SYLLABUS - Spec Notes · Web viewSYLLABUS UNIT – I Electric Drives: Type of electric drives,...

SYLLABUS

UNIT – IElectric Drives: Type of electric drives, choice of motor, starting and running characteristics, speed control, temperature rise, particular applications of electric drives, types of industrial loads, continuous, intermittent and variable loads, load equalization.

UNIT—IIElectric Heating & Welding: Electric Heating: Advantages and methods of electric heating, resistance heating induction heating and dielectric heating. Electric welding: resistance and arc welding, electric welding equipment, comparison between A.C. and D.C. Welding.

UNIT — IIIIllumination: Introduction, terms used in illumination, laws of illumination, polar curves, photometry, integrating sphere, sources of light. Discharge lamps, MV and SV lamps — comparison between tungsten filament lamps and fluorescent tubes, Basic principles of light control, Types and design of lighting and flood lighting.

UNIT –IVElectric Traction-I : System of electric traction and track electrification Review of existing electric traction systems in India. Special features of traction motor, methods of electric braking-plugging rheostatic braking and regenerative braking. Mechanics of train movement. Speed-time curves for different services — trapezoidal and quadrilateral speed time curves.

UNIT – VElectric Traction-II: Calculations of tractive effort, power, specific energy consumption for given run, effect of varying acceleration and braking retardation, adhesive weight and braking retardation adhesive weight an coefficient of adhesion.

TEXT BOOK Utilization of Electrical Power, Er. R. K. Rajput, Laxmi Publications.

M & Science of Utilization of electrical Energy, Paab, Dhanpat Rai & Sons.

REFERENCE BOOKS Utilization of Electric Energy, E. Openshaw Taylor, University press.

Generation, Distribution and Utilization of electrical Energy, C.L. Wadhwa, New Age International (P) Limited,

Utilization of Electrical Power including Electric drives and Electric traction, N.V.Suryanarayana, New Age International (P) Limited.

Utilization of Electric Energy, VVL Rao, University Press,

CONTENTS PAGE NO.S

Unit-1 :

1.1 Electrical Drives

1.1.1 Classification of Electric Drives

1.2 Selection of a Motor

1.2.1 Electrical Characteristics

1.2.2 Bearings

1.3 Types of Electric Drives

1.3.1 Group Electric Drive

3.3.2 Individual Electric Drive

3.3.3 Multimotor Electric Drive

1.4 Classification of Load Torques

1.5 Components of Load Torques

1.6 Characteristics of Different types of Loads

1.6.1 Constant Torque Characteristics

1.6.2 Torque Proportional to Speed

1.6.3 Torque Proportional to Square of the Speed

1.6.4 Torque Inversely Proportional to Speed

1.7 Motors for Different Industrial Drives

1.8 Size and Rating

1.9 Different Types of Industrial Loads

1.10 Load Equalization

1.10.1 Use of Flywheels

1.10.2 Choice of Flywheels

1.11 Modern Methods of Speed control of Industrial Drives

1.11.1 The Various Types of Variable Speed Drives

1.11.2 D.C Variable Speed Drive

1.11.3 Speed Control of Induction Motor

1.12 Slip Power Recovery System

UNIT – 2

2.1 Electric Heating

2.1.1 Advantages of Electric Heating

2.2 Different Methods of Heat Transfer

2.3 Requirement of a Good Heating Element

2.4 Design of Heating Element

2.5 Methods of Heating

2.5.1 Resistance Heating

2.5.2 Direct Resistance Heating

2.5.3 Indirect Resistance Heating

2.5.4 Core Type Furnace

2.6 Welding

2.6.1 Use of Electricity in Welding

2.6.2 Arc Welding Machines

2.6.3 V-I Characteristics of Arc Welding DC Machines

2.6.4 DC Welding Machines with Motor Generator Set

2.6.5 AC Rectified Welding Unit

2.6.6 AC Welding Machines

UNIT- 3

3.1 Introduction

3.2 Definitions

3.3 Nature of Radiation

3.4 Laws of Illumination

3.5 Polar Curves

3.5.1 Determination of M.S.C.P and M.H.C.P from polar diagrams

3.6 Design of Illumination Systems

3.6.1 Direct Lighting

3.6.2 Indirect Lighting

3.6.3 Semi-direct Lighting

3.6.4 Semi-Indirect Lighting

3.6.5 General Diffusing System

3.7 Types of Lamps

3.7.1 Discharge Lamps

3.7.1.1 Incandescent Lamp

3.7.1.2 Sodium Vapour Lamp

3.7.1.3 Mercury Vapour Lamp

UNIT-4

4.1 Introduction

4.1.1 Traction Systems

4.1.2 Advantages of Electric Traction

4.1.3 Disadvantages of Electric Traction

4.2 Systems of Railway Electrification

4.2.1 Direct current system

4.2.2. Single-phase ac system

4.2.3 Three-phase ac system

4.2.4 Composite system

4.3 Types of Railway Services

4.3.1 Train Movement

4.3.2 Typical Speed/Time Curve

4.3.3 Speed/Time Curves for Different Service

4.4. Quantities Involved in Traction Mechanics

4.4.1 Relationship between Principal Quantities in Trapezoidal Diagram

4.4.2 Relationship between Principal Quantities in Quadrilateral Diagram

4.7 Mechanism of Train Movement

4.8 Braking in Traction

4.8.1 Rheostatic Braking

4.8.2 Regenerative Braking with D.C. Motors

UNIT-5

5.1 Introduction5.2. Calculations of tractive effort

5.2.1. power, 5.2.2. specific energy consumption for given run, 5.3. effect of varying acceleration 5.3.1 braking retardation, 5.3.2 adhesive weight 5.3.3 braking retardation adhesive weight an coefficient of adhesion.

UNIT-I

ELECTRIC DRIVES

1.1 ELECTRICAL DRIVES

Motion control is required in large number of industrial and domestic

applications like transportation systems, rolling mills, paper machines, textile mills,

machine tools, fans, pumps, robots, washing machines etc. Systems employed for

motion control are called DRIVES, and may employ any of prime movers such as

diesel or petrol engines, gas or steam turbines, steam engines, hydraulic motors and

electric motors, for supplying mechanical energy for motion control. Drives

employing electric motors are known as electrical drives. An electric drive can be

defined as an electromechanical device for converting electrical energy into

mechanical energy to impart motion to different machines and mechanisms for various

kinds of process control.

1.1.1 Classification of Electric Drives

According to Mode of Operation

Continuous duty drives

Short time duty drives

Intermittent duty drives

According to Means of Control

Manual

Semi automatic

Automatic

According to Number of machines

Individual drive

Group drive

Multi-motor drive

According to Dynamics and Transients

Uncontrolled transient period

Controlled transient period

According to Methods of Speed Control

Reversible and non-reversible uncontrolled constant speed control

Reversible and non-reversible step speed control.

Variable position control

Reversible and non-reversible smooth speed control.

Advantages of Electrical Drive

They have flexible control characteristics.

The steady state and dynamic characteristics of electric drives can be shaped to satisfy the load

requirements.

Drives can be provided with automatic fault detection systems.

Programmable logic controller and computers can be employed to automatically control the drive

operations in a desired sequence.

They are available in wide range of torque, speed and power.

They are adaptable to almost any operating conditions such as explosive and radioactive

environments.

It can operate in all the four quadrants of speed-torque plane.

They can be started instantly and can immediately be fully loaded.

Control gear requirement for speed control, starting and braking is usually simple

and easy to operate.

1.2 Selection of a Motor

The selection of a driving motor depends primarily on the conditions under

which it has to operate and the type of load it has to handle. Main guiding factors for

such a selection are as follows:

(a) Electrical characteristics

1. Starting characteristics

2. Running characteristics

3. Speed control

4. Braking

(b)Mechanical considerations

1. Type of enclosure

2. Type of bearings

3. Method of power transmission

4. Type of cooling

5. Noise level

(c) Size and rating of motor

1. Requirement for continuous, intermittent or variable load cycle

2. Overload capacity

(d) Cost

1. Capital cost

2. Running cost

In addition to the above factors, one has to take into consideration the type of

current available whether alternating or direct. However, the basic problem is one of

matching the mechanical output of the motor with the load requirement i.e. to select a

motor with the correct speed/torque characteristics as demanded by the load. In fact,

the complete selection process requires the analysis and synthesis of not only the load

and the proposed motor but the complete drive assembly and the control equipment

which may include rectification or frequency changing.

1.2.1 Electrical Characteristics

Types of Enclosures

The main function of an enclosure is to provide protection not only to the

working personnel but also to the motor itself against the harmful ingress of dirt,

abrasive dust, vapours and liquids and solid foreign bodies such as a spanner or screw

driver etc. At the same time, it should not adversely affect the proper cooling of the

motor. Hence, different types of enclosures are used for different motors depending

upon the environmental conditions.

Some of the commonly used motor enclosures are as under:

1. Open Type

In this case, the machine is open at both ends with its rotor being supported on

pedestal bearings or end brackets. There is free ventilation since the stator and rotor

ends are in free contact with the surrounding air. Such, machines are housed in a

separate neat and clean room. This type of enclosure is used for large machines such

as d.c. motors and generators.

2. Screen Protected Type

In this case, the enclosure has large openings for free ventilation. However,

these openings are fitted with screen covers which safeguard against accidental

contacts and rats entering the machine but afford no protection from dirt, dust and

falling water. Screen protected type motors are installed where dry and neat conditions

prevail without any gases or fumes.

3. Drip Proof Type

This enclosure is used in very damp conditions. i.e. for pumping sets. Since

motor openings are protected by over-hanging cowls, vertically falling water and dust

are not able to enter the machine.

4. Splash-proof Type

In such machines, the ventilating openings are so designed that liquid or dust

particles at an angle between vertical and 100° from it cannot enter the machine. Such

type of motors can be safely used in rain.

5. Totally Enclosed (TE) Type

In this case, the motor is completely enclosed and no openings are left for

ventilation. All the heat generated due to losses is dissipated from the outer surface

which is finned to increase the cooling area. Such motors are used for dusty

atmosphere i.e. sawmills, coal-handling plants and stone-crushing quarries etc.

6. Totally-enclosed Fan-cooled (TEFC) Type

In this case, a fan is mounted on the shaft external to the totally enclosed

casing and air is blown over the ribbed outer surfaces of the stator and end shields.

Such motors are commonly used in flour mills, cement works and sawmills etc. They

require little maintenance apart from lubrication and are capable of giving years of

useful service without any interruption of production.

Figure: 3.1 A Three Phase Motor

7. Pipe-ventilated Type

Such an enclosure is used for very dusty surroundings. The motor is totally

enclosed but is cooled by neat and clean air brought through a separate pipe from

outside the dust-laden area. The extra cost of the piping is offset by the use of a

smaller size motor on account of better cooling.

8. Flame-proof (FLP) Type

Such motors are employed in atmospheres which contain inflammable gases

and vapours i.e. in coal mines and chemical plants. They are totally enclosed but their

enclosures are so constructed that any explosion within the motor due to any spark

does not ignite the gases outside. The maximum operating temperature at the surface

of the motor is much less than the ignition temperature of the surrounding gases.

1.2.2 Bearings

These are used for supporting the rotating parts of the machines and are of two

types:

1. Ball or roller bearings 2. Sleeve or bush bearings

(a) Ball Bearings

Up to about 75kW motors, ball bearings are preferred to other bearings

because of their following advantages:

1. They have low friction loss

2. They occupy less space

3. They require less maintenance

4. Their use allows much smaller air-gap between the stator and rotor of an induction motor

5. Their life is long.

Their main disadvantages are with regard to cost and noise particularly at high motor speeds

(b)Sleeve Bearings

These are in the form of self-aligning porous bronze bushes for fractional kW

motors and in the form of journal bearings for larger motors. Since they run very

silently, they are fitted on super-silent motors used for driving fans and lifts in offices

or other applications where noise must be reduced to the absolute minimum.

1.3 Types of Electric Drives

1.3.1 Group Electric Drive

This drive consists of a single motor, which drives one or more line shafts

supported on bearings. The line shaft may be fitted with either pulleys and belts or

gears, by means of which a group of machines or mechanisms may be operated. It is

also sometimes called as Shaft drives.

Advantages

A single large motor can be used instead of number of small motors

Disadvantages

There is no flexibility. If the single motor used develops fault, the whole process will be stopped.

1.3.2 Individual Electric Drive

In this drive each individual machine is driven by a separate motor. This motor

also imparts motion to various parts of the machine.

1.3.3 Multi Motor Electric Drive

In this drive system, there are several drives, each of which serves to actuate

one of the working parts of the drive mechanisms.

E.g.: Complicated metal cutting machine tools, Paper making industries, Rolling machines etc.

1.4 Classification of Load Torques

Various load torques can be classified into broad categories.

Active load torques

Passive load torques

Load torques which has the potential to drive the motor under equilibrium conditions are called active

load torques. Such load torques usually retains their sign when the drive rotation is changed (reversed).

Eg: Torque due to force of gravity , Torque due tension, Torque due to compression and torsion etc.

Load torques which always oppose the motion and change their sign on the

reversal of motion are called passive load torques

Eg: Torque due to friction, cutting etc.

1.5 Components of Load Torques

The load torque can be further divided into following components

(i) Friction Torque (TF)

Friction will be present at the motor shaft and also in various parts of the load.

TF is the equivalent value of various friction torques referred to the motor shaft.

(ii) Windage Torque (TW)

When motor runs, wind generates a torque opposing the motion. This is known

as windage torque.

(iii) Torque required to do useful mechanical work

Nature of this torque depends upon particular application. It may be constant

and independent of speed. It may be some function of speed, it may be time invariant

or time variant, its nature may also change with the load’s mode of operation. Value of

friction torque with speed is shown in figure below.

Figure: 3.2 Torque With Speed Curve

Its value at stand still is much higher than its value slightly above zero speed.

Friction at zero speed is called stiction or static friction. In order to start the drive the

motor should at least exceed stiction.

Friction torque can also be resolved into three components.

Figure: 1.3 Torque with Speed Curve

Component Tv varies linearly with speed is called VISCOUS friction and is given by

Tv = B ωn

Where B is viscous friction co-efficient.

Another component TC, which is independent of speed, is known as COULOMB friction.

Third component Ts accounts for additional torque present at stand still. Since Ts is present only at stand

still it is not taken into account in the dynamic analysis. Windage torque, TW which is proportional to

speed squared is given by

Tw = Cω2m

Where C is Constant.

From the above discussions, for finite speed

Tl = TL + Tc + Cω2m+ B ωn

1.6 Characteristics of Different types of Loads

One of the essential requirements in the section of a particular type of motor for

driving a machine is the matching of speed-torque characteristics of the given drive

unit and that of the motor. Therefore the knowledge of how the load torque varies with

speed of the driven machine is necessary. Different types of loads exhibit different

speed torque characteristics. However, most of the industrial loads can be classified

into the following four categories.

Constant torque type load

Torque proportional to speed (Generator Type load)

Torque proportional to square of the speed (Fan type load)

Torque inversely proportional to speed (Constant power type load)

1.6.1 Constant Torque Characteristics

Most of the working machines that have mechanical nature of work like

shaping, cutting, grinding or shearing, require constant torque irrespective of speed.

Similarly cranes during the hoisting and conveyors handling constant weight of

material per unit time also exhibit this type of characteristics.

Figure: 3.4 Constant Torque Characteristics

1.6.2 Torque Proportional to Speed

Separately excited dc generators connected to a constant resistance load, eddy

current brakes have speed torque characteristics given by T=k

Figure: 3.5 Torque Proportional to Speed

1.6.3 Torque Proportional to Square of The Speed

Another type of load met in practice is the one in which load torque is

proportional to the square of the speed. Eg Fans rotary pumps, compressors and ship

propellers.

Figure: 3.6 Torque proportional to square of the speed

1.6.4 Torque Inversely Proportional to Speed

Certain types of lathes, boring machines, milling machines, steel mill coiler

and electric traction load exhibit hyperbolic speed-torque characteristics.

Figure: 3.7 Torque Inversely proportional to speed

1.7 Motors for Different Industrial Drives

1. D.C. Series Motor. Since it has high starting torque and variable speed, it is used for heavy duty

applications such as electric locomotives, steel rolling mills, hoists, lifts and cranes.

2. D.C. Shunt Motor. It has medium starting torque and a nearly constant speed. Hence, it is used for

driving constant-speed line shafts, lathes, vacuum cleaners, wood-working machines, laundry washing

machines, elevators, conveyors, grinders and small printing presses etc.

3. Cumulative Compound Motor. It is a varying-speed motor with high starting torque and is used for

driving compressors, variable-head centrifugal pumps, rotary presses, circular saws, shearing machines,

elevators and continuous conveyors etc.

4. Three-phase Synchronous Motor. Because its speed remains constant under varying loads, it is used

for driving continuously-operating equipment at constant speed such as ammonia and air compressors,

motor-generator sets, continuous rolling mills, paper and cement industries.

5. Squirrel Cage Induction Motor. This motor is quite simple but rugged and possesses high over-load

capacity. It has a nearly constant speed and poor starting torque. Hence, it is used for low and medium

power drives where speed control is not required as for water pumps, tube wells, lathes, drills, grinders,

polishers, wood planers, fans, blowers, laundry washing machines and compressors etc

6. Double Squirrel Cage Motor. It has high starting torque, large overload capacity and a nearly

constant speed. Hence, it is used for driving loads which require high starting torque such as compressor

pumps, reciprocating pumps, large refrigerators, crushers, boring mills, textile machinery, cranes,

punches and lathes etc.

7. Slip-ring Induction Motor. It has high starting torque and large overload capacity. Its speed can be

changed up to 50% of its normal speed. Hence, it is used for those industrial drives which require high

starting torque and speed control such as lifts, pumps, winding machines, printing presses, line shafts,

elevators and compressors etc.

8. Single-phase Synchronous Motor. Because of its constant speed, it is used in teleprinters, clocks, all

kinds of timing devices, recording instruments, sound recording and reproducing systems.

9. Single-phase Series Motor. It possesses high starting torque and its speed can be controlled over a

wide range. It is used for driving small domestic appliances like refrigerators and vacuum cleaners etc.

10. Repulsion Motor. It has high starting torque and is capable of wide speed control. Moreover, it has

high speed at high loads. Hence, it is used for drives which require large starting torque and adjustable

but constant speed as in coil winding machines.

11. Capacitor-start Induction-run Motor. It has fairly constant speed and moderately high starting

torque. Speed control is not possible. It is used for compressors, refrigerators and small portable hoists.

12. Capacitor-start-and-run Motor. Its operating characteristics are similar to the above motor except

that it has better power factor and higher efficiency. Hence, it is used for drives requiring quiet

operations.

1.8 Size and Rating

The factors which govern the size and rating of motor for any particular

service are its maximum temperature rise under given load conditions and the

maximum torque required. It is found that a motor which is satisfactory from the point

of view of maximum temperature rise usually satisfies the requirement of maximum

torque as well. For class-A insulation, maximum permissible temperature rise is 40oC

whereas for class – B insulation, it is 50oC. This temperature rise depends on whether

the motor has to run continuously, intermittently or on variable load.

Different ratings for electrical motors are as under:

1. Continuous Rating. It is based on the maximum load which a motor can deliver for an indefinite

period without its temperature exceeding the specified limits and also possesing the ability to take 25%

overload for a period of time not exceeding two hours under the same conditions.

For example, if a motor is rated continuous 10 KW, it means that it is capable of giving an output of 10

KW continuously for an indefinite period of time and 12.5 KW for a period of two hours without its

temperature exceeding the specified limits.

2. Continuous Maximum Rating. It is the load capacity as given above but without overload capacity.

Hence, these motors are a little bit inferior to the continuous-rated motors.

