb/in

Gram per square g/cm

Kilogram per kg/m

Newton per square N/m

centimetre square metre metre

14 Energy, Work Foot.pound ft.lb Gram centimetre g.cm Kilogram metre kg.m joule J(Nm)

15 Heat British thermal unit Btu calorie Cal joule J joule J(Nm)

16 Torque Pound force foot lbf.ft Newton millimetre N mm Kilogram metre kg.m Newton metre Nm

17 Temperature Degree Fahrenheit °F Degree Centigrade °C Kelvin K Kelvin K

18 Specific heat BTU per pound degree Btu/lb°F Calorie per gram Cal/g°C Joule per kilogram J/(kgK) Joule per J/(kgK)

fahrenheit degree Celsius kelvin kilogram kelvin

Fundamental and Derived units F.P.S, C.G.S, M.K.S and SI units Exercise 1.1.02

S.No Physical quantity British units Metric units International units

FPS Symbol CGS Symbol MKS Symbol SI Units Symbol

19 Frequency Cycle per second 1/s Hertz Hz Hertz Hz Hertz Hz

20 Moment of inertia Pound force foot lbf.ft.s

Gram square g.cm

Kilogram kg.m

Kilogram per square Kg.m

square second centimetre square metre metre

21 Momentum Pound second lb.s Gram centimetre g.cm/sec Kilogram metre kg.m/sec Kilogram metre Kg.m/

per second per second per second sec

22 Moment of force Pounds foot lbs/ft Gram centimetre g.cm Kilogram metre kg.m Newton metre Nm

23 Angle degree deg degree deg degree deg Radian rad

24 Specific volume Cubic foot per pound ft

/lbs Cubic centimetre per Cm

/g Cubic metre per m

/kg Cubic metre per m

/kg

gram kilogram kilogram

25 Specific resistance Ohm foot Ω ft Ohm centimetre Ω cm Ohm meter Ω m Ohm meter Ω m

26 Specific weight Pound per cubic foot lbf/ft

Gram per cubic g/cm

Kilogram per cubic kg/m

Newton per cubic N/m

centimetre metre metre

27 Fuel consumption Miles per gallon m/gal Centimetre per cubic cm/cm

Kilometre per litre km/l Metre per cubic metre m/m

centimetre

28 Dynamic viscosity Pound force per lbf/ft

Centi poise CP pascal second P

a.s

pascal second P

a.s

square foot

29 Surface tension Poundal per foot pdl/ft dyne per centimetre dyn/cm Newton per metre N/m Newton per metre N/m

30 Entropy

British thermal unit Btu/

F Calorie per degree Cal/

c Joule per kelvin J/K Joule per kelvin J/K

per degree Fahrenheit centigrade

31 Electric current Columb per second C/s Biot Bi Ampere A Ampere A

32 Electric voltage Volt V Volt V Volt V Volt V

33 Electric resistance Ohm Ω Ohm Ω Ohm Ω Ohm Ω, (V/A)

34 Electric Mho, Siemens ,s Mho ,s Siemens s Siemens s

conductance

35 Light intensity Candela Cd Candela Cd Candela Cd Candela Cd

36 Specific gravity No unit - No unit - No unit - No unit -

Workshop Calculation And Science : (NSQF) Exercise 1.1.02

Decimal multiples and parts of unit

Decimal power Value Prefixes Symbol Stands for

1000000000000 tera T billion times

1000000000 giga G thousand milliotimes

1000000 mega M million times

1000 kilo K thousand times

100 hecto h hundred times

10.10

deca da ten times

-1

0.1 10

-1

deci d tenth

-2

0.01 centi c hundreth

-3

0.001 milli m thousandth

-6

0.000001 micro μ millionth

-9

0.000000001 nano n thousand millionth

-12

0.000000000001 pico p billionth

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Measurement units and conversion Exercise 1.1.03

Units and abbreviations

Quantity Units Abbreviation of unit

Calorific value kilojoules per kilogram kJ/kg

Specific fuel kilogram per hour per newton kg/hr/N

consumption

Length millimetre, metre, kilometre mm, m, km

Mass kilogram, gram kg, g

Time seconds, minutes, hours s, min, h

Speed centimetre per second, cm/s, m/s

metre per second

kilometre per hour, miles km/h, mph

per hour

Acceleration metre-per-square second m/s

Force newtons, kilonewtons N,kN

Moment newton-metres Nm

Work joules J

Power horsepower, watts, kilowatts Hp, W, kW

Pressure newton per square metre N/m

kilonewton per square metre kN/m

Angle radian rad

Angular speed radians per second rad/s

radians-per-square second rad/s

revolutions per minute Rpm

revolutions per second rev/s

SI units and the British units:

Quantity SI unit

→ →

→ British unit British unit

→→

→ SI unit

Length 1 m = 3.281 ft 1 ft = 0.3048 m

1 km = 0.621 mile 1 mile = 1.609 km

Speed 1 m/s = 3.281 ft/s 1 ft/s = 0.305 m/s

1 km/h = 0.621 mph 1 mph = 1.61 km/h

Acceleration 1 m/s

= 3.281 ft/s

1 ft/s

= 0.305 m/s

Mass 1 kg = 2.205 lb 1 lb = 0.454 kg

Force 1 N = 0.225 lbf 1 lbf = 4.448 N

(1 million newtons)

Torque 1 Nm = 0.738 lbf ft 1 lbf ft = 1.355 Nm

Pressure 1 N/m

= 0.000145 lbf/in

1 lbf/in

= 6.896 kN/m

1 Pa = 1 N/m

1 bar = 14.5038 lbf/in

1 lbf/in

= 6.895 kN/m

Energy, work 1 J = 0.738 ft lbf 1 ft lbf = 1.355 J

1 J = 0.239 calorie 1 calorie = 4.186 J

1 kJ = 0.948 Btu 1 Btu = 1.055 kJ

(1 therm = 100 000 Btu)

1 kJ = 0.526 CHU 1 CHU = 1.9 kJ

Power 1 kW = 1.34 hp 1 hp = 0.7457 kW

Fuel consumption 1km/L = 2.82 mile/gallon 1 mpg = 0.354 km/L

Specific fuel 1 kg/kWh = 1.65 lb/bhp h 1 lb/bhp h = 0.606 kg/kWh

consumption 1 litre/kWh=1.575 pt/bhp h 1 pt/bhp h = 0.631 litre/kWh

Calorific value 1 kJ/kg = 0.43 Btu/lb 1 Btu/lb = 2.326 kJ/kg

1 kJ/kg = 0.239 CHU/lb 1 CHU/lb = 4.188 kJ/kg

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

Hμ

kwh

Units in measuring practice with definitions

Quantity Unit Explanation

Force F Newton N 1 Newton is equal to the force which imports an

acceleration of 1m/s

to a body of mass 1 kg

1N = 1 kg m/s

Pressure P Newton 1 Newton per square metre (1 pascal)

per square is equal to the pressure with which

metre the force of 1 N is exercised perpendicular

to the area of 1 m

Pascal Pa 1Pa = 1 N/m

. 1 Bar (bar) is the special name

for 100 000 Pa.

Normal stress Newton per 1 Newton per square metre (1 pascal)

tensile or square to the mechanical stress with which the

compressive metre force of 1 n is exercised on the area of 1 m

stress, Shear stress In many branches of engineering the mechani-

cal stress and strength are specified in N/m

1 N/m

= 1000 000 Pa = 1 MPa

Heat Energy W Joule J 1 Joule is equal to the work that is done when

Quantity of heat the point of application of the force of 1 N is

shifted by 1 m in the direction of the force.

1 J = 1 Nm = 1 Ws = 1 kgm

3600 000 J = 1 kWh

Moment of a force Newton Nm 1 Newton is equal to the moment of a force

(torque) metre which results from the product of the force

joule J of 1 N and the lever arm of 1 m.

1 Nm = 1 J = 1 Ws = 1 kgm

Power P Watt W 1 Watt is equal to the power with which the

Energy flow energy of 1 J is converted during the time of 1s.

Heat flow ø The unit watt is also called volt ampere in

the specification of apparent electric power

1 W = 1 J/s = 1 Nm s = 1 VA

Specific Joule per 1 Joule per kilogram is equal to the quantity of

heat value kilogram heat which on complete burning of the mass of

1 kg releases the energy of 1 J

Fuel gram per 1 gram per kilowatt-hour is equal to the fuel

consumption kilowatt- consumption of the mass of 1 g for the work

hour of 1 kWh.

Temperature T Kelvin K The kelvin is defined as the fraction of

the thermodynamic temperature of the triple

point of water.

Electric current I Ampere A 1 Ampere is the strength of a current which

would bring about an electrodynamic force of

0.2.10 N per 1 m length between two parallel

conductors placed at a distance of 1 m.

273.16

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

Electric voltage V Volt V 1 Volt is equal to the electric voltage between

two points of a metallic conductor in which a

power of 1 W is expended for a current of 1 A

strength.

Electric resistance R Ohm Ω 1 Ohm is equal to the electric resistance be-

tween two points of a metallic conductor in

which an electric current of 1 A flows at a

voltage of 1 V.

Electric conductance G Siemens S 1 Siemens is equal to the electric conductance

of a conductor of electric resistance of 1ohm

Quantity Q Coulomb C 1 Coulomb is equal to the quantity

of electricity ampere-second As of electricity which flows through the conductor

cross-section during the time of 1 s at an

electric current of 1A.

Prefixes for decimal multiples and submultiples

Use

1 Megapascal = 1 MPa = 1000000 Pa

1 Kilowatt = 1 kW = 1000 W

1 Hectolitre = 1 hL= 100 L

Decanewton = 1 daN = 10 N

Decimetre = 1 dm = 0.1 m

1 Centimetre = 1 cm = 0.01 m

1 Millimetre = 1 mm = 0.001 m

1 Micrometre = 1 um = 0.000001 m

Conversion factors

1 inch = 25.4 mm

1 mm = 0.03937 inch

1 metre = 39.37 inch

1 micron = 0.00003937"

1 kilometre = 0.621 miles

1 pound = 453.6 gr

1 kg = 2.205 lbs

1 metric ton = 0.98 ton

Units of physical quantities

Units of length

Micron 1 μ = 0.001 mm

Millimetre 1 mm = 1000 μ

Centimetre 1 cm = 10 mm

Decimetre 1 dm = 10 cm

Metre 1 m = 10 dm

Kilometre 1 km = 1000 m

Inch 1" = 25.4 mm

Foot 1" = 0.305 m

Yard 1 Yd = 0.914 m

Nautical mile 1 NM = 1852 m

Geographical mile 1 = 1855.4 m

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

Units of area

Square millimetre 1 mm

Square centimetre 1 cm

= 100 mm

Square decimetre 1 dm

= 100 cm

Square metre 1 m

= 100 dm

Are 1 a = 100 m

Hectare 1 ha = 100 a

Square kilometre 1 km

= 100 ha

Square inch 1 sq.in = 6.45 cm

Square foot 1 sq.ft = 0.093 m

Square yard 1 sq.yd = 0.84 m

Square metre 1 m

= 10.76 ft

Acre 1 = 40.5 a

1 Acre = 100 cent 1 Hectare = 2.47 acres

1 Cent = 436 Sq. ft. 1 acre = 0.4047 Hec

1 Ground = 2400 Sq.ft. tare

1 Hectare = 10000 sq.

metre

Units of weight

Milligram - force 1 mgf

Gram-force 1 gf 1000 mgf

Kilogram-force 1 kgf = 1000 gf

Tonne 1 t = 1000 kgf

Ounce 1 = 28.35 gf

Pound 1 lbs = 0.454 kgf

Long ton 1 = 1016 kgf

Short ton 1 = 907 kgf

Time

Second 1 s

Minute 1 min 60 s

Hour 1 hr = 60 min

Units of volume and capacity

Cubic millimetre 1 mm

Cubic centimetre 1 cm

= 1000 mm

Cubic decimetre 1 dm

= 1000 cm

Cubic metre 1 m

= 1000 dm

Litre 1 l = 1 dm

Hectolitre 1 hl = 100 l

Cubic inch 1 cu. in = 16.387 cm

Cubic foot 1 cu. ft = 28317 cm

Gallon (British) 1 gal = 4.54 l

1cubic metre 1 m

= 1000 litres

1000 Cu.cm 1000 cm

= 1 litre

1 cubic foot 1 ft

= 6.25 Gallon

1 litre 1It = 0.22 Gallon

Angle

1 Centessimal unit

1 Right Angle = 100 grade (100

)

1 grade (1

) = 100 Minute (100’)

1 minute (1’) = 100 second (100”)

2 Sexagesimal unit

1 Right angle = 90 Degree (90

)

1 Degree (1

) = 60 minutes (60’)

1minute (1’) = 60 seconds (60”)

3 Circular unit

Radian

Relationship between Radian and Degree

1 Radian =

180

= π Radian;

1 Degree = Radian

180

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

Work

Kilogram-force 1 kgfm = 9.80665 J

Metre 1 kgfm = 9.80665 Ws

Joule 1 J = 1 Nm

Watt-second 1 Ws = 0.102 kgfm

Kilowatt hour 1 kWh = 3.6 x 10

= 859.8456 kcal

I.T.Kilocalorie 1 kcal

= 426.kgfm

Pressure

Pascal 1 Pa = 1 N/m

1 atm = 101325 Pa

Bar 1 bar = 10N/cm

= 100000 Pa–Torr 1 torr = ≈ 133.32 pa

Atmosphere 1 atm = 1 kgf/cm

1 kgf/cm

= 735.6 mm of mercury

Power

Kilogram-force metre/second

1 kgfm/s = 9.80665 W

Kilowatt 1 kW = 1000 W = 1000 J/s

= 102 kgfm/s (approx.)

Metric horse power 1 HP = 75 kgfm/s

= 0.736 kW

1 Calorie = 4.187J

I.T.Kilocalorie/hour = 1 kcal

IT/h

= 1.163 W

101325

760

Geometrical quantities

Symbol Physical quantity Conventional Units S.I.Units Symbol

S.I. units

l Length m Metre m

h Height m Metre m

b Width, breadth m Metre m

r Radius m Metre m

d Diameter m Metre m

d,δ Wall thickness m Metre m

S Length of path m Metre m

A (S) Area m

Square metre m

V (v) Volume m

Cubic metre m

α,β,γ etc Angle ° Radian (1 rad = 57.3°) rad

λ Wave length km Kilometre km

l,la Second moment of area cm

Metre to the fourthpower m

MASS

m Mass kg Kilogram kg

ρ Density g/cm

Kilogram per cubicmetre kg/m

l,J Moment of inertia kg, m

Newton metre

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

TIME

T Time or time interval s Second s

nu Rotational frequency l/min Reciprocal second l/s

u,v,w,c Velocity speed m/min Metre per second m/s

ω Angular velocity rad/s Radian per second rad/s

g Acceleration of freefall m/s

Metre per second squared m/s

a Acceleration m/s

Metre per second squared m/s

Retardation m/s

Metre per second squared m/s

FORCE AND PRESSURE

F Force kgf Newton (1kgf = 9.80665N) N

G(P,W) Weight kgf Newton N

γ Specific weight kgf/m

Newton per cubic metre N/m

M Moment of force kgf.m Newton metre N.m

(force x distance)

p Pressure (force/ area) kgf/cm

pascal, Newton per Pa,N/m

square metre

p Normal stress kgf/mm

bar (1 bar = 10 N/m)

τ p Shear stress kgf/mm

bar

E Modulus of elasticity kgf/mm

Newton per square metre N/m

G Shear modulus kgf/mm

Newton per square metre N/m

μ Co-efficient of friction No Unit

TEMPERATURE

Scale Freezing point Boiling point

Centigrade (

C) 0°C 100°C

Faranheit (°F) 32°F 212°F

Kelvin (K) 273K 373K

Reaumur (°R) 0°R 80°R

°R

°C

100

K- 273

100

°F- 32

180

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

HEAT, WORK, ENERGY,FORCE

A,W Work kgfm Joule (1 Joule=1 N.m) J (Nm)

P Power kgfm/s Watt W (J/s)

E,W Energy kgfm Joule J (Nm)

η Efficiency - - -

W,A,E,Q Quantity of heat kcal Joule J

C Specific heat kcal/kgf°C Joule per newton per J/N.°K

degree Kelvin

Thermal conductivity kcal/mh°C Joule per metre per J/ms°K

second per degree

Kelvin

Force In C.G.S. System : Force (Dyne) = Mass (gm)XAcceleration (cm/sec

)

In F.P.S. System : Force (Poundal) = Mass (Ib) X Acceleration (ft./sec

)

In M.K.S System : Force (Newton) = Mass (Kg) x Acceleration (mtr./sec

)

1 Dyne = 1 gm x1 cm/sec

1 Poundal = 1 Ib x 1 ft/sec

1 Newton = 1 kg x 1 mtr/sec

= 10

dynes

1gm weight = 981 Dynes

1 Ib weight = 32 Poundals

1 kg weight = 9.81 Newtons

ELECTRICAL QUANTITIES

V Electric potential V Volt V(W/A)

E Electromotive force V Volt V(W/A)

I Electric current A Ampere A

R Electric resistance Ω Ohm Ω (V/A)

e Specific resistance Ω m Ohm metre Vm/A

G Conductance Ω

-1

Siemens S

Workshop Calculation And Science : (NSQF) Exercise 1.1.03

1 Convert the following as indicated

a 5 yards into metres ______

b 15 miles into kilometres ______

c 7 metres into yards ______

d 320 kilometres into miles ______

2 Convert

a 5 pounds into kilograms ______

b 8.5 kilograms into pounds ______

c 5 ounces into grams ______

d 16 tons into kilograms ______

3 Convert

a 40 inches into centimetres ______

b 12 feet into metres ______

c 5 metres into inches ______

d 8 metres into feet ______

4 Convert

a 234 cubic metres into gallons ______

b 2 cubic feet into litres ______

c 2.5 gallons into litres ______

d 5 litres into gallons ______

5 Answer the following questions

a 120°C = ______ °F.

b 8 mm = ______ inches

12 mm = ______ inches

6 Convert and find out

A car consumes fuel at the rate of one gallon for a travel

of 40 miles.

The same car travels a distance of 120 kilometer. what

is the consumption of fuel in litres.

7 Write equivalent British units for the given metric units

a Seconds, minutes, Hours

b Grams, Kilograms

c Litres, Cubic meters

d Square centimeter, Square kilometer

8 Expand the abbreviations of the following

a km/l

b N/m

cKW

d m/s

e RPM

9 Convert the following S.I. units as required.

a Length

i 3.4 m = ______ mm

ii 1.2 m = ______ cm

iii 0.8 m = ______ mm

iv 0.02 km = ______ cm

v 10.2 km = ______ mile

vi 6 m = ______ km

vii 18 m = ______ mm

viii 450 m = ______ km

ix 85 cm = ______ km

x 0.06 km = ______ mm

b Mass

i 650 g = ______ kg

ii 300 cg = ______ g

iii 8 g = ______ dg

iv 120 mg = ______ g

v 8 dag = ______ mg

vi 2.5 g = ______ mg

vii 2.5 g = ______ kg

viii 350 mg = ______ mg

ix 20 cg = ______ mg

x 0.05 Mt = ______ kg

c Force

i 1.2 N = ______ kg

ii 2.6 N = ______ kg

iii 800 N = ______ KN

iv 14.5 kg = ______ N

v 25 kg = ______ N

d Work, energy, amount of heat

i 2 Nm = ______ Ncm

ii 50 Ncm = ______ Nm

iii 120 KJ = ______ J

iv 40 J = ______ KJ

v 40 J = ______ KJ

vi 300 wh = ______ kwh

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Conversions of length, mass, force,work, power and energy Exercise 1.1.04

e Power

i 200 mW = ______ W

ii 0.2 kW = ______ W

iii 300 kW = ______ mW

iv 2.10

W = ______ mW

v 6.10

-4

kW = ______ W

vi 2 W = ______ KW

vii 350 W = ______ kW

viii 0.08 W = ______ kW

ix 2 x 10

-3

kW =______ W

x 0.04 W = ______ mW

f Convert as required.

i 3 Nm = ______ J

ii 2 J = ______ Ws

iii 12 J = ______ KJ

iv 3 Nm/s = ______ J/s

v 8 J/s = ______ J/s

vi 5 N = ______ KN

vii 5 Ws = ______ Ws

viii 3 KJ = ______ Nm

ix 18 J/s = ______ W

x 12 W = ______ J/s

xi kJ/s = ______ Nm/s

Workshop Calculation And Science : (NSQF) Exercise 1.1.04

Factors, HCF, LCM and problems Exercise 1.2.05

Prime Numbers and whole Numbers

Factor

A factor is a small number which divides exactly into a

bigger number.e.g.

To find the factors of 24, 72, 100 numbers

24 = 2 x 2 x 2 x 3

72 = 2 x 2 x 2 x 3 x 3

100 = 2 x 2 x 5 x 5

The numbers 2,3,5 are called factors.

Definition of a prime factor

Prime factor is a number which divides a prime number into

factors.e.g.

57 = 3 x 19

The numbers 3 and 19 are prime factors.

They are called as such, since 3 & 19 also belong to prime

number category.

Definition of H.C.F

The Highest Common Factor

The H.C.F of a given group of numbers is the highest

number which will exactly divide all the numbers of that

group.e.g.

To find the H.C.F of the numbers 24, 72, 100

24 = 2 x 2 x 2 x 3

72 = 2 x 2 x 2 x 3 x 3

100 = 2 x 2 x 5 x 5

The factors common to all the three numbers are

2 x 2 = 4. So HCF = 4.

Definition of L.C.M

Lowest common multiple

The lowest common multiple of a group of numbers is the

smallest number that will contain each number of the given

group without a remainder.e.g.

• Factorise the following numbers

7,17,20,66,128

7,17 - These two belong to Prime numbers. Hence no

factor except unity and itself.

Factors of 20 = 2 x 2 x 5

• Select prime numbers from 3 to 29

3,5,7,11,13,17,19,23,29

• Find the HCF of the following group of numbers HCF of

78, 128, 196

78 = 2 x 3 x 13

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

196 = 2 x 2 x 49

HCF = 2

• Find LCM of 84,92,76

LCM =

Factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

220

210

266

333

2 128

264

232

216

278

339

2 128

2 6 4

2 3 2

2 1 6

2 8

2 4

2 196

298

2 84, 92, 76

2 42, 46, 38

3 21, 23, 19

7, 23, 19

LCM = 2 x 2 x 3 x 7 x 23 x 19 = 36708

• To find out the LCM of 36, 108, 60

Factors of 66 = 2 x 3 x 11

2 36, 108, 60

2 18, 54, 30

3 9, 27, 15

3 3, 9, 5

1, 3, 5

LCM of the number

36, 108, 60 = 2 x 2 x 3 x 3 x 3 x 5 = 540

The necessity of finding LCM and HCF arises in subtraction

and addition of fractions.

Fractions Exercise 1.2.06

Description

A minimal quantity that is not a whole number. For e.g. .

A vulgur fraction consists of a numerator and denomi-

nator.

Numerator/Denominator

The number above the line in a vulgar fraction showing how

many of the parts indicated by the denominator are taken

is the numerator. The total number of parts into which the

whole quantity is divided and written below the line in a

vulgar fraction is the denominator. e.g.

1,3,7 - numerators 4,12 - denominators

Fraction: Concept

Every number can be represented as a fraction.e.g.

1 

, A full number can be represented as an apparent

fraction.e.g. (Fig 1)

Fraction: Value

The value of a fraction remains the same if the numerator

and denominator of the fraction are multiplied or divided by

the same number.(Fig 2)

Multiplication

When fractions are to be multiplied, multiply all the

numerators to get the numerator of the product and multiply

all the denominators to form the denominator of the

product. (Fig 3)

Division

When a fraction is divided by another fraction the dividend

is multiplied by the reciprocal of the divisor. (Fig 4)

Addition and Subtraction

The denominators of the fractions should be the same when

adding or subtracting the fractions. Unequal denominators

must first be formed into a common denominator. It is the

lowest common denominator and it is equal to the product

of the most common prime numbers of the denominators of

the fractions in question.(Fig 5)

Examples

•

Multiply



•

Divide



•

and

Add





•

from

sub



1417







Types of fractions

• Proper fractions are less than unity. Improper fractions

have their numerators greater than the denominators.

• A mixed number has a full number and a fraction.

Addition of fraction

Add 

To add these fractions we have to find out L.C.M of

denominators 2,8,12.

Find L.C.M of 2,8,12

Step 1 L.C.M

2 2,8,12

2 1,4,6

1,2,3

Factors are 2,2,2,3

Hence L.C.M = 2 x 2 x 2 x 3 = 24

Step 2

10312









Subtraction of fraction

or(17

from17

9subtract 

)

Step 1: Subtract whole number first 17 - 9 = 8

Step 2: L.C.M of 16,32 = 32

Since number 16 divides the number 32

Subtracting fractions =

Adding with whole number from Step 1

8 get we 

Common fractions

Problems with plus and minus sign

Example

3solve 

Rule to be followed

1 Add all whole numbers

2 add all + Numbers

3 Add all - Numbers

4 Find L.C.M of all denominators

Solution

Step 1: Add whole numbers = 3 + 6 - 4 = 5

Step 2: Add fractions =



L.C.M of 4,8,16,32 is 32

1952

9102824









Step 3: Adding again with the whole number

we get

15 

Brackets and their solution

Sometimes fractions are added and subtracted with brack-

ets, to express two or more expressions. The problems

with brackets are solved by using a rule called ‘BODMAS’

rule.

