POSTGRADUATE DIPLOMA IN ELECTRICAL ENGINEERING
ELECTRICAL POWER SYSTEMS LECTURE NOTES PROFESSOR JAMES KATENDE
DEPARTMENT OF ELECTRICAL ENGINEERING FIRST EDITION JULY 2005
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CHAPTER ONE INTRODUCTION The ability of the human race to develop sources of energy needed to accomplish useful work has played an essential role in the continual improvement in the standard of living of societies around the globe. The use of energy can be seen in everyday devices such as home appliances and machinery. Energy consumption in homes and industries increased beyond the point where useful forms of energy could be produced at the locations at which energy was being used. Hence, centralised energy processing and generating stations, together with elaborate transmission and distribution systems, were developed and the electric power system emerged as a tool for converting and transmitting energy. Energy is converted from its basic source, such as fossil fuel or hydro, into electric energy at places remote from population centres. Then the electric energy is sent over transmission and distribution systems to various locations, where it is converted to light, heat, and mechanical energy. An electric power system is designed to generate, control, dispatch and distribute electrical energy to consumers economically and with minimum ecological disturbance and to transfer this energy over transmission lines and distribution networks with maximum efficiency and reliability to deliver to consumers at virtually fixed voltage and frequency. Chapter two describes the typical structure of power systems as well the principles of bulk power generation. In chapter three an overview of the principles of power systems analysis is presented. Chapter four explains the need for power factor correction and the principle techniques available for doing so. Computations of the required rating of power factor correction gear are illustrated with several examples. Chapter five presents electrical load characteristics and some aspects of power station operating economics. Chapter six presents some of the common techniques for pricing the cost of generating and delivering electrical energy to consumers of electricity. Chapter seven expounds on the essentials of power flow analysis. Finally chapter eight introduces the essential features of power system protection
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CHAPTER TWO POWER SYSTEM STRUCTURE AND POWER GENERATION 2.1 Introduction Electric energy is produced in large quantities at various electric power plants by converting different forms of energyfossil fuels, nuclear energy, water power, etc. Electric energy is transformed by the use of transformers to different voltage levels most suitable for transmission, distribution and consumption. Electric power is transmitted using overhead or cable lines to customers at varied distances from its sources. Electric energy is utilized by various conversion devices such as electric motors, electric ovens, lighting systems, air condition units, etc. The need for power transmission lines arises from the fact that bulk electric power generation is done at electric power plants remote from consumers. However, consumers require small amounts of energy and they are scattered over wide areas. Thus the transmission of energy over a distance offers a number of advantages such as the following: 1. Use of remote energy sources. 2. Reduction of the total power reserve of generations 3. Utilization of the time difference between various time zones when the peak demands are not coincidence. 4. Improved reliability of electric power supply. The different power stations located in different geographical locations are interconnected by transmission lines thereby forming a power system network usually referred to as the GRID. This chapter presents an overview of the power system structure and principles of power generation. 2.2 The Structure Of Power Systems Generating stations, transmission lines and the distribution systems are the main components of an electric power system. Generating stations and a distribution station are connected through transmission lines, which also connect one power system (grid, area) to another. A distribution system connects all the loads in a particular area to the transmission lines. For economical and technological reasons, individual power systems are organized in the form of electrically connected areas or regional grids (also called power pools). Each area or regional grid operates technically and economically independently, but these are eventually interconnected* to form a national grid (which may even form an international grid) so that each area is contractually tied to other areas in respect to certain generation and scheduling features. Nigeria has a 330kV national grid. The major advantages of interconnecting power systems include the following: • Increased reliability. In the event of a forced or planned outage of a power station, the affected system can be fed from other stations. River flow, storage facilities, floods, and draughts are the factors that may affect hydrogeneration, for example. Outages can easily be met by load transfer once systems are interconnected. • Reduction in total installed capacity. In an isolated system reserve units must be maintained separately in power station. However, the reduction in total installed capacity depends on the characteristics of the interconnected system and the desired degree of service reliability. • Economic operation. The location of hydro power stations is determined by the natural water power sources. The choice of site for fossil-fuel fired thermal stations is more flexible. The following two alternatives are possible. 1. Power stations may be built close to sources of fossil fuel (coal mines or petroleum refineries) and electric energy is evacuated over transmission lines to the load centres. 2. Power stations may be built close to the load centres and coal is transported to them from the mines by rail road. In practice, however, power station location will depend upon many factors—technical, economical and environmental. As it is considerably cheaper to transport bulk electric energy over extra high voltage (EHV) transmission lines than to transport equivalent quantities of gas or oil over rail road, the recent trend is to build super (large) thermal power stations near sources of natural gas. Bulk power can be transmitted to fairly long distances over transmission lines of 400kV and above. However, the Nigeria’s gas resources are located mainly in the southern belt and some thermal power stations will continue to be sited in distant western and southern regions.
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As nuclear stations are not constrained by the problems of fuel transport and air pollution, a greater flexibility exists in their location, so that these stations are located close to load centres while avoiding high density pollution areas to reduce the risks, however remote, of radio-activity leakage. In Nigeria, as of now, the largest fraction of electric power used is generated in thermal plants. The remaining smaller fraction comes from hydro stations. Oil/natural gas is the fuel for most of the steam plants; the rest depends upon hydro and coal. Electric power is generated at a voltage of 11 to 25kV which then is stepped up to the transmission levels in the range of 66 to 330kV (or higher). As the transmission capability of a line is proportional to the square of its voltage, research is continuously being carried out to raise transmission voltages. Some countries are already employing 765kV. The voltages are expected to rise to 1200kV in the near future. For very long distances (over 600km), it is economical to transmit bulk power by DC transmission. It also obviates some of the technical problems associated with very long distance AC transmission. The DC voltages used are 400kV and above, and the line is connected to the AC systems at the two ends through a transformer and converting/inverting equipment (silicon controlled rectifiers are employed for this purpose). Several DC transmission lines are in use in Congo, India, Europe and the U.S.A. Figure 2.1 depicts schematically the structure of a power system.
Fig. 2.1 Schematic diagram depicting power system structure The first stepdown of voltage from transmission level is at the bulk power substation, where the reduction is to a range of 33 to 132kV, depending on the transmission line voltage. Some industries may require power at these voltage levels. This stepdown is from the transmission and grid level to subtransmission level. The next stepdown in voltage is at the distribution substation. Normally, two distribution voltage levels are employed: 1. The primary or feeder voltage (11kV) 2. The secondary or consumer voltage (415 V three phase/ 230 V single phase). The distribution system, fed from the distribution transformer stations, supplies power to the domestic, or industrial and commercial consumers. Thus, the power system operates at various voltage levels separated by transformers. Fig 2.2 illustrates the Nigerian power system structure.
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Fig 2.2 The Nigerian power system structure 2.3 Conventional Sources of Electric Energy Thermal (coal, oil, nuclear) and generations are the main conventional sources of electric energy. The necessity to conserve fossil fuels has forced scientist and technologists across the world to search for unconventional sources of electric energy. Some of the sources being explored are solar, wind and tidal sources. The conventional and some of the unconventional sources and techniques of energy generation are briefly surveyed. 2.3.1 Thermal (Coal, Oil/Natural Gas Fired) Power Stations The chemical energy stored in coal is transformed into electric energy in thermal power plants. The heat released by the combustion of coal or oil/natural gas produces steam in a boiler at high pressure and temperature, which when passed through a steam turbine gives off some of its internal energy as mechanical energy. The axial-flow type of turbine is normally used with several cylinders on the same shaft. The steam turbine acts as a prime mover and drives the electric generator (alternator). A simple schematic diagram of a coal fired thermal plant is shown in Fig. 2.3. The efficiency of the overall conversion process is poor and its maximum value is about 40% because of the high heat losses in the combustion gases and the large quantity of heat rejected to the condenser which has to be given off in cooling towers or into a stream/lake in the case of direct condenser cooling. The team power station operates on the Rankine cycle, modified to include superheating, feed-water heating and steam reheating. The thermal efficiency (conversion of heat to mechanical energy) can be increased by using steam at the highest possible pressure and temperature. With steam turbines of this size, additional increase in efficiency is obtained by reheating the steam after it has been partially expanded by an external heater. The reheated steam is then returned to the turbine where it is expanded through the final states of bleeding. To take advantage of the principle of economy of scale (which applies to units of all sizes), the present trend is to go in for larger sizes of units. Larger units can be installed at much lower cost per kilowatt. They are also cheaper to operate because of higher efficiency. They require lower labour and maintenance expenditure. There may be a saving of as high as 15% in capital cost per kilowatt by going up from a 100 to 250MW unit size and an additional saving in fuel cost of about 8% per kWh. Since larger units consume less fuel per kWh, they produce less air, thermal and waste pollution, and this is a significant advantage in our concern for environment. The only trouble in the case of a large unit is the tremendous shock to the system when outage of such a large capacity unit occurs. This shock can be tolerated so long as this unit size does not exceed 10% of the on-line capacity of a large grid.
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Fig. 2.3 Schematic diagram of a coal fired steam plant Today’s maximum generator unit size is (nearly 1200MW) limited by the permissible current densities used in rotor and stator windings. Efforts are on to develop super conducting machines where the winding temperatures will be nearing absolute zero. Extreme high current and flux densities obtained in such machines could perhaps increase unit sizes to several GWs which would result in better generating economy. Air and thermal pollution is always present in a coal fired steam plant. The air polluting agents (consisting of particulates and gases such as NOX, oxides of sulphur, etc.) are emitted via the exhaust gases and thermal pollution is due to the rejected heat transferred from the condenser to cooling water. Cooling towers are used in situations where the stream/lake cannot withstand the thermal burden without excessive temperature rise. The problem of air pollution can be minimized through scrubbers and precipitators and by resorting to minimum emission dispatch and passing Clean Air Act like that passed in USA in the 1970s. Fluidized-bed Boiler The main problem with coal is its high ash content (up to 40% max). To solve this, fluidized-bed combustion technology is being developed and perfected. The fluidized-bed boiler is undergoing extensive development and is being preferred due to its lower pollutant level and better efficiency. Direct ignition of pulverized coal is being introduced but initial oil firing support is needed. Cogeneration Considering the tremendous amount of waste heat generated in thermal power generation, it is advisable to save fuel by the simultaneous generation of electricity and steam (or hot water) for industrial use or space heating. Now called cogeneration, such systems have long been common, here and abroad. Currently, there is renewed interest in these because of the overall increase in energy efficiencies which are claimed to be as high as 65% Cogeneration of steam and power is high energy efficient and is particularly suitable for chemicals, paper, textiles, food, fertilizer and petroleum refining industries. Thus, these industries can solve energy shortage problem in a big way. Further, they will not have to depend on the grid power which is not so reliable. Of course they can sell the extra power to the government for use in deficient areas. They may also sell power to the neighbouring industries, a concept called wheeling power. There are two possible ways of cogeneration of heat and electricity: (i) Topping cycle (ii) Bottoming cycle. In the topping cycle, fuel is burnt to produce electrical and mechanical power and the waste heat from the power generation provides the process heat. In the bottoming cycle, fuel first produces process heat and the waste heat from the processes is then used to produce power. Coal-fired plants share environmental problems with some other types of fossil-fuel plants; these include “acid rain” and the “greenhouse” effect. 2.3.2 Hydroelectric Power Generation The oldest and cheapest method of power generation is that of utilizing the potential energy of water. The energy is obtained almost free of running cost and is completely pollution free. Of course, it involves high capital cost because of
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the heavy civil engineering construction works involved. Also it requires a long gestation period of about five to eight years as compared to four to six years for steam plants. Hydroelectric stations are designed, mostly, as multipurpose projects such as river flood control, storage of irrigation and drinking water, and navigation. A simple block diagram of a hydro plant is given in Fig. 2.4. The vertical difference between the upper reservoir and tail race is called the head.
Fig. 2.4 A typical layout for a storage type hydro plant Hydro plants are of different types such as run-of-river (use of water as it comes), pondage (medium head) type, and reservoir (high head) type. The reservoir type plants are the ones which are employed for bulk power generation. Often, cascaded plants are also constructed, i.e., on the same water stream where the discharge of one plant becomes the inflow of a downstream plant. Different types of turbines such as Pelton, Francis and Kaplan are used for storage, pondage and run-of-river plants, respectively. Hydroelectric plants are capable of starting quickly—almost in five minutes. The rate of taking up load on the machine is of the order of 20 MW/min. Further, no losses are incurred at standstill. Thus, hydro plants are ideal for meeting peak loads. The time from start up to the actual connection to the grid can be as short as 2 min. The power available from a hydro plant is given by P = 981ρWH W where W = discharge m3/s through turbine ρ = density 1000 kg/m2 H = head (m) Problems peculiar to hydroplant which inhabit expansion are: 1. Silting—reportedly Bhakra dead storage has silted fully in 30 years 2. Seepage 3. Ecological damage to region 4. These cannot provide base load, must be used for peak shaving and energy saving in coordination with thermal plants. In areas where sufficient hydro generation is not available, peak load may be handled by means of pumped storage. This consists of an upper and lower reservoirs and reversible turbine-generator sets, which can also be used as motor-pump sets. The upper reservoir has enough storage for about six hours of full load generation. Such a plant acts as a conventional hydro plant during the peak load period, when production costs are the highest. The turbines are driven by water from the upper reservoir in the usual manner. During the light load period, water in the lower reservoir is pumped back into the upper one so as to be ready for use in the cycle of the peak load period. The generators in this period change to synchronous motor action and drive the turbines which now work as pumps. The electric power is supplied to the sets from the general power network or adjoining thermal plant. The pumped storage scheme, in fact, is analogous to the charging and discharging of a battery. It has the added advantage that the synchronous machine can be used as synchronous condensers for VAR compensation of
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the power network, if required. In a way, from the point of the thermal sector of the system, the pumped storage scheme “shaves the peaks” and “fills the troughs” of the daily load-demand curve. 2.3.3 Nuclear Power Stations There are two ways in which energy can be released in a nuclear power plant: • By combining light nuclei like hydrogen or helium. This process is known as fusion. • By breaking up heavy nuclei into nuclei of intermediate atomic number with resultant release of energy. This process is known as fission. When Uranium-235 is bombarded with neutrons, fission reaction takes place releasing neutrons and heat energy. These neutrons then participate in the chain reaction of fissioning more atoms of 235U. In order that the freshly released neutrons be able to fission the uranium atoms, their speeds must be reduced to a critical value. Therefore, for the reaction to be sustained, nuclear fuel rods must be embedded in neutron speed reducing agents (like graphite, heavy water, etc.) called moderators. For reaction control, rods made of neutron-absorbing material (boron-steel) are used which, when inserted into the reactor vessel, control the amount of neutron flux thereby controlling the rate of reaction. However, this rate can be controlled only within a narrow range. The schematic diagram of a nuclear power plant is shown in Fig 2.5. The heat released by the nuclear reaction is transported to a heat exchanger via primary coolant (CO2, water, etc.).
Fig. 2.5 Schematic view of a nuclear power plant Steam is then generated in the heat exchanger, which is used in a conventional manner to generate electric energy by means of a steam turbine. Various types of reactors are being used in practice for power plant purposes, viz., advanced gas reactor (AGR), boiling water reactor (BWR), heavy water moderated reactor, etc. The associated merits and problems of nuclear power plants as compared to conventional thermal plants are mentioned below: Merits 1. A nuclear power plant is totally free from Air pollution. 2. It requires little fuel in terms of volume and weight, and therefore poses no transportation problems and may be sited, independently of nuclear fuel supplies, close to load centres. However, safety consideration requires that they be normally located away from populated areas. Problems 1. Nuclear reactors produce radioactive fuel waste, the disposal of which poses serious environmental hazards. 2. The rate of nuclear reaction can be lowered only by a small margin, so that the load on a nuclear power plant can only be permitted to be marginally reduced below its full load value. Nuclear power stations must, therefore, be reliably connected to a power network, as tripping of the lines connecting the station can be quite serious and may require shutting down of the reactor with all its consequences. 3. Because of relatively high capital cost as against running cost, the plant should operate continuously as a base load station. Wherever possible, it is preferable to support such a station with a pumped storage scheme mentioned earlier.
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The world uranium resources are quite limited, and at the present rate may not last much beyond 50 years. However, there is a redeeming feature. During the fission of 235U, some of the neutrons are absorbed by the more abundant uranium isotope 238U (enriched uranium contains only about 3% of 235U while most of it is 238U) converting it to plutonium (239U), which in itself is a fissionable material and can be extracted from the reactor fuel waste by a fuel reprocessing plant. Plutonium would then be used in the next generation reactors (fast breeder reactors—FBRs), thereby considerably extending the life of nuclear fuels. The FBR technology is been intensively developed it will extend the availability of nuclear fuels at predicted rates of energy consumption to several centuries. Figure 2.5 shows the schematic diagram of an FBR. It is essential that for breeding operation, conversion ratio (fissile material generated/fissile material consumed) has to be more unity. This is achieved by fast moving neutrons so that no moderator is needed. The neutrons do slow down a little through collisions with structural and fuel elements. The energy density/kg of fuel is very high and so the core is small. It is therefore necessary that the coolant should possess good thermal properties and hence liquid sodium is used. The fuel for an FBR consists of 20% plutonium plus 8% uranium oxide. The coolant liquid sodium leaves the reactor at 650oC at atmospheric pressure. The heat so transported is led to a secondary sodium circuit which transfers it to a heat exchanger to generate steam at 540oC.
Fig. 2.6 Fast breeder reactor (FBR) With a breeder reactor release of plutonium, an extremely toxic material, would make the environmental considerations most stringent. Typical power densities (MW/m3) in fission reactor cores are: gas cooled 0.53, high temperature gas cooled 7.75, heavy water 18.0, boiling water 29.0, pressurized water 54.75, fast breeder reactor 760.0. Fusion Energy is produced in this process by the combination of two light nuclei to form a single heavier one under sustained conditions of extremely high temperatures (in millions of degree centigrade). Fusion is futuristic. Generation of electricity via fusion would solve the long term energy needs of the world with minimum environmental problems. A commercial reactor is expected by 2000 AD. Considering radioactive wastes, the impact of fusion reactors would be much less than the fission reactors. In case of success in fusion technology sometime in the distance future or a break through in the pollution-free solar energy, FBRs would become obsolete. However, there is an intense need today to develop FBR technology as an insurance against failure to develop these two technologies. In the past few years, serious doubts have been raised about the safety claims of nuclear power plants. There have been as many as 150 near disaster nuclear accidents from the Three-mile accident in USA to the recent Chernobyl accident in the former USSR. There is a fear that all this may put the nuclear energy development in reverse gear. If this happens
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there could be serious energy crisis in the third world countries which have pitched their hopes on nuclear energy to meet their burgeoning energy needs. France and Canada are possibly the two countries with a fairly clean record of nuclear generation. 2.3.4 Magnetohydrodynamic (MHD) Generation In thermal generation of electric energy, the heat released by the fuel is converted to rotational mechanical energy by means of a thermocycle. The mechanical energy is then used to rotate the electric generator. Thus two stages of energy conversion are involved in which the heat to mechanical energy conversion has inherently low efficiency. Also, the rotating machine has its associated losses and maintenance problems. In MHD technology, electric energy is directly generated by the hot gases produced by the combustion of fuel without the need for mechanical moving parts. In a MHD generator, electrically conducting gas at a very high temperature is passed in a strong magnetic field, thereby generating electricity. High temperature is needed to ionize the gas, so that it has good electrical conductivity. The conducting gas is obtained by burning a fuel and injecting a seeding material such as potassium carbonate in the product of combustion. The principle of MHD power generation is illustrated in Fig. 2.6. About 50% efficiency can be achieved if the MHD generator is operated in tandem with a conventional steam plant.