3. Intermittent Rating. It is based on the output which a motor can deliver for a specified period; say

one hour or ½ hour or ¼ hour without exceeding the temperature rise. This rating indicates the maximum

load of the motor for the specified time followed by a no-load period during which the machine cools

down to its original temperature

Estimation of Motor Rating

Since primary limitation for the operation of an electric motor is its

temperature rise, hence motor rating is calculated on the basis of its average

temperature rise. The average temperature rise depends on the average heating which

itself is proportional to the square of the current and the time for which the load

persists.

For example, if a motor carries a load L1 for time t1 and load L2 for time t2

and so on, then

In fact, heating is proportional to square of the current but since load can be

expressed in terms of the current drawn, the proportionality can be taken for load

instead of the current.

Generally, load on a motor is expressed by its load cycle. Usually, there are periods of no-load in the

cycle. When motor runs on no-load, heat generated is small although heat dissipation continues at the

same rate as long as the machine is running. Hence, there is a difference in the heating of a motor running

at no-load and when at rest. It is commonly followed practice in America to consider the period at rest as

one - third while calculating the size of motor. It results in giving a higher motor rating which is

advantageous and safe.

1.9 Different Types of Industrial Loads

The three different types of industrial loads under which electric motors are required to work are as

under:

(i) continuous load (ii) intermittent load and (iii) variable or fluctuating load

(i) Continuous Load. In such cases, the calculation of motor size is simpler because the loads like

pumps and fans require a constant power input to keep them operating. However, it is essential to

calculate the KW rating of the motor correctly. If the KW rating of the motor is less than what is

required, the motor will overheat and consequently burn out. If, on the other hand, KW rating is more

than what is needed by the load, the motor will remain cool but will operate at lower efficiency and

power.

(ii) Intermittent Loads. Such loads can be of the following two types:

(a) In this type of load, motor is loaded for a short time and then shut off for a sufficient by

long time, allowing the motor to cool down to room temperature. In such cases, a motor with a short time

rating is used as in a kitchen mixer.

Figure: 3.8 Intermittent Loads

(b) In this type of load, motor is loaded for a short time and then it is shut off for a short time. The shut

off time is so short that the motor cannot cool down to the room temperature. In such cases, a suitable

continuous or short-time rated motor is chosen which, when operating on a given load cycle, will not

exceed the specified temperature limit.

(iii) Variable Loads. In the case of such loads, the most accurate method of selecting a suitable motor is

to draw the heating and cooling curves as per the load fluctuations for a number of motors. The smallest

size motor which does not exceed the permitted temperature rise when operating on the particular load

cycle should be chosen for the purpose. However, a simpler but sufficiently accurate method of selection

of a suitable rating of a motor is to assume that heating is proportional to the square of the current and

hence the square of the load. The suitable continuous rating of the motor would equal the r.m.s. value of

the load current.

1.10 Load Equalization

If the load fluctuates between wide limits in space of few seconds, then large

peak demands of current will be taken from supply and produce heavy voltage drops

in the system. Large size of conductor is also required for this. Process of smoothing

out these fluctuating loads is commonly referred to as load equalization and involves

storage of energy during light load periods which can be given out during the peak

load period, so that demand from supply is approximately constant. Tariff is also

affected as it is based on M.D.(Maximum Demand).For example, in steel rolling mill,

when the billet is in between the rolls it is a peak load period and when it comes out it

is a light load period, when the motor has to supply only the friction and internal

losses.0

Figure: 3.9 Load Equalization

1.10.1 Use of Flywheels

The method of Load Equalization most commonly employed is by means of a flywheel. During peak

load period, the flywheel decelerates and gives up its stored kinetic energy, thus reducing the load

demanded from the supply. During light load periods, energy is taken from supply to accelerate flywheel,

and replenish its stored energy ready for the next peak. Flywheel is mounted on the motor shaft near the

motor. The motor must have drooping speed characteristics, that is, there should be a drop in speed as the

load comes to enable flywheel to give up its stored energy. When the Ward -Leonard system is used with

a flywheel, then it is called as Ward – Leonard llgner control.

Flywheel Calculations

The behavior of flywheel may be determined as follows.

Fly wheel decelerating :- (or Load increasing)

Let

TL- Load torque assumed constant during the time for which load is applied in kg-m

Tf- Torque supplied by flywheel in kg-m

To- Torque required on no load to overcome friction internal losses etc., in kg-m

Tm- Torque supplied by the motor at any instant, in kg-m

ωo-No Load speed of motor in rad/sec.

ω- Speed of motor at any instant in rad/sec.

S-motor slip speed in rad/sec.

I- Moment of inertia of flywheel in kg-m2

g- Acceleration due to gravity in m/sec2

t- time in sec.

Tm= TL - Tf or TL = Tm + Tf (1)

Energy stored by Flywheel when running at speed ‘ω’ is ½ I ω2/g

If speed is reduced from ωo to ω

The energy given up by flywheel is

= ½ I/g (ω2o -ω2)

= ½ I/g (ωo - ω) (ωo+ ω) (2)

(ωo+ ω)/2 = mean speed. Assuming speed drop of not more than 10%., this may be assumed

equal to ω.

(ωo+ ω)/2 = ω . Also (ωo - ω) = s

From equation (2), Energy given up = I/g ωs

Power given up = I/g ω ds/dt

But Torque = Power / ω

When the flywheel decelerates, it gives up its stored energy.

Torque supplied by flywheel is

Tf = I/g ds/dt

From equation (1),

Tm= TL - I/g ds/dt

For values of slip speed upto 10% of no-load speed, Slip is proportional to torque or

s = K Tm

Tm= TL - I/g K d Tm /dt

This equation is similar to the equation for heating of the motor W – Aλθ = G.S. dθ/dt

i.e.,( TL - Tm ) = I/g K d Tm /dt

g dt/IK = d Tm/( TL - Tm )

by integrating on both sides,

-ln( TL - Tm ) = tg/IK +C1 (3)

At t=0 , When load starts increasing from no load i.e., Tm = T0

Hence, at t=0 Tm = T0

C1= - ln( TL – T0 )

By substituting the value of C1 in equation (3) we get

-ln( TL - Tm ) = tg/IK – ln ( TL – T0 )

ln[( TL - Tm )/ ( TL – T0 )] = - tg/IK

[( TL - Tm )/ ( TL – T0 )] = e- tg/IK

( TL - Tm ) = ( TL – T0 ) e- tg/IK

Tm= TL - ( TL – T0 ) e- tg/IK

If the load torque falls to zero between each rolling period , then

Tm= TL (1-e- tg/IK) (T0=0)

Load Removed (Flywheel Accelerating)

Slip speed is decreasing and therefore ds/dt is negative

Tm= T0 +Tf = T0 – I/g ds/dt

g dt/ IK= d Tm/ (T0 – Tm)

After integrating on both sides

-ln(T0 – Tm) = tg/IK + C

Figure: 3.10 Rolling mill drives with flywheel

1.10.2 Choice of Flywheel

There are two choices left for selecting a flywheel to give up its maximum stored energy:

1. Large drop in speed and small flywheel (But with this the quality of production will suffer, since a

speed drop of 10 to 15% for maximum load is usually employed)

2. Small drop in speed and large flywheel. (This is expensive and creates additional friction losses. Also

design of shaft and bearing of motor is to be modified.) So compromise is made between the two and a

proper flywheel is chosen.

1.11 Modern Methods of Speed control of Industrial Drives

1. Constant torque Applications

2. Variable torque applications

1.11.1 The various types of variable speed drives

D.C variable speed drive

A.C variable speed drive

Variable frequency drive for induction motor drive

Slip power recovery drives for HT slip ring induction motor

1.11.2 D.C Variable Speed Drive

Phase angle control

Silicon-controlled rectifiers (SCR) are solid state semiconductor devices that

are usually used in power switching circuits. SCR controls the output signal by

switching it ‘on’ or ‘off’, thereby controlling the power to the load in context. The two

primary modes of SCR control are phase angle fired-where a partial waveform is

passed to regulate the power. In the phase-angle controller, the firing pulse is delayed

to turn on the SCR in middle of every half cycle. This means that every time a part of

an AC cycle is cut, the power to the load also gets cut. To deliver more or less power

to the load, the power to the load also gets cut, to deliver more or less power to the

load. The phase angle is increased or decreased, thereby controlling the throughput

power.

Figure: 1.11 SCR /Triac Connections for Various Methods of Phase Control

Phase control is the most common form of Thyristor power control. The

Thyristor is held in the off condition -- that is, all current flow in the circuit is blocked

by the Thyristor except a minute leakage current. Then the Thyristor is triggered into

an “on” condition by the control circuitry. For full-wave AC control, a single Triac or

two SCRs connected in inverse parallel may be used. One of two methods may be

used for full-wave DC control a bridge rectifier formed by two SCRs or an SCR

placed in series with a diode bridge.

Figure illustrates voltage waveform and shows common terms used to describe Thyristor operation.

Figure: 3.12 Sine Wave Showing Principles of Phase Control

Delay angle is the time during which the Thyristor blocks the line voltage. The

conduction angle is the time during which the Thyristor is on. It is important to note

that the circuit current is determined by the load and power source. For simplification,

assume the load is resistive; that is, both the voltage and current waveforms are

identical.

1.11.3 Speed control of Induction Motor

a) Variable Voltage Speed Control

In this method, the speed of a SCIM is varied by varying the stator voltage

with the help of three sets of SCRs connected back-to back. The stator voltage is

reduced by delaying the firing (or triggering) of the thyristors. If we delay the firing

pulses by 100°, the voltage obtained is about 50% of the rated voltage which decreases

the motor speed considerably.

Figure: 1.12 Variable Voltage Speed Control

Unfortunately, I2R losses are considerable due to distortion in voltage.

Moreover, p.f. is also low due to large lag between the current and voltage. Hence,

this electronic speed control method is feasible for motors rated below 15 kW but is

quite suitable for small hoists which get enough time to cool off because of

intermittent working. Of course, p.f. can be improved by using special thyristors

called gate turn-off thyristors (GTOs) which force the current to flow almost in phase

with the voltage (or even lead it).

b) Variable-frequency Speed Control

A 3-phase SCIM connected to the outputs of three 3-phase cycloconverters. As seen, each cycloconverter

consists of two 3-phase thyristor bridges, each fed by the same 3-phase, 50-Hz line. The +R bridge

generates the positive half cycle for R-phase whereas −R generates the negative half. The frequency of

the cycloconverter output can be reduced to any value (even up to zero) by controlling the application of

firing pulses to the thyristor gates

Figure: 1.13 Variable-frequency Speed Control

. This low frequency permits excellent speed control. For example, the speed

of a 4-pole induction motor can be varied from zero to 1200 rpm on a 50-Hz line by

varying the output frequency of the cycloconverter from zero to 40 Hz. The stator

voltage is adjusted in proportion to the frequency in order to maintain a constant flux

in the motor. This arrangement provides excellent torque/speed characteristics in all 4-

quadrants including regenerative braking. However, such cycloconverter-fed motors

run about 10°C hotter than normal and hence require adequate cooling. A small part of

the reactive power required by SCIM is provided by the cycloconverter, the rest being

supplied by the 3-phase line. Consequently, power factor is poor which makes

cycloconverter drives feasible only on small and medium power induction motors.

1.12 Slip Power Recovery System

Slip Power Recovery System (SPRS) is a variable speed drive for slip ring

induction motors. It recovers and delivers the slip* dependent rotor power from the

motor to the grid. At changeover speed, SPRS is connected to the rotor and the rotor

resistance is disconnected. The diode rectifier converts the rotor voltage to DC

voltage. This rectified rotor voltage is counter-balanced by a line commutated inverter.

By controlling the ‘counter-balancing’ inverter voltage, the rotor current, hence rotor

speed is regulated. The slip power collected at the slip rings is fed back to the grid

through the inverter.

Figure: 1.14 Slip Power Recovery System

The slip power recovery (SPR) drive is an external system connected to the

rotor circuit in place of the external resistors. The SPR provides speed and torque

control like the resistors but can also recover the power taken off the rotor and feed it

back into the power system to avoid energy waste. In usual practice, an SPR drive

consists of two interconnected power converters .The rotor converter is connected to

the three-phase rotor winding. The feedback power converter is connected to the

power system, usually through a transformer that matches the output voltage of the

converter to the power system. The regenerative or feedback power converter is

controlled to modulate the amount of power put back into the power system, allowing

control of the motor speed. All the rotor energy previously lost as heat in the rheostat

is now saved, and for large motors, this amounts to significant cost savings.

UNIT II

ELECTRIC HEATING & WELDING


Electric heating is extensively used both for domestic and industrial applications.

Domestic applications include (i) room heaters (ii) immersion heaters for waterheating(iii) hot plates for

cooking (iv) electric kettles (v) electric irons (vi) popcornplants(vii) electric ovens for bakeries and (viii)

electric toasters etc.


(i) Cleanliness. Since neither dust nor ash is produced in electric heating, it is a clean

system of heating requiring minimum cost of cleaning. Moreover, the material to be

heated does not get contaminated.

(ii) No Pollution. Since no flue gases are produced in electric heating, no provisionhas to be made

for their exit.

(iii) Economical. Electric heating is economical because electric furnaces are cheaper in

their initial cost as well as maintenance cost since they do not require big space for

installation or for storage of coal and wood. Moreover, there is no need to construct

anychimney or to provide extra heat installation.

(iv) Ease of Control. It is easy to control and regulate the temperature of an electric

furnace with the help of manual or automatic devices. Temperature can be controlled

within ± 5°C which is not possible in any other form of heating.

(v) Special Heating Requirement. Special heating requirements such as uniformheating of a material

or heating one particular portion of the job without affecting its other parts or heating with no

oxidation can be met only by electric heating.

(vi) Higher Efficiency. Heat produced electrically does not go away waste through thechimney and

other by products. Consequently, most of the heat produced isutilised for heating the material

itself. Hence, electric heating has higher efficiency ascompared to other types of heating.

(vii) Better working Conditions. Since electric heating produces no irritating noisesand alsothe

radiation losses are low, it results in low ambient temperature. Hence,working with electric

furnaces is convenient and cool.

(viii) Heating of Bad Conductors. Bad conductors of heat and electricity like wood,plastic and

bakery items can be uniformly and suitably heated with dielectric heatingprocess.

(ix) Safety-Electric heating is quite safe because it responds quickly to the controlledsignals.

(x) Lower Attention and Maintenance Cost. Electric heating equipment generally willnot

requiremuchattention and supervision and their maintenance cost is almostnegligible.

Hence, labour charges arenegligibly small as compared to other forms ofheating.


The different methods by which heat is transferred from a hotbody to a cold body are as under:

1. Conduction

In this mode of heat transfer, one molecule of the body gets heatedand transfers some of the heat to the

adjacent molecule and so on.There is a temperature gradient between the two ends of the bodybeing

heated. Consider a solid material of cross-section Asq.m. andthickness x metre as shown

Figure: 1.22 Solid Material representing Heat Transfer Phenomenon

If T1 and T2 are the temperatures of the two sides of the slab in °K, where K is thermal conductivity of

the material.

2. Convection

In this process, heat is transferred by the flow of hot and cold air currents. Thisprocess is applied in the

heating of water by immersion heater or heating of buildings. The quantity of heat absorbed by the body

by convection process depends mainly on the temperature of the heating element above the surroundings

and upon the size of the surface of the heater. It also depends, to some extent, on the positionof the

heater. The amount of heat dissipated is givenH = a (T1 - T2),where a and b are constants and T1 and T2

are the temperatures of the heatingsurface and the fluid in °K respectively. In electric furnaces, heat

transferred by convection is negligible.

3. Radiation

It is the transfer of heat from a hot body to a cold body in a straight line withoutaffecting the intervening

medium. The rate of heat emission is given by Stefan’s lawaccording to which heat dissipated

where K is radiating efficiency and e is known as emissivity of the heating element.If d is the diameter of

the heating wire and l its total length, then its surface areafrom which heat is radiated = πd×l . If H is the

power radiated per m2 of the heatingsurface, then total power radiated as heat = H ×πd l . If P is the

electrical power input to the heating element, then P=πdl×H.

2.3 Requirement of a Good Heating Element(1) High Specific Resistance.When specific resistance of the material of the wireishigh, only short length

of it will be required for a particular resistance (and henceheat) or for the same length of the wire and the

currrent, heat produced will be more.

(2) High Melting Temperature. If the melting temperature of the heating elementishigh, it would be

possible to obtain higher operating temperatures.

(3) Low Temperature Coefficient of Resistance. In case the material has lowtemperature coefficient of

resistance, there would be only small variations in itsresistance over its normal range of temperature.

Hence, the current drawn by theheating element when cold (i.e., at start) would be practically the same

when it is hot.

(4) High Oxidising Temperature. Oxidisation temperature of theheating element should be high in order

to ensure longer life.

(5) Positive Temperature Coefficient of Resistance. If the temperature coefficient ofthe resistance of

heating element is negative, its resistance will decrease with rise intemperature and it will draw more

current which will produce more wattage and henceheat. With more heat, the resistance will decrease

further resulting in instability ofoperation.

(6) Ductile. Since the material of the heating elements has to have convenientshapes and sizes, it should

have high ductility and flexibility.

(7) Mechanical Strength. The material of the heating element should posses highmechanical strength of

its own. Usually, different types of alloys are used to getdifferentoperatingtemperatures. For example

maximum working temperature ofconstant an (45% Ni, 55% Cu) is 400°C, that ofnichrome (50%, Ni

20% Cr) is 1150°C, that of Kantha(70% Fe, 25% Cr, 5% Al) is 1200° C and that of silicon carbide is

1450°C. With the passage of time, every heating element breaks open and becomes unserviceable. Some

of the factors responsible for its failure are :

(1) Formation of hot spots which shine brighter during operation, (2) Oxidation (3)

Corrosion (4) Mechanical failure

2.4 Design Of Heating ElementNormally, wires of circular cross-section or rectangular conducting ribbons are usedas heating elements.

Under steady-state conditions, a heating element dissipates asmuch heat from its surface as it receives the

power from the electric supply. If P is the power input and H is the heat dissipated by radiation, then P =

H under steady-state conditions.As per Stefan’s law of radiation, heat radiated by a hot body is given by

where T1 is the temperature of hot body in °K and T2 that of the cold body (or coldsurroundings) in °K

Total surface area of the wire of the element = (πd) × l

If H is the heat dissipated by radiation per second per unit surface area of the wire,then heat radiated per

second=(πd) × l × H

2.5 Methods of Heating2.5.1 Resistance Heating

2.5.1.1 Direct Resistance Heating

In this method the material (or charge) to be heated is treated as a resistance

andcurrent is passed through it. The charge may be in the form of powder, small

solidpieces or liquid. The two electrodes are inserted in the charge and connected to

either a.c. or d.c. supply .

Figure: 2.1 Direct Resistance Heating

Obviously, two electrodes will be required in the case of d.c. or single-phase a.c. supply but there would

be three electrodes in the case of 3- phase supply. When the charge is in the form of small pieces, a

powder of high resistivity material is sprinkled over the surface of the charge to avoid direct short circuit.

Heat is produced when current passes through it. This method of heating has high efficiency because the

heat is produced in the charge itself.

1.12.1.2 Indirect Resistance Heating

In this method of heating, electric current is passed through a resistance element which is placed in an

electric oven. Heat produced is proportional to I2Rlosses in the heating element.

Figure: 1.24 Indirect Resistances Heating

The heat so produced is delivered to the chargeeither by radiation or convection or by a combination of

the two. Sometimes,resistance is placed in a cylinder which is surrounded by the charge placed in

thejacket as shown above. This arrangement provides uniform temperature.Moreover, automatic

temperature control can also be provided.

