UNESCO

EDUCATION REVITALISATION PROJECT

NATIONAL

ELECTRICAL ENGINEERI

ELECTRICAL

UNESCO-NIGERIA TECHNICAL & VOCATIONAL

EDUCATION REVITALISATION PROJECT-PHASE II

YEAR I- SEMESTER II

THEORY

Version 1: December 2008

NATIONAL DIPLOMA IN

ELECTRICAL ENGINEERING TECHNOLOGY

ELECTRICAL POWER SYSTEM (I)

COURSE CODE: EEC 122

NG TECHNOLOGY

TABLE OF CONTENTS

Department Electrical Engineering Technology

Subject Electrical power system I

Year 2

Semester 2

Course Code EEC 122

Credit Hours 3

Theoretical 2

Practical 3

Week 1: Principle of Electrical Energy Generation ……………………………1

1.1 Introduction……………………………………………………………1

1.2 Importance of Electrical Energy………………………………………2

1.3 Generation of Electrical Energy……………………………………….2

1.4 Sources of Energy……………………………………………………..2

1.5 Generating power Station……………………………………………...6

1.6 Steam power Station……………………………………………………8

Week 2: …………………………………………………………………………..10

1.7 Gas power Station……………………………………………………...10

1.8 Advantages…………………………………………………………….11

1.9 Disadvantages………………………………………………………….11

1.10 Main Component of Gas power Plant……………………………….12

Week 3:………………………………………………………………………….13

1.11Wind Energy Power Plant…………………………………………….13

1.12 Solar Energy Power Plant…………………………………………….13

1.13 Component of Solar Energy Plant…………………………………….14

1.14 Types of Photovoltaic (PV) Solar System…………………………….14

1.15 Solar Energy…………………………………………………………..15

1.16 Solar cells power Generation Unit…………………………………….16

Week 4: …………………………………………………………………………..17

1.17 Various Voltage Levels………………………………………………..17

1.18 Transmission Lines…………………………………………………….17

1.19 Classification of overhead Transmission lines………………………...18

Week 5:……………………………………………………………………………20

1.20 Principle of protection System and Devices……………………………20

1.21 Fuses……………………………………………………………………20

1.22 High Breaking Capacity………………………………………………..21

1.23 Semi enclose Rewirable Fuses…………………………………………21

Week 6:…………………………………………………………………………...22

1.24 Conductors……………………………………………………………..22

1.25 Commonly Used Conductors Materials……………………………….22

1.26 Types of Conductors…………………………………………………..23

Week 7: Principle of distribution system.........................................................26

2.1 Introduction.........................................................................................26

2.2 Distribution system ...........................................................................26

2.3 Feeders...............................................................................................27

2.3 Distributors.........................................................................................27

2.4 Service mains......................................................................................28

2.4 Classification of Distribution System.................................................29

Week 8:....................................................................................................................31

2.5 Direct current (DC) Distribution.........................................................31

2.6 Methods of feeding a distributor.........................................................31

2.8 Uniformly loaded distribution ............................................................34

2.9 D.C distributor feed at one end...........................................................35

2.10 Distributor feed at both ends...............................................................36

2.11 Uniformly loaded fed at both ends......................................................37

Week 9:....................................................................................................................39

2.12 Alternating current distribution (AC) ................................................39

2.13 classification of A.C distribution system............................................39

2.14 Power factors referred to receiving end..............................................40

Week 10: principle of protections in power system................................................43

3.1 Fuses and if components....................................................................43

3.2 Current Rating....................................................................................45

3.2 fusing current......................................................................................45

3.4 fusing factor........................................................................................45

Week 11:.................................................................................................................48

3.5 Moulded case current breaker.............................................................48

3.6 current breakers..............................................................................48

3.7 Maintenance moulded case circuit breaker.....................................49

3.8 Circuit Breaker Ratings...................................................................49

Week 12:........................................................................................................51

3.9 Functions Circuit Breaker...............................................................51

3.10 Principle of operation......................................................................51

3.11 Arc Phenomenon................................................................................52

3.12 Principle of Arc Extinction.................................................................53

3.13 Methods of Arc Extinction..................................................................54

3.14 Resistance the arc Increase...................................................................54

Week 13:.................................................................................................................55

3.15 Isolators..............................................................................................55

3.16 different isolator & Circuit Breaker........................................................

Week 14: Types of insulators & supports...........................................................57

3.7 Overhead & underground system insulators.......................................57

3.18 requirement of distribution system.....................................................58

Week 15:..................................................................................................................59

3.19 insulators.............................................................................................59

3.20 properties of insulators........................................................................59

3.21 Types of Insulators..............................................................................59

3.22 Advantages ........................................................................................61

Principle of Electrical Energy Generation Week 1

1.1 INTRODUCTION

Energy is the basic necessity for the economic development of a country due to it importance in

human life. Most of our day to day activities make use of electrical energy because it makes the activity

much easier, simple within a limited time. It is practically impossible to estimate the actual magnitude

of the part that electrical energy has played in the building up of present day civilization. With the

advance or the availability of huge amount of energy in the modern times has resulted in a shorter

working day, higher agricultural and industrial production, a healthier and more balanced diet and

better transportation facilities. As a matter of fact, there is a close relationship between the energy

used per person and his standard of living. The greater the per capital consumption of energy in a

country, the higher is the standard of living of its people. An example of energy generation is shown in

figure 1.1

Figure 1.1

The energy exist in different forms in native but t he most important form is the

electrical energy. The modern society is so much dependent upon the use of

electrical energy in the sense that, it has become a part and parcel of our life.

1.2 IMPORTANCE OF ELECTRICAL ENERGY

Energy may be needed as heat, light or as motive power etc. The present-day

advancement in science and technology has made it possible to convert electrical

energy into any desired form. This has given electrical energy a place of pride in

the modern hold. The survival of industrial undertakings and our social structures

depends primarily upon low cost and uninterrupted supply of electrical energy. In

fact, the advancement of a country is measured in terms of per capital

consumption of electrical energy. Electrical energy is higher to all other form of

energy due to the fact that electrical energy is;

• Convenient in form

• Easy control

• Greater flexibility

• Cheapness

• Cleanliness and

• Higher transmission efficiency.

1.3 GENERATION OF ELECTRICAL ENERGY (ELECTRICITY GENERATION)

Electricity generation is the process where energy available in different forms in

nature is being converted into electrical energy. The electrical energy must be

produced and transmitted tot eh point of use at the instant it is needed. The

entire process takes only a fraction of a second. This instantaneous production of

electrical energy introduces technical and economical considerations unique to

the electrical power industry.

1.4 SOURCES OF ENERGY

Electrical energy is produced from energy available in various forms in nature.

The various sources of energy are:

Sun

Wind

Water

Fuels

Nuclear energy.

SUN: The sun is the primary source of energy. The heat energy radiated by the

sun can focused over a small area by means of reflectors. This heat can be used

to raise steam and electrical energy can be produced. But this system has some

limitations such as:

It requires a large area for the generation of even a small amount of electric

power

It cannot be used in cloudy days or at night

It is an uneconomical method.

