Outcome 1 : Power Supply Network.
Electrical ac power systems consist of three-phase generation systems, transmission, distribution networks, and loads. The networks supply large three-phase industrial loads at various distribution and transmission voltages as well as single phase residential and commercial loads.
Generated voltages are up to 35kV for generators used in large electrical power stations. Primary Transmission voltages are High voltages typically 200-400kV, Secondary transmission voltages are typically 132–66kV. Primary Distribution voltages are typically 33–11kV and Secondary
Distribution voltages are typically 400V.*
*The general structure and values mentioned are indicative only, it varies depending on the requirement of specific power system network
Electrical Power Systems by D. Das, Publisher : New Age International Ltd
Fig:1.1 Schematic Diagram of Power Supply Network*
*The general structure and values in the above picture is indicative only, it varies depending on the requirement of specific power system network
Electrical Power Systems by D. Das, Publisher : New Age International Ltd
Fig:1.2 Single Line Diagram of Power Supply Network* *Emphasize on the symbols of Generating stations, Transformers, Circuit Breaker, Substation and Load
Electric power supply based on the type of power transmission may be classified as 1. AC system
2. DC system
Electric power supply network based on the method of carrying power from one point to another may be classified as
1. Overhead system
2. Underground system
Generation and Transmission of Electric power adopts 3 phase 3 wire system
Distribution of Electric power is done by 3 phase 3 wire to industrial loads and 3 phase 4 wire to small consumer loads.
How Electrical Power is Generated?
The working method of most Electrical power plant is by employing a method to rotate a turbine which in turn is connected to an alternator.
Alternator is a synchronous machine which is three-phase machine, able to exchange power with a three-phase network.
The vast majority of power stations generate electricity in three-phase form, using synchronous machines.
The most common fuel is fossil fuel viz., Coal, petrol, diesel, water, gas, Uranium 235.
Outcome2: Conventional and Non Conventional Power Plants.
Power can be generated from Conventional and Non-Conventional Sources Examples for Conventional Sources of Energy are
1. Hydro Electric Power Plant 3. Thermal Power Plant
Examples for Non Conventional Sources of Energy are
5. Tidal Power
7. Geothermal Power 9. Fuel Cell
2. Nuclear Power Plant 4. Diesel Power Plant
6. Wind Power
8. Magneto Hydro Dynamic (MHD) 10. Solar Power
Conventional Energy Resources (1-4)
1. Hydro Electric Power Plant
Water stored at a certain height has potential energy which is converted into kinetic energy and this kinetic energy is converted to the mechanical energy by allowing the water to flow through the hydraulic turbine runner. This mechanical energy is utilized to run an electric generator which is coupled to the turbine shaft.
Hydro projects are developed for the following purposes:
1. To control the floods in the rivers.
2. Generation of power.
3. Storage of irrigation water.
4. Storage of the drinking water supply.
Fig 2.1 Hydro Electric Power Plant
Selection of site for Hydro Electric Power Plant
Essential characteristics of a good site are: large catchment areas, high average rainfall and a favorable place for constructing the storage or reservoir.
The following factors should be given careful consideration while selecting a site for a hydro-electric power plant:
1. Availability of water
2. Facility for storing water
3. Head available
4. Distance from Load centres
5. Access to site
The essential features of a Hydro power plant are as below:
1. Catchment area (The catchment area of a hydro plant is the whole area behind the dam)
2. Reservoir (Whole of the water available from the catchment area is collected in a reservoir behind the dam) 3. Dam (A dam is built across a river for storing water)
4. Surge Tank (The use of the surge tank is to avoid water hammer in the penstock)
5. Penstock (Water passage which leads water from tunnel to power house)
6. Valve House (To control the water flow in the penstocks. Valves are electrically driven)
7. Power house (The power house is a building in which the turbines, alternators are housed)
8. Tail race (Passage for discharging the water leaving the turbines, into the river)
Classification of Hydro Electric Power Plant
The hydro-power plants can be classified as : l. Storage plant
(a) High head plants (about 300 m and above) (b) Medium head plants (about 50 to 300 m) (c) Low head plants. (upto about 50 m)
2. Run-of-river power plants
(a) With storage
(b) Without storage.
3. Pumped storage power Plants.
2. Nuclear Power Plant
Nuclear reactions of importance in energy production are fusion, fission, and radioactivity.
Fuel of a nuclear reactor should be fissionable material which can be defined as an element whose nuclei can be caused to undergo nuclear fission by nuclear bombardment and to produce a fission chain reaction.
Natural uranium found in earth crust contains three isotopes namely U234, U235 and U238.
U235 is most unstable and is capable of sustaining chain reaction and has been given the name as primary fuel. U233 and Pu239 are used as secondary fuels.