2.5.1.2 Core Type Furnace

(a)Core-type Furnaces which operate just like a two wind ing transformer. These canbe further sub-

divided into (i) Direct core-type furnaces (ii) Vertical core-typefurnaces and (iii) Indirect core-type

furnaces.

(b) Coreless-type Furnaces — in which an inductively-heated element is madetotransfer heat to the

charge by radiation.

Core Type Induction Furnace

It is essentially a transformer in which the charge to beheated forms a single-turn short-circuited

secondary and is magnetically coupled to the primary by an iron core. The furnace consists of a circular

hearth which contains the charge to be melted in the form of an annular ring. When there is no molten

metal in the ring, the secondary becomes open-circuited there-by cutting off the secondary current.

Figure: 1.25 Core Type Induction Furnace

Hence, to start the furnace, molted metal has to be poured in the annular hearth. Since, magnetic

coupling between the primary and secondary is very poor, it results in high leakage and low power factor.

In order to nullify the effect of increased leakage reactance, low primary frequency of the order of 10 Hz

is used. If the transformer secondary current density exceeds 500 A/cm2 then, due to the interaction of

secondary current with the alternating magnetic field, the moltenmetal is squeezed to the extent that

secondary circuit is interrupted.This effect isknown as “pinch effect”.

2.6 Welding

It is the process of joining two pieces of metal or non-metal at faces rendered

plastic or liquid by the application of heat or pressure or both. Filler material may be

used to effect the union.

All welding processes fall into two distinct categories:

1. Fusion Welding—It involves melting of the parent metal. Examples are:

(i) Carbon arc welding, metal arc welding, electron beam welding, electro slagwelding and electrogas

welding which utilize electric energy and(ii) Gas welding and thermit welding which utilize chemical

energy for the meltingpurpose.

2. Non-fusion Welding—It does not involve melting of the parent metal. Examples are:

(i) Forge welding and gas non-fusion welding which use chemical energy.

(ii) Explosive welding, friction welding and ultrasonic welding etc.,which use mechanical energy.

(iii) Resistance welding which uses electrical energy. Proper selection of the weldingprocess depends on

the (a) kind of metals to be joined (b) cost involved (c) nature ofproducts to be fabricated and (d)

production techniques adopted.


Electricity is used in welding for generating heat at the point of welding in

order tomelt the material which will subsequently fuse and form the actual weld joint.

Thereare many ways of producing this localised heat but the two most common

methods are as follows

1. Resistance welding—here current is passed through the inherent resistance of thejoint to be welded

thereby generating the heat as per the equation I2 Rt/J kilocalories.

2. Arc welding—here electricity is conducted in the form of an arc which isestablished between the two

metallic surfaces


Welding is never done directly from the supply mains. Instead, special

weldingmachines are used which provided currents of various characteristics. Use of

suchmachines is essential for the following reasons

1. To convert a.c. supply into d.c. supply when d.c. welding is desired.

2. To reduce the high supply voltage to a safer and suitable voltage for weldingpurposes.

3. To provide high current necessary for arc welding withoutdrawing a corresponding high

current from the supply mains.

4. To provide suitable voltage/current relationships necessary forarc welding at minimum

cost.

There are two general types of arc welding machines:

(a) d.c. welding machines

(i) motor-generator set

(ii) a.c. transformers with rectifiers

(b) a.c. welding machines


It is found that during welding operation, large fluctuations in current andarc

voltage result from the mechanism of metal transfer and other factors. Thewelding

machine must compensate for such changes in arc voltage in order tomaintain an even

arc column. There are three major voltage/ current characteristicsused in modern d.c.

welding machines which help in controlling these currentfluctuations:

1. Drooping arc voltage (DAV).

2. Constant arc voltage (CAV).

3. Rising arc voltage (RAV).

Figure: 1.26 V-I characteristics of Arc Welding process

The machines with DAV characteristics have high open-circuit voltage

whichdrops to a minimum when arc column is started. The value of current rises

rapidly asshown in Figure 1.26 (a). This type of characteristic is preferred for manual

shield metal arc welding. The CAV characteristic shown in Figure.1.26 (b) is suitable

for semiautomatic or automatic welding processes because voltage remains

constantirrespective of the amount of current drawn . Because of its rising

voltagecharacteristic, RAV has an advantage over CAV because it maintains a

constant arc gap even if short circuit occurs due to metal transfer by the arc.

Moreover, it is welladopted to fully automatic process. DC welding machines can be

controlled by a simple rheostat in the exciter circuit or by a combination of exciter

regulator and series of field taps. Some arcs welding are equipped with remote-

controlled current units enabling the operator to vary voltage amperage requirement

without leaving themachines.


Such a welding plant is a self-contained single-operator motor-generator

setconsisting of a reverse series winding d.c generator driven by either a d.c. or an

a.c.motor (usually 3-phase). The series winding produces a magnetic field

whichopposes that of the shunt winding. On open-circuit, only shunt field isoperative

and provides maximum voltage for striking thearc. After the arc hasbeen established,

current flows through the series winding and sets up a fluxwhich opposes the flux

produced by shunt winding. Due to decreases in the net flux,generator voltage is

decreased With the help of shunt regulator,generator voltage and current values can be

adjusted to the desired level. Matters are so arranged that despite changes in arc

voltage due to variations in arc length, current remains practically constant. Figure.

1.27 shows the circuit of a d.c. motor-generator type of welding machine.

Figure: 1.27 DC motor-generator welding machine

Advantages

1. It permits portable operation.

2. It can be used with either straight or reverse polarity.

3. It can be employed on nearly all ferrous and non-ferrous materials.

4. It can use a large variety of stick electrodes.

5. It can be use for all positions of welding.


Figure: 1.28 AC Rectified Welding Machine

It consists of a transformer (single-or three-phase) and a rectifier unit asshown in Figure. 1.28. Such a

unit has no moving parts, hence it has long life. The onlymoving part is the fan for cooling the

transformer. But this fan is not the basic part ofthe electrical system. Figure. 1.28 shows a single-phase

full-wave rectified circuit ofthe welder. Silicon diodes are used for converting a.c. into d.c. These diodes

arehermetically sealed and are almost ageless because they maintain rectifyingcharacteristics indefinitely

Such a transformer-rectifier welder is most adaptablefor shield arc welding because it provides both d.c.

and a.c. polarities. It is veryefficient and quiet in operation. These welders are particularly suitable for

thewelding of (i) pipes in all positions (ii) non-ferrous metals (iii) low-alloy andcorrosion-heat and creep-

resisting steel (iv) mild steels in thin gauges


It consists of a step-down trans-former with a tapped secondary having an

adjustable reactor in series with it for obtaining drooping V/I characteristics. The

secondary is tapped to give differentvoltage/ current settings.

Advantages

(i) Low initial cost

(ii) Low operation and maintenance cost

(iii) Low wear

(iv) No arc blow

Disadvantages

(i) its polarity cannot be changed

(ii) it is not suitable for welding of cast iron andnon-ferrous metals

Question BankPART - A

1. What are the Modes of Heat transfer?

2. What are the Properties of heating element Materials?

3. What are the Types of Electric Heating?

4. What are the Reasons for failure of heating element?

5. 11 What are the Application of Dielectric Heating?

6. What are the Advantages of Dielectric Heating?

7. What are the Advantages of Coated Electrodes?

PART - B

1. Explain coreless type induction furnace

2. Explain the construction and working principle of dielectric heating.

3. What are the types of ARC furnace? Describe the operation of them. (8)

4. Discuss the characteristic requirement of welding generator sets

(both AC and DC).

UNIT II

ELECTRIC HEATING & WELDING


Electric heating is extensively used both for domestic and industrial applications.

Domestic applications include (i) room heaters (ii) immersion heaters for waterheating(iii) hot plates for

cooking (iv) electric kettles (v) electric irons (vi) popcornplants(vii) electric ovens for bakeries and (viii)

electric toasters etc.


(i) Cleanliness. Since neither dust nor ash is produced in electric heating, it is a clean

system of heating requiring minimum cost of cleaning. Moreover, the material to be

heated does not get contaminated.

(ii) No Pollution. Since no flue gases are produced in electric heating, no provisionhas to be made

for their exit.

(iii) Economical. Electric heating is economical because electric furnaces are cheaper in

their initial cost as well as maintenance cost since they do not require big space for

installation or for storage of coal and wood. Moreover, there is no need to construct

anychimney or to provide extra heat installation.

(iv) Ease of Control. It is easy to control and regulate the temperature of an electric

furnace with the help of manual or automatic devices. Temperature can be controlled

within ± 5°C which is not possible in any other form of heating.

(v) Special Heating Requirement. Special heating requirements such as uniformheating of a material

or heating one particular portion of the job without affecting its other parts or heating with no

oxidation can be met only by electric heating.

(vi) Higher Efficiency. Heat produced electrically does not go away waste through thechimney and

other by products. Consequently, most of the heat produced isutilised for heating the material

itself. Hence, electric heating has higher efficiency ascompared to other types of heating.

(vii) Better working Conditions. Since electric heating produces no irritating noisesand alsothe

radiation losses are low, it results in low ambient temperature. Hence,working with electric

furnaces is convenient and cool.

(viii) Heating of Bad Conductors. Bad conductors of heat and electricity like wood,plastic and

bakery items can be uniformly and suitably heated with dielectric heatingprocess.

(ix) Safety-Electric heating is quite safe because it responds quickly to the controlledsignals.

(x) Lower Attention and Maintenance Cost. Electric heating equipment generally willnot

requiremuchattention and supervision and their maintenance cost is almostnegligible.

Hence, labour charges arenegligibly small as compared to other forms ofheating.


The different methods by which heat is transferred from a hotbody to a cold body are as under:

1. Conduction

In this mode of heat transfer, one molecule of the body gets heatedand transfers some of the heat to the

adjacent molecule and so on.There is a temperature gradient between the two ends of the bodybeing

heated. Consider a solid material of cross-section Asq.m. andthickness x metre as shown

Figure: 1.22 Solid Material representing Heat Transfer Phenomenon

If T1 and T2 are the temperatures of the two sides of the slab in °K, where K is thermal conductivity of

the material.

2. Convection

In this process, heat is transferred by the flow of hot and cold air currents. Thisprocess is applied in the

heating of water by immersion heater or heating of buildings. The quantity of heat absorbed by the body

by convection process depends mainly on the temperature of the heating element above the surroundings

and upon the size of the surface of the heater. It also depends, to some extent, on the positionof the

heater. The amount of heat dissipated is givenH = a (T1 - T2),where a and b are constants and T1 and T2

are the temperatures of the heatingsurface and the fluid in °K respectively. In electric furnaces, heat

transferred by convection is negligible.

3. Radiation

It is the transfer of heat from a hot body to a cold body in a straight line withoutaffecting the intervening

medium. The rate of heat emission is given by Stefan’s lawaccording to which heat dissipated

where K is radiating efficiency and e is known as emissivity of the heating element.If d is the diameter of

the heating wire and l its total length, then its surface areafrom which heat is radiated = πd×l . If H is the

power radiated per m2 of the heatingsurface, then total power radiated as heat = H ×πd l . If P is the

electrical power input to the heating element, then P=πdl×H.

2.3 Requirement of a Good Heating Element(1) High Specific Resistance.When specific resistance of the material of the wireishigh, only short length

of it will be required for a particular resistance (and henceheat) or for the same length of the wire and the

currrent, heat produced will be more.

(2) High Melting Temperature. If the melting temperature of the heating elementishigh, it would be

possible to obtain higher operating temperatures.

(3) Low Temperature Coefficient of Resistance. In case the material has lowtemperature coefficient of

resistance, there would be only small variations in itsresistance over its normal range of temperature.

Hence, the current drawn by theheating element when cold (i.e., at start) would be practically the same

when it is hot.

(4) High Oxidising Temperature. Oxidisation temperature of theheating element should be high in order

to ensure longer life.

(5) Positive Temperature Coefficient of Resistance. If the temperature coefficient ofthe resistance of

heating element is negative, its resistance will decrease with rise intemperature and it will draw more

current which will produce more wattage and henceheat. With more heat, the resistance will decrease

further resulting in instability ofoperation.

(6) Ductile. Since the material of the heating elements has to have convenientshapes and sizes, it should

have high ductility and flexibility.

(7) Mechanical Strength. The material of the heating element should posses highmechanical strength of

its own. Usually, different types of alloys are used to getdifferentoperatingtemperatures. For example

maximum working temperature ofconstant an (45% Ni, 55% Cu) is 400°C, that ofnichrome (50%, Ni

20% Cr) is 1150°C, that of Kantha(70% Fe, 25% Cr, 5% Al) is 1200° C and that of silicon carbide is

1450°C. With the passage of time, every heating element breaks open and becomes unserviceable. Some

of the factors responsible for its failure are :

(1) Formation of hot spots which shine brighter during operation, (2) Oxidation (3)

Corrosion (4) Mechanical failure

2.4 Design Of Heating ElementNormally, wires of circular cross-section or rectangular conducting ribbons are usedas heating elements.

Under steady-state conditions, a heating element dissipates asmuch heat from its surface as it receives the

power from the electric supply. If P is the power input and H is the heat dissipated by radiation, then P =

H under steady-state conditions.As per Stefan’s law of radiation, heat radiated by a hot body is given by

where T1 is the temperature of hot body in °K and T2 that of the cold body (or coldsurroundings) in °K

Total surface area of the wire of the element = (πd) × l

If H is the heat dissipated by radiation per second per unit surface area of the wire,then heat radiated per

second=(πd) × l × H

2.5 Methods of Heating2.5.1 Resistance Heating

2.5.1.1 Direct Resistance Heating

In this method the material (or charge) to be heated is treated as a resistance

andcurrent is passed through it. The charge may be in the form of powder, small

solidpieces or liquid. The two electrodes are inserted in the charge and connected to

either a.c. or d.c. supply .

Figure: 2.1 Direct Resistance Heating

Obviously, two electrodes will be required in the case of d.c. or single-phase a.c. supply but there would

be three electrodes in the case of 3- phase supply. When the charge is in the form of small pieces, a

powder of high resistivity material is sprinkled over the surface of the charge to avoid direct short circuit.

Heat is produced when current passes through it. This method of heating has high efficiency because the

heat is produced in the charge itself.

1.12.1.2 Indirect Resistance Heating

In this method of heating, electric current is passed through a resistance element which is placed in an

electric oven. Heat produced is proportional to I2Rlosses in the heating element.

Figure: 1.24 Indirect Resistances Heating

The heat so produced is delivered to the chargeeither by radiation or convection or by a combination of

the two. Sometimes,resistance is placed in a cylinder which is surrounded by the charge placed in

thejacket as shown above. This arrangement provides uniform temperature.Moreover, automatic

temperature control can also be provided.

2.5.1.2 Core Type Furnace

(a)Core-type Furnaces which operate just like a two wind ing transformer. These canbe further sub-

divided into (i) Direct core-type furnaces (ii) Vertical core-typefurnaces and (iii) Indirect core-type

furnaces.

(b) Coreless-type Furnaces — in which an inductively-heated element is madetotransfer heat to the

charge by radiation.

Core Type Induction Furnace

It is essentially a transformer in which the charge to beheated forms a single-turn short-circuited

secondary and is magnetically coupled to the primary by an iron core. The furnace consists of a circular

hearth which contains the charge to be melted in the form of an annular ring. When there is no molten

metal in the ring, the secondary becomes open-circuited there-by cutting off the secondary current.

Figure: 1.25 Core Type Induction Furnace

Hence, to start the furnace, molted metal has to be poured in the annular hearth. Since, magnetic

coupling between the primary and secondary is very poor, it results in high leakage and low power factor.

In order to nullify the effect of increased leakage reactance, low primary frequency of the order of 10 Hz

is used. If the transformer secondary current density exceeds 500 A/cm2 then, due to the interaction of

secondary current with the alternating magnetic field, the moltenmetal is squeezed to the extent that

secondary circuit is interrupted.This effect isknown as “pinch effect”.

2.6 Welding

It is the process of joining two pieces of metal or non-metal at faces rendered

plastic or liquid by the application of heat or pressure or both. Filler material may be

used to effect the union.

All welding processes fall into two distinct categories:

1. Fusion Welding—It involves melting of the parent metal. Examples are:

(i) Carbon arc welding, metal arc welding, electron beam welding, electro slagwelding and electrogas

welding which utilize electric energy and(ii) Gas welding and thermit welding which utilize chemical

energy for the meltingpurpose.

2. Non-fusion Welding—It does not involve melting of the parent metal. Examples are:

(i) Forge welding and gas non-fusion welding which use chemical energy.

(ii) Explosive welding, friction welding and ultrasonic welding etc.,which use mechanical energy.

(iii) Resistance welding which uses electrical energy. Proper selection of the weldingprocess depends on

the (a) kind of metals to be joined (b) cost involved (c) nature ofproducts to be fabricated and (d)

production techniques adopted.


Electricity is used in welding for generating heat at the point of welding in

order tomelt the material which will subsequently fuse and form the actual weld joint.

Thereare many ways of producing this localised heat but the two most common

methods are as follows

1. Resistance welding—here current is passed through the inherent resistance of thejoint to be welded

thereby generating the heat as per the equation I2 Rt/J kilocalories.

2. Arc welding—here electricity is conducted in the form of an arc which isestablished between the two

metallic surfaces


Welding is never done directly from the supply mains. Instead, special

weldingmachines are used which provided currents of various characteristics. Use of

suchmachines is essential for the following reasons

1. To convert a.c. supply into d.c. supply when d.c. welding is desired.

2. To reduce the high supply voltage to a safer and suitable voltage for weldingpurposes.

3. To provide high current necessary for arc welding withoutdrawing a corresponding high

current from the supply mains.

4. To provide suitable voltage/current relationships necessary forarc welding at minimum

cost.

There are two general types of arc welding machines:

(a) d.c. welding machines

(i) motor-generator set

(ii) a.c. transformers with rectifiers

(b) a.c. welding machines


It is found that during welding operation, large fluctuations in current andarc

voltage result from the mechanism of metal transfer and other factors. Thewelding

machine must compensate for such changes in arc voltage in order tomaintain an even

arc column. There are three major voltage/ current characteristicsused in modern d.c.

welding machines which help in controlling these currentfluctuations:

1. Drooping arc voltage (DAV).

2. Constant arc voltage (CAV).

3. Rising arc voltage (RAV).

Figure: 1.26 V-I characteristics of Arc Welding process

The machines with DAV characteristics have high open-circuit voltage

whichdrops to a minimum when arc column is started. The value of current rises

rapidly asshown in Figure 1.26 (a). This type of characteristic is preferred for manual

shield metal arc welding. The CAV characteristic shown in Figure.1.26 (b) is suitable

for semiautomatic or automatic welding processes because voltage remains

constantirrespective of the amount of current drawn . Because of its rising

voltagecharacteristic, RAV has an advantage over CAV because it maintains a

constant arc gap even if short circuit occurs due to metal transfer by the arc.

Moreover, it is welladopted to fully automatic process. DC welding machines can be

controlled by a simple rheostat in the exciter circuit or by a combination of exciter

regulator and series of field taps. Some arcs welding are equipped with remote-

controlled current units enabling the operator to vary voltage amperage requirement

without leaving themachines.


Such a welding plant is a self-contained single-operator motor-generator

setconsisting of a reverse series winding d.c generator driven by either a d.c. or an

a.c.motor (usually 3-phase). The series winding produces a magnetic field

whichopposes that of the shunt winding. On open-circuit, only shunt field isoperative

and provides maximum voltage for striking thearc. After the arc hasbeen established,

current flows through the series winding and sets up a fluxwhich opposes the flux

produced by shunt winding. Due to decreases in the net flux,generator voltage is

decreased With the help of shunt regulator,generator voltage and current values can be

adjusted to the desired level. Matters are so arranged that despite changes in arc

voltage due to variations in arc length, current remains practically constant. Figure.