Fig. 2.6 The principle of MHD power generation Though the technological feasibility of MHD generation has been established, its economic feasibility is yet to be demonstrated. India had started a research and development project in collaboration with the former USSR to install a pilot MHD plant based on coal and generating 2 MW power. In Russia, a 25 MW MHD plant which uses natural gas as fuel has been in operation for some years. 2.3.5 Gas Turbines Compressed air is fed to the combustion chamber where continuous combustion of fuel oil is maintained. The hot gases so produced are made to run a gas turbine. In normal running, a gas turbine is less economical compared to a conventional steam plant; but it is used as a prime mover for its ability to start and take up load quickly. Thus it is capable of meeting sudden peaks of system load. Fuel economy can be improved by using hot exhaust gases of the gas turbine in a conventional steam generation cycle. There are currently many installations in the world with 100 MW generators. Recently a 6 x 30 MW gas turbine station has been put up in Delhi. A gas turbine unit can also be used as synchronous compensator to help maintain flat voltage profile in the system. 2.4 UNCONVENTIONAL ENERGY SOURCES Because of the limited availability of oil/natural gasl, there is considerable international effort into the development of alternative/new/unconventional/renewable/clean sources of energy. Most of the new sources (some of them in fact have been known and used for centuries now!) are nothing but the manifestation of solar energy, e.g., wind, sea waves, ocean thermal energy conversion (OTEC) etc. In this section, we shall discuss the possibilities and potentialities of various methods of using solar energy. 2.4.1 Geothermal Power Plants In a geothermal power plant, heat deep inside the earth acts as a source of power. There has been some use of geothermal energy in the form of steam coming from underground in the USA, Italy, New Zealand, Mexico, Japan, Philippines, India, and some other countries.
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The present installed geothermal plant capable in the world is about 500 MW and the total estimated capacity is only about 2000 MW. Since the pressure and temperatures are low, the efficiency is even less than the conventional fossil fuelled plants, but the capital cost are less and the fuel is available free of cost. 2.4.2 Wind Power Stations Winds are essentially created by the solar heating of the atmosphere. Several attempts have been made since 1940 to use wind to generate electric energy and development is still going on. However, techno-economic feasibility has yet to be satisfactorily established. Wind as a power source is attractive because it is plentiful, inexhaustible and non-polluting. Further, it does not impose extra heat burden on the environment. Unfortunately, it is non-steady and undependable. Control equipment has been devised to start the wind power plant whenever the wind speed reaches 30km/h. Methods have also been found to generate constant frequency power with varying wind speeds and consequently varying speeds of wind mill propellers. Wind power may prove practical for small power needs in isolated sites. But for maximum flexibility, it should be used in conjunction with other methods of power generation to ensure continuity. For wind power generation, there are three types of operations: 1. Small, 0.5—10 kW for isolated single premises 2. Medium, 10—100 kW for communities 3. Large, 1.5 MW for connection to the grid. The theoretical power in a wind stream is given by P = 0.5 ρ A V 3 W where ρ = density of air (1201 g/m3 at NTP) V = mean air velocity (m/s) and A = swept area (m2). For a rotor of 17m diameter and a velocity of 48 km/h the theoretical power is 265kW and the practical would be roughly half of this value. There are some distinctive energy end-use features of wind power systems: 1. Most wind power sites are in remote rural, island or marine areas. 2. Rural grid systems are likely to be ‘weak’ in these areas, since they carry relatively low voltage supplies (e.g. 33kV). 3. There are always periods without wind. 2.4.3 Solar Energy The average accident solar energy received on earth surface is about 600 W/m2, but the actual value varies considerably. It has the advantage of being free of cost, non-exhaustible and completely pollution-free. On the other hand, it has several drawbacks—energy density per unit area is very low, it is available for only a part of the day, and cloudy and hazy atmospheric conditions greatly reduce the energy received. Therefore, in harnessing solar energy for electricity generation, challenging technological problems exist, the most important being that of the collection and concentration of solar energy and its conversion to the electrical form through efficient and comparatively economical means. At present, two technologies are being developed for conversion of solar energy to the electrical form. In one technology, collectors with concentrators are employed to achieve temperatures high enough (700oC) to operate a heat engine at reasonable efficiency to generate electricity. However, there are considerable engineering difficulties in building a single tracking bowl with a diameter exceeding 30m to generate perhaps 200kW. The scheme involves large and intricate structures involving huge capital outlay and as of today is far from being competitive with conventional electricity generation. The solar power tower generates steam for electricity production. There is a 10 MW installation of such a tower by the Southern California Edison Co. in USA using 1818 plane mirrors, each 7m x 7m reflecting direct radiation to the raised boiler.
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Electricity may be generated from a Solar Pond by using a special ‘low temperature’ heat engine coupled to an electric generator. A solar pond at Ein Borek in Isreal produces a steady 150kW from 0.74 hectare at a busbar cost of about $0.10/kWh. Solar power potential is unlimited. Direct Conversion to Electricity (photovoltaic generation) This technology converts solar energy to the electrical form by means of silicon wafer photoelectric cells known as “Solar Cells”. Their theoretical efficiency is about 25% but the practical value is only about 15%. But that does not matter as solar energy is basically free of cost. The chief problem is the cost and maintenance of solar cells. With the likelihood of a breakthrough in the large scale production of cheap solar cells with amorphous silicon, this technology may compete with conventional fuels become scarce. Solar energy could, at the most, supplement up to 5-10% of the total energy demand. It has been estimated that to produce 1012kWh per year, the necessary cells would occupy about 0.1% of US land area as against highways which occupy 1.5% (in 1975) assuming 10% efficiency and a daily insolation of 4kWh/m2. In all solar thermal schemes, storage is necessary because of the fluctuating nature of sun’s energy. This is equally true with many other unconventional sources as well as sources like wind. Fluctuating sources with fluctuating loads complicate still further the electricity supply.
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CHAPTER THREE FUNDAMENTAL THEORY 3.1 INTRODUCTION The main components that make up a power system are generating sources, transmission and distribution networks, and loads. For the purpose of analysis, the power system is treated as a network of circuit elements contained in branches radiating from nodes to from closed loops or meshes. The system variables are current and voltage, and in steady state analysis, they are regarded as time varying quantities at a single and constant frequency. The network parameters are impedance and admittance; these are assumed to be linear, bilateral (independent of current direction) and constant for a constant frequency. The analysis of electric power systems involves the study of the performance of the system under both normal and abnormal conditions. Such analysis considers single phase and three phase circuits. This chapter deals with steady state ac (alternating current) analytical techniques. 3.2 VECTOR ALGEBRA A vector represents a quantity in both magnitude and direction. In figure 3.1 the vector OP has a magnitude Z at an angle θ with the reference axis OX.
Fig 3.1 Vector OP It may be resolved into two components at right angles to each other, in this case x and y. The magnitude or scalar value of vector
Z is known as the modulus or amplitude, Z and the angle θ is the argument, and is written as arg Z . The
conventional method of expressing a vector
Z is to write simply: Z ∠ θ.
This form completely specifies a vector for graphical representation or conversion into other forms. For vectors to be useful, they must be expressed algebraically. In figure 3.1, the vector adding its components x and y; algebraically this vector may be written as:
Z is the resultant of vectorially
(3.1) Z = x + jy where the operator j indicates that the component y is perpendicular to component x. In electrical nomenclature, the axis OX is the ‘real’ or ‘in-phase’ axis, and the vertical axis OY is called the ‘imaginary’ or ‘quadrature’ axis. The operator j rotates a vector anti-clockwise through 90o. If the vector is made to rotate anti-clockwise through 180o, then the operator j has performed its function twice, and since the vector has reversed its sense, then: j x j or j2 = -1 Hence, j = − 1 . When a vector quantity is expressed in terms of its rectangular co-ordinates it is called a ‘complex quantity’. Therefore, x + jy is a complex quantity and is the rectangular form of the vector Z ∠ θ where:
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Z = (x 2 + y 2 ) ⎫ ⎪ y ⎪ θ = tan ⎪ x ⎬ ⎪ x = Z cos θ ⎪ ⎪⎭ y = Z sin θ
(3.2)
From equations (3.1) and (3.2):
Z = Z (cos θ + j sin θ )
(3.3)
and since cosθ and sinθ may be expressed in exponential form by the identities:
e jθ − e − jθ sinθ = 2j it follows that
e jθ − e − jθ and cosθ = 2
Z may also be written as: Z = Z e jθ
(3.4)
Therefore, a vector quantity may also be represented trigonometrically and exponentially. 3.3 MANIPULATION OF COMPLEX QUANTITIES Complex quantities may be represented in any of the four co-ordinate systems given below:
Z ∠θ
a.
Polar
b.
Rectangular
c.
Trigonometric
Z (cosθ + jsinθ)
d.
Exponential
Z ejθ
x + jy
The modulus Z and the argument θ are together known as ‘polar co-ordinates’, and x and y are described as ‘Cartesian co-ordinates’. Conversion between co-ordinate systems is easily achieved. As the operator j obeys the ordinary laws of algebra, complex quantities in rectangular form can be manipulated algebraically, as can be seen by the following:
Z 1 + Z 2 = (x1 + x2) + j(y1+ y2) Z 1 − Z 2 = (x1 – x2) + j(y1-y2) Z 1 Z 2 = Z 1 Z 2 ∠(θ1 + θ 2 )⎫ ⎪ Z1 ⎬ Z1 = ∠(θ 1 − θ 2 ) ⎪ Z2 Z2 ⎭
(3.5) (3.6)
(3.7)
Fig. 3.2 Addition of vectors
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Given a complex quantity Z=X+jY the complex conjugate of Z is denoted by Z* and is obtained by changing the sign of the imaginary part of Z, i.e, Z*=X-jY. If Z is given in polar form as;
Z = Z ∠θ then
Z* = Z∠ − θ In other words the complex conjugate in polar form is obtained by simply changing the sign of the argument 3.3.1 Complex Quantities Some complex quantities are variable with, for example, time; when manipulating such variables in differential equations it is expedient to write the complex quantity in exponential form. When dealing with such functions it is important to appreciate that the quantity contains real and imaginary components. If it is required to investigate only one component of the complex variable, separation into components must be carried out after the mathematical operation has taken place. EXAMPLE: Determine the rate of change of the real component of vector Z ∠ (ωt) with time. Solution The vector is given as:
Z ∠(ωt ) = Z (cosωt + jsinωt) = Z ejωt The real component of the vector is Z cosωt. Differentiating Z ejωt with respect to time:
d Z e j ωt = j ω Z e j ωt dt =jω Z (cosωt + jsinωt) Separating into real and imaginary components:
(
)
d Z e jωt = Z (-ωsinωt + jωcosωt) dt Thus, the rate of change of the real component of the vector Z ∠ωt is: - Z ωsinωt 3.3.2 Complex Numbers A complex number may be defined as a constant that represents the real and imaginary components of a physical quantity. The impedance parameter of an electric circuit is a complex number having real and imaginary components, which are described as resistance and reactance, respectively. Confusion often arises between vectors and complex numbers. A vector, as previously defined, may be complex number. In this context, it is simply a physical quantity of constant magnitude acting in a constant direction. A complex number that is a physical quantity relating stimulus and response in a given operation is known as a ‘complex operator’. In this context, it is distinguished from a vector by the fact that it has no direction of its own. Because complex numbers assume a passive role in any calculation, the form taken by the variables in the problem determines the method of representing them. 3.3.3 Mathematical Operators
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Mathematical operators are complex numbers that are used to move a vector through a given angle without changing the magnitude or character of the vector. An operator is not a physical quantity; it is dimensionless. The symbol j, which has been compounded with quadrature components of complex quantities, is an operator that rotates a quantity anti-clockwise through 90o. Another useful operator is one which moves a vector anti-clockwise through 120o, commonly represented by the symbol a. Operators are distinguished by one further feature; they are the roots of unity. Using De Moivre’s theorem, the nth root of unity is given by solving the expression:
11 / n = (cos 2 π m + j sin 2 π m)1 / n where m is any integer. Hence:
11 / n = cos
2π m 2π m + j sin n n
where m has values 1,2,3,… (n-1) From the above expression j is found to be the 4th root and a the 3rd root of unity, as they have four and three distinct values, respectively. Some useful functions of the a operator are listed below: 2π
j 1 3 a=− + j =e 3 2 2 4π j 1 3 a2 = − − j =e 3 2 2
1=1 + j0 = ej0 1 + a + a2 = 0
1 – a = j 3a
2
1 – a2 = - j 3a a – a2 = j 3 j=
a − a2 3
3.4 CIRCUIT QUANTITIES AND CONVENTIONS Circuit analysis may be described as the study of the response of a circuit to an imposed condition, for example, a short circuit. The circuit variables are current and voltage. Conventionally, current flow results from the application of a driving voltage, but there is complete duality between the variables and either may be regarded as the cause of the other. When a circuit exists, there is an interchange of energy; a circuit may be described as being made up of ‘sources’ and ‘sinks’ for energy. The parts of a circuit are described as elements; a ‘source’ may be regarded as an ‘active’ element and a ‘sink’ as a ‘passive’ element. Some circuit elements are dissipative, that is, they are continuous sinks for energy, for example, resistance. Other circuit elements may be alternately sources and sinks, for example, capacitance and inductance. The elements of a circuit are connected together to form a network having nodes (terminals or junctions) and branches (series groups of elements) that form closed loops (meshes). In steady state ac circuit theory, the ability of a circuit to accept a current flow resulting from a given driving voltage is called the impedance of the circuit. Since current and voltage are duals the impedance parameter must also have a dual, called admittance. 3.4.1 Circuit Variables As current and voltage are sinusoidal functions of time, varying at a single and constant frequency, they are regarded as rotating vectors and can be drawn as plane vectors that is, vectors defined by two co-ordinates) on a vector diagram. Such rotating vector is called a phasor and the diagram is said to be a phasor diagram.
16
For example, the instantaneous value, e, of a voltage varying sinusoidally with time is expressed mathematically as: (3.8) e = E m sin( wt + δ ) where: Em is the maximum amplitude of the waveform; ω = 2πf, the angular velocity in rad/s. δ is the argument defining the amplitude of the voltage at a time t = 0 f is the frequency in hertz (Hz) At t = 0, the actual value of the voltage is Emsinδ. So if Em is regarded as the modulus of a vector, whose argument is δ, then Emsinδ is the imaginary component of the vector E m ∠δ . Figure 3.3 illustrates this quantity as a vector and as a sinusoidal function of time.
Fig. 3.3 Representation of a sinusoidal function The current resulting from applying a voltage to a circuit depends upon the circuit impedance. If the voltage is a sinusoidal function at a given frequency and the impedance is constant the current will also vary harmonically at the same frequency, so it can be shown on the same vector diagram as the voltage vector, and is given by the equation
i=
Em
(3.9)
Z sin(wt + δ − φ )
where:
⎫ Z = R2 + X 2 ⎪ ⎪ 1 ⎞⎪ ⎛ X = ⎜ ωL − ⎟⎬ ωC ⎠⎪ ⎝ ⎪ X φ = tan −1 ⎪ R ⎭
(3.10)
From equations (3.9) and (3.10) it can be seen that the angular displacement φ between the current and voltage vectors and the circuit magnitude I m = E m ⁄ Z is dependent upon the impedance Z . In complex form the impedance may be written as Z = R + jX . The ‘real component’, R, is the circuit resistance, and the ‘imaginary component’, X, is the circuit reactance. When the circuit reactance is inductive (that is, ωL angle φ, and when it is capacitive (that is, ωL <
>
1 ), the current ‘lags’ the voltage by an ωC
1 ) it ‘leads’ the voltage by an angle φ. ωC
When drawing phasor diagrams, one phasor is chosen as the ‘reference’ and all other phasors are drawn relative to the reference phasor in terms of magnitude and angle. The circuit impedance Z is a complex operator and is distinguished from a phasor only by the fact that it has no direction of its own.
17
A further convention is that sinusoidally varying quantities are described by their ‘effective’ or ‘root-mean-square’ (rms) values; these are usually written using the relevant symbol without a suffix. Thus,
Im ⎫ ⎪ 2 ⎪ ⎬ Em ⎪ E= 2 ⎪⎭ I=
(3.11)
The ‘root-mean-square’ value is that value which has the same heating effect as a direct current quantity of that value in the same circuit, and this definition applies to non-sinusoidal as well as sinusoidal quantities. 3.4.2 Sign Conventions In describing the electrical state of a circuit, it is often necessary to refer to the ‘potential difference’ existing between two points in the circuit. Since wherever such a potential difference exists, current will flow and energy will either be transferred or absorbed, it is obviously necessary to define a potential difference in more exact terms. For this reason, the terms voltage rise and voltage drop are used to define more accurately the nature of the potential difference. Voltage rise is a rise in potential measured in the direction of current flow between two points in a circuit. Voltage drop is the converse. A circuit element with a voltage rise across it acts as a source of energy. A circuit element with a voltage drop across it acts as a sink of energy. Voltage sources are usually active circuit elements, while sinks are usually passive circuit elements. The positive direction of energy flow is from sources to sinks. Kirchhoff’s first law states that the sum of the driving voltages must equal the sum of the passive voltages in a closed loop. This is illustrated by the fundamental equation of a series RLC circuit:
iR + L
di 1 + idt = e dt C ∫
(3.12)
where the terms on the left-hand side of the equation are voltage drops across the circuit elements and the term on the right-hand side is the driving voltage (i.e. source voltage). Expressed in steady state terms equation (3.12) may be written as:
∑E = ∑IZ
(3.13)
It is the equation most usually adopted in electrical network calculations, since it equates the driving voltages, which are known, to the passive voltages, which are functions of the currents to be calculated. In describing circuits and drawing phasor diagrams, for formal analysis or calculations, it is necessary to adopt a notation which defines the positive direction of assumed current flow, and establishes the direction in which positive voltage drops and voltage rises act. Two methods are available; one, the double suffix method, is used for symbolic analysis, the other, the single suffix or diagrammatic method, is used for numerical calculations. In the double suffix method the positive direction of current flow is assumed to be from node a to node b and the current is designated Iab.With the diagrammatic method, an arrow indicates the direction of current flow. The voltage rises are positive when acting in the direction of current flow. It can be seen from figure 3.4 that E 1 and
E an are positive voltage rises and E 2 and E bn are negative voltage rises. In the diagrammatic method their direction of action is simply indicated by an arrow, whereas in the double suffix method, E an and E bn indicate that there is a potential rise in directions na and nb.
18
Fig 3.4 Methods of representing a circuit Voltage drops are also positive when acting in the direction of current flow. From figure 3.4 (a) it can be seen that
( Z 1 + Z 2 + Z 3 ) I is the total voltage drop in the loop in the direction of current flow, and must equate to the total voltage rise, E 1 − E 2 . In figure 3.4 (b), the voltage drop between nodes a and b, designated V ab , indicates that point b is at a lower potential than a, and is positive when current flows from a to b. Conversely V ba is a negative voltage drop. Symbolically:
V ab = V an − V bn ⎫⎪ ⎬ V ba = V bn − V an ⎪⎭
(3.14)
where n is a common reference point.
