Principal of Power System PDF

CONTENTS CONTENTS CHAPTER  Introduction General E nergy is the basic necessity for the eco-...

CONTENTS CONTENTS CHAPTER  Introduction General E nergy is the basic necessity for the eco- nomic development of a country. Many functions necessary to present-day 1.1 Importance of Electrical Energy living grind to halt when the supply of energy 1.2 Generation of Electrical Energy stops. It is practically impossible to estimate the actual magnitude of the part that energy has 1.3 Sources of Energy played in the building up of present-day 1.4 Comparison of Energy Sources civilisation. The availability of huge amount of energy in the modern times has resulted in a 1.5 Units of Energy shorter working day, higher agricultural and in- 1.6 Relationship Among Energy Units dustrial production, a healthier and more balanced diet and better transportation facilities. As a 1.7 Efficiency matter of fact, there is a close relationship be- 1.8 Calorific Value of Fuels tween the energy used per person and his stan- dard of living. The greater the per capita con- 1.9 Advantages of Liquid Fuels Over sumption of energy in a country, the higher is the Solid Fuels standard of living of its people. 1.10 Advantages of Solid Fuels Over Energy exists in different forms in nature but Liquid Fuels the most important form is the electrical energy. The modern society is so much dependent upon the use of electrical energy that it has become a part and parcel of our life. In this chapter, we shall focus our attention on the general aspects of elec- trical energy. 1 CONTENTS CONTENTS 2 Principles of Power System 1.1 Importance of Electrical Energy Energy may be needed as heat, as light, as motive power etc. The present-day advancement in science and technology has made it possible to convert electrical energy into any desired form. This has given electrical energy a place of pride in the modern world. The survival of industrial undertakings and our social structures depends primarily upon low cost and uninterrupted supply of electrical energy. In fact, the advancement of a country is measured in terms of per capita consumption of electrical energy. Electrical energy is superior to all other forms of energy due to the following reasons : (i) Convenient form. Electrical energy is a very convenient form of energy. It can be easily converted into other forms of energy. For example, if we want to convert electrical energy into heat, the only thing to be done is to pass electrical current through a wire of high resistance e.g., a heater. Similarly, electrical energy can be converted into light (e.g. electric bulb), mechanical energy (e.g. electric motors) etc. (ii) Easy control. The electrically operated machines have simple and convenient starting, control and operation. For instance, an electric motor can be started or stopped by turning on or off a switch. Similarly, with simple arrangements, the speed of electric motors can be easily varied over the desired range. (iii) Greater flexibility. One important reason for preferring electrical energy is the flexibility that it offers. It can be easily transported from one place to another with the help of conductors. (iv) Cheapness. Electrical energy is much cheaper than other forms of energy. Thus it is overall economical to use this form of energy for domestic, commercial and industrial purposes. (v) Cleanliness. Electrical energy is not associated with smoke, fumes or poisonous gases. Therefore, its use ensures cleanliness and healthy conditions. (vi) High transmission efficiency. The consumers of electrical energy are generally situated quite away from the centres of its production. The electrical energy can be transmitted conveniently and efficiently from the centres of generation to the consumers with the help of overhead conductors known as transmission lines. 