1.27 shows the circuit of a d.c. motor-generator type of welding machine.

Figure: 1.27 DC motor-generator welding machine

Advantages

1. It permits portable operation.

2. It can be used with either straight or reverse polarity.

3. It can be employed on nearly all ferrous and non-ferrous materials.

4. It can use a large variety of stick electrodes.

5. It can be use for all positions of welding.


Figure: 1.28 AC Rectified Welding Machine

It consists of a transformer (single-or three-phase) and a rectifier unit asshown in Figure. 1.28. Such a

unit has no moving parts, hence it has long life. The onlymoving part is the fan for cooling the

transformer. But this fan is not the basic part ofthe electrical system. Figure. 1.28 shows a single-phase

full-wave rectified circuit ofthe welder. Silicon diodes are used for converting a.c. into d.c. These diodes

arehermetically sealed and are almost ageless because they maintain rectifyingcharacteristics indefinitely

Such a transformer-rectifier welder is most adaptablefor shield arc welding because it provides both d.c.

and a.c. polarities. It is veryefficient and quiet in operation. These welders are particularly suitable for

thewelding of (i) pipes in all positions (ii) non-ferrous metals (iii) low-alloy andcorrosion-heat and creep-

resisting steel (iv) mild steels in thin gauges


It consists of a step-down trans-former with a tapped secondary having an

adjustable reactor in series with it for obtaining drooping V/I characteristics. The

secondary is tapped to give differentvoltage/ current settings.

Advantages

(i) Low initial cost

(ii) Low operation and maintenance cost

(iii) Low wear

(iv) No arc blow

Disadvantages

(ii) its polarity cannot be changed

(ii) it is not suitable for welding of cast iron andnon-ferrous metals

UNIT- III

ILLUMINATION

3.1 INTRODUCTION

The aim of artificial lighting is to supplement the day light or to replace it in

modern offices, homes, industries, work places etc.Good illumination ensures

increased production, effectivity of work and reduced accidents. Before entering into

the field of illumination technology, a glimpse of light energy may be helpful.In the

beginning of the nineteenth century it was not possible to do day time work after

sunset, due to lack of adequate light.During those days crude system of lighting was

use but during the middle of the 19th century a gas mantle was used as a source of

light.In the early 1900, the electric filament lamps came into the field as a source of

light.The electric lamps are preferred to other sources of illumination for reason of

cleanliness, convenience, steady light output and reliability.

Radiant Energy

Light is a form of radiant energy, which produces the sensation of sight in µs.

According to quantum theory of max plank, light energy is emitted in the form of small

‘bundless’ or pockets’s called quanta or photons.

Velocity of light is 3 x 108 metre/ sec

Velocity of light wave propagation v= λf

Where λ = wave length

F = frequency

Wave length of high frequency wave can be measured in micron or Angstrom

1 micron (µ) = 10-6 m

1 Angstrom (Aº) = 10-10 m

Radiations of different wavelengths produce different colour sensation on the eyes

Visible light can have a wave length between 4000 Aº 7500 Aº.

Illumination

Illumination is the result of that light on surfaces on which it falls.

3.2Definitions

Light

Light is defined as that radiant energy in the form of waves which produces a sensation of vision upon the human eye.

Luminous flux

Luminous flux is defined as the energy in the form of light waves radiated per second

from a luminous body.(eg.: incandescent lamp).

Luminous efficiency

It is defined as the output in lumens per watt of the power consumed by the

source of light. It is measured in lumens per wattage.

Lumens emitted by the source

Luminous efficiency = --------------------------------------

Wattage of source per light

Luminous flux

Luminous flux is defined as the energy in the form of light waves radiated per

second

from a luminous body.(eg.: incandescent lamp).

Luminous Intensity

Luminous intensity is defined as the flux emitted by the source per unit solid

angle.

Lumen

It is a unit of flux and is defined as the luminous flux per unit angle from a source

of one candle power.

Lumens = Candle power X Solid angle.

= C.P. X ω.

Candle Power

Candle power is the light rendering capacity of a source in a given direction and is

defined as the number of lumens given out by the source in a unit solid angle in a given

direction.

LumensC.P. = ----------

ωPlane Angle

When two straight lines lying in the same plane meet at a point, there will be an angle

between these converging lines at the meeting point. This angle is termed as plane angle

ArcPlane angle = -----------

RadiusUnit of plane angle is radians.

Solid Angle

The angle subtended at a point in space by an area is termed as solid angle. In

plane angle it is the area which is enclosed by two lines, but in case of solid angle. It

is thevolume enclosed by numerous lines lying on the surface and meeting at a point.

Solid angle is denoted by ω.Unit of solid angle is steradian.

Figure:1.1Solid Angle representation

Illumination

When the light falls on a surface it is illuminated. The illuminance is defined as the

luminous flux received per unit area.Illuminance is denoted by the symbol E and is

measured in lumens/m2or lux or metre-candle

Brightness

It is defined as the flux emitted per unit area or the luminous intensity per unit

projected area of the source in a direction perpendicular to the surface.

Mean Hemispherical Candle Power

The mean hemispherical candle power of a source of light is the mean or averageof

the candle power in all directions within the hemisphere either above the horizontalplane

or below the horizontal plane.

Mean Spherical Candle Power

The mean spherical candle power of a source of light is the mean or average of

thecandle power in all directions in all the planes.

Mean Hemispherical Candle Power

The mean hemispherical candle power of a source of light is the mean or averageof the

candle power in all directions within the hemisphere either above the horizontalplane

or below the horizontal plane.

Reduction Factor

Reduction factor of a source of light is defined as the ratio of its mean

sphericalcandle power to its mean horizontal candle power.

Reduction factor = M.S.C.P / M.H.C.P.

Reflection Factor

It is defined as the ratio of the ratio of reflected light to the incident light. It

isalways less than unity.

Reflected lightReflection factor = ----------------------

Incident light

Lamp Efficiency

Lamp efficiency is defined as the ratio of luiminous flux to the power input. It

is expressed in lumens/watt.

Depreciation Factor

Depreciation factor is defined as , illumination under normal working

condition to illumination when everything is clean. So this occurs when the source is

not clean.(eg. lamps covered with dust, dirt or smoke).

Utilization factor (UF)

This is the proportion of the luminous flux emitted by the lamps, reaching the

working plane. It is a measure of the effectiveness of the lighting scheme.

3.3Nature of Radiation

The usual method of producing artificial light consists in raising a solid body or

vapour to

incandescence by applying heat to it. It is found that as the body is gradually heated above

roomtemperature, it begins to radiate energy in the surrounding medium in the form of electromagnetic

waves of various wavelengths. The nature of this radiant energy depends on the temperature of the hot

body. Thus, when the temperature is low, radiated energy is in the form of heat wave so only but when a

certain temperature is reached, light waves are also radiated out in addition to he waves and the body

becomes luminous. Further increase in temperature produces an increase in the amount of both kinds of

radiations but the colour of light or visible radiation changes from bright red to orange, to yellow and

then finally, if the temperature is high enough, to white. As temperature is increased, the wavelength of

the visible radiation goes on becoming shorter. It should be noted that heat waves are identical to light

waves except that they are of longer wavelength and hence produce no impression on the retina.

Obviously, from the point of view of light emission, heat energy represents much wasted energy.

3.4 Laws of Illumination

There are 2 important laws in illumination.

Inverse Square Law

Cosine Law

Inverse Square Law

The illumination of a surface is inversely proportional to the square of the distance

between the surface and the light source provided that the distance between the

surface and the source is sufficiently large so that the source can be regarded as a

point source.

This law assumes that the illumination received on a surface from a light source is

inversely proportional to the square of it’s distance from the source

In other words, the further away, the less illumination

Figure: 1.2 Illumination level

The inverse square law can be calculated by

Where E is illuminance

I is Luminous intensity

d is distance

Example

The illuminance on a surface directly below a point source is 400 lux. If the distance

between the light source and the surface is 2m, what is the intensity of the light

source?

The solution assumes that the lamp is a filament or energy saver type

A long fluorescent type does not use this law

I= 400 X 4

I = 1600 cd

Cosine Law

According to this law the illumination at any point on a surface is proportional to

the cosine of the angle between the normal at that point and the direction of luminous

flux.

This method allows us to calculate the illuminance with one or more lamps or with

reflection from surroundings.

Figure: 1.3 Illumination for cosine law representation

ab 1------ = ----------ac cos θ

cos θand the illumination decreases in the ratio -----

1The expression for the illumination then becomes

3.5Polar Curves

In most lamps or sources of light the luminous intensity is not the same in all

directions. If the luminous intensity, ie the candle power is measured in a horizontal

plane about a vertical axis and a curve is plotted between candle power and the

angular position, a curve is obtained is called horizontal polar curve.

The luminous intensity in all the directions can be represented by polar curves. If the

luminous intensity in a vertical plane is plotted against the angular position, a curve

known as vertical polar curve is obtained. The typical polar curves for an

ordinaryfilament lamp are shown in Figure.

Figure: 1.4Horizontal &Figure: 1.5 Vertical plane curve

Uses:

The polar curves are used to determine the MHCP,MSCP and the actual illumination of a

surface by employing the following methods.

Rousseau’s Construction

All our calculations so far were based on the tacit assumption that the light source

was of equal luminous intensity or candle-power in all directions. However, lamps and

other sources of light, as a rule, do not give uniform distribution in the space

surrounding them. If the actual luminous intensity of a source in various directions be

plotted to scale along lines radiating from the centre of the source at corresponding

angles, we obtain the polar curve of the candle power. Suppose we construct a Figure

consisting of large number of spokes radiating out from a point —the length of each

spoke representing to some scale the candle power or luminous intensity of the source

in that particular direction. If now we join the ends of these spokes by some suitable

material, say, by linen cloth, then we get a surface whose shape will represent to scale

the three dimensional candle power distribution of the source placed at the centre. In

the ideal case of a point source having equal distribution in all directions, the surface

would be spherical.

Figure: 1.6 &Figure: 1.7 -Two polar curves of c.p. distribution in a vertical plane

It is difficult to give a graphic representation of such a 3- dimensional model in

a plane surface. Therefore, as with engineering drawings, it is usual to draw only one

or more elevations and a plan of sections through the centre of the source. Elevations

represent c.p. distribution in the vertical plane and the plans represent c.p. distribution

in horizontal plane. The number of elevations required to give a complete idea of the

c.p. distribution of the source in all directions depends upon the shape of the plan i.e.

on the horizontal distribution. If the distribution is uniform in every horizontal plane

i.e. if the polar curve of horizontal distribution is a circle, then only one vertical curve

is sufficient to give full idea of the space distribution.

Figure: 1.8 Distribution of luminous intensity by the source

In Figure.1.6are shown two polar curves of c.p. distribution in a vertical plane.Curve 1 is for vacuum

type tungsten lamp with zig-zag filament whereas curve 2 is for gas filled tungsten lamp with filament

arranged as a horizontal ring. If the polar curve issymmetrical about the vertical axis as in the Figures

given below, then it is sufficient togive only the polar curve within one semicircle in order to completely

define thedistribution of c.p. as shown in Figure. 1.7 The curves 1 and 2 are as in Figure. 1.6, a curve 3 is

for d.c. open arc with plain carbons and curve 4 is for a.c. arc with plain carbons. However, if the source

and/or reflector are not symmetrical about vertical axis, it is impossible to represent the space distribution

of c.p. by a single polar diagram and even polar diagrams for two planes at right angles to each other give

no definite idea as to the distribution in the intermediate planes. Consider a filament lamp with a helmet-

type reflector whose axis is inclined and crosssection elliptical—such reflectors are widely used for

lighting shop windows.Figure.1.8 represents the distribution of luminous intensity of such source and its

reflector in two planes at right angles to each other.

The importance of considering the polar curves in different planes when the

c.p.distribution in asymmetrical is even more strikingly depicted by the polar curves in

Y Yplane and X X plane of a lamp with a special type of reflector designed for street

lightingpurposes (Figure. 1.9). It would be realized from above that the polar

distribution of light from any source can be given any desired form by using reflectors

and/or refractors of appropriate shape. In Figure.1.10 is shown the polar curve of c.p.

distribution of a straight type of lamp in a horizontal plane.

Figure: 1.9 Polar curves depicting the c.p.distribution in asymmetrical mode

Figure: 1.10 Polar curve of c.p. distribution of a straight type of lamp in a horizontal plane.

Figure: 3.11 (a) &Figure: 1.11(b) Polar distribution curve of a filament lamp representing the c.p distribution in a horizontal plane & vertical plane.

Determination of M.S.C.P. and M.H.C.P. from Polar Diagrams

Figure.1.11 (a) shows the polar distribution curve of a filament lamp in a

horizontal plane and the polar curve in Figure. 1.11(b) represents the c.p. distribution

in a vertical plane. It will be seen that the horizontal candle-power is almost uniform

in all directions in that plane. However, in the vertical plane, there is a large variation

in the candle power which falls to zero behind the cap of the lamp. The curve in

Figure.1.11 (a) has been drawn with the help of a photometer while the lamp is rotated

about a vertical axis, say, 10° at a time. But the curve in Figure.1.11 (b) was drawn

while the lamp was rotated in a vertical plane about a horizontal axis passing through

the centre of the filament. The M.H.C.P. is taken as the mean of the readings in

Figure. 1.11 (a). However, a more accurate result can be obtained by plotting candle

power on an angular base along the rectangular axes and by determining the mean

height of the curve by the mid-ordinate or by Simpson’s rule.The M.S.C.P. of the

lamp can be obtained from the vertical polar curve of Figure. 1.11 (b) by Rousseau’s

construction as explained below.

Only half of the vertical polar curve is shown in the Figure (Figure. 1.12) since it is symmetrical about

the vertical axis. With O is the centre and radius OR equal to the maximum radius of the polar curve, a

semi-circle LRM is drawn. A convenient number of points on this semi-circle (say 10° points) are

projected onto any vertical plane as shown. For example, points a,b,c etc. are projected to d,e,f and so on.

From point d, the horizontal line dg is drawn equal to the intercept OA of the polar diagram on the radius

oa. Similarly, eh = OB, fk = OC and so on. The point’s g, h, k etc., define the Rousseau Figure. The

average width w of this Figure represents the M.S.C.P. to the same scale as that of the candle powers in

the polar curve. The average width is obtained by dividing the Rousseau area by the base of the Rousseau

Figure i.e. length lm which is the projection of the semi-circle LM on the vertical axis. The area may be

determined by Simpson’s rule or by using a planimeter.

M.S.C.P. = area of Rousseau Figure

length of the baseSpherical reduction factor f = M.S.C.P

M.H.C.P. ∴M.S.C.P. = f × M.H.C.P.

Figure: 1.12 Half-side of the Vertical Polar Curve

3.6 Design of Illumination Systems

Design of Lighting Schemes and Lay-outs

A well-designed lighting scheme is one which

(i) provides adequate illumination (ii) avoids glare and hard shadows (iii) provides sufficiently uniform

distribution of light all over the working plane.

It is the ratio of the lumens actually received by a particular surface to the total lumens emitted by a

luminous source.

Different lighting schemes may be classified as (i) direct lighting (ii) indirect lighting and (iii) semi-

direct lighting (iv) semi-indirect lighting and (v) general diffusing systems.

3.6.1 Direct Lighting

As the name indicates, in the form of lighting, the light from the source falls directly on the object or the

surface to be illuminated

Figure: 1.13 Direct Lighting

With the help of shades and globes and reflectors of various types, most of the

light is directed in the lower hemisphere and also the brilliant source of light is kept

out of the direct line of vision. Direct illumination by lamps in suitable reflectors can

be supplemented by standard or bracket lamps on desk or by additional pendant

fittings over counters.

The fundamental point worth remembering is planning any lighting installation is that sufficient and

sufficiently uniform lighting is to be provided at the working or reading plane. For this purpose, lamps of

suitable size have to be so located and furnished with such fittings as to give correct degree and

distribution of illumination at the required place. Moreover, it is important to keep the lamps and fittings

clean otherwise the decrease in effective illumination due to dirty bulbs or reflectors may amount to 15 to

25% in offices and domestic lighting and more in industrial areas as a result of a few weeks neglect.

Direct lighting, though most efficient, is liable to cause glare and hard shadows.

3.6.2 Indirect Lighting

In this form of lighting, light does not reach the surface directly from the source but indirectly by diffuse

reflection.

Figure: 1.14 Indirect Lighting system

The lamps are either placed behind a cornice or in suspended opaque bowls. In

both cases, a silvered reflector which is corrugated for eliminating striations is placed

beneath the lamp. In this way, maximum light is thrown upwards on the ceiling from

which it is distributed all over the room by diffuse reflection. Even gradation of light

on the ceiling is secured by careful adjustment of the position and the number of

lamps. In the cornice and bowl system of lighting, bowl fittings are generally

suspended about three fourths the height of the room and in the case of cornice

lighting, a frieze of curved profile aids in throwing the light out into the room to be

illuminated. Since in indirect lighting whole of the light on the working plane is

received by diffuse reflection, it is important to keep the fittings clean.

One of the main characteristics of indirect lighting is that it provides shadow

less illumination which is very useful for drawing offices, composing rooms and in

workshops especially where large machines and other obstructions would cast

troublesome shadows if direct lighting were used. However, many people find purely

indirect lighting flat and monotonous and even depressive.Most of the users demand

50 to 100% more light at their working plane by indirect lighting than with direct

lighting. However, for appreciating relief, a certain proportionof direct lighting is

essential.

3.6.3 Semi-direct System

This system utilizes luminaries which send most of the light downwards directly on the working plane

but a considerable amount reaches the ceilings and walls also.

Figure: 1.15 Semi-direct System

The division is usually 30% upwards and 45% downwards. Such a system is best suited to rooms with

high ceilings where a high level of uniformly-distributed illumination is desirable. Glare in such units is

avoided by using diffusing globes which not only improve the brightness towards the eye level but

improve theefficiency of the system with reference to the working plane.

3.6.4 Semi-indirect Lighting

In this system which is, in fact, a compromise between the first two systems, the light is partly received

by diffuse reflection and partly direct from the source.

Figure: 1.16 Semi-Indirect Lighting

Such a system, therefore, eliminates the objections of indirect lighting mentioned above. Instead of using

opaque bowls with reflectors, translucent bowls without reflector are used. Most of the light is, as before,

directed upwards to the ceiling for diffuse reflection and the rest reaches the working plane directly

except for some absorption by the bowl.

3.6.5 General Diffusing System

Figure: 1.17 General Diffusing Lighting

In this system, luminaries are employed which have almost equal light distribution downwards and

upwards.

3.7 Types of Lamps

1. By temperature incandescence - In this method, an electric current is passedthrough a filament of thin

wire placed in vacuum or an inert gas. The current generates enough heat to raise the temperature of the

filament to luminosity. Incandescent tungsten filament lamps are examples of this type and since their

output depends on the temperature of their filaments, they are known as temperature radiators.

2. By establishing an arc between two carbon electrodes. The source of light, in theircase, is the

incandescent electrode.

3. By Discharge Lamps.

3.7.1 Discharge Lamps

In these lamps, gas or vapour is made luminous by an electricdischarge

through them. The colour and intensity of light i.e. candle-power emitteddepends on

the nature of the gas or vapour only. It should be particularly noted thatthese discharge

lamps are luminiscent-light lamps and do not depend on temperature for higher

efficiencies. In this respect, they differ radically from incandescent lamps whose

efficiency is dependent on temperature. Mercury vapour lamp, sodiumvapour lamp,

neon-gas lamp and fluorescent lamps are examples of light sources based on discharge

through gases and vapours.