Voltage-Current relationships for circuit elements When an alternating current, i=Imsinωt passes through a resistor R, the voltage drop across the resistor is given by v=Ri =R Imsinωt =Vmsinωt where Vm=RIm is the amplitude of the voltage. Consequently for a resistor the voltage and current waveforms are in phase and of same frequency. They may only differ in amplitude depending on the value of resistance R. If the same current I passes through a pure inductance L, the voltage drop across the inductor is given by
19
di dt = ωLI m cos ωt
v=L
= ωLI m sin(ωt + 90 0 ) Thus for an inductor, the current phasor lags the voltage phasor by 90o. If the same current passes through a pure capacitor, the voltage drop across the capacitor is given by
1 idt C∫ 1 = ∫ I m sin ωtdt C 1 =− I cos ωt ωC m 1 = I m sin(ωt − 90 0 ) ωC v=
Thus the current phasor leads the voltage phasor by 90ofor a capacitor 3.4.3 Electric Power Power is an indication of how much work (conversion of energy from one form to another) can be accomplished in a specified amount of time. i.e. power is rate of doing work). Work is done when energy is converted from one form to another. Power =
work joules/second time
In electrical terms Power =
QV = IV watts t
where Q is the charge transferred due to potential difference V and t is time interval during which the transfer takes place. Note that electric current I =Q/t, and I = V/R where R is the resistance to flow of current. Hence Power P=
V2 = I 2 R watts. R
In general the power, Po, delivered by an electrical device is less or equal to the power Pi supplied to its input. Power is lost in the energy conversion process. Consequently, Pi = Po + PL and efficiency of the device is given by
η=
Po Po = Pi Po + PL
WORKED EXAMPLE: A 2.5hp motor operates on an efficiency of 75%. What is the power drawn from the supply in watts. If the input current is 10.8A what is the supply voltage? 1hp = 746W Solution
20
η=
Po 2.5 × 746 × 100 ⇒ 0.75 = ⇒ Pi = 2486.7 Pi Pi
V=
P 2486.7 = = 230V I 10.8
The product of the potential difference across and the current through a branch of a circuit is a measure of the rate at which energy is exchanged between that branch and the remainder of a circuit. If the potential difference is a positive voltage drop, the branch is passive and absorbs energy. Conversely, if the potential difference is a positive voltage rise, the branch is active and supplies energy. The rate at which energy is exchanged is known as power, and by convention, the power is positive when energy is being absorbed and negative when being supplied. With ac circuits the power alternates, so, to obtain a rate at which energy is supplied or absorbed, it is necessary to take average power over one whole cycle. If the voltage v = Vmcos(ωt + θv) and the current i = Imcos(ωt + θi), then the instantaneous power is given by:
p = vi = Vm I m cos(ωt + θ v ) cos(ωt + θ i ) 1 Using the trigonotry identity, cos A cos B = [cos( A − B ) + sin( A + B ) , results in 2 1 p = Vm I m [cos(θ v − θ i ) + cos(2ωt + θ v − θ i )] 2 1 = Vm I m {cos(θ v − θ i ) + cos[2(ωt + θ v ) − (θ v − θ i )]} 2 1 = Vm I m [cos(θ v − θ i ) + cos 2(ωt + θ v ) cos(θ v − θ i ) + sin 2(ωt + θ v ) sin(θ v − θ i )] 2 Thus the instantaneous power is p = P[1+cos2(ωt + θv)] + Qsin2(ωt + θv) (3.15) where: P = VI cos θ , is the real or active power absorbed by the circuit. Q = VI sin θ , is the reactive power borrowed and returned by the circuit. θ =θv-θi, is the angle between voltage and current, or the impedance angle. V =Vm/ 2 is the rms voltage and I is the rms current
From equation (3.15) it can be seen that the quantity P varies from 0 to 2P and quantity Q varies from –Q to +Q in one cycle, and that the waveform is of twice the periodic frequency of the current voltage waveform. The average value of the power exchanged in one cycle is a constant, equal to quantity P, and as this quantity is the product of the voltage and the component of current which is ‘in phase’ with the voltage it is known as the ‘real’ or ‘active’ power. The average value of quantity Q is zero when taken over a cycle, suggesting that energy is stored in one half-cycle and returned to the circuit in the remaining half-cycle. Q is the product of voltage and quadrature component of current, and is known as ‘reactive power’. Expressing the rms voltage and current phasors in polar form as: V = V ∠θ v expression VI* yields
21
and
I = I ∠θ i then the
VI * = V I ∠(θ v − θ i ) = V I ∠θ = V I cos φ + j V I sin θ
(3.16)
∴ S = VI * = P + jQ
The quantity S is described as the ‘apparent power’, and is the term used in establishing the rating of a circuit or power equipment. The unit of apparent power, S , is the volt-ampere (VA) and the larger units are kVA (103VA) or MVA(106VA). The SI unit for real power P is the watt (W) and the larger units are kW (103W) or MW (106W). The unit for reactive power Q is the volt-ampere reactive (VAr). Its larger versions are kVAr (103VAr) or MVAr (106VAr). The reactive power Q is positive when the phase angle θ between voltage and current (impedance angle) is positive (i.e., when the load impedance is inductive and I lags V). Q is negative when θ is negative (i.e., when the load impedance is capacitive and I leads V). Based on eqn (3.16) it is convenient to think of P, Q, and S as forming the sides of a right triangle, referred to as the power triangle. This is illustrated in Fig 3.5
Fig 3.5 Power triangle In this case the voltage is leading the current by ϕ degrees (radians) or the current is lagging the voltage by (radians). In terms of power factor (pf) we say that: 1. pf is lagging when current is lagging behind voltage (as in inductive ccts) 2. pf is leading when current is leading voltage (as in capacitive ccts).
ϕ
degrees
From the power triangle we have
P 2 + Q 2 = V 2 I 2 cos 2ϕ + V 2 I 2 sin 2ϕ The power to an inductive load, Z=R+jXL, for example, is: Real power, P = VIcosϕ = I R 2
Reactive power, Q = VIsin ϕ = I X L 2
Apparent power, S = VI =
P2 + Q2 = I 2Z
Hence, power factor
pf = cos ϕ
⎡ ⎛ Q ⎞⎤ = cos ⎢ tan −1 ⎜ ⎟⎥ ⎝ P ⎠⎦ ⎣ ⎡ ⎛ X ⎞⎤ = cos ⎢ tan −1 ⎜ L ⎟⎥ ⎝ R ⎠⎦ ⎣
3.4.4 Single Phase and Polyphase System A system is single or polyphase depending upon whether the sources feeding it are single or polyphase. A source is single or polyphase according to whether there is one or several driving voltages associated with it. For example, a three-
22
phase source is a source containing three alternating driving voltages that are assumed to reach a maximum in phase sequence; red (R), yellow (Y), blue (B). The positive phase sequence is RYB, whereas the negative phase sequence is RBY.
Phase sequence Each phase driving voltage is associated with a phase branch of the system network as shown in figure 3.5(a). If a polyphase system has balanced voltages, that is, equal in magnitude and reaching a maximum at equally displaced time intervals, and the phase branch impedances are identical, it is called a ‘balanced’ system. It will become ‘unbalanced’ if any of the above conditions are not satisfied. Calculations using a balanced polyphase system are simplified, as it is only necessary to solve for a single phase, the solution for the remaining phases being obtained by symmetry. The power system is normally operated as a three-phase, balanced, system. For this reason the phase voltages are equal in magnitude and can be represented by three vectors spaced 120o (or 2π ⁄ 3 radians) apart, as shown in figure 3.5(b).
Figure 3.5 Three-phase systems Since the voltages are symmetrical, they may be expressed in terms of one, that is:
⎫ ⎪⎪ E y = a 2 E r = E sin(ωt − 120 o )⎬ ⎪ E b = aE r = E sin(ωt + 120 o ) ⎭⎪ E r = E r = E sin ωt
(3.17)
where a is the vector operator ej2π ⁄ 3. Further, if the phase branch impedances are identical in a balanced system, it follows that the resulting currents are also balanced. There are two types of 3-φ circuit connections. 1. Star connection (Y) is as illustrated below
23
In a Y-connected circuit the phase voltages are measured between the lines and the neutral or star point: so we have VRN (red phase voltage), VYN (yellow phase voltage) and VBN (blue phase voltage). The line voltages are measured between the lines R, Y, and B: so we have VRY (voltage between red and yellow lines), VYB (voltage between yellow and blue lines) and VBR (voltage between blue and red lines). The phase currents are IR, IB and IY 2. Delta connection (∆) Here the component connections form the sides of a ∆. The neutral wire or star-point is not available in a ∆connected circuit. The line currents are I1 (in red line), I2 (in yellow line) and I3 in blue line. The phase currents are IR (directed from Y to R nodes of the ∆), IB (from R to B nodes) and IY (from B to Y nodes). A phasor diagram for these quantities is as follows.
Fig. Line current, phase current and phase voltage phasors in a ∆-connected circuit Let VL denote line voltage, VP phase voltage, IL line current and IP phase current. It is easy to show from appropriate phasor diagrams that: For a star-connected 3-phase circuit • The VP lags VL by 30o • VL= 3 VP • IL=IP For a ∆-connected circuit • IL lags IP by 30o • VL=VP •
IL=
3 IP
POWER Power in a ∆- or Y-connected system is given by P=
3 VLIL cosφ 3 VLIL sinφ
(Real power)
(Reactive power) Q= where φ = angle between line voltage and current phasors
24
S=
P 2 + Q 2 = 3 VLIL
(Apparent power)
cosφ is the power factor. Rating of 3-φ equipment such as generators, motors, transformers or transmission lines are normally given as total or 3-φ power in MW, or as total apparent power in MVA, along with the line-line voltage in kV. WORKED EXAMPLE A generator supplies a load through a feeder whose impedance is Zfdr = 1 + j2Ω, the load impedance is ZL = 8 + j6Ω, the voltage across the load is 220V. Find the real power and the reactive power supplied by the generator take the load voltage VL as the reference phasor.
Solution Zfdr = 1 + 2j = 2.24 ∠ 63.40 Ω ZL = 8 + j6 = 10 ∠ 36.90 Ω VL = 220 ∠ 00 V
VL 220∠0 0 = ∴ IL = = 22∠ − 36.9 0 A 0 Z L 10∠36.9 Kirchoff’s voltage law yields, Eg = VL + IL Zfdr = 220 ∠ 00 + (22 ∠ -36.90)(2.24 ∠63.4 ) = 264 ∠ 4.80 The complex power supplied by the generator is given by S = Eg Ig * = (264 ∠ 4.8) x (22 ∠ 36.90) = 4336.5+j3863.6 0
= 5,808 ∠41.7 kVA 0
MEASUREMENT OF 3-φ POWER 1. Single wattmeter method: This is employed when the neutral is available and the system is balanced. The total power taken by the load is 3 times the wattmeter reading W.
2.
Two-wattmeter method: This is employed for both balanced and unbalanced loads.
25
The wattmeter readings are W1= VLIL cos (30o +φ) W2= VLIL cos (30o+φ) P= W1+W2 P= VLIL [cos (30o+φ) + cos (30o-φ)] P= 3 VLIL cosφ The difference between the wattmeter readings is W2-W1= ILVL [cos (30-φ)-cos (30+φ] = VLIL Sinφ Which is, 1 / 3 times the total 3-φ reactive power. Thus, reactive power, is given by Q = 3 [W2 – W1] The power factor is given by
Q ] P ⎡ 3 ( w2 − w1 ) ⎤ -1 ⎢ ⎥ = tan ⎣⎢ w2 + w1 ) ⎦⎥
φ = tan-1 [
The power factor is given by pf = cos(φ) Let Wh be the higher wattmeter reading and WL be the lower reading. a)
If the ratio
WL > 0.5, both wattmeter readings are positive and the total power is WT = Wn + WL whether Wh
power factor is leading or lagging. b) If c)
WL < 0.5 is indicative of the smaller wattmeter reading negatively and the total power WT = Wh – WL for Wh
both leading and lagging power factor. For power factor = 1(cos0o = 1) we have a purely resistive load and Wh = WL i.e. both wattmeters have the same reading.
d) When power = 0(cos90o = 0) we have a purely reactive load with indicate same reading but with opposite signs. Worked Example: The phase sequence of the Y-connected generator in fig θ1 is RYB. a) Find the phase angles θ2 and θ3 b) Find the magnitudes of the line voltages c) Find the line currents
26
WL = 1 or WL = -Wh. Both wattmeter Wh
d) Verify that the current in the wire is zero
Solution a)
The phase angles are: θ2 = -120o θ3 = +120o or – 240o
b) EL =
3E P
= 3 x 120 = 208V ∴ ERY = EYB = EBR = 208V c) EP = VP ∴ Vrn = ERN Vyn = EYN Vbn = EBN Irn =
Iyn =
Ibn =
Vrn 120∠0 o 120∠0 o = = = 24∠ − 53.13 o ∆ 3 + j4 Z rn 5∠53.13 o
Vyn Z yn
=
120∠ − 120 o 120∠ − 120 o = = 24∠ − 173.13o ∆ o 3 + j4 5∠53.13
Vbn 120∠120 o 120∠120 o = = = 24∠66.87 o ∆ o Z bn 3 + j4 5∠53.13
d) KCL yields IN = Irn + Iyn + Ibn Irn = 24 ∠ − 53.13 = 14.4 – j19.20 o
Iyn = 24 ∠ − 173.13 = -23.83 – j2.87 o
Ibn = 24 ∠66.87 = o
9.43 + j22.07 0 + j0
∴ For a balanced system the neutral current is zero. Worked Example: For the 3-θ system in fig. below the phase sequence is RYB a) Find the phase angles θ2 and θ3 b) Find the current in each phase of the load
27
c)
Find the magnitude of the line currents
Solution (For RYB phase sequence) a) The phase angles are θ2 = -120o θ3 = 120o or – 240o b) VP = EL, therefore Vry = ERY, Vyb = EYB AND Vbr = EBR The phase currents are Iy =
c)
Vry Z ry
150∠0 o 150∠0 o = = = 15∠ − 53.13o ∆ o 6 + j8 10∠53.13
Ib =
150∠ − 120 o = 15∠ − 173.13 o ∆ o 10∠53.13
Ir =
150∠120 o = 15∠66.87 o ∆ o 10∠53.13
IL = =
3I P 3 × 15
= 25.95∆ Since it is a balanced load system, IR = IY = IB = 25.95∆ WORKED EXAMPLE: A 500KVA, 6.6kV alternator is ∆-connected. Calculate the current in each phase of the alternator when it delivers its full output to a balanced 3-φ load at a power factor of 0.8 lagging. Solution: Power supplied by the alternator P = 3V L I L cos ϕ Power supplied, P = 500 x 0.8 = 400kW Thus,
400,000 =
3 × 6600 I L × 0.8
28
400,000
∴ IL =
3 × 6600 × 0.8
= 43.8 A
For a ∆ connection IP =
IL
3
=
43.8 3
= 25.3 A
WORKED EXAMPLE: A balanced 3-φ load is connected to 2.2 kV feeder. The load draws a line current of 60A at a pf of 0.9 lagging. Calculate the real, reactive, and apparent power absorbed by the load. Solution: VL = 2200V IL = 60A P.F = 0.9 lag ⇒ φ = -cos-1(0.9) = -25.8o P=
3 VLIL cosφ = 205.8kW
Φ=
3 VLIL sinφ = 99.7kVAR 3 VLIL = 228kVA
S=
WORKED EXAMPLE: A 3-φ, 1.5, kW, unity power factor; heating unit, and a 5 hp induction motor with a full load efficiency of 80% and pf of 0.85 are served from the same 3-ф, 3-wire 208V system. Find the magnitude of the line current for rated output from the 5hp motor. Note that 1 hp = 746W. Solution: Motor Output = 5x746=3730W Motor Input =
3730
η
=
3730 = 4662W 0.8
The motor is a balanced 3-φ load so that P=
3 VLILmcos φ
3 x 208 x ILm x 0.85 4662 ILm = = 15.25 A 3 × 208 × 0.85
4662 =
Ф = -cos-1 (0.85) = -31.7o Hence, the motor line current is ILm = 15.25 ∠ -31.70A For the heater P = 1500 = ILh =
3 VLILh 3 x 208 x ILh 1500 = 4.16∠0 o A 3 × 208
Therefore, total line current is the phasor sum of motor and the heater line current. IL = 15.25 ∠ -31.7o + 4.16 ∠ 0o IL = 18.9 ∠ -25.1oA WORKED EXAMPLE: A ∆- connected load consists of three identical impedances of 8+j6Ω each and supplied from a three phase 200V source. Calculate (a) the power factor (b) the phase current and line current (c) the real, reactive and apparent power taken by the load. Solution (a)
Z∆ = 8 + j6 = 10 ∠ 36.9oΩ
29
VL = Vph. = 200 ∠ 0o Iph =
Vph
=
200∠0 o 10∠36.9 o
Z∆ Iph = 20 ∠ - 36.9o IL = 3 Iph ∠ (-36.9o-30o) = 34.6 ∠ -66.9oA
(b) (c)
PF = cos36.9o = 0.8 lagging P = 3VphIph cosφ = 3 x 20 x 200 x 0.8 = 9600W QL = 3Vph x Iph sin = 3 x 20 x 200 x sin(36.9o) QL = 7200 VAR SL = 3VphIph = 3 x 200 x 20 = 7200 PL =
3 VLIL cosφ = 96KW
3 VLIL sinφ = 7.2KVAR S = 3 VLIL = 12KVA
Q =
If the impedance is Y- connected we have ZY = 8 + j6Ω VL = 200 ∠ 0oV Vph =
200 3
∠0 o = 115.5V∠0 o V
V ph
115.5 = 11.53∠ − 36.9 0 A ZY 8 + j6 IL = IP = 11.53 ∠ -36.90A
(a) Iph =
=
(b) PF = cos36.9o = 0.8 lagging
3 VLIL cosφ = 3 x 200 x 11.55 cos36.90 = 3.2kW QL = 3 VLIL sinφ = 2.4kVAR
(c) PL =
SL = 3 VLIL = 4kVA WORKED EXAMPLE: An unbalanced Y – connected load is supplied by a 3-φ 3- wire 208V RYB system. Find the line current at the voltage across the load impedance.
30
The mesh equation by inspection is
⎡10∠0 o + 15∠ − 3o ⎢ o ⎣⎢− 15∠30 I1 I2
⎤ ⎡ I 1 ⎤ ⎡208∠120 o ⎤ ⎥ ⎢ ⎥ =⎢ ⎥ 15∠30 o + 10∠ − 30⎦⎥ ⎣ I 2 ⎦ ⎢⎣208∠0 o ⎦⎥ − 15∠30 o
5210∠90 o = 14.15 ∠ 86.1A 367.5∠3.9 3730∠56.6 o = = 10.15 ∠ 52.7A 367.5∠3.9 o =
Line Currents IR = I1 = 14.5 ∠ 86.1o A IY = I2 – I1 = 8.0 ∠ -49.5o A IB = - I2 = - 10.15 ∠ 52.7o or 10.15 ∠ (52.7 – 180o) A Load Voltage VRN = IRZR = (14.5 ∠ 86.1)(10 ∠ 0o) 141.5 ∠ 86.1o V VYN = IYZY = (8 ∠ -49.5o)(15 ∠ 30o) = 120 ∠ -19.5o V VBN = IBZB = (10.15)(-127.3o)(10 ∠ -30o) = 101.5 ∠ -157.3o V EXERCISE 1. Each arm of a ∆-connected load consists of resistance of 20Ω in series with an inductive reactance of 34.6Ω. Two wattmeters are used to measure the power supply to the load from a balanced 415V 3-Φ supply. Calculate the magnitude of the line and phase currents and the wattmeter readings. Taking as the reference the voltage between the supply neutral and the line without a wattmeter current coil (line B) .Give the phase angles of all line and phase currents. Answer IL = 18A ∠ -30 ∠ -150 ∠ -90. Iph = 10.4A, ∠ -60o ∠ -180o ∠ -60o wattmeter reading WR = 6490W, WY = 0W 2. (a) (b) (c) (d) 3. 4.
A 345kV, 3-φ transmission line delivers 500MVA at 0.868 pf lagging to a 3-φ load connected to the receiving end terminals. Assume that the load is delta-connected and the voltage at the receiving end is 345kV. Find the complex load impedance per phase. Calculate the line and phase currents. Find the real and reactive power per phase. Find the total real and reactive power. Repeat problem 2 assuming the load is Y-connected. A 3-φ sub-station bus supplies two Y-connected loads that are connected in parallel through a 3-φ feeder with an impedance of 0.5 + j0.2 Ω per phase. Load 1 draws 50kW at 0.868 lagging pf and load 2 draws 32 kVA at 0.9 leading PF. The line to line voltage of the load is 460V. Find the following: (a) The impedance of each load per phase. (b) Total line current flowing through the feeder. (c) Line to line voltage at the sub-station bus. (d) Total real and reactive power supplied by the bus.
3.5 PER UNIT NOTATION
31
In the per-unit (pu) notation, any electrical quantity may be expressed in pu as the ratio of the actual quantity and the chosen base value for the quantity expressed either in decimal form or percent. Four basic quantities must be considered, namely, power Sb, voltage Vb, current Ib, and impedance Zb. In single phase systems, the relationships among these quantities are: Sb = VbIb Vb = IbZb With only two equations relating the four basic quantities, it is necessary to specify two base values. The power and voltage bases are usually chosen equal to the rated value and the other two are computed from the above equations as follows. Ib =
Sb Vb
Zb =
( KVb ) 2 Vb (Vb ) 2 = = Ib Sb MVAb
The specified power base is applicable to all parts of the power system. The voltage base varies across a transformer and so do the current base and impedance base. The pu electrical quantities are calculated as follows: Su =
P + jQ = Pu + jQu Sb
pu
V pu Vb I pu Iu = Ib Z Zu = pu Zb Vu =
It can be seen by inspection of any power system diagram that: a. Several voltage levels exist in a system b. It is common practice to refer to plant MVA in terms of per unit or percentage values c. Transmission line and cable constants are given in ohms/km. Before any system calculation can take place, the system parameters must be referred to ‘base quantities’ and represented as a unified system of impedances in the either ohmic, percentage, or per unit values. The base quantities are power and voltage. Normally, they are given in terms of the three-phase power in MVA and the line voltage in kV. The base impedance resulting from the above base quantities is:
Zb =
(kVb ) 2 ohms MVA b
(3.18)
and, provided the system is balanced, the base impedance may be calculated using either single-phase or three-phase quantities. The per unit or percentage value of any impedance in the system is the ratio of actual to base impedance values. Hence:
MVA b ⎫ ⎪ (kVb ) 2 ⎬ ⎪ Z(%) = Z(pu) × 100 ⎭
Z(pu) = Z(ohms) ×
(3.19)
where, MVAb = base MVA kVb = base kV Simple transposition of the above formulae will refer the ohmic value of impedance to the per-unit or percentage values and base quantities. Having chosen base quantities of suitable magnitude all system impedances may be converted to those base quantities by using the equations given below:
32
MVA b2 ⎫ MVA b1 ⎪⎪ 2⎬ ⎛ kVb1 ⎞ ⎪ ⎟⎟ = Z b1 × ⎜⎜ ⎪ ⎝ kVb2 ⎠ ⎭
Z b2 = Z b1 × Z b2
(3.20)
where suffix b1 denotes the value to the original base and suffix b2 denotes the value to the new base The choice of impedance notation depends upon the complexity of the system, plant impedance notation and the nature of the system calculations envisaged. If the system is relatively simple and contains mainly transmission line data, given in ohms, then the ohmic method can be adopted with advantage. However, the per-unit method of impedance notation is the most common for general system studies since: 1. Impedances are the same referred to either side of a transformer if the ratio of base voltages on the two sides of a transformer is equal to the transformer turns ratio 2. Confusion caused by introduction of powers of 100 in percentage calculations is avoided 3. By a suitable choice of bases, the magnitudes of the data and results are kept within a predictable range, and hence errors in data and computations are easier to spot. Most power system studies are carried out using software in per unit quantities. Irrespective of the method of calculation, the choice of base voltage, and unifying system impedances to this base, should be approached with caution, as shown in the following example (Fig. 3.6).