BODMAS rule:

• B - Brackets - 1

• O - of - 2

• D - Division - 3

• M - Multiplication - 4

• A - Addition - 5

• S - Subtraction - Last to be done

Problem with brackets

solve







































91456

Steps to be followed

( ) 1 Solve inside bracket ( )

2 Solve multiplication and division together

3 Solve addition and subtraction together

{ } 4 Solve curly bracket before the final stage

[ ] 5 Solve square bracket at the final stage.

Workshop Calculation And Science : (NSQF) Exercise 1.2.06

Examples

Common fractions

• Multiply





• Division



4 

• Addition

124

82,4,8L..C.M











• Subraction











Bracket problem

1 solve by use ‘BODMAS’ rule







































2116810

1230

123

810

15810

1580









































 1

2 solve by use ‘BODMAS’ rule

367

1470

1256

19456





























































































Workshop Calculation And Science : (NSQF) Exercise 1.2.06

Decimal fractions Exercise 1.2.07

Description

Decimal fraction is a fraction whose denominator is 10 or

powers of 10 or multiples of 10 (i.e.) 10, 100, 1000, 10000

etc. Meaning of a decimal number:-

12.3256 means

10000

1000

100

1)X(210)X(1 

Representation

The denominator is omitted. A decimal point is placed

at different positions of the number corresponding to the

magnitude of the denominator

3.648

1000

3648

0.0127,

10000

127

0.35

100

0.5,

.Ex

Addition and subtraction

Arrange the decimal fractions in a vertical order, placing the

decimal point of each fraction to be added or subtracted, in

succession one below the other, so that all the decimal

points are arranged in a straight line. Add or subtract as

you would do for a whole number and place the decimal

point in the answer below the column of decimal points.

Decimal fractions less than 1 are written with a zero before

the decimal point. Example: 45/100 = 0.45 (and not simply

.45)

Add 0.375 + 3.686

0.375

3.686

4.061

Subtract 18.72 from 22.61

22.61

18.72

3.89

Multiplication

Ignore the decimal points and multiply as whole numbers.

Find the total number of digits to the right of the decimal

point. Insert the decimal point in the answer such that the

number of digits to the right of the decimal point equals to

the sum of the digits found to the right of the decimal points

in the problem.

Multiply 2.5 by 1.25

= 25 x 125 = 3125. The sum of the figures to the right

of decimal point is 3. Hence the answer is 3.125.

Division

Move the decimal point of the divisor to the right to make

it a full number. Move the decimal point in the dividend to

the same number of places, adding zeroes if necessary.

Then divide.

Divide 0.75 by 0.25

37525

100

0.25

0.75

0.750.25



Move the decimal point in the multiplicand to

the right to one place if the multiplier is 10, and

to two places if the multiplier is 100 and so on.

When dividing by 10 move the decimal point

one place to the left, and, if it is by 100, move

them point by two places and so on.

Example

Allowing 3 mm for cutting off each pin how many pins, can

be made from a 900 mm long bar? How much material will

be left out?

Length of pin

= 2.25 + 55.36 + 12.18

= 69.79 mm

Length of the bar = 900 mm

Step 1

Let the number of pins to be made =

Length of number of pins = x 69.79 mm

Step 2

Waste for each pin = 3 mm

Waste for number of pins = 3 x mm=3 mm

Adding step (1) + step (2) and equating to length of bar

69.79 mm + 3 mm = 900 mm

(69.79mm + 3mm) = 900mm

(72.79mm) = 900mm

= 900



72.79

Hence Number of pins to be made = 12

Secondly

Left out materials

= Total length of bar - Length for 12 pins+ wastage of

cutting

= 900mm - (12 x 69.79 + 12 x 3)mm

= 900 - (837.48 + 36)mm

= 900 - 873.48mm

=26.52mm

Left out material 26.52 mm

Conversion of Decimals into fractions and vice-versa

• Convert decimal into fractions

Example

Convert 0.375 to a fraction

Now place 1 under the decimal point followed by as many

zeros as there are numbers

• Convert fraction into decimal

Example

• Convert to a decimal

Proceed to divide in the normal way of division but put

zeros (as required) after the number 9 (Numerator)

=0.5625

• Convert

to a decimal

0.875

70008

0.875



Recurring decimals

While converting from fraction to decimals, some fractions

can be divided exactly into a decimal. In some fractions the

quotient will not stop. It will continue and keep recurring.

These are called recurring decimals.

Examples

•

convert

Recurring0.3333

10000



Recurring0.666

20000















 Recurring20.14285714

10000

These are written as below with a dot over the figure

0.3333 ——> 0.3

0.6666 ——> 0.6

. .

0.14857142 —> 0.14857

Note the dots marked over numbers.

We normally carry the decimal points upto 4 places in

Engineering calculations.

Approximations in Measured Value calculations

In Measured Value calculations 4 places of decimals are

sufficients and in many dimensions of parts even 3 decimal

places are near enough to complete the maintenance job

operations.

Workshop Calculation And Science : (NSQF) Exercise 1.2.07

Method of writing approximations in decimals

1.73556 = 1.7356 Correct to 4 decimal places

5.7343 = 5.734 Correct to 3 decimal places

0.9345 = 0.94 Correct to 2 decimal places

Multiplication and division by 10,100,1000

Multiplying decimals by 10

A decimal fraction can be multiplied by 10,100,1000 and so

on by moving the decimal point to the right by as many

places as there are zeros in the multiplier.

• 4.645 x 10 = 46.45 (one place)

• 4.645 x 100 = 464.5 (two places)

• 4.645 x 1000 = 4645 (three places)

Dividing decimals by 10

A decimal fraction can be divided by 10,100,1000 and so

on, by moving the decimal point to the left by as many

places as required in the divisor by putting zeros

Examples

• 3.732 ÷ 10 = 0.3732 (one place)

• 3.732 ÷ 100 = 0.03732 (two places)

• 3.732 ÷ 1000 = 0.003732 (three places)

Examples

• Rewrite the following number as a fraction

453.273

1000

273

453

100

1)(310)(5100)(4





• Write the representation of decimal places in the given

number 0.386

3 - Ist decimal place

8 - IInd decimal place

6 - IIIrd decimal place

• Write approximations in the following decimals to 3

places.

a 6.9453 ——> 6.945

b 8.7456 ——> 8.746

• Convert fraction to decimal

0.875



• Convert decimal to fraction

10000

625

0.0625 

ASSIGNMENT

1 Write down the following decimal numbers in the

expanded form.

a 514.726

b 902.524

2 Write the following decimal numbers from the expansion.

1000

100

570500 

1000

100

9200 

3 Convert the following decimals into fractions in the

simplest form.

a 0.72

b 5.45

c 3.64

d 2.05

4 Convert the following fraction into decimals

1000

Workshop Calculation And Science : (NSQF) Exercise 1.2.07

Pocket calculator and its applications Exercise 1.2.08

A pocket calculator allows to spend less time in doing

tideous calculations. A simple pocket calculator enables

to do the arithmatical calculations of addition substraction,

multiplication and division, while a scientific type of calcu-

lator can be used for scientific and technical calculations

also.

No special training is required to use a calculator. But it is

suggested that a careful study of the operation manual of

the type of the calculator is essential to become familiar

with its capabalities. A calculator does not think and do.

It is left to the operator to understand the problem, interpret

the information and key it into the calculator correctly.

Constructional Details (Fig 1)

• Function keys

• Memory keys

M Store the display number

M+ The displayed value is added to the memory

M- The displayed value is subtracted from the

memory

MR RCL The stored value is recalled on to the

display

Further functional keys included in Scientific calculators

are as shown in Fig 2.

Addition key

Subtraction key

Multiplication keyx

Division key

Equals key to display the result

Pi key

Square root key

Percentage key

Sign change key

Square key

Reciprocal key

+/-

The key board is divided into five clear and easily recogniz

able areas and the display

• Data entry keys

The entry keys are from ..............to

and a key for the decimal point .

• Clearing keys

These keys have the letter ‘C’

Clear totally

Clear entry only

, Clear memory

CLR

CM MC

For trigonometric functions and

for brackets

Sin Cos

Ta n

(

)

Exponent key

Some of the keys have coloured lettering above or

below them. Toi use a function in coloured lettering,

press INV key. INV will appear on the display.

Then press the key that the coloured lettering

identifies. INV will disappear from the display.

, to obtain the logarithm of the dis-

played number and the antilogarithm of the displayed value.

to convert displayed rectangular coordinates

into polar coordinates.

to convert displayed polar coordinates into

rectangular coordinates.

• The display

The display shows the input data, interim results and

answers to the calculations.

The arrangement of the areas can differ from

one make to another. Keying in of the numbers

is done via. an internationally agreed upon set

of ten keys in the order that the numbers are

written.

Exp

INV

log

INV

R–P

INV

P–R

Rules and Examples:

• Addition: Example 18.2 + 5.7

Sequence Input Display

Input of the 1st term 1 8 . 2 18.2

of the sum

Press + key + 18.2

Input 2nd term of the

sum. the first term 5 . 7 5.7

goes into the register

Press the = key = 23.9

• Subtraction: Example 128.8 - 92.9

Sequence Input Display

Enter the subtrahend 1 2 8 . 8 18.2

Press - key - 128.8

Enter the minnend.

The subtrahend goes 9 2 . 9 92.9

into the register

Press the = key = 35.9

• Multiplication: Example 0.47 x 2.47

Sequence Input Display

Enter multiplicand . 4 7 0.47

Press x key x 0.47

Enter multiplier,

multiplicand goes 2 . 4 7 2.47

to register

Press = key = 1.1609

• Division: Example 18.5/2.5

Sequence Input Display

Enter the dividend 1 8 . 5 18.5

Press ÷ Key ÷ 18.5

Enter the divisor

goes to the register 2 . 5 2.5

Press = key = 7.4

Workshop Calculation And Science : (NSQF) Exercise 1.2.08

• Multiplication & Division:

Example : 2.5 x 7.2 / 4.8 x 1.25

Sequence Input Display

Enter 2.5 2 . 5 2.5

Press x key x 2.5

Enter 7.2 7 . 2 7.2

Press ÷ key ÷ 18

Enter 4.8 4 . 8 4.8

Press x key x 3.75

Remember:Before

input of thefirst value

under the fraction

line, the x keymust

be operated

Enter 1.25

Press = key

• Store in memory Example (2+6) (4+3)

Sequence Input Display

Workout for the first 2 2

bracket

+ 2

6 6

= 8

Store the first result in STO , M 8

x or M+

Workout for the 4 4

2nd bracket

+ 4

3 3

= 7

Press x key x 7

Recall memory RCL or MR 8

Press = key = 56

1.25

3.0

• Percentage: Example 12% of 1500

Sequence Input Display

Enter 1500 1 5 0 0 1500

Press x key x 1500

Enter 12 1 2 12

Press INV % INV % 12

Press = key = 180

• Square root: Example

53 2 

Sequence Input Display

Enter 2 2 2

Press

key 2

Press + key + .

Press bracket key ( .

Enter 3 3 3

Press

key .

Press x key x .

Enter 5 5 5

Press

key

Press bracket close key ) .

Press = key = 5.2871969

553 2 

• Common logaritham: Example log 1.23

Sequence Input Display

1 . 2 3 log = 0.0899051

• Power: Example 123 + 30

Sequence Input Display

1 2 3 + 3 0 INV X

= 1023

2 +(

)

5.2871969

Workshop Calculation And Science : (NSQF) Exercise 1.2.08

ASSIGNMENT

1 Using calculator solve the following

a 625 + 3467 + 20 + 341 + 6278 = ______

b 367.4 + 805 + 0.7 + 7.86 + 13.49 = ______

c 0.043 + 1.065 + 13.0 + 34.76 + 42.1 = ______

d 47160 + 1368.4 + 0.1 + 1.6901 + 134.267 =

_______

2 Using calculator simplify the following

a 24367 - 4385 = ______

b 9.643 - 0.7983 = ______

c 4382.01 - 381.3401 = ______

d 693.42 - 0.0254 = ______

3 Using calculator find the values of the following

a 23 x 87 = ______

b 1376 x 0.81 = ______

c 678 x 243 = ______

d 0.75 x 0.24 = ______

4 Using calculator solve the following

a 22434 ÷ 3 = ______

b 4131 ÷ 243 = ______

c 469890 ÷ 230 = ______

d 3.026 ÷ 0.89 = ______

5 Solve the following

13x215

1170x537.5

= ________



8.031000

3500182.28

________

6 Solve the following





8x0.3

0.52)128)x(384(634

________

• Before starting the calculations be sure to

press the ‘ON’ key and confirm that ‘0’ is

shown on the display.

• Do not touch the inside portion of the calcu-

lator. Avoid hard knocks and unduly hard

pressing of the keys.

• Maintain and use the calculator in between

the two extreme temperatures of 0° and 40°

• Never use volatile fluids such as lacquer,

thinner, benzine while cleaning the unit.

• Take special care not to damage the unit by

bending or dropping.

• Do not carry the calculator in your hip pocket.







17.04).10.021)x(28(3.89

2.6)0.0512.2)x(842(389

_________

7 2a = 450 mm(major axis)

2b = 250mm(minor axis)

Perimeter of the ellipse

c = _____metre

Hint C =

)(2

ba 



8 ø = 782 mm

α = 136°

Area of the sector

A = ______





360

d x π

AHint

9 d = 1.25 metre

V = ______ dm

πr

VHint 

10 L = 1.2 metres

B = 0.6 metre

H = 0.5 metre

'r' of steel

= 7.85 kg/dm

m = ______ kg

(mass ‘m = V x r )

Workshop Calculation And Science : (NSQF) Exercise 1.2.08

1 Convert the following into improper fractions.

__________

1 

__________

4 

__________

3 

__________

5 

_________

3 

______

5 

__________

7 

__________

182 

2 Convert the following into mixed numbers.

__________



__________



__________



__________



__________



__________



__________



__________



3 Place the missing numbers.

__________



__________



__________



4 Simplify.

_________



__________



__________



__________



__________



__________



5 Multiply.

__________

x5 

__________2x



__________



__________3x

3 

__________

2 

__________

6x5 

Solving problems by using calculator Exercise 1.2.09

6 Divide

_________



________

6 

__________



__________4

3 

________ 

_________

38 

7 Place the missing numbers.

x_________



x________



x_________



x_________



x_________



x_________

3 

x_________



x_________



8 Add the followings:

_________



_________



_________



_________

6 

9 Subtract

_________



_________



10 Simplify

_________

2 

________8

2 

_________

3 

11 Express as improper fractions

12 Reduce to mixed number or whole number

163

144

13 Reduce to lowest terms

144

Workshop Calculation And Science : (NSQF) Exercise 1.2.09

14 Addition of decimals

a 4.56 + 32.075 + 256.6245 + 15.0358

b 462.492 + 725.526 + 309.345 + 626.602

15 Subtract the following decimals

a 612.5200 – 9.6479

b 573.9246 – 215.6000

c 968.325 – 16.482

d 5735.4273 – 364.2342

16 Add and subtract the following

a 56.725 + 48.258 – 32.564

b 16.45 + 124.56 + 62.7 - 3.243

17 Multiplication of decimals

a By 10,100,1000

i 3.754

ii 8.964 x 100

iii 2.3786 x 1000

iv 0.005 x 1000

b By whole numbers

i 8.4 x 7

ii 56.72 x 8

c By another decimal figure (use calculator)

i 15.64 x 7.68

ii 2.642 x 1.562

18 Divide the following

62.5

14.4

64.56

100

0.42

1000

48.356

25.5

19 Division

1.2

16.8

1.2

1.68

1.2

0.168

1.1

1.54

1.6

27.2

f 31.5 ÷ 10.5

g 1.54 ÷ 1.1

h 4.41 ÷ 2.1

20 Change the fraction into a decimal

21 Find the value

20.5 x 40 ÷ 10.25 + 18.50

A = 12.613 mm

X = __________mm.

B = __________mm.

Workshop Calculation And Science : (NSQF) Exercise 1.2.09

Square and square root Exercise 1.3.10

a basic number

2 exponent

radial sign indicating the square root.

square root of 'a' squared

radicand

Square number

The square of a number is the number multiplied by itself.

Basic number x basic number = Square number

a x a = a

4 x 4 = 4

= 16

Splitting up

A square area can be split up into sub-areas. The largest

square of 36 is made up of a large square 16, a small

square 4 and two rectangles 8 each.

Large square 4 x 4 = 16 a

Two rectangles 2 x 4 x 2 = 16 2ab

Small square 2 x 2 = 4 b

Sum of sub-areas = 36 = a

+ 2ab + b

bab2a 36 

Extracting the square root procedure

• Starting from the decimal point form groups of two

figures towards right and left. Indicate by a prime

symbol.

00.24,46

• Find the root of the first group, calculate the difference,

bring down the next group.

• Multiply the root by 2 and divide the partial radicand.

• Enter the number thus calculated in the divisor for the

multiplication.

If there is a remainder, repeat the procedure.

6 46,24

36 36

128 1024

1024

6846,24 

Basic number x basic number = Square

number basic number Square 

Example

The cross-sectional of a rivet is 3.46 cm

. Calculate the

diameter of the hole.

46,246

1024

1024128

Result: In order to find the square root, we split up the

square numbers.

Rivet cross-section is the hole cross-section.

To find d

Given that Area = 3.46 cm

Area = 0.785 x d

(formula)

3.46 cm

= d

x 0.785

21mm (or) cm 2.1 d

0.785

3.46

0.785

cm 3.46



Simple problems using calculator Exercise 1.3.11

2916 

57964 

.29448 

3.8456 

2 A = 2025 mm

a = mm

3 A = 176.715 mm

d = mm

4 A = 65031 mm

d = 140 mm

D = mm

5 I = 58 cm

b = 45 cm

= A

a = cm

6 A = 807.77 cm

d = 140 mm

D = mm

7 a x a = 543169 mm

a = mm

8 d : l = 1:1.5

A = 73.5 mm

d = mm

9 increase in area

A = 12.7%

A = 360 mm

d = mm

(d = diameter after the

increase in area)

Applications of pythagoras theorem and related problems Exercise 1.3.12

1 What is the side AC if AB = 15 cm, BC = 25 cm.

= AB

+ BC

= 15

+ 25

= 225 + 625 = 850

AC =

850 = 29.155 cm

2 What is the side BC if AB = 10 cm, AC = 30 cm.

= AB

+ BC

= 10

+ BC

900 = 100

+ BC

= 900 - 100 = 800

BC = 28.284 cm

3 What is the side AB if BC = 20 cm, AC = 35 cm.

= AB

+ BC

= AB

+ 20

1225 = AB + 400

= 1225 - 400 = 825

AB = 28.72 cm

4 What is the value of side BC if AB = 8 cm, AC = 24 cm.

= AB

+ BC

= 8

+ BC

576 = 64

+ BC

= 576 - 64 = 512

BC = 572 = 22.63 cm

5 What is the value side AC if AB = 6.45 cm, BC = 8.52

cm.

= AB

+ BC

= 6.45

+ 8.52

= 41.60

+ 72.59

= 114.19

AC = 114.19 = 10.69 cm

6 What is the value of side AB if BC = 3.26 cm, AC = 8.24

cm.

= AB

+ BC

8.24

= AB

+ 3.26

67.9 = AB + 10.63

= 67.9 - 10.63

= 57.27

AB = 57.27 = 7.57 cm

7 What is the value of side AB if AC = 12.5 cm, BC = 8.5

cm.

= AB

+ BC

12.5

= AB

+ 8.5

156.25 = AB + 72.25

= 156.25 - 72.25

= 84

AB = 84 = 9.17 cm

ASSIGNMENT

1 What is the value of side AB, if right angled triangled of

side AC = 12.5 cm and BC = 7.5 cm.

2 What is the value of side AC, if right angled triangled of

side AB = 6.5 cm and BC = 4.5 cm.

3 What is the value of side BC, if right angled triangled of

side AC = 14.5 cm and AB = 10.5 cm.

4 What is the value of side AC, if right angled triangled of

side AB = 7 cm and BC = 5 cm.

5 What is the value of side BC, if right angled triangled of

side AC = 13.25 cm and AB = 8.75 cm.

Ratio and proportion Exercise 1.4.13

Ratio

Introduction

It is the relation between two quantities of the same kind

and is expressed as a fraction.

Expression

a, b two quantities of the same kind.

or a:b or a



b or

a in b is the ratio.

Ratio is always reduced to the lowest terms.

Example

Proportion

It is the equality between the ratios, a : b is a ratio and c:

d is another ratio. Both ratios are equal. Then

a :b :: c : d or

Example

250 : 2000 :: 1 : 8

Proportion fundamentals

If then

• ad = bc

•

3:4::6:8 or

• 3 x 8 = 6 x 4

•

Ratio - relation of two quantities of the same kind.

Proportion - equality between two ratios.

Example

• A steel plate of 800 x 1400 mm is to be drawn to a scale

of 1:20. What will be the lengths in the Fig 1

The reduction ratio is .

B is reduced from 800 to 800 x = 40 mm.

L is reduced from 1400 x = 70 mm.

• Find the number of teeth of the larger gear in the gear

transmission shown in the Fig 2.

Speed ratio = 400 : 300

Teeth ratio = 24:T

ASSIGNMENT

: l

= 2:3

L = 2.75 metres

______ metres

= ______ metres

2 d: L of shaft = 2:7

d = 40 mm

L = ______ mm

3 D : L=1:10

L=150mm

D=______ mm

I = 140 mm

h = _____ mm

5 D : d = 1.75: 1

D = 35 mm

d = ______ mm

6 a:s = 5:1

s = 1.5mm

a =______ mm

7 A:B=9:12

B:C=8:10

Then A:B:C=___________

8 A:B=5:6

B:C=3:4

Then A:B:C= ______

9 A:55=9:11

A = __________

10 15:9.3=40:x

x = __________

Find the ratio of A:B:C

If A:B= 2:3 and B:C=4:5

A:B = 2:3

B:C = 4:5

A:B = 8 :12 (Ratio 2:3 multiply by 4)

B:C = 12:15 (Ratio 4:5 multiply by 3)

∴ A:B:C = 8:12:15

Workshop Calculation And Science : (NSQF) Exercise 1.4.13

Direct and indirect proportions Exercise 1.4.14

Proportion

Description

It is the equality between the ratios, a:b is a ratio and c:d

is another ratio. Both ratios are equal. Then

a : b::c : d or e.g. 250 : 2000::1 : 8

Rule of three

A three step calculation

statement

single

multiple.

Direct proportion

The more in one the more in the other - An increase in one

denomination produces an increase in the other. (Fig 1)

Example

4 turners earn 300 Rupees. How much will 6 Turners earn?

Statement

4 turners = 300 Rupees

Single

1 Turner = 75 Rupees

Multiple

6 Turners = 6 x 75 = 450 Rupees

Result - The more the more.

Indirect or inverse proportion

The more in one the lesser other - Increase in one quantity

will produce a decrease in the other. (Fig 2)

Example

Four turners finish a job in 300 hours. How much time will

6 turners take to do the same job?

Solution procedure in three steps:

Statement

4 turners taken = 300 hours

The time will reduce if 6 turners to do the same job.

Therefore this is inverse propotion.

Multiple fraction

6 Turners = 200 hours

Result - The more the less.

Problems involving both

Example

Two turners need three days to produce 20 pieces. How

long does it take for six turners to produce 30 such pieces?

Statement

2 turners, 20 piece = 3 days

6 turners, 30 pieces = how many days.

First step (Fig 3)

Statement 2 turners for 20 pieces = 3 days

1 turners for 20 pieces = 3 x 2 = 6 days

Multiple 6 turners for 20 pieces = = 1 day

Inverse proportion - The more the less.

Second step (Fig 4)

Compound and Inverse proportions

• Compound proportions

Example

If 5 Fitter take 21 days to complete overhauling of 6 vehicles

how long 7 Fitters will take to over haul 8 vehicles (Assume

time of overhauling each vehicle is constant)

In this both direct and indirect proportions are used.

• 1 Fitter will over haul 1 vehicle in days (shorter time).

• Quantities (No. of days) are taken in last as that is

the answer required in this case.

Ans: 7 Fitters will overhaul 8 vehicles in 20 days.

Inverse proportion

Some times proportions are taken inversely.

Examples

• If one water pump fills the fuel tank in 12 minutes, two

pumps will take half the time taken.

The time should not be doubled.

• If two pumps take 30 minutes to fill up a tank how long

will 6 similar pumps take this to fill this tank.

Ans: Time taken by 6 pumps =

minutes 10

230





Proportional parts in combustion equation

Introduction

Proportion of quantities form an important factor in the

combustion process of a fuel. The following happens during

the combustion process.

Fuel is a hydro carbon substance. The combustion air is

supplied from atmosphere and contains oxygen and nitro-

gen. Now the following chemical changes take place during

combustion of a fuel.

• Carbon burns with oxygen and forms Co and Co

(Carbon monoxide and there carbon dioxide.)