3.7.1.1 Incandescent Lamp

An incandescent lamp essentially consists of a fine wire of a high-resistancemetal

placed in an evacuated glass bulb and heated to luminosity by the passage ofcurrent

through it. Such lamps were produced commercially for the first time by Edison in

1879. His early lamps had filaments of carbonized paper which were, later on,

replaced by carbonized bamboo. They had the disadvantage of negativetemperature

coefficient of resistivity. In 1905, the metallized carbon-filament lampswere put in the

market whose filament shad a positive temperature coefficient ofresistivity(like

metals).Such lamps gave 4 lm/W.

Figure: 1.18 Coiled coil filament & Gas filled Lamp

At approximately the same time, osmium lamps were manufactured by

whichfilaments had made of osmium which is very rare and expensive metal. Such

lamps had a very fair maintenance of candle-power during their useful life and an

average efficiency of 5 lm/W. However, osmium filaments were found to be very

fragile. In 1906 tantalum lamps having filaments of tantalum were produced which

had an initial efficiency of 5 lm/watt. All these lamps were superseded by tungsten

lamps which were commercially produced in about 1937 or so. The superiority of

tungsten lies mainly in its ability to withstand a high operating temperature without

undue vaporisation of the filament. The necessity of high working temperature is due

to the fact that the amount of visible radiation increases with temperature and so does

the radiant efficiency of the luminous source. The melting temperature of tungsten is

3655°K whereas that of osmium is 2972°K and that of tantalum is 3172°K. Actually,

carbon has a higher melting point than tungsten but its operating temperature is

limited to about 2073°K because ofrapid vaporization beyond this temperature.

In fact, the ideal material for the filament of incandescent lamps is one which has thefollowing properties

1.A high melting and hence operating temperature

2.A low vapour pressure

3.A high specific resistance and a low temperature coefficient

4.Ductility and

5.Sufficient mechanical strength to with-stand vibrations.

Since tungsten possesses practically all the above mentioned qualities, it is usedin almost all modern

incandescent lamps. The earlier lamps had a square-cagetype filament supported from a central glass

stem enclosed in an evacuated glassbulb.

The object of vacuum was two fold :

(a) To prevent oxidation and

(b) Tominimize loss of heat by convection and the consequent lowering of filamenttemperature.

However, vacuum favoured the evaporation of the filament with theresulting blackening of the lamp so

that the operating temperature had to be kept aslow as 2670º K with serious loss in luminous efficiency.

It was, later on, found that this difficulty could be solved to a great extent by inserting a chemically inert

gas like nitrogen or argon. The presence of these gases within the glass bulb decreased the evaporation of

the filament and so lengthened its life. The filament could now be run at a relatively higher temperature

and hence higher luminous efficiency could be realized. In practice, it was found that an admixture of

85% argon and about 15 percent nitrogen gave the best results.

However, introduction of gas led to another difficulty i.e. loss of heat due to convection which offsets the

additional increase in efficiency. However, it was found that for securing greater efficiency, a

concentrated filament having a tightly-wound helical construction was necessary. Such a coiled filament

was less exposed to circulating gases, its turns supplying heat to each other and further the filament was

mechanically stronger. The latest improvement is that the coiled filament is itself ‘coiled’ resulting in

‘coiled-coil’ filament Figure.1.18 (a) which leads to further concentrating the heat, reducing the effective

exposure to gases and allows higher temperature operation, thus giving greater efficiency. The

construction of a modern coiled- coil gas-filled filament lamp is shown in Figure. 1.18 (b). The lamp has

a ‘wreath’ filament i.e. a coiled filament arranged in the form of a wreath on radial supports

Incandescent Lamp Characteristics

The operating characteristics of an incandescent lamp are materially affectedby

departure from its normal working voltage. Initially, there is a rapid heating up ofthe

lamp due to its low thermal capacity, but then soon its power intake becomes steady.

If the filament resistance were not dependent on its temperature, the rate of generation

of heat would have been directlyproportional to the square of voltage applied across

the lamp.

Figure: 1.19 Gas filled Lamp

However, because of

(i) Positive temperature coefficient of resistance and

(ii) Complex mechanism of heat transfer from filament to gas, the relations between the lamp

characteristics and its voltage are mostly experimental.

3.7.1.2 Sodium Vapour Lamp

Figure:1.20 Sodium Vapour Lamp

One type of low-pressure sodium-vapour lamp along with its circuit connection is shown above. It

consists of an inner U-tube A made of a special sodium-vapour-resisting glass. It houses the two

electrodes and contains sodium together with the small amount of neon-gas at a pressure of about 10 mm

of mercury and one per cent of argon whose main function is to reduce the initial ionizing potential. The

discharge is first started in the neon gas (which gives out redishcolour). After a few minutes, the heat of

discharge through the neon gas becomes sufficient to vaporise sodium and then discharge passes through

the sodium vapour. In this way, the lamp starts its normal operation emitting its characteristic yellow

light. The tungsten-coated electrodes are connected across auto-transformer T having a relatively high

leakage reactance. The open-circuit voltage of this transformer is about 450 V which is sufficient to

initiate a discharge through the neon gas. The leakage reactance is used not only for starting the current

but also for limiting its value to safe limit. The electric discharge or arc strikes immediately after the

supply is switched on whether the lamp is hot or cold. The normal burning position of the lamp is

horizontal although two smaller sizes of lamp may be burnt vertically. The lamp is surrounded by an

outer glass envelope B which serves to reduce the loss of heat from the inner discharge tube A. In this

way, B helps to maintain the necessary high temperature needed for the operation of a sodium vapour

lamp irrespective of draughts. The capacitor C is meant for improving the power factor of the circuit

3.7.1.3 Mercury Vapour Lamp

Figure: 1.21 High Pressure Mercury Vapour Lamp

Like sodium-vapour lamp, this lamp is also classified as electric discharge

lamp inwhich light is produced by gaseous conduction. Such a lamp usually consists

of twobulbs — an arc-tube containing the electric discharge and an outer bulb which

protects the arc-tube from changes in temperature. The inner tube or arc tube A is

made of quartz (or hard glass) the outer bulb B of hard glass. As shown in Figure.

1.21, the arc tube contains a small amount of mercury and argon gas and houses three

electrodes D, E and S. The main electrodes are D and E whereas S is the auxiliary

starting electrode. S is connected through a high resistance R (about 50 kS) to the

main electrode situated at the outer end of the tube. The main electrodes consist

oftungsten coils with electron-emitting coating or elements of thorium metal. When

the supply is switched on, initial discharge for the few seconds is established in the

argon gas between D and S and then in the argon between D and E. The heat produced

due to this discharge through the gas is sufficient to vaporize mercury. Consequently,

pressure inside A increases to about one or two atmospheres and the p.d. across D and

E grows from about 20 to 150 V, the operation taking about 5-7 minutes. During this

time, discharge is established through the mercury vapours which emit greenish-blue

light. The choke serves to limit the current drawn by the discharge tube A to a safe

limit and capacitor C helps to improve the power factor of the circuit. True colour

rendition is not possible with mercury vapour lamps since there is complete absence of

red-light from their radiations. Consequently, red objects appear black, all blues

appear mercuryspectrum blue and all greens the mercury-spectrum green with the

result that colour values are distorted. Correction for colour distortion can be achieved

by

1. Using incandescent lamps (which are rich in red light) in combination with themercury

lamps.

2. Using colour-corrected mercury lamps which have an inside phosphor coat to addred

colour to the mercury spectrum. Stroboscopic (Flickering) effect in mercuryvapour

lamps is caused by the 100 on and off arc strikes when the lamps are used on the 50-Hz

supply. The effect may be minimized by

Using two lamps on lead-lag transformer

Using three lamps on separated phases of a 3-phase supply and

Using incandescent lamps in combination with mercury lamps.

In the last few years, there has been tremendous improvement in the construction and operation of

mercury-vapour lamps, which has increased their usefulness and boosted their application for all types of

industrial lighting, floodlighting and street lighting etc. As compared to an incandescent lamp, a mercury-

vapour lamp smaller in size and has 5 to 10 times longer operating life and has 3 times higher efficiency

i.e. 3 times more light output for given electrical wattage input.

Typical mercury-vapour lamp applications are:

1. High-bay industrial lighting — where high level illumination is required and colourrendition is not

important.

2. Flood-lighting and street-lighting

3. Photochemical applications — where ultra-violet output is useful as in chlorination,

water sterilization and photocopying etc.

4. For a wide range of inspection techniques by ultra-violet activation of fluorescentand

phos-phorescent dyes and pigments.

5. Sun-tan lampsfor utilizing the spectrum lines in the erythemal region ofultraviolet energy for

producing sun-tan.

UNIT- IV

ELECTRIC TRACTION-I

4.1 INTRODUCTIONBy electric traction is meant locomotion in which the driving (or tractive) force

is obtained from electric motors. It is used in electric trains, tramcars, trolley buses

and diesel-electric vehicles etc. Electric traction has many advantages as compared to

other non-electrical systems of traction including steam traction.

4.1.1 Traction Systems

a) Non-electric traction systems

They do not involve the use of electrical energy at any stage. Examples are : steam engine drive used in

railways and internal-combustion-engine drive used for road transport.

(b) Electric traction systems

They involve the use of electric energy at some stage or the other.

They may be further subdivided into two groups:

1. First group consists of self-contained vehicles or locomotives. Examples

are: battery electric drive and diesel-electric drive etc.

2. Second group consists of vehicles which receive electric power from a

distribution network fed at suitable points from either central power stations or

suitably-spaced substations.

Examples are: railway electric locomotive fed from overhead ac supply and

tramways and trolley buses supplied with dc supply.

Requirements of an Ideal Traction System

1. High adhesion coefficient, so that high tractive effort at the start is possible to have rapid

acceleration.

2. The locomotive or train unit should be self-contained so that it can run on any route.

3. Minimum wear on the track

4. It should be possible to overload the equipment for short periods.

5. The equipment required should be minimum, of high efficiency and low maintenance cost

6. It should be pollution free.

7. Speed control should be easy

8. Braking should be such that minimum wear is caused on the brake shoes, and if possible energy

should be regenerated and returned to the supply during braking period.

9. There should be no interference to the communication lines running near the track.

4.1.2 Advantages of Electric Traction

1. Cleanliness. Since it does not produce any smoke or corrosive fumes, electric traction is most suited

for underground and tube railways. Also, it causes no damage to the buildings and other apparatus due to

the absence of smoke and flue gases.

2. Maintenance Cost. The maintenance cost of an electric locomotive is nearly 50% of that for a steam

locomotive. Moreover, the maintenance time is also much less.

3. Starting Time. An electric locomotive can be started at a moment's notice whereas a steam

locomotive requires about two hours to heat up.

4. High Starting Torque. The motors used in electric traction have a very high starting torque. Hence, it

is possible to achieve higher accelerations of 1.5 to 2.5 km/h/s as against 0.6 to 0.8 km/h/s in steam

traction. As a result, we are able to get the following additional advantages:

(i) High schedule speed

(ii) increased traffic handling capacity

(iii) Because of (i) and (ii) above, less terminal space is required—a factor of

great importance in urban areas.

5. Braking. It is possible to use regenerative braking in electric traction system. It leads to the

following advantages :

(i) About 80% of the energy taken from the supply during ascent is returned to it during

descent.

(ii) Goods traffic on gradients becomes safer and speedier.

(iii) Since mechanical brakes are used to a very small extent, maintenance of brake shoes, wheels,

tyres and track rails is considerably reduced because of less wear and tear.

6. Saving in High Grade Coal. Steam locomotives use costly high-grade coal which is not so abundant.

But electric locomotives can be fed either from hydroelectric stations or pithead

thermal power stations which use cheap low-grade coal. In this way, high-grade coal can be save for

metallurgical purposes.

7. Lower Centre of Gravity. Since height of an electric locomotive is much less than that of a steam

locomotive, its centre of gravity is comparatively low. This fact enables an electric locomotive to

negotiate curves at higher speeds quite safely.

8. Absence of Unbalanced Forces. Electric traction has higher coefficient of adhesion since there are no

unbalanced forces produced by reciprocating masses as is the case in steam traction. It not only reduces

the weight/kW ratio of an electric locomotive but also improves its riding quality in addition to reducing

the wear and tear of the track rails.

4.1.3 Disadvantages of Electric Traction

1. the most vital factor against electric traction is the initial high cost of laying out overhead electric

supply system. Unless the traffic to be handled is heavy, electric traction becomes uneconomical.

2. Power failure for few minutes can cause traffic dislocation for hours.

3. Communication lines which usually run parallel to the power supply lines suffer from electrical

interference. Hence, these communication lines have either to be removed away from the rail track or

else underground cables have to be used for the purpose which makes the entire system still more

expensive.

4. Electric traction can be used only on those routes which have been electrified. Obviously, this

restriction does not apply to steam traction.

5. Provision of a negative booster is essential in the case of electric traction. By avoiding the flow of

return currents through earth, it curtails corrosion of underground pipe work and interference with

telegraph and telephone circuits.

4.2 Systems of Railway Electrification

Direct current system

Single-phase ac system

Three-phase ac system

Composite system

4.2.1 Direct Current System

Direct current at 600-750 V is universally employed for tramways in urban

areas and for many suburban railways while 1500-3000 V dc is used for main line

railways. The current collection is from third rail (or conductor rail) up to 750 V,

where large currents are involved and from overhead wire for 1500 V and 3000 V,

where small currents are involved. Since in majority of cases, track (or

running) rails are used as the return conductor, only one conductor rail is required.

Both of these contact systems are fed from substations which are spaced 3 to 5 km for

heavy suburban traffic and 40-50 km for main lines operating at higher voltages of

1500 V to 3000 V. These substations themselves receive power from 110/132 kV, 3-

phase network (or grid). At these substations, this high-voltage 3-phase supply is

converted into low-voltage 1-phase supply with the help of Scott-connected or V-

connected 3-phase transformers. Next, this low ac voltage is converted into the

required dc voltage by using suitable rectifiers or converters (like rotary converter,

mercury-arc, metal or semiconductor rectifiers). These substations are usually

automatic and are remote-controlled.

The dc supply so obtained is fed via suitable contact system to the traction

motors which are either dc series motors for electric locomotive or compound motors

for tramway and trolley buses where regenerative braking is desired.

It may be noted that for heavy suburban service, low voltage dc system is undoubtedly superior to 1-

phase ac system due to the following reasons:

1. DC motors are better suited for frequent and rapid acceleration of heavy trains than ac motors.

2. DC train equipment is lighter, less costly and more efficient than similar ac equipment.

3. When operating under similar service conditions, dc train consumes less energy than a1-phase ac train.

4. The conductor rail for dc distribution system is less costly, both initially and in maintenance than the

high-voltage overhead ac distribution system.

5. DC system causes no electrical interference with overhead communication lines.

The only disadvantage of dc system is the necessity of locating ac/dc conversion sub-stations at relatively

short distances apart.

4.2.2 Single-Phase Low-frequency AC System

In this system, ac voltages from 11 to 15 kV at1623or 25 Hz are used. If

supply is from a generating station exclusively meant for the traction system, there is

no difficulty in getting the electric supply of1623or 25 Hz. If, however, electric supply

is taken from the high voltage transmission lines at 50 Hz, then in addition to step-

down transformer, the substation is provided with a frequency converter. The

frequency converter equipment consists of a 3-phase synchronous motor which drives

a I-phase alternator having or 25 Hz frequency. The 15 kV1623or 25 Hz supply is fed

to the electric locomotor via a single over-head wire (running rail providing the return

path).

A step-down transformer carried by the locomotive reduces the 15-kV voltage

to 300-400 V for feeding the ac series motors. Speed regulation of ac series motors is

achieved by applying variable voltage from the tapped secondary of the above

transformer. Low-frequency ac supply is used because apart from improving the

commutation properties of ac motors, it increases their efficiency and power factor.

Moreover, at low frequency, line reactance is less so that line impedance drop

and hence line voltage drop is reduced. Because of this reduced line drop, it is feasible

to space the substations 50 to 80 km apart. Another advantage of employing low

frequency is that it reduces telephonic interference.

4.2.3Three-phase Low-frequency AC System

It uses 3-phase induction motors which work on a 3.3 kV, 1623 Hz supply.

Substations Receive power at a very high voltage from 3-phase transmission lines at

the usual industrial frequency of 50 Hz. This high voltage is stepped down to 3.3 kV

by transformers whereas frequency is reduced from 50 Hz to 1623 Hz by frequency

converters installed at the sub-stations. Obviously, this system employs two overhead

contact wires, the track rail forming the third phase (of course, this leads to insulation

difficulties at the junctions).

Induction motors used in the system are quite simple and robust and give

trouble free operation. They possess the merits of high efficiency and of operating as a

generator when driven at speeds above the synchronous speed. Hence, they have the

property of automatic regenerative braking during the descent on gradients. However,

it may be noted that despite all its advantages, this system has not found much favour

and has, in fact, become obsolete because of it’s certain inherent limitations given

below:

1. The overhead contact wire system becomes complicated at crossings and junctions. 2. Constant-

speed characteristics of induction motors are not suitable for traction

work.

3. Induction motors have speed/torque characteristics similar to dc shunt

motors. Hence, they are not suitable for parallel operation because, even with little

difference in Rotational speeds caused by unequal diameters of the wheels, motors

will becomes loaded very unevenly.

4.2.4 Composite System

Such a system incorporates good points of two systems while ignoring their bad points.

Two such composite systems presently in use are :

phase to 3-phase system also called Kando system

1-phase to dc system.

Kando System

In this system, single-phase 16-kV, 50 Hz supply from the substation is picked

up by the locomotive through the single overhead contact wire. It is then converted

into 3-phase ac supply at the same frequency by means of phase converter equipment

carried on the locomotives. This 3-phase supply is then fed to the 3-phase induction

motors. As seen, the complicated overhead two contact wire arrangement of ordinary

3-phase system is replaced by a single wire system. By using silicon controlled

rectifier as inverter, it is possible to get variable-frequency 3-phase supply at 1/2 to 9

Hz frequency. At this low frequency, 3-phase motors develop high starting torque

without taking excessive current. In view of the above, Kando system is likely to be

developed further.

Single-phase AC to DC System

This system combines the advantages of high-voltage ac distribution at

industrial frequency with the dc series motors traction. It employs overhead 25-kV,

50-Hz supply which is stepped down by the transformer installed in the locomotive

itself. The low-voltage ac supply is then converted into dc supply by the rectifier

which is also carried on the locomotive. This dc supply is finally fed to dc series

traction motor fitted between the wheels. The system of traction employing 25-kV,

50-Hz, 1-phase ac supply has been adopted for all future track electrification in India.

4.3 Types of Railway Services

There are three types of passenger services offered by the railways:

1. City or Urban Service. In this case, there are frequent stops, the distance between stops being nearly

1 km or less. Hence, high acceleration and retardation are essential to achieve moderately high schedule

speed between the stations.

2. Suburban Service. In this case, the distance between stops averages from 3 to 5 km over a

distance of 25 to 30 km from the city terminus. Here, also, high rates of acceleration and retardation are

necessary.

3. Main Line Service. It involves operation over long routes where stops are infrequent.

Here, operating speed is high and accelerating and braking periods are relatively unimportant.

On goods traffic side also, there are three types of services (i) main-line freight service (ii)

local or pick-up freight service and (iii) shunting service.