Fig 2.2 Nuclear Power Plant
Advantages of Nuclear Power Plant
1. Space requirement of nuclear power plant is less compared to conventional power plants.
2. A nuclear power plant consumes very small quantity of fuel.
3. Nuclear power plants are not effected by weather conditions.
4. It does not require large quantity of water.
Disadvantages of Nuclear Power Plant
1. Initial cost of nuclear power plant is higher.
2. Radioactive wastes should be disposed carefully.
3. Maintenance cost of the plant is high.
Selection of site for Nuclear Power Plant
1. Availability of water.
2. Distance from load center
3. Distance from populated area. 4. Accessibility to site.
5. Waste disposal.
3. Thermal Power Plant (Steam Power Plant)
Steam is used to drive steam engines, steam turbines etc. Steam power station is most suitable where coal is available in abundance. Thermal electrical power generation is one of the major method.
A steam power plant must have following equipment's
1. A furnace to burn the fuel.
2. Steam generator or boiler - Heat generated in furnace is used to convert water to steam. 3. Main power unit - Engine or turbine to use the heat energy of steam and perform work. 4. Piping system to convey steam and water.
Fig 2.3a Thermal Power Plant
Fig 2.3b Thermal Power Plant
4. Diesel Power Plant
The oil engines and gas engines are called Internal Combustion Engines. In IC engines fuels burn inside the engine and the products of combustion form the working fluid that generates mechanical power. Whereas, in Gas Turbines the combustion occurs in another chamber and hot working fluid containing thermal energy is admitted in turbine.
Operating principle of diesel power plant can be stated as
1. Cylinder is charged (Fuel filling)
2. Cylinder contents (Fuel) are compressed
3. Combustion (Burning) creates high pressure pushing the piston, thereby converting linear motion to rotatory motion. This rotatory motion will rotate alternator generating power.
4. Exhaust of spent products of combustion is released to atmosphere.
Fig 2.4 Diesel Power Plant
Advantages of Diesel Power Plants
1. Very simple design and simple installation.
2. Limited cooling water requirement.
3. Standby losses are less as compared to other Power plants.
4. Low fuel cost.
Disadvantages of Diesel Power Plants
1. High Maintenance and operating cost.
2. The plant cost per kW is comparatively more.
3. The life of diesel power plant is small due to high maintenance. 4. Noise is a serious problem in diesel power plant.
Non Conventional Energy Resources (5-10)
5. Tidal Power
Tidal systems would use the rise and fall of the water along a coastal area as a source of energy for producing electrical power.
6. Wind Power
Wind systems have also been considered for producing electrical energy. However, winds are variable in most parts of our country. This fact causes wind systems to be confined to being used with storage systems, such as batteries. However, large amounts of power would be difficult to produce by this method.
7. Geothermal Power
These systems utilize the heat of molten masses of material in the interior of the earth. Thus, heat from the earth is a potential source of energy for power generation. The principle of geothermal systems is similar to other steam turbine-driven systems.
8. Magneto Hydro Dynamics
The operation of an MHD system relies upon the flow of a conductive gas through a magnetic field, thus causing a direct current (DC) voltage to be generated. The electrical power developed depends upon the strength of the magnetic field that surrounds the conductive gas, and on the speed and conductivity of the gas.
9. Fuel Cell
This type of cell converts the chemical energy of fuels into direct current electrical energy. The chemical reactions of the electrodes with the electrolyte release electrons to an external circuit.
10. Solar Power
The largest energy source available today is the sun, which supplies practically limitless energy. The energy available from the sun far exceeds any foreseeable future need. Solar Cells directly converts solar energy to Electrical Energy.
Possible solar-energy systems include home heating, orbiting space systems, and steam driven electrical power systems. Each of these systems uses solar collectors that concentrate the light of the sun so that a large quantity of heat will be produced. Potentially, this heat could be used to drive a steam turbine in order to generate additional electrical energy.
Fundamental of Transmission of Electrical Power, Under ground HT and LT Power Cables, Over Head line conductors, Efficiency of transmission line, the line drop, types of transmission circuits
Fundamentals of transmission of electrical Power:
The purpose of the electric transmission system is interconnection of power plants or generating stations with the loads. RMS voltage between the conductors is called line-to-line voltage and RMS voltage between conductor and ground is called Phase Voltage.
Operating Frequencies: 50Hz (Followed by majority countries) and 60Hz.
The location of overhead transmission lines is limited by zoning laws (ex:-safety regulation in some areas may demand an underground transmission line) and by populated areas, highways, railroads, and waterways, as well as other topographical and environmental factors. Power transmission lines ordinarily operate at voltage levels from 12 kV to 500 kV for AC.
Underground HT and LT Power Cables
All electric cables consist of three essential points.
(a)The conductor (For transmitting electrical power)
(b)The insulation (To insulate the conductor from direct contact with earth or other objects)
(c)External protection (To prevent from mechanical damage, chemical or electro-chemical attack, fire) Underground Transmission and Distribution:
The use of underground cable is ordinarily confined to the short lengths required in congested urban areas. The cost of underground cable is much more than that of overhead conductor. Cryogenic or superconductor cables (Cooled conductors) have more power handling capability.
Insulating materials used for cable should posses the following properties
1. High insulation resistance. (To resist the flow of current through the insulation)
2. High dielectric strength. (To make the breakdown voltage high)
3. Good mechanical properties (Ex:- tenacity (for gripping the cable) and elasticity (flexibility to work)) 4. It should not be affected by chemicals around it.