Wind: This method can be used where wind flows for a considerable length of

time. Wind energy is used to run the wind mill continuously as shown in figure

1.2, the generator is arranged to charge the batteries. This batteries supply t he

energy when the wind stops. This method has the advantages that maintenance

and generation costs are negligible. However, the draw backs or disadvantages of

this method are (i) variable output (ii) unreliable because of uncertainty about

wind pressure and (iii) power generated is quite small.

Figure 1.2

Water: When water is stored at a suitable place, it possesses potential energy

because of the head created. This water energy can be converted into

mechanical energy with the help of water turbines as shown in figure 1.3. The

water turbines drive the alternator which converts mechanical energy into

electrical energy. This method or generation of electrical energy has become very

popular because it has low production and maintenance costs.

Figure 1.3

Fuels: This can be further classifying into solid fuel (coal), liquid, fuel (oil) and gas

fuel (Gas). The heat energy of these fuels is converted into mechanical energy by

suitable prime movers such as steam engines, steam turbines, internal

combustion engine etc. the prime mover drives the alternator which converts

mechanical energy into electrical energy as can be seen in figure 1.4.

Figure 1.4

Nuclear energy: It has been discovered that large amount of heat energy is

liberated by the fission of uranium and other fissionable materials. It is estimated

that heat produced by 1kg of nuclear fuel is equal to that produced by 4500 tones

of coal. T he heat produced due to nuclear fission can be utilized to raise steam

with suitable arrangements as shown in figure1.5. The steam can run the steam

turbine which in two can drive the alternator to produce electrical energy. The

disadvantages of this system are (a) High cost of nuclear plant (b) Disposal of

radioactive waste and death of trained personnel to handle the plant.

Figure 1.5

1.5 GENERATING STATION

Figure 1.6

Bulk electric power is produced by special plants known as generating stations or

called power plants as shown in figure 1.6.

A generating station essentially employs a prime mover coupled to an alternator

for the production of electric power. The prime mover (e.g. steam turbine, water

turbine etc) that converts energy from some other form into mechanical energy.

The alternator converts mechanical energy o f the prime mover into electrical

energy. The electrical energy produced by t he generating station is transmitted

and distributed with the help of conductors to various consumers.

But in modern generating station, several auxiliary equipment and instruments

are used, apart from prime mover – alternator combination, in order to ensure

cheap, reliable and continuous service.

Depending upon the form of energy being converted into electrical energy, the

generating stations are classified as:

Steam power plants

Hydro power plant

Diesel power plant/Gas power plant

Nuclear power plants

wind power plant

Solar power plant

MHD P-plants.

1.6 STEAM POWER STATION (THERMAL STATION)

Figure 1.7a

A generating station which converts heat energy of coal combustion into electrical

energy is known as a steam power station. The steam is produced in the boiler by

utilizing the heat of coal combustion as shown in figure 1.7a & 17b. The steam is

then expanded in the prime mover (i.e steam turbine) and is condensed in a

condenser to be fed into the boiler again. The steam turbine drives the alternator

which converts mechanical energy of the turbine into electrical energy. This type

of power station is suitable where coal and water are available in abundance and

a large amount of electric power is to be generated.

Figure 1.7b

Generation of Electric Energy Week 2

1.7 GAS TURBINE POWER PLANT

A generating station which employs gas turbine as the prime mover for the

generation of electrical energy is known as gas turbine power plant.

In a gas turbine power plant, air is used as the working fluid.

Figure 1.8

Figure 1.9

The air is combustion chamber where heat is added to air, thus raising its

temperature.

Heat is added to the compressed air either by burning fuel in the chamber or by

the use of air heaters. The hot and high pressure air from the combustion

chamber is then passed to the gas turbine where it expands and does the

mechanical work. The gas turbine drives the alternator which converts

mechanical energy into electrical energy as shown in figure 1.9 & 1.10.

Figure 1.10

1.8 ADVANTAGES

• It is simple in design as compared to steam power station since no bitters

and auxiliaries are required.

• It is much smaller in sizes as compared to steam station of the same

capacity.

• The initial and operating cost are much lower than that of equivalent steam

power station.

• It requires comparatively less water as no condenser is used

• The maintenance charges are quite small.

• Gas turbines are much simpler in construction and operation.

• It can be started quickly from cold conditions

• There are no standby losses.

1.9 DISADVANTAGES

• There is a problem for starting the unit. This is because before starting the

turbine, t he compressor has to be operated for which power is required

from external source. But once it start, the external power will be no

longer required as the turbine itself supplies the necessary power to the

compressor.

• The overall efficiency of such plant is how (about 20%)

• The temperature of combustion chamber is quite high (300o)

1.10 THE MAIN COMPONENTS OF THE PLANT ARE:

(i) Compressor (ii) Regenerator

(iii) Combustion chamber (iv) Gas turbine

(v) Alternator (vi) Starting motor

Generation of Electric Energy Week 3

1.11 WIND POWER PLANT

Figure 1.11

This method can be used where wind flows for a considerable length of time.

Wind energy is used to run the wind mill continuously as shown in figure 1.11, the

generator is arranged to charge the batteries. This batteries supply t he energy

when the wind stops. This method has the advantages that maintenance and

generation costs are negligible. However, the draw backs or disadvantages of this

method are

variable output

Unreliable because of uncertainty about wind pressure and

Power generated is quite small.

1.12 SOLAR ENERGY POWER PLANT

This is the energy receives from the sun as a result of the sun rays known as

radiation from the sun to the earth surface. This energy is trapped through the

use of photovoltaic cell and converted into DC power output and DC output can

further be converted into AC power, and this output can be use into many

applications such as water pumping for irrigation, lighting, and refrigeration of

vaccine etc.

These arrangements is shown in figure 1.12

1.13 THE BASIC COMPONENTS OF SOLAR ENERGY POWER PLANT INCLUDE :

The photovoltaic cell (PV Solar Panel)

The Storage facilities (Batteries)

The charger Controller

The Inverter

The load

Figure 1.12

Solar Cell Array

Subsystem

DC to AC

Conversion ( )

Battery for Storage Control unit

Power Distribution

Unit

Loads

1.14 TYPES OF PV SOLAR SYSTEMS There are various types of PV systems configurations used for different

applications. PV system could be use in the stand alone, integrated and grid

connected mode. It could also be used as directly connected systems without

storage battery or with storage battery. The output from the system could be

used as a DC or AC systems. The different modes of usage provides for flexibility.

1.15 SOLAR ENERGY

The sun is a good source of energy. Heat energy in the suns

Rays can be used for a number of advantages particularly solar energy

applications. The solar energy is very versatile as it has limitless potential in

transforming our lives. Studies have shown that endowed with availability of this

resource as well as it’s viability for practical use.

Nigeria receives 5.80 x 106 MWh of electricity can be obtained from solar energy.