4.3.1 Train Movement

The movement of trains and their energy consumption can be conveniently

studied by means of speed/time and speed/distance curves. As their names indicate,

former gives speed of the train at various times after the start of the run and the later

gives speed at various distances from the starting point. Out of the two, speed/time

curve is more important because

1. its slope gives acceleration or retardation as the case may be.

2. area between it and the horizontal (i.e. time) axis represents the distance

travelled.

3. energy required for propulsion can be calculated if resistance to the motion of train is known.

4.3.2 Typical Speed/Time Curve

Figure: 4.1 speed / Time Curve

1. Constant Acceleration Period (0 to t1)

It is also called notching-up or starting period because during this period, starting resistance of the motors

is

gradually cut out so that the motor current (and hence, tractive effort) is maintained nearly constant

which produces constant acceleration alternatively called ‘rheostatic acceleration’ or ‘acceleration while

notching.

2. Acceleration on Speed Curve (t1 to t2)

This acceleration commences after the starting resistance has been all cut-out

at point t1 and full supply voltage has been applied to the motors. During this period,

the motor current and torque de-crease as train speed increases. Hence, acceleration

gradually decreases till torque developed by motors exactly balances that due to

resistance to the train motion. The shape of the portion AB of the speed/time curve

depends primarily on the torque/speed characteristics of the traction motors.

3. Free-running Period (t2 to t3)

The train continues to run at the speed reached at point t2. It is represented by

portion

BCand is a constant-speed period which occurs on level tracks.

4. Coasting (t3 to t4)

Power to the motors is cut off at point t3 so that the train runs under its

momentum, the speed gradually falling due to friction, windage etc. (portion CD).

During this period, retardation remains practically constant. Coasting is desirable

because it utilizes some of the kinetic energy of the train which would, otherwise, be

wasted during braking. Hence, it helps to reduce the energy consumption of the train.

5. Braking (t4 to t5)

At point t4, brakes are applied and the train is brought to rest at point t5. It may

be noted that coasting and braking are governed by train resistance and allowable

retardation respectively.

4.3.3 Speed/Time Curves for Different Service

The graph (a) is representative of city service where relative values of

acceleration and retardation are high in order to achieve moderately high average

speed between stops. Due to short distances between stops, there is no possibility of

free-running period though a short coasting period is included to save on energy

consumption.

Figure:4.2 Speed/Time Curves

In suburban services (b), again there is no free-running period but there is

comparatively longer coasting period because of longer distances between stops. In

this case also, relatively high values of acceleration and retardation are required in

order to make the service as attractive as possible. For main-line service (c), there are

long periods of free-running at high speeds. The accelerating and retardation periods

are relatively unimportant.

4.4 Quantities Involved in Traction Mechanics

Following principal quantities are involved in train movement

D = distance between stops

Me = effective mass of the train

We = effective weight of the train

βc = retardation during coasting

Va = average speed

t = total time for the run

t2 = time of free running = t − (t1 + t3) Ft = tractive effort

M=dead mass of the train

W = dead weight of the train

α = acceleration during starting period β = retardation during braking

Vm = maximum (or crest) speed.

t1 = time of acceleration t3 = time of braking

T = torque

4.4.2 Relationship between Principal Quantities in Trapezoidal Diagram

α = Vm /t1 or t1 = Vm /α

β = Vm /t3 or t3 = Vm /β

As we know, total distance D between the two stops is given by the area of trapezium

OABC.

D = area OABC

= area OAD + area ABED + area BCE

1 1= --- vm t1 + vm t2 + -- vm t3 2 2

1 1= --- vm t1 + vm [t- (t1+t3)] + -- vm t3 2 2

1= vm [ t1 /2+ t- (t1+t3)] + -- vm t3 2

1= vm [ t - -- (t1+t3)] 2

Vm 1 1= vm [ t - -- ( ---- + ----- ) ] 2 α β

Let,

D = Vm (t − KVm)

or KV 2 − Vmt + D = 0

Obviously, if Vm, Va and D are given, then value of K and hence of _ and _ can be

found.

4.4.3 Relationship between Principal Quantities in Quadrilateral Diagram

Let βc represent the retardation during coasting period. As before,

Figure: 4.3 Quadrilateral Diagram

4.4 Mechanism of Train Movement

The essentials of driving mechanism in an electric vehicle are illustrated. The armature of the driving

motor has a pinion which meshes with the gear wheel keyed to the axle of the driving wheel. In this way,

motor torque is transferred to the wheel through the gear.

Let,

T = torque exerted by the motor

F1= tractive effort at the pinion

Ft= tractive effort at the wheel

ϒ= gear ratio

Here,d1, d2= diameters of the pinion and gear wheel respectively

D = diameter of the driving wheel

ƞ = efficiency of power transmission from the motor to driving axle

Now,T = F1 × d1/2 or F1 = 2T/d1

Tractive effort transferred to the driving wheel is

For obtaining motion of the train without slipping, Ft ≤µa Wa where µa is the coefficient of

adhesion and Wa is the adhesive weight.

Control of D.C. Motors

The starting current of motor is limited to its normal rated current by starter during starting. At the instant

of switching on the motor, back e.m.f. Eb = 0

Supply voltage = V = IR + Voltage drop across the starting current of motor is

limited to its normal rated current by starter during starting. At the instant of switching

on the motor, back e.m.f. Eb = 0

Supply voltage = V = IR + Voltage drop across Rs

At any other instant during starting

V = IR + Voltage across Rs + Eb

At the end of accelerating period, when total Rs is cut-off

V = Eb + IR

Figure: 4.4 Voltage Curve

If T is the time in sec.

For starting and neglecting IR drop, total energy supplied = V.I.T. watt-sec

Energy wasted in Rs = Area of triangle ABC × I = ½. T.V.I. watt -sec. = ½ VIT watt - sec.

But total energy supplied = V.I.T watt - sec.

Half the energy is wasted in starting

ƞstarting = 50%

Series - Parallel Starting

With a 2 motor equipment ½ the normal voltage will be applied to each motor at starting as (a) (Series

connection) and they will run upto approximate ½ speed, at which instant they are switched on to parallel

and full voltage is applied to each motor. Rs is gradually cutout, with motors in series connection and

then reinserted when the motors are connected in parallel, and again gradually cut-out.

Figure: 4.5 Series Connection

In traction work, 2 or more similar motors are employed. Consider 2 series motors started by series

parallel method, which results in saving of energy.

(a) Series operation. The 2 motors, are started in series with the help of Rs. The current during

starting is limited to normal rated current ‘I’ per motor. During series operation, current ‘I’ is

drawn from supply. At the instant of starting OA = AB = IR drop in each motor. OK = Supply

voltage ‘V’. The back e.m.f. of

(b) motors jointly develop along OM. At point. E, supply voltage V = Back e.m.f of 2 motors + IR

drops of 2 motor. Any point on the line BC represents the sum of Back e.m.f. of 2 motors + IR

drops of 2 motors + Voltage across resistance Rs of 2 motors OE = time taken for series running.

At pt ‘E’ at the end of series running period, each motor has developed a back e.m.f.

(b) Parallel operation. The motors are switched on in parallel at the instant ‘E’, with Rs reinserted.

Current drawn is 2I from supply. Back e.m.f. across each motor = EL. So the back e.m.f. now develops

along LG. At point ‘H’ when the motors are in full parallel,(Rs = 0 and both the motors are running at

rated speed)

Supply voltage= V = HF = HG + GF

= Normal Back e.m.f. of each motor + IR drop in each motor

Series Parallel Control by Shunt Transition Method

The various stages involved in this method of series – parallel control are shown in Figure. In steps 1, 2,

3, 4 the motors are in series and are accelerated by cutting out the Rs in steps. In step 4, motors are in full

series. During transition from series to parallel, Rs is reinserted in circuit- step 5. One of the motors is

bypassed -step 6 and disconnected from main circuit - step 7. It is then connected in parallel with other

motor -step 8, giving 1st parallel position. Rs is again cut-out in steps completely and the motors are

placed in full parallel.

Figure: 4.6 Series Parallel Connection

The main difficulty with series parallel control is to obtain a satisfactory

method of transition from series to parallel without interrupting the torque or allowing

any heavy rushes of current.

In shunt transition method, one motor is short circuited and the total torque is

reduced by about 50% during transition period, causing a noticeable jerk in the motion

of vehicle. The Bridge transition is more complicated, but the resistances which are

connected in parallel with or ‘bridged’ across the motors are of such a value that

current through the motors is not altered in magnitude and the total torque is therefore

held constant and hence it is normally used for railways. So in this method it is seen

that, both motors remain in circuit through-out the transition. Thus the jerks will not

be experienced if this method is employed.

Series Parallel Control by Bridge Transition

(a) At starting, motors are in series with Rs i.e. link P in position = AA’

(b) Motors in full series with link P in position = BB’ (No Rs in the circuit)

The motor and Rs are connected in the form of Wheatstone Bridge. Initially motors are in series with full

Rs as shown in Fig. 43.32 (a). A and A’ are moved in direction of arrow heads. In position BB’ motors

are in full series, with no Rs present in the circuit.

Figure: 4.7 Series Parallel Control by Bridge Transition

In transition step the Rs is reinserted. In Ist parallel step, link P is removed and motors are connected in

parallel with full Rs. Advantage of this method is that the normal acceleration torque is available from

both the motors, through - out starting period. Therefore acceleration is smoother, without any jerks,

which is very much desirable for traction motors.

4.5 Braking in Traction

Both electrical and mechanical braking is used. Mechanical braking provides holding torque. Electric

Braking reduces wear on mechanical brakes, provides higher retardation, thus bringing a vehicle quickly

to rest. Different types of electrical braking used in traction are discussed.

4.5.1 Rheostatic Braking

(a) Equalizer Connection (b) Cross Connection

(a) Equalizer Connection

For traction work, where 2 or more motors are employed, these are connected in parallel for braking,

because series connection would produce too high voltage. K.E. of the vehicle is utilized in driving the

machines as generators, which is dissipated in braking resistance in the form of heat. To ensure that the 2

machines share the

load equally, an equalizer connection is used. If it is not used, the machine whose acceleration builts-up

first would send a current through the 2nd machine in opposite direction, causing it to excite with reverse

voltage. So that the 2 machines would be short circuited on themselves. The current would be

dangerously high. Equalizer prevents such conditions. Hence Equalizer connection is important during

braking in traction.

Figure: 2.8 Equalizer Connection

(b) Cross Connection

In cross connection the field of machine 2 is connected in series with armature of machine 1 and the field

of machine 1 is connected in series with armature of machine 2. Suppose the voltage of machine 1 is

greater than that of 2. So it will send greater current through field of machine 2, causing it to excite to

higher voltage. At the same time machine 1 excitation is low, because of lower voltage of machine 2.

Hence machine 2 will produce more voltage and machine 1 voltage will be reduced. Thus automatic

compensation is provided and the 2 machines operate satisfactorily.

Because of cross - connection during braking of traction motors, current in any of the motor will not go

to a very high value.

4.5.2 Regenerative Braking with D.C. Motors

In order to achieve the regenerative braking, it is essential that (i) the voltage generated by the machine

should exceed the supply voltage and (ii) the voltage should be kept at this value, irrespective of machine

speed. The following figure shows the 4 series motors connected in parallel during normal running i.e.

motoring.

One method of connection during regenerative braking is to arrange the machines as shunt machines,

with series fields of 3 machines connected across the supply in series with suitable resistance. One of the

field winding is still kept in series across the 4 parallel armatures. The machine acts as a compound

generator. (with slight differential compounding) Such an arrangement is quiet stable; any change in line

voltage produces a change in excitation which produces corresponding change in e.m.f. of motors, so that

inherent compensation is provided e.g. let the line voltage tends to increase beyond the e.m.f. of

generators. The increased voltage across the shunt circuit

increases the excitation thereby increasing the generated voltage. Vice versa is also true. The

arrangement is therefore self compensating.

Figure:4.9 Regenerative Braking with D.C. Motors

D.C. series motor can’t be used for regenerative braking without modification for obvious reasons.

During regeneration current through armature reverses; and excitation has to be maintained. Hence field

connection must be reversed.

UNIT- 5

ELECTRIC TRACTION-IIINTRODUCTION

The movement of trains and their energy consumption can be most conveniently studied by means of the speed–distance and the speed–time curves. The motion of any vehicle may be at constant speed or it may consist of periodic acceleration and retardation. The speed–time curves have significant importance in traction. If the frictional resistance to the motion is known value, the energy required for motion of the vehicle can be determined from it. Moreover, this curve gives the speed at various time instants after the start of run directly.

TYPES OF SERVICES

There are mainly three types of passenger services, by which the type of traction system has

to be selected, namely:

1. Main line service.2. Urban or city service.3. Suburban service.

Main line services

In the main line service, the distance between two stops is usually more than 10 km. High balancing speeds should be required. Acceleration and retardation are not so important.

Urban service

In the urban service, the distance between two stops is very less and it is less than 1 km. It requires high average speed for frequent starting and stopping.

Suburban service

In the suburban service, the distance between two stations is between 1 and 8 km. This service requires rapid acceleration and retardation as frequent starting and stopping is required.

SPEED–TIME AND SPEED–DISTANCE CURVES FOR DIFFERENT SERVICES

The curve that shows the instantaneous speed of train in kmph along the ordinate and time in seconds along the abscissa is known as ‘speed–time’ curve.

The curve that shows the distance between two stations in km along the ordinate and time in seconds along the abscissa is known as ‘speed–distance’ curve.

The area under the speed–time curve gives the distance travelled during, given time internal and slope at any point on the curve toward abscissa gives the acceleration and retardation at the instance, out of the two speed–time curve is more important.

Speed–time curve for main line service

Typical speed–time curve of a train running on main line service is shown in Fig.10.1. It mainly consists of the following time periods:

1. Constant accelerating period.2. Acceleration on speed curve.3. Free-running period.4. Coasting period.5. Braking period.

Fig. 10.1 Speed–time curve for mainline service

Constant acceleration

During this period, the traction motor accelerate from rest. The curve ‘OA’ represents the constant accelerating period. During the instant 0 to T1, the current is maintained approximately constant and the voltage across the motor is gradually increased by cutting out the starting resistance slowly moving from one notch to the other. Thus, current taken by the motor and the tractive efforts are practically constant and therefore acceleration remains constant during this period. Hence, this period is also called as notch up accelerating period or rehostatic accelerating period. Typical value of acceleration lies between 0.5 and 1 kmph. Acceleration is denoted with the symbol ‘α’.

Acceleration on speed-curve

During the running period from T1 to T2, the voltage across the motor remains constant and the current starts decreasing, this is because cut out at the instant ‘T1’.

According to the characteristics of motor, its speed increases with the decrease in the current and finally the current taken by the motor remains constant. But, at the same time, even though train accelerates, the acceleration decreases with the increase in speed. Finally, the acceleration reaches to zero for certain speed, at

which the tractive effort excreted by the motor is exactly equals to the train resistance. This is also known as decreasing accelerating period. This period is shown by the curve ‘AB’.

Free-running or constant-speed period

The train runs freely during the period T2 to T3 at the speed attained by the train at the instant ‘T2’. During this speed, the motor draws constant power from the supply lines. This period is shown by the curve BC.

Coasting period

This period is from T3 to T4, i.e., from C to D. At the instant ‘T3’ power supply to the traction, the motor will be cut off and the speed falls on account of friction, windage resistance, etc. During this period, the train runs due to the momentum attained at that particular instant. The rate of the decrease of the speed during coasting period is known as coasting retardation. Usually, it is denoted with the symbol ‘βc’.

Braking period

Braking period is from T4 to T5, i.e., from D to E. At the end of the coasting period, i.e., at ‘T4’ brakes are applied to bring the train to rest. During this period, the speed of the train decreases rapidly and finally reduces to zero.

In main line service, the free-running period will be more, the starting and braking periods are very negligible, since the distance between the stops for the main line service is more than 10 km.

Speed–time curve for suburban service

In suburban service, the distance between two adjacent stops for electric train is lying between 1 and 8 km. In this service, the distance between stops is more than the urban service and smaller than the main line service. The typical speed–time curve for suburban service is shown in Fig. 10.2.

Fig. 10.2 Typical speed–time curve for suburban service

The speed–time curve for urban service consists of three distinct periods. They are:

1. Acceleration.2. Coasting.3. Retardation.

For this service, there is no free-running period. The coasting period is comparatively longer since the distance between two stops is more. Braking or retardation period is comparatively small. It requires relatively high values of acceleration and retardation. Typical acceleration and retardation values are lying between 1.5 and 4 kmphp and 3 and 4 kmphp, respectively.

Speed–time curve for urban or city service

The speed–time curve urban or city service is almost similar to suburban service and is shown in Fig. 10.3.

Fig. 10.3 Typical speed–time curve for urban service

In this service also, there is no free-running period. The distance between two stop is less about 1 km. Hence, relatively short coasting and longer braking period is required. The relative values of acceleration and retardation are high to achieve moderately high average between the stops. Here, the small coasting period is included to save the energy consumption. The acceleration for the urban service lies between 1.6 and 4 kmphp. The coasting retardation is about 0.15 kmphp and the braking retardation is lying between 3 and 5 kmphp. Some typical values of various services are shown in Table. 10.1.

Table 10.1 Types of services

SOME DEFINITIONS

Crest speed

The maximum speed attained by the train during run is known as crest speed. It is denoted with ‘Vm’.

Average speed

It is the mean of the speeds attained by the train from start to stop, i.e., it is defined as the ratio of the distance covered by the train between two stops to the total time of rum. It is denoted with ‘Va’.

where Va is the average speed of train in kmph, D is the distance between stops in km, and T is the actual time of run in hours.

Schedule speed

The ratio of the distance covered between two stops to the total time of the run including the time for stop is known as schedule speed. It is denoted with the symbol ‘Vs’.

where Ts is the schedule time in hours.

Schedule time

It is defined as the sum of time required for actual run and the time required for stop.

, Ts = Trun + Tstop.

FACTORS AFFECTING THE SCHEDULE SPEED OF A TRAIN

The factors that affect the schedule speed of a train are:

1. Crest speed.2. The duration of stops.3. The distance between the stops.4. Acceleration.5. Braking retardation.

Crest speed

It is the maximum speed of train, which affects the schedule speed as for fixed acceleration, retardation, and constant distance between the stops. If the crest speed increases, the actual running time of train decreases. For the low crest speed of train it running so, the high crest speed of train will increases its schedule speed.

Duration of stops

If the duration of stops is more, then the running time of train will be less; so that, this leads to the low schedule speed.

Thus, for high schedule speed, its duration of stops must be low.

Distance between the stops

If the distance between the stops is more, then the running time of the train is less; hence, the schedule speed of train will be more.

Acceleration

If the acceleration of train increases, then the running time of the train decreases provided the distance between stops and crest speed is maintained as constant.Thus, the increase in acceleration will increase the schedule speed.

Breaking retardation

High breaking retardation leads to the reduction of running time of train. These will cause high schedule speed provided the distance between the stops is small.

SIMPLIFIED TRAPEZOIDAL AND QUADRILATERAL SPEED TIME CURVES

Simplified speed–time curves gives the relationship between acceleration, retardation average speed, and the distance between the stop, which are needed to estimate the performance of a service at different schedule speeds. So that, the actual speed–time curves for the main line, urban, and suburban services are approximated to some from of the simplified curves. These curves may be of either trapezoidal or quadrilateral shape.

Analysis of trapezoidal speed–time curve

Trapezoidal speed–time curve can be approximated from the actual speed–time curves of different services by assuming that:

o The acceleration and retardation periods of the simplified curve is kept same as to that of the actual curve.

o The running and coasting periods of the actual speed–time curve are replaced by the constant periods.