5. It should be non-hygroscopic.(The electrical property of cable degrades with the amount of moisture)
Construction of cables:
Fig: 3.1 Figure 3.2 Schematic diagram of 3 conductor 3 phase Cable (a) Solid conductor
(b) Stranded conductor
(c) Cross section of stranded conductor
The different layers of Cables are
1. Conductors or Cores :
A cable may have one or more than one core (conductor) depending upon the type of service for which it is intended. The conductors are made of tinned copper or aluminum and are usually stranded in order to provide flexibility to the cable.
Each core or conductor is provided with a suitable thickness of insulation, the thickness of layer depending upon the voltage to be withstood by the cable. The commonly used materials for insulation are impregnated paper, varnished cambric or rubber mineral compound.
3. Metallic sheath:
Protect the cable from moisture, gases or other damaging liquids (acids or alkalies) in the soil and atmosphere, a metallic sheath of lead or aluminum is provided over the insulation.
Over the metallic sheath is applied a layer of bedding which consists of a fibrous material like jute or hessian tape. The purpose of bedding is to protect the metallic sheath against corrosion and from mechanical injury due to armouring.
Over the bedding, armouring is provided which consists of one or two layers of galvanized steel wire or steel tape. Its purpose is to protect the cable from mechanical injury while laying it and during the course of handling.
In order to protect armouring from atmospheric conditions, a layer of fibrous material (like jute) similar to bedding is provided over the armouring.
Classification of Cables (Voltage level based)
(i) Low-tension (L.T.) cables — upto 1000 V
(ii) High-tension (H.T.) cables — upto 11,000 V
(iii) Super-tension (S.T.) cables — from 22 kV to 33 kV
(iv) Extra high-tension (E.H.T.) cables — from 33 kV to 66 kV (v) Extra super voltage cables — beyond 132 kV
Types of Cables Single core low tension cable
Figure 3.3 Schematic diagram of single core low tension cable
Cables for 3-Phase Service
1. Belted cables - upto 11 kV
2. Screened cables - from 22 kV to 66 kV 3. Pressure cables - beyond 66 kV.
Three Phase cables
Figure 3.4 Schematic diagram of 3-phase belted cable
Figure 3.5 Schematic diagram of 3-phase screened cable
Oil filled cables
Gas Pressure cables
Figure 3.7 Schematic diagram of three phase gas pressure cable
Figure 3.6 Schematic diagram of three phase Oil filled cable
Overhead Line Conductors:
AC overhead transmission voltages have increased to levels in the range of 765 kV, with research now dealing with voltages of over 1000 kV. Overhead conductors can dissipate heat better.
The portions of the electrical distribution system that carry current are known as conductors. Conductors may be in the form of solid or stranded wires, cable assemblies, or large metallic bus- bar systems. A conductor may have insulation, or, in some cases, it may be bare metal. Conductor Characteristics:
Majority of the conductors are made of either copper or aluminum as they possess necessary flexibility, current-carrying ability, and are economical. Copper is a better conductor; however, aluminum is 30 percent lighter in weight. Aluminum-conductor steel-reinforced (ACSR) type is used for long-distance power transmission. This type of conductor has stranded aluminum wires.
The major components of a High Voltage Transmission Line
High-voltage and Extra high-voltage (EHV) transmission lines interconnect power plants and loads, and form an electric network. The major components are
1. Tower (The supporting structure)
2. Insulator (To insulate the live conductor from the tower)
3. Conductor (Each conductor is stranded, steel-reinforced aluminum cable)
4. Foundation and grounding (Steel-reinforced concrete foundation & grounding electrodes placed in ground)
5. Shield conductors: (Two grounded shield conductors protect the phase conductors from lightning)
Overhead lines Versus Underground cables:
The inductance is more predominant.
The capacitance is more predominant.
Used for long distance transmission.
Large charging current on very high voltage cables limits the cables to use for long length.
Conductors used are less expensive.
Conductors used are more expensive.
The size of the conductor is small because of the better heat dissipation.
The size of the conductor for the same power is large.
The insulation cost is less.
The insulation cost is more.
The erection cost is less.
The erection cost is more.
Less safety and more interference.
Underground cables give greater safety to the public, less interference with amenities and better outlook to the city.
Transmission line Performance parameters: The performance parameters of transmission line are determination of efficiency and regulation of lines.
for 1-phase, Loss = I2R
for 3-phase phase, Loss = 3I2R
When a transmission line is carrying current, there is a voltage drop in the line due to resistance and inductance of the line. The result is that receiving end voltage (VR) of the line is generally less than the sending end voltage (VS). This voltage drop (VS - VR) in the line is expressed as a percentage of receiving end voltage VR and is called voltage regulation.
% regulation =
1. Short transmission line:
Types of transmission Circuits:
When the length of the line is less than 80KM or voltage level is less than 66KV, it is usually considered as short transmission line. Capacitance may be ignored without much error.