• Hydrogen burns with oxygen and becomes water (H

• Sulphur burns with oxygen and becomes sulphur

dioxide.

• Nitrogen is an inert gas and does not take part in

combustion.

Method of finding proportional parts in one lb of a

substance

To be found out now

• Proportion of oxygen and hydrogen in one lb/Kg of

water.

Statement 6 turners for 20 pieces = 1 day

Single 6 turners for 1 piece = days

Multiple 6 turners for 30 pieces =

x 30 = 1.5 days

Direct proportion - The more the more.

Solve the probÎem by first writing the statement

and proceed to single and then to the multiple

according to the type of proportion that is

involved.

Introduction

Proportional fundamentals, as applicable to motor vehicle

calculations are discussed below.

Simple Proportion

• Proportion

This is an equality between two ratios

Examples

• If one vehicle fleet uses 30 litres of petrol per day how

much petrol is used by 6. Vehicles operating under

similar condition.

One vehicle uses petrol = 30 litres per day.

Then six vehicles will use = 6 Times as much

= 6 x 30 = 180 litres/day.

• If 4 vehicles of a fleet use 120 gallons of petrol per day

how much petrol will be used by 12 vehicles operating

under the same condition.

4 vehicles use 120 gallons per day

1 Vehicle will use

120

= 30 gallons/day



12 vehicles will use

Both examples are called simple proportion

because only two quantities were used and the

day is common for both ratios.

Workshop Calculation And Science : (NSQF) Exercise 1.4.14

• Proportion of hydrogen and carbon in one lb/kg of fuel.

Examples

• The chemical formula for water is H

O. This means 2

atoms of hydrogen and one atom of oxygen combined

to make one molecule of water. If oxygen atom weighs

16 times as much as hydrogen find out the proportions

in one kg of water.

Solution

Parts by weight of water are as below

Oxygen = 16/2 = 8kg

Hydrogen = 1/1 = 1kg

Total = 8+1= 9kg

• A hydrocarbon fuel has formula C

. This shows one

molecule of fuel contain 6 atoms of carbon and 14

atoms of hydrogen. If the carbon atom weighs 12 times

as the hydrogen atom, find the proportionate parts of

hydrogen and carbon in one kg of fuel.

Solution

Parts of carbon by weight

= 6 x 12 = 72

Parts of hydrogen by weight = 14.

Total No. of parts = 72 + 14 = 86.

Weight of Carbon = 72/86 = 0.8372 kg

Weight of Hydrogen 14/86 = 0.1628 kg

Ratio and Proportion

Proportion of air quantity required for combustion

process

Mass of air required for complete combustion of fuel

depends on the following factors and is called Air - fuel

Ratio

• Carbon, Hydrogen, Sulphur are to burn with oxygen

in the combustion process.

• It has been found that the following quantities of air

(by wt) are required for this purpose to supply sufficient

quantity of oxygen.

• For complete combustion of 1kg of carbon =

Kgs

of oxygen

• For complete combustion of 1 kg of hydrogen = 8kgs

of oxygen

• For complete combustion of 1 kg of sulphur = 1 kg

of oxygen

• Formula for calculation of mass of air for

complete combustion.

Air contains 23% oxygen and 77% nitrogen

Mass of air = Mass of oxygen x for each constituent

air kg.of 4.35

100

1sulphur For

air 34.8kg.of

100

8hydrogen For

air 11.6kg.of

100

2Carbon For



Total 50.75 Kg

Hence 50.75 kg of air is to be supplied to the engine for

combustion of 1 kg of fuel.

As the combustion process is not even more

quantity of air than 50.7 kg is to be supplied to

the engine.

The calculations involved in the combustion

equations is beyond the scope of ITI students

as it involves chemistry and physics for comput-

ing the proportions of different elements.

Workshop Calculation And Science : (NSQF) Exercise 1.4.14

ASSIGNMENT

1 Length = 6.1 metre

Weight = 32 kgf

Weight of 1 metre of

the same rod =

_______ kgf

= 120 mm

= 720 mm

= 1200 rpm

= ______ rpm

= 42 T

= 96 rpm

= 224 rpm

= _______ T

D = 50 mm

H = 80 mm

h = 36 mm

d = ______ mm

5 If a mechanic assembles 8 machines in 3 days. How

long he will take to assemble 60 machines.

6 In an autoshop the grinding wheel makes 1000 rpm and

the driven pulley is 200 mm dia. If the driving pulley is

150 mm dia. Find out the rpm of the driving pulley.

7 In a gearing of a vehicle the following facts are found.

A 180 mm dia of gear meshes with 60mm dia gear. If the

bigger gear makes 60 rpm. What will be the rpm of

smaller gear.

8 A vehicular job is completed by 5 mechanics in 4 days.

If only 3 mechanics are available, in how many days the

work can be completed.

9 In a gearing arrangement of a vehicle a gear having 26

teeth is meshing with a gear of 52 teeth. The dia of 52

teeth gear is 200mm. Find out the diameter of 26 teeth

gear wheel.

10 If two water pumps take 45 minutes to fill up a tank how

long will 4 similar pumps will take to this tank.

11 In a belt-pulley drive the driving pulley is of 12 cm

diameter and rotates at 360 rpm. Find the rpm of driven

pulley whose dia meter is 20 cm diameter.

12 To overhaul a gear box, 12 mechanics are needed to

complete the work in 5 days. If only 7 mechanics are

available, how many days they will be able to complete

the overhauling work.

13 Express in simple ratios the following

6045 

Rs4.00 paise 40 

metres 4

20mm

d C100C4 

14 Air contains 24% oxygen and 78% nitrogen by mass

(weight). Calculate the quantity of air (mass of air)

required for complete combustion of unit mass fuel (The

main constitutents that take part in combustion proc-

ess are carbon, hydrogen and sulphur)

Note: Given the following data (Solve the problem)

a 1 kg of carbon requires kg of oxygen.

b 1 kg of hydrogen requires 8 kg of oxygen.

c 1 kg of sulphur requires 1 kg of oxygen.

15 A fuel is a hydro carbon substance of C

. This shows

each molecule of fuel contains 7 atoms of carbon and

14 atoms of hydrogen. If carbon atomic weight is 12

times greater than hydrogen atom, find out the propor-

tionate parts of hydrogen and carbon in one kg of fuel.

16 A vehicle worth Rs.20,000/- can be insured at a cost of

Rs.150/-. How much wired it cost to insure a vehicle

worth Rs.24000/- for one year and 3 month at the same

rate. (Compound proportion)

Workshop Calculation And Science : (NSQF) Exercise 1.4.14

Percentage Exercise 1.5.15

Ex.

0.16

100



In decimal form, it is 0.16. Percentage calculation also

involves rule of three. The statement (the given data), for

unit, and then to multiple which is for calculating the

answer.

Example

The amount of total raw sheet metal to make a door was 3.6

metre

and wastage was 0.18 metre

. Calculate the % of

wastage.

Solution procedure in three steps.

Statement:

Area of door (A) =3.6 m

= 100 %.

Wastage = 0.18 m

Single:

3.6

100

1 m

Analyse the given data and proceed to arrive at

the answer through the unit.

Example

A fitter receives a take-home salary of 984.50 rupees.

If the deduction amounts to 24%, what is his total salary?

(Fig 3)

Total pay 100%

Percentage

Percentage is a kind of fraction whose denominator is

always 100. The symbol for percent is %, written after the

number 16%.

Multiple: for 0.18 m

3.6

100

x 0.18.

Wastage = 5%.

Conclusion

The three steps involved are,

step one : describe the situation (availability)

step two : decide for unit

step three : proceed for the multiple.

Deduction 24%

Take home salary 76%

If the take home pay is Rs.76, his salary is 100.

For 1% it is

For Rs.984.50, it is

x 984.50.

For 100% it is

1295.39X100

984.50



100% i.e. gross pay = Rs.1295.40.

Example 1

75 litres of oil is taken out from a oil barrel of 200 litres

capacity. Find out the percentage taken in this.

Solution

% of oil taken = Oil taken out (litres) / Capacity of Barrel

(litres) x 100

%37 100 x

200



Example 2

A spare part is sold with 15%. Profit to a customer, to a

price of Rs.15000/-. Find out the following (a) What is the

purchase price (b) What is the profit.

Solution: CP =

CP = cost price

SP = sale price

SP=CP+15%of CP

15000= +

Profit = SP-CP = 15000-13043.47 = 1956.53

Purchase price = Rs.13,043/,Profit = Rs. 1957

Example 3

Out of 80000 cars, which were tested on road, only 16000

cars had no fault. What is the percentage in this accep-

tance.

Example 4

The price of a motor cycle dropped to 92% of original price

and now sold at Rs.18000/- What was the original price.

Solution

Present price of Motor cycle Rs.18000

This is the value of 92% of original price

Original Price = 18000 x

= Rs.19565

Example 5

A Motor vehicle uses 100 litres of Petrol per day when

travelling at 30 kmph. After top overhauling the consump-

tion falls to 90 litres per day. Calculate percentage of

saving.

Solution

Percentage of saving = Decrease in consumption/Original

consumption x 100

= (100 – 90) x100

= 10% Saving in fuel.

ASSIGNMENT

1 a = 400mm (side of

square)

d = 400 mm

Wastage = ______ %.

2 d = 26mm

'a' depth of u/cut =

2.4mm

reduction of area at

cross-section

= ______ %

3 Percentage of increase

= 36%

Value of increase

= 611.2 N/mm

Original tensile strength

= ______ N/mm

4 Copper in alloy = 27 kg

Zinc in alloy = 18 kg

% of Copper

= _______ %

% of Zinc = _______%.

5 Weight of alloy = 140

Kgf

Weight of Sn 40%

Pb = ______ Kgf

Sn = ______ Kgf.

6 Shaded portion

= ______ %.

7 Compression length =

______ %.

8 d = 360 mm

a = 0.707 x d

Wastage = ______ %.

Workshop Calculation And Science : (NSQF) Exercise 1.5.15

Workshop Calculation And Science : (NSQF) Exercise 1.5.16

9 Cu = 36 Kg

Zn = 24 Kg

Cu = ______ %

Zn = ______ %

10 Cu = 42.3 Kg

Sn = 2.7 Kg

Cu ______ %

Sn = ______ %.

11 What is the selling price, If a trader buys a spare part

for Rs.195/- This is 65% of selling price.

12 What is the purchase price if 25% profit is added to it,

If a Motor cycle tyre is sold for Rs.300/-.

13 How many m

of elements of air present in 120 m

of air,

If the composition of Air is 23% of Oxygen and 77% of

Nitrogen.

14 How many kg of each of these elements are found. If an

Engine bearing made of alloy of 40 kg consists of the

following constitutents.

a) Copper (Cu) - 86%

b) Tin (Sn) - 10%

c) Zinc (Zn) - 4%

15 How much weight of these elements are found to exist.

If a solder consists of 35%. Tin and 65% Lead. In a

solder of 40 kg.

16 Find out the following:

1 Average consumption per journey.

2 Average consumption per mile.

3 Using the average consumption express maximum

consumption as a percentage of the average correct

to two decimal places. If the total petrol consump-

tion of car on 4 different journeys each of 200 miles

are found to be as 6.65, 7.5, 6.85,7.05 gallons

respectively.

17 In a Transport workshop, the following expenditure was

found to be occuring on the capital income.

1 40% income spent on tyres

2 30% income spent on fuel and lubricants

3 10% income spent on spare parts

If the month end saving comes to Rs.2000/- what is the

total income?

18 What is the final weight of the machined job if a casting

weighs 80kg. During preliminary machining weight is

reduced by 4% and final machining by 5%.

19 What is the weight of zinc, copper and tin, casting

weight is 25kg. If a casting has 35% zinc, 40% copper

and 25% tin.

20 What is the total weight of a solder, if solder consists

of 35% tin 65% lead, and tin consists 14 grams.

21 What is the total annual income of the salesman, if a

salesman gets a monthly pay of Rs.1000 and a com-

mission of 2.5% on his sale. In one year sale amount

is Rs.60,000.

22 What is the total income of a man, if he spends 15% of

his income on agriculture, 21% on family, 24% on

education of children and he saves Rs.360.

23 What is the percentage of his savings, if a person’s

monthly salary is Rs.450 and saves Rs.90 every months.

Changing percentage to decimal and fraction Exercise 1.5.16

Conversion of Fraction into Percentage

1 Convert

into percentage.

Solution:

x 100

= 50%

2 Convert

into percentage

Solution:

x 100 =

100

= 9.01%

Convert the following fraction into percentage.

Conversion of Percentage into Fraction

1 Convert 24% into fraction.

Solution:

100

2 Convert

% into fraction.

Soluction:

100



Convert the following percentage into fraction

1 15%

3 80%

4 125%

Conversion of Decimal Fraction into Percentage

1 Convert 0.35 into percentage.

Soluction: 0.35 x 100

= 35%

2 Convert 0.375 into percentage.

Soluction: 0.375 x 100

= 37.5%

Convert the following Decimal Fraction into Percentage

1 0.2

2 0.004

3 0.875

4 0.052

Conversion of Percentage into Decimal fraction

1 Convert 30% into decimal fraction.

Soluction:

100

= 0.3

2 Convert

into decimal fraction.

Soluction:

100



= 0.333

Convert the following percentage into decimal fraction

1 15%

27%

4 90%

Algebraic symbols and fundamentals Exercise 1.6.17

Introduction

Algebra is a form of mathematics in which letters may be

used in place of unknown. In this mathematics numbers

are also used in addition to the letters and the value of

number depends upon its place. For example in 3x and x

the place of x is different. In 3x=3 is multiplied with x,

whereas in x

- 3 is an Index of x.

Positive and negative numbers

Positive numbers have + sign in front of them, and negative

numbers have – sign in front of them. The same applies to

letters also.

Example +x – y

+8 or simply 8 positive number.

– 8 negative number.

Addition and subtraction

Two positive numbers are added, by adding their absolute

magnitude and prefix the plus sign.

To add two negative numbers, add their absolute magni-

tude and prefix the minus sign.

To add a positive and a negative number, obtain the

difference of their absolute magnitudes and prefix the sign

of the number having the greater magnitude.

+7 + 22 = +29

–8 – 34 = – 42

–27 + 19 = –8

44 + (–18) = +26

37 + (–52) = –15

Multiplication of positive and negative numbers

The product of two numbers having like signs is positive and

the product of two numbers with unlike signs is negative.

Note that, where both the numbers are negative, their

product is positive.

Ex. –20 x –3 = 60

5 x 8 = 40

4 x –13 = –52

–5 x 12 = –60

Division

The number that is divided is the dividend, the number by

which we are dividing is the divisor and the answer is the

quotient. If the signs of the dividend and the divisor are the

same then the quotient will have a + sign. If they are unlike

then the quotient will have a negative sign.

28+

=+7

4–

56+

= –14

72–

=–8

6–

96–

= +16

When an expression contains addition,

subtraction, multiplication and division,

perform the multiplication and division

operations first and then do the addition and

subtraction.

Example

12 x 8 – 6 + 4 x 12 = 96 – 6 + 48 = 138

102 ¸ 6 – 6 x 2 + 3 = 17 – 12 + 3 = 8

Parantheses and grouping symbols

( ) Brackets

{ } Braces

7 + (6–2) = 7 + 4 = 11

6 x (8–5) = 6 x 3 = 18

Parentheses

These are symbols that indicate that certain addition and

subtraction operations should precede multiplication and

division. They indicate that the operations within them

should be carried out completely before the remaining

operations are performed. After completing the grouping,

the symbols may be removed.

In an expression where grouping symbols immediately

preceded or followed by a number but with the signs of

operation omitted, it is understood, that multiplication

should be performed.

Grouping symbols are used when subtraction and multipli-

cation of negative numbers is done.

To remove grouping symbols which are preceded by

negative signs, the signs of all terms inside the grouping

symbols must be changed (from plus to minus and minus

to plus).

Parentheses which are preceded by a plus sign may be

removed without changing the signs of the terms within the

parentheses.

When one set of grouping symbols is included within

another set, remove the innermost set first.

When several terms connected by + or – signs contain a

common quantity, this common quantity may be placed in

front of a parentheses.

8 + 6(4–1) = 8 + 6 x 3 = 26

(6+2)(9–5) = 8 x 4 = 32

Plus 4 less negative 7 is written as 4 – (–7).

Plus 4 times negative 7 is written as 4(–7).

4 –(–7) = 4 + 7 = 11

8 –(7–4) = 8 – 3 = 5

3 +(–8) = 3 – 8 = –5

7 +(4 – 19) = 7 + (–15) = 7 – 15 = – 8

3{40 +(7 + 5)(8–2)} = 3 {40 + 12 x 6} = 3 x 112 = 336.

8x + 12 - quantity 4 may be factored out giving the

expression 8x + 12 as 4(2x + 3).

The innermost set in a grouping symbols of an expres-

sion is to be simplified first.

Algebraic symbols and simple equations

Algebraic symbol

An unknown numerical value of a quantity is represented by

a letter which is the algebraic symbol.

Factor

A factor is any one of the numbers or letters or groups which

when multiplied together give the expression. Factors of 12

are 4 and 3 or 6 and 2 or 12 and 1.

8x + 12 is the expression and this may be written as

4(2x + 3), 4 and (2x + 3) are the factors.

Algebraic terms

If an expression contains two or more parts separated by

either + or –, each part is known as the term.

y – 5 x is the expression. y and – 5 x are the terms.

The sign must precede the term.

Coefficient

When an expression is formed into factors whose product

is the expression, then each factor is the coefficient of the

remaining factors.

48x = 4 x 12 x x.

4 is the coefficient of 12 x . x is the coefficient of 48.

Equation

It is a statement of equality between numbers or numbers

and algebraic symbols.

12 = 6 x 2, 13 + 5 = 18.

2x + 9 = 5, y – 7 = 4y + 5.

Simple equation

Equations involving algebraic symbols to the first power are

simple equations.

2x + 4 = 10. 4x + 12 = 14.

Addition and subtraction

Quantities with algebraic symbols are added or subtracted

by considering those terms involving same symbols and

powers.

10x + 14 – 7y

– 11a + 2x – 4 – 3y

– 4a + 8

= 10x + 2x – 7y

– 3y

– 11a – 4a + 14 – 4 + 8

= 12x – 10y

– 15a + 18

2x = 10, 2x + 6 = 10 + 6

y + 12 = 20, y + 12 – 8 = 20 – 8

x + 10 = 12,

x + 10 – 10 = 12 – 10

3x = 6, 2 x 3x = 2 x 6, 6 x = 12

5y = 20,

The same number may be added or subtracted to both

members of an equation without changing its equality.

Each member of an equation may be multiplied or divided

by the same number or symbol without changing its

equality.

The equality of an equation is not altered when the numbers

or symbols are added or subtracted from both sides.

Multiplication and division by the same numbers or sym-

bols on both sides also will not affect the equality.

Transposition of the terms of the equations

= equals to

+ plus

– minus

x multiply

÷ divided by

Concept of equality (Fig 1)

Workshop Calculation And Science : (NSQF) Exercise 1.6.17

An equation can be compared to a pair of scales which

always remain in equilibrium. The two sides of the equation

can fully be transposed. 9 = 5 + x may also be written as

5 + x = 9.

We must always perform the same operation on both sides

of the equation to keep the equilibrium. Add or subtract the

same amount from both sides. 5 + x = 9 By adding 3 on

both sides, the equation becomes 5 + x + 3 = 9 + 3 or x

+ 8 = 12.

5 + x = 9 Subtract 5 from both sides then 5 + x – 5 = 9– 5.

x = 4.

5 is transposed from left side to the right side by changing

its sign from + to –.

= 20. Multiply both sides by 4. Then

x 4 = 20 x 4.

x = 80,

5x = 25.

Divide both sides by 5 then

x = 5.

When transposing numbers or letter symbols from one

side to the other side multiplication becomes division and

the division becomes multiplication.

The equality of an equation remains unchanged

when both sides of the equation are treated in

the same way. When transposing from one

side to the other side,

a plus quantity becomes minus quantity.

a minus quantity becomes a plus quantity

a multiplication becomes a division

a division becomes a multiplication.

To solve simple equations isolate the unknown

quantity which is to be found on the left side of

the equation.

Example

• Solve for x if 4x = 3(35 – x )

4x = 105 – 3x (brackets removed)

4x + 3x = 105 (By transposing –3x on the right side

to the left side)

7x = 105

x = 15 (dividing both sides by 7)

Workshop Calculation And Science : (NSQF) Exercise 1.6.17

Addition,subtraction,multiplication and division of algebra Exercise 1.6.18

Addition

Example 1

Add 2x, 3x, 7x

Solution: Since this is only addition, put ‘+’ symbol &

each bracket

= (+2x) + (+3x) + (+7x) = 12x

Add

-3a, -2a, -5a

= (-3a) + (-2a) + (-5a) = -10a

Example 2

Add 10xy, -8x

, -7xy, -7x

= (10xy) + (-8x

) + (-7xy) + (-7x

)

= (10xy) + (-7xy) + (-8x

) + (-7x

)

= (3xy) + (-15x

)

Example 3

Add



a4a2a













































Subtraction

Example 1

Subtraction 8m from 2m

Solution : (2m) - (8m)

To be subtracted from +8m after changed + 2m

the symbol - 8m

after added - 6m

Example 2

Subtraction (-5x + 3y) from (7x + 8y)

Solution: (7x + 8y) - (-5x + 3y)

+ 7x + 8y

+ 5x - 3y

+ 12x + 5y

(Multiply all the terms which should be subtracted with

‘-’ symbol subsequently add or subtract the similar terms

as atated in the steps)

Solving of Simple Equations

Rules Related to signs in Algebra

1 Multiplication Rules

(i) + x + = +

(ii) - x - = +

(iii) + x - = -

(iv) - x + = -

2 Division Rules

(i)

(ii)

−

(iii)

−=

−

(iv)

−=

−

Examples

1) a + 4 = 7

a = 7 - 4

a = 3

2) M - 5 = 3

m = 3 + 5

m = 8

3) 7P = 17

P =

= 9

x = 9 x 6

x = 54

5) 7x + 3 = 15 + x

7x -x = 15 - 3

6x = 12

x =

= 2

Y = 0 x

y = 0

2XM

68M

86M

−=

36x

6 x 30

6 x 305x

3x 2x

144

x36x

2x -5x -4x 6x

=−−+

+=−+

30- a

8 x 0 30 a

30 a

30 4a - 5a

=+−

10)

Multiplication

While dealing with multiplication, first of all multiply the

symbols, subsequently coefficient, afterwards letters.

Example 1

Multiply (2x-5) x (x-1)

(2x-5) (x-1)

= 2x (x-1) - 5(x-1)

= 2x

- 2x -5x +5

= 2x

- 7x + 5

Example 2

Multiply (4x

+ 3x -12) by (2x +3)

(4x

+ 3x -12) (2x +3)

= 2x(4x

+ 3x -12) + 3 (4x

+ 3x -12)

= 8x

+6x

- 24x + 12x

+ 9x - 36

= 8x

+6x

+ 12x

- 24x

+ 9x - 36

= 8x

+18x

- 15x -36

Division

Division is nothing but reverse operation of multiplication.

Example 1

Division (x

+ 11x + 30) by (x + 6)

+ 11x + 30) ÷ (x + 6)

+ 11x + 30 ÷ x + 6 = x + 5

Workshop Calculation And Science : (NSQF) Exercise 1.6.18

Addition

1 Add 7y, 8y, 5y , 3y

2 Add 5x, 6x, 7y - 9y, 10x

3 Add 9a

, -7m, 5, -14a

, 15m, -6

4 Add

Substraction

1 Subtract 7x from 9x

2 Subtract 8a from 3a

3 (-5 -3b) - (3a + 5b)

4 Subtract (5m - 6n) from (3m + 5n)

Multiplication

1 Multiply 6a x (- 2ab)

2 Multiply 3x (4x - 2y + 5)

3 Multiply (3x + 2) (3x - 2)

Division

525x 

15a60ab 

3 77y14x 

Assignment

Workshop Calculation And Science : (NSQF) Exercise 1.6.18

Metal:

Metal is a mineral used in all types of engineering works

such as machineries, bridges, aero planes etc . so we

must have basic knowledge about the metals.

Understanding the physical and mechanical properties of

metals has become increasingly important for a machinist

since he has to make various components to meet the

designed service requirements against factors, such as

the raise of temperature, tensile, compressive and impact

loads etc. A knowledge of different properties of materials

will help him to do his job successfully. If proper material/

metal is not used it may cause fracture or other forms of

failures, and endanger the life of the component when it is

put into function.

Fig 1 shows the way in which the metals get deformed

when acted upon by the same load.

Note the difference in the amount of deformation.

Physical properties of metals

– Colour

– Weight/specific gravity

– Structure

– Conductivity

– Magnetic property

– Fusibility

Colour

Different metals have different colours. For example,

copper is distinctive red colour. Mild steel is blue/black

sheen.

Weight

Metals may be distinguished, based on their weights for

given volume. Metals like aluminium lighter weigh (Specific

gravity 2.7) and metals like lead have a higher weight.

(Specific gravity 11.34)

Structure (Figs2 and 3)

Generally metals can also be differentiated by their internal

structures while seeing the cross-section of the bar through

a microscope. Metals like wrought iron and aluminium

have a fibrous structure and metals like cast Iron and

bronze have a granular structure.