Wrong selection of base voltage 11.8kV
132kV
11kV
141kV
141 x 11 = 11.7kV 132
Right selection 11.8kV
Fig 3.6 Selection of base voltages WORKED EXAMPLE A 500KVA, 6.6kv alternator is ∆ connected. Calculate the current in each phase of the alternator when it delivers its full output to a balanced 3-φ load at a power factor of 0.8 lagging. Solution. Power supplied by the alternator P = 3V L I L cos ϕ Power supplied, P = 500 x 0.8 = 400kw Thus,
3 × 6600 I L × 0.8 400,000 ∴ IL = = 43.8 A 3 × 6600 × 0.8
400,000w =
For a ∆ connection IP =
IL
3
=
43.8 3
= 25.3 A
WORKED EXAMPLE A balanced 3-φ load is connected to 2.2 kV feeder, the load draws a line current of 60A at a P.f of 0.9 lagging. Calculate the real, reactive, and apparent power absorbed by the load. Solution
33
VL = 2200V IL = 60A P.F = 0.9 lag ⇒ φ = -cos-1(0.9) = -25.80 P=
3 VLIL cosφ = 205.8KW
Φ=
3 VLIL sinφ = 99.7KVAR
S=
3 VLIL = 228KVA
WORKED EXAMPLE A 3-φ 1.5 KW, unity power factor; rating unit, and a 5 horse power induction motor with a full load efficiency of 80% and P.f of 0.85 are served from the same 3Q, 3 wire 208v system. Find the magnitude of the line current for rated O/P from the 5HP motor. Note that 1 HP = 746W. Solution Motor Output = 5x746=3730w Motor Input =
3730
η
=
3730 = 4662W 0.8
The motor is a balanced 3-φ load so that
3 VLIL cos φ
P=
3 x 208 x IL x 0.85 4662 = 15.25 A 3 × 208 × 0.85
4662 = IL =
Ф = cos-1 (0.85) = 31.70 Hence, the motor line current is IL = 15.25 ∠ -31.70A For the heater P = 1500 = ILH =
3 VLIL 3 x 208 x IL 1500 = 4.16∠0 0 A cosφ = 1 3 × 208
Therefore, total line current is the phasor sum of motor and the later line current. IL = 15.25 ∠ -31.70 + 416 ∠ 00 IL = 18.9 ∠ -25.10A WORKED EXAMPLE A ∆- connected load consists of three identical impedance of 8+j6w each and supplied from a three phase 200V source. Calculate (a) the power factor (b) phase current and line current (c) the real, reactive and apparent power taken by the load. Solution VL = Vph. = 200 ∠ 00 Z ∆ = 8 + j6 =10 ∠ 36.90 Ω 0 (a) pf = cos36.9 = 0.8 lagging (b) Iph = IL =
V ph ZA
=
200∠0 0 10∠36.9 0
Iph = 20 ∠ -36.90
3 Iph ∠ -300 = 34.6 ∠ -66.90A
(c) P = 3VphIph cosφ = 3 x 20 x 200 x 0.8 = 9600W QL = 3Vph x Iph sin = 3 x 20 x 200 x sin(36.90) QL = 7200VAR SL = 3VphIph = 3 x 200 x 20 = 7200 kVA
34
PL =
3 VLIL cosφ = 96 kW
Q =
3 VLIL sinφ = 7.2 kVAR
S =
3 VLIL = 12kVA
If the impedance is Y- connected we have Zy = 8 + j6Ω VL = 200 ∠ 0o V Vph =
200 3
(a)
∠0 0 = 115.5V∠0 0 V
pf = cos36.9 0.8 lagging
V ph
115.5 = 11.53∠ − 36.9 0 A Z y 8 + j6 IL = IP = 11.53 ∠ -36.90A (c) PL = 3 VLIL cosφ = 3 x 200 x 11.55 cos36.90 = 3.2kW
(b) Iph =
=
QL =
3 VLIL sinφ = 2.4 kVAR
SL =
3 VLIL = 4kVA
WORKED EXAMPLE A generator supplied a load through a feeder whose impedance is Zf = 1+j202. the load impedance ZL = 8+j602 the voltage across the load is 120V. find the real power and reactive power supplied by the generator using per unit representation. Take the load voltage as the reference phasor and choose Sb = 1500VA, Vb = 120V Solution Ib =
S b 1500 = = 12.5A Vb 120 2
V (120) 2 Zb = b = = 9.6 Sb 1500 Per unit value (PU)
120∠0 10∠36.9 0 = 1.0 ∠ 00 ZL = = 1.04 ∠ 36.90PU 120 9.6 2.24∠63.4 1.0∠0 0 0 = 0.96∠ − 36.9 Zfd = = 0.233 ∠ 63.40PUm IL = 0 9.6 1.04∠36.9 VL=
Generator voltage Eg = VL + IL Zfdr = 1 ∠ 00 + (0.96 ∠ -36.90)(0.233 ∠ 63.10) = 1.204 ∠ 4.8PU V The real voltage for generator is 1.204 ∠ 4.80 x 120-----base voltage = 144.5 ∠ 4.8v Complex power Sg = EgIg* =Eg IL* ∴ Sg = (1.204 ∠ 4.80) (0.96 ∠ -36.90) = 1.156 ∠ 41.70 = 0.863 + j0.969 pu Pg = 0.863 x 1500 = 1295W Qg = 0.769 x 1500 = 1154VAR. ADVANTAGES OF THE PER UNIT SYSTEM 1. It eliminates the need for conversion of the voltages, current and impedances across every xmer in a given network. This because the P.U value of this quantities are the same on both sides of the xmer. Moreover the P.U impedance are
35
unchanged when referred from primary to secondary and verser. Furthermore, in comparing xmer of different ratings, comparison of P.U impedance is to half full than comparison of actual impedance i.e. if 2 xmers have the same P.U leakage reactance and resistances, there regulation are the same for all load P.F, although their actual reactance and resistances may be very different. 2. The need to transform from 3φ to single phase equivalent vice verser is avoided with hence there is less confusion in handling and in handling various parameters in 3φ system.
the use of P.U quantities
EXERCISE A single phase source is connected to an electrical load, the load draws 0.6 pu current at 1.1 pu voltage while taking a real power of 0.4 pu at a lagging pf. Choose a base voltage of 8kV and a base current of 125A. Calculate the following. (a) Real power in kW (b) Reactive power in kVAR (c) pf (d) Ohmic values of resistance and reactance of the load.
36
CHAPTER FOUR POWER FACTOR CORRECTION 4.1 Introduction The power factor (pf) of a given circuit is equal to the cosine of the phase angle between the circuit voltage and current phasors. The voltage is always taken as reference. An ordinary consumer of electrical energy is not interested in the phase angle between the voltage and the current applied to the equipment (e.g. electric motor). Concern is usually not with the power factor but with the actual real power output (kW) of the equipment. However, for constant voltage and output power, the current, I, through the supply feeder depends on the power factor according to the relation, I=
Pi n
(4.1)
3V L cos φ
where Pin is the power supplied by the feeder, VL is the line voltage, and cosΦ is the power factor. Note that from eqn. (4.1), a reduction in the power factor (cosΦ) from unity to 0.8 will result in a 25% increase in current I. Conversely, an increase (or improvement) in the pf will result in a reduction in the current, which will reduce the following: a) The losses in the feeder (I2R). b) The required size of feeder conductors (conductor size depends on current carrying capacity). c) The required rating of the transformer and other station equipment (e.g., a lower rated transformer can serve) d) The losses in the transformer. e) The voltage drop in the feeder. This results in improved voltage regulation. In general, the efficiency of power generation, transmission and distribution equipment is improved when operating near unity power factor. Consequently, utilities encourage consumers of electric energy to improve or correct the power factor of their electrical load. An improved power factor implies reduced reactive power supplied by the source. Equipment that compensate for the reactive power demand of the load are often used: e.g. static capacitors, synchronous condensers, and phase advancers. This chapter deals with principles of power factor correction. 4.2 Concept of Power Factor Correction To introduce the concept of power factor improvement, we consider three types of load connected to a 2.3kV power line: a pure capacitive load, an inductive load and a combination of capacitive and inductive loads. 4.2.1 Pure Capacitive Load For the pure capacitive load shown in Fig 4.1, the current drawn from the supply source is given by:
Ic =
2300∠0 o 230 o = =230 ∠ 90oA − j10 10∠ − 90 o
Fig 4.1 Pure capacitive load The current phase angle is positive indicating that the current phasor leads the voltage phasor by 90o. The apparent power delivered to the capacitive load is Sc = VI* = 2300 ∠ 0o x 230 ∠ -90o = 529 ∠ -90o kVAR Hence, we observe the following: 1. The capacitor does not draw real power, i.e. Pc = 0 kVA 2. The capacitor draws energy as it is charged, but returns the same energy during the discharge part of each cycle. A wattmeter would read zero. In this case the reactive power Qc = -529 kVAR. This can be costly to a power company since the supply system must be sized to carry the current without unreasonable voltage drop or power loss. For this reason power factor meters are installed to monitor large industrial consumers. 3. The power factor, pfc = cos φ = cos 90o =0
37
4.2.2. Inductive Load Fig 4.2 shows an inductive load connected to the 2.3kV power line.
The load impedance is: ZL = 3 + j4 = 5 ∠ 53o Ω.
Fig 4.2 Inductive load
4⎞ ⎛ cos⎜ tan −1 ⎟ = 0.6 3⎠ ⎝ o 2300∠0 The load current, IL = = 460 ∠ -53oA 5∠53o The load power factor is, pf =
The current phase angle is negative indicating that the current phasor lags behind the voltage phasor. The power delivered to the load is given by : SL = VI* ∴ SL = (2300 ∠ 0o)(460 ∠ -53o) SL = 635 + j846kVA = PL+QL where, PL = 635kW, QL = 846 kVAR. If this load is an industrial plant, PL corresponds to the “useful” power purchased from the power company to light and heat the plant and to run power tools, etc. The reactive power QL, has only nuisance value. The apparent power dawn by the load is: SL =
(846 2 + 635 2 ) = 1058kVA
The power triangle for this load is sketched in Fig. 4.3. The apparent power dawn by the load is SL =
(846 2 + 635 2 )
= 1058 kVA
Fig 4.3 Power triangle for inductive load Note that: (1) (2)
The real or active power absorbed by the load is that actually the one taken by the resistive component of the load in form of I2R, i.e. PL=4602x3 =635 kW. The pure inductive component of the load is responsible for the reactive power, QL = I2XL = 4602x4= 846 kVAR. The larger the inductor reactance XL becomes, the larger the load impedance angle and the larger the reactive power the source will supply and hence the smaller the power factor. The reactive power taken from the source is not utilized by the load but rather is returned to the source.
4.2.3. Combined Capacitive and Inductive Loads Fig. 4.4 shows a combination of the capacitive and inductive loads.
38
Fig. 4.4 Combined capacitive and inductive loads Since the current through each of the elements remains unchanged the resulting active and reactive power drawn from the source is: P = 0+635 = 635 kW, Q = -529+846 = 317 kVAR. The power factor of the combined load is given by
pf = cos(tan −1
317 Q ) = cos(tan −1 ) = 0.89 635 P
The new phase angle is Φ=cos-1(0.89) = 27o The total current drawn by the combined load is
I T = I c + I L = 230∠90 0 + 460∠ − 530 = j 230 + 276.2 − j 367.9 276.2 − j137.9 = 308.7∠27 0 A The current drawn by the inductive load has been reduced from 460 A to 308 A as a result of connecting the capacitor in parallel with the load. Moreover, the overall phase angle has reduced from 53o to 27o. The smaller the phase angle, the better the system power factor. The addition of the capacitor, which is common practice in large power systems, reduces the reactive power seen by the incoming feeder, and has brought the power factor much closer to unity. Example 4.1 An industrial plant draws 250kW with pf of 0.707 lagging, a capacitor is to be added in parallel with the plant to improve the overall pf to 0.866 lagging. Find the value of the capacitance required. The source is 2.3kV, 60Hz Solution Plant: pf = 0.707 lagging, Pp = 250kW Now, pf =
∴ SP =
PP SP
PP 2.50 = = 353.6∠φ o kVA pf 0.707
Qp = Sp sinф = Sp sin(cos-10707) = Sp sin(45o) Qp = 250kVAR Sp = 250 + j250 kVA For the plant capacitor combination Q2 = cos-1(0.866) = 30o S2 = 250 + jQ2 Now, tan300 =
Q2 and therefore, Q2 = 250 tan300 = 144KVAR 250
Thus the required capacitor rating is: Qc = 144 – 250 = -106 kVAR
∴ Qc = VIc sin(-900) = -106 x 103 =
V2 1 /(ωC )
f = 60Hz ω = 2πf = 2π x 60 rad/S
39
∴C=
106 × 10 3 = 53.1µf (2300) 2 (2π × 60)
EXAMPLE 4.2 A three phase motor draws 20kVA at 0.707 pf lagging from a 220V source. It is desired to improve the pf to 0.9 lagging by connecting a capacitor bank across the terminals of the motor a) Calculate the line current before and after the addition of the capacitor bank. b) Determine the required kVA rating of the capacitor bank. Solution The motor real and reactive powers are Pm = 20 x 0.707 = 14.14 kW Qm = 20 sin [cos-1 (0.707)] = 14.14 kVAR a) The current of the motor is Im =
Sm
3VL
=
20,000 3 × 220
= 52.5 A
Thus before power factor correction the line current is 52.5 A The corrected power factor is, pfcorr = 0.90 The new line current is, Icorr =
Pm
3VL × Pf corr
=
14140 3 × 220 × 0.9
= 41.2A
That is, after power factor correction the line current is 41.2A b) The corrected value of reactive power is Qcorr = Pm tan(cos-10.9) = 14.14 tan(25.80) = 6.85kVAR. The kVAR rating of the capacitor bank required to improve the pf from 0.707 to 0.9 lagging is found as Qcap = Qcorr - Qm = 6.85 – 14.14 = -7.29 kVAR Thus the capacitor bank rated 7.29kVA will correct the power factor to 0.9 lagging 4.3 USE OF STATIC CAPACITORS The least expensive way to achieve near unity pf is with the use of power-factor correction capacitors. Capacitors provide a static source of leading reactive current and can be installed close to the load. The leading current drawn by the capacitors may partly or completely neutralize the lagging reactive component of the load current. This raises the pf of the load. Four of the most common capacitor connections are illustrated below. a) 3-Φ grounded-star b) 3-Φ ungrounded-star c) 3-Φ delta d) 1-Φ Grounded or ungrounded Y-connections are usually made on high-voltage circuits (primary) while ∆ and 1-Φ connections are usually made on low-voltage circuits (secondary) through protective fuses. Majority of power capacitor equipment installed on primary feeders is connected grounded-Y due to the following benefits it offers. i) With grounded-Y connections switch tanks and frames are at ground potential. This provides for increased personnel safety. ii) It provides for faster operation of the series fuse in case of capacitor failure. iii) Grounded capacitors can by-pass some line surges to ground and therefore exhibit a degree of self protection from transient voltages and lightning surges. iv) It provides a low impedance path for harmonics.
40
4.3.1 Advantages of static capacitors • Low losses (almost negligible) • Require little maintenance as there are no moving parts • Can be easily installed as they are light and require no foundation • Operate under ordinary atmospheric conditions 4.3.2 Disadvantages static capacitors • Short service life ranging from 8 to 10 years • Easily damaged when the voltage exceeds the rated value • It is un-economical to repair damaged capacitors 4.4 USE OF SYNCHRONOUS CONDENSERS A synchronous motor draws a leading current when over-excited and therefore behaves like a capacitor - hence the name synchronous condenser. When connected to the power supply the leading current it draws may partially or completely neutralize the lagging reactive component of the load current – thereby improving the pf. 4.4.1 Advantages of synchronous condensers • The motor can provide output torque and leading or lagging KVAR to the power network. • The operation is very smooth since by varying the field excitation the magnitude of current drawn by the motor can be changed by any amount – hence step-less control of pf is achievable. • Faults can easily be removed. • Effects of harmonics and switching problems are less severe than is the case with capacitors. 4.4.2 Disadvantages of synchronous condensers • There are nominal losses in the motor • High maintenance cost (bearings, ventilation, etc) • Produces noise • Except for sizes above 500 kVA the cost is greater than that of sane rated static capacitors • Since it has no self-starting torque, auxiliary equipment has to be provided to run it to synchronous speed It should be noted that neither the consumer nor supply authority benefits by over-correction of power factor. An optimal value of power factor can be computed based on economic considerations. In any case corrective appliances should be switched on or off with the loads which they correct. Example 4.3 A single phase source is connected to an electrical load. The load draws a 0.6 pu current at 1.10 pu voltage while taking a real power of 0.4 pu at a lagging power factor. Choose a base voltage of 8 kV and a base current of 125 A. Calculate the following: a) Real power in kW b) Reactive power in kVAR c) Power factor d) Ohmic values of the resistance and reactance e) The capacitor kVA rating required to improve the power factor to 0.9 lagging. Solution Vb = 8kV, Ib = 125A ∴ Sb = VbIb = 8 x 125 = 1000 kVA Iu = 0.6pu, Vu = 1.1pu and Pu = 0.4 pu pf lagging a) Real power P = 0.4 x 1000 = 400kW b) Apparent power Su = 1.1*0.6 = 0.66pu Reactive power Qu =
S u2 − Pu2 = 0.66 2 − 0.4 2 = 0.525pu
∴ Q = 0.525 x 1000 = 525kVAR ⎛ −1 Q ⎞ c) Power factor, pf = cosΦ = cos ⎜ tan ⎟ P⎠ ⎝
41
⎡ −1 525 ⎤ = 0.606 lagging. 400 ⎥⎦ ⎣ 8000 d) Base impedance Zb = = 64Ω 125 V 1.1 New Zu = u = = 1.833 pu ⇒ Z = Zu x Zb = 1.8 33 x 64 0.6 Iu = cos ⎢ tan
ф = cos-10.606 = 52.70
e)
∴ Z = 117.33 ∠52.7 0 = 71.1+ j93.33Ω ∴ R = 71.1Ω and X = 93.33Ω
Φnew = cos-10.8 = 36.870 Qnew = P tanΦnew = 400 tan36.870 = 400 x 0.75 = 300kVAR Qnew = Qold +Qc ⇒ 300 = 525 + Qc ⇒ Qc = -225kVAR Qc = 225kVA
EXAMPLE 4.4 An over-excited synchronous motor is connected across a 250kVA inductive load of 0.6 lagging power factor. The motor takes 20kW while running on no-load. Calculate the kVA rating of the motor in order to raise the overall power factor of the motor plus inductive load combination to 0.95 lagging. Solution SL = 250kVA, ФL = - cos-1(0.6) = -53.1o, PL=250x0.6=150 kW QL = PL tan(cos-10.6) = 150 tan(53.1o) = 200kVAR PM = 20kW, PT = PL + PM = 150 + 20 = 170kW QT = PT tan(cos-10.95) = 170 tan(18.2o) = 55.9kVAR QM = QT - QL = 55.9 – 200 = -144.1kVAR SM =
PM2 + Q 2M = 20 2 + (−144.1) 2 = 145.5kVA
Hence a150kVA synchronous motor will do.