Figure 3.8 Short transmission line (a) Equivalent circuit, (b) Phasor diagram
From the phasor diagram
VS cosΦS = Vr cosΦr+ Ir R --------- (1)
VS sinΦS = Vr sinΦr+ Ir X ---------- (2)
Squaring and adding the above two equations and simplifying , we get Sending end voltage is given by,
VS=Vr +IrRcosΦr+IrXsinΦr Sending end power factor is given by,
Voltage regulation is given by
Efficiency is given by,
2. Medium transmission line:
When the length of the line is more than 80 KM and below 250 KM, it is usually considered as medium transmission line. Line charging currents becomes appreciable and shunt capacitance must be considered. For medium transmission line, half of the shunt capacitance may be considered to be lumped at each end of the line. This is referred to as nominal Π network as shown.
Figure 3.9 Medium transmission line nominal Π network
3. Long transmission line:
When the length of the line is more than 250 KM, it is usually considered as long transmission line.
1. A load of 1000 kW at 0•8 power factor lagging is received at the end of a 3-phase line 20 km long. The resistance and reactance of conductor are 0•25 Ω per km and 0•28 Ω per km. If the receiving end line voltage is maintained at 11 kV, calculate:
(i) Sending end voltage (line-to-line) (ii) percentage regulation and (iii) Transmission efficiency
2. A single phase overhead transmission line delivers 1100 kW at 33 kV at 0•8 p.f. lagging. The total resistance and inductive reactance of the line are 10 Ω and 15 Ω respectively. Determine:
(i) sending end voltage (ii) sending end power factor and (iii) transmission efficiency.
3. An overhead 3-phase transmission line delivers 5000 kW at 22 kV at 0•8 p.f. lagging. The resistance and reactance of each conductor is 4 Ω and 6 Ω respectively. Determine:
(i) sending end voltage (ii) percentage regulation (iii) transmission efficiency.
Electrical design of O/H lines : Mechanical Design of over head lines : Sag calculation
Electrical design of overhead lines
a) The selection of the most appropriate conductor size must take into account both technical and economic criteria as listed below:
1. The conductor cross-sectional area should be such as to minimize the initial capital cost.
2. Conductor should conform to standard sizes already used.
3. The conductor thermal capacity must be adequate.
b) When the electric field intensity reaches the critical value of about 30 kV/cm, the air in the immediate vicinity of conductors no more remains a dielectric and at this intensity, the air becomes conducting, hence a complete electric breakdown occurs and an arc is established. This phenomenon is named corona.
Mechanical design of overhead lines:
The energized conductors of transmission and distribution lines must be placed to totally eliminate the possibility of injury to people. Overhead conductors, however, elongate with time, temperature, and tension, thereby changing their original positions after installation.
Despite the effects of weather and loading on a line, the conductors must remain at safe distances from buildings, objects, and people or vehicles passing beneath the line at all times.
The mechanical design of overhead lines is determined by
1. Minimum distances of the conductors to earth, to other conductors and to the tower
2. Minimum clearances and minimum length of insulators,
3. Sag, taking account of conductor and ambient conditions
4. Tensile stress of the conductor
5. Wind forces, Ice loading and temperature on the conductor, insulators and towers.
The level differences between the supporting structures affect the shape of the conductor catenary. The catenary’s shape affect sag and tension of the conductor, and therefore, must be determined using mathematical equations. Sag and Tension are inversely related.
Supporting structures may be at same level or at different levels. The shape of a catenary is a function of conductor weight per unit length ( w (kg/m) ), horizontal component of tension ( T (Kg) ) which acts tangentially is same at any point on the catenary , span length ( l (m) ) , and the maximum sag of the conductor ( S (m) ).
Figure 4a: Towers at level support Figure 4b: Towers at different level supports
The sag of the conductor depends on: 1. Conductor unit weight
2. Span Length
3. Tension in the conductor
4. Weather Conditions (effect of wind, ice) 5. Temperature
6. Elasticity of the conductor materials
For a towers at level supports, the lowest point is in the center, and the sag S, is found by
S = 𝑤𝑙2 8𝑇
For a level span, the total conductor length required between level supports (L) can be found by
𝐿 = 𝑙 + 8𝑆2 3𝑙
Tension in a conductor 𝑇 = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡h 𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟
Ex 1:- A transmission line conductor has been suspended freely from two towers and has taken the form of a catenary that has ultimate strength of 1200Kg with a safety factor of 2. The distance between the two towers is 152 m, and the weight of the conductor is 1160 kg/km. Calculate the following: ( a ) Sag and ( b ) Length of the conductor
Ex 2:- A 132 KV transmission line has the following data
Weight of conductor = 680 kg/km : length of span = 260 m
Ultimate strength = 3100kg : safety factor = 2
Calculate the height above ground at which the conductor should be supported. Ground clearance required is 10 m.
Ice and Wind Loading in Conductors
When a conductor is covered with ice or is exposed to wind, the effective conductor weight per unit length increases. During occasions of heavy ice or wind load, the conductor catenary tension increases dramatically along with the loads on angle and structures. Both the conductor and its supports can fail unless these high-tension conditions are considered in the line design.