Solar energy technologies can be classified into two;-

Solar-thermal; and

Solar photovoltaic

Solar thermal; here the solar radiation is converted to thermal energy.

This heat energy can be directly used or indirectly by using the heat to boil water

and generate steam, which would in turn be used to generate steam turbine for

electricity generation. In other words, it may be used in application such as

drying, cooking, refrigeration and air conditioning.

Solar photovoltaic. Here the photovoltaic (pv) devices converts sunlight

directly into direct current (DC) electrical energy. This is done through the use of

silicon solar cells. Several solar cells are linked together to form a solar module.

Because they are modular in form, adding one or more cells can expand them. Or

they can be dismantled and used for other applications. The solar PV modules are

light and easily installed. They require small amount of maintenance. The

modules produce DC electricity which can be used directly or even stored in

batteries to be used later. The PV modules have been used for the following

applications;-

Photo-voltaic pumps for pumping water

Photo –voltaic refrigeration for preserving vaccines;

1.16 SOLAR CELL POWER GENERATION UNIT

For power generation the system consist of arrays, which are made up of

photovoltaic devices, the inverter to convert the DC into AC; the battery to store

the energy during daylight, as well as controller unit to manage the automatic

operation of the system.

One of the main of solar photovoltaic electricity generation is the high cost of

module. A 12V module cost about #30,000 (at exchange rate of 1 US $)

Generation of Electric Energy Week 4

1.17 VARIOUS VOLTAGE LEVELS

Generating voltages: 6.6KV, 11KV, 13.2KV or 33KV

High voltage transmission: 330KV, 132KV, 66KV, 6.6KV 3.3KV

Low voltage distribution: A.C 415/240V, 3 -,φ 4 wires

Standard frequency: Nigeria : 50Hz + 1% and – 1%

1.18 TRANSMISSION LINES

Figure 1.13 Transmission Lines

The important consideration in the design and operation of a transmission line

are the determination of voltage drop, line losses and efficiency of transmission.

These values are greatly influenced by the lines constants R, L and C of

transmission line as in figure 1.13 above. For instance, the voltage drop in the line

depends upon the values of above three line constants. Similarly, the resistance

of transmission line conductors is the most important cause of power loss in the

line and determines the transmission efficiency. In this chapter, we shall develop

formular by which we can calculate voltage regulation, line losses and efficiency

of transmission lines. These formular are important for two principal reasons.

Firstly, they provide an opportunity to understand the effects of the line on bus

voltages and the flow of power. Secondly, they help in developing an overall

understanding of what is occurring on electric power system.

1.19 CLASSIFICATION OF OVERHEAD TRANSMISSION LINES

A transmission line has three constant R, L and C distributed uniformly along

the whole length of the line. The resistance and inductance from the series

impedance. The capacitance existing between conductors for 1-phase line or from

a conductor to neutral for a 3-phase line forms a shunt path throughout the

length of the line. Therefore, capacitance effects introduce complications in

transmission line calculations. Depending upon the manner in which capacitance

is taken into account; the overhead transmission lines are classified as:

(i) Short transmission lines: when the length of an overhead transmission line

is up to about 50km and the line voltage is corporately low (<20kV), it is

usually considered as a short transmission line. Due to smaller length and

lower voltage, the capacitance effects are small and hence can be

neglected. Therefore, while studying the performance of a short

transmission line, only resistance and inductance of the line are taken into

account.

(ii) Medium transmission lines: When the length of an overhead transmission

line is about 50-150km and the line voltage is moderately high (>20KV<

100kV), it is considered as a medium transmission line. Due to sufficient

length and voltage of the line, the capacitance effects are taken into

account. For purposes of calculations, the distribution capacitance of the

line is divided and lumped in the form of condensers shunted across the

line and at one or more points.

(iii) Long transmission lines: When the length of an overhead transmission

line is more than 150 km and line voltage is very high (>100kV), it is

considered as a long transmission line. For the treatment of such a line,

the line constants are considered uniformly distributed over the whole

length of the line and rigorous methods are employed for solution.

It may be emphasized here that exact solution of any transmission

line must consider the fact that the constants of the line are not lumped

but are distributed uniformly throughout the length of the line. However,

reasonable accuracy can be obtained by considering these constants as

lumped for short

WEEK 5

1.20 Principle of Protection system and Devices

Circuit protection would be unnecessary if overloads and short circuits could be

eliminated. Unfortunately, overloads and short circuits do occur. To protect a circuit

against these currents, a protective device must determine when a fault condition

develops and automatically disconnect the electrical equipment from the voltage

source. An over current protection device must be able to recognize the difference

between over currents and short circuits and respond in the proper way. Slight over

currents can be allowed to continue for some period of time, but as the current

magnitude increases, the protection device must open faster. Short circuits must be

interrupted instantly. Several devices are available to accomplish this.

1.21 Fuses

A fuse is a one-shot device (Figure1). The heat produced by overcurrent causes the

current carrying element to melt open, disconnecting the load from the source voltage.

There are three types of fuses, namely

Semi-enclosed (Rewireable) fuse

Cartridge fuses

High Breaking Capacity(HBC)

Generation of Electrical Energy Week 5

Figure 1.15 Plug fuse

Figure 1.14 Cartridges

The cartridge type has fuses which look similar to those you would find in a standard

household plug. This type is improvement of the rewirable fuse type. It is main

advantages, is easy to replace, totally enclosed and its current rating is very accurate

1.22 HIGH BREAKING CAPACITY (HBC)

HBC stands for "high blow current (sometimes described as HRC = high rupture

current). HBC fuses are designed not to explode when

failing under currents many times their normal working

current (e.g. 1500 amps in a 10 amp circuit). They are

therefore to be preferred for the protection of main

voltage circuits where the power source may be

capable of providing very high currents. HBC types can

usually be recognized by being sand filled though they

may have a thick ceramic body.

Figure1.16 A HBC fuse

1.23 SEMI-ENCLOSED (REWIREABLE) FUSES

As the name indicates, the rewireable type has a fuse wire

held at both ends by a small retaining screw. Once the fuse is

blown, the fuse wire is the only pieces to be replaced. It is

cheap, but replacing a wrong size of element can cause

catastrophic consequences.

Figure1.17 Rewireable fuses

Generation of Electrical Energy Week 6

1.24 CONDUCTORS

The conductor is one of the important items as most of the capital outlay is invested for it. Therefore,

proper choice of material and size of conductor is of considerable importance. The conductor material

used for transmission and distribution of electric power should have the following properties:

(i) High electrical conductivity.

(ii) High tensile strength in order to withstand mechanical stresses.

(Iii) Low cost so that it can be used for long distances.

(iv) Cross arm which provide support to the insulators.

(v) Miscellaneous items such as phase plates, danger plate, lightning arrestors, anti-climbing wires

etc.

All above requirements are not found in a single material. Therefore while selecting a conductor

material for a particular case, a compromises made between the cost and the required electrical and

mechanical properties.