42
CHAPTER FIVE STATION LOAD CHARACTERISTICS AND ECONOMICS 5.1 Introduction The ultimate objective of any power system is to deliver electrical energy to the consumer safely, reliably, economically and with good quality. Operation of the power system requires that proper attention be given to the safety not only of the utility personnel but also of the general public. At the load centre, electrical energy is converted to other more desirable and useful forms of energy. This implies that electricity supply should be available where, when and in whatever amount the consumer requires. The quality of the supplied electrical energy is partially dependent on energy usage by the consumers. When there is high demand on the limited capabilities of the power system, voltage may deviate from its acceptable levels. Switching of large machinery could cause fluctuation of the voltage as well as the frequency. Usually high and prolonged demands may lead to overloaded equipment, which may cause tripping of protective devices to prevent further damage to the equipment. Finally, the power system must be operated such that the overall costs of producing electricity, including all attendant losses in the generation and delivery systems, are minimized. The most economical conditions derive not only from proper operating procedures but also from efficient system planning and design. 5.2 Categories of Electrical Loads Electrical loads are commonly grouped into four categories - Residential, commercial, industrial, and others as described in chapter six. These are sometimes subdivided into subgroups depending on their usage levels, for example, residential A, B or C. Residential loads are private homes and apartments. They include lighting, cooking, comfort heating and cooling, refrigerators, water heaters, washers and dryers, electronic gadgets for entertainment, pressing irons and many more different appliances. Commercial loads include office buildings, department stores, grocery stores and shops. Industrial loads consist of factories, manufacturing plants, steel, lumber, paper, mining, textile and other industrial factories. In both commercial and industrial groups, loads include lighting, comfort heating and cooling, and various types of office equipment. In addition, industrial loads contain various types and sizes of motors, fans, presses, furnaces, etc. Electrical load refers to the amount of electrical energy or electrical power consumed by a particular device or by a whole community. It is also referred to as electrical demand. At the individual consumer level, the electrical demand is quite unpredictable. However, as the demands of the various users are accumulated and added at a feeder or a substation, they begin to exhibit a definite pattern. 5.3 Load-Duration Curve A typical plot of the electrical load is shown in Fig 5.1, and it is referred to as a load curve. The period of interest is normally one day; thus, it is called a daily load curve. The daily load curve shows the kilowatt demand from midnight to noon to midnight (i.e, over a 24-hour period).
Fig 5.1 Daily load curve
43
The daily load curves for Monday through Friday are similar in shape and peak values. Weekend curves are generally different, particularly for industrial and commercial customers because of shutdown of operations on Saturdays and Sundays. Load curves may be constructed for feeders, substations, generating plants, or for the whole system. Load curves are also drawn for different periods or seasons of year. Thus, the system peak load for the hot season, or rain season, or cold season (e.g. Harmattan), may be read as the highest ordinate from the corresponding load curve. The daily load curves may be accumulated for the whole year and presented in another curve, the annual load duration curve (LDC). A typical LDC is shown in Fig 5.2.
Fig 5.2 Load duration curve The LDC shows the 8760 hourly loads during the whole year, although not in the order in which they occurred. Rather the LDC shows the number of hours during which the load exceeded a certain kilowatt demand. If the area under the LDC curve is calculated and divided by the total number of the hours, the average demand is determined. Several factors used in the electricity supply industry may be defined in terms of parameters of load and load-duration curves: 5.3.1 Load Factor The characteristics of a daily load curve on a gross basis are indicated by peak load and the time of its occurrence and load factor defined as: Load factor, m =
Average load over a given interval Peak load during the same time interval
(5.1)
The average load determines the energy consumption over the day while the peak load along with considerations of standby capacity determines plant capacity for meeting the load. A high load factor helps in drawing more energy from a given installation. As individual load centres have their own characteristics, their peaks in general have a time diversity, which when utilized through transmission interconnection, greatly aids in jacking up load factors at an individual plant - excess power of a plant during light load periods is evacuated through long distance high voltage transmission lines, while a heavily loaded plant receives power. Thus, if a plant’s peak load is 1000MW but the average annual load is 350MW, its annual load factor would be only 0.35. Most of its capacity remains un-utilized for the major part of the year and so the cost would be high. The average load is estimated by dividing the area under the daily load curve by the time period considered. Average load =
Area under load curve (kWh) 24 (h)
(5.2)
5.3.2 Diversity Factor This is defined as the sum of individual maximum demands by the consumers, divided by the maximum load on the system, i.e., Sum of individual consumer groups maximum demand (5.3) Diversity factor = Actual peak load of the system
44
This factor gives the time diversification of the load and is used to decide the installation of sufficient generating and transmission plant. If all the demands came at the same time, i.e. unity diversity factor, the total installed capacity required would be much more. Luckily, the factor is much higher than unity, especially for domestic loads. A high diversity factor could be obtained by: 1. Giving incentives to farmers and/or some industries to use electricity in the night or during light load periods 2. Using day-light saving as in many other countries. 3. Staggering the office timings. 4. Having different time zones in the country like USA, Australia etc. 5. Having two-part tariff in which a consumer has to pay an amount dependent on the maximum demand he makes, plus a charge for each unit of energy consumed. Sometimes a consumer is charged on the basis of kVA demand instead of kW to penalize load of low power factor. Diversity helps to improve the load factor and the economic operation of the power plant. A high load factor is, in general, an indication of balanced load curve with relatively small load changes. High values of demand factor, load factor, diversity factor and capacity factor (define below) are desired for economic operation of the plant and to produce electricity at less cost.
5.3.3 Other Factors Other factors frequently used in connection with power plants are: Plant capacity factor =
Actual kWh produced Installed plant kW capacity Average load = Installed capacity
Actual kWh produced Plant capacity (kW ) × Time (in hours ) in operation kWh generated in a year = 8760xkWinstal
(5.4)
Plant use factor =
(5.5)
Each consumer has a “connected load” which is the sum of the continuous ratings of all the equipment and outlets on the consumer’s circuits. The maximum demand is the maximum load which a consumer uses at any time. It is always less than or equal to the connected load. It is the ratio of the energy produced in a given time to the maximum possible energy that could have been produced during the same time of operation. If the operating time is one year (or 8760) hours, the plant use factor is equal to the capacity factor. If the rated capacity of the plant is equal to the peak load, then the load factor and capacity factor will be numerically equal. The difference between load factor and capacity factor is an indication of the reserve capacity.
Reserve factor =
kWinstal kWmax
=
load factor capacity factor
(5.6)
If the daily load curve of a power station shows a peak demand that is greater than the installed capacity of the station, then the peak load is shared by other stations interconnected with it. The load duration curve is usually divided into three parts 1. Base load 2. Intermediate load 3. Peaking load.
45
The base load is the load below which the demand never falls and is supplied 100% of the time. The peaking load occurs for about 15% of the time. The intermediate load represents the remaining region of the LDC.
5.4 Worked Examples EXAMPLE 1: A power station supplies the following loads to the consumers. Time (hour) 0-6 6-10 10-12 12-16 16-20 20-22 22-24 Load (MW) 30 70 90 60 100 80 60 a) Draw the load curve and estimate the load factor for the plant b) What is the load factor of a standby equipment of 30MW capacity if it takes up all loads above 70MW? What is plant the use factor? SOLUTION: a) The load curve is sketched below:
Energy generated = area under load curve = 30 x 6 + 70 x 4 + 90 x 2 + 60 x 2 + 100 x 4 + 80 x 2 + 60 x 2 = 1560MWh
1560 MW 24 Average load 65 Load factor = = = 0.65 100 peak load
Average load =
b) If the load above 70MW is supplied by a standby unit of 30MW capacity, the energy generated by it is = 20 x 2 + 30 x 4 + 10 x 2 = 180 MWh. Time during which standby unit remains in operation = 2 + 4 + 2 = 8 hours.
180MWh = 22.5MW 8h 22.5 Load factor = = 0.75 30 energy generated 180 × 10 3 kWh = = 0.75 Use factor = plant capacity × operating hrs 30 × 10 3 kW × 8 Average load =
EXAMPLE 2 A substation supplies power via four feeders to its consumers. Feeder No 1 supplies six consumers, whose individual daily maximum demands are 70kW, 90kW, 20kW, 50kW, 10kW and 20kW, while the maximum demand on the feeder is 200kW. Feeder No 2 supplies four consumers, whose daily maximum demands are 60kW, 40kW, 70kW and 30kW, while the maximum demand on the feeder 2 is 160kW. Feeders number 3 and 4 have daily maximum demand of 150kW and 200kW respectively, while the maximum demand on the station is 600kW. Determine the diversity factors for the consumers of feeder No 1, feeder No 2 and for the four feeders. SOLUTION:
⎡ Maximum demand of the Group ⎤ Diversity = ⎢ ⎥ ⎣ Sum of individual Maximum demands ⎦
−1
46
70 + 90 + 20 + 50 + 10 + 20 1 = = 1.3 200 0.77 60 + 40 + 70 + 30 = 1.25 Diversity factor for feeder 2: Div = 160 200 + 160 + 150 + 200 1 Diversity factor for the feeders: Div = = = 1.183 600 0.845
Diversity factor for feeder 1: Div =
EXAMPLE 3 The peak load on a power station is 60MW. The loads having max demands of 30MW, 20MW, 10MW and 14MW are connected to the station. The station capacity is 80MW and the annual load factor is 0.50. Estimate; (a) the average station load, (b) the energy supplied per year, (c) the demand factor, and (d) the diversity factor. SOLUTION : a)
Load factor =
Average load peak load
Average load 60MW ∴ Average load = 30MW Energy supplied per year = average load x 8760 = 30 x 8760MWh = 262.8 x 106kWh Max Demand 60 60 = = 0.811 c) Demand factor = = Connected load 30 + 20 + 10 + 14 74
0.5 =
d) Diversity factor =
Sum of individual Max demand 74 30 + 20 + 10 + 14 = = = 1.233 60 60 Simultaneous Max demand
EXAMPLE 4 A power plant has the following annual factors: load factor = 0.75, capacity factor = 0.60, use factor = 0.65. Maximum demand is 60MW. Estimate: a) The annual energy production b) The reserve capacity over and above the peak load, and c) The hours during which the plant is not in service per year. SOLUTION: Average load Load factor = Peak load ∴ Average load = 0.75 x 60 = 45MW Annual energy production = 45000 x 8760 = 394,200,000kWh = 394.2 x 106kWh Average load Capacity factor = Plant capacity Plant capacity =
45 = 75MW 0.6
Reserve capacity over and above the peak load = 75 – 60 = 15MW Energy generated per year Use factor = Plant capacity × hours in operation
47
∴ Hours in operation =
394.2 × 10 6 kWh
= 8086 hours 75 × 10 3 kW × 0.65 Hours not in service in service in a year = 8760 – 8086 = 674 hours
EXAMPLE 5 A power station has to supply load as follows Time (h) 0-6 6-12 12-14 14-18 18-24 Load (MW) 30 90 60 100 50 a) Draw the load curve b) Draw the load duration curve c) Select suitable generating units to supply the load d) Calculate the load factor e) Calculate the capacity of the plant and the plant capacity factor SOLUTION: a) Draw the load curve b) Draw the load duration curve c) Energy generated = 30 x 6 + 90 x 6 + 60 x 2 + 100 x 4 + 50 x 6 = 1540MWh = 1540 x 103kWh. 1540 × 10 3 kW = 64.17MW 24 3 Max demand = 100 x 10 Kw
Average load =
Load factor =
1540 × 10 3
= 0.64 24 × 100 × 10 3 To supply the load, three generating units, each of 30MW capacity and one generating unit of 10MW capacity will be selected. One additional unit will be kept standby. Its capacity will be equal to that of the largest unit, i.e. 30MW. Load duration curve will indicate the operational schedule of different generating units, which will be as follows One 30MW unit will run for 24 hours i) ii) Second 30MW unit will run for 18 hours Third 30MW unit to run for 10 hours iii) Fourth 30MW unit to run for 4 hours iv) Plant capacity = 30 x 4 + 10 x 1 = 130Kw Energy generated 1540 × 10 3 kWh = 0.494 = Capacity factor = Capacity × operating time 130 × 10 3 × 24kWh
5.4 LOAD FORECASTING
Of all the information, which, collectively, defines the task of supplying electric power to the consumer, none is more important than the estimate of the load. Indeed, the electric load is the most important input in the utility's budget and the most vital for the study of transmission and distribution system, as rates. Load or demand estimation is the starting point for planning the future electric power supply. The consistency of demand growth over the years has lead to numerous attempts to fit mathematical curves to this trend. One of the simplest curves is (5.7) P = P0 e α( t − t 0 ) where α is the average per unit growth rate, P is the demand in year t, and P0 is the given demand at year t0.
Example In a certain country, the energy consumption is expected to double in ten years. Assuming a simple exponential growth given by P = P0 e αt calculate the growth α. Solution Let the starting demand be P0 in year 1. After 10 years the demand double, i.e P=2P0. Thus the demand growth equation yields Or 2 = e10α 2P0 = P0 e10α
48
Taking natural logarithms on both sides gives
λn( 2 ) = 0.0693 10 The tool often used by engineers in making long-term forecasts is termed the method of least squares. The method can be used to establish the trend of random or scattered data accurately. Although the method of least squares is mathematically correct, the accuracy of predictions based on its use is less reliable, as the trend is projected continually further into the future. In practice, a line might be extended into the future with reasonable accuracy for a period of time equal to that of the data used to calculate the trend.
λn( 2 ) = 10α
and
∴ α=
In the electric utility industry, two variations of the method are used. These are the straight line and the compoundgrowth methods. The, compound-growth method is more popular, but the choice of the straight-line method is a matter of individual judgement in any particular situation.
Method of Least Squares Plots of typical historical maximum-hour loads are not usually uniform, because of several factors such as wild variations in weather, industrial shut-downs (especially where industrial load is a substantial proportion of the system load), changes in service area, economic conditions (such as might affect industrial capacity utilization), etc. Table 2 shows a list of annual peak loads, which were obtained from a study of a substation. Table 5.1 Substation annual peak loads %Annual Year Maximum Annual Growth+ Load Growth* 22200 1 17.1 3800 26000 2 23.1 6000 32000 3 21.2 6800 38800 4 1.0 400 38400 5 * Annual growth is the arithmetic difference between the load of two successive years. + Percentage annual growth is the annual growth divided by the maximum load of the preceding year, expressed as a percentage. Based on the above historical data, the load can be predicted five years ahead using the least-squares method. The principle is as follows. Suppose we are given a set of m data points: (x1, y1), (x2, y2),….(xm, ym). We would like to determine a function F(x) such that (5.8) yi = F(xi) + µi For i = 1, 2, ….., m, where the approximation errors µi are small. The form of the function F(x) depends on the problem at hand. Here we assume that it has the form of a linearly weighted sum: n
F(x) =
∑c f ( x )
(5.9)
j j
j=1
where the number of summands n and the specific basis functions fi are chosen based on knowledge of the problem at hand. A common choice is fj = xj-1 which means that F(x) = c1 + c2x + c3x2 + …….+ cnxn
(5.10)
is a polynomial of degree n-1 in x. Let
49
⎡ f1 ( x1 ) f 2 ( x1 ) ⎢ f (x ) f (x ) 2 2 A=⎢ 1 2 ⎢ . . ⎢ ⎣ f1 ( x m ) f 2 ( x m )
f n ( x1 ) ⎤ . f n ( x 2 ) ⎥⎥ = [aij ] ⎥ . . ⎥ . f n ( x m )⎦
.
denote the matrix of values of basis functions at the given points; aij = fj(xi) Let c = ck denote the desired n-vector of coefficients of eqn. (5.10). Then
⎡ f1 ( x1 ) . f 2 ( x1 ) ⎢ f (x ) f (x ) 2 2 Ac = ⎢ 1 2 ⎢ . . ⎢ ⎣ f1 ( xm ) f 2 ( x m )
f n ( x1 ) ⎤ ⎡c1 ⎤ . f n ( x 2 ) ⎥⎥ ⎢⎢c 2 ⎥⎥ ⎥⎢ . ⎥ . . ⎥⎢ ⎥ . f n ( x m ) ⎦ ⎣c n ⎦
.
⎡ F ( x1 ) ⎤ ⎢F (x ) ⎥ 2 ⎥ =⎢ ⎢ . ⎥ ⎥ ⎢ ⎣ F ( x m )⎦
is the vector of “predicted values” of y. thus µ = Ac – y is the m-vector of approximation errors. The least squares solution that minimizes µ is given by c = [(ATA)-1AT]y = A+y where matrix A+ = (ATA)-1AT is called the pseudoinverse of matrix A.
Worked Example Based on the annual peak load data given in table 1 (a) Find the function of the form y = c1 + c 2 x that is the best least-squares fit to the data points. (b) Hence estimate the annual peak load and growth for five more years. Solution: (a) From the given data we form matrix A as follows ⎡1 ⎢1 ⎢ A = ⎢1 ⎢ ⎢1 ⎢1 ⎣
x1 ⎤ ⎡1 x2 ⎥⎥ ⎢1 ⎢ x3 ⎥ = ⎢1 ⎥ ⎢ x4 ⎥ ⎢1 x5 ⎥⎦ ⎢⎣1
1⎤ 2⎥⎥ 3⎥ ⎥ 4⎥ 5 ⎥⎦
50
⎡1 ⎢1 ⎡1 1 1 1 1⎤ ⎢ T A A=⎢ ⎥ ⎢1 ⎣1 2 3 4 5⎦ ⎢1 ⎢ ⎢⎣1
1⎤ 2⎥⎥ ⎡5 15 ⎤ 3⎥ = ⎢ ⎥ ⎥ 15 55⎦ 4⎥ ⎣ 5 ⎥⎦
(ATA)-1 = 1 ⎡ 55 − 15⎤ 50 ⎢⎣− 15
5 ⎥⎦
A+
(ATA)-1AT
=
=
1 ⎡ 55 − 15⎤ ⎡1 1 1 1 1⎤ 1 ⎡ 40 25 10 − 5 − 20⎤ = 5 10 ⎥⎦ 50 ⎢⎣− 15 5 ⎥⎦ ⎢⎣1 2 3 4 5⎥⎦ 50 ⎢⎣− 10 − 5 0 c= A+y ⎡22200⎤ ⎢26000⎥ ⎥ 1 ⎡896000⎤ ⎡17920⎤ 1 ⎡ 40 25 10 − 5 − 20⎤ ⎢ ⎢32000⎥ = = ⎥ ⎢ ⎥ ⎥=⎢ ⎢ 5 10 ⎦ ⎢ 50 ⎣− 10 − 5 0 ⎥ 50 ⎣226000⎦ ⎣ 4520 ⎦ 38800 ⎢ ⎥ ⎢⎣38400⎥⎦
i.e. c1 = 17920 and c2 = 4520 Thus the mathematical model for the annual peak load for the station is: y= 17920 + 4500x kW (b) Based on this model we can estimate the annual peak load for the subsequent five in Table 3.
years. The results are shown
Table 3: Estimated annual peak loads for years 6 to 10 Year X 6 7 8 9 10
Peak Load Y 45040 49560 54080 58600 63120
Annual Growth 6640 4520 4520 4520 4520
51
% Annual Growth 17.30 10.04 9.12 7.71 7.71
CHAPTER SIX TARIFFS 6.1 Introduction Tariffs are prices charged by a utility for supplying the energy needs of consumers. The costs involved in the generation, transmission and distribution of electrical energy fall into two major categories. i) The cost incurred due to maximum demand on the power system (i.e. the capital expenditure of providing sufficient generating plant to supply the maximum needs of all consumers). ii) The cost incurred in operating the power system (i.e. fuel costs for generation of power, maintenance costs, personnel emoluments, etc). Tariffs may be structured such as to influence the load curve and improve the load factor. Some of the tariff structures are: a) Two-part tariff b) Block rate tariff c) Power factor tariff 6.2 Two-Part Tariff In this tariff structure electrical energy charges are divided in two parts: a) A fixed charge on the basis of the maximum demand in kVA b) A running charge based on energy consumed in kWh. For the fixed charge either a maximum demand (MD) meter is installed in the consumer premises to indicate the maximum kVA demanded in a prescribed period of time (e.g. a month) or it is assessed based on installed capacity. Whether or not an MD meter is installed usually depends on the size of demand. When maximum demand is not likely to exceed 50kVA an assessment is made. The cost of metering is of necessity quite high because of the accuracy required to avoid errors which can amount to large cost to the consumer. Consumers of electrical energy may be divided into classes depending on their load characteristics. Typical classes include the following: • Residential • Commercial (shops, offices, etc) • Industrial • Street lighting • Special Class (e.g. Agro-allied enterprises involved in farming; crop cultivation; religious houses; Government and Teaching Hospitals; Government Research Institutes; Education and Secondary Schools; Water Boards). Table 1.1 shows hypothetical two-part tariff codes for the above categories of consumers. Advantages of 2-part tariff • It is easily understood by consumers. • It caters for the fixed charges that depend on maximum demand but are independent of units consumed. Disadvantages • Fixed charges have to be paid whether energy is consumed or not • Requires additional MD meter cost or else there is always error in assessing the maximum demand.