Weight of ice acts vertically downwards. (wi)
Force due to wind is assumed to act horizontally. (ww)
Total weight of conductor per unit length = Wt= √((𝑤 + 𝑤 )2+ 𝑤2 ) 𝑖𝑤
Ex 3:- A transmission line has a span of 150m between level supports. The conductor has a weight of 1.98kg. The weight additional due to ice loading is 0.6kg. Weight due to effect of wind is 1.2kg. (All weights given are for unit length).The tension in the conductor is 2100 Kg. If the height of the tower is 12.5m. Find a) the sag and b) the clearance from ground
Calculation of Resistance, Inductance and Capacitance of transmission lines
An overhead transmission line consists of a group of conductors running parallel and can be represented with parameters such as resistance(R), inductance(L), capacitance(C) and shunt conductance(G).
The series resistance causes a real power loss in the conductor.
Power transmission capacity of the transmission line is mainly governed by series inductance.
The capacitance together with conductance forms the shunt admittance of a transmission line. The shunt conductance accounts for leakage currents flowing across insulators and ionized
pathways in the air.(Leakage current flows even when the circuit is open). The leakage currents
are negligible as compared to the current flowing in the transmission lines.
These parameters are uniformly distributed throughout but can be lumped for the purpose of
analysis on approximate basis.
Line Resistance Calculation:
The dc resistance of a solid round conductor is given by R = 𝜌×𝑙
Where ρ = Resistivity of conductor l = Length of conductor
A = Area of cross section of conductor
The dc resistance of a stranded conductor is greater than the value given by above equation, because spiraling of the strands makes them longer than the conductor itself.
Problem:- Determine Resistance of a 180m long wire, If the resistivity is 2.65×10-8 Ωm, Area of cross section of the conductor is 4mm2.
The conductor resistance increases with the increase of temperature. For small changes in temperature, the resistance increases linearly as temperature increases and the resistance at a temperature T is given by
RT=R0 (1+α0T) α0 = Temperature coefficient of resistance at 00 Celsius
Where RT = Resistance at T0 Celsius R0 = Resistance at 00 Celsius
Problem:- Resistance of a wire at 0° Celsius is measured to be 15Ω. If the temperature increased to 35° Celsius, determine the change of resistance for the wire. Assuming the use of copper wire (α0 = 0.004041)
Inductance of a single phase two-wire (1-phase) transmission line:
Inductance is the flux linkage per ampere. There are two flux linkages: (i) due to internal flux, and (ii) due to external flux.
Loop Inductance Lab = 4×10-7 ( ln e1/4 + ln 𝐷 )
Where, D = distance between conductors & R = Radius of the conductor. R’ = 0.7788R
The multiplying factor of 0.7788 to adjust the radius in order to account for internal flux linkages.
Lab = 4×10-7 ( ln 𝐷 𝑅×𝑒−1
Lab = 4×10-7 ln 𝑫 𝑹′
Inductance of three phase transmission line: (i) Symmetrical spacing:
For symmetrical spacing a=b=c=D
La = 2×10-7 ln 𝐷 H/m 𝑅′
(Inductance of a conductor)
(ii) Unsymmetrical spacing:
La = 2×10-7 ln 𝐷′ H/m
(Inductance of a conductor) Unsymmetrical placing of wires on a horizontal plane
1. A single phase line has two parallel conductors 2m apart. The diameter of each conductor is 1.2 cm. Calculate the loop inductance / phase / km of the line.
2. A single phase line has two parallel conductors 3 m apart. The radius of each conductor is 1 cm. Calculate the loop inductance / phase / km of the line.
3. Find the inductance / phase / km of a three phase transmission line using 1.24 cm diameter conductors when they are placed at the corners of an equilateral triangle of each side 2m.
4. The three conductors of a three phase line are arranged at the corners of a triangle of sides 2m , 2.5 m and 4.5 m. Calculate the inductance / phase / km of the line . The diameter of each conductor is 1.24 cm.
5. Calculate the inductance / phase / km in a three phase, three wire system when the conductors are arranged in a horizontal plane with spacing such that d13 = 4m; d12=d23= 2m. The diameter of each conductor is 2.5 cm.
Capacitance of a 1-phase transmission line:
Let V be the voltage applied between the conductors so that the charge
per unit length of each conductor is 𝜌L coulomb per meter. The length
of the line is very large as compared with the distance of separation D of the conductors, and radius R of each conductor is very small as compared to the distance of separation. Capacitance between the wires is
C =𝜌L=𝜋𝜀0F/m ab 𝑉 𝑙𝑛𝐷
The capacitance of one conductor with respect to the neutral plane is two times the capacitance of the single-phase line
C = 2C = 2𝜋𝜀0 F/m an ab 𝑙𝑛𝐷
Capacitance of a 3-phase transmission line: (i) Symmetrical spacing:
For symmetrical spacing a=b=c=D
C = 2𝜋𝜀0 F/m an 𝑙𝑛𝐷
(ii) Unsymmetrical spacing:
ForunsymmetricalspacingD’=3 𝑎×𝑏×𝑐 C = 2𝜋𝜀0 F/m
Charging current of a line per phase = 𝑉 𝑋𝐶𝑎𝑏
Where, Cab = line-to-line Capacitance Xc= Capacitive reactance in ohms V = line voltage in volts
an 𝑙𝑛𝐷′ 𝑅
Unsymmetrical placing of wires on a horizontal plane
1. A single phase line has two parallel conductors 3m apart. The radius of each conductor is 1cm. Calculate the capacitance / phase / km of the line.