Conductivity (Figs 4 and 5)

Thermal conductivity and electrical conductivity are the

measures of ability of a material to conduct heat and

electricity. Conductivity will vary from metal to metal.

Copper and aluminium are good conductors of heat and

electricity.

Magnetic property

A metal is said to possess a magnetic property if it is

attracted by a magnet.

Almost all ferrous metals, except some types of stainless

steel, can be attracted by a magnet, and all non-ferrous

metals and their alloys are not attracted by a magnet.

Physical and mechanical properties of metals Exercise 1.7.19

Fusibility (Fig 6)

It is the property possessed by a metal by virtue of which

it melts when heat is applied. Many materials are subject

to transformation in the shape (i.e) from solid to liquid at

different temperatures. Lead has a low melting temperature

while steel melts at a high temperature.

Tin melts at 232°C.

Tungsten melts at 3370°C.

Mechanical properties

– Ductility

– Malleability

– Hardness

– Brittleness

– Toughness

– Tenacity

– Elasticity

Ductility (Fig 7)

A metal is said to be ductile when it can be drawn out into

wires under tension without rupture. Wire drawing depends

upon the ductility of a metal. A ductile metal must be both

strong and plastic. Copper and aluminium are good

examples of ductile metals.

Malleability (Figs 8 and 9)

Malleability is that property of a metal by which it can be

extended in any direction by hammering, rolling etc.

without causing rupture. Lead is an example of a malleable

metal.

Hardness (Fig 10)

Hardness is a measure of a metal's ability to withstand

scratching, wear and abrasion, indentation by harder

bodies. The hardness of a metal is tested by marking by

a file etc.

Workshop Calculation And Science : (NSQF) Exercise 1.7.19

Brittleness (Fig 11)

Brittleness is that property of a metal which permits no

permanent distortion before breaking. Cast iron is an

example of a brittle metal which will break rather than bend

under shock or impact.

Toughness (Fig 12)

Toughness is the property of a metal to withstand shock or

impact. Toughness is the property opposite to brittleness.

Wrought iron is an example of a tough metal.

Tenacity

The tenacity of a metal is its ability to resist the effect of

tensile forces without rupturing. Mild steel, Wrought Iron

and copper are some examples of tenacious metals.

Elasticity

Elasticity of a metal is its power of returning to its original

shape after the applied force is released. Properly heat-

treated spring is a good example for elasticity.

Workshop Calculation And Science : (NSQF) Exercise 1.7.19

Introduction of iron and cast iron Exercise 1.7.20

Ferrous Metals

Metals which contain iron as a major content are called

ferrous metals. Ferrous metals of different properties are

used for various purposes.

Introduction of Iron, Cast Iron, wrought Iron and steel

The ferrous metals and alloys used commonly are:

– Pig-iron

– Cast Iron

– Wrought Iron

– Steels and Alloy steels

Different processes are used to produce iron and steel.

Pig-iron (Manufacturing process)

Pig-iron is obtained by the chemical reduction of iron ore.

This process of reduction of the iron ore to Pig-iron is known

as SMELTING.

The main raw materials required for producing Pig-iron are:

– Iron ore

– Coke

– Flux

Iron ore

The chief iron ores used are:

– magnetite

– hematite

– limonite

– carbonite.

These ores contain iron in different proportions and are

naturally available.

Coke

Coke is the fuel used to give the necessary heat to carry

on the reducing action. The carbon from the coke in the

form of carbon monoxide combines with the iron ore to

reduce it to iron.

Flux

This is the mineral substance charged into a blast furnace

to lower the melting point of the ore, and it combines with

the non-metallic portion of the ore to form a molten slag.

Limestone is the most commonly used flux in the blast

furnace.

Properties and use of Pig-iron

Pig-iron is, therefore, refined and remelted and used to

produce other varieties of iron and steel.

Cast Iron (Manufacturing process)

The pig-iron which is tapped from the blast furnace is the

crude form of raw material for the cupola, and should be

further refined for making castings. This refining is carried

out in the cupola furnace which is a small form of a blast

furnace.

Generally cupolas are not worked continuously like blast

furnaces but are run only as and when required.

Cast Iron (Types)

Cast iron is an alloy of iron, carbon and silicon. The carbon

content ranges from 2 to 4%.

Types of cast iron

The following are the types of cast iron.

– Grey cast iron

– White cast iron

– Malleable cast iron

– Nodular cast iron

Grey cast iron

This is widely used for the casting of machinery parts and

can be machined easily.

Machine base, tables, slideways are made of cast iron

because it is dimensionally stable after a period of aging.

Because of its graphite content, cast iron provides an

excellent bearing and sliding surface.

The melting point is lower than that of steel and as grey cast

iron possesses good fluidity, intricate casting can be

made.

Grey cast iron is widely used for machine tools because of

its ability to reduce vibration and minimize tool chatter.

Grey cast iron, when not alloyed, is quite brittle and has

relatively low tensile strength. Due to this reason it is not

used for making components subjected to high stress or

impact loads.

Grey cast iron is often alloyed with nickel, chromium,

vanadium or copper to make it tough.

Grey cast iron is weldable but the base metal needs

preheating.

White cast iron

This is very hard and is very difficult to machine, and for this

reason, it is used in components which should be abrasion-

resistant.

White cast iron is produced by lowering the silicon content

and by rapid cooling. When cooled in this manner, it is

called chilled cast iron.

White cast iron cannot be welded.

Malleable cast iron

Malleable cast iron has increased ductility, tensile strength

and toughness when compared with grey cast iron.

Malleable cast iron is produced from white cast iron by a

prolonged heat-treatment process lasting for about 30

hours.

Nodular cast iron

This is very similar to malleable cast iron. But this is

produced without any heat treatment. Nodular cast iron is

also known as: Nodular Iron - Ductile Iron - Spheroidal

Graphite Iron

This has good machinability, castability, resistance to

wear, low melting point and hardness.

Mealleable and nodular castings are used for machine

parts where there is a higher tensile stress and moderate

impact loading. These castings are less expensive and are

an alternative to steel castings.

Wrought Iron (Manufacturing process) (Fig 1)

Wrought iron is the purest form of iron. The analysis of

Wrought iron shows as much as 99.9% of iron. (Fig 1)

When heated, wrought iron does not melt, but only be-

comes pasty and in this form it can be forged to any shape.

Modern methods used to produce wrought iron in large

quantities are the

– puddling process

– aston or Byers process

Steel

This is pure iron. Carbon content is more. Due to

excessive carbon it is harder and tougher. Carbon content

is from 0.15 to 1.5%. Besides there are other impurities

like sulphur, phosphorous etc. Are there which cannot be

separated. This is hardened and tempered by heating it to

a definite temperature and cooling it in oil or water.

The following methods are adopted for making different

types of steel:

1 Cementation process 2 Crucible process

3 Bassemer process 4 Open hearth process

5 Electro thermo process 6 High frequency process.

11.13 Main types of steel

Mainly steel is the following two types:

1 Plain steel

2 Alloy steel

1 Plain steel. In this carbon and iron are mixed. Accoring

to the percentage of carbon. plain steel is the following

types:

A Low carbon steel

B Medium carbon steel

C High carbon steel

A Low carbon steel: It is also called mild steel. In

this. the percentage of carbon is from 0.15%to0.25%.

Due to less quantity of carbon is sufficiently soft and

tolerates the strain. It can be put in different shapes

through forging and rolling. This is not very hard or

strong. This cannot be hardened or tempered by

ordinary methods. Nuts, bolts, rivets, sheets,

wires, T-iron and angle iron etc. Are made out of it.

B Medium carbon steel: The carbon content is from

0.25% to 0.5%. Due to excess of carbon, it is harder

and tougher than mild steel. The tenacity is more.

This is used to hardened or tempered. Various

things are made by forging and rolling. This is used

for making high tensile tubes, wires, agricultural

implements, connecting rods, cam shafts, span-

ners, pulleys etc.

C High carbon steel: It has carbon content from

0.5% to 1.5%. It is very hard and wears least. This

can be hardened by heat treatment. This can

neither be cast nor rolled. This is very hard and

tough. It acquires permanent magnetic properties.

This is used for making pointed tools, springs,

pumps, files, cutleries, cold chisels press die etc.

2 Alloy Steel

When the steel is mixed with other metals like vinoleum,

manganese tungsten etc. It is called an alloy steel. Alloy

steel has properties of its ingredients.

Types of Alloy Steel

Alloy steel is mainly of two types:

A Low alloy steel

B High alloy steel

A Low Alloy steel: Besides carbon other metals are

in lesser quantity. Its tensile strength is more. The

welding can work on it. This can also be hardened

and tempered. It is used in manufacturing various

parts of an aeroplane and cam shaft etc.

B High Alloy Steel: Besides carbon it has a high

percentage of the metals higher than low steel alloy.

This is following types:

Workshop Calculation And Science : (NSQF) Exercise 1.7.20

a High Speed Steel: It is also called high tungsten

alloy steel because it has more quantity of tungsten.

According to the quantity of tungsten it is classified

in three types:

1 Tungsten 22%, Chromium4%, Vanadium 1%

2 Tungsten 18%, Chromium 4%, Vanadium 1%

3 Tungsten 14%, Chromium 4%, Vanadium 1%

Cutting tools are made out of it because it is very

hard but becomes soft at low critical temperature.

This temperature is raised out of cutting process of

tool, then the cutting tool becomes useless and is

unfit for work. But due to high percentage of

tungsten it keeps working upto high temperature. It

is used for cutting tools, drilling, cutter, reamers,

hacksaw blades etc.

b Nickel Steel: In this 0.3% carbon and 0.25 to

0.35% nickel is pressent. Due to nickel its tensile

strength, elastic limit and hardness is increased. It

does not catch rust. Its cutting resistance in-

creases 6 times more than plain carbon and steel

due to 0.35% nickel present in it. This is used for

making rivets, pipes, axle shafting, parts of buses

and aeroplanes. If 5% of cobalt is mixed with 30-

35% nickel, it becomes invar steel. It is mainly used

for making precious instruments.

c Vanadium Steel: It contains 1.5% carbonm 12.5%

tungsten, 4.5% chromium, 5% vanadium and 5%

cobalt. Its elastic limit, tensile strength and ductility

is more. It has strength to bear sharp jerks. It is

mainly used to manufacture of tools.

d Manganese Steel: It is also called special high

alloy steel. It contains 1.6 to 1.9% of manganese

and 0.4 to 0.5% carbon. It is hard and less wear. It

is not affected by magnet. It is used in grinders and

rail points etc.

e Stainless Steel: Along with iron it contains 0.2 to

90.6% carbnon, 12 to 18% chromium, 8% nickel

and 2% molybdenum. It is used for making knives,

scissors, utensils, parts of aeroplane, wires, pipes

and gears etc.

Properties of stainless steel:

1 Higher corrosion resistance

2 Higher cryogenis toughness

3 Higher work hardening rate

4 Higher hot strength

5 Higher ductility

6 Higher strength and hardness

7 More attractive appearance

8 Lower maintenance

f Silicon Steel: It contains 14% of silicon. Its uses

are multiferrous according to the percentage of

silicon. 0.5% to 1% silicon, 0.7 to 0.95% manganese

mixture is used for construction work. 2.5 to 4%

silicon content mixture is used for manufacturing

electric motors, generators, laminations of

transformers. In chemical industries 14% silicon

content mixture is used.

g Cobalt Steel: High carbon steel contains 5 to 35%

cobalt. Toughness and tenacity is high. It has

magnetic property therefore used to made perma-

nent magnets.

Workshop Calculation And Science : (NSQF) Exercise 1.7.20

Types of ferrous and non ferrous Exercise 1.7.21

Ferrous and Non ferrous alloys

Alloying metals and ferrous alloys

An alloy is formed by mixing two or more metals together

by melting.

For ferrous metals and alloys, iron is the main constituent

metal. Depending on the type and percentage of the

alloying metal added, the property of the alloy steel will

vary.

Metals commonly used for making alloy steels

Nickel (Ni)

This is a hard metal and is resistant to many types of

corrotion rust.

It is used in industrial applications like nickel, cadmium

batteries, boiler tubes, valves of internal combustion en-

gines), engine spark plugs etc. The melting point of nickel

is 1450°C. Nickel can be magnetised. In the manufacture

of permanent magnets a special nickel steel alloy is used.

Nickel is also used for electroplating. Invar steel contains

about 36% nickel. It is tough and corrosion resistant.

Precision instruments are made of Invar steel because it

has the least coefficient of expansion.

Nickel-steel alloys are available containing nickel from 2%

to 50%.

Chromium (Cr)

Chromium, when added to steel, improves the corrosion

resis-tance, toughness and hardenability of steel. Chro-

mium steels are available which may contain chromium

up to 30%.

Chromium, nickel, tungsten and molybdenum are alloyed

for making automobile components and cutting tools.

Chromium is also used for electroplating components.

Cylinder liners are chrome-plated inside so as to have wear

resistance properties. Stainless steel contains about 13%

chromium. Chromium-nickel steel is used for bearings.

Chrome-vanadium steel is used for making hand tools like

spanners and wrenches.

Manganese (Mn)

Addition of manganese to steel increases hardness and

strength but decreases the cooling rate.

Manganese steel can be used to harden the outer surface

for providing a wear resisting surface with a tough core.

Manganese steel containing about 14% manganese is

used for making agricultural equipment like ploughs and

blades.

Silicon (Si)

Addition of silicon for alloying with steel improves resis-

tance to high temperature oxidation.

This also improves elasticity, and resistance against

corrosion. Silicon alloyed steels are used in manufacturing

springs and certain types of steel, due to its resistance to

corrosion. Cast iron contains silicon about 2.5%. It helps

in the formation of free graphite which promotes the

machine-ability of cast iron.

Tungsten (W)

The melting temperature of tungsten is 3380° C. This can

be drawn into thin wires.

Due to this reason it is used to make filaments of electric

lamps.

Tungsten is used as an alloying metal for the production of

high speed cutting tools. High speed steel is an alloy of

18% tungsten, 4% chromium and 1% vanadium.

Stellite is an alloy of 30% chromium, 20% tungsten, 1 to

4% carbon and the balance cobalt.

Vanadium (Va)

This improves the toughness of steel. Vanadium steel is

used in the manufacture of gears, tools etc. Vanadium

helps in providing a fine grain structure in tool steels.

Chrome-vanadium steel contains 0.5% to 1.5% chromium,

0.15% to 0.3% vanadium, 0.13% to 1.10% carbon.

This alloy has high tensile strength, elastic limit and

ductility. It is used in the manufacture of springs, gears,

shafts and drop forged components.

Vanadium high speed steel contains 0.70% carbon and

about 10% vanadium. This is considered as a superior high

speed steel.

Cobalt (Co)

The melting point of cobalt is 1495°C. This can retain

magnetic properties and wear- resistance at very high

temperatures. Cobalt is used in the manufacture of

magnets, ball bearings, cutting tools etc. Cobalt high

speed steel (sometimes known as super H.S.S.) contains

about 5 to 8% cobalt. This has better hardness and wear

resistance properties than the 18% tungsten H.S.S.

Molybdenum (Mo)

The melting point of molybdenum is 2620°C. This gives

high resistance against softening when heated. Molybde-

num high speed steel contains 6% of molybdenum, 6%

tungsten, 4% chromium and 2% vanadium. This high

speed steel is very tough and has good cutting ability.

Cadmium (cd)

The melting point of cadmium is 320°C. This is used for

coating steel components.

Alloying Metals and Non Ferrous Alloys

Non-ferrous Metals And Alloys

Copper and its alloys

Metals without iron are called non-ferrous metals. Eg.

Copper, Aluminium, Zinc, Lead and Tin.

Copper

This is extracted from its ores ‘MALACHITE’ which con-

tains about 55% copper and ‘PYRITES’ which contains

about 32% copper.

Properties

Reddish in colour. Copper is easily distinguishable be-

cause of its colour.

The structure when fractured is granular, but when forged

or rolled it is fibrous.

It is very malleable and ductile and can be made into sheets

or wires.

It is a good conductor of electricity. Copper is extensively

used as electrical cables and parts of electrical apparatus

which conduct electric current.

Copper is a good conductor of heat and also highly

resistant to corrosion. For this reason it is used for boiler

fire boxes, water heating apparatus, water pipes and

vessels in brewery and chemical plants. Also used for

making soldering iron.

The melting temperature of copper is 1083

The tensile strength of copper can be increased by ham-

mering or rolling.

Copper Alloys

Brass

It is an alloy of copper and zinc. For certain types of brass

small quantities of tin or lead are added. The colour of brass

depends on the percentage of the alloying elements. The

colour is yellow or light yellow, or nearly white. It can be

easily machined. Brass is also corrosion-resistant.

Brass is widely used for making motor car radiator core and

water taps etc. It is also used in gas welding for hard

soldering/brazing. The melting point of brass ranges from

880 to 930

Brasses of different composition are made for various

applications. The following table-1 gives the commonly

used brass alloy compositions and their application.

Bronze

Bronze is basically an alloy of copper and tin. Sometimes

zinc is also added for achieving certain special properties.

Its colour ranges from red to yellow. The melting point of

bronze is about 1005

C. It is harder than brass. It can be

easily machined with sharp tools. The chip produced is

granular. Special bronze alloys are used as brazing rods.

Bronze of different compositions are available for various

applications.

Lead and its alloys

Lead is a very commonly used non-ferrous metal and has

a variety of industrial applications.

Lead is produced from its ore ‘GALENA’. Lead is a heavy

metal that is silvery in colour when molten. It is soft and

malleable and has good resistance to corrosion. It is a

good insulator against nuclear radiation. Lead is resistant

to many acids like sulphuric acid and hydrochloric acid.

It is used in car batteries, in the preparation of solders etc.

It is also used in the preparation of paints.

Lead Alloys

Babbit metal

Babbit metal is an alloy of lead, tin, copper and antimony.

It is a soft, anti-friction alloy, often used as bearings.

An alloy of lead and tin is used as ‘soft solder’.

Zinc and its alloys

Zinc is a commonly used metal for coating on steel to

prevent corrosion. Examples are steel buckets, galvanized

roofing sheets, etc.

Zinc is obtained from the ore-calamine or blende.

Its melting point is 420

It is brittle and softens on heating; it is also corrosion-

resistant. It is due to this reason it is used for battery

containers and is coated on roofing sheets etc.

Galvanized iron sheets are coated with zinc.

Tin and tin alloys

Tin

Tin is produced from cassiterite or tinstone. It is silvery

white in appearance, and the melting point is 231

C. It is

soft and highly corrosion-resistant.

It is mainly used as a coating on steel sheets for the

production of food containers. It is also used with other

metals, to form alloys.

Example: Tin with copper to form bronze. Tin with lead to

form solder. Tin with copper, lead and antimony to form

Babbit metal.

Aluminium

Aluminium is a non-ferrous metal which is extracted from

‘BAUXITE’. Aluminium is white or whitish grey in colour. It

has a melting point of 660

C. Aluminium has high electrical

and thermal conductivity. It is soft and ductile, and has low

tensile strength. Aluminium is very widely used in aircraft

industry and fabrication work because of its lightness. Its

application in the electrical industry is also on the increase.

It is also very much in use in household heating appliances.

Workshop Calculation And Science : (NSQF) Exercise 1.7.21

Difference between iron & steel, alloy steel & carbon steel Exercise 1.7.22

S.No Basic Iron Steel

distinction

1 Formation Pure substance Made up of

iron and

carbon

2 Types Cast iron, Carbon steel

Wrought and alloy

iron and steel steel

3 Rusting Quickly gets Have different

oxidised and elements that

result in rust protect from

rusting

4 Surface Its surface Its surface

is rusty stays shiny

5 Usage Used in Used in

buildings,tools buildings,

and automobiles cars,

railways

and

automobiles

6 Existence Available in Has to be

nature formed

Steel Plants in India

S.No Name of the Steel plant State

1 Tata Iron Bihar

2 Indian Iron Steel West Bengal

3 Visweshvaraiah Iron Steel Karnataka

4 Bhilai Steel Plant Madya Pradesh

5 Durgapur Steel Plant West Bengal

6 Alloy Steel Plant (Durgapur) West Bengal

7 Bokaro Steel Plant Bihar

8 Rourkela Steel Plant Orissa

9 Salem Steel Plant Tamilnadu

10 Vishakapatnam Steel Plant Andhra Pradesh

Difference between iron and steel:

Comparison of the Properties of Cast Iron, Mild Steel and steel

Property Cast Iron Mild Steel Steel

Composition Carbon contents Carbon contents Carbon contents

from 2 to 4.5% from 0.1 to 0.25% from 0.5 to 1.7%

Strength – High – Moderate – High

compressive strength compressive strength compressive strength

– Poor tensile strength – Moderate tensile strength – High tensile strength

– Poor shearing strength – High shearing strength – High shearing strength

Malleability Poor High High

Ductility Poor High High

Hardness Moderately hard and can be Mild Hard

hardened by heating to

hardening temperature and

quenching

Toughness Possesses poor toughness Very tough Toughness varies

with carbon content

Brittleness Brittle Malleable Malleable

Forgeability Cannot be forged Can be forged Can be forged

Weldability Cannot be welded with Can be welded very easily Can be welded

difficulty

Casting Can be easily cast Can be cast but not easily Can be cast

Elasticity Poor High High

Ferrous metals Non Ferrous metals

1 Iron content is more 1 Iron content is missing

2 The melting point is 2 The melting point is low.

high

3 This is of brown and 3 This is of different colours

black colour

4 This catches rust 4 This doesn’t catch rust.

5 This can be 5 This cannot be

magnetised magnetised

6 This is brittle in cold 6 This becomes brittle in hot

state. state.

Difference between cast Iron and steel

Cast Iron Steel

1 Carbon content is high Carbon content is less

2 Carbon is in free state Carbon is mixed

3 Melting point is low Melting point is high

4 It cannot be magnetised It can be magnetised

5 Because it is brittle, In can be forged

it cannot be forged

6 It rusts with difficulty It rusts quickly

7 It cannot be welded It can be welded

Difference between metals and non-metals

Metals Non Metals

Shiny dull

Usually good conductors Usually poor conductors of

of heat and electricity heat and electricity

Most are ductile not ductile

Opaque (opposite of Transparent when as a thin

‘transparent’) sheet

Most are malleable Usually brittle when solid

Form alkaline oxides Form acidic oxides

Sonorous (make a bell not sonorous

-like sound when struck)

Usually have 1-3 valency Usually have 4-8 valency

electrons electrons

Most corrode easily

Usually high melting

points (usually solid at

room temperature

except for mercury)

Workshop Calculation And Science : (NSQF) Exercise 1.7.22

Properties and uses of rubber, timber and insulating materials Exercise1.7.23

Properties and uses of rubber

Rubber

Rubber is an elastic material. It can be classified into

• Natural rubber

• Hard rubber

• Synthetic rubber

Natural rubber

It is obtained from the secretion of plants. It softens on

heating, becomes sticky at 30°C and hardens at about

5°C.

Sulphur is added to rubber and the mixture is heated. This

process is called vulcanising. By this process, stronger,

harder and more rigid rubber is obtained. Further, it

becomes less sensitive to changes of temperature and

does not dissolve in organic solvents. Its oxidisation is also

minimised by increasing its weathering properties.

By adding carbon black, oil wax, etc, the deformation

properties are minimised. Rubber is moisture-repellent and

possesses good electrical properties. The main disadvan-

tages of the rubber are as given under.

• Low resistance to petroleum oils.

• Cannot be exposed to sunlight.

• Cannot be used for high-voltage insulation.

• Low operating temperature (as it becomes brittle and

develops cracks at a temperature of 60°C)

• Sulphur in rubber reacts with copper. Hence, copper

wires are to be tinned.

Hard rubber

By increasing the sulphur content and prolonged

vuicanization, a rigid rubber product called hard rubber or

ebonite is obtained. It possesses good electrical and

mechanical properties.

Uses

It is used for battery containers, panel boards, bushing,

ebonite tubes, etc.

Synthetic rubber

This is similar to natural rubber and is obtained from

thermoplastic vinyl high polymers. Some of the important

synthetic rubbers are:

• Nitrite butadience rubber

• Butyl rubber

• Hypalon rubber

• Neoprene rubber

• Silicon rubber

Sl.No. Name Properties Uses

1 Nitrite Goods resilience, wear resistance, flexibility at low Automobile tyre inner tubes.

butadiene temperature, resistance to ageing, oxidation, low

rubber tensile strength, high thermal conductivity, low

hygroscopicity

2 Butyl It is attacked by petroleum oils, gases and alcoholic Used as insulation in hot and wet

solvents. It has thermal and oxidation stability and conditions, used as tapes in repair

high resistance to ozone. work.

3 Hypalon Resistance to deterioration when exposed to sunlight Used in jacketing of electric wires

rubber and temperature (up to 150°C). and cables

4 Neoprene Better resistance to ageing, oxidation and gas Used for wire insulation and cable

rubber diffusion, better thermal conductivity and flame sheating.

resistance, poor mechanical properties.

5 Silicon High operating temperature (200°C) flexibility, moisture Insulation for power cables and

and corrosion resistance, resistance to oxidation, control wires of blast furnace coke

ozone, arcing, good insulating properties and thermal ovens, steel mills and nuclear

conductivity. It is a good insulator. power stations high frequency

enerators, boiler, airport lighting

cranes.