6.3 Block Rate Tariff The rates are fixed for blocks of energy consumed. For example, it may be N 400 per unit for the first 100 units N 300 per unit for the next 200 units N 200 per unit for the rest. The tariff structure has the advantage of encouraging energy consumption and the utility needs only one meter for the purpose. It is usually convenient for domestic or residential consumers. Table 1.1 Sample two-part tariff codes S/N Tariff code Customer’s Fixed Demand Monthly
Meter Maintenance
52
Minimum Charge
Energy Charge
? (N)
1.0 1.1 1.2 1.3 1.4 2.0 2.1 2.2 2.3 2.4 3.0 3.1 3.2 3.3 3.4 3.5 4.0 4.1 5.0 5.1 5.2 5.3 5.4
Residential R1 R2 R3 R4 Commercial C1 C2 C3 C4 Industrial D1 D2 D3 D4 D5 Street lighting S1 Special tariff A1 A2 A3 A4
Levels
Charge (N)
Charge per kVA (N)
per Month (N)
per kWh (N)
< 5kVa 5-15kVA 15-45kVA 45-500kVA
20 30 120 120
100 100 500 1500
-
20 30 120 5000
1.20 4.00 6.00 8.50
5-15kVA 15-45kVA 45-500kVA 0.5-2MVA
90 120 240 -
100 500 1500 2200
230 250
90 120 5000 31250
6.50 8.50 8.50 8.50
5-15kVA 15-45kVA 45-500kVA 0.5-2MVA >2MVA
90 120 240 -
100 500 1500 2200 2200
230 250 270
90 120 5000 31250 1500000
6.50 8.50 8.50 8.50 8.50
1-φ, 3-φ
-
5.00
-
240
6.50
15-45kVA 45-500kVA 0.5-2MVA >2MVA
120 240 -
500 1500 2200 2200
-
120 5000 31250 31250
5.80 5.80 5.80 5.80
6.4 Power Factor Tariff This is devised to make a distinction between overall charge to be made to types of consumer: those with good pf and those with poor power factor. PF tariff is divided into three main classes a) Maximum demand (kVA) tariff b) Average power factor tariff c) kWh and kVARh tariff: kWh and kVARh are charged separately. A poor power factor means more kVARh and consequently more charge per unit of energy consumed. Example 6.1 A factory to be set up is to have a fixed load of 760 kW at 0.8 pf. The electricity board offers to supply energy at the following alternate rates: a) LV supply at N 32/kVA max demand/annum + 10 kobo/kWh b) HV supply at N 30/kVA max demand/annum + 10kobo/kWh. The HV switchgear costs N 60/kVA and switchgear losses at full load amount to 5%. Interest, depreciation charges for the switchgear are 12% of the capital cost. If the factory is to work for 48 hours/week, determine the more economical tariff. Solution 760 = 950kVA 0. 8 Loss in switchgear = 5% 950 = 1000kVA ∴ Input demand = 0.95 Cost of switchgear = 60 x 1000 = N 60,000 Annual charges on depreciation = 0.12 x 60,000 = N 7,200 Annual fixed charges due to maximum demand corresponding to tariff (b)
Maximum demand =
53
= 30 x 1,000 = N 30,000 Annual running charges due to kWh consumed = 1000 x 0.8 x 48 x 52 x 0.10 = N 199,680 Total charges/annum = N 236,880 Maximum demand corresponding to tariff (a) = 950kVA Annual fixed charges = 32 x 950 = N 30,400 Annual running charges for kWh consumed = 950 x 0.8 x 48 x 52 x 0.10 = N 189,696 Total charges/annum = N 220,096 Therefore, tariff (a) is economical.
54
CHAPTER SEVEN POWER FLOW STUDIES 7.1 INTRODUCTION Successful operation of power system under normal balanced three-phase steady-state conditions requires that: • Generation supplies the demand (load) plus losses. • Bus voltage magnitudes remain close to rated values. • Generators operate with specified real and reactive power limits. • Transmission lines and transformers are not overloaded Power flow (also commonly called load flow) studies routinely investigate these requirements with the aid of a power flow computer program. Such a program computes: ¾ The voltage magnitude and angle at each bus in a power system under balanced 3-Φ steady-state conditions, ¾ Real and reactive power flow for all equipment interconnecting the buses, and the ¾ Technical losses in the power system Power flow studies investigate both existing power systems and proposed changes including new generation and transmission to meet projected load growth. Power flow studies also allow determination of: o The best size and the most favourable locations for power capacitors both for power factor improvement or raising bus voltages. o The best location as well as the optimal capacity of proposed generating stations, substations or new power lines. o The state of the power system under a given operational condition so that operators detect overloads and evaluate transfer limits.
7.2 TYPES OF BUSES FOR LOAD-FLOW STUDIES There are generally three categories of buses in an electric power network, namely: • Generation bus (also called ‘voltage-controlled bus) • Load bus • Slack bus Two of the following quantities are specified at each bus 1.
Magnitude of voltage, V
2. 3. 4.
Phase angle of the voltage, Φ Active or real power, P Reactive power, Q
The quantities specified at each bus type are as follows. a)
Generation bus (or voltage controlled bus): Voltage magnitude V and real power P are specified. Hence it
is also referred to as the P-V bus. b) Load bus: Real power P and reactive power Q are specified. Hence, it is also called the P-Q bus. c)
Slack (or swing) bus: the voltage magnitude V and phase angle Φ are specified. This bus takes up the slack in the power system, that is, it also provides additional real and reactive power to supply the transmission losses, since these are not known until the final solution is obtained
The voltage angle of the slack bus generator is set to zero or any arbitrary value, so its voltage becomes the reference voltage for other bus voltages in the system.
Table 7.1 summarizes the status of the variables at each bus type. It is assumed that the system has N buses out of which G are generator buses.
55
Table 7.1 Typical power flow bus specifications Bus type Known Unknown quantities quantities Slack (swing) bus V, Φ = 0 P, Q Generator bus Q, Φ P, V Load bus
P, Q
V, Φ
Number of Unknowns 2 2(G-1) 2(N-G)
7.3 POWER FLOW PROBLEM FORMULATION The overall power flow problem can be divided into the following sub-problems. (i) Draw a single-line diagram of the power system (ii) Assuming a balanced three phase system, the transmission system is represented by its positive phase sequence network of linear lumped series and shunt branches. The line impedances and shunt admittances in per-unit values are then found, including transformer impedances, shunt capacitor ratings and transformer taps. (iii) Formulation of system power flow equations relating bus voltages and powers. (iv) Specification of static operating conditions in terms of the power and voltage constraints applicable to the various buses. (iii) Numerical solution of the power flow equations, subject to specified constraints, to obtain values of all bus voltages. (iv) Actual power flows in all transmission links are computed based on the determined bus voltages. 7.4. FORMULATION OF THE POWER FLOW EQUATIONS The power flow equation (PFE) is a mathematical model of the power system used to predict power flows in the network. PFE formulation in this course is based on the nodal admittance matrix approach: I = YbusV (7.1) V is the Nx1 vector of bus voltages where I is the Nx1 vector of bus currents Ybus= [Yik] is the N x N bus admittance matrix Hence the current at bus i is given by; N
Ii =
∑Y V ik
(7.2)
k
k =1
where N = no of buses in the network. • The diagonal elements Yii of Ybus are obtained as the algebraic sum of admittances of all lines connected to bus i, including any fixed impedances connected from that bus to ground • The off-diagonal elements Yik = Yki are obtained as the negative of the admittances of lines directly between buses i and k. The complex power at bus i is given by the relation Si = Vi I*i = Pi + jQ i (7.3) It is usually the case that powers rather than currents are known, hence, eqn (7.3) is rearranged as: S P + jQ i I i* = i = i Vi Vi *
Taking the complex conjugate of I i yields
Ii = Substituting for
S*i Vi*
=
Pi − jQ i
(7.4)
Vi*
I i from eqn (7.4) into eqn (7.2) yields Pi − jQ i Vi*
N
=
∑Y V ik
k
i = 1,2, …., N.
k =1
56
Or
Pi − jQ i = Vi*
N
∑Y V ik
i = 1,2, …., N
k
(7.5)
k =1
where Vi is the voltage at bus i Vk is the voltage at bus k Yik is element (i,k) in the bus admittance matrix (Ybus) It should be noted that the N equations in eqn (7.5) are complex. These equations represent 2N real equations which may be expressed in rectangular, polar or hybrid form (a combination of rectangular and polar). For the rectangular form the voltages Vi = ei + jfi and the bus admittance matrix elements Yik = Gik + jBik are substituted into eqn. (7.5). After simplification we obtain the rectangular form: ⎡N ⎤ ⎡ N ⎤ (7.6a) Pi = ei ⎢ ( G ik e i − B ik f i )⎥ + fi ⎢ ( G ik f k + B ik e i )⎥ ⎣⎢ k =1 ⎦⎥ ⎣⎢ k =1 ⎦⎥
∑
⎡
∑
N
⎤
⎡
⎦⎥
⎣⎢ k =1
⎤
N
Qi = fi ⎢∑ ( G ik e k − B ik f k )⎥ - ei ⎢∑ ( G ik f k + B ik e k )⎥ ⎣⎢ k =1
For the polar form the voltages Vi = Vi
⎦⎥
(7.6b)
∠σ i or Vi = Vi e jσ i and the bus admittance elements Yjk = Y jk ∠Ψik are
substituted into eqn (7.5). After simplification the equations in polar form are: N
∑Y
Pi = Vi
ik
Vk cos( σ i − σ k − ψ ik )
(7.7a)
Vk sin( σ i − σ k − ψ ik )
(7.7b)
k =1 N
Q i = Vi
∑Y
ik
k =1
Example 7.1 Write the power flow equations for the 3-bus network in Fig. 7.1 in polar form.
Fig. 7.1 Three bus network for example 7.1 0 − 4 + j5 ⎤ ⎡Y11 Y12 Y13 ⎤ ⎡4 − j5 ⎥ ⎢ Step 1: Form Ybus = ⎢Y13 Y22 Y23 ⎥ = ⎢⎢0 4 − j10 − 4 + j10⎥⎥ ⎢⎣Y13 Y23 Y33 ⎥⎦ ⎢⎣− 4 + j5 − 4 − j10 8 − j15 ⎥⎦ Given data: Bus 1: V1 = 1.0 pu, σ1 = 0o (Slack bus) Bus 2: P2 = 1.7 pu and V2 = 1.1249 (PV bus) Bus 3: P3 = -2.0 and Q3 = -1 (PQ bus) Step 2: The power flow equations are obtained using *
Pi – jQi = V i
3
∑Y V ik
(7.8)
k
k =1
Converting the elements of Ybus into polar form yields: Y12=Y21=0 Y11 = 6.4031 ∠ − 51.3 0 0 Y23 = 10.77 ∠111.8 0 Y22 = 10.77 ∠ − 68.2
Y13 = 6.4031 ∠128.7 0 Y32=Y23
57
Y31=Y13 Y33 = 17.00 ∠ − 61.9 Converting the bus voltages into polar form yields: V1 = 1 ∠0 pu, 0
V2 = 1.124 ∠σ 2 ,
0
and V3 = V3 ∠σ 3
For bus 1 P1 – jQ1 = V1* [ Y11V1 + Y12 V2 + Y13V3 ] = ( 1∠0 0 )* [( 6.4031∠ − 51.3 0 )( 1∠0 0 ) + ( 0 )( 1.124∠σ 2 ) + + ( 6.4031∠128.7 0 )( V3 ∠σ 3 )] ∴ P1 − jQ1 = 6.4031 ∠ − 51.3 0 + 0 + 6.4031 V3 ∠( 128.7 0 + σ 3 )
(7.9)
For bus 2 P2 – jQ2 = V2* [ Y21V1 + Y22V2 + Y23V3 ] = ( 1.124∠ − σ 2 )[( 0 )( 1∠0 0 ) + ( 10.77∠ − 68.2 0 )( 1.124∠σ 2 ) + + ( 10.77∠111.8 0 )( V3 ∠σ 3 )] Thus the nodal equation at bus 2 is: 1.7 - jQ2 = 0 + 13.606 ∠ − 68.2 0 + 12.105 V3 ∠( 111.8 0 + σ 3 − σ 2 )
(7.10)
For bus 3 P3 – jQ3 = V3* [ Y31V1 + Y32V2 + Y33V3 ] = ( V3 ∠ − σ 3 )[( 6.4031∠128.7 0 )( 1∠0 0 ) + ( 10.77∠111.8 0 )( 1.124∠σ 2 ) + + ( 17∠ − 61.9 0 )( V3 ∠σ 3 )] Thus the nodal equation for bus 3 is: − 2 + j1 = 6.4031 V3 ∠( 128 .7 0 − σ 3 ) + 12 .105 V3 ∠( 111 .8 0 + σ 2 − σ 3 ) + 2
+ 17 V3 ∠ − 61 .9 0
(7.11)
Simultaneous equations (7.9)-(7.10) relate the powers in the network to the bus voltages. These equations are typically complex and nonlinear. In a power flow study the equations are solved for the unknown quantities using a numerical iterative technique such as the Gauss method, the Gauss-Siedel method, the Newton-Raphson method, etc. The Gauss and Gauss-Siedel iterative methods are outlined in the next section.
7.5 GAUSS AND GAUSS-SIEDEL ITERATIVE METHODS The load flow equations can be written in general vector form as: f(x) =0 where x is the vector of bus powers and voltages and f (.) is a vector-valued function.
(7.12)
The solution vector to eqn. (7.12) can be obtained by iterative techniques to estimate the numerical solution as accurately as desired. The steps in solving the problem by iteration methods are: (i) Guess the initial solution approximately. (ii) This solution is used in conjunction with the original equation to compute a new and more accurate estimate. (iii) The second estimate is used in finding out the third estimate and so on. Any process of successive approximations used to obtain a numerical solution of algebraic equations, differential equations, etc. is known as iterative method. Digital computers are used for obtaining the solution quickly. The list of computer instructions stating the sequence of operations is an algorithm. The quality of an algorithm is decided by the speed of convergence to the acceptable solution. The solution is said to have converged if the largest change between successive iterates is less than a given tolerance value. The Gauss and Gauss-Siedel methods of iterative solution of linear and nonlinear algebraic equations are illustrated by the following examples.
58
Example 7.2 Solve the following set of linear simultaneous equations using the: (a) Gauss iterative method (b) Gauss-Siedel iterative method 4 x1 − x 2 − x3 − 7 = 0 4 x1 − 8 x2 + x3 + 21 = 0 − 2 x1 + x 2 + 5 x3 − 15 = 0
Solution These equations can be re-written in the form
x1 =
7 + x 2 − x3 ; 4
x2 =
21 + 2 x1 + x3 ; 8
and
x3 =
15 + 2 x1 − x 2 5
(a) The Gauss iterative process is as follows:
x1( k +1) =
7 + x 2( k ) − x3( k ) 4
x 2( k +1) =
21 + 2 x1( k ) + x3( k ) 8
x3( k +1) =
15 + 2 x1( k ) − x 2( k ) 5
(7.13)
In eqn (7.13) k is the iteration count. Suppose the guess to the solution at iteration k=0 is P0 =
( x1( 0 ) , x 2( 0) , x3( 0 ) ) = (1, 2, 2). Then at iteration k=1 eqns.(7.13)
yield:
x1(1) =
7+2−2 = 1.75 4
x 2(1) =
21 + 4 + 2 = 3.375 8
x3(1) =
15 + 2 − 2 =3 5
Subsequent iterations yield the results shown in following table
k
x1( k )
x 2( k )
x3( k )
0 1 2 3 4 5
1 1.75 1.84375 1.9625 1.99062500 1.99414063
2 3.375
2 3
3.875
3.025 3.925 3.97656250 3.99531250
2.9625 3.000000 3.00093750
Μ
Μ
Μ
Μ
15
1.99999993
3.99999985
2.99999993
Μ
Μ
Μ
Μ
19
2.00000000
4.00000000
3.00000000
59
After 19 steps the iteration has converged to the nine-digit machine approximation. The solution to the above system of linear equations is actually P = (2, 4, 3). (b) In the Gauss-Siedel iteration method new values of each variable are used as soon as they are available. Hence the iteration equations become x1( k +1) =
7 + x 2( k ) − x3( k ) 4
x 2( k +1) =
21 + 2 x1( k +1) + x3( k ) 8
x3( k +1) =
15 + 2 x1( k +1) − x 2( k +1) 5 (0)
(0)
(0)
Starting with Po = ( x1 , x 2 , x3 ) = (1, 2, 2) then
x1(1) =
7+2−2 = 1.75 4
x 2(1) =
21 + 4 x1.75 + 2 = 3.75 8
x3(1) =
15 + 2 x1.75 − 3.75 = 2.95 5
The results of the iteration process are given in the following Table.
k
x1( k )
x 2( k )
x3( k )
0 1 2 3 . . . 8 9 10
1 1.75 1.95 1.995625 . . . 1.99999983 1.99999998 2.00000000
2 3.75 3.96875 3.99609375 . . . 3.99999988 3.99999999 4.00000000
2 2.95 2.98625 2.99903125 . . . 2.99999996 3.00000000 3.00000000
It has converged to the solution in 10 iterations only which is a considerable improvement on the Gauss method. Example 7.3 Consider the nonlinear equations 3x1 + x1 x 2 − 2 = 0 3x 2 − x1 x 2 + 2 = 0 Solve this set using the a) Gauss iterative method. b) Gauss-Siedel iterative method
Solution: Using the usual method, it is seen that the solution to the equations is x1 = 1, x2 = -1. Let the equations be re-arranged as:
60
x1 = 2 - x 1 x 2 and x2 = - 2 + x 1 x 2 3 3 (Note that there are other possible re-arrangements for the two equations. Some of them may not necessarily result in solutions that converge to the actual solution. This is a subject of numerical analysis which is beyond the scope of this course.) a) The Gauss iteration equations are
x1( k +1) =
2 − x1( k ) x 2( k ) 3
and
x 2( k +1) =
− 2 + x1( k ) x 2( k ) 3 ( 0)
Initially, start iteration by assuming values for x1 ( 0)
Let for k=0, x1
( 0)
and x 2
= 0 and x 2( 0) = 0
Then, iteration k = 1 yields:
x1(1) = 0.6666 – 0 = 0.6666 x 2(1) ) = - 0.6666 + 0 = - 0.6666 Iteration k=2 yields: x1( 2 ) =
2 − (0.6666) × ( −0.6666) = 0.8147 3
x 2( 2 ) =
− 2 + (0.6666) × ( −0.6666) = −0.8147 3
Iteration k=3 yields: x1(3) =
2 − (0.8147 ) × (−0.8147) = −0.8878 3
x1( 3) =
− 2 − (0.8147 ) × ( −0.8147 ) = −0.8878 3 ( 4)
( 4)
( 5)
( 5)
Substitute the values obtained of x1 and x2 and continue steps for x1 , x 2 , x1 , x 2 , etc. till the final results obtained are steady. b) Gauss-Siedel method. In the Gauss-Siedel method, the convergence is obtained faster by making use of the upgraded iterates as soon as they are available, i.e. use x(k+1) and not x(k), as soon as it is available. (k ) (k ) − 2 + x1( k +1) x 2( k ) So the algorithm becomes: x1( k +1) = 2 − x1 x 2 and x 2( k +1) = 3 3 ( 0)
Let for k=0, x1
= 0 and x 2( 0) = 0
(1)
Iteration k=1: x1 = 0.6666 – 0 = 0.6666
x 2(1) ) = - 0.6666 + 0 = - 0.6666 Iteration k=2:
x1( 2 ) =
2 − (0.6666) × (−0.6666) = 0.8147 3
x 2( 2 ) =
− 2 + (0.8147 ) × (−0.6666) = −0.8476 3
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Iteration k=3: x1(3) =
2 − (0.8147 ) × (−0.8476) = 0.8476 3
x 2( 3) =
− 2 + (0.8476) × (−0.8476) = −0.9200 3 ( 4)
( 4)
( 5)
( 5)
Substitute the values obtained of x1 and x2 and continue steps for x1 , x 2 , x1 , x 2 , etc. till the final results obtained are steady. It will be observed that the speed of convergence in this method is increased or the number of iterations needed is less.