1.25 COMMONLY USED OF CONDUCTOR MATERIALS.

The most commonly used conductor materials for over head lines are copper, aluminum, steel-cored

aluminum, galvanized steel and cadmium copper. The choice of a particular material will depend upon

the cost, the required electrical and mechanical properties and the local conditions.

All conductors used for over head lines are preferably stranded in order to increase flexibility. In

stranded conductors, there is generally one central wire and round this, successive layers of wires

containing 6, 12, 18, 24…… wires. Thus, if there are n layers, the total number of individual wires is 3n

(n+1) +1. In the manufacture of stranded conductors, the consecutive layers of wires are twisted or

spiraled in opposite direction so that layers are bound together.

TYPES OF CONDUCTORS

1. Copper. Copper is an ideal material for over head lines owing to its high electrical conductivity

and greater tensile strength. It is always in the hard drawn form as stranded conductors. Although hard

drawing decrease the electrical conductivity slightly yet it increases the tensile strength considerably.

Copper has high current density i.e., the current carrying capacity of copper per unit of X-

sectional area id quite large. This leads to two advantages. Firstly, smaller X-sectional area of conductor

is required and secondly, the area offered by the conductor to wind load is reduced. Moreover, this

metal is quite homogeneous, durable and high scrap value.

There is hardly any doubt that copper is an ideal material for transmission and distribution of electric

power. However, due to its higher cost and non- availability, it is rarely used for these purpose. Now– a

– days the trend is to use aluminium in place of copper.

2. Aluminium. Aluminium is cheap and light as compared to copper but it has much smaller

conductivity and tensile strength. The relative comparison of the two materials is briefed

below:

i The conductivity of aluminium is 60% that of copper. The smaller conductivity of aluminium

means that for any particular transmission efficiency, the X - sectional area of conductor must be

lager in aluminium than in copper. For the same resistance, the diameter of aluminium conductor is

about 1.26 times the diameter of the copper conductor.

The increase X- section of Aluminium exposes a greater surface to wind pressure and, therefore,

supporting towers must be design for greater transverse strength. This often requires the used of

higher towers with consequence of greater sag.

ii The specific gravity of aluminium(2.71gm/cc) is lower than that of copper (8.9gm/cc). Therefore,

an aluminium conductor has a most one-half the weigh of equivalent copper conductor. For this

reason, the supporting strictures for aluminium need not be made so strong as that of copper

conductors.

iii Aluminum conductor being light, is liable to greater swings and hence larger cross- arms are

required.

iv Due to lower tensile strength and higher co - efficient of linear expansion of aluminium, the sag

is greater in aluminium conductors.

Considering the combined properties of cost, conductivity, tensile strength, weight etc., aluminium has

an edge over copper. Therefore, it is being used as a conductor material. It is particularly profitable to

use aluminium for heavy-current transmission where the conductors’ size is large and its cost forms a

major proportion of the total cost of complete installation.

3. Steel cored aluminium. Due to low tensile strength, aluminium conductors produce greater sag.

This prohibit their used for larger span and makes them unsuitable for distance transmission. in order to

increase the tensile strength , the aluminium conductor is reinforced with a core of galvanized steel

wires. The composite conductor thus obtained is known as steel core aluminium and is abbreviated as

A.C.S.R.(aluminium conductor reinforced).

Steel – cored aluminium conductors consists of central core of galvanized steel wires

surrounded by a number of aluminium strands. Usually diameter of both steel and aluminium wires is

the same. The X- section of the two metal is generally in the ratio of 1:6but can be modified to 1:4in

order to get more tensile strength for the conductor. Fig 8.1 shows steel cored aluminium conductor

having one steel wire surrounded by six wires of aluminium. The result of this composite conductors is

that steel cored takes greater percentage of mechanical strength while the aluminium strand carry the

bulk of current. The steel cored aluminum conductors have the following advantages:

(i) The reinforcement with steel increases the tensile strength but at the same time keeps the composite

conductors light. Therefore steel cored aluminium conductors produce smaller sag and hence longer

span can be used.

(ii) Due to smaller sag with steel cored aluminium conductors. Towers of smaller height can be used.

4. Galvanized steel. Steel have high tensile strength. Therefore, galvanized steel conductors can be

used for extremely long span or short line section exposed to abnormally high stresses due to

climatic conditions. They have been found very suitable in rural areas where cheapness is the main

consideration. Due to poor conductivity and high resistance of steel, such conductors are not

suitable for transmitting high large power over a long distance. However, they can be used to

advantage for transmitting a small power over a small distance were the size of the copper

conductor desirable from economic considerations would be too small and thus unsuitable for used

because of poor mechanical strength.

5. Cadmium copper. The conductor material being employed in certain cases is copper alloyed

with cadmium. An addition of 1%or2% cadmium to copper increases the tensile strength by

about 50% and the conductivity is only reduced by 15% below that of pure copper. Therefore,

cadmium copper conductor can be used for exceptionally long spans. However, due to high cost

of cadmium, such conductors will be economical only for lines of small X- section i.e., where the

cost of conductor material is comparatively small compared with the cost of supports.

2.0 Principle of Distributions System Week 7 2.1 INTRODUCTION

Electrical power is usually generated and transmitted in 3-phase. It is distributed

in three - phase or single – phase depending on the need of the consumer. The

figure 2.1below shows a typical power distribution system. Power is supplied

from the generator (for example at 11KV) which is stepped – up by the step – up

transformer to a higher voltage (about 132KV). This high voltage is used to

transmit electricity over long distances so as to minimize power losses at a far end

of the line. The overhead high – voltage transmission line terminates in step –

down transformers in a substation where the voltage is stepped – down for

distribution.

Fig.2. 1 & 2.2 A typical electric power Distribution system

2.2 DISTRIBUTION SYSTEM

The distribution system is that part of the power system which distributes electric

power for local use (to the consumer). It is the electric system between the

substation fed by the transmission system and the consumer’s meters. The

distribution system consists of feeders, distributors and the service mains.

Figure 2.3

2.3 FEEDERS:

Feeders are conductors which connect the source (the substation or

localized generating station) to the distributors serving a particular area.

Current loading on a feeder is the same throughout its entire length as no

tapings are taken from the feeder. A feeder is designed on the basis of its

current carrying capacity. The voltage drop of a feeder is relatively

unimportant during design as it can be compensated by means of voltage

regulating equipment at the substation.

2.4 DISTRIBUTORS:

A distributor is a conductor that receives power directly from the feeder. It

is a conductor from which tapings are taken for supply to the consumer. It

has distributed loading which gives rise to variations of current along its

entire length. A distributor is designed from the point of view of the voltage

drop in it.

Figure 2.4 Figure 2.5

2.5 SERVICE MAINS:The connecting wires or connecting link between the

distributors and the consumer’s terminal are the service mains.

Figure 2.6

Fig 2.23,2.24 & 2.25 above shows a typical A.C. distribution system. In the figure,

feeders connect the substation to the distributors. Power is tapped from the

distributors through the sub-distributors via the service mains to the consumer’s

premises.