Workshop Calculation And Science : (NSQF) Exercise 1.7.23

Properties and uses of timber

General properties

Timber should have the following properties

• Straight fibres.

• Siliky lustre when planed.

• Uniform colour.

• Regular annual rings.

• Heaviness.

• Firm adhesion of fiber and compact modulary rays.

• Sweet smell.

• It should be free from loose or dead knots and shakes.

• The surface should not clog the teeth of the saw on

cutting but should remain bright.

Classification

• Timbers are classified as

a Softwood

b hardwood

Softwood timber

• Usually all trees with needle leaves of softwood and

those with broad leaves are of hard-wood.

• The wood contains resins and turpentines.

• The wood has a fragrant smell.

• Fibres are straight.

• Texture is soft and regular.

• Tough for resisting tensile stresses.

• Weak across the fibres.

• Annual rings are distinct, having one side soft, porous

and light coloured. The other side is dense and dark.

• The general colour of the wood is pale tinted or light

such as pine spruce, fir, ash, kail, deodar etc.

Properties of hardwood

• The wood generally contains a large percentage of acid.

• It is brightly coloured.

• Annual rings are not distinct.

• It is difficult and hard to work with.

• It resists shearing stress.

• Fibre are overlapped.

• The general colour is dark brown uch as oak, walnut,

teak, mahagony, sishim, babul, sal etc.

Uses

Soft timber

• Because of its cheapness it is used for low grade

furniture, doors and windows for cheap tyes of houses.

• Used as fuel.

• Some timbers are used for baskets and mat making.

• The bark is used as garment is some places.

Hard timber

• Used for high quality furniture such as chairs, tables,

sofas, dewans, beds, etc.

• Used for door, window frames for high quality houses as

they can take good polish and painting finish.

• Used for manufacturing katha.

Wood as an electrical insulator

Wood is impregnated with oil or other substance, for use

as insulator.

Example

It is used in electrical machine windings, as slot wedges.

Insulating materials

Description

These are the materials which offer very high resistance to

the flow of current and make current flow very negligible or

nil. These materials have very high resistance - usually of

may megohms (1 Megohm = 10

ohms) oer centimetre

cubed. The insulators should also posseses high dielectric

strength. This means that the insulating material should

not break down or puncture even on application of a high

voltage (or high electrical pressure) to a given thickness.

Properties of insulators

The main requirements of a good insulating material are:

• High specific resistance (many megohms/cm cube) to

reduce the leakage currents to a negligible value.

• Good dielectric strength i.e. high value of breakdown

voltage (expressed in kilovolts per mm).

• Good mechanical strength, in tension or compression

(It must resist the stresses set up during erection and

under working conditions.)

• Little deterioration with rise in temperature (The insulating

properties should not change much with the rise in

temperature i.e. when electrical machines are loaded.)

• Non-absorption of moisture, when exposed to damp

atmospheric condition. (The insulating properties,

specially specific resistance and dielectric strength

decrease considerably with the absorption of even a

slight amount of moisture.)

Workshop Calculation And Science : (NSQF) Exercise 1.7.23

Products and insulators

Insulators Uses in electric field

1 Mica In elements or winding (Slot insulation)

2 Rubber Insulation in wires

3 Dry cotton Winding

4 Varnish Winding

5 Asbestos In the bottom of irons and kettles, etc.

6 Gutta Submarine cables

parcha

7 Porcelain Overhead lines insulators

8 Glass -do-

9 Wood dry Cross arms in overhead lines

10 Platic Wires insulation or switches body

11 Ebonite Bobbin of transformer

12 Fibre Bobbin making and winding insulation

13 Empire Winding insulation

cloth

14 Leathroid -do-

paper

15 Millimax -do-

paper

16 P.V.C. Wire insulation

17 Bakelite Switch etc. making, for insulation

18 Shellac -do-

19 Slate Making panel board

20 Paraffin Sealing

Wax

Mass, units of mass, density and weight Exercise 1.8.24

m - mass of a body

g - acceleration due to gravity in metre/sec

= 9.81 m/

sec

V - volume of the body

ρ - density (pronounced as `rho')

W or FG - weight or weight force

Mass (Fig 1)

Mass of a body is the quantity of matter contained in a

body. The unit of mass in F.P.S system is pound (lb), in

C.G.S. system gram (gr) and in M.K.S and S.I systems

kilogram (kg). 1ton which is 1000 kg is also used sometimes.

The conversion factor is 1000. Three decimal places are

shifted during conversion.E.g.1 ton =1000 kg 1g = 1000 mg

Density

Density is the mass of a body per unit volume. Hence its

unit will be gr/cm

or kg/dm

or ton/m

Density =

System Absolute Derived Conversion

unit unit

F.P.S. system 1 poundal 1 Lbwt 32.2 poundals

(1 lb x 1 ft/sec

= 1 pound)

C.G.S. system 1 dyne 1 Gr.wt 981 dynes

1 gr x 1

cm/sec

M.K.S. Newton 1 kg.wt 1 Newton =

S.I.system Newton Newton 1 kg x 1 m/sec

1 kg.wt = 9.81 Newton 1 Newton = 10

dynes.

( approximately 10N)

Weight (Fig 2)

Weight is the force with which a body is attracted by the

earth towards its centre. It is the product of the mass of the

body and the acceleration due to gravity. The weight of a

body depends upon its location.

weight = W or FG = mass x gravitational force

= m x g

Mass and weight are different quantities.

Mass of a body is equal to volume x density.

Weight force is equal to mass x acceleration

due to gravity.

Weight , Density and Specific gravity

It is now seen that the mass of a substance is measued by

its weight only without any reference to volume. But if equal

weights of lead & aluminium, are compared the volume of

lead is much smaller than volume of aluminium. So we can

now say that lead is more dense than aluminium,. i.e In

other words the density of lead is greater than aluminium.

(Fig 1)

Unit

The density is measured as below

MKS/SI= Kg/m

, CGS - 1 gm/cm

FPS–lbs/c ft

Difference between mass and weight

S. No Mass Weight

1 Mass is the quantity Weight is measure

of matter in a body (ie) of amount of force

measurement of acting on mass due

matter in a body to acceleration

due to gravity

2 It does not depend on It depends on the

the position or space position, location

and space

3 Mass of an object will Weight of an object

not be zero will be zero if gravity

is absent

4 It is measured using It is measured using

by physical balance by spring balance

5 It is a scalar quantity It is a vector quantity

6 When immersed in When immersed in

water mass does not water weight will

change change

7 The unit is in grams The unit is in

and kilogram kilogram weight,

a unit of force

Difference between mass & weight, density & specific gravity Exercise 1.8.25

The relation of mass and volume is called density.

The density expresses the mass of volume E.g. 1 dm

water has the mass of 1kg - thus the density of 1kg/dm

Solids gm/cc

Liquids gm/cc

1 Aluminum 2.7 Water 1.00

2 Lead 11.34 Petrol 0.71

3 Cast iron 6.8 to 7.8 Oxygen 1.43

4 Steel 7.75 to 8.05 Diesel Oil 0.83

The specific gravity of a substance is also called its relative

density.

Formula

Specific gravity

(or) Relative density

From the above table, we can calculate the weight of any

given volume of a substance (say Diesel oil) in any units

provided we know the specific gravity of the substance.

Also vice-versa for volume of density is known.

Comparision Between Density And Specific Gravity

(Relative Density)

Density Relative density or

Specfiic gravity

Mass per unit volume of The density of substance

a substance is called its to denisty of water at 4°C

density is its relative density

Its unit is gm per cu cm; It has no unit of measure-

lbs per cu.ft and kg/cubic ment simply expressed in

meter a number

Density = Relative density

Solids Sp.gy Liquids Sp.gy

1 Aluminium 2.72 Petrol 0.71

2 Lead 11.34 Battery acid 1.2 to 1.23

3 Cast iron 6.8 to 7.8 Water 1.00

4 Steel 7.82 Diesel Oil 0.83

Related problems with assignment of mass, weight & density Exercise 1.8.26

1 Calculate the mass in kg of a rectangular steel plate of

dimensions 220 x 330 x 15 mm (Fig 1) (density of

steel = 7.82 gm/cm

Mass = Volume x density

= 22 x 33 x 1.5cm

x 7.82 gm/cm

= 1089 cm

x 7.82 gm/cm

mass = 8.516 kg

2 A storage container holds 250 litres of water. What

weight in N will this amount of water exert on the surface

on which it is standing?(Fig 2)

( 1 litre of water = 1 kg of water )

Density of water 1 gm/cm

or 1 kg/dm

Acceleration due to gravity is taken as 10

metre/sec

(approximation).

Capacity = 250 litres = 250 dm

in volume.

Mass of water = volume x density of water

= 250 dm

x 1 kg/dm

= 250 kg

Weight extended = mass x acceleration due to gravity

= 250 kg x 10 metre/sec

= 2500 kg.metre/sec

= 2500 N( 1 kg.m/sec

=1N)

3 A force of 15 dynes acting on a mass of `m' produces

an acceleration of 2.5 cm/sec

. Find the mass.

1 Gr. wt. = 981 dynes

15 dynes = Gr.wt

Force = m x acceleration produced by the force

Gr.wt = mass x 2.5 cm/sec

gr.cm/sec

= mass x 2.5 cm/sec

mass = grams =

mass = 0.00612 gram

4 A force of 2 N acts on a mass of 10 kg. Find the

acceleration produced by the force on the mass.

Force = 2 N ( 1 N = 1kg.m/sec

)

Force = mass x acceleration

2 kg.metre/sec

= 10 kg x acceleration produced

2 x 1 kg.metre/sec

= 10 kg x acceleration

produced

acceleration produced = metre/sec

= 0.2 metre/sec

5 Calculate the weight of a body having a mass of 1 kg if

the acceleration due to gravity is 9.81 metres/sec

Weight force = mass x acceleration due to gravity

= 1 kg x 9.81 metres/sec

(1 kg.metre/sec

= 1 N)

9.81 kg metre/sec

= 9.81 N

In the examples solved the value of `g' is taken

as 10 metre/sec

, unless specifically men-

tioned otherwise.

• The outside and inside diametres of a hollow sphere are

150 & 70mm respectively. Calculate its mass if the

density of material is 7.5 gm/cm

(Fig 3)

Mass = Volume x Density

= Volume x 7.5 gm/cm

D= 150 mm = 15cm R= 7.5 cm

Workshop Calculation And Science : (NSQF) Exercise 1.8.26

d = 70mm = 7 cm r = 3.5 cm

=1587.5 cm

Mass

= 1587.5 cm

X 7.5 gm/cm

= 11906.6 gm=11.9kg say 12kg

6 A car has a mass of 800 kg. Find out its weight force

(Take 9.81 m/sec

)

( 1n = 1kg.m/sec

)

The Wt. force of a car=Mass of car x gravitational

acceleration

= 800 x 9.81 N

= 7848 Newtons

7 A cylindrical tank 2m dia x 3.5 m deep is filled with

petrol. Find the weight of petrol in Tonnes, Assume

density of petrol 720 Kg/m

.(Fig 4)

Volume of Tank

3.14 x 3.5 m

= 10.99 m

Since 1 m

= 1000 litres

Volume of Tank = 10.99 x 1000 litres

Density of petrol = 720 Kg/m

Weight of Petrol in Kg =10.99x1000 litresx720Kg

= 720 x 10990 Kg

Weight of Petrol in Tonnes

(Metric Units)

Weight of Petrol = 7912.8 Tonnes

8 If the battery acid specific gravity is 1.3, and this is

being filled up into a cylindrical tank. Find out its

density.

(Density of water = 1000 gm/cm

)

Specific gravity or Relative density

Now, density of battery acid

= Specific gravity x Density of water

= 1.3 x 1000 gm/cm

= 1300 gm/cm

Determination of specific gravity of a substance

The specific gravity of a substance may be determined by

1 Arche medies principle

2 Hydrometer

Archemedes Principle

Archemedes principle states that when a body is fully or

partially immersed in a liquid, the amount of liquid dis-

placed by the body is equal to the loss of weight of the body

in the liquid.

Weight of a body in a liquid = total weight of the body

- weight of the liquid displaced by the body

This quantity if it is zero then the body will float. It is

negative the body will rise up till the weight of liquid

displaced by the immersed portion of the body is equal and

equal to the weight of the body. If it is positive the body will

sink. Specific gravity of solids in soluble in water

specific gravity of solids soluble in water

specific gravity of a liquid

The solid chosen should be such that it is

insoluble in both water and the liquid whose

specific gravity is to be determined.

Example

1 An iron piece weighs 160 kgf in air and 133 kgf when

it is fully immersed in water. Determine the volume and

specific gravity of the iron piece.

Weight of the solid in air = 160 kgf

Weight of the solid in water = 133 kgf

∴Loss of weight in water = 27 kgf

By Archemedies principle the loss of weight of a solid

in water = volume of water displaced.

∴Volume of water displaced = 27 cm

Workshop Calculation And Science : (NSQF) Exercise 1.8.26

Sl.No Metal Density gm/cc

1 Aluminium 2.7

2 Cast Iron 6.8 - 7.8

3 Copper 8.92

4 Gold 19.32

5 Iron 7.86

6 Lead 11.34

7 Nickel 8.912

8 Silver 0.5

9 Steel 7.75 - 8.05

10 Tin 7.31

11 Zinc 7.14

12 Diamond 3.51

13 Bismuth 9.78

14 Brass 8.47

15 Phosphrous 8.7 - 8.9

Bronze

16 Ice 0.93

17 Air 0.0013

18 Mercury 13.56

19 Petrol 0.71

20 Diesel 0.83

21 Kerosene 0.78 - 0.81

22 Water 1.0

∴Volume of the solid= 27 cm

Specific gravity of iron piece = 5.93

2 A metal piece weighs 6.5 kgf in air and 3.5 kgf in water.

Find its weight when it is fully immersed in a liquid

whose specific gravity is 0.8 and also the S.G of the

metal.

Weight of metal piece in air = 6.5 kgf

Weight of metal piece in water= 3.5 kgf

∴Loss of weight in water = 3.00 kgf (6.5 - 3.5)

∴Specific gravity of metal

By applying the principle of Archemedes the above

results are derived.

By using a hydrometer also, the specific gravity of a

liquid is determined. The most common type of

hydrometer is the Nicholson’s hydrometer which is a

variable weight but constant immersion type.

Specific gravity of a liquid

weight of hydrometer+ weight required to sink the =

in water up to the same mark.

Let the weight of the metal piece in the liquid = W

loss of weight of the metal in the liquid = 6.5 kgf_W

w = 6.5 kgf - 3 kgf x 0.8 = 4.1 kgf

loss of weight of the metal in the liquid = 4.1 kgf.

3 A solid of wax weighs 21 kgf in air. A metal piece

weighing 19 kgf in water is tied with the wax solid and

both are immersed in water and the weight was fund to

be 17 kgf. Find the specific gravity of wax.

Weight of wax in air = 21 kgf

Weight of metal and wax in water = 17 kgf

Weight of metal piece only in water = 19 kgf

weight of wax in water = (17 -19) kgf

= - 2 kgf

loss of weight of wax in

water = (21 - (-2)) kgf

= 23 kgf

specific gravity of wax

ASSIGNMENT

1 l = 1800 mm

b = 65 mm

h = 12 mm

ρ = 7.85 g/cm

m = ______ kg

2 Capacity = 36 litres

d = 32 cm

H = ______ cm

3 D = 74 mm

d = 68 mm

l = 115 mm

ρ = 8.6 gm/cm

m = ______ gms

= 80 mm

= 61 mm

d = 39 mm

L = 112 mm

l = 90 mm

ρ = 7.85 gm/cm

m = ______ kg

5 D = 44 mm

d = 20 mm

L = 120 mm

= 60 mm

= 40 mm

ρ = 7.85 gm/cm

m = ______ kg

6 L = 120 mm

B = 90 mm

= 60 mm

= 30 mm

d = 55 mm

H = 42 mm

h = 18 mm

ρ = 7.85 gm/cm

m = ______ kg

7 L = 200 mm

= 75 mm

= 50 mm

B = 80 mm

H = 110 mm

h = 45 mm

ρ = 2.7 gm/cm

m = ______ kg

8 V = 320 cm

= 8.9 gm/cm

g = 9.80665

metre/ sec

m = ______ kg

FG = ______ N

9 Capacity = 35 litres

g = 10 metres/sec

FG = ______ N

10 (m

) mass of chain

= 150 kg

Total FG = 8 KN

Load = ______ N

mass m

= _____kg

11 W (FG) = 22.5 N

V (volume) = ______

12 F = 250 d N

side of cube

= ______ mm

(cubical counter

weight balances `F')

Workshop Calculation And Science : (NSQF) Exercise 1.8.26

13 unbalanced load in

the set up = 16 cN

ø of balancing weight

= 20 mm

l of balancing weight

= ______ mm

14 d

= 40 mm

= 9 x 10

–2

= r

= 60 mm

= ______ N

15 l x b = 1 m

FG = 7.85 x 10

–2

s = ______ mm

F = 400 N

m = ______ kg

17 m

= 200 gms

FG = 16 N

F = ______ dN

18 R = 14 kN

m = ______ kg

19 V = 4 dm

FG = 10.8 daN

ρ = ______ gm/cm

20 l = 500 mm

b = 300 mm

H = 250 mm

ρ of oil = 0.9 gm/cm

m = 2.5 kg

h = ______ mm

21 Engine cooling

Data given

Water in Radiator = 10

litres

Find

Mass of water =

__________ kg

(Assume 1 litre = dm

in volume)

Density of water = 1 kg/

22 Cylinder Liner Dimen-

sion

Data given

OD = 111 mm

ID = 103 mm

Length = 240 mm

Material = C.I

Density of C.I = 7.259

gm/cm

Find its mass___ in kg

23 Gudgeon Pin (Solid)

Data given

Dia = 200 mm

Length = 70 mm

Material = M.S

Density = 7.85 gm/cm

Find out its

Workshop Calculation And Science : (NSQF) Exercise 1.8.26

24 Data given

Dia = 80 mm

Length = 100 mm

Density of Aluminum =

2.7 g/cm

Find its Mass _______

in kg

25 Hollow sphere (Cast

Brass)

Data given

O.D = 150 mm

I.D = 120 mm

Density of Brass = 6.89

gm/cc

Use Vol = ( (R

)

Find

Mass of Hollow sphere

= _______ kg

26 Diesel Tank

Data given

Diameter = 400 mm

Depth of filling (h)

= 600 mm

Spongy of oil = 0.8

Density of water = 1000

kg/m

Find

Mass of oil in Tank =

______ in kg

27 Definition

Define the following term

a Mass

b Weight

c Density

d Specific gravity

28 Conversion of vehicle weights

Take g = 10 m/sec

Weight force Mass

a 480 Newton _______

b 14800 N _______

c 2000 N _______

d 7000 N _______

29 Covnersion of mass of vehicle

Take g = 9.81 m/sec

Mass of Vehicle Its weight

a 1200 kg ________ N

b 800 kg ________ N

c 700 kg ________ N

d 900 kg ________ N

30 Fill up the blanks

Comparison of Metals & Liquids

Material Sp.gy Density

a Lead 11.34 _______

b Copper 8.92 _______

c Cast Iron 7.20 _______

d Petrol 0.71 _______

e Diesel 0.83 _______

f Sulphuric

Acid 1.84 _______

31 Fill in the blanks with correct statement in a & b

a The density of water - 1000 kg/m3 specific gravity of

nitric acid = 1.2. The density of nitric acid = _______

b Material Density Specific gravity

i Water 1000 kg/m

_______

ii Aluminium 2.7 g/cm

_______

iii Iron 8 g/cc _______

iv Copper 8.7 g/cc _______

c Mass of a body = Volume x _______

d Weight force = Mass x _______

e Give abbreviation for

i Mega newton _______

ii Kilo newton per square metre _______

f 1 litre of water = _______kg.

Workshop Calculation And Science : (NSQF) Exercise 1.8.26

Rest, motion, speed, velocity, difference between speed & velocity,

acceleration & retardation Exercise 1.9.27

Body at rest

When a body does not change its position, with respect to

its surroundings, it is said to be at rest.

Body at motion

When a body changes its position, with respect to its

surroundings, it is said to be in motion. The motion may

be linear if the body moves in a straight line or it may be

circular when it moves in a curved path.

Terms relating to motion

Displacement

When a body is in motion from one place to another, the

displacement is the distance from the starting position to

the final position.

Speed

It is the rate of change of displacement of a body in motion.

It has got no direction and it is a scalar quantity.

Speed = distance travelled per unit time

Unit = m/s, km/Hr.mile/Hr.

Velocity

It is the rate of change of displacement of a body in motion

in a given direction. It is a vector quantity and can be

represented both in magnitude and direction by a straight

line. Velocity may be linear or angular. The unit of linear

velocity is metre/sec,

Velocity =

Unit = m/s, km/Hr,mile/Hr.

Difference between speed & velocity

unit = m/s

(metre per square second)

u = Initial velocity in metre per second(m/sec)

v = Final velocity in metre per second(m/sec)

s = Distance in metre (m)

t = Time in second (sec)

a = Acceleration m/sec

(positive value)

R = Retardation m/sec

(negative value of acceleration)

Equations of motion

Then v = u + at

s = ut + at

and v

– u

= 2as

= u

+ 2as

Retardation

When the body has its initial velocity lesser than its final

velocity it is said to be in acceleration. When the final

velocity is lesser than the initial velocity the body is said to

be in retardation. Then the three equation of motion will be

v = u – at

s = ut – at

–v

= 2as

Average speed

Vm - Average speed in metre/min, (metre/sec)

n - Revolutions per minute or number of strokes per

minute

s - Distance travelled, length of stroke.

Stroke speed (Fig 1)

For one revolution of the point k, of the crank pin the

distance the power saw blade moves = 2 x s

Therefore ‘n’ revolutions in a minute the distance = 2 x s x

n. Since the stroke of the blade will be given in metre to

determine the average speed

Vm = 2 x s x n

Piston speed (Fig 2)

Acceleration

Rate of change of velocity is known as acceleration or it is

the change of velocity in unit time. Its unit is metre/sec

It is a vector quantity.

m/sec

Speed

The rate of change

place of an object is

its speed.

In the speed, direc-

tion is not indicated.

Only the magnitude is

expressed.

Speed

Velocity

The speed in a definite direc-

tion is called velocity.

Both the magnitude and di-

rection are expressed.

Velocity

As the piston moves backward and forward, its speed

constantly changes between the upper and lower dead

centres. Hence in this case also the average speed Vm =

2 x s x n. Since s is expressed in mm and n in number of

revolutions/per minute and since Vm is given in metre/sec,

we have

Vm = 2 x s x

metre/min.

If s is given in metres then

Vm = 2 x s x = s x metre/sec.

2 x s denotes a double stroke.

In case of the reciprocating motion the average

speed is taken into account for calculations.

Vm = 2 x s x n metre/min if s is given in metres

Example (Fig 3)

An extrusion press has a crank radius of 20 cm and an rpm

of 30/min. Calculate the average speed in metre/min,

metre/sec.

s = The diameter = 40 cm.

One crank revolution makes the piston to travel in 2s=80cm

Vm = 2 x 400 x metre/min.

= 24 metre/min = 0.4 metre/sec

NEWTON’S LAWS OF MOTION

Motion under gravity

A body falling from a height, from rest, has its velocity goes

on increasing and it will be maximum when it hits the

ground. Therefore a body falling freely under gravity has a

uniform acceleration. When the motion is upward, the

body is subjected to a gravitational retardation. The

acceleration due to gravity is denoted with ‘g’.

Momentum

It is the quantity of motion possessed by a body and is

equal to the product of its mass, and the velocity with which

it is moving. Unit of momentum will be kg metre/sec.

Momentum = mass x velocity

Newton’s laws

First law

Every body continues to be in a state of rest or of uniform

motion in a straight line unless it is compelled to change

that state of rest or of uniform motion by some external

force acting upon it.

Second law

The rate of change of momentum of a moving body is

directly proportional to the external force acting upon it and

takes place in the direction of the force.

Third law

To every action there is always an equal and opposite

reaction.

In the rivet joint equal forces act on the strap and they

opposite force F

. (Fig 4)

Law of conservation of momentum

When two moving bodies have an intentional or uninten-

tional impact, then sum of the momentum of the bodies

before impact = sum of the momentum after impact, or the

change in momentum after the impact is zero.

- mass of one body and

- velocity with which it moves

- mass of second body

- velocity with which it moves

Momentum = m x v= mass of the body x its velocity

Rate of change of momentum = force acting on the body

force = mass x acceleration

Workshop Calculation And Science : (NSQF) Exercise 1.9.27

Equations of motions under gravity

Upward Downward

V = u – gt v = u + gt

s = ut – gt

s = ut + gt

–v

= 2gs v

–u

= 2gs

Momentum of two bodies before impact = momentum after

impact

x v

+ m

x v

= (m

+ m

Terms - Some Examples in vehicles

Displacement

The piston displacement is the space between 2 dead

centres (TDC and BDC) where in the piston moves in the

cylinder. (Fig 5)

Speed

This is reckoned in 2 ways in a vehicle

– Vehicle speed in kmh/mph

– Engine speed in rpm

Velocity

A motor vehicle, normally changes its speed and direction

on road. Hence used in velocity calculation.