Use of Acceleration Factors. The convergence process can be considerably increased in the above methods by the application of acceleration factor. If (k )
(k )
∆ x1 and ∆ x 2 indicate the variable change in the (k – 1)th and kth iteration, the difference variables can be written as:
∆x1( k ) = x1( k ) − x1( k −1) ∆x 2( k ) = x 2( k ) − x 2( k −1) Each difference is multiplied by the acceleration factor, α. The accelerated iterates are obtained as follows: k) k −1) x1(acc = x1(acc + α∆x1( k )
) −1) x 2( kacc = x 2( kacc + α∆x 2( k )
These accelerated values, x1acc and x2acc, are used in the iteration process instead of x1 and x2. The value of α is taken between 1.5 and 1.7 for the solution of load-flow equations.
7.6 GAUSS-SIEDAL METHOD OF SOLVING POWER FLOW EQUATIONS This is an interactive method of solving the nonlinear algebraic PFEs. Depending on bus type the equations are solved to obtain the unknown quantities. For example, for a load bus the phasor voltage may be computed using ⎡ ⎤ Vk( i +1) = 1 ⎢ Pk − jQ k − N Y V ( β ) ⎥ k = G + 1, G+2, G+3, ….., N ∑ km m ⎥ Ykk ⎢ V k(i )* m =1 ⎢⎣ ⎥⎦ m≠ k
β β
=i
for m>k
= i +1 for m
∆Q ki +1 = Qki +1 − Qki ≤ 0.01 pu i +1 i ∆σ ki +1 = σ k − σ k ≤ 0.01 pu
After the solution has converged, the phasor voltages at all buses are known and may be used to find other information about the steady-state operating point characteristics of the power system such as: Real and reactive power generation at
62
slack bus; Reactive power generation at PV buses; Real and reactive power flows in the lines; Real and reactive power losses between buses. The slack bus power is found using, S1 =P1 + jQ1 = V1 ⎡
⎤ ⎢∑ Y1k Vk ⎥ ⎣ k =1 ⎦ N
*
The real and reactive power flowing between bus k and m may be computed using
[
]
Skm = Pkm + jQkm = Vk (Vk − Vm )Ykm + Yshk Vk * where Yshk is the shunt admittance in the π model of the transmission line. Specifically, the line power Skm measured at bus k and defined positive in direction k to m is given by, *
* Skm = Pkm + jQkm = VkI km = Vk [Vk* − Vm* ]Ykm + Vk Vk*Yshk Real and reactive power losses are computed using the line flows as follows
loss
S km = Skm + Smk P km + jQ km = [Pkm + Pmk ] + j (Qkm + Qmk ) loss
loss
Example 7.4 For the 3-bus system in Fig 7.2, the line per-unit reactances are: x12=0.75, x13=0.25, and x23=0.50. Bus 1 is selected as the swing bus, and its voltage set at V1 = 1.0∠0 pu. 0
Bus 2 is a generator bus with voltage magnitude V2 = 1.02 pu and real injected power, P2=0.6 pu. Bus 3 is a load bus with injected power: P3=-0.8 pu and Q3=-0.6 pu. a) Use the Gauss-Seidel method to determine the power flow solution for power system within a tolerance of 0.01 pu. b) Calculate the power flow through line 1-3 at both ends c) Calculate the power loss on line 1-3.
Fig. 7.2 Three-bus network of Example 7.4 Solution: Transmission line reactances are: X12 = j0.75pu, X13 = j0.25pu, and X23 = j0.5pu ⎡Y11 Y12 Y13 ⎤ ⎥ The bus admittance matrix is Y = ⎢Y bus ⎢ 21 Y22 Y23 ⎥ ⎢⎣Y31 Y32 Y33 ⎥⎦ Where,
⎡ 1 1 1 1 ⎤ = − j5.333 + =⎢ + jx12 jx13 ⎣ j 0.75 j 0.25 ⎥⎦ 1 1 Y12 = Y21 = − =− = j1.333 jx12 j 0.75
Y11 =
Y22 =
⎡ 1 1 1 1 ⎤ + = + = − j 3.333 jx12 jx 23 ⎢⎣ j 0.75 j 0.5 ⎥⎦
Y13 = Y31 = −
1 1 =− = j 4. 0 jx13 j 0.25
63
⎡ 1 1 1 1 ⎤ + = + = − j 6.0 jx13 jx 23 ⎢⎣ j 0.25 j 0.5 ⎥⎦ Known quantities: Bus 1: Slack bus: V1 = 1 ∠ 0opu Y33 =
Bus 2:
PV bus: V2 = 1.02pu, P2 = 0.6pu
Bus 3: PQ bus: P3 = -0.8pu, Q3 = -0.6pu Unknown quantities to be determined Bus 2: σ 2 and Q 2 Bus 3:
V3
and σ 3
Gauss-Siedel equations:
Im (V2(i)* (Y21 V1 + Y22 V2(i) + Y23 V3(i) ) ) Q (i) 2 =⎡ ⎤ V2(i) = 1 ⎢ P2 − jQ 2 − Y21V1 − Y23 V3(i−1) ⎥ * (i)
[ ]
Y22 ⎢ V2 ⎥ ⎣ ⎦ ⎡ ⎤ V3(i) = 1 ⎢ P3 − jQ 3 − Y31 V1 − Y32 V2(i −1) ⎥ Y33 ⎢ V3(i) * ⎥ ⎣ ⎦
[ ]
Constant terms: Y21 V1 = j1.333(1∠0 0 ) and Y31 V1 = j4.0 Using a flat start:
V
(0) 2
(0)
= V3 = 1∠0 0
Thus at iteration i=0, the reactive power injected by bus 2 is given by: Q2( 0 ) = − Im (1.02∠0 0 ) * j1.333 + (− j 3.333)(1.02∠0 0 ) + ( j 2)(1∠0 0 ) = 0.07 Iteration 1 ⎡ 0.6 − j 0.07 ⎤ 1 − j1.333 − ( j 2)(1∠0 0 )⎥ = 1.04∠9.8 0 V2(1) = − j 3.333 ⎢⎣ (1.02∠0 0 ) * ⎦ Now the magnitude of V2 is fixed to be 1.02 hence the above calculated value of 1.04 is reset to 1.02 while the phase angle is retained. Hence V2(1) = 1.02∠9.8 0
[
}]
{
⎤ 1 ⎡ − 0.8 + j0.6 − j4 − (j2)(1.02∠9.8 o )⎥ = 0.91∠ − 4.8 o ⎢ − j6 ⎣ (1∠0 o ) * ⎦ Test for convergence: V31 =
∆V3(1) = V3(1) − V3( 0 ) = 0.09
[
]
Q2(1) = −Im (1.02∠9.8 o ) * {j1.333 + (− j3.333)(1.02∠9.8 o ) + (j2)(0.91∠ − 4.8 o )} = 0.33 Iteration 2
V2( 2 ) =
⎡ 0.6 − j0.33 ⎤ 1 − j1.333 − (j2)(0.91∠ − 4.8 o )⎥ = 1.02∠8.2 o ⎢ o − j3.333 ⎣ (1.02∠9.8 ) * ⎦
V3( 2) =
⎤ 1 ⎡ − 0.8 + j0.6 − j4 − (j2)(1.02∠8.2 o )⎥ = 0.89∠ − 5.7 o ⎢ o − j6 ⎣ (0.91∠ − 4.8 ) * ⎦
∆V3(2) = V3( 2 ) − V3(1) = 0.02
[
]
Q2( 2) = − Im (1.02∠8.2 0 ) * { j1.333 + (− j 3.333)(1.02∠8.2 0 ) + ( j 2)(0.89∠ − 5.7 0 ) = 0.36 Iteration 3 V2( 3) =
⎡ 0.6 − j0.36 ⎤ 1 − j1.333 − (j2)(0.89∠ − 5.7 o )⎥ = 1.02∠7.7 o ⎢ − j3.333 ⎣ (1.02∠8.2 o ) * ⎦
64
V3(3) =
⎤ 1 ⎡ − 0.8 + j 0.6 − j 4 − ( j 2)(1.02∠7.7 0 )⎥ = 0.88∠ − 6 0 ⎢ 0 * − j 6 ⎣ (0.89∠ − 5.7 ) ⎦
∆V2(3) = V3( 3) − V3( 2 ) = 0.01
[
]
Q (23) = − Im (1.02∠7.7 o ) * {j1.333 + (− j3.333)(1.02∠7.7 o ) + (j2)(0.88∠ − 6 o )} = 0.38 The solution has converged and the system bus voltages are V2 = 1.02 ∠ 7.7o and V3 = 0.88 ∠ -6o pu V1 = 1 ∠ 0o (ii) Power flows on line 1-3 S13 = V1 ⎛⎜ V1 − V3 ⎞⎟ ⎜ ⎝
jx 13
*
S31 = V3 ⎛⎜ V3 − V1 ⎞⎟ ⎜ ⎝
jx 13
*
o o o ⎡1∠0 − 0.88∠ − 6 ⎤ o 1 0 = ∠ ⎥ = 0.62∠53.5 ⎢ ⎟ j0.25 ⎦ ⎣ ⎠
*
pu
*
⎡ 0.88∠ − 6 o − 1∠0 o ⎤ o ⎥ = 0.62∠ − 126 ⎟ = 0.88∠ − 6 ⎢ j0.25 ⎣ ⎦ ⎠
Power loss on line 1-3 = S13 + S31 = 4.4 x 10-3 ∠ -0.0032 o
pu
pu
EXERCISES 7.1 The one-line diagram of an unloaded power system is shown in fig. 7.3. The reactance of the transmission line is X= 30Ω. The generators and transformers are rated as follows: G1: 20 MVA, 12kV, X = 1.20 pu G2: 60 MVA, 13.8kV, X = 1.40 pu G3: 20 MVA, 13.2kV, X = 1.20 pu T1: 25 MVA, 12/69kV, X = 0.08 PU T2: 60 MVA, 69/13.8kV, X = 0.14 pu T3: 75 MVA, 13.2/69kV, X = 0.16 pu
Fig 7.3 Choose a power base of 50 MVA and a voltage base of 12kV in the circuit of generator 1. Form the bus admittance matrix.
7.2 The power system shown in fig. 7.4 has the following generator and transformer data. G1: 50 MVA, 13.2kV, X = 1.20 pu G2: 20 MVA, 13.2kV, X = 1.10 pu T1: 60 MVA, 13.2/115kV, X = 0.16 pu T2: 25 MVA, 34.5/13.2kV, X = 0.10 pu T3: 60 MVA, 115/34.5kV, X = 0.16 pu T4: 10 MVA, 4.16/34.5kV, X = 0.05 pu
65
Fig. 7.4 The 115-kV transmission line is 50 km long, and it has a reactance of 0.8 Ω/km. The 34.5-kV line is 10 km long, and it has a reactance of 1.2 Ω/km. Choose a power base of 50 MVA and a voltage base of 115kV for the high-voltage transmission line. a. Choose the voltage bases in the other parts of the system in order that per-unit turns ratios of all transformers are 1:1. b. Draw the one-line impedance diagram showing all impedance values in per unit. c. Form the bus admittance matrix.
7.3 For the power system shown in fig. 7.5, bus 1 is selected as the slack or swing bus and its voltage is set to V1 = 1.0 ∠ 0o pu. The chosen power base is 100 MVA. Generator 2 delivers a real power of 0.75 pu at a voltage of 1.02 pu. The loads on buses 3 and 4 are SD3 = (0.40 +j0.30) pu and SD4 = (0.80 + j0.60) pu, respectively. The impedance parameters for the transmission lines referred to the given power base and a voltage base of 115kV are Line 1-2: Z = (0.01 + j0.05) pu; Y = j0.30 pu Line 2-3: Z = (0.03 + j0.15) pu; Y = j0.90 pu
Fig 7.5 The transformer is connected between buses 3 and 4, and its reactance is 0.10 pu. From a flat start, perform one iterate of the Gauss-Siedel iterative technique to find the voltages at buses 2, 3, and 4.
66
CHAPTER EIGHT POWER SYSTEM PROTECTION 8.1 Introduction The operation of a power system is affected by disturbances that could be due to natural occurrences such as lightning, wind, trees, animals, and human errors or accidents. These disturbances could lead to abnormal system conditions such as short circuits, overloads, and open circuits. Short circuits, which are also referred to as faults, are of the greatest concern because they could lead to damage to equipment or system elements and other operating problems including voltage drops, decrease in frequency, loss of synchronism, and complete system collapse. There is, therefore, a need for a device or a group of devices that is capable of recognizing a disturbance and acting automatically to alleviate any ill effects on the system element or on the operator. Such capability is provided by the protection system. The protection system is designed to disconnect the faulted system element automatically when the short circuit currents are high enough to present a direct danger to the element or to the system as a whole. When the fault results in overloads or short-circuit currents that do not present any immediate danger, the protection system will initiate an alarm so that measures can be implemented to remedy the situation.
8.2 Component of Protection Systems There are three principal components of a protection system: 1. Transducer 2. Protective relay 3. Circuit breaker These components are described briefly in the following paragraphs. 8.2.1 Transducers. The transducer serves as a sensor to detect abnormal system conditions and to transform the high values of short-circuit current and voltage to lower levels. The main sensors used are the current transformer (CT) and the potential transformer (PT). The current transformer is designed to provide a standard continuous secondary current of 5 A. Standard CT ratios available include 50/5, 100/5, 150/5, 200/5, 250/5, 300/5, 400/5, 600/5, 800/5, 900/5, 1000/5, 1200/5, 1500/5, 1500/5, and 2000/5. During fault conditions, the short-circuit currents could reach over 10 times normal for short periods of time without damaging the CT windings. The current transformer has a primary winding that usually consists of one turn and a secondary winding of several turns. It is, therefore, unsafe to open-circuit the secondary of a CT whose primary is energized. The potential transformer is designed to operate at a constant standard secondary voltage of 120 V. For low-voltage applications, the VT is just like any other two-winding voltage transformer. For primary voltages in the HV and EHV levels, a capacitor voltage-divider circuit is used together with the VT, forming the capacitor voltage transformer (CVT). The primary voltage is impressed across the series-connected capacitors. The VT is used to measure the voltage of a few kilovolts across the capacitor of the smaller capacitance value.
8.2.2 Protective Relays. A protective relay is a device that processes the signals provided by the transducers, which may be in the form of a current, a voltage, or a combination of current and voltage. These signals arise as a result of a faulted condition such as a short circuit, defective equipment or lines, lightning strikes, or surges. The protective relay can initiate or permit the opening of various interrupting devices or sound an alarm. There are two main classifications of protective relays based on their construction: electromechanical and solid state. The electromechanical relay develops an electromagnetic force or torque from the signal provided by the transducer; this force or torque is used to physically open, or close a set of contacts to permit or initiate the tripping of circuit breakers or actuate an alarm.
67
The solid-state, or static, relay is energized by the same signals as in an electromechanical relay. However, there is no physical opening, or closing, of the relay contacts. Instead, the switching of the relay contacts is simulated by causing a solid state device to change its status from conducting (closed position) to non-conducting (open position). Electromechanical relays predate the solid-state relays. A majority of power system installations still use electromechanical relays. The improved reliability, versatility, and faster response (as low as one-quarter cycle) of solidstate relays have made them more attractive. Some electromechanical relays have been replaced by solid-state relays, and in newer installations a mixture of both types would usually be found.
8.2.3 Circuit Breakers. A circuit breaker (CB) is a mechanical device used to energize and interrupt an electric circuit. It should be able to open and close quickly, maybe in the order of a few milliseconds. It should be able to carry the rated current continuously at the nominal voltage, and it must be able to withstand the large short-circuit current (called its momentary rating) that flows during the first cycle after a fault occurs. The circuit breaker must be able to interrupt a large short-circuit current called its interrupting rating. The momentary rating is about 1.6 times the interrupting rating because the former includes the effect of the DC component of the transient short-circuit current. The actual value of the current interrupted by the circuit breaker depends on its speed, which could be ½, 3, 5, or 8 cycles. When the current-carrying contacts of the circuit breaker are opened, an electric field appears across the contacts that ionizes the medium between them, and an arc is established between the contacts. The circuit breaker must be able to extinguish, or interrupt, this arc as quickly as possible. The arc is made to take an elongated path, cooled, and finally extinguished when the AC current feeding the arc passes through its zero value. Sometimes, the arc is extinguished in air, oil, sulphur hexafluoride (SF6), or a vacuum.
8.3 Types of Protection Systems The fundamental principle in designing protection systems is to divide the power system into zones that can be provided with the appropriate protection and can be disconnected or isolated as quickly as possible in order to minimize the effect on the rest of the power system. Each zone of protection is provided with two types of protection: primary and backup protection. 8.3.1 Primary or Main Protection. Primary protection is provided to ensure fast and reliable tripping of the circuit breakers to clear faults occurring within the boundary of its own zone of protection. In general, primary protection is provided for each transmission line segment, major piece of equipment, and switchgear. A separate zone of protection is established around each system element as in Fig. 8.1. There is overlapping of adjacent zones of protection around the circuit breakers. Thus, if a fault occurs within the overlap region, more breakers are tripped than is actually necessary. On the other hand, if there is no overlap, a fault that occurs in the region between two zones of protection will not be in either zone, and no protective relay will operate and no circuit breaker will be tripped. Although each major piece of equipment and transmission line segment is provided with primary protection and the primary zones of protection are made to overlap, it is still possible for the protection to fail. Hence, there is a need for backup protection.
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Fig 8.1 Zones of primary protection in a power system 8.3.2 Backup Protection. Backup protection is provided in case the primary protection fails to operate or is under repair or maintenance. The protective relays used for backup protection have longer time delays; that is, they are slower acting than the relays used for primary protection to give the latter the opportunity to perform their function. The backup protection should be designed in such a way that the cause of failure of the primary protection is not going to cause the same failure in the backup protection. Therefore, the backup protection should be located at a different station from the primary protection. Consider a fault on line segment 3-4 of the power system shown in Fig. 8.2. If circuit breaker 3 fails to trip, the backup relay would trip circuit breaker 1; or, if the circuit breaker 4 fails to trip, the backup relays would trip circuit breakers 7 and 8. Because circuit breaker 1 is located differently from 3 and circuit breakers 7 and 8 are located differently from 4, the cause of failure of the primary protection would probably not cause failure of the backup protection, which might happen if the backup protection were provided at 2 or 5 and 6 instead. If a fault occurred at station C, primary protection would be provided by circuit breakers at 4, 5 and 6. Backup protection would be located at 3, 7 and 8.
Fig 8.2 Location of backup protection 8.4 Requirements of Protection Systems In order to perform its functions properly, the protection system must have the following characteristics: speed, selectivity, reliability and sensitivity. 8.4.1 Speed. The speed of the protection system refers to the operating times of the protective relays. The potential damage to the faulted element depends on the length of time the short-circuit currents are allowed to flow. The speed of clearing, or isolating, the faulted system component also affects the stability of the whole system. Protective relays may be characterized as instantaneous with an operating time of about 0.10s, or as high speed with an operating time of less than 0.05s. Solid-state, or static, relays can have operating times as low as one quarter of a cycle. 8.4.2 Selectivity. Selectivity is the ability of the protection system to detect a fault, identify the point at which the fault occurred, and isolate the faulted circuit element by tripping the minimum number of circuit breakers. Selectivity of the protection system is obtained by proper coordination of the operating currents and time delays of the protective relays.
8.4.3 Reliability. The reliability of the protection system is its ability to operate upon the occurrence of any fault for which it was designed to protect. In other words, the protection system should operate when it is supposed to and not operate when it is not supposed to. 8.4.4 Sensitivity.
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Sensitivity refers to the characteristic of a protective relay that it operates reliably, when required, in response to a fault that produces the minimum short-circuit current flowing through the relay.