2.6 CLASSIFICATION OF DISTRIBUTION SYSTEMS

Distribution systems are classified based on three main aspects; nature of current,

type of construction and scheme of connection.

NATURE OF CURRENT – Based on nature of current, distribution

systems are grouped into two – alternating current (A.C.) distribution system

and direct current (D.C.) distribution system.

TYPE OF CONSTRUCTION – Based on type of construction, distribution

systems are divided into overhead systems or underground systems. In the

overhead system, bare aluminum or copper conductors are strung between

wooden, steel or concrete poles. These conductors are connected to the poles

by insulators and cross arms. The underground system uses insulated cables

to convey power from one system to the other. Underground cables have

better voltage regulation than overhead cables which is as a result of low

inductance and low inductive drops due to small spacing between the

conductors. Overhead conductors have considerably higher current carrying

capacity than underground conductors of the same material and cross-

section.

SCHEME OF CONNECTION – Under scheme of connection distribution

systems are further classified as radial system, ring main system or

interconnected system.

i) Radial System – In this system feeders branch out radially from a common

source and feed the distributors at one end only. In this type of system if a feeder

fails due to a fault, the supply to the consumer is interrupted until repairs are

done. It is the simplest distribution circuit and has the lowest initial cost but has

some drawbacks such as,

• Any fault on the feeder or distributor cuts off supply to the consumers on the

side of the fault away from the substation as they are dependent on a single

feeder and distributor.

• The consumer at the farthest end of the distributor would be subject to

voltage fluctuations when the load on the distributor changes.

ii) Ring Main System – In the ring main distribution system the feeder

branches out in the form of a loop or ring. The loop circuit starts from the

substation bus bars makes loop through the area to be served and returns to the

substation. This makes a complete loop and has isolating switches provided at the

poles at strategic points for isolating a particular section in case of a fault. Thus

failure of one interconnecting feeder does not interrupt the supply.

2.0 Principle of Distributions System Week 8 2.7 DIRECT CURRENT (D.C.) DISTRIBUTION

Electrical power is mostly generated, transmitted and distributed as alternating

current. Direct current however is necessary for certain applications such as for

the operation of variable speed machinery, for electrochemical work and electric

traction. A.C. power is converted into D.C. power by use of mercury-arc rectifier,

rotary converters and motor generator sets. D.C. supply can be obtained as either

2-wire system or 3-wire system for distribution.

1) 2-wire D.C. System – This system has 2 wires; the outgoing or positive

wire and the return or negative wire. Due to its low efficiency, it is not used

for transmission purposes but for distribution of D.C. power.

2) 3-wire D.C. System – This system has 3 wires, the middle wire which is

the neutral is earthed. The voltage between either of the outer wires and

neutral is half that between the negative and positive wire making two

voltages available at the consumer terminal.

2.8 METHODS OF FEEDING A DISTRIBUTOR

There are various methods of feeding a distributor

i) Distributor fed at one end

ii) Distributor fed at both ends

iii) Distributor fed at the center

iv) Ring mains

The nature of loading on the distributor also varies such as

a) Concentrated loading

b) Uniform loading

c) Combination of both concentrated and uniform loading

i) Distributor fed at one end:

A C D E B

I1 I2 I3

The distributor AB above is connected to supply at one end with loads I1, I2 and I3

taken at different points along its length. In this type of distributor when a fault

occurs on any section of the distributor, the whole distributor will have to be

disconnected from supply. Voltage across loads decreases away from feeding

point (point E will therefore have the lowest voltage). Current also decreases

along various sections of the distributor.

In fig 5 above is shown a distributor fed at one end. The voltage drop in the

distributor is

V = IACRAC + ICDRCD + IDERDE + …

IAC is the current in section AC of the distributor which is the sum of load currents

I1, I2, I3 …

IAC = I1 + I2 + I3 + …

ICD = I2 + I3 + …

IDE = I3 + …

ii) Distributor fed at both ends:

A C D E B

I1 I2 I3

The distributor AB in fig. 6 above is connected to the supply mains at both ends

with loads taken at different points. Voltage at the supply ends A and B may or

may not be equal. The load voltage decreases away from one feeding point

reaches minimum value then increases towards the other feeding point. In this

type of distributor continuity of supply is maintained in case of faults along the

distributor as there are two feeding points.

iii) Distributor fed at the center:

A C B

I1 I2 I3 I4

In this type of feeding, the center of the distributor is connected to the supply

making it two singly fed distributors having a common feeding point.

iv) Ring Mains: I1

feeder I2

The distributor is in the form of a closed ring. It is the same as a straight

distributor fed at both ends with equal voltages and the two ends brought

together to form a closed ring.

Looking at the various types of loading (uniform loading, concentrated loading

and a combination of both) on the distributors above we have.

2.9 UNIFORMLY LOADED DISTRIBUTOR FED AT ONE END

A B A C B

i i i x dx

In fig. a above conductor AB is fed at one end A and uniformly loaded with i

amperes per unit length. Let,

i = current tapped off per unit length

l = total length of distributor

r = resistance per unit length of the distributor

Finding the voltage drop at a point C fig. b which is at a distance of x units from

feeding end A.

Current at point C = (il – ix) = i(l – x)

Consider a small section of length dx near point C

resistance = rdx

Voltage drop over length dx is

dv = i(l – x)(rdx) = (ilr – ixr)dx

Total drop up to point x is

=

V = ilrx – ½ irx2

= ir (lx – x2/2) ----- (1)

At point B voltage drop can be calculated by taking, x = l

ir (l2 – l

2/2) = irl2/2 = ½ IR ----- (2)

Where I = il = total current entering at point A

R = rl = total resistance of distributor AB

Thus total drop in the distributor AB = ½ IR

Example 1: A 250m 2-wire D.C. distributor fed from one end is loaded uniformly

at the rate of 1.6A/meter. The resistance of each conductor is 0.0002Ω per meter.

Find the voltage necessary at feed point to maintain 250V (a) at the far end (b) at

the midpoint of the distributor.

Solution:

Current entering the distributor I = il = 1.6 x 250 = 400A

Resistance of distributor per meter run r = 2 x 0.0002 = 0.0004Ω

Total resistance of distributor R = r x l

R = 0.0004 x 250 = 0.1Ω

Voltage drop over entire distributor = ½ IR

= ½ x 400 x 0.1 = 20V

At feeding point voltage drop = 250 + 20 = 270V

At a distance x meters from feeding point the voltage drop is taken from eqn (1)

above.

V = ir (lx – x2/2)

At mid point x = l/2 = 250/2 = 125m

Voltage drop = 1.6 x 0.0004 ((250 x 125) - 1252/2) = 15V

At feeding point voltage = 250 + 15 = 265V

2.10 DC DISTRIBUTOR FED AT ONE END – CONCENTRATED LOADING

Example 2: A 2-wire dc distributor AB is 300m long. It is fed at point A with loads

of 30A, 40A, 100A and 50A at distances of 40m, 100m, 150m and 250m from A. If

the maximum permissible voltage drop is not to exceed 10V, find the cross-

sectional area of the distributor. Take ρ = 1.78 x 10-8

Ωm.