1.2 Generation of Electrical Energy The conversion of energy available in different forms in nature into electrical energy is known as generation of electrical energy. Electrical energy is a manufactured commodity like clothing, furniture or tools. Just as the manufacture of a commodity involves the conversion of raw materials available in nature into the desired form, similarly electrical energy is produced from the forms of energy available in nature. However, electrical energy differs in one important respect. Whereas other commodities may be produced at will and consumed as needed, the electrical energy must be produced and transmitted to the point of use at the instant it is needed. The entire process takes only a fraction of a second. This instantaneous production of electrical energy introduces technical and economical considerations unique to the electrical power industry. Energy is available in various forms from different natural sources such as pressure head of water, chemical energy of fuels, nuclear energy of radioactive substances etc. All these forms of energy can be converted into electrical energy by the use of suitable arrangements. The arrangement essentially employs (see Fig. 1.1) an alternator coupled to a prime mover. The prime mover is driven by the energy obtaimed from various sources Introduction 3 such as burning of fuel, pressure of water, force of wind etc. For example, chemical energy of a fuel (e.g., coal) can be used to produce steam at high temperature and pressure. The steam is fed to a prime mover which may be a steam engine or a steam turbine. The turbine converts heat energy of steam into mechanical energy which is further converted into electrical energy by the alternator. Similarly, other forms of energy can be converted into electrical energy by employing suitable machinery and equipment. 1.3. Sources of Energy Since electrical energy is produced from energy available in various forms in nature, it is desirable to look into the various sources of energy. These sources of energy are : (i) The Sun (ii) The Wind (iii) Water (iv) Fuels (v) Nuclear energy. Out of these sources, the energy due to Sun and wind has not been utilised on large scale due to a number of limitations. At present, the other three sources viz., water, fuels and nuclear energy are primarily used for the generation of electrical energy. (i) The Sun. The Sun is the primary source of energy. The heat energy radiated by the Sun can be focussed over a small area by means of reflectors. This heat can be used to raise steam and electrical energy can be produced with the help of turbine-alternator combination. However, this method has limited application because : (a) it requires a large area for the generation of even a small amount of electric power (b) it cannot be used in cloudy days or at night (c) it is an uneconomical method. Nevertheless, there are some locations in the world where strong solar radiation is received very regularly and the sources of mineral fuel are scanty or lacking. Such locations offer more interest to the solar plant builders. (ii) The Wind. This method can be used where wind flows for a considerable length of time. The wind energy is used to run the wind mill which drives a small generator. In order to obtain the electrical energy from a wind mill continuously, the generator is arranged to charge the batteries. These batteries supply the energy when the wind stops. This method has the advantages that maintenance and generation costs are negligible. However, the drawbacks of this method are (a) variable output, (b) unreliable because of uncertainty about wind pressure and (c) power generated is quite small. (iii) Water. When water is stored at a suitable place, it possesses potential energy because of the head created. This water energy can be converted into mechanical energy with the help of water turbines. The water turbine drives the alternator which converts mechanical energy into electrical energy. This method of generation of electrical energy has become very popular because it has low production and maintenance costs. (iv) Fuels. The main sources of energy are fuels viz., solid fuel as coal, liquid fuel as oil and gas fuel as natural gas. The heat energy of these fuels is converted into mechanical energy by suitable prime movers such as steam engines, steam turbines, internal combustion engines etc. The prime mover drives the alternator which converts mechanical energy into electrical energy. Although fuels continue to enjoy the place of chief source for the generation of electrical energy, yet their reserves are diminishing day by day. Therefore, the present trend is to harness water power which is more or less a permanent source of power. (v) Nuclear energy. Towards the end of Second World War, it was discovered that large amount of heat energy is liberated by the fission of uranium and other fissionable materials. It is estimated that heat produced by 1 kg of nuclear