2. Find the capacitance / phase / km of a three phase transmission line using 1.25 cm diameter conductors when these are placed at the corners of an equilateral triangle of each side 2m.
3. The three conductors of a three phase line are arranged at the corners of a triangle of sides 2m , 2.5 m and 4.5 m. Calculate the capacitance / phase / km and capacitance between lines / km. The diameter of each conductor is 1.24 cm.
4. A three phase 50 Hz, 66KV overhead line conductors are placed in a horizontal plane. The distance between lines are 2m , 2.5 m and 4.5 m respectively. If the length of the line is 100km and the diameter of each conductor is 1.25 cm . Calculate i) capacitance / phase / m ii) charging current per phase .
5. A three phase 50 Hz, 132 KV overhead line conductors are placed 4m apart in a horizontal plane. The length of the line is 100km. The diameter of each conductor is 2 cm . Calculate i) capacitance / phase / km ii) charging current per phase.
O/H lines insulators:
Types of insulators
The insulators for overhead lines provide insulation to the power conductor from the ground. These insulators are mainly made of either glazed porcelain or toughened glass. The materials used for porcelain are silica -20%, feldspar-30% and clay 50%.
Types of insulators used are
1. Pin type
2. Suspension type (Disc Insulator) 3. Strain type (Disc Insulator)
4. Shackle type
5. Post type
I. Pin type insulators :
• The pin type insulators are normally used up to 33 kV.
• The insulators and its pin should be sufficiently mechanically strong to withstand the resultant force due to combined effect of the weight of the conductor, wind pressure and ice loading if any per span length.
• The pin type of insulators are uneconomical beyond 33 kV operating voltage. Also the replacement of these insulators is expensive.
II. Suspension insulators :
• For insulating overhead lines against higher voltages beyond 33 kV, suspension insulators are used.
• These insulators consist of one or more insulator units flexibly connected together and adapted to be hung for the cross arm of the supporting structure and to carry a power conductor at its lowest extremity. Such composite units are known as string insulators.
• The cap at the top is recessed so that it can take the pin of another unit and in this way a string of any required number of insulators can be built. The cap and the pin are secured to the insulator by means of cement. The standard unit is 10′′ × 5 3/4 ′′ in size.
• Suspension insulators being free to swing, the clearances required between the power conductor and the supporting structure are more as compared to pin type insulators. This means the length of the cross arm for suspension insulators is more as compared with the pin type.
• The suspension insulators, in addition to being economical as compared to pin type for voltages more than 33 kV, have the following further advantages:
1. Each insulator is designed for 11 kV and hence for any operating voltage a string of insulators can be used. For example, for 132 kV transmission, the number of insulators required is 12.
2. In case of failure of one of the units in the string, only that particular unit needs replacement rather than the whole string.
III. Strain insulators :
• The strain insulators are exactly identical in shape with the suspension insulators. These strings are placed in the horizontal plane rather than the vertical plane as is done in case of suspension insulators (discs are in vertical plane in case of string insulators).
• These are used to take the tension of the conductors at line terminals, at angle towers, at road crossings, long river crossing and at junction of overhead lines with cables. These insulators are, therefore, known as tension or strain insulators.
• For low voltages of the order of 11 kV, shackle insulators are used. But for higher voltages a string of insulators is used.
IV. Shackle Insulators:
• In earlier days it was used as strain insulators.
• Shackle or spool type insulator is usually used on L.T lines.
• It can be directly fixed to the pole with a bolt or to the cross arm.
• May be used either in the horizontal (or) vertical position.
• The conductor in the groove is fixed with a soft binding wire.
Both low voltage conductors and the house wires are attached to shackle insulator.
The tapered hole of this insulator distributes the load more evenly reduces the possibility of breakage when heavily loaded.
V. Post Insulators:
• Post-type insulators are used for medium and low-voltage transmission lines, where insulators replace the cross-arm. However, the majority of post insulators are used in substations where insulators support conductors, bus bars, and equipment.
Potential distribution over string of suspension Insulator
• The voltage across the discs of the string is not uniformly distributed. This is because of the capacitance's formed between the metal parts of the insulators and the tower structure.
• These capacitance's could be made negligibly small by increasing the distance between the insulators and the tower structure which requires larger lengths of cross arms. This will result into bigger size of the towers and hence uneconomical.