Solution:

Total voltage drop over the distributor is

V = i1R1 + i2R2 + i3R3 + …

For 2-wire distributor voltage drop is

V = 2(i1R1 + i2R2 + i3R3 + …)

Cross-sectional area A =

Resistance of section AC, RAC = ρl/A = 1.78 x 10-8

x 40/A

RAD = 1.78 x 10-8

x 100/A

RAE = 1.78 x 10-8

x 150/A

RAF = 1.78 x 10-8

x 250/A

V = 10V

10 = 2 x 1.78 x 10-8/A [(40 x 30) + (100 x 40) + (150 x 100) + (250 x 50)]

A = 116.34 x 10-6

m2

2.11 DISTRIBUTOR FED AT BOTH ENDS – CONCENTRATED LOADING

Example 3: A 2-wire dc distributor AB is fed from both ends. At feeding point A,

the voltage is maintained at 230V and at B 235V. The total length of the

distributor is 200m and loads of 25A, 50A, 30A and 40A are tapped at distances of

50m, 75m, 100m and 150m from A respectively. The resistance per kilometer of

one conductor is 0.3Ω. Calculate (a) the currents in various sections of the

distributor (b) minimum voltage and the point at which it occurs.

Solution:

Resistance of 1000m length of distributor of both wires = 2 x 0.3 = 0.6Ω

At section AC resistance RAC is

RAC = 0.6 x 50/1000 = 0.03Ω

RCD = 0.6 x 25/1000 = 0.015Ω

RDE = 0.6 x 25/1000 = 0.015 Ω

REF = 0.6 x 50/1000 = 0.03 Ω

RFB = 0.6 x 50/1000 = 0.03 Ω

Voltage at B = voltage at A – voltage drop over AB

VB = VA – [IARAC + (IA – 25)RCD + (IA – 75)RDE + (IA – 105)REF + (IA – 145)RFB]

235 = 230 – [0.03IA + 0.015(IA – 25) + 0.015(IA – 75) + 0.03(IA – 105) + 0.03(IA –

145)]

235 = 230 – (0.12IA – 9)

IA = 33.33A

Current in AC, IAC = IA = 33.33A

ICD = IA – 25 = 8.33A

IDE = IA – 75 = - 41.67A ------ this shows that current flows in the opposite

direction that is E to D.

IEF = IA – 105 = - 71.67A

IFB = IA – 145 = - 111.67A

From the current at various sections calculated above, the currents are coming to

load point D from both sides making it the point of minimum potential.

VD = VA – (IACRAC + ICDRCD)

= 230 – [(33.33 x 0.03) + (8.33 x 0.015)] = 228.875V

2.12 UNIFORMLY LOADED DISTRIBUTOR FED AT BOTH ENDS

Example 4: (i) A uniformly loaded distributor is fed at the center. Show that

maximum voltage drop = IR/8 where I is the total current fed to the distributor

and R is the total resistance of the distributor. (ii) A 2-wire dc distributor 1000m

long is fed at the center and is loaded uniformly at the rate of 1.25A/m. If the

resistance of each conductor is 0.05 Ω/km find the maximum voltage drop in the

distributor.

Solution:

The distributor is fed at center C and uniformly loaded with loads of i A/m. Taking

the resistance per meter run of the distributor as r Ω. Maximum voltage drop

occurs at either end of the distributor.

Maximum voltage drop = voltage drop in half distributor

= ½ (il/2) (rl/2) = 1/8 (il) (rl) = 1/8 IR

Where I = il = total current fed to distributor

R = rl = total resistance of distributor

(ii) I = il = 1.25 x 1000 = 1250A

R = rl = 2 x 0.05 x 1 = 0.1 Ω

Maximum voltage drop = 1/8IR

= 1/8 (1250)0.1 = 15.62V

2.0 Principle of Distributions System Week 9

2.13 ALTERNATING CURRENT DISTRIBUTION (A.C)

Electricity was initially generated, transmitted and distributed as direct current

(D.C.). The main disadvantage of direct current system was that voltage levels

could not be easily changed. Now-a-days electrical energy is generated,

transmitted and distributed in the form of alternating current. One important

reason for the widespread use of alternating current in preference to D.C. is the

fact that alternating voltage can be conveniently changed in magnitude by means

of a transformer. Transformer has made it possible to transmit A.C. power at high

voltage and utilize it at a safe potential.

2.14 CLASSIFICATION OF A.C. DISTRIBUTION SYSTEM

The A.C. distribution system is classified into two; primary distribution system and

secondary distribution system.

PRIMARY DISTRIBUTION SYSTEM

The primary distribution system is that part of A.C. distribution system which

operates at voltages higher than general utilization and handles large blocks of

electrical energy than the average low-voltage consumer uses. Voltage used for

primary distribution depends upon the amount of power to be conveyed. The

most commonly used primary distribution voltages are 11KV, 6.6KV and 3.3KV. It

is carried out by 3-phase, 3-wire system.

SECONDARY DISTRIBUTION SYSTEM

That part of A.C. distribution system which includes the range of voltages at which

the ultimate consumer utilizes the electrical energy delivered to him. It employs

400/230V, 3-phase, 4-wire system.

In A.C. systems, voltage drops are due to the combined effects of resistance,

inductance and capacitance. Power factor has to be taken into account as loads

tapped off from the distributor are generally at different power factors. The

power factors of load currents may be referred to receiving end voltage or to the

respective load voltages.

2.15 POWER FACTORS REFERRED TO RECEIVING END VOLTAGE

A R1 + jX1 R2 + jX2 B

I1 cosØ1 I2 cosØ2

Consider an A.C. distributor AB (fig. a) with concentrated loads of I1 and I2 tapped

off at points C and B as shown above. Let the receiving end voltage VB be the

reference vector, lagging power factors at C and B be cosØ1 and cosØ2 with

respect to VB. Let R1, X1 and R2, X2 be the resistance and reactance of sections AC

and CB of the distributor. Then,

Impedance of section AC, Z AC = R1 + jX1

Impedance of section CB, Z CB = R2 + jX2

Load current at point C, I 1 = I1(cosØ1 – jsinØ1)

Load current at point B, I 2 = I2(cosØ2 – jsinØ2)

Current in section CB, I CB = I 2 = I2(cosØ2 – jsinØ2)

Current in section AC, I AC = I 1 + I 2

= I1(cosØ1 - jsin Ø1) + I2(cosØ2 - jsinØ2)

Voltage drop in section CB, V CB = I CB Z CB

= I2(cosØ2 - jsinØ2)(R2 + jX2)

Voltage drop in section AC, V AC = I AC Z AC = ( I 1 + I 2) Z AC

= [I1(cosØ1 - jsinØ1) + I2(cosØ2 - jsinØ2)](R1 + jX1)

Sending end voltage, V A = V B + V CB + V AC

Sending end current, I A = I 1 + I 2

Example:

1) A single phase A.C. distributor AB 300m long is fed from end A and is loaded as

under (i) 100A at 0.707 p.f. lagging 200m from point A. (ii) 200A at 0.8 p.f. lagging

300m from point A. The load resistance and reactance of the distributor is 0.2Ω

and 0.1Ω per km. Calculate the total voltage drop in the distributor. The load

power factors refer to the voltage at the far end.