fuel is equal to that produced by 4500 tonnes of coal. The heat produced due to nuclear fission can be utilised to raise steam with suitable arrangements. The steam 4 Principles of Power System can run the steam turbine which in turn can drive the alternator to produce electrical energy. However, there are some difficulties in the use of nuclear energy. The principal ones are (a) high cost of nuclear plant (b) problem of disposal of radioactive waste and dearth of trained personnel to handle the plant. Coal Crude oil Natural gas Hydro-electric power Nuclear power Renewables Energy Utilisation 1.4 Comparison of Energy Sources The chief sources of energy used for the generation of electrical energy are water, fuels and nuclear energy. Below is given their comparison in a tabular form : S.No. Particular Water-power Fuels Nuclear energy 1. Initial cost High Low Highest 2. Running cost Less High Least 3. Reserves Permanent Exhaustable Inexhaustible 4. Cleanliness Cleanest Dirtiest Clean 5. Simplicity Simplest Complex Most complex 6. Reliability Most reliable Less reliable More reliable 1.5 Units of Energy The capacity of an agent to do work is known as its energy. The most important forms of energy are mechanical energy, electrical energy and thermal energy. Different units have been assigned to various forms of energy. However, it must be realised that since mechanical, electrical and thermal energies are interchangeable, it is possible to assign the same unit to them. This point is clarified in Art 1.6. (i) Mechanical energy. The unit of mechanical energy is newton-metre or joule on the M.K.S. or SI system. The work done on a body is one newton-metre (or joule) if a force of one newton moves it through a distance of one metre i.e., Mechanical energy in joules = Force in newton × distance in metres (ii) Electrical energy. The unit of electrical energy is watt-sec or joule and is defined as follows: One watt-second (or joule) energy is transferred between two points if a p.d. of 1 volt exists between them and 1 ampere current passes between them for 1 second i.e., Introduction 5 Electrical energy in watt-sec (or joules) = voltage in volts × current in amperes × time in seconds Joule or watt-sec is a very small unit of electrical energy for practical purposes. In practice, for the measurement of electrical energy, bigger units viz., watt-hour and kilowatt hour are used. 1 watt-hour = 1 watt × 1 hr = 1 watt × 3600 sec = 3600 watt-sec 1 kilowatt hour (kWh) = 1 kW × 1 hr = 1000 watt × 3600 sec = 36 x 10 watt-sec. 5 (iii) Heat. Heat is a form of energy which produces the sensation of warmth. The unit* of heat is calorie, British thermal unit (B.Th.U.) and centigrade heat units (C.H.U.) on the various systems. Calorie. It is the amount of heat required to raise the temperature of 1 gm of water through 1ºC i.e., 1 calorie = 1 gm of water × 1ºC Sometimes a bigger unit namely kilocalorie is used. A kilocalorie is the amount of heat required to raise the temperature of 1 kg of water through 1ºC i.e., 1 kilocalorie = 1 kg × 1ºC = 1000 gm × 1ºC = 1000 calories B.Th.U. It is the amount of heat required to raise the temperature of 1 lb of water through 1ºF i.e., 1 B.Th.U. = 1 lb × 1ºF C.H.U. It is the amount of heat required to raise the temperature of 1 lb of water through 1ºC i.e., 1 C.H.U. = 1 lb × 1ºC 1.6 Relationship Among Energy Units The energy whether possessed by an electrical system or mechanical system or thermal system has the same thing in common i.e., it can do some work. Therefore, mechanical, electrical and thermal energies must have the same unit. This is amply established by the fact that there exists a definite relationship among the units assigned to these energies. It will be seen that these units are related to each other by some constant. (i) Electrical and Mechanical 1 kWh = 1 kW × 1 hr = 1000 watts × 3600 seconds = 36 × 10 watt-sec. or Joules 5 ∴ 1 kWh = 36 × 10 Joules 5 It is clear that electrical energy can be expressed in Joules instead of kWh. (ii) Heat and Mechanical (a) 1 calorie = 4·18 Joules (By experiment) (b) 1 C.H.U. = 1 lb × 1ºC = 453·6 gm × 1ºC = 453·6 calories = 453·6 × 4·18 Joules = 1896 Joules ∴ 1C.H.U. = 1896 Joules (c) 1 B.Th.U. = 1 lb × 1ºF = 453·6 gm × 5/9 ºC = 252 calories = 252 × 4·18 Joules = 1053 Joules ∴ 1 B.Th.U. = 1053 Joules It may be seen that heat energy can be expressed in Joules instead of thermal units viz. calorie, B.Th.U. and C.H.U. * The SI or MKS unit of thermal energy being used these days is the joule—exactly as for mechanical and electrical energies. The thermal units viz. calorie, B.Th.U. and C.H.U. are obsolete. 