Acceleration (Fig 6)

When the speed of the vehicle is increased on road, it is

said to be accelerated.

Deceleration (Fig 6)

Deceleration or Retardation (this is further explained)

During the application of brakes of a vehicle the speed of the

vehicle is decreased. Then it is said to be decelerated or

retarded.

Example

In circular motion bodies (like shafts, axles, gear-wheels,

pulleys, flywheels, grinding wheels) turn with constant

speed around its axis.

The angular of circular motion is also called Angular

velocity or Peripheral speed.

Expressed in Metre/sec or Radians per second.

Bodies at rest and in motion

Terms related to brake system

Every vehicle has a brake system. When brakes are

applied on a moving vehicle (with certain velocity) its

velocity is reduced and vehicle is decelerated and it stops

at a certain distance. So the definition of the terms related

to Brake application are set forth below.

Deceleration (a) (Fig 8)

Workshop Calculation And Science : (NSQF) Exercise 1.9.27

Circular or Angular motion (Fig 7)

When a body rotates about an axis, it is said to have

angular motion or circular motion.

This is the decrease in velocity within a certain time. e.g

A car travelling at 90 kmph stops after 10 Sec.

The deceleration = 90 x x 1/10

= 25 m/s/10 sec

= 2.5 m/sec

Deceleration time

The time 10 seconds is called the above time to stop the

vehicle.

Stopping distance

During the deceleration time the car travels a distance

called i.e Stopping distance ‘d’.

But the total stopping distance is reckoned as equal to

normal stopping distance and distance travelled by the car

during reaction time of the driver.

The reaction time is explained as below

During the application of brakes, the driver takes sometime

to recognise the danger and then to apply the brakes. The

time (thus elapsed) is called reaction time. During this time

the vehicle travels some more distance before coming to a

stop. So the total stopping distance actually varies due to

the reaction time of the driver and it is longer than the

normal stopping distance. The reaction time varies be-

tween driver to driver.

Example

A car is travelling with a speed of 72 kmph and its

aceleration (a) = 5 m/sec

. The reaction time of driver to

apply brakes is 1.5 seconds. calculate the total stopping

distance.

Solution

Velocity of car = 72 kmph

= 20 m/sec

aceleration = 5 m/sec

Normal stopping distance S = (m)

Total stopping distance

= 40 metre + Velocity x Reaction time

= 40 m + (20 x 1.5) m

= 70 metres.

Newon's Law of Motion

Some Examples in vehicles

First law (with examples) (Fig 9)

Bodies at rest or in Uniform motion

The diesel engine piston remains at rest at TDC or BDC due

to its inertia. Expansion of gas pressure or flywheel

momentum moves the piston from TDC or BDC.

Second law (with examples) (Fig 10)

The rate of change of momentum of a moving body (say

Engine part or Vehicle)is directly proportinal to external

force acting take place in the direction of force.

– A connecting rod in motion is brought to rest at BDC.

– The direction of movement of a vehicle is altered by force

of wind.

– When a vehicle travels in a down gradient its speed

increases.

– The speed of vehicle is decreased when travelling up

gradient.

Workshop Calculation And Science : (NSQF) Exercise 1.9.27

Third law (with examples) (Fig 11&12)

To every action there is always an equal and opposite

reaction.

All upward force = All downward forces

– Jack is lifting a differential

– Crane rope is lifting an engine.

0.3m/sec

sec 45

m/sec 13.88

m/sec 13.88 m/sec

x50km/hr





Related problems with assignment of speed & velocity Exercise 1.9.28

Examples

• A body travels a distance of 168 metres in a straight line

in 21 secs. What velocity the body is travelling.

Velocity = distance travelled/time

= 168 metres/21 secs

= 8m/sec

• A train covers a distance of 150 kilometres, between

two stations, in 2 1/2 hours. Determine the average

velocity with which the train is moving.

Average velocity = Distance travelled/time taken

= 150 Km/2 1/2 hrs =

150



Km/hr

= 60 Km/hr

• A vehicle accelerates uniformly from a velocity of 8 km/

hr to 24 km/hr in 4 secs. Determine the acceleration

and the distance travelled by it during that time.

Initial velocity = 8 km/hr (u)

Final velocity = 24 km/hr (v)

time = 4 sec (t) acceleration (a)



v = u + at

24 km/hr = 8 km/hr + a x 4 sec

(24km/hr - 8km/hr - 16km/hr)

4a sec = 16 km/hr = 16000 metre/3600 sec

acceleration (a) = 16000 metre/3600 x 4 sec

v-u = 4.44 m/sec a=

Acceleration (a) = 1.1 metre/sec

• A car moving with a velocity of 50 km/hr is brought to

rest in 45 secs. Find out the retardation.

Initial velocity = 50 km/hr (1km= 1000 metres)

Final velocity = 0 km/hr (1 Hour = 3600 seconds)

Time = 45 secs

v = u – at

0 = u – at

u = at

50000/3600 metre/sec = a x 45 sec



Retardation = 50000/3600 x 45 metre/sec

= 0.30 metre/sec

• A body falling freely under the action of gravity reaches

the ground in one second. Determine the height from

which the body fell. Take g = 9.81 metre/sec

Initial velocity = 0 metre/sec (U)

Acceleration due to gravity = 9.81 metre/sec

(g)

Time taken = 1 sec (t)

= ut + gt

0 x 1 sec= x 9.81 m/sec

x 1

sec

= 0 x 1 sec + x 9.81 metre/sec

x 1 sec

1 Sec

= 4.905 metres. s = 4.905 metres

• A force of 30 N acts on a body at rest. The mass

of the body is 50 kg. Determine the velocity of the body

after 4 secs, the distance it covers during that period

and the acceleration

F = m x a

30 N = 50 kg x a

3 kg x metre/sec

= 50 kg x a

∴ acceleration = 3/50 metre/sec

= 0.06 metre/sec

a = 0.06 m/sec

v = u + at

= 0 + 0.06 metre/sec

x 4 sec = 0.24 metre/sec

s = ut + 1/2 at

= 0 + 1/2 x 0.06 metre/sec

x 16 sec

= 0.48 metre s = 0.48 metre

• A stone is thrown vertically upwards with a velocity

of 120 metre/sec. Determine (a) the maximum height

to which it travels before starting to return to earth. (b)

The total time taken by the stone to go up and come

down. (c) The velocity with which it will strike the

ground.

Initial velocity of throw = 120 metre/sec (u)

Final velocity = 0 metre/sec (v) (taken g = 10 m/sec

)

Retardation due to gravity = 10 metre/sec

–v

= 2g.s

∴ 120

metre

/sec

– 0 = 2 x 10 metre/sec

x s

∴ s = 120 x120/2 x 10 metre

= 720 metre

when it comes down its velocity at start = 0 metre/sec.

The acceleration due to gravity = 10 metre/sec

and the

distance travelled = 720 metre

∴ v

– u

= 2as v

-0 = 2x 10 m/sec

x 720 m

– 0 = 2 x 10 x 720 metre

/sec

sec/m 14400v

∴ v = 120 metre/sec

Time taken to go up and reach a velocity of 0 metre/sec

= u/g = 120 metre/sec/10 metre/sec

= 12 sec.

Time taken to start from rest and attain a velocity of 120

metre/sec = v/g = 12 sec.

∴ Total time taken = 24 sec.

Workshop Calculation And Science : (NSQF) Exercise 1.9.28

• Calculate the Angular velocity in radian/second of

an engine flywheel when it is rotating at 2800 rpm.

(Fig 1 & 2)

Angular velocity (W) = This is the rate of change of

displacement or angle turned through per unit time.

Solution

Angular velocity of flywheel W= 2πN/60 rad/sec.

[N = 2800 rpm]

= 2π x 2800/60 radian/sec.

= 293.3 radian/sec.

• A motor car road wheel of dia 540 mm turns through an

angle of 120°. Calculate the distance moved by a point

on tyre thread of the wheel.

Solution

There are 2π radians in one turn of wheel. i.e 2π radians

= 360°

Since wheel turns 120° angle, 120° = 120 x 2π/360

= 2.094 radians

Distance moved by a point on tyre S – re

[where r = 270 mm

θ = 2.094 radian]

S = 270 x 2.094 mm

= 565.38 mm

Circumferential distance moved by the point = 565.38

• The rear wheels of a car have diameter of 600 mm. The

rear axle makes 250 rpm. Find out the peripheral speed

of rear wheels in m/sec.

Solution

Peripheral speed V =

(m/s)

1000

πdN



m/sec7.85

1000

250

6003.14







• Calculate the stopping distance of a car travelling with

a speed of 72 km/h and being accelerated with

a - 5 m/ sec

Solution

Va (initial speed of a car) = 72 kmph

m/sec

72 m/sec)

3600

1000

kmph (1 

= 20 metres/sec

Stopping distance S =

(metre)

40metre

400







ASSIGNMENT

1 S = 180 mm

n = 65 (double stroke)

Vm = _____metre/min

Vm is average cutting

speed)

2 V = 16 metre/min

s = 210 mm

n = ______

(V is the cutting speed)

3 n = 22 strokes (Double

stroke)/min

V = 18 metre/min

s = ______ mm

4 s = 240 mm

n = 30 (working stroke)

V = ______ metre/min

5 n = 50 cutting strokes

V = 32 metre/min

d = ______ mm

6 s = 64 mm

n = 3600 rpm

Vm = _____metre/sec

Vm is the average

piston speed)

7 Vm = 0.35 metre/sec

s = 200 mm

n = ______ rpm

s = 650 mm

Vm = 90 metre/min

n = ______ rpm

9Vm

= 5.2 metre/sec

Increased to

= 6.3 metre/sec

Increase in n (rpm) = ______ %

10 s = 250 mm

n = 45 (double

strokes)

V = ______metre/min

11 Is : Vm = 25 : 1

n =_____(double

strokes)

Is = rack travel

Vxm = stroke speed/

min

12 Vm = 10 metre/min.

n = 12.5 / min.

Rack travel = ______

13 dia of crank = 100 mm

Rack

speed = 12 metre/min

Crank disc 'n'' = _____

rpm

Workshop Calculation And Science : (NSQF) Exercise 1.9.28

14 spindle 'n' = 250 rpm

Average stroke speed

= 30 metre/min

stroke length =

________mm

15 Car Speed = 90 km/hr

Time to stop = 10 sec

Deceleration

= ________ metre/

sec

16 Car speed = 80 km/hr

Distance stopped = 60

metre

Deceleration of car

=________ metre/sec

17 Deceleration = 4.5 m/

sec

Stopping distance = 50

metres

Velocity of car

=________ km/hr

18 Distance travelled by

car = 600 km

Time = 8 hrs 20 min

Average velocity

= ________ km/hr

19 Average velocity

= 56.3 km/hr

Distance travelled

= 464.475 km

Travelling time

= ________ hrs

20 Car wheel

n = 720 rpm

Peripheral speed

= 18.84 m/sec

d = ________

21 Angular speed

n = 2000 rpm

Angular velocity

= _______ radians/sec

Use

W = 2πN/60 rad/sec

22 Piston Velocity/Speed

S = 74 mm

n = 4500 rpm

Mean velocity

= _______ m/sec

Maximum velocity

5= _______ m/sec

(Average Speed of

Piston)

23 Total Stopping Distance

time reactionvelocity



(Use = V

/2a)

V = Vehicle speed = 80 km/hr

Deceleration = 5m/sec

Reaction Time of driver = 2 seconds

Total Stopping Distance = _______ meter

Workshop Calculation And Science : (NSQF) Exercise 1.9.28

Units of work, power and energy, horse power of engines and

mechanical efficiency Exercise 1.10.29

F - force or weight force in N

s - distance the body on which force acts is moved in

metres

t - time in seconds

v - speed in metre/sec

w - work done by the force in joules

P - Power in Watts

out

- Power output

- Power input

Force

A Force is that which changes or tends to change the state

of rest or motion of a body.

Force = Mass x Acceleration

F= Ma

Unit

F = M x a

= kg x m/sec

= 1 Newton (SI unit)

(Newton: If 1 kg of mass accelerates at the rate

of 1m/sec

then the force exerted on the mass

is 1 newton)

FPS = 1 pound x 1 Feet/second

= 1 pound

CGS = 1 gm x 1cm/second

= Dyne

Mks = 1 kg x 1m/second

= Newton.

1 Newton = 10

dynes

1kg wt = 9.81N

1 pound = 4.448N,

Newton = 0.225 pound.

Work(Fig 1)

Work is said to be done by a force, when it moves, its point

of application through a distance. Applied for ‘F’ moves a

body through a distance’s.

Work done ‘W’ = F x s.

The S.I. unit of work is 1 joule which is the work done by a

force of moving the body through a distance of 1 metre.

Therefore joule = 1 N x 1 metre = 1 Nm

Also 1 joule = 1 Nm = 10

dynes x 100 cm = 10

dynes cm

= 10

ergs.

Absolute units (Fig 2)

In C.G.S. system unit of work = 1 erg = 1 dyne x 1 cm

In F.P.S system unit of work = 1 Foot poundal = 1 poundal

x 1 foot

In M.K.S. system unit of work = 1 joule = 1 Newton x 1

metre

Derived units

C.G.S. system 1 Gm Wt x 1 cm = 981 ergs.

F.P.S. system 1 ft LB = 981 foot poundal

M.K.S. system 1 kgf metre = 981 joule.

Power(Fig 2)

It is the work done in unit time.

which is equal to 1 Watt. Power in watts = w/t = F.s/t

= F x V

In M.K.S. system the unit is 1 kgf meter/sec. One horse

power is = 75 kg metre/sec or 4500 kgf metre/min.

1HP (metric) = 735.5 Watts

1HP (British) = 746 Watts = 0.746 KW

1 KW = 1.34 HP

Power input is the power given to a machine to do work.

Power output is what we get out of the machine. Power

output is always less than power input due to friction in the

machine. The ratio between power output to power input

is efficiency of the machine and it is expressed

percentage.(Fig 3)

100%

input power

output power

efficiency 

Indicated Horse Power and Brake Horse Power

The power actually generated by the engine or generator is

the indicated horse power which is indicated on the plate.

The Brake horse power is the power available to do useful

work. B.H.P is always less than I.H.P. due to losses to

overcome frictional resistance.

100%

I.H.P

B.H.P

efficiency mechanical 

Work done by a force = Magnitude of the force x distance

moved by the body

Power = Total work done / total time taken

100%

input power

output power

efficiency 

Energy

The energy of a body is its capacity to do work. It is equal

to power x time. Hence the unit of energy is the same as

the unit of work in all systems.

Forms of energy

Mechanical energy, Electrical energy, Atomic energy,

Heat energy, Light energy, Chemical energy, sound en-

ergy. Energy of one form can be transformed into energy

of another form.

Law of conservation of energy

– The energy can neither be created nor destroyed.

– Total energy possessed by a body remains the

same.(Fig 4)

Depending upon the position of the body or body in motion,

mechanical energy possessed by the body may be

potential energy or kinetic energy respectively.

Potential energy

Potential energy is the energy possessed by a body by

virtue of its position (Fig 4). A body of mass ‘m’ kept at a

height ‘h’ from a datum possesses a potential energy of

mgh or Wh or Fh; where W or F are the Weight force. When

the body is allowed to fall it will be able to do a useful work

of Fh.

Example

• Water stored in a Tank

• Coil Spring.

Kinetic energy

It is the energy possessed by a body by virtue of its motion.

If a body of mass ‘m’ starting from rest attains a velocity of

‘v’ after covering a distance of ‘s’, by the action of an applied

force ‘F’, then work done on the body=F x s But F= m x a.

Therefore work done on the body = m x a x s.

But a x s =

because the body is starting from rest.

Therefore Work done on the body =

Since work done on the body = The energy possessed by

the body

Kinetic Energy =

Energy possessed by a body = work done on the body

Potential energy = mgh

Kinetic energy =

If friction is neglected potential energy = Kinetic energy

Example

• Rolling vehicle

• Rotating fly wheel

• Flowing water

• Falling weight

Workshop Calculation And Science : (NSQF) Exercise 1.10.29

Potential energy, kinetic energy and related problems with assignment

Exercise 1.10.30

Potential energy

Hammer head drops from height ‘h’ . m = 10 kg.

h = 1.4 m.

sec

metre

0u 

= 2 gs

= 2 x 9.81 x 1.4

= 27.468

= 5.24 m/sec

= 10 kg x 9.81 metre/sec

x 1.4 metre (Fig 5)

= 137.3 N metre (



1N = 1kg.m/sec

)

K.E = x 10 kg x 5.24

sec

metre

= 137.3 N metre.

Examples

• A pulley is used to lift a mass with a force of 900 N to

a height of 10 metres in 2 minutes. Find the work done

by the force and also the power.(Fig 6)

Work done = F x s = 900N x 10 metre

= 9000 Nm = 9000 joules.

Power

120sec

joules 9000



watts75

sec

joules75



• Determine the horse power required to drive a lift in

raising a load of 2000 kgf at a speed of 2 metre/sec, if

the efficiency is 70%.

Useful workdone to raise the lift in 1 sec

Force = 2000 kgf

Work = F x d

22000

Power









= 4000 w

Power output = 4000 w

Power input = Power output

w5714

0.7

4000



100%

Input

Output





HP7.67.659

746

5714

HP 

Power input = 7.6 HP

• A mass of 100 gm is allowed to fall from a height of 10

metres. Determine the amount of Kinetic energy

gained by the body. (Take the value of g as 10 metre/

sec

Since inital velocity is 0 and distance travelled is 10

metres. final velocity

= 2 x g x s = 2 x 10x10 metre

/sec

Joules. 10

ergs10 10

/secmetre gm 10000

/secmetres 200 gm 100

K.E

222









K.E. developed by the vehicle at a constant speed

• A motor vehicle of one tonne is travelling at 60 km/hr.

Calculate K.E of the vehicle at this speed.

K.E of the vehicle =

Where m = one tonne or 1000 kg

v = 60 km/hr

Workshop Calculation And Science : (NSQF) Exercise 1.10.30

Solution

Changing v into meter/sec we get,

( 1km = 1000m)

( 1hour = 3600sec)

K.E.developed by a vehicle during acceleration

• A motor vehicle of 1200 Kg mass is being accelerated

36 km to 48 km/hr speed. Calculate the increase in K.E

during its acceleration.

Solution

Mass of motor vehicle= 1200 kg

K.E. of the vehicle at 36 km/hr speed

= x 1200 x 36

J KE= mv

v= 36km/hr =36 x = 10m/sec

K.E of the vehicle at 48 km/hr speed

= x 1200 x 48

J ( 1kg.m/sec

= 1N)

( 1Nm = 1J)

v= 48 km/hr =48 x= m/sec

KE = x1200 x 10 x10=60000J

KE = x1200 x x =106666.67 J

Increase in K.E of the vehicle = 106666.67 J - 60000J

= 46666.67 J

= 46.666 KJ.

Workdone in vehicle operation

The Mechanical Work performed by the motor vehicle for its

propellsion on road can generally be classified into two

major categories of work done.

– Workdone by the IC engine in developing full power

under all condition of speed and load.

– Workdone by the motor vehicle in performing different

operations on road like hill climbing/acceleration/brak-

ing/ towing and reversing operation

ASSIGNMENT

1 m= 55 kg

a) s = 1.82 metres

W = ______ joules

b) s = 1.40 metres

W= ____ joules

c) s = 0.85 metres

W =Joules

2 t = 8 secs

a) P = _____ Watts

b) P = _____ Watts

c) P = _____ Watts

3 W = 1312.5 Joules

m = 350 kg

s = _____ metres

4 m = 75 kg

s = 100 metres

t = 12 secs

W = ______ Nm

P = ______ Watts

5 V = 1 m

/min

H = 2 m

η = 0.75

Power input = ______

6 P = 12 kw

s = 4 metres

t = 20 secs

m = ______ kg

7 d = 3 metre

H = 2 metre

t = 20 minutes

s = 6 metres

P = ______ kW

Water filled in the

tank. s is the pumping

height

8 d = 200 mm

n = 750 rpm

F = 700 N

P = _____ kW

9 P input = 4 kW

P output

= 3450 Joules/sec

η = ______ %

10 Volume of water

‘V’ = 10 metre

H = 18 metres

t = 20 sec

η = 70 %

P output

= ______ kW

11 d = 225 mm

s = 450 mm

Piston pressure

‘P’ = 4.5 bar

V = 2.5 metre/sec

(piston speed)

η = 70 %

Power input

= _____ kW

‘V’ of water pumped

= 3 metre3/min

H = 6 metre

η = 0.8

Power input

= ______ kW

Workshop Calculation And Science : (NSQF) Exercise 1.10.30

Heat

It is a form of energy. Heat energy can be transformed into

other forms of energies. Heat flows from a hotter body to a

colder body. (Fig 1)

Calorific value: The amount of heat released by the

complete combustion of unit quantity of the fuel (Mass or

volume) is known as calorific value of fuels.

Water equivalent

It is the mass of water which will absorb the same amount

of heat as the given substance for the same temperature

rise. Water equivalent = Mass of the substance x specific

heat of the substance.

Therefore water equivalent = ms

Types of heat

1 Sensible heat and

2 Latent heat

1 Sensible heat

Sensible heat is the heat absorbed or given off by a

substance without changing its physical state. It is

sensible and can be obsorbed by the variation of tempera-

ture in the thermometers.

2 Latent heat

The heat gained or given by the substance during a change

of state (from solid to liquid to gas) is called latent heat or

hidden heat. The heat absorbed or given off does not cause

and temperature change in the substance.

Types, 1. Latent heat of fusion of solid

2. Latent heat of vaporisation of solid.

1 Latent heat of fusion of solid

The amount of heat required per unit mass of a substance

at melting point to convert it from the solid to the liquid state

is called latent heat of fusion of solid. Its unit is cal/gram.

Latent heat of fusion of ice

The amount of heat required to convert per unit mass of the

ice into water at 0

C temperature is called latent heat of

fusion of ice.

Latent heat of fusion of ice(L) = 80 cal/gram

2 Latent heat of vapourisation of liquid

The amount of heat required to vaporise a unit mass of liquid

at its boiling point is called latent heat of vapourisation.

Latent heat of vaporisation of water or latent heat of

steam

The amount of heat required to convert into steam of a unit

mass of water at its boiling point (100

C) is called latent

heat of vaporisation of water or latent heat of steam.

Latent heat of steam(L) = 540 cal/gram

Units of heat

Calorie: It is the quantity of heat required to raise the

temperature of 1 gram of water through 1°C.

BTHU: It is the quantity of heat required to raise 1 lb of water

through 1°F. (British thermal unit).

C.H.U; It is the quantity of heat required to raise 1 lb of

water through 1°C.

Joule : S.I. Unit (1 Calorie = 4.186 joule)

Effects of heat

• Change in temperature

• Change in size

• Change in state

• Change in structure

• Change in Physical properties

Specific heat

The quantity of heat required to raise the temperature of one

gm of a substance through 1

C is called specific heat. It

is denoted by the letter ‘s’.

Specific heat of water = 1

Aluminium = 0.22

Copper = 0.1

Iron = 0.12

Thermal capacity:

It is the amount of heat required to raise the temperature of

a substance through 1

C is called the thermal capacity of

the substance.

Thermal capacity = ms calories.

Concept of heat and temperature, effects of heat, thermometer scale, celcius,

fahrenheit, reaumer, kelvin and difference between heat and temperature

Exercise 1.11.31

Temperature

It is the degree of hotness or coldness of a body. The

temperature is measured by thermometers.

Temperature Scales

Temperatures are caliberated between two fixed reference

points namely the freezing point of water, and the boiling

point of water. These two fixed points on different tempera-

ture scales are:

Scale Freezing point Boiling point

Centigrade (°C) 0°C 100°C

Fahrenheit (°F) 32°F 212°F

Kelvin (K) 273°K 373°K

Reaumur (°R) 0°R 80°R

Workshop Calculation And Science : (NSQF) Exercise 1.11.31

Difference between heat and temperature

Heat Temperature

1 It is a form of energy. This tells the state of heat.

2 Its unit is calorie. Its unit is degree.

3 Heat is measured by calorimeter. Temperature is measured by thermometer.

4 By adding quantity of heat of two substances their By adding two temperatures we cannot find the

total heat can be calculated. temperature of the mixture.

5 By heating a substance the quantity of heat is Two substances may read the same temperature though

increased regardless of increase in temperature. they might be having different amount of heat in them.

Heat is a form of energy. Temperature is the

degree of hotness or coldness of a body. The

relationship for conversion from one tempera-

ture scale to the others is

180

32F

100

273K

100

0000









Boiling point

Any substance starts turning into a gas shows the tem-

perature at which it boils this is known as the boiling point.