This known as trapezoidal approximation, a simplified trapezoidal speed–time curve is shown in fig,

Fig. Trapezoidal speed–time curve

Calculations from the trapezoidal speed–time curve

Let D be the distance between the stops in km, T be the actual running time of train in second, α be the acceleration in km/h/sec, β be the retardation in km/h/sec, Vm be the maximum or the crest speed of train in km/h, and Va be the average speed of train in km/h. From the Fig. 10.4:

Area under the trapezoidal speed–time curve gives the total distance between the two stops (D).

∴ The distance between the stops (D) = area under triangle OAE + area of rectangle ABDE + area of triangle DBC

= The distance travelled during acceleration + distance travelled during free- running period + distance travelled during retardation.

Now:

The distance travelled during acceleration = average speed during accelerating period × time for acceleration

The distance travelled during free-running period = average speed × time of free running

The distance travelled during retardation period = average speed × time for retardation

The distance between the two stops is:

Solving quadratic Equation (10.5), we get:

By considering positive sign, we will get high values of crest speed, which is practically not possible, so negative sign should be considered:

Analysis of quadrilateral speed–time curve

Quadrilateral speed–time curve for urban and suburban services for which the distance between two stops is less. The assumption for simplified quadrilateral speed–time curve is the initial acceleration and coasting retardation periods are extended, and there is no free-running period. Simplified quadrilateral speed–time curve is shown in Fig. 10.5.

Fig. Quadrilateral speed–time curve

Let V1 be the speed at the end of accelerating period in km/h, V2 be the speed at the end of coasting retardation period in km/h, and βc be the coasting retardation in km/h/sec.

Time for acceleration,

Time for coasting period,

Time period for braking retardation period,

Total distance travelled during the running period D:

= the area of triangle PQU + the area of rectangle UQRS + the area of triangle TRS.

= the distance travelled during acceleration + the distance travelled during coastingretardation + the distance travelled during breaking retardation.

But, the distance travelled during acceleration = average speed × time for acceleration

The distance travelled during coasting retardation =

The distance travelled during breaking retardation = average speed × time for breaking retardation

∴ Total distance travelled:

Example 10.1: The distance between two stops is 1.2 km. A schedule speed of 40 kmph is required to cover that distance. The stop is of 18-s duration. The values of the acceleration and retardation are 2 kmphp and 3 kmphp, respectively. Then, determine the maximum speed over the run. Assume a simplified trapezoidal speed–time curve.

Solution:

Acceleration α = 2.0 kmphp.

Retardation β = 3 kmphp.

Schedule speed Vs = 40 kmph.

Distance of run, D = 1.2 km.

Actual run time, T = Ts – stop duration

= 108 – 18

= 90 s.

where

Example 10.2: The speed–time curve of train carries of the following parameters:

1. Free running for 12 min.2. Uniform acceleration of 6.5 kmphp for 20 s.3. Uniform deceleration of 6.5 kmphp to stop the train.4. A stop of 7 min.

Then, determine the distance between two stations, the average, and the schedule speeds.

Solution:

Acceleration (α) = 6.5 kmphps.

Acceleration period t1 = 20 s.

Maximum speed Vm = αt1

= 6.5 × 20 = 130 kmph.

Free-running time (t2) = 12 × 60

= 720 s.

The distance travelled during the acceleration period:

The distance travelled during the free-running period:

The distance travelled during the braking period

The distance between the two stations:

D = D1 + D2 + D3

= 0.36 + 26 + 0.362

= 26.724 km.

Example 10.3: An electric train is to have the acceleration and braking retardation of 0.6 km/hr/sec and 3 km/hr/sec, respectively. If the ratio of the maximum speed to the average speed is 1.3 and time for stop is 25 s. Then determine the schedule speed for a run of 1.6 km. Assume the simplified trapezoidal speed–time curve.

Solution:

Acceleration α = 0.6 km/hr/s.

Retardation β = 3 km/hr/s.

Distance of run D = 1.6 km.

Let the cultural time of run be ‘T ’ s.

Example 10.4: The distance between two stops is 5 km. A train has schedule speed of 50 kmph. The train accelerates at 2.5 kmphps and retards 3.5 kmphps and the duration of stop is 55 s. Determine the crest speed over the run assuming trapezoidal speed–time curve.

Solution:


Retardation (β) = 3.5 kmphps.

By using the equation:

Example 10.5: A train is required to run between two stations 1.5 km apart at an average speed of 42 kmph. The run is to be made to a simplified quadrilateral speed–time curve. If the maximum speed is limited to 65 kmph, the acceleration to

, kmphps, and the casting and braking retardation to 0.15 kmphs and 3 kmphs, respectively. Determine the duration of acceleration, costing, and braking periods.

Solution:

Distance between two stations D = 1.5 km.

Average speed Va = 42 kmph.

Maximum speed Vm = 65 kmph.


Coasting retardation βc = 0.15 kmphps.

Barking retardation β = 3 kmphps.

Before applying brakes; let the speed be V2.

The actual time of run, T = t1 + t2 + t3

Example 10.6: A train has schedule speed of 32 kmph over a level track distance between two stations being 2 km. The duration of stop is 25 s. Assuming the braking retardation of 3.2 kmphps and the maximum speed is 20% grater than the average speed. Determine the acceleration required to run the service.

Solution:


Distance D = 2 km.

Duration of stop = 25 s.

Braking retardation = 3.2 kmphps.

Example 10.7: A suburban electric train has a maximum speed of 75 kmph. The schedule speed including a station stop of 25 s is 48 kmph. If the acceleration is 2 kmphps, the average distance between two stops is 4 km. Determine the value of retardation.

Solution:


The distance of run (D) = 4 km.

Schedule speed (Vs) = 48 kmph.

Acceleration (α) = 2 kmphps.

The duration of stop = 25 s.

Example 10.8: An electric train is accelerated at 2 kmphps and is braked at 3 kmphps. The train has an average speed of 50 kmph on a level track of 2,000 min between the two stations. Determine the following:

1. Actual time of run.2. Maximum speed.3. The distance travelled before applying brakes4. Schedule speed.

Assume time for stop as 12 s. And, run according to trapezoidal.

Solution:


Retardation (β) = 3 kmphps.

Average speed (Va) = 50 kmph.

Distance D = 2,000 min = 2 km.


The distance travelled before applying brakes

D1 + D2 = D - D3

= 2 – 0.17 = 1.83 km.

Example 10.9: An electric train has an average speed of 40 kmph on a level track between stops 1,500 m apart. It is accelerated at 2 kmphps and is braked at 3 kmphps. Draw the speed–time curve for the run.

Solution:


The distance of run (D) = 1,500 m = 1.5 km.


Retroaction (β) = 3 kmphps.

Using the equation (Fig. P.10.1):

where

Fig. P.10.1

Example 10.10: An electric train has quadrilateral speed–time curve as follows:

1. Uniform acceleration from rest at 1.5 kmphps for 25 s.2. Coasting for 45 s.3. The duration of braking 20 s.

If the train is moving a uniform up gradient of 1.5%, the reactive resistance is 45 N/ton, the rotational inertia effect is 10% of dead weight, the duration of stop is 15 s, and the overall efficiency of transmission gear and motor is 80%. Find schedule speed.

Solution:

Time for acceleration t1 = 25 s.

Time for coasting t2 = 45 s.

Time for braking t3 = 20 s.


Maximum speed Vm = α t1

= 1.5 × 25 = 37.5 kmph.

According to the equation:

TRACTIVE EEFFORT (FT)

It is the effective force acting on the wheel of locomotive, necessary to propel the train is known as ‘tractive effort’. It is denoted with the symbol Ft. The tractive effort is a vector quantity always acting tangential to the wheel of a locomotive. It is measured in newton.

The net effective force or the total tractive effort (Ft) on the wheel of a locomotive or a train to run on the track is equals to the sum of tractive effort:

1. Required for linear and angular acceleration (Fa).2. To overcome the effect of gravity (Fg).3. To overcome the frictional resistance to the motion of the train (Fr).

Mechanics of train movement

The essential driving mechanism of an electric locomotive is shown in Fig. 10.6. The electric locomotive consists of pinion and gear wheel meshed with the traction motor and the wheel of the locomotive. Here, the gear wheel transfers the tractive effort at the edge of the pinion to the driving wheel.

Fig. Driving mechanism of electric locomotives

Let T is the torque exerted by the motor in N-m, Fp is tractive effort at the edge of the pinion in Newton,Ft is the tractive effort at the wheel, D is the diameter of

the driving wheel, d1 and d2 are the diameter of pinion and gear wheel, respectively, and η is the efficiency of the power transmission for the motor to the driving axle.

The tractive effort at the edge of the pinion transferred to the wheel of locomotive is:

10.7.2 Tractive effort required for propulsion of train

From Equation (10.8), the tractive effort required for train propulsion is:

Ft = Fa + Fg + Fr,

where Fa is the force required for linear and angular acceleration, Fg is the force required to overcome the gravity, and Fr is the force required to overcome the resistance to the motion.

Force required for linear and angular acceleration (Fa)

According to the fundamental law of acceleration, the force required to accelerate the motion of the body is given by:

Force = Mass × acceleration

F = ma.

Let the weight of train be ‘W ’ tons being accelerated at ‘α’ kmphps:

Equation (10.12) holds good only if the accelerating body has no rotating parts. Owing to the fact that the train has rotating parts such as motor armature, wheels, axels, and gear system. The weight of the body being accelerated including the rotating parts is known as effective weight or accelerating weight. It is denoted with ‘We’. The accelerating weight ‘(We)’ is much higher (about 8–15%) than the dead weight (W) of the train. Hence, these parts need to be given angular acceleration at the same time as the whole train is accelerated in linear direction.

∴ The tractive effort required-for linear and angular acceleration is:

Tractive effort required to overcome the train resistance (Fr)

When the train is running at uniform speed on a level track, it has to overcome the opposing force due to the surface friction, i.e., the friction at various parts of the rolling stock, the fraction at the track, and also due to the wind resistance. The magnitude of the frictional resistance depends upon the shape, size, and condition of the track and the velocity of the train, etc.

Let ‘r’ is the specific train resistance in N/ton of the dead weight and ‘W’ is the dead weight in ton.

Tractive effort required to overcome the effect of gravity (Fg)

When the train is moving on up gradient as shown in Fig. 10.7, the gravity component of the dead weight opposes the motion of the train in upward direction. In order to prevent this opposition, the tractive effort should be acting in upward direction.

∴ The tractive effort required to overcome the effect of gravity:

Now, from the Fig. 10.7:

Fig. 10.7 Train moving on up gradient

From Equations (10.15) and (10.16):

+ve sign for the train is moving on up gradient.

–ve sign for the train is moving on down gradient.

This is due to when the train is moving on up a gradient, the tractive effort showing Equation (10.17)will be required to oppose the force due to gravitational force, but while going down the gradient, the same force will be added to the total tractive effort.

∴ The total tractive effort required for the propulsion of train Ft = Fa + Fr ± Fg:

Power output from the driving axle

Let Ft is the tractive effort in N and ν is the speed of train in kmph.

∴ The power output (P) = rate of work done

If ‘ν’ is in m/s, then P = Ft × ν W.

If ‘η’ is the efficiency of the gear transmission, then the power output of motors,

:

SPECIFIC ENERGY CONSUMPTION

The energy input to the motors is called the energy consumption. This is the energy consumed by various parts of the train for its propulsion. The energy drawn from the distribution system should be equals to the energy consumed by the various parts of the train and the quantity of the energy required for lighting, heating, control, and braking. This quantity of energy consumed by the various parts of train per ton per kilometer is known as specific energy consumption. It is expressed in watt hours per ton per km.

Determination of specific energy output from simplified speed–time curve

Energy output is the energy required for the propulsion of a train or vehicle is mainly for accelerating the rest to velocity ‘Vm’, which is the energy required to overcome the gradient and track resistance to motion.

Energy required for accelerating the train from rest to its crest speed ‘Vm'

Energy required for overcoming the gradient and tracking resistance to motion

Energy required for overcoming the gradient and tracking resistance:

where Ft′ is the tractive effort required to overcome the gradient and track resistance, W is the dead weight of train, r is the track resistance, and G is the percentage gradient.

Factors affecting the specific energy consumption

Factors that affect the specific energy consumption are

given as follows.

Distance between stations

From equation specific energy consumption is inversely proportional to the distance between stations. Greater the distance between stops is, the lesser will be the specific energy consumption. The typical values of the specific energy consumption is less for the main line service of 20–30 W-hr/ton-km and high for the urban and suburban services of 50–60 W-hr/ton-km.

Acceleration and retardation

For a given schedule speed, the specific energy consumption will accordingly be less for more acceleration and retardation.

Maximum speed

For a given distance between the stops, the specific energy consumption increases with the increase in the speed of train.

Gradient and train resistance

From the specific energy consumption, it is clear that both gradient and train resistance are proportional to the specific energy consumption. Normally, the coefficient of adhesion will be affected by the running of train, parentage gradient,

condition of track, etc. for the wet and greasy track conditions. The value of the coefficient of adhesion is much higher compared to dry and sandy conditions.

IMPORTANT DEFINITIONS

1 Dead weight

It is the total weight of train to be propelled by the locomotive. It is denoted by ‘W’.

2 Accelerating weight

It is the effective weight of train that has angular acceleration due to the rotational inertia including the dead weight of the train. It is denoted by ‘We’.

This effective train is also known as accelerating weight. The effective weight of the train will be more than the dead weight. Normally, it is taken as 5–10% of more than the dead weight.

3 Adhesive weight

The total weight to be carried out on the drive in wheels of a locomotive is known as adhesive weight.

4 Coefficient of adhesion

It is defined as the ratio of the tractive effort required to propel the wheel of a locomotive to its adhesive weight.

Ft ∝ W= μW,

where Ft is the tractive effort and W is the adhesive weight.

Example 10.11: A 250-ton motor coach having four motors each developing 6,000 N-m torque during acceleration, starts from rest. If the gradient is 40 in 1,000, gear ration is 4, gear transmission efficiency is 87%, wheel radius is 40 cm, train resistance is 50 N/ton, the addition of rotational inertia is 12%. Calculate the time taken to attain a speed of 50 kmph. If the line voltage is 3,000-V DC and the efficiency of motors is 85%. Find the current during notching period.

Solution:

The weight of train W = 250 ton.

Gear ratio r = 4.

Wheel diameter D = 2 × 40 = 80 cm.

Or, D = 0.8 m.

Train resistance r = 50 N/ton.

Rotational inertia = 12%.

Accelerating weight of the train We = 1.10 × 250 = 275 ton.

Total torque developed T = 4 × 6,000 = 24,000 Nm.

But,

Ft = 277.8 We α + 98.1 WG + Wr

208,800 = 277.8 × 275 α + 98.1 × 250 × 4 + 250 × 50

∴ α = 1.285 kmphps.

The time taken for the train to attain the speed of 50 kmph:

Power output from the driving axles:

Example 10.12: An electric train of weight 250 ton has eight motors geared to driving wheels, each is 85 cm diameter. The tractive resistance is of 50/ton. The effect of rotational inertia is 8% of the train weight, the gear ratio is 4–1, and the gearing efficiency is 85% determine. The torque developed by each motor to accelerate the train to a speed of 50 kmph in 30 s up a gradient of 1 in 200.

Solution:

The weight of train W = 250 ton.

The diameter of driving wheel D =0.85 m.

Tractive resistance, r = 50N/ton.

Gear ratio r = 4.

Gearing efficiency η = 0.85.

Accelerating weight of the train:

We = 1.10 × W

= 1.10 × 250 =275 ton.


Tractive effort Ft = 277.8 We α + 98.1 WG + Wr

= 126,815.7+12,262.5+12,500

= 151,578.2 N.

Example 10.13: A tram car is equipped with two motors that are operating in parallel, the resistance in parallel. The resistance of each motor is 0.5 Ω. Calculate the current drawn from the supply mains at 450 V when the car is running at a steady-state speed of 45 kmph and each motor is developing a tractive effort of 1,600 N. The friction, windage, and other losses may be assumed as 3,000 W per motor.

Solution:

The resistance of each motor = 0.5 Ω.

Voltage across each motor V = 450 V.

Tractive effort Ft = 1,600 N.


Losses per motor = 3,000 W.

Copper losses = I2Rm = I2× 0.5

Motor input = motor output + constant loss + copper losses 450

× I = 20,000 + 3,000 + 0.5I2

0.5 I2– 450I + 23,000 = 0.

After solving, we get I = 54.39 A.

Total current drawn from supply mains = 2 × 54.39

= 108.78 A.

Example 10.14: A locomotive exerts a tractive effort of 35,000 N in halting a train at 50 kmph on the level track. If the motor is to haul the same train on a gradient of 1 in 50 and the tractive effort required is 55,000 N, determine the power delivered by the locomotive if it is driven by (i) DC series motors and (ii) induction motors.

Solution:

Tractive effort Ft= 35,000 N.


The power delivered by the locomotive on up gradient track with the DC series motors:

Since the power output the power delivered by the locomotive on up gradient with the induction motors is:

Example 10.15: A train weighting 450 ton has speed reduced by the regenerative braking from 50 to 30 kmph over a distance of 2 km on down gradient of 1.5%.Calculate the electrical energy and the overage power returned to the line tractive resistance is 50 N/ton. And, allow the rotational inertia of 10% and the efficiency conversion 80%.

Solution:

The accelerating weight of the train We = 1.1 W

= 1.1 × 450 = 495 ton.

The distance travelled D = 2 km.

Gradient G = 1.5%

Track resistance r = 50 N/ton.

Efficiency η = 0.8.

The energy available due to the reduction in the speed is:

The tractive effort required while going down the gradient:

Ft = Wr – 98.1 WG

= 450 × 50 – 98.1 × 450 × 1.5

= -43,717.5 N.

The energy available while moving down the gradient a distance of 2 km is:

The total energy available = 8.49 + 24.2875

= 32.7775 kW-hr.

Example 10.16: A train weighing 450 ton is going down a gradient of 20 in 1,000, it is desired to maintain train speed at 50 kmph by regenerative braking. Calculate the power fed into the line and allow rotational inertia of 12% and the efficiency of conversion is 80%. Traction resistance is 50 N/ton.

Solution:

The dead weight of train W = 450 ton.

The maximum speed Vm = 50 kmph.

Tractive resistance r = 50 N/ton.


The efficiency of conversion = 0.8




The efficiency of conversion = 0.8


= Wr – 98.1 WG

= 450 × 50 – 98.1 × 450 × 2

= –65,790 N.

The power fed into the line = power available × efficiency of conversion

= 913.75 × 0.8

= 731 kW.

Example 10.17: The speed–time curve of an electric train on a uniform raising gradient of 10 in 1,000 comprise of:

1. Uniform acceleration from rest at 2.2 kmphps for 30 s.2. Wasting with power off for 30 s.3. Braking at 3.2 kmphps to standstill the weight of the train is 200 ton. The tractive

resistance of level track being 4 kg/ton and the allowance for rotary inertia 10%. Calculate the maximum power developed by traction motors and the total distance travelled by the train. Assume the transmission efficiency as 85%.

Solution:


Braking (β) = 3.2 kmph.

The dead weight of train W = 200 ton.

Track resistance r = 4 kg/ton = 4 × 9.81 = 39.24 N/ton.

Maximum velocity Vm = αt1 = 2.2 × 30 = 66 kmph.

Tractive effort required:

Ft = 277.8 Weα + 98.1 WG + Wr

= 277.8 × 8 × 1.1 × 200 × 2.2 + 98.1 × 200 × 1 + 200 × 39.24

= 161,923.2 N.

The maximum power developed by the traction motor = = 3492.46kW.