6 Principles of Power System (iii) Electrical and Heat 1 kWh = 1000 watts × 3600 seconds = 36 × 10 Joules 5 (a) 36 × 10 5 calories = 860 × 10 calories 3 = 4 ⋅18 ∴ 1 kWh = 860 × 10 calories or 860 kcal 3 1 kWh = 36 × 10 Joules = 36 × 10 /1896 C.H.U. = 1898 C.H.U. 5 5 (b) [Œ 1 C.H.U. = 1896 Joules] ∴ 1 kWh = 1898 C.H.U. 36 × 10 5 1 kWh = 36 × 10 Joules = 5 (c) B.Th.U. = 3418 B.Th.U. 1053 [Œ 1 B.Th.U. = 1053 Joules] ∴ 1 kWh = 3418 B.Th.U. The reader may note that units of electrical energy can be converted into heat and vice-versa. This is expected since electrical and thermal energies are interchangeable. 1.7 Efficiency Energy is available in various forms from different natural sources such as pressure head of water, chemical energy of fuels, nuclear energy of radioactive substances etc. All these forms of energy can be converted into electrical energy by the use of suitable arrangement. In this process of conversion, some energy is lost in the sense that it is converted to a form different from electrical energy. Therefore, the output energy is less than the input energy. The output energy divided by the input energy is called energy efficiency or simply efficiency Measuring efficiency of compressor. of the system. Output energy Efficiency, η = Input energy As power is the rate of energy flow, therefore, efficiency may be expressed equally well as output power divided by input power i.e., Output power Efficiency, η = Input power Example 1.1. Mechanical energy is supplied to a d.c. generator at the rate of 4200 J/s. The generator delivers 32·2 A at 120 V. (i) What is the percentage efficiency of the generator ? (ii) How much energy is lost per minute of operation ? Introduction 7 Solution. (i) Input power, Pi = 4200 J/s = 4200 W Output power, Po = EI = 120 × 32·2 = 3864 W P 3864 × ∴ Efficiency, η = o × 100 = 100 = 92 % Pi 4200 (ii) Power lost, PL = Pi − Po = 4200 − 3864 = 336 W ∴ Energy lost per minute (= 60 s) of operation = PL × t = 336 × 60 = 20160 J Note that efficiency is always less than 1 (or 100 %). In other words, every system is less than 100 % efficient. 1.8 Calorific Value of Fuels The amount of heat produced by the complete combustion of a unit weight of fuel is known as its calorific value. Calorific value indicates the amount of heat available from a fuel. The greater the calorific value of fuel, the larger is its ability to produce heat. In case of solid and liquid fuels, the calorific value is expressed in cal/gm or kcal/kg. However, in case of gaseous fuels, it is generally stated in cal/litre or kcal/litre. Below is given a table of various types of fuels and their calorific values along with composition. S.No. Particular Calorific value Composition 1. Solid fuels (i) Lignite 5,000 kcal/kg C = 67%, H = 5%, O = 20%, ash = 8% (ii) Bituminous coal 7,600 kcal/kg C = 83%, H = 5·5%, O = 5%, ash = 6·5% (iii) Anthracite coal 8,500 kcal/kg C = 90%, H = 3%, O = 2%, ash = 5% 2. Liquid fuels (i) Heavy oil 11,000 kcal/kg C = 86%, H = 12%, S = 2% (ii) Diesel oil 11,000 kcal/kg C = 86·3%, H = 12·8%, S = 0·9% (iii) Petrol 11,110 kcal/kg C = 86%, H = 14% 3. Gaseous fuels 3 (i) Natural gas 520 kcal/m CH4 = 84%, C2H6 = 10% Other hydrocarbons = 5% (ii) Coal gas 7,600 kcal/m3 CH4 = 35%, H = 45%, CO= 8%, N = 6% CO2 = 2%, Other hydrocarbons = 4% 1.9 Advantages of Liquid Fuels over Solid Fuels The following are the advantages of liquid fuels over the solid fuels : (i) The handling of liquid fuels is easier and they require less storage space. (ii) The combustion of liquid fuels is uniform. (iii) The solid fuels have higher percentage of moisture and consequently they burn with great difficulty. However, liquid fuels can be burnt with a fair degree of ease and attain high temperature very quickly compared to solid fuels. (iv) The waste product of solid fuels is a large quantity of ash and its disposal becomes a problem. However, liquid fuels leave no or very little ash after burning. (v) The firing of liquid fuels can be easily controlled. This permits to meet the variation in load demand easily. 