UNESCO
UNESCO-NIGERIA TECHNICAL & VOCATIONAL
EDUCATION REVITALISATION PROJECT
PROJECT-PHASE II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERI
ENGINEERING
NG TECHNOLOGY
ELECTRICAL POWER SYSTEM (I)
COURSE CODE: EEC 122
YEAR I- SEMESTER II
THEORY
Version 1: December 2008
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:
2.1
Principle of distribution system.........................................................26
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
Solar Cell Array
Subsystem
DC to AC
Conversion ( )
Battery for Storage
Control unit
Power Distribution
Unit
Figure 1.12
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
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%
Week 4
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
Generation of Electrical Energy
Week 5
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)
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 Xsectional 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 crosssection.
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
I1
D
I2
E
B
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
I1
I2
E
B
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
I1
I2
B
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:
feeder
I1
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
i
i
A
i
C
x
B
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 – l2/2) = 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 crosssectional area of the distributor. Take ρ = 1.78 x 10-8Ωm.
Solution:
Total voltage drop over the distributor is
V = i 1R 1 + i 2R 2 + i 3R 3 + …
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
B
R2 + jX2
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 2001000
Z AC = (0.04 + j0.02) Ω
Impedance of section CB, Z CB = (0.2 + j0.1) x 1001000
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
PORCELAIN
FIXING HOLE
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
30A RED
60a PURPLE
15A-BLUE
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 shortcircuit 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
PORCELAIN
BODY
SILVER
END CAP
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.