Eec 122 Power System i Thoery

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< 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.