• In practice the insulators are not very far from the tower structure and hence these capacitances affect the voltage distribution across the string. The capacitance of each unit is known as mutual capacitance. Fig. represents an equivalent circuit for a string of 3 insulator discs. C is self capacitance and kC is shunt capacitance.𝐼𝑛and 𝑖𝑛 are charging currents.
• Here capacitance to ground is the capacitance of metal part of the insulator disc to the tower structure. Since the insulator discs are identical, each disc is represented by self capacitance C.
• Let V be the operating line voltage and V1, V2, and V3 the voltage drops across the units starting from the cross arm towards the power conductor. 𝑉 = V1 + V2 + V3. voltage across each disc
as a multiple of the operating voltage and to compare the voltage across each unit. Its shows that V1 < V2 < V3.
• This means the voltage drop across the unit nearest the cross arm is minimum and the voltage drop across the unit nearest the power conductor is maximum.
• It is clear that the lowermost unit in a string of insulators is fully stressed or utilized. As we go towards the cross arm the units are less stressed as compared to their capacity and hence they are not utilized fully. String efficiency is a measure of the utilization of material in the string.
𝑘 = 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑒𝑎𝑟𝑡h = 𝑘𝐶 𝑆𝑒𝑙𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝐶
𝐼2 = 𝐼1 + 𝑖1
𝑉2 =𝑉1+𝑉1 111
2𝛱𝑓𝐶 2𝛱𝑓𝐶 2𝛱𝑓𝑘𝐶
2𝛱𝑓𝐶𝑉 = 2𝛱𝑓𝐶𝑉 +2𝛱𝑓𝑘𝐶𝑉 211
𝑉 = 𝑉 (1+k) 21
Similarly𝑉 = 𝑉 1+3𝑘+𝑘2 31
and𝑉 = 𝑉 +𝑉 + 𝑉 =𝑉 +𝑉(1+k)+𝑉 1+3𝑘+𝑘2 =𝑉 (3+4k+𝑘2) 𝑝h123111 1
Methods of improving string efficiency
1. Selection of k:
One of the methods for equalising the potential drop across the various units of the string is to have a smaller value of k which needs longer cross arms and hence taller supporting structures and hence it is uneconomical to go beyond certain value of the length of cross arm. It has been found that the value of k = 1/10 is about the minimum which is normally obtained.
2. Grading of Units:
Unequal distribution of voltage is due to the leakage current from the insulator pin to the tower structure. This current can’t be eliminated. The other possibility is that disc of different capacities could be used. This means that in order to carry out unit grading, units of different capacities are required. This requires large stocks of different sized units, which is uneconomical and impractical. Therefore, this method is normally not used except for very high voltage lines.
3. Static Shielding:
In static shielding the idea is to cancel exactly the pin to tower charging currents so that the same current flows through the units of identical capacities to produce equal voltage drops across each unit. In this method a guard ring or grading ring is connected round to the power conductor such that this surrounds the bottom unit.
1. In a three phase overhead system , each line is suspended by a string of three insulator. The voltages across top unit and the middle unit are 8KV and 11KV respectively. Calculate a) ratio of shunt capacitance to self capacitance b) line voltage c) string efficiency
2. In a 33KV overhead line, there are 3 units in the string of insulators. If the capacitance between each insulator to pin and earth is 11% of self capacitance of each insulator. Find distribution of voltage over three insulators and string efficiency
3. Each line of a three phase system is suspended by a string of three similar insulators. If the voltage across the line unit is 17.5 KV, calculate the line to neutral voltage. Assume that the shunt capacitance between each insulator and earth is 1/8th of the capacitance of the insulator itself. Also find the string efficiency
4. The three bus bar conductors in an outdoor substation are supported by units of post type insulators . each unit consists of a stack of 3 post insulators fixed one on the top of the other. The voltage across the top insulator is 13.1 KV and that across the next unit is 11KV . find the bus bar voltage of the station.
Corona : power losses due to Corona in
Corona phenomenon is the ionization of air surrounding the power conductor. Free electrons are normally present in free space because of radioactivity and cosmic rays. As the potential between the conductors is increased, the gradient around the surface of the conductor increases.
When the electric field intensity or potential gradient reaches the critical value of about 30 kV/cm, the air in the immediate vicinity of conductors no more remains a dielectric and at this intensity, the air becomes conducting, hence a complete electric breakdown occurs and arc is established between the two electrodes.
This phenomenon is called Corona.
Main Effects of Corona Discharges on Overhead Lines
1. Corona Loss (Generally negligible under fair-weather conditions but can reach values of several kilowatts per kilometer of line during foul weather.)
2. Electromagnetic Interference (Television and radio Interference) (The frequency spectrum of corona discharges has frequencies around a few tens of megahertz. As a result, the interference levels at the television and radio frequencies are attenuated.)
3. Audible Noise (A pulsating sound wave is generated from the discharge site, propagates through the surrounding ambient air, and is perfectly audible in the immediate vicinity of the HV lines.)