The boiling point of water is 100

Melting point

The temperature at which any solid melts into liquid or

liquid freezing to solid is called the melting point of

substance. ` The melting point of ice is 0

Conversion between centigrade, fahrenheit, reaumer, kelvin scale of

temperature Exercise 1.11.32

1 Convert 0

C into

100

180

32F





180

100

32F



180

100

32F



°F = 0 + 32

= 32

C = 32

2 Convert -40

C into

100

180

32F





180

100

32F



180

100

32F







F - 32 = -72

°F = -72 + 32

= -40

-40

C = -40

3 Convert 37

C into k

100

273K

100





°K-273 = C

°K = C + 273

°K = 37 + 273

= 310 K

C = 310K

4 Convert 70

C into Reaumer

100



100



5680

100



C= 56

5 Convert -25

F into

180

32F

100





180

3225

100

C 





100

180

57-



31.66

285







- 25

F = -31.7

6 Convert 98.6

F into

100

180

32F







100

180

3298.6







100

180

66.6



C37

180

6660



98.6

F = 37

ASSIGNMENT

Convert the following

1 10.5

C = ___

240

C = ___

360

C = ___

480

C = ___

5 105

C = ___

6 -100

C = ___

777

F = ___

820

F = ___

9 428

F = ___

10 -210

F = ___

11 72

R = ___

12 143

C = ___K

13 373

K = ___

14 746

K = ___

15 At what temperature will the reading of a fahrenheit

thermometer be double of a centigrade one.

Temperature measuring instruments, types of thermometer, pyrometer and

transmission of heat Exercise 1.11.33

Measuring heat energy

Energy can be released in chemical reactions as light,

sound or electrical energy. But it is most often released as

heat energy. This allows us to easily measure the amount

of heat energy transferred.

The apparatus used to measure the amount of heat by

mixer method is called calorimeter. It is nothing but

cylindrical shaped vessel and a stirrer made out of mostly

copper.

In a calorimeter when the hotter solid/liquid substance are

mixed with the cooler solid/liquid substances, heat transfer

takes place until both substances reach the same tem-

perature. By the same time calorimeter also reaches the

same temperature. By mixing rule,

⎥

⎦

⎤

⎢

⎣

⎡

⎥

⎦

⎤

⎢

⎣

⎡

⎥

⎦

⎤

rcalorimeteby

edHeat absorb

substance

idsolid / liqu

ed byHeat absorb

liquid

by solid/

Loss of heat

Measurement

Temperature is generally measured in degrees Celsius. In

this system the freezing point of water is defined as 0°C and

the boiling point of water is defined as 100°C. The Kelvin

temperature scale begins from absolute 0. i.e.— 273°. The

temperature intervals are the same.

∴ 273K = 0°C, 20°C = 273K + 20°C = 293K.

Instruments

The instruments used to measure and read temperature

takes into account changes in the properties of materials,

electrical phenomena incandescence, radiation and

melting.

Thermometer

They are based on the principle that liquids and solids

expand when they are subjected to heat. Mercury and

alcohol expand uniformly. When heat is applied the volume

of the liquid increases and the liquid rises in the capillary

tube integral with the container. Mostly mercury is used in

this type of thermometers because of its properties (Shiny

Pyrometer

Thermoelectric pyrometer is based on the principle that the

soldering point between the wires of different metals, when

heated a contact voltage is generated. The voltage depends

upon the temperature difference between the hot measuring

point and the cold end of the wire. Thermocouple elements

are constructed of copper and Constant (up to 600°C) or of

platinum and platinum-rhodium (up to 1600°C)

Radiation pyrometers are used to measure temperatures

of red hot metals up to 3000°C. These concentrate thermal

rays through an optical lens and focus them on to a thermo

element. The scale of the ammeter is calibrated in degrees

Celsius or Kelvin.

and will not adhere to the glass tubes and we can measure

up to 300

The bimetal thermometer consists of metals with different

coefficient of expansion. The bimetal is twisted into a spiral

which curls when the temperature rises.

Workshop Calculation And Science : (NSQF) Exercise 1.11.33

Transmission of Heat

Heat is a form of energy and is capable of doing work. Heat

flows from a hot body to a colder body or from a point of high

temperature to a point of low temperature. The greater is

the temperature difference the more rapidly will be the heat

flow. Heat is transmitted in three ways.

• By Conduction

• By Convection

• By Radiation

Conduction

Conduction is the name given to the transmission of heat

energy by contact. The heat source is in contact with the

Conductor. (metal rod). The rod is in contact with a

thermometer. Due to Conduction heat is transferred from

the heated end to the free end. In general good electrical

conductors are also good heat conductors and good

electrical insulators are also good heat insulators. A good

heat insulator does not necessarily withstand high

temperature.

Convection

Convection is the name given to the transmission of heat

energy by the up-ward flow. When heated, the fluid (liquid/

gas) becomes less dense and because of its mobility, is

displaced upwards, by a similar but colder and more dense

fluid. e.g., The domestic hot water system, The cooling

system in motor cars.

Radiation

Heat is radiated or transmitted from one object to the other

in space without actually being in contact, by means of

electro-magnetic waves. These waves are similar to light

waves and radio waves. They can be refracted by lenses

and reflected by mirrors. This radiation is called infrared. It

requires no medium to carry the radiation. (e.g) The heat of

the sun travels through the space.

Transmission of heat takes place in three ways

Conduction, Convection and Radiation.

Expansion due to heat

When a solid, liquid or gaseous substance is heated, it

expands and volume is increased. Similarly when it is

cooled, it contracts (shrinks) and volume is decreased.

E.g : small gaps are left in between the lines of railway track

to allow for expansion during summer. If this is not done,

the rails would expand and bend there by causing derail-

ment of trains.

Except a few substances, all solids, liquids and gases

expand. For the same amount of heat given, the expansion

of liquids is greater than solid and expansion of gas is more

than liquid.

Volume of water is reducing while heating from 0

C to 4

After that volume is increasing. The data at 4

C of water will

be taken as reference point for any calculations relating

with water.

Co-efficient of linear expansion and related problems with assignment

Exercise 1.11.34

Expansion of solids

A solid substance shows the following types of expansion

when heated.

1 Linear expansion

2 Superficial expansion and

3 Cubical expansion

1 Linear expansion

When a solid is heated, its length increases. This is called

linear expansion. It depends upon the material, original

length and change in temperature.

Co-efficient of linear expansion

The co-efficient of linear expansion is the change in length

per unit original length per degree rise in temperature. It is

represented by α (Alpha).

Length of the solid at t

C = l

Length of the solid at t

C = l

Change in Temperature = t

- t

Change in length = l

- l





ttt

ttl

121















etemperatur in change x length Original

length in Change

expansion linear

of efficientCo

−

⎭

⎬

⎫

Increased length l

-l

= αl

Final length l

= l

(1 + αt)

2. Superficial expansion

When a solid is heated, its area increases is called su-

perficial expansion.

Co-efficient of superficial expansion

The increase in area per unit original area per degree rise

in temperature is called co-efficient of superficial expan-

sion. It is represented by β (Beta).

Co-efficient of superficial

Expansion = 2 x linear expansion

β = 2α

3. Cubical expansion

When a solid is heated, its volume increases is called

cubical expansion.

Co-efficient of cubical expansion

The increase in volume per unit original volume per degree

rise in temperature. It is represented by γ (Gama).

Co-efficient of cubical expansion

= 3 x linear expansion

γ = 3α

Examples

Find the co-efficient of linear expansion. If an 8 metre

long metal rod is heated from 30

C to 80

C. So that it

may produce an elongation of 0.84 mm.

Initial length (l) = 8m

Increased length = 0.84 mm

Increased temperature(t) = 80 - 30 = 50







)expansion(

linear of effiecient-Co



temp Increasedlength Initial

length Increased





508000

0.84





400000

0.84



= 2.1 x 10

-6

If iron bridge is 100 metre long at 0

C. What will be

the length of bridge if the temperature is 40

C and

the co-efficient of linear expansion is 12 x 10

-6

per

degree.

Initial length of iron bridge = 100 m

Increased temperature = 40 - 0 = 40







)expansion(

linear of effiecient-Co



temp Increasedlength Initial

length Increased





40100

length Increased

1012







40100

1000000

length Increased 

= 0.048 m

Iron bridge at 40

C = 100 + 0.048 = 100.048 m

The length of a metal rod is 100 cm at 30

C and 100.14

cm at 100

C. Calculate the co-efficient of linear ex-

pansion and the rod length in 0

Initial length at 30

C = 100 cm

Final length at 100

C = 100.14 cm

Increased length = 0.14 cm

Increased temperature = 100 - 30 = 70

Workshop Calculation And Science : (NSQF) Exercise 1.11.34







)expansion(

linear of effiecient-Co



temp Increasedlength Initial

length Increased





70100

0.14





10070100





100000



= 2 x 10

-5

To find the length at 0

= l

(1 + αt)

100 = l

(1 + 2 x 10

-5

x 30)

100 = l

(1 + 0.0006)

0.00061

100





Length at 0°C = 99.94 m

Find the change in length of metallic rod 100 cm

long, when its temperature is increased from 25

C to

C and the co-efficient of linear expansion is 10 x

-6

Initial length = 100 cm

Increased temperature = 40 - 25 = 15

Co-efficient of linear = 10 x 10

-6

expansion (α)







)expansion(

linear of effiecient-Co



temp Increasedlength Initial

length Increased





15100

length Increased

1010







Increased length = 10 x 10

-6

x 100 x 15

1000000

1510010 



0.015cm

1000



Find out the temperature that the rod will extend by

0.54 mm in linear direction piece of metal rod is 2.5

metre long and the co-efficient of linear expansion

is 10.4 x 10

-6

per degree centigrade.

Initial length = 2.5 m = 2500 mm

Increased length = 0.54 mm

Initial temperature = 20

Co-efficient of linear = 10.4 x 10

-6

expansion (α)







)expansion(

linear of effiecient-Co



temp Increasedlength Initial

length Increased





temp Increased2500

0.54

1010.4







Increased temperature =

1010.42500

0.54





10.42500

10000000.54





C20.77

260

5400



Final temperature = 20 + 20.77

= 40.77

ASSIGNMENT

Co-efficient of linear expansion

1 Calculate the co-efficient of linear expansion of rod. If

rod is found to be 100m long at 20

C and 100.14m long

at 100

2 Find the change in length if the co-efficient of linear

expansion of rod is 0.00024/

C and the temperature of

a rod of 3.6m length is raised by 120

3 find the change in length if the co-efficient of linear ex-

pansion of rod is 0.00024/

C. If the temperature of a

rod of 6m length is raised by 120

4 Find the increase in length 100 cm iron rod if the tem-

perature raise from 40

C to 90

C. The co-efficient of

linear expansion of the iron is 10x10

-6

5 If micrometer reading is standardised at 15

C. What

will be the true reading of the micrometer if the reading

taken at 35

C is 20.20 mm?

The co-efficient of linear expansion of material of mi-

crometer is 11 x 10

-6

Problems of heat loss and heat gain with assignment Exercise 1.11.35

Mixing of heat

- Mass of first substance

- specific heat of first substance

- mass of 2nd substance

- specific heat of 2nd substance

- temperature of mixture

m - mass

Q - Quantity of heat

δt/Δt - temperature difference

- temperature of the mixture.

Unit of amount of heat

The derived unit for the amount of heat is S.I. unit is 1

joule (j).

Specific heat

It is also expressed as the amount of heat required to raise

the temperature of unit mass of a substance through 1°C.

In S.I. unit in order to heat a mass of 1 kg of water through

1°C,

the amount of heat needed or the

mechanical equivalent of heat = 4186 joules

= 4.2 kj/kg°C.

Quantity of heat needed for a substance to rise the

temperature

The amount of heat needed for heating 1 kg of the

substance through 1°C is equal to the specific heat of the

substance ‘s’. For heating a mass of ‘m’ kg of the

substance to attain a temperature difference of t,

the quantity of heat needed = m x s x Δt

Therefore Q = m x s x Δt.

Mixing

When there is an exchange of temperatures, there is an

exchange in the amount of heat. When hotter bodies

involve with colder substances, heat transference takes

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

place from hotter substances to the colder substances

until the mixture or both the substances acquire the same

temperature.

Heat lost by the bodies at higher temperature

= Heat gained by the bodies at lower temperature and

hence the total amount of heat of the component

substances

= amount of heat in the mixture.

Heat loss by hot substance =

Heat gained by colder substance

S of the component amounts of heat =

amount of heat in the mixture

x s

x t

+ m

x s

x t

= (m

+ m

)tm.

Example

A bath tub contains 40 litres of water at 15°C and 80 litres

of water at 60°C is poured to it. What is the temperature

of the mixture.

x s

x t

+ m

x s

x t

= (m

+ m

)tm.

C60 x

Ckg

kg 4.2

x kg 80 C15 x

Ckg

kj 4.2

x kg 40 









t60

Ckg

4.2kj

80kg15

Ckg

4.2kj

40kg















C45C

4.2120

22680







Examples

A container contains 25 kg of water. Initial tempera-

ture of container and water is 25

C. Calculate the

heat required to heat the water to the boiling

temperature of water. Assume water equivalent of

container = 1 kg.

Mass of the water (m) = 25 Kg.

Initial temperature of water and container = 25

Final temperature of water and container = 100

Increased temperature (t) = 100 - 25

= 75

Water equivalent (m s) = 1 Kg.

Required amount of heat to container = m s t

= 25 x 1 x 75

= 1875 K.cal.

Required amount of heat to container = m s t

= 1 x 75

= 75 K.cal.

Total required amount of heat = 1875 + 75

=1950 K.cal.

300 gram of water 25°C is mixed with 200 gram of

water at 85

C. Find out the final temperature of the

mixture assuming that no heat escapes.

i) Weight of water = 300 gram

Initial temperature = 25

Final temperature = Assume ‘X’

Temperature gained = x - 25

ii) Mass of water = 200 gram

Initial temperature = 85

Temperature lost = 85

C - x

Heat gained by 300 gram water = m s t

= 300 x 1 x (x - 25)

= 300 x -7500 cal.

Heat lost by 200 gram water = m s t

= 200 x 1 x (85 - x)

= 17000 - 200 x cal.

Heat gained = Heat lost

300 x -7500 = 17000 - 200 x

300 x + 200 x = 17000 + 7500

500 x = 24500

x =

C49

500

24500



Final temperature = 49

20gm of common salt at 91

C immersed in 250 gram

of turpentine oil at 13

C. The final temperature is

found to be 16

C. If the specific heat of turpentine oil

is 0.428. Calculate the specific heat of common salt.

Mass of the salt(m) = 20 gram

Initial temperature(t) = 91

Mass of the turpentine(m) = 250 gram

Initial temperature(t) = 13

Specific heat of turpentine(s) = 0.428

Final temperature of mixture = 16

Heat gained by turpentine(Q) = m s t

= 250 x 0.428 x (16-13)

= 250 x 0.428

= 321 calories.

Heat lost by salt (Q) = m s t

= 20 x s x (91-16)

= 20 x s x 75

= 1500 s calories

Heat lost = Heat gained

1500 s = 321

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

1500

321

s 

Specific heat of salt = 0.214

If copper calorimeter contains 80 gram of water at

C. The water equivalent of calorimeter is 20 gm.

What will be be resultant temperature of the mix-

ture, when 100 gm of water at 40

C is added to the

mixture?

Mass of the water in calorimeter = 80 gram

Temperature = 20

Final temperature of the mixture = Assume ‘x’

Temperature raised in water = x - 20

Specific heat of calorimeter(ms) = 20 gram

Mass of water added = 100 gram

Temperature = 40

Temperature lost = 40 - x

Heat gained

Heat gained by water in calorimeter = m s t

= 80 x 1 x (x - 20)

= 80 x (-1600)

Heat gained by calorimeter = m s t

= 20 x (x - 20)

= 20 x (-400)

Heat lost

Heat lost by added water = m s t

= 100 x 1 x (40 - x)

= 4000 - 100x

Heat gained = Heat lost

80x (-1600) + 20x (-400) = 4000 - 100x

100x -2000 = 4000 - 100x

100x+100x = 4000 + 2000

200x = 6000

200

6000

x 

= 30

Final temperature = 30

Find the amount of heat required to boil 15 gram of

ice at -8

C. Latent heat of ice = 336 joule/gm. Latent

heat of steam = 2268 J/gm. Relative specific heat of

ice = 0.5

Heat of ice cube

-8

C to 0

C Ice Q = mct kJ

= m x s x 4.2 x t kJ

= 0.015 x 0.5 x 4.2 x 8 kJ

= 0.252 kJ

C Ice to 0

C water = m x h s f kJ

= 0.015 x 336 kJ

= 5.04 kJ

C water to 100

C water = m c t kJ

= 0.015 x 4.2 x 100 kJ

= 6.3 kJ

100

C water to 100

C steam Q= m x h

sfg

= 0.015 x 2268 kJ

= 34.02 kJ

Total amount of heat Q = 0.252 + 5.04 + 6.3 + 34.02 kJ

Answer = 45.612 kJ

ASSIGNMENT

Mixing of Heat

1. m = 120 litres

= 20°C

= 85°C

s = 4.2

Q = ______ kj

2. m

=80 litres of water

=40 litres of water

= 10°C

= 70°C

= ______ °C

3. m

= 25 litres of water

= 12°C

= 70°C

= 33.75°C

= ______ litres

4. m

= 100 kg of water

= 12°C

= 50 kg of steel

= 600°C

CKg

4.2





CKg

0.46





Rise in temperature

=_____°C of water

5. m

= 250 litres of water

= 150 kg of steel

= 15°C

= 70°C

= ______ °C

6. m

= 20 litres of

machine oil

= 30 kg of steel

= 160°C

= 60°C

density of oil

= ______ °C = 0.91

7. m

= 10 litres

CKg

4.2

CKg

0.46

= 0.5 kg

= 18°C

= 780°C

Rise in

temperature =___°C

of water

8. m

= 60 gms of water

m = 70 gms of

calorimeter

= 80 gms of metal

= 20

= 95

= 25

s = 0.2

= 1

=_____

9. m

= 250 gms of oil

= 15

= 150 gms of brass

= 90

= 25°C

= 0.09 water

equivalent of = 3 gms

calorimeter

= ______ gms

Heat loss and heat gain

1 Calculate the amount of heat required to raise the

temperatureof 85.5 g. of sand from 20

C to 35

C. Spe-

cific heat of sand = 0.1

2 How much quantity of heat will be rejected in one hour,

if the rate of flow of water is 11 kg/min and the raise of

temperature of water is 12

3 Find out its specific heat. If we require 510 calories to

raise the temperature of 170/ gm of material 50

C to

4 Calculate the specific heat of metal piece. If 500gm

metal piece at 300

C is dropped in 5 kg of water. Its

temperature raises from 30

C to 75

C are no heat losses.

5 Find out the final temperature of mixture, assuming

that no heat escapes. If 300gm of water at 25

C is

mixed with 200gm of water at 85

6 What will be the resultant temperature of the mixture,

when 100gm of water at 40

C is added to the mixture.

If copper calorimeter contains 80 gm of water at 20

The water equivalent of calorimeter is 20 gm.

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Concept of pressure and its units in different system Exercise 1.11.36

Pressure

F - Force, Thrust

P - Pressure

Concept of pressure

The action of a force on unit area of the body on which it

acts is termed as pressure.

Pressure ‘P’ = Force or Thrust F per unit area ‘A’ (Fig 1)

A - Area of surface on which force acts

Example

A liquid gives force of 100 N over an area of 2m

. What is

the pressure?

Force = 100N

Area = 2m

Pressure = ?

= 50 N/m

Pressure

British unit Pounds per square inch lb/in

FPS

Metric units Gram per square contribute g/cm

CGS

MKS Kilogram per square metre kg/m

International Newtons per square metre N/m

circuits SI

Unit of pressure N/m

, 1 N/m

= 1 pascal.

This unit is too small (pressure of a fly on a area of

1cm

). Hence ‘bar’ is introduced as the unit of pressure.

1 bar = 10

pascal.

1 bar = 1000 mbar. [SI unit of pressure is pascal (Pa) and

Metric unit of pressure is bar].

Properties of pressure

1 The pressure in a liquid increases with increase in

depth.

2 The pressure at a point increases with the density of

the liquid.

3 The pressure is same in all directions about a point in

liquid at rest.

4 Upward pressure at a point in a liquid is equal to

downward pressure.

Saturation temperature of a fluid (liquid/vapour)

The temperature at which a liquid changes to vapour by

heating further at a given pressure. If this pressure is kept

constant, the heat removal process changes the vapour

into liquid at constant temperature. It is the saturated con-

dition of a fluid (Liquid/vapour).

Saturation pressure of a fluid

By adding or removing heat at a givcen temperature, there

is physical change is fluid. The given constant pressure is

known as saturation-pressure.

Critical temperature and pressure

Each fluid (liquid/gas) has its highest temperature and

pressure; below which the removal or addition of heat

changes physical state. If the conditions of temperature /

pressure exceeds the saturation level, the vapour cannot

be condensed into liquid. This vapour which cannot be

condensed into liquid, is called gas.

Super heated vapour

If the temperature of the vapour is raised above its satu-

rated temperature, it is called super heated vapour, for a

given pressure.

Insulating materials: Heat will flow from high tempera-

ture to low temperature. Heat flow by radiation, conduc-

tion and convection method through the wall, door, ceiling

and glass door to the refrigerated space.

The material which restricts such heat flow is called insu-

lating materials

Properties of insulating materials

- It is low conductivity

- Resistance to fire

- Less moisture absorption

- Good rigidity

- Odourless

- Vapour permeability

- Light in weight

Selection of insulating material: The following factors

are the prime importance in the selection of a proper insu-

lating material.

- Low thermal conductivity: Thermal conductance

value of a material is a measure of its effectiveness to

allow the flow of heat through it by conduction, obvi-

ously an insulating material should have a very low

thermal conductivity.

- Resistance to fire.

- Mechanical strength

- Low moisture absorption capacity

- Easy to lay

- Cost

- Easy of handling

- Low cost

Types of insulating materials

Glass wool, PUF, Cork sheet, Thermocole, Insulating foil,

fiber glass.

Types of insulating materials: Basic types of insulating

materials are inorganic fibrous or cellular materials. Ex-

ample, glass wool, slag wool ceramic products, as bestos,

etc. Organic fibrous materials, cork, cotton, rubber foam,

saw dust, rice husk, polystyrene, polyurethane,

phenotherm, etc. The type and form available as the appli-

cations of various insulations as follows.

Glass wool: Available as semi-rigid, resin bonded slabs/

sheets of different densities -higher density gives strength

and lower conductivity but allows vapour transmission.

Available with foil or other coverings.

Cork: Compressed and moulded into a rigid block, light

but strong, can be cut easily with a saw, resists water but

allows relatively high rate of water vapour transmission.

Expended polystyrene(Thermocole): it is available as

a rigid board, beads, moulded into shape for pipe/curved

surface, can be cut easily with a saw, light weight allows

relatively low vapour transmission.

Polyurethane: available as a rigid board, flexible board,

liquid can be sprayed on surfaces and allowed to foam,

can be used for in site applications.

Wood shaving/Saw dust: It needs good supporting com-

partment, can easily settle down. Fairly high conductivity

absorbs moisture/water.

Phenotherm: Available slabs with different facings, and

as performed pipe sections, can be easily cut with a saw.

Insulting materials and properties/specifications:

There are many insulating materials used in refrigeration

and air conditioning field. For our water tank only few of

them were in use.

Now-a-days the following insulating materials were in broad

use.

- Thermocole

- Glasswool/Tar felt

- Puf

- Fiberglass

Thermocole: It is one of the insulting materials in normal

use. It is available in low and high density. This is avail-

able in various thicknesses ranging 0.25" to 5".

Thermocole is available in various shapes (moulded) of

necessity.

Thermocole allows (Characteristically) low transmission

of vapour, thereby heat entry through is cut short. This

may vary with its low/high density.

It can be cut very easily even with knife to a required shape.

Thermocole withstands cool/heat for a longer time.

The 'K' factor of an insulation material follows (thermocole).

Thermocole -0.20 btu/hr Ft2 deg.f°/inch

Fibreglass: Also one of the insulating materials used for

its manufactured from inorganic materials (sand, dolomite,

limestone). Glass fibre insulation does not shrink due to

temperature variation.

This insulation materials used for higher temperatures also

upto 450°C (842°F).

Fibreglass products does not absorb moisture from the

amblent air.

Glass wool: Normally glass wool material is heavily thin

weighted object in layers, soft (touching). It comes off in

various sizes (thickness from 0.5" to 2.5". it comes in

white, yellow colours mixed up with broken glass pieces.

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Thermal conductivity and insulations Exercise 1.12.37

Handling glass wool is hazardous and harmful (if it is

breathed). Always it is advisable to handle glass wool with

gloves and goggles (eye) while working on it. It also comes

off in various densities.

Glass wools are of two types of uses. One type of glass

wool used for low temperature refrigeration/air condition-

ing purpose. The other type is used for boiler materials

(heat prevention) purposes.

The 'K' factor of insulation material:

Glasswool: 0.230-.27 Btu/Hr ft2 deg. F°/inch.

Puf: The other mode of insulating materials used in water

cooler at the evaporator tank's external body.

For this kind of insulation two chemicals used namely

isocyanide-R11., Both available in liquid form in bottles

(for lesser capacities) and cans (for higher capacities).

Both the liquids (chemicals) should always kept cool. When

both of them added in a container and stirred in few min-

utes it becomes foamy (initially with thin and becomes

thicker and becomes hard (sticks with the unit).