8.5 Types of Protective Relays The protective relay is the device that responds to signals from the transducers by quickly initiating or allowing a control action to be implemented in order to prevent damage to the faulted equipment and to restore service as soon as possible. The operating characteristics of the more commonly used protective relays are described in this section. A relay is said to pick up when it operates to open its normally closed (NC) contact or to close its normally open (NO) contact in response to a disturbance to produce a desired control action. The smallest value of the actuating quantity for the relay to operate is called its pickup value. A relay is said to reset when it operates to close an open contact that is normally closed (NC) or to open a closed contact that is normally open (NO). The largest value of the actuating quantity for this to happen is called the reset value.
8.5.1 Over-current Relays. The actuating quantity of an over-current relay is a current. The relay is designed to operate when the actuating quantity equals, or exceeds, its pickup value. An over-current relay can be either of two types: instantaneous or time-delay type. Both relay types are frequently provided in one relay case and are actuated by the same current; however, their individual pickup values can be adjusted separately. The pickup values may be adjusted by varying the tap settings in the input. The instantaneous relay element has no intentional time delay, and it operates quickly from ½ to 3 cycles depending on the value of the fault current. A typical operating characteristic of this relay type is shown in Fig. 8.3. The time-delay relay element is characterized by having an operating time that varies inversely as the fault current flowing through the relay. A typical inverse time characteristic is shown in Fig. 8.3. The time-delay characteristic may be shifted up or down by adjusting the time-dial setting so that the relay operates with a different time delay for the same value of fault currents. The difficulty in using over-current relays is that they are inherently non-selective. They detect over-current (faults) not only in their own zones of protection but also in adjacent zones. The selectivity can be improved by proper coordination of the relay pickup values and time-delay settings. As the electric load grows and the power system configuration changes, operating conditions and magnitudes of short-circuit currents will vary. The pickup values of the over-current relays have to be readjusted continually in response to these changes. Over-current relays are popular especially on low-voltage circuits because of their low cost. They are also used in specific applications on high-voltage systems.
Fig. 8.3 Time characteristics of over-current relays 8.5.2 Directional Relays. A directional relay is able to distinguish between current flowing in one direction and current flowing in the opposite direction. The relay responds to the phase angle difference between the actuating current and a reference current (or voltage) called the polarizing quantity.
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Directional relaying is typically used in conjunction with some other relay, usually the over-current relay to improve its selectivity.
8.5.3 Differential Relays. The operation of a differential relay is based on the vector difference of two or more similar electrical actuating quantities. The most common application is current differential relaying, in which the current entering and the current leaving the protected element are compared. If the difference exceeds the pickup value of the relay, it operates to trip the breakers to isolate the element. Typical differential relaying employing an over-current relay is shown in Fig. 8.4. Identical current transformers are placed at both ends of the protected element, and the CT secondaries are connected in parallel with an over-current relay. The directions of current flows shown in Fig. 8.4 are those corresponding to normal load conditions or to a fault external to the protected element. Thus, it is seen that the CT secondary currents merely circulate between the CTs, and no current flows in the over-current relay.
Fig, 8.4 Differential relaying Suppose a fault occurs on the protected element as shown in Fig 8.5. The short-circuit currents flow into the fault, and the CT secondary currents no longer circulate. The vectorial sum of the CT secondary currents flows through the overcurrent relay and causes the relay to disconnect the element from the system. Even though the current transformers used for the differential relay are identical, the secondary currents may not be identical because of CT transformation inaccuracies. Thus, the secondary currents will no longer merely circulate for normal load conditions or for external faults. The differential current that will flow through the over-current relay may be sufficient for the relay to pick up and cause false tripping of the circuit breakers.
Fig 8.5 Fault currents in a differential relay 8.5.4 Percentage-Differential Relays.
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The difficulty encountered in differential relaying due to CT errors is eased by the use of a percentage-differential relay. This type of relay has an operating coil and two restraining coils. The operating current is proportional to (IA-IB) and must exceed a certain percentage of the restraining current, which is proportional to ½(IA+IB), before the relay will operate.
8.5.5 Distance Relays. In a distance relay, a voltage and a current are balanced against each other and the relay responds to the ratio of the voltage to the current, which is the impedance of the transmission line from the relay location to the point of interest. The impedance may be used to measure distance along a transmission line, hence the name distance relay. This distance relay is useful because it is able to differentiate between a fault and normal operating conditions and to differentiate between faults in a specific area and a fault in a different part of the system. The operation of the distance relay is limited to a certain range of pick up values of impedance. The distance relay picks up whenever the measured impedance is less than or equal to the selected pickup value of impedance. There are several types of distance relays, including impedance, reactance and mho relays. The mho relay has an inherent directional characteristic; that is, it responds to or “sees” faults in only one direction. On the other hand, the impedance and the reactance relays see faults in all directions. Thus, a directional relay is commonly used together with impedance relay, and a mho relay is used as a starting unit for the reactance relay.
8.5.6 Pilot Relays. Pilot relaying is a means of communicating information from the end of a protected line to the protective relays at both line terminals. The relay determines whether a fault is internal or external to the protected line. The communication channel, or pilot, is used to transmit this information between line terminals if the fault is internal to the protected line, all the circuit breakers at the terminals of the line are tripped in high speed. If the fault is external to the protected line, the tripping of the circuit breakers is prevented, or blocked. Three types of pilot are commonly used for protective relaying: wire, power line carrier, and microwave pilot. A wire pilot consists of a twisted pair of copper wires of the telephone line type. It may be leased from the telephone company, or it may be owned by the electric utility. The power line carrier is the most commonly used pilot for protective relaying. In this type of pilot, a low-voltage, highfrequency current (30 to 300 kHz) is transmitted along one phase of the high-voltage power line to a receiver at the other end of the protective line. Line traps, located at both line terminals, serve to contain a carrier signal inside the zone of the protected line. The microwave pilot is an ultra-high-frequency radio system operating above 900 MHz. In this pilot, transmitters and receivers operate the same way as in power line carrier pilot; however, line traps are replaced by a line-of-sight antenna.
8.6 Applications of Protection Systems The applications of the different types of protection systems for the protection of various types of equipment and transmission lines are described in this section. These discussions are confined to protection for the high-voltage, bulk power system components. 8.6.1 Generator Protection. The protection system provided to the synchronous generator must be able to detect any abnormal condition immediately and act quickly to prevent damage to the generator and minimize the effect on the power system. Synchronous generators are provided with protection against various disturbances, including short circuits in the stator windings, loss of field excitation, stator and rotor overheating, and over-speed. Short-circuit protection of the stator windings is of primary concern and is discussed in this section. When a short circuit occurs between stator windings of a synchronous generator, or between a stator winding and ground, the protection system should quickly trip the main circuit breaker to disconnect the machine from the rest of the system and at the same time disconnect the field winding from the exciter. The best stator winding short-circuit protection is provided by a percentage-differential relay, which is shown in Fig 8.6. It may be noted that the relaying protect against a two-phase fault, or line-to-line fault, in the stator windings.
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When the synchronous generator neutral is grounded through a high resistance, the short-circuit current for a single lineto-ground fault is much less than the short-circuit current for a fault involving the phase windings. Thus, the ground fault current may not be detected by the differential protection. An over voltage connected across the grounding resistor would be able to detect the increased voltage across the resistor in the presence of a ground fault, and the over-voltage relay will operate.
Fig 8.6 Generator protection 8.6.2 Transformer Protection. The protection system provided for a power transformer depends primarily on its capacity and its voltage rating. Thus, for small transformers with capacities up to about 2 MVA, power fuses are deemed to be adequate. For larger transformers, with capacities greater than 10 MVA, percentage-differential relays with harmonic restraint are recommended. The one-line diagram of the protection of a three-winding transformer using a three-winding percentagedifferential relay is illustrated in Fig. 8.7.
Fig 8.7 Three winding transformer protection The differential relaying protection must satisfy two basic requirements: 1. The protection must not operate for normal load conditions and faults external to the transformer. 2. The relays must operate to trip the circuit breakers for an internal fault that is severe enough to cause direct damage to the transformer. T hree-phase transformers with Y-∆-connected windings require further consideration. The primary and secondary currents of such transformers normally differ not only in magnitude but also in phase angles because of the inherent phase shifts
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in Y-∆ or ∆-Y connections. The current transformers must, therefore, be connected in such a manner that the CT secondary line currents as seen by the protective relays are equal under normal operating conditions or for external faults. The correct magnitude relationship is obtained by proper choice of CT ratios and, if necessary, the use of an autotransformer in the CT secondary circuit. The correct phase-angle relationship is obtained by connecting the CTs on the Y-connected side of the transformer in delta ∆ and the CTs on the ∆-connected side in Y. In this way, the CT conditions are able to compensate for the phase shift introduced by the Y-∆ or ∆-Y connection. The design of the protection for a ∆-Y transformer using percentage-differential relays is illustrated in Fig. 8.8 and the following example.
Example 8.1 Design the protection of a three-phase, 50-MVA, 230/34.5 kV power transformer using available standard CT ratios. The high-voltage side is Y-connected and the low-voltage side is ∆-connected. Specify the CT ratios, and show the threephase wiring diagram indicating the CT polarities. Determine the currents in the transformer and the CTs. Specify the rating of an autotransformer, if one is needed.
Fig 8.8 Y- ∆ transformer protection Solution: When the transformer is carrying rated load, the line currents on the high-voltage side and low-voltage side are IHV = ILV =
50,000 3 (230) 50,000 3 (34.5)
= 125.5 A = 836.7 A
The CTs on the low-voltage side are Y-connected, and the CT ratio selected for this side is 900/5. The current in the leads flowing to the percentage-differential relay on this side is equal to the CT secondary current and is given by
⎛ 5 ⎞ ⎟ = 4.65 A ⎝ 900 ⎠
ILV lead = 836.7 ⎜
The current in the leads to the relay from the low-voltage side must be balanced by an equal current in the leads connected to the ∆-connected CTs on the high-voltage side. This requires a CT secondary current equal to ICT sec =
4.65
= 2.68 A
3
To obtain a CT secondary current of 2.68 A, the CT ratio of the high-voltage CTs is chosen as CT ratio =
125.5 = 46.8 2.68
The nearest available standard CT ratio is 250/5. If this CT ratio is selected, the CT secondary currents will actually be
⎛ 5 ⎞ ⎟ =2.51 A ⎝ 250 ⎠
ICT sec = 125.5 ⎜
Therefore, the currents in the leads to the ∆-connected CTs from the percentage-differential relays will be IHV lead =
3 (2.51) = 4.35 A 74
It is seen that the currents in the leads on both sides of the percentage-differential relay are not balanced. This condition cannot just be ignored because it could lead to improper tripping of the circuit breaker for an external fault. This problem can be solved by using an autotransformer as shown in Fig. 8.9. The autotransformer should have a turns ratio of Nautotransformer =
4.65 = 1.07 4.35
Fig 8.9 Y- ∆ Protection of Example 8.1 In the design of the transformer protection of Example 8.1, the magnetizing current of the transformer has been assumed to be negligible. This is a reasonable assumption during normal operating conditions because the magnetizing current is a small percentage of the rated load current. However, when a transformer is being energized, it may draw a large magnetizing inrush current that soon decays with time to its normal value. The inrush current flows only in the primary, causing an unbalance in current, and the differential relay will interpret this an internal fault and will pick up to trip the circuit breakers. To prevent the protection system from operating and tripping the transformer during its energization, percentagedifferential relaying with harmonic restraint is recommended. This is based on the fact that the magnetizing inrush current has high harmonic content, whereas the fault current consists mainly of fundamental frequency sinusoid. Thus, the current supplied to the restraining coil consists of the fundamental and harmonic components of the normal restraining current of (IA + IB)/2, plus another signal proportional to the harmonic content of the differential current (IA – IB)/2. Only the fundamental frequency of the differential current is supplied to the operating coil of the relay.
8.6.3 Bus Protection. The sum of the currents flowing through the lines connected to the same bus is equal to zero for normal operating conditions or for a fault external to the bus. Hence, differential relaying using over-current relays can be used to provide protection against bus faults. This is illustrated in the one-line diagram shown in Fig. 8.10.
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Fig 8.10 Bus protection using an over-current relay All of the CTs are connected in parallel and an over-current relay is connected across their output. When there is no fault on the bus, the line currents are indicated by the solid arrows, and the current through the over-current relay is given by IR =
1 (I1 + I2 – I3 – I4) = 0 CT ratio
Therefore, no current flows through the relay. On the other hand, when there is a fault at the bus, currents I3 and I4 reverse directions as shown by the broken arrows; thus, the current through the relay is now given by IR =
1 (I1 + I2 + I3 + I4) > 0 CT ratio
Hence, the over-current relay picks up to trip all circuit breakers.
8.6.4 Transmission Line Protection. Transmission lines provide the means for bringing the electric energy from the generating plants to the major substations, from these substations to the load centre substations, and ultimately to the individual consumers. Transmission lines travel over wide-open spaces, as well as thickly populated metropolitan areas. Because of these long distances and extensive exposure to nature and human accidents, most of the short circuits that occur in power systems are on overhead transmission and distribution lines. For the low-voltage distribution circuits, the protection system usually provided consists of circuit reclosers and power fuses that acts as relays and circuit breakers combined. For the protection of medium-voltage and high-voltage transmission lines, separate relays and circuit breakers are employed. Since HV and EHV transmission lines are the major means of bulk-power transmission, the protection of these transmission lines is designed to be more reliable and more selective, and it is also more expensive.
8.6.4.1 Transmission Line Protection by Over-current Relays. Consider that the portion of the power system to be protected is radial as shown in Fig. 8.11.
Fig 8.11 A radial power system
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The generator connected to bus 1 represents the rest of the power system, and it supplies loads at buses 1, 2, 3, 4 and 5 through four transmission lines. Since the short-circuit current comes from the left side of each line, it is sufficient to provide only one circuit breaker at the sending end for each line. So that service disruption is minimized, for a fault on line 4-5, only circuit breaker CB4 should be tripped; for a fault on the line 3-4, only breaker CB3 should be opened; for a fault on the line2-3, only CB2 should be tripped; and so on. The short-circuit current due to a fault on any of the lines depends on the fault location and the type of fault. Since the total impedance increases with the distance from the generator to the fault, the short-circuit current is inversely proportional to this distance. Over-current relays are used mainly to provide protection for subtransmission and distribution lines. Two forms of overcurrent protection are provided: primary protection for the line itself and backup protection for an adjacent line. Two types of overcurrent relay units are used: time overcurrent relay and instantaneous relay. The time overcurrent relays at each of the four buses 1, 2, 3 and 4 provide primary protection for their own line segment and provide remote backup protection to adjacent line downstream from the relay location. Thus, the relay at bus 1 provides primary protection for line 1-2 and also provides backup protection for line 2-3; the relay at bus 2 provides primary protection for line 2-3 and backup protection for line 3-4; and so forth. When the relay at bus 1 provides backup protection for line 2-3, it must be adjusted to be selective with the primary relaying at bus 2. The relay at bus 2 is expected to operate first for a fault on line 2-3 before the relay at bus1 operates. The operating times of the relays at the different buses are shown in Fig. 8.12.
Fig 8.12 Overcurrent protection of radial power system Thus, for the indicated fault, the relay at bus 4 picks up to trip circuit breaker CB4. If CB4 fails to open for any reason the fault current persists, and after a certain time delay the relay at bus 3 picks up to trip CB3. The relay at bus 2 is selective with the relay at bus 3; thus, if both CB3 and CB4 failed to open, the relay at bus 2 would pick up and trip CB2 after a longer time delay. For setting the pickup values and the selectivity clearances between the time-delay overcurrent relays for backup protection, there are four criteria to consider: 1. The relay must be able to pick up for the minimum short-circuit currents for which the relay is designed to protect. 2. The relay should not pick up for normal load conditions. A common practice is to set the minimum pickup value equal to twice the peak load current in order to carry emergency peak loads and pick up cold load. 3. The relay nearer the source providing backup protection for the downstream relay must pick up for one third of the minimum current seen by the latter. 4. The relay nearer the source providing backup protection for the downstream relay should pick up for the maximum current seen by the latter but no sooner than (a selectivity clearance of) at least 0.3 s. The instantaneous overcurrent relay unit provides primary protection for its own protected line segment to supplement the time-delay overcurrent relay unit. The instantaneous unit operates quickly with no intentional time delay, so it should
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be adjusted such that it does not operate for a fault on neighbouring lines. The pickup value is usually adjusted so that the instantaneous relay is able to detect a fault occurring at a distance of up to 80% of the line segment. The operating times of the instantaneous relay and the time-delay units are shown in Fig. 8.13. The complete protection system for a line consists of three overcurrent relays for phase fault protection and one overcurrent relay for ground fault protection. This protection system is shown in Fig. 8.14.
8.6.5 Transmission Line Protection by Distance Relays. To improve reliability, the generating plants, transmission lines, and substations of a high-voltage power system are interconnected to form a network. Overcurrent relay protection can no longer be used because coordinating the relays becomes a difficult, if not impossible, task.
Multiples of pickup current Figure 8.13 Typical time characteristics of an overcurrent relay.
Figure 8.14
Complete overcurrent protection of a radial system
Protection of transmission lines connected as a network can be provided by distance relays. These distance relays provide phase fault protection for the line, while an overcurrent relay provides ground fault protection. Distance relays provide primary protection for a line section and backup protection for an adjacent line. The value of impedance at the farthest fault location for which a distance relay picks up is called its “reach.” One distance relay contains three distance relay units responding to three independently adjustable pickup impedance values (or zones of protection) with three independently adjustable time delays.
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The primary impedance corresponding to a particular fault location, or relay unit reach, is converted to a secondary value that is used to adjust the phase or ground distance relay. This secondary impedance value is given by
⎛ CT ratio ⎞ ⎜ VT ratio ⎟⎟ ⎠ ⎝
Zsec = Zpri ⎜
The first-zone, or high-speed zone, unit of the distance relay has a reach of up to 80% to 90% of the length of the transmission line. This relay unit operates with no time delay. The second-zone unit of the distance relay provides protection for the rest of the line beyond the reach of at least 20% of the adjoining line section. Because this unit “sees” faults in the adjoining line, it must be selective with the first-zone unit of the adjacent line. Therefore, this second-zone unit is provided with a time delay of about 0.2s to 0.5s. The third-zone unit of the distance relay provides backup protection for the rest of the adjoining line section. This unit is adjusted to reach beyond the end of the adjoining line to ensure that backup protection is provided for the full line section. The time delay provided for this unit is usually about 0.4s to 1.0s. The protection of a portion of a power system by using distance relays is illustrated in Fig. 8.15.
Figure 8.15 Line protection by distance relays.
EXERCISES 8.1 For the power system of Fig. 8.2, specify the location of the backup protection so that the cause of failure in the primary protection of the following system components will not cause the same failure in the backup. a. Line 5-7 b. Line 6-8 c. Station B. 8.2 Consider the power system of Fig. 8.2. In response to a fault, the protection system tripped breakers 3, 7, and 8. Determine all possible locations of the fault. Determine what single failure had occurred, if any. 8.3 Repeat Problem 8.2 if the breakers tripped are 4, 6, and 7. 8.4 Repeat Problem 8.2 if the breakers tripped are 4, 6, and 8. 8.5 For the three-phase high-voltage bus shown in Fig. 8.16, sketch the developed three-phase wiring diagram for the bus differential protection using overcurrent relays.
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Fig 8.16 High-voltage bus of Problem 8.5 8.6 A three-phase step-down transformer bank is rated 10 MVA and 69/13.8kV. The high-voltage side is Y-connected, and the low-voltage side is ∆-connected. Sketch the developed three-phase wiring diagram for the protection of the transformer bank using percentage-differential relays. Show all CT ratings, connections, and polarities. Also show the values of the current in the lines, leads, relay windings, and transformer windings. Indicate the connections and ratings of any autotransformer that may be needed. 8.7 Repeat Problem 8.6 for the protection of a three-phase power transformer rated 100 MVA and 230/69kV. Assume that the transformer windings are Y-connected in both the primary and secondary sides. 8.8 Repeat Problem 8.6 when the transformer windings are connected in Y at the primary and in ∆ at the secondary. 8.9 A portion of a power system is shown in Fig. 8.17. Stations A and B have distance relays that are each adjusted for a first-zone reach of 100 ohms and a third-zone reach of 125 ohms. A fault occurs at F. a. What is the impedance seen by the relay at A and by the relay at B? b. Can the relay at A see the fault before the breaker at B has tripped? c. Can the relay at A see the fault after the breaker at B has tripped?
Fig 8.17 Portion of a power system for Problem 8.9 8.10 Repeat Problem 8.9 if the fault occurs at G and the breaker D2 fails to open. Assume that the fault current magnitudes are the same as those shown in Fig. 8.17.
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