Factors Affecting Corona Loss
The following are the factors that affect corona loss on overhead transmission lines:
(i) Electrical factors (Frequency and waveform of supply) (ii) Atmospheric factors (Pressure and temperature effect)
(iii) Factors connected with the conductors (Diameter of the Conductor, Surface Conditions of the Conductors)
Methods of Reducing Corona Loss
These losses can be reduced by using
(i) large diameter conductors, (ii) hollow conductors, and (iii) bundled conductors.
Voltage regulation in Power systems
(Important methods of voltage regulations in power systems)
When power is supplied to a load through a transmission line keeping the sending end voltage constant, the receiving end or load voltage undergoes variations depending upon the magnitude of the load and the power factor of the load. The higher the load with smaller power factor the greater is the voltage variation.
Practically all the equipment's on the power systems are designed to operate satisfactorily only when the voltage levels on the system correspond to their rated voltages or at the most the variations are within 5%.
Thus the necessity of controlling the voltage on the system is very much strong.
The methods for voltage control are
(i) Shunt capacitors (Shunt capacitors are used across an inductive and capacitive loads so as to supply part of the reactive VARs required by the load)
(ii) Series capacitors (Static capacitor connected in series with the line, it reduces the inductive reactance between the load and the supply point)
(iii) Synchronous capacitors (Synchronous capacitor supplies VARs when over-excited, i.e. during peak load conditions and it consumes VARs when under-excited during light load conditions.)
(iv) Tap changing transformers (The tap changing transformers control the voltage by changing the transformation ratio, the voltage in the secondary circuit is varied and voltage control is obtained. This method is the most popular as it can be
used for controlling voltages at all levels)
(v) Booster transformers. (Used for boosting the voltage only. Can be used at any point in the system)
Different bus-bar systems, Generating stations and substation earthing.
When a number of generators or feeders operating at the same voltage have to be directly connected electrically, bus-bars are used. Bus-bars are copper rods or thin walled tubes and operate at constant voltage.
Bus Bar Classification
1. Single Bus Bar System
The single bus-bar system has the simplest design and is used for
power stations. It is also used in small outdoor stations having relatively
few outgoing or incoming feeders and lines.
The chief advantages of this type of arrangement are low initial cost, less maintenance and simple operation.
2. Single Bus Bar System with Sectionalisation
In large generating stations where several units are installed, it is a common practice to sectionalize the bus so that fault on any section of the bus-bar will not cause complete shut down.
Here bus bar is divided into two sections connected by a circuit breaker and isolators. If a fault occurs on any section of the bus-bar, that section can be isolated without affecting the supply to other sections.
3. Duplicate Bus Bar System
In large generating stations breakdown and maintenance should not affect the continuity of supply. For that Main bus bar and a spare bus bar is used. Each generator and feeder may be connected to either bus bar with the help of circuit breaker and isolator.
3. Earth Resistance calculation of Substation and Generating station
The purpose of a earthing system is to ensure safe conditions for personnel and plant in and around the site during normal and earth fault conditions. An earth grid with minimum resistance can provide a passage for fault current to flow to ground.
A simplified formula for approximating the resistance, R ohms, of a earth grid is:
ρ=soil resistivity (Ωm),
r=equivalent circular plate radius (m) L=total length of buried conductor (m)
𝑅=ρ1+1 4𝑟 𝐿
Example 1:-Find the Resistance of earthing grid made-up of sand and graval with resistivity 2000 Ωm, Grid area 1408𝑚2, buried length of conductor 950m.(IEEE80 standard data)
Example 2:-Find the Resistance of earthing grid made-up of sand and clay with resistivity 200 Ωm, Grid area 1750𝑚2, buried length of conductor 540m. .(IEEE80 standard data)
Basics ideas of AC DC power Transmission
The terminal converter stations of a DC scheme are more expensive than AC substations but DC transmission using overhead line is cheaper after a particular distance. The choice of whether to use AC or DC is usually made on cost. With the increasing use of high-voltage, high-current semiconductor devices, converter stations and their controls are becoming cheaper and more reliable, which may make DC more attractive at shorter distances in the near future.
Block diagram of AC DC power Transmission
Transformer Step up
Transformer Step down
Four terminal DC grid with Converters (Rectifier and Inverter)
Vs1 Zs1, Vs4 Zs4 represents the ac input voltage model, Xl1,Xl4 represents step-up transformers
Vs2 Zs2, Vs3 Zs3 represents the ac output voltage model, Xl2,Xl3 represents step-down transformers Only the Resistance between grids are denoted being DC interconnection.
Advantages of HVDC transmission:
1. Two conductors, positive and negative to ground, are required instead of three, thereby reducing tower or cable costs;
2. The voltage stress at the conductor surface can be reduced with DC, thereby reducing corona loss, audible emissions, and radio interference;
Disadvantages of HVDC transmission:
1. The higher cost of converter stations compared with an AC transformer substation;
2. The need to provide filters and associated equipment to ensure acceptable waveform and power factor on the AC networks.
For DC transmission,
HVDC usually refers to transmission voltages near to 660 kV; (HVDC – High Voltage DC)
UHVDC refers to transmission voltages near to 800 kV or higher. (UHVDC – Ultra High Voltage DC)