We should be careful that there is no air gap in the tank

covered. It foams out with high density and uneven finish

at the outer level.

Puf (materials) insulations are widely used by our

manufacturer's for their products as it keeps the tempera-

ture for a longer period.

The main disadvantage of the insulation is as soon as the

chemicals are mixed and stirred it should be poured over

the evaporator coil (or) outside the evaporator tank within

the shortest period. If the time exceeds the solution starts

framing at the container itself and becomes useless.

The evaporator tank should be covered well with wooden/

steel boards with required gaps for insulation tightened all

the corners well giving small gaps to pour the solution.

Method of laying duct insulation: when there is no

chance of moisture condensation on the duct, glass wool

can be used. Since it is economical and fire resistant.

However if moisture condensation can occur greater care

should be exercised in case of glass wool. First a uniform

coat of bitumen is applied to the duct surface and the

wool is stuck to the bitumen. The insulation is then cov-

ered with a polythene sheet which acts as a vapour bar-

rier. The surface can be plastered after spreading chicken

wire mesh as reinforcement.

Expanded polystyrene can be laid easily as it is rigid.

Bitumen is applied on the duct and the insulation is stuck

joints are also sealed with bitumen. No separate vapour

barrier is needed other than a coat of bitumen. The insula-

tion can be finished with cement and plaster or metal clad-

ding.

Purpose of false ceiling: The conditioned air arrives

through the ducts at the supply air diffusers and enters

the conditioned space. Most diffusers are attached to the

false ceiling and a variety of diffusers are available for dif-

ferent air spreading needs. The return air grills will be fixed

to the false ceiling. The false ceiling prevents mixing of

conditioned air and return air.

Return air usually flow into the plenum or return air box

through grill placed in the false ceiling. Since substantial

amount of energy goes into the air in the first place. It is

a practice to recycle to air. The air is therefore brought

back to the air conditioning. Plant room it is common to

route the return air through the gap between the false ceil-

ing and the main ceiling. A space referred to as a plenum,

the false ceiling is also known as a return air duct.

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Electricity is a kind of energy. It is the most useful sources

of energy which is not visible but its presense can be felt

by its effects. Electricity is obtained by conversion of

other forms of energy like heat energy, chemical energy,

nuclear energy, mechanical energy and energy stored in

water etc.,

To understand electricity, one must understand the struc-

ture of an atom.

Basically an atom contains electrons, protons and neu-

trons. The protons and neutrons are located in the centre

of an atom and the electrons, a negative electric charge

particle revolving around the nucleus in an atom. The pro-

ton has a positive charge. Neutrons are neutral and have

no charge.

Sources of electricity

Battery

Battery stores electrical energy in the form of chemical

energy and it gives power when required. Battery is used

in automobiles and electronics, etc.,

Generator

It is a machine which converts the mechanical enregy into

electrical energy.

When a conductor rotates between a magnetic field using

prime mover an emf will be induced. By using this method

all types of AC and DC generator - generates power.

E.g. Thermal power station

Hydro power station

Nuclear power station

Wind power station

Solar power station

Uses of Electricity

1 Lighting - Lamps

2 Heating - Heaters, ovens

3 Power - Motor, fan

4 Traction - Electromotives, lift, crane

5 Communication - Telephone, telegraph, radio,

wireless

6 Entertainment - Cinema, radio, T.V.

7 Medical - x-rays, shock traeatment

8 Chemical - Battery charging, electroplating

9 Magnetic - Temporary magnets

10 Engineering - Magnetic chucks, welding,

x-rays of welding

Molecule

All substances are made by tiny molecules. The smallest

part on any matter which has the properties of the parent

substance, is called molecule. The molecules of a sub-

stance are alike in shape and properties. Each molecules

possesses the physical and chemical properties of the

mother substance.

Atom

The are three particles in a atom. They are:

i Proton: The electron present in the middle of the atom

is called proton. It is in the nucleus of the atom. The

proton liquid-equilent is 1840 times heavier than an elec-

tron liquid equivalent and is equal to the atom of hydro-

gen. It has a positive charge.

ii Neutron: It is a basic particle which is in the nucleus

of atom. It is held fast with proton in the nucleus. It has

no charge. It increases the liquid equivalent of the cou-

lomb.

iii Electron: It is unstable or moving electron. It keeps

moving round the nucleus on different orbits. It carries

negative charge. It is 1/1240 of hydrogen atom liquid

equivalent. It charge is 1.6 x 10

-19

coulomb.

How electricity is produced

In the middle of each atom there is electricity. This is in

form of electrons. Those parts which are stable known as

protons and those moving are called electrons. The pro-

tons are of positive and electrons are negative charges.

By external force only the moving electrons are reduced

or increased as compared to protons and then only both

the charges combine. When the number of electrons pro-

duced are less than the number of protons, then it be-

comes a positive charge. Conversely when the number of

electrons produced are more tan the number of protons it

becomes negative. In this way electricity is produced by

atom.

Electrical terms and units

Quantity of electricity

The strength of the current in any conductor is equal to the

quantity of electrical charge that flows across any section

of it in one second. If ‘Q’ is the charge and ‘t’ is the time

taken

then I =

Q = I x t

The SI unit of current is coulomb. Coulomb is equivalent to

the charge contained in nearly 6.24 x 10

electrons.

Coulomb

In an electric circuit if one Ampere of current passes in one

second, then it is called one coulomb. It is also called

ampere second (As). Its larger unit is ampere hour (AH)

1 AH = 3600 As (or) 3600 coulomb

Introduction, use of electricity, molecule, atom, electricity is produced,

electric current, voltage, resistance and their units Exercise 1.13.38

Electro motive force (EMF)

It is the force which causes to flow the free electrons in any

closed circuit due to difference in electrical pressure or

potential. It is represented by ‘E.’ Its unit is Volt.

Potential difference (P.D)

This is the difference in electrical potential measured

across two points of the circuit. Potential difference is

always less than EMF. The supply voltage is called

potential difference. It is represented by V.

Voltage

It is the electric potential between two lines or phase and

neutral. Its unit is volt. Voltmeter is used to measure

voltage and it is connected parallel between the supply

terminals.

Volt

It is defined as when a current of 1 ampere flows through a

resistance of 1 ohm, it is said to have potential difference

of 1 volt.

Current

It is the flow of electrons in any conductor is called current.

It is represented by I and its unit is Ampere. Ammeter is

used to measure the current by connecting series with the

circuit.

Ampere

When 6.24 x 10

electrons flow in one second across any

cross section of any conductor, the current in it is one

ampere.(or) If the potential difference across the two ends

of a conductor is 1 volt and the resistance of conductor is

1 ohm then the current through is 1 ampere.

Resistance

It is the property of a substance to oppose to the flow of

electric current through it, is called resistance. Symbol: R,

Unit : Ohm (W), Ohm meter is used to measure the

resistance.

Ohm

If the potential difference across the two ends of conductor

is 1 volt and the current through it is 1 ampere, then the

resistance of the conductor is 1 Ohm.

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Ohm’s law, relation between V.I.R & problems, series & parallel circuits

problem Exercise 1.13.39

Ohm’s law

V - Voltage in volts

l - Current in Ampere

R - Resistance in ohms.

In any closed circuit the basic parametres of electricity

(Voltage, Current and resistance) are in a fixed relationship

to each other.

Basic values

To clarify the basic electrical values, they can be compared

to a water tap under pressure

Water pressure - electron pressure - Voltage

Amount of water - electron flow - Current

throttling of tap - obstruction to - Resistance

electron flow

Relationships

If the resistance is kept constant and the voltage is

increased, the current is increased

I ∞ V

If voltage is constant and the resistance is increased,

current is decreased

Ohm’s law

From the above two relationships we obtain Ohm’s law,

which is conveniently written as V = R.I.

Ohm’s law states that at constant temperature

the current passing through a closed circuit is

directly proportional to the potential differ-

ence, and inversely proportional to the resis-

tance.

By Ohm’s law

EXAMPLE

A bulb takes a current of 0.2 amps at a voltage of 3.6 volts.

Determine the resistance of the filament of the bulb to find

R. Given that V = 3.6 V and l = 0.2 A.

To find ‘R’. Given that V = 3.6V and I = 0.2 A

Therefore V = l x R

3.6 V = 0.2 A x R

Therefore

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Voltage drop = 12V – 10.8 = 1.2V

The supply voltage is called Potential differ-

ence.

Example

The Internal resistance of a dynamo is 0.1 ohm. The

voltage of dynamo is 12V. What is the Voltage dynamo

when a current of 20 amps being supplied to an outside

circuit.

Solution

Voltage drop = Current x Internal resistance

= 20 x 0.1 volts

= 2 volts

Example (Fig 6)

The Internal resistance of a Battery is 2 ohms. When a

resistance of 10 ohms is connected to a battery it draws 0.6

amps. What is the EMF of the battery.

P.D = Current flowing x Resistance

= 0.6 A x 10Ω

= 6 volts

V.D = Current flowing x Internal resistance of battery

= 0.6 x 2 volts

= 1.2 volts

EMF of the Battery = (6.00 + 1.2)V

= 7.2 volts

Example

The voltage supply to a filament lamp is 10.8V. The voltage

should be 12V. Find out loss of voltage.(Fig 5)

The total resistance is equal to the sum of all the resis-

tances. In a series connection the end of the first load is

connected to the beginning of the second load and all loads

are connected end to end.

Features of series connection:

• The same current flows through all the loads.

• The voltage across each load is proportional to the

resistance of the load.

• The sum of the voltages across each load is equal to the

applied voltage.

• The Total resistance is equal to the sum of all the

resistances.

l=l

= l

= ...

V= V

+ V

+ ...

R= R

+ R

+ ...

Parallel connection

In a parallel connection the beginning and the ends of the

loads are connected together.

Resistance connections

V - Voltage (in volts)

R - Resistance (in ohms)

I - Current intensity (in Amperes)

Series Connection

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Features of parallel connection:

• The current flowing through each load depends upon

the resistance of the load.

• The voltage across each load is the same and is equal

to the voltage applied to the circuit.

• The total resistance of a parallel connection is always

smaller than the smallest resistance in the circuit.

• In parallel connection the reciprocal of the total

resistance is equal to the sum of the reciprocals of all

resistances in the circuit.

l= l

+ l

+ ...

V= V

= V

...

.......... 

Example

Two resistances of 4 ohms and 6 ohms are connected in

parallel. Determine the total resistance.



(since parallel connection)

Therefore



Therefore R



ohms = 2.4 ohms

Example

Two resistors of 2 and 4 ohms are switched in parallel

to a 6V battery

– Calculate the total resistance

– Find the total current and partial current.

Solution

Total resistance

1+2

tot

21tot

I Total = I

+ I

current

Amp4.5 =

1.5A +3A = total I

1.5

3A=

But

Assume the given resistors in the assignment

as bulb with filaments and other current con-

suming devices like Horn, Wiper etc of the

vehicle.

100

Table of analogies between mechanical and electrical quantities

Mechanical quantity Unit Electrical quantity Unit

Force 'F’ N Voltage ‘V’ V

Velocity v =

Time

ntDisplaceme

m/s Current I A

Time t seconds Time t seconds

Power P = F x v N

sec

Power P = V x iW = V x A

Energy = F x v x tj = Nm Energy W = V x i x tj = W x s

Electrical power, energy and their units, calculation with assignment

Exercise 1.13.40

V - Voltage (Volts) V

i - Current Intensity (Ampheres) A

P - Power (Watts, Kilowatts) W, kW

W - Work, Energy (Watt hour, Kilowatt hour) wh, Kwh

t - time (hours) h

Electric Power

In mechanical terms we defined power as the rate of doing

work. The unit of power is Watt. In an electrical circuit also

the unit of electrical power is 1 Watt. In mechanical terms

1 Watt is the work done by a force of 1 N to move the body

through 1 metre in one second. In an electrical circuit, the

electromotive force overcomes the resistance and does

work. The rate of doing work depends upon the current

flowing in the circuit in amperes. When an e.m.f of one volt

causes a current of 1 ampere to flow the power is 1 Watt.

Hence Power = Voltage x Current

P= V x l

Power in Watts = Voltage in Volts x Current in Amperes

Electric work, energy

Electrical work or energy is the product of electrical power

and time

Work in Watt seconds = Power in Watts x time in sec-

onds

W = P x t

Since 1 joule represents 1 Watt x 1 sec, which is very

small, larger units such as 1 Watt hour and 1 kilowatt hour

are used.

1 W.h = 3600 Watt sec.

1 Kwh = 1000 Wh = 3600000 Watt sec

Note: The charge for electric consumption is

the energy cost per Kwh and it varies according

to the country and states.

W=V I

V=I R

101

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Example

1 Calculate the power rating of the lamp in the

circuit, if 0.25 amperes of current flows and the

voltage is 240 volts.

P = V x I

V = 240 Volts

I = 0.25 Amperes

Therefore Power = 240 Volts x 0.25 Amperes

= 60 Volts Ampere

But 1 Watt = 1 Volt x 1 Amphere

Therefore Power = 60 Watts

2 A current of 15 amperes flow through a resistance

of 10 Ohms. Calculate the power in kilowatts

consumed.

Given that R = 10 and I = 15A

Power = V x I = I x R x I = I

x R

Therefore Power = 15

x 10 = 2250 Watts = 2.25 kW

3 At a line voltage of 200 Volts a bulb consumes a

current of 0.91 ampheres. If the bulb is on for 12

hour calculate the work in Wh to find the work

given that V = 200 Volts.

I = 0.91 Amps.

t = 12 hours

Therefore Power=V x I = 200 Volts x 0.91 Amps

= 182 Watts

Therefore Work = P x t = 182 Watts x 12 hours

= 2184 Watt hour.

4 An adjustable resistor bears the following label:

1.5 k Ohms/0.08 A. What is its rated power?

Given: R = 1.5 k Ohms; I = 0.08 A

Find: P

V = R x I = 1500 Ohms.0.08 A = 120 volts

P = V x I = 120 volts.0.08 A = 9.6 W alternatively:

P = 1

.R = (0.08 A)

.1500 Ohms = 9.6 W.

5 Find the current and power consumed by an

electric iron having 110

ΩΩ

Ω resistance when feed

from a 220 v supply

Resistance of electric iron (R) = 110 ohms

Voltage (V) = 220 volts

Current (I) =

110

220

2 amperes

Power (w) = V x I

= 220 x 2

= 440 watts

6 Find the total power if four 1000 W, 180 volt

heaters are connected in series across 240 V

supply and current carrying capacity is 15 amp.

Find the total power.

Connection = Series

No. of heaters = 4

Heater power (W) = 1000 watts

Heater voltage = 180 V

Supply voltage = 240 V

Heater resistance (R) =

324

1000

180 x 180

= 32.4 ohms

Total resistance = 32.4 x 4 = 129.6 ohms

Total current (I) =

129.6

240

= 1.85 amperes

Total Power (W) = V x I

= 240 x 1.85 = 444 watts

7 If a 40 watt fluorescent lamp draws a current of

0.10 ampere. How much voltage will be required

to illuminate it?

Lamp power (W) = 40 watt

Current (I) = 0.10 ampere

Voltage (V) =

0.1

= 400 volts

8 Find the cost if running 15 HP motor for 15 days @

6 hrs per day and the cost of energy is Rs. 3 per unit.

Motor power (w) = 15 HP

= 15 x 746 = 11,190 watts

Consumption per day = 11,190 x 6

= 67140 = 67.14 KWH

Consumption for 15 days = 67.14 x 15

= 1007 KWH (or) unit

Cost per unit = Rs. 3

Cost for total energy = 3 x 1007 = Rs. 3021

102

parallel

2 ASSIGNMENT

1 ASSIGNMENT

1 R = 40 Ohms

I = 6.5 Amps

V = ______ Volts

2 V = 6 Volts

I = 0.5 Amps

R = ______ Ohms

3 V = 220 Volts

R = 820 Ohms

I = ______ Amps

4 I = 4.5 Amps

V = 220 Volts

R = ______ Ohms

5 R = 50 Ohms

V = 220 Volts

I = _____ Amps

6 V = 110 Volts

I = 4.55 Amps

R =_____ Ohms

7 R = 250 Ohms

I = 0.44 Amps

V = ______ Volts

8 I = 11.5 Amps

V = 380 Volts

R =______ Ohms.

9 R = 22 Ohms

= 7.8 Amps

(Voltage drop)

V = ______ Volt

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

= 12 ohms

= 22 ohms

= 24 ohms in series

R = ______ ohms

= 15 ohms

= 25 ohms

V = 220 V

= ______ V

3 V = 220 V

= 40 ohms

= 100 V

(in series)

= ohms

4 V = 80 V

I = 2 A

= 30 ohms

(in series)

= ______ ohms

5 R

= 6 ohms

= 12 ohms

= 18 ohms

R = ______ ohms

103

6 R = 6 ohms

are in parallel

= 12 ohms

= 16 ohms

= ______ ohms

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

7 R

= 40 ohms

= 60 ohms

V = 220 V

I = ______ A

= ______ A

3 ASSIGNMENT

1 Current Consumed

I = 0.136 A

Voltage ‘V’ = 220 V

P = _____ Watts

2 P = 500 Watts

I = 2.27 A

V = ______ v

3 P = 750 W

V = 220 v

I = ______ A

4 P = 60 W

V = 200 v

R = ______ W

5 I = 0.455 A

R = 484 ohms

P = ______ Watts

6 P = 550 W

R = 22 ohms

I = ______ A

7 P consumed = 1.8 kW

R = 8 ohms

V = ______ v

8 I consumed = ____ A

P = 2 kW

= 220 v-Heating

element voltage

R = ______ W

9 P = 100 W

t = 1 hour

Energy consumption

= ______ kWh

10 Energy consumed

‘W’ = 1 kWh

Power ‘P’ = 100 W

t = ______ hr

11 W = 1.5 kWh

t = 45 min.

P = ______ kW.

12 Energy metre reading

= 6755.3 kWh

Increases to W

= 6759.8 kWh

t = 45 min.

P = ______ kW.

13 Power consumed

‘P’ = 6.2 kW

t = 8 hours

Charge per kwh

= 1.25 Rupees

Total cost

= ______ Rupees

14 I = 5.45 A

V = 220 v

Energy consumed

= 1 kWh

t = ______ hr.

104

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

Magnetic induction, self and mutual inductance and EMF generation

Exercise 1.13.41

Magnetic induction

When a magnet is brought near to an iron bar is brought

near to a magnete, a magnetism is produced in the iron bar.

The phenomenon is know as magnetic induction. Actually,

before attracting an iron bar towards it, a magnet induces

an opposite polarity in the iron bar and then due to

attraction between unlike poles, magnet attracts the iron

bar. The magnet need not to touch the iron bar for magnetic

induction.

In various electrical measuring instruments, soft iron pole

pieces are used along with bar magnets in order to given the

desired shape to the magnet used, such pole piece work

on the principle of magnetic induction.

Intensity of magnetic field

The force acting on a unit pole placed in a magnetic field

(attractive or repulsive force) is called the intensity of

magnetic field. It is denoted by letter H and its unit is

Wb/m.

Principles and laws of electromagnetic induction

Faraday’s laws of electromagnetic induction are also

applicable for conductors carrying alternating current.

Faraday’s laws of elctromagnetic induction

Faraday’s first law states that whenever the magnetic

flux is linked with a circuit changes, an emf is always

induced in it.

The second law states that the magnitude of the induced

emf is equal to the rate of change of flux linkage.

Dynamically induced EMF

Accordingly induced emf can be produced either by moving

the conductor in a stationery magnetic field or by changing

magnetic flux over a stationery conductor. When conductor

moves and produces emf, the emf is called as dynamically

induced emf Example: Generators.

Statically induced EMF

When changing flux produces emf the emf is called as

statically induced emf as explained below.

Example:Transformer.

Statically induced emf: When the induced emf is pro-

duced in a stationery conductor due to changing magnetic

field, obeying Faraday’s laws of electro magnetism, the

induced emf is called as statically induced emf.

There are two types of statically induced emf as stated

below:

1 Self induced emf produced with in the same coil.

2 Mutually induced emf produced in the neighbouring

coil.

Self-induction: The production of an electromotive force in

a circuit, when the magnetic flux linked with the circuit

changes as a result of the change in a current inducing in

the same circuit.

At any instant, the direction of the magnetic field is

determined by the direction of the current flow.

With one complete cycle, the magnetic field around the

conductor builds up and then collapses. It then builds up

in the opposite direction, and collapses again. When the

magnetic field begins building up from zero, the lines of

force or flux lines expand from the centre of the conductor

outward. As they expand outward, they can be thought of

cutting through the conductor.

Self induction

According to Faraday’s Laws, an emf is induced in the

conductor. Similary, when the magnetic field collapses,

the flux lines cut through the conductor again, and an emf

is induced once again. This is called self-induction (Fig 1).

Mutual induction

When two or more coils one magnetically linked together

by a common magnetic flux, they are said have the

property of mutual inductance. It is the basic operating

principal of the transformer, motor generators and any

other electrical component that interacts with another

magnetic field. It can define mutual induction on the current

flowing in one coil that induces as voltage an adjacent coil.

In the Fig 2 current flowing in coil L1 sets up a magnetic field

around it self with some of its magnetic field line passing

through coil L2 giving in mutual inductance coil one L on

has a current of I, and N, turns while coil two L2, has N2

turns therfore mutual inductance M, of coil two that exists

with respect to coil one L, depend on their position with

inspect to each other.

The mutual inductance M that exists between the two coils

can be greately measured by positioning them on a

common soft iron cone or by measuring the number of turns

of either coil on would he found in a transformer.

105

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

The two coils are tightly would one on top of the other over

a common soft iron core unity said to exist between them

as any losses due to the leakage of flux will be extremely

small. Then assuring a perfect flux leakage between the

two coils the mutual inductance M.

Dynamically induced EMF

Generator: An electrical generator is a mahine which

converts mechanical energy into electrical energy.

Principle of the Generator: To facilitate this energy

conversion, the generator works on the principle of Faraday’s

Laws of Electromagnetic induction.

Faraday’s laws of electromagnetic induction: There

are two laws

The first law states:

• whenever the flix linking to a conductor or circuit

changes, an emf will be induced.

The second law states:

• the magnitude of such induced emf (e) depends upon

the rate of change of the flix linkage.

Types of emf: According to Faraday’s Laws, an emf can

be induced, either by the relative movement of the conduc-

tor and the magnetic field or by the change of flux linking

on a stationary conductor.

Dynamically induced emf: In case, the induced emf is

due to the movement of the conductor in a stationary

magnetic field as shown in Fig 1a or by the movement of the

magnetic field on a stationary conductor as shown in Fig

1b, the induced emf is called dynamically induced emf.

As shown in Fig 1a & 1b, the conductor cuts the lines of

force in both cases to induce an emf, and the presence of

the emf could be found by the deflection of the needle of the

galvanometer ‘G’. This principle is used in DC and AC

generators to produce electricity.

Statically induced emf: In case, the induced emf is due

to change of flux linkage over a stationary conductor as

shown in Fig 2, the emf thus induced is termed as statically

induced emf. The coils 1 and 2 shown in Fig 2 are not

touching each other, and there is not elctrical connection

between them.

According to Fig 2, when the battery (DC) supply is used

in coil 1, an emf will be induced in coil 2 only at the time of

closing or opening of the switch S. If the switch is

permanently closed or opened, the flux produced by coil 1

becomes static or zero respectively and no emf will be

induced in coil 2. EMF will be induced only when there is

a change in flux which happens during the closing or

opening of the circuit of coil 1 by the switch in a DC circuit.

Alternatively the battery and switch could be removed and

coil 1can be connected to an AC supply as shown in Fig

2. Then an emf will be induced in coil 2 continuously as long

as coil 1 is connected to an AC source which produces

alternating magnetic flux in coil 1 and links with coil 2. THis

principle is used in transformers.

Production of dynamically induced emf: Whenever a

conductor cuts the magnetic flux, a dynamically induced

emf is produced in it. This emf causes a current to flow if

the circuit of the conductor is closed.

For producting dynamically induced emf, the requirements

are:

• magnetic field

• conductor

• relative motion between the conductor and the mag-

netic field.

If the conductor moves with a relative velocuty ‘v’ with

respect to the field, then the induced emf ‘e’ will be

= BLV Sinθ Volts

106

Workshop Calculation And Science (NSQF - Level 4 & 5) - 1st Semester

where

B = magnetic flux density, measured in tesla

L = effective length of the conductor in the field in metres

V = relative velocity between field and conductor in

metre/second.

θ = the angle at which the conductor cuts the magnetic

field.

pattern of induced emf in a conductor when it rotates under

N and S poles of uniform magnetic field.

The emf induced by this process is basically alternating in

nature, and this alternating current is converted into direct

current in a DC generator by the commutator.

Fleming’s right hand rule: The direction of dynamically

induced emf can be identifued by this rile. Hold the thumb,

forefinger and middle finger of the righ hand at right angles

to each other as shown in Fig 4 such that the forefinger is

in the direction of flux and the thumb is in the direction of

the motion of the conductor, then the middle finger indi-

cates the direction of emf induced, i.e. towards the ob-

server or away from the observer.

Likewise for every position of the remaining conductors in

the periphery, the enf induced could be calculated. If these

values are plotted on a graph, it will represent the sine wave