Let, the coasting retardation be βc:

Ft = 277.8 We(-βc) + 98.1 WG + Wr

0 = -277.8 × (1.1 × 200) × βc + 98.1 × 200 × 1 + 200 × 39.24

βc = 0.449 kmphps

V2 = Vm – βcV2

= 66 – 0.449 × 65

= 36.815 kmph.

The total distance travelled by the train:

Example 10.18: A 2,300-ton train proceeds down a gradient of 1 in 100 for 5 min, during which period, its speed gets reduced from 40 to 20 kmph by the application of the regenerative braking. Find the energy returned to the lines if the tractive resistance is 5 kg/ton, the rotational inertia 10%, and the overall efficiency of the motors during regeneration is 80%.

Solution:

The dead weight of the train W = 2,300 ton.

The accelerating weight of the train We = 1.1 × 2,300 s

= 2,530 ton.

Tractive resistance r = 5×9.81= 49.05 N/ton.

Regenerative period t = 5 × 60

= 300 s.

Overall efficiency η = 0.8.

The energy available due to the reduction in speed:

= 0.01072 We

= 0.01072 × 2,530 × (402-202)

= 32,545.92

= 32.54 kW-hr.


= Wr – 98.1 WG

= 2,300 × 49.05–98.1 × 2,300 × 1

= –112,815.

The distance moved during regeneration:

The energy available on the account of moving down the gradient over a distance of 2,500 m:

The total energy available = 32.54 + 78.34

= 88.707 kW-hr.

The energy returned to the line = 0.8×11.08

= 88.707 kW-hr.

Example 10.19: An electric train has an average speed of 50 kmph on a level track betweenstops 1,500 m a part. It is accelerated at 2 kmphs and is braked at 3 kmphs. Estimate the energy consumption at the axle of the train per ton-km. Take the reactive resistance constant at 50 N/ton and allow 10% for rotational inertia.

Solution:

Acceleration (α) = 2 kmphs.

Retardation (β) = 3 kmphs.

The distance of run (D) = 1.5 km.


Using the equation:

The distance travelled during braking:

The energy consumption at the axle of the train per ton-km:

Example 10.20: An electric train has quadrilateral speed–time curve as follows.

1. The uniform acceleration for rest at 2.2 kmphs for 30 s.2. Coasting for 45 s.3. The braking period of 20 s.

The train is moving in a uniform up gradient of 1%, the tractive resistance is 50 N/ton, the rotational inertia effect 10% of the dead weight the duration of the station stop 20 s and overall efficiency of transmission gear and motor as 80%. Determine the value of is schedule speed and specific energy consumption of run.

Solution:

Time of acceleration t1 = 30 s.

Time of coasting t2 = 45 s.

Time of braking t3 = 20 s.


Maximum speed Vm = α t1 = 2.2 × 30 = 66 kmph.

Gradient G = 1%.

Let the coasting retardation be βc:

Ft = 277.8 We(–βc) + 98.1 WG + Wr.

0 = 277.8 × 1.1 W βc + 98.1 × W × 1 + 50 W

=–305.58 W βc + 98.1 W + 50 W.

βc = 0.4846 kmphps.

V2 = Vm – βct2

= 66 – 0.4846 × 45

= 44.193 kmph.

When power is on, the distance travelled is:

D1 = distance travelled during acceleration period

The specific energy output:

Example 10.21: A train weighing 200-ton accelerates uniformly from rest to a speed of 40 kmph up a gradient of 1 in 100, the time taken being 30 s. The power is then cut off and train coasts down a uniform gradient of 1 in 1,000 for period of 40 s. When brakes are applied for period of 20 s so as to bring the train uniformly to rest on this gradient determine:

1. The maximum power output from the driving axles.2. The energy taken from the conductor rails in kW-hr assuming an efficiency of 70%

assume tractive resistance to be 45 N/ton at all speeds and allow 10% for rotational inertia.

Solution:


Accelerating period t1 = 30 s.

Tractive effort required:

Ft = 27.88 We α + 98.1 WG + Wr

= 277.8 × 220 × 1.33 + 98.1 × 200 × 1 + 200 × 45

= 109,904.28 N.

1. The maximum power output from driving axle:

Total energy required for the run:

2.

Example 10.22: Calculate the energy consumption if a maximum speed of 12 m/sec and for a given run of 1,500 m, an acceleration of 0.36 m/s2 desired. The tractive resistance during acceleration is 0.052 N/kg and during the coasting is 6.12

N/1,000 kg. Allow a 10% of rotational inertia, the efficiency of the equipment during the acceleration period is 60%. Assume quadrilateral speed–time curve.

Solution:

Accelerating weight of the train We = 1.1 W.

Maximum speed Vm = 12 m/s.

The distance of run D = 1,500 m.

Acceleration α = 0.36 m/s2.

The tractive resistance during acceleration r = 0.52 N/kg. The

tractive effort required during acceleration Ft = Weα + Wr

= 1.1 W × 0.36 + W × 0.052

= 0.448 W N.

Example 10.23: A 100-ton weight train has a rotational inertia of 10%. This train has to be run between two stations that are 3 km a part and has an average speed of 50 km/hr. The acceleration and the retardation during braking are 2 kmphps and 3 kmphps, respectively. The percentage gradient between these two stations is 1% and the train is to move up the incline the track resistance is 50 N/ton, then determine:

1. Maximum power at the driving axle.2. Total energy consumption.3. Specific energy consumption.

The combined efficiency of the alembic train is 70%. Assume simplified trapezoidal speed–time curve.

Solution:

The dead weight of the train, W = 100 ton.

The accelerating weight of the train, We = 1.1 × W = 1.1 × 100 = 110 ton.

The distance of run (D) = 3 km.




Gradient (G) = 1%.


Using the equation, the maximum speed:

Tractive effort required during free running is :

Example 10.24: An electric train has quadrilateral speed–time curve as follows:

1. Uniform acceleration from rest 2 kmphps for 30 s.2. Coasting for 40 s.3. Braking period of 25 s.

The train is moving a uniform down gradient of 1% and the tractive resistance of 50 N/ton. The rotational resistance is 10% of the dead weight, the duration of the stop is 20 s and the overall efficiency of the transmission the gear and the motor as 80%. Calculate its schedule speed and specific energy consumption.

Solution:


Acceleration period (t1) = 30 s.

Gradient (G) = 1%.

The tractive of resistance (r) = 50 N/ton.


Overall efficiency (η) = 80%.

Maximum speed Vm = αt1

= 2 × 30 = 60 kmph.

Let the coasting retardation be βc:

Tractive effort:

Ft = 277.8 Wc (–βc) – 98.1 × WG + Wr

0 = –277.8 × 1.1 W βc – 98.1 × W × 1 + 50 W

βc = – 0.157 kmphps

V2 = Vm – βct2

= 60 – (–0.517 × 40)

= 66.28 kmph.


Example 10.25: The schedule speed of a electric train is 40 kmph. The distance between two stations is 3 km with each stop is of 30 s duration. Assuming the acceleration and the retardation to be 2 and 3 kmphps, respectively. The dead weight of the train is 20 ton. Assume the rotational inertia is 10% to the dead weight and the track resistance is 40 N/ton. Calculate:

1. The maximum speed.2. The maximum power output from driving axles.3. The specific energy consumption is watt-hours per ton-km. The overall efficiency is

80%, assume simplified speed–time curve.

Solution:


The distance between the two stations (D) = 3 km.




The dead weight of the train (w) = 20 ton.

The track resistance (r) = 40 N/ton.

The overall efficiency (η) = 80%.

where:

The tractive effort during acceleration:

Ft = 277.8We × α + Wr

= 277.8 × 1.1 × 20 × 2 + 20 × 40

= 13,023.2 N.

The maximum power output = 177.958 kW.

(i) The distance travelled during braking:

The distance travelled with power is on:

D1 = 3 – 0.112

= 2.88 km.


CALCULATION OF ENERGY RETURNED TO THE SUPPLY DURING REGENERATIVE BRAKING

When the train is accelerating, it acquires kinetic energy corresponding to that speed. During the coasting period, some of the kinetic energy is wasted, to propel the train against the friction and windage resistance.

While the train is moving on the down gradients or level track, the KE acquired by the rotating parts is converted into the electrical energy, which is fed back to the supply system. The amount of energy fed back to the system is depending on the following factors.

1. The initial and final speeds during the regenerative braking.2. The train resistance and the gradient of the track.3. The efficiency of the system.

Consider the initial and final speeds of the train during regenerative braking are V1

and V2 in KMPH, and the effective weight of the train is We tons.

The kinetic energy stored by the train at a speed of V1 kmph:

Thus, the kinetic energy at speed V2 kmph:

Therefore, the energy available during the regeneration:

If D is the distance in km covered during the regenerative braking, then the energy fed back to the supply during the braking while the train is moving on down gradient:

If r is the train resistance in N/ton, then the energy lost to overcome the resistance to the motion and friction, windage losses:

Hence, the total energy available during regeneration:

The energy returned to the supply system:

where η is the efficiency of the system.

Advantages of regenerative breaking

1. In regenerative breaking, a part of the energy stored by the rotating parts is converted into the electrical energy and is fed back to the supply. This will lead to the minimum consumption of energy, thereby saving the operating cost.

2. High breaking retardation can be obtained during regenerative breaking.3. Time taken to bring the vehicle to rest is less compared to the other breakings; so that,

the running time of the vehicle is considerably reduced.4. The wear on the brake shoes and tyre is reduced, which increases the life of brake

shoe and tyre.

Disadvantages

In addition to the above advantages, this method suffers from the following disadvantages.

1. In addition to the regenerative breaking, to bring the vehicle to standstill, mechanical breaking is to be employed.

2. In case of DC traction, additional equipment is to be employed for regenerative breaking, which increases the cost and sometimes, substation are equipped with mercury arc rectifiers to convert AC to DC supply.

3. The electrical energy returned to the supply will cause the operation of substations complicated.

Example 10.26: A 450-ton train travels down gradient of 1 in 75 for 110 s during which its speed is reduced from 70 to 55 kmph. By the regenerative braking, determine the energy returned to the lines if the reactive resistance is 4.5 kg/ton and the allowance for the rotational inertia is 7% and the overall efficiency of the motor is 80%.

Solution:

Accelerating weight Wa = 1.075 × 450

= 483.75 ton.

Track resistance r = 9.81 × 4.5

= 44.145 N-m/ton

= 9,904.782 W-hr

= 9.904 kW-hr.

Factiva effort required during retardation:

= Wr – 98.1 WG

= 450 × 44.45 – 98.1 × 450 × 4/3

= 19,865.25 – 58,860

= –38,994.75 N.

The distance travelled during the retardation period:

As the train is moving in downward gradient, so that the tractive effort will provide additional energy to the system. The energy available when the train moves over a gradient is given as:

The period of regeneration = 110 s. Overall

efficiency (η) = 80%.

The kinetic energy of the train at a speed of 70 kmph is:

The kinetic energy of the train at the speed of 55 kmph is:

The energy available due to the retardation by the regenerative braking:

= 2,588.495 – 15,979.713

= 20.68 kW-hr.

The energy returned to the supply system:

= 0.8 × total energy available

= 0.8 × (20.68+9.904)

= 24.467 kW-hr.INTRODUCTION

The movement of trains and their energy consumption can be most conveniently studied by means of the speed–distance and the speed–time curves. The motion of any vehicle may be at constant speed or it may consist of periodic acceleration and retardation. The speed–time curves have significant importance in traction. If the frictional resistance to the motion is known value, the energy required for motion of the vehicle can be determined from it. Moreover, this curve gives the speed at various time instants after the start of run directly.

TYPES OF SERVICES

There are mainly three types of passenger services, by which the type of traction system has to be selected, namely:

1. Main line service.2. Urban or city service.3. Suburban service.

Main line services

In the main line service, the distance between two stops is usually more than 10 km. High balancing speeds should be required. Acceleration and retardation are not so important.

Urban service

In the urban service, the distance between two stops is very less and it is less than 1 km. It requires high average speed for frequent starting and stoppig

In the suburban service, the distance between two stations is between 1 and 8 km. This service requires rapid acceleration and retardation as frequent starting and stopping is required.

SPEED–TIME AND SPEED–DISTANCE CURVES FOR DIFFERENT SERVICES

The curve that shows the instantaneous speed of train in kmph along the ordinate and time in seconds along the abscissa is known as ‘speed–time’ curve.

The curve that shows the distance between two stations in km along the ordinate and time in seconds along the abscissa is known as ‘speed–distance’ curve.

The area under the speed–time curve gives the distance travelled during, given time internal and slope at any point on the curve toward abscissa gives the acceleration and retardation at the instance, out of the two speed–time curve is more important.

Speed–time curve for main line service

Typical speed–time curve of a train running on main line service is shown in Fig.10.1. It mainly consists of the following time periods:

1. Constant accelerating period.2. Acceleration on speed curve.3. Free-running period.4. Coasting period.5. Braking period.

SHORT ASNWERS QUESTIONS:

Unit –I

ELECTRIC DRIVES

1. What are all the types of drives?

2. What are all the electrical characteristics used to select a motor?

3. What are all the Mechanical characteristics used to select a motor?

4. What are the types of Load?

5. Define Load Equalization.

6. What are advantages of Thyristor control?

7. Mention two advantage of electric drive

8. What is meant by composite load?

9. Size of motors depends upon what factors

10. Define continuous rating of motor

Unit -IIELECTRIC HEATING & WELDING

1. What are the Modes of Heat transfer?

2. What are the Properties of heating element Materials?

3. What are the Types of Electric Heating?

4. What are the Reasons for failure of heating element?

5. 11 What are the Application of Dielectric Heating?

6. What are the Advantages of Dielectric Heating?

7. What are the Advantages of Coated Electrodes?

Unit –III

ILLUMINATION

1. Define Mean Horizontal candle Power (MHCP):

2. Define Mean spherical Candle power (MSCP):

3. Define Mean Hemispherical candle power (MHSCP):

4. Define Reduction Factor:

5. What are the types of Lightning?

6. What are the Factors affecting the design of Illumination?

Unit-IVELECTRIC TRACTION-I

1. What are various traction systems you know of?

2. Discuss merits and demerits of steam engine drive.

3. What are the different systems of railway electrification?

4. What are the merits and demerits of D.C system of track electrification?

5. What are the functions of a D.C. substation?

6. What are the factors affecting the schedule speed of a train?

7. What is the difference between dead weight and accelerating weight of a locomotive?

8. What do you meant by tractive effort?

9. What are the advantages of electric braking?

10.Why three phase traction system employing induction motors is now obsolete give reasons?

11. What are the various types of braking methods?

12. Discuss the current trends in AC traction systems?

13. What are the advantages of pantographs?

14. What are various current collection systems?

UNIT-V

ELECTRIC TRACTION-II1. With the help of a complete Speed-Time curve, discuss how different parameters. Of this curve

change with the type of train service.

2. Deriven expression for the distance traveled by an electric train using trapezoidal speed-time

curve.

3. Draw the speed-time curves for different services and explain them in detail.

4. Assuming a quadrilateral speed-time curve, develop a method of determining the specific energy

consumption of a train.

5. Discuss how different parameters of speed-time curve will vary with the type

of train service.

6. With the help of trapezoidal speed-time curve, derive an expression for the maximum speed and

hence estimate the values of acceleration and retardation.

7. Derive the expressions for the speed-Torque characteristics of dc shunt motor under the

following conditions:

a)Without control

b) External resistance in the armature circuit

c) External resistance in the field circuit

d) Armature shunted with resistance R

8. Draw the typical characteristics for all the conditions

9. Derive an expression for specific energy output on level track using a simplified speed–time

curve.

10. Derive the relationship between acceleration, retardation, maximum speed, running time and

distance between two stops assuming a trapezoidal Speed- Time curve.

What are the different methods of approximation of speed-Time curves? Derive expression for

distance travelled using quadrilateral approximation method of V (t) curve

11. What steps are taken to reduce the interference in the telecommunication circuits?

12. Compare Ac and DC systems of traction?

13. What types of train services railways have to cater for and what are their

distinguishing features?

14. What are the factors affecting the schedule speed of a train?

15. What are the different types of functions performed by the tractive effort developed by traction

unit

LONG ASNWERS QUESTIONS:

Unit –I

ELECTRIC DRIVES

1. Explain steady state characteristics and transient characteristics of drives

2. What type of motors is suitable for i) textile mill ii) pumps iii) lifts? Explain the reason for the

choice

3. What do you understand by the term load equalization? Derive the expression for motor torque

incase of motor fitted with the flywheel when load is increasing.

4. Discuss the various factors determines the choice of an electric motor for an

application.

5. Briefly explain the modern speed control of induction motors.

Unit -IIELECTRIC HEATING & WELDING

6. Explain coreless type induction furnace

1. Explain the construction and working principle of dielectric heating.

2. What are the types of ARC furnace? Describe the operation of them. (8)

3. Discuss the characteristic requirement of welding generator sets

(both AC and DC).

Unit –III

ILLUMINATION

4. Explain the method of working of a Neon lamp with a neat sketch.

5. State the Lambert’s cosine law of illumination.

6. Explain the working of a sodium vapour lamp with in a neat sketch.

7. What is Halogen lamp?

8. Explain the types of lamps and lighting fitments you should select for

(i)A large machine shop with rows of drilling machines

(ii) A drawing office and lathes.

9. Explain the operation principle and working of a mercury vapour lamp and

compareitsperformance with a fluorescent lamp.

10. Describe with a neat sketch the principle ofoperation of a fluorescent lamp.

11. Mention the function of each component.

12. Explain the principle of street lighting? Show different types of lighting with

neat sketches.

13. Discuss about Diffusion principle of street lighting.

Unit-IVELECTRIC TRACTION-I

1. Sketch the typical speed-time curve for (1) Main line service and to sub – urban services in electric

Traction.

2. Explain the mechanics of train movement?

3. What is multiple unit control in electric train and explain in details each one of them?

4. What are different braking systems and explain them in details?

5. What is the speed controls of different system of motors used in electric train?

6. Define co-efficient of adhesion “ and explain the factors on which it depends?

7. Discuss the various arrangement of current collection used in electric traction.

8. Write short notes on the recent trends in electric traction.

9. State the principle of regenerative braking. Explain regenerative braking in respect of

DC motors, b) Induction motors.

10.What are the various methods of speed control of series motors and their scope of speed range?

11. Discuss the merits and demerits of the induction motor for traction duties?

12. What is the main advantage of series parallel control of motors over rheostatic method of starting

and speed control?

13. What is multiple unit control and for what application will you suggest this?

UNIT-V

ELECTRIC TRACTION-II

1. With the help of a complete Speed-Time curve, discuss how different parameters. Of this curve

change with the type of train service.

2. Derivean expression for the distance traveled by an electric train using

trapezoidal speed-time curve.

3. Draw the speed-time curves for different services and explain them in detail.

4. Assuming a quadrilateral speed-time curve, develop a method of determining the specific energy

consumption of a train.

5. Discuss how different parameters of speed-time curve will vary with the type of train service.

6. With the help of trapezoidal speed-time curve, derive an expression for the maximum speed and

hence estimate the values of acceleration and retardation

Derive an expression for specific energy output on level track using a simplified speed–time curve.

7. Derive the relationship between acceleration, retardation, maximum speed, running time and

distance between two stops assuming a trapezoidal Speed- Time curve.

8. What are the different methods of approximation of speed-Time curves? Derive expression for

distance travelled using quadrilateral approximation method of V (t) curve.

9. What are the major equipments of a DC substation?

10. What are the different systems of current collection and give their merits and demerits?

11. What are the advantages of automatic weight tensioning and temperature compensation

arrangements in the OHE?

12. What are the advantages and disadvantages of linear induction motor?

13. What is the function of a motor starter?

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