Solution:

Impedance of distributor/km = (0.2 + j0.1) Ω

Impedance of section AC, Z AC = (0.2 + j0.1) x 1000200

Z AC = (0.04 + j0.02) Ω

Impedance of section CB, Z CB = (0.2 + j0.1) x 1000100

Z CB = (0.02 + j0.01) Ω

Taking voltage at the far end B as the reference vector,

Load current at point B, I 2 = I2(cosØ2 - jsinØ2)

= 200(0.8 – j0.6)

I 2 = (160 – j120) A

Load current at point C, I 1 = I1(cosØ1 - jsinØ1)

= 100(0.707 – j0.707)

I 1 = (70.7 – j70.7) A

Current in section CB, I CB = I 2 = (160 – j120) A

Current in section AC, I AC = I 1 + I 2 = (70.7 – j70.7) + (160 – j120)

I AC = (230.7 – j190.7) A

Voltage drop in section CB, V CB = I CB Z CB = (160 – j120) (0.02 + j0.01)

V CB = (4.4 – j0.8) V

Voltage drop in section AC, V AC = I AC Z AC

= (230.7 – j190.7) (0.04 + j0.02)

V AC = (13.04 – j3.01) V

Voltage drop in the distributor = V AC + V CB

= (13.04 – j3.01) + (4.4 – j0.8) = (17.44 – j3.81) V

Magnitude of drop = √(17.44)2 + (3.81)

2

= 17.85 V

2.0 Principle of Distributions System Week

10

3.1 FUSES AND ITS COMPONENTS

A fuse is defined in the I.E.E. Regulation as: “A device for opening a circuit by

means of a conductor designed to melt when an excessive current flows. The

fuse comprises all the parts that form the complete device”.

There are three types of fuses:

1. Rewirable fuse

2. Cartridge fuse

3. High braking capacity (H.B.C) fuse, formerly termed the high rupturing

capacity (H.R.C) fuse.

Rewirable Fuse: This consist (Fig. 6.14) of a porcelain bridge and base. The

bridge has two sets of copper contacts which fit into contacts in the base.

BRIDGE BASE

CABLE ENTRY

ASBESTOR TUBE

FUSE ELEMENT FIXING HOLE

PORCELAIN PROCELAIN

BRASS CONTACT WITH

CONNECTING TERMINAL

FIG. Rewirable Fuse

The fuse element, for example, tinned copper wire, is connected between the

terminals of the bridge. An asbestos tube, or pad, is generally fitted in the fuse tp

minimize the effect of arcing when the fuse element melts.

This type of fuse is termed a ‘semi-enclosed fuse’ to distinguish it form the older

type of fuse which consisted simply of a piece of wire connected between two

terminals.

That 3 of the I.E.E Regulations gives approximate sizes of tinned copper wire to be

used for elements in semi-enclosed fuse.

Example: 0.2, (standard wire gauge) – 5A current rating

0.35mm - 10A current rating

0.50mm - 15A current rating

3.2 CURRENT RATING

is the current which the fuse element will carry continuously without

deterioration.

3.3 FUSING CURRENT

is the current at which the fuse element will melt. This is approximately twice the

current rating of the fuse element (fusing factor = 2).

3.4 FUSING FACTOR

is the ratio

Advantages

1. Cheap

2. Easy to replace fuse element.

Disadvantages

1. Fuse elements deteriorate in use

2. Any size of fuse wire can be fitted, thus defeating the purpose of the

fuse.

Note: The fuse must be capable of protecting the smallest conductor in

the circuit.

3. Lacking in discrimination. It is possible that a 15A fuse element may

melt before a 10A fuse element, depending largely on the condition of

the wire. Further, the rewritable fuse is not capable of discriminating

between a momentary high starting current and fault current.

4 Easily damage, particularly with short-circuit currents.

Cartridge Fuse. This type has come into common use with

FUSE ELEMENT

END END

CAP CAP

PORCELAIN TUBE

FIG. Cartridge fuse

The fused 13A plug used on the domestic ring circuit. The diagram above shows

the construction of a cartridge fuse. The fuse element is contained in a porcelain

tube fitted with two connecting caps, and has a fusing factor of 1.5.

The colour rode for these fuses is as follows:

5A-WHITE 13A-BROWN 15A-BLUE

30A RED 60a PURPLE

High Breaking Capacity fuse. (H.B.C). This type of fuse is designed to protect

circuits against heavy overloads and is capable of opening a circuit under short-

circuit conditions without damaging surrounding equipment.

The diagram below shows the construction of a high breaking capacity fuse. This

consists of the following:

1. Porcelain tube

2. Silver element

3. Indicating element which ignites powder under the label to show when

the fuse element has opened.

4. INDICATING ELEMENT AND

INDICATING POWER CONNECTING LUG

SILICON BODY SILVER END CAP

PORCELAIN ELEMENT

FIG. High breaking capacity fuse

5. End caps

6. Silica (fine sand) filling used to quench the arc.

NOTE. The fuse must always be placed in the phase or non-earthed

conductor of the installation, never in the neutral (earthed) conductor.

2.0 Principle of Distributions System Week

11

3.5 MOLDED - CASE CIRCUITS BREAKER

The molded case of a CB provides the physical means of positioning the breaker components, and it protects the working parts from damage and contamination. The molded case also protects people from contact with energized components in the breaker.

Molded - case circuit breakers can be used in any electrical circuits where protection is required, including main service and feeders as well as branch circuits. They are found in switchboards, panel boards, control centres and individual enclosures.

3.6 CIRCUIT BREAKERS

Figure 4.7: Miniature circuit breakers with different poles

The problem with fuses is they only work once. Every time you blow a fuse, you have to

replace it with a new one. A circuit breaker (Figure 4.7) does the same thing as a fuse .It

opens a circuit as soon as current climbs to unsafe levels, but you can use it over and

over again.

The basic circuit breaker consists of a simple switch,(see figure 4.8) connected to either

a bimetallic strip or an electromagnet. The diagram below shows a typical

electromagnet design.

Figure4.8: Cut view of a miniature circuit breaker

3.7 MAINTENANCE OF MOLDED-CASE CIRCUIT BREAKER

Molded case breakers are relatively trouble - free devices, requiring little maintenance. The only maintenance required is to see that all conductor terminals are tight and free from corrosion, and that the breaker is dry and free from accumulated dirt and dust.

3.8 CIRCUIT BREAKER RATINGS

A circuit’s breaker is selected for a particular duty taking the following factors into

consideration. :

i. The normal current it will have to carry

ii. The amount of current the supply system will feed into the circuit

Fault, which is the current the circuit breaker, will have to interrupt

Without damage to itself.

Ratings of circuit breakers are specified according to certain standards and recommendations. The ratings are the same for one pole, three pole, and four pole circuit breakers.

2.0 Principle of Distributions System Week

12 3.9 FUNCTIONS CIRCUIT BREAKERS

A circuit breaker is a piece of equipment which can

(i) Make or break a circuit either manually or by remote control

under normal conditions.

(ii) Break a circuit automatically under fault conditions

(iii) Make a circuit either manually (or by remote control) as well

automatic control for switching functions. The latter control

employs relays and operates only under fault conditions.

The mechanism of opening of the circuit breaker under fault

conditions has already been briefed in the previous chapter.

OPERATING PRINCIPLE: A circuit breaker essentially consists of fixed

and moving contacts, called electrodes. Under normal operating

conditions, these contacts remain close and will not open

automatically until and unless the system becomes faulty. Or course,

the contacts can be opened manually or by remote control

whenever desired, when a fault occurs on any part of the system,

the trip coils of circuit breaker get energize and the moving contacts

are pulled apart by some mechanism, thus opening the circuit.

When the contacts of a circuit breaker are separated under fault

conditions, an arc is struck between them, the current is thus able to

continue until the discharge ceases. The production of arc not only

delays the current interruption process but it also generates

enormous heat which may cause damage to the system or to the

circuit breaker itself. Therefore, the main problem in a circuit

breaker is to extinguish the arc within the shortest possible time so

that heat generated by it may not reach a dangerous value.

2.0 Principle of Distributions System Week

13

3.10 ISOLATORS

This is an electrical manual device used to protect electrical circuit in a

power system of a transmission and distribution network.

The operation of an isolator serves as a protection to the circuit, to and

safe guard the use of the circuit.

It also contents a lock that is used fixed or make contact between the

two terminals. And the terminals can be separated by removing the

lock.

While A circuit breaker is a piece of equipment which can

(i) Make or break a circuit either manually or by remote control

under normal conditions.

(ii) Break a circuit automatically under fault conditions

(iv) Make a circuit either manually (or by remote control) as well

automatic control for switching functions. The latter control

employs relays and operates only under fault conditions.

The mechanism of opening of the circuit breaker under fault

conditions has already been briefed in the previous chapter.

2.0 Principle of Distributions System Week

14

3.13 OVERHEAD AND UNDERGROUND SYSTEMS

Distribution systems can be either overhead or underground. Underground

systems use conduits, cables and manholes under the surface of streets while

overhead systems consist of lines mounted on wooden, concrete or steel poles.

They are arranged to carry distribution transformers as well as conductors. Some

of the merits and demerits of overhead and underground systems are:

1) FAULTS – Chances of faults in underground systems are very rare as

cables are laid underground and have better insulation.

2) INITIAL COST – Underground system is more expensive than

overhead system due to high cost of trenching, conduits, cables e.t.c.

3) FLEXIBILITY – Overhead system is more flexible than underground

system as poles and transformers can be easily shifted to meet

changes in load conditions.

4) FAULT LOCATION AND REPAIRS – There are little chances of

faults in an underground system, but if they do occur it is difficult to

locate and repair. In overhead systems, conductors are visible thus

faults are easily traced and repaired.

5) CURRENT CARRYING CAPACITY & VOLTAGE DROP –

Overhead conductor has a higher current carrying capacity than

underground conductor of the same material and cross section.

6) MAINTENANCE COST – Maintenance cost of underground system

is very low compared with that of overhead system because of fewer

chances of faults.

3.14 REQUIREMENTS OF A DISTRIBUTION SYSTEM

Some of the requirements of a good distribution system are:

I. PROPER VOLTAGE – Voltage variations at consumer’s terminals

should be as low as possible. Changes in voltage are generally caused

due to the variation of load on the system. A good distribution system

should ensure that the voltage variations at consumer’s terminals are

within permissible limits which are 6± % of rated value at consumer’s

terminals.

II. AVAILABILITY OF POWER ON DEMAND – Power must be

available to consumers’ in any amount that they may require from

time to time.

III. RELIABILITY – Modern industry is almost dependent on electric

power for its operation. Reliability of the system can be improved by

interconnected systems, providing additional reserve facilities.

2.0 Principle of Distributions System Week

15

3.15 INSULATORS

Insulators are materials that do not allow the flow of current through them.

Overhead line conductors should be supported on the poles in such a way that

currents from conductors do not flow to earth through supports that is line

conductors must be properly insulated from supports. This is achieved by securing

line conductors to supports with the help of insulators. The insulators provide

necessary insulation between line conductors and supports and thus prevent any

leakage current from conductors to earth.

3.16 PROPERTIES OF INSULATORS

i. High mechanical strength in order to withstand conductor load, wind load.

ii. High electrical resistance of insulator material in order to avoid leakage

currents to earth.

iii. High relative permittivity of insulator material in order that dielectric

strength is high.

iv. The insulator material should be non-porous; free from impurities and cracks

otherwise the permittivity will be lowered.

v. High ratio of puncture strength to flashover.

3.17 TYPES OF INSULATORS

The most common used material for insulators of overhead lines is porcelain but

glass, steatite and special composition materials are also used to a limited extent.

The successful operation of an overhead line depends to a considerable extent

upon the proper selection of insulators. The most commonly used types of

insulators are; pin type, suspension, strain insulator and shackle insulator.

1) PIN TYPE INSULATORS:

The pin type insulator is secured to the cross arm on the pole. There is a groove

on the upper end of the insulator for housing the conductor. The conductor

passes through this groove and is bound by the annealed wire of the same

material as the conductor.

Pin type insulators are used for transmission and distribution of electric power at

voltages up to 33KV. Beyond operating voltage of 33KV, the pin type insulators

become too bulky and hence uneconomical.

2) SUSPENSION TYPE INSULATORS:

Suspension type insulators are used for high voltages above 33KV. They consist of

a number of porcelain discs connected in series by metal links in the form of a

string. The conductor is suspended at the bottom end of this string while the

other end of the string is secured to the cross-arm. Each disc is designed for low

voltage say 11KV. The number of discs in series would depend upon the working

voltage.

ADVANTAGES:

i. Suspension type insulators are cheaper than pin type insulators

for voltages beyond 33KV.

ii. If any one disc is damaged, the whole string does not become

useless because the damaged disc can be replaced.

iii. Suspension arrangement provides greater flexibility to the line.

iv. Suspension type insulators are generally used with steel towers.

v. Each unit or disc of suspension type insulator is designed for

low voltage usually 11KV.

3) SHACKLE INSULATORS

Shackle insulators are frequently used for low voltage distribution lines. They can

be used either in a horizontal position or in a vertical position. They can be

directly fixed to the pole with a bolt or to the cross arm.