1.10 Advantages of Solid Fuels over Liquid Fuels The following are the advantages of solid fuels over the liquid fuels : 8 Principles of Power System (i) In case of liquid fuels, there is a danger of explosion. (ii) Liquids fuels are costlier as compared to solid fuels. (iii) Sometimes liquid fuels give unpleasant odours during burning. (iv) Liquid fuels require special types of burners for burning. (v) Liquid fuels pose problems in cold climates since the oil stored in the tanks is to be heated in order to avoid the stoppage of oil flow. SELF-TEST 1. Fill in the blanks by inserting appropriate words/figures. (i) The primary source of energy is the...................... (ii) The most important form of energy is the......................... (iii) 1 kWh =..................... kcal (iv) The calorific value of a solid fuel is expreessed in...................... (v) The three principal sources of energy used for the generation of electrical energy are........................ and......................... 2. Pick up the correct words/figures from the brackets and fill in the blanks. (i) Electrical energy is....................... than other forms of energy. (cheaper, costlier) (ii) The electrical, heat and mechanical energies......................... be expressed in the same units. (can, cannot) (iii)......................... continue to enjoy the chief source for the generation of electrical energy. (fuels, radioactive substances, water) (iv) The basic unit of energy is......................... (Joule, watt) (v) An alternator is a machine which converts......................... into.......................... (mechanical energy, electrical energy) ANSWERS TO SELF-TEST 1. (i) Sun, (ii) electrical energy, (iii) 860, (iv) cal/gm or kcal/kg, (v) water, fuels and radioactive substances. 2. (i) Cheaper, (ii) can, (iii) fuels, (iv) Joule, (v) mechanical energy, electrical energy. CHAPTER REVIEW TOPICS 1. Why is electrical energy preferred over other forms of energy ? 2. Write a short note on the generation of electrical energy. 3. Discuss the different sources of energy available in nature. 4. Compare the chief sources of energy used for the generation of electrical energy. 5. Establish the following relations : (i) 1 kWh = 36 × 10 Joules 5 (ii) 1 kWh = 860 kcal (iii) 1 B.Th.U. = 1053 Joules (iv) 1 C.H.U. = 1896 Joules 6. What do you mean by efficiency of a system ? 7. What are the advantages of liquid fuels over the solid fuels ? 8. What are the advantages of solid fuels over the liquid fuels ? DISCUSSION QUESTIONS 1. Why do we endeavour to use water power for the generation of electrical energy ? 2. What is the importance of electrical energy ? 3. What are the problems in the use of nuclear energy ? 4. Give one practical example where wind-mill is used. 5. What is the principal source of generation of electrical energy ? GO To FIRST CONTENTS CONTENTS CHAPTER Generating Stations 2.1 Generating Stations 2.2 Steam Power Station (Thermal Station) 2.3 Schematic Arrangement of Steam Power Introduction Introduction Station I n this modern world, the dependence on 2.4 Choice of Site for Steam Power Stations electricity is so much that it has become a 2.5 Efficiency of Steam Power Station part and parcel of our life. The ever increas- 2.6 Equipment of Steam Power Station ing use of electric power for domestic, commer- 2.7 Hydro-electric Power Station cial and industrial purposes necessitates to pro- 2.8 Schematic Arrangement of Hydro- vide bulk electric power economically. This is electric Power Station achieved with the help of suitable power produc- 2.9 Choice of Site for Hydro-electric Power ing units, known as Power plants or Electric Stations power generating stations. The design of a power plant should incorporate two important aspects. 2.10 Constituents of Hydro-electric Plant Firstly, the selection and placing of necessary 2.11 Diesel Power Station power-generating equipment should be such so 2.12 Schematic Arrangement of Diesel Power that a maximum of return will result from a mini- Station mum of expenditure over the working life of the 2.13 Nuclear Power Station plant. Secondly, the operation of the plant should 2.14 Schematic Arrangement of Nuclear be such so as to provide cheap, reliable and Power Station continuous service. In this chapter, we shall 2.15 Selection of Site for Nuclear Power focus our attention on various types of generat- Station ing stations with special reference to their advan- tages and disadvantages. 2.16 Gas Turbine Power Plant 2.17 Schematic Arrangement of Gas 2.1 Generating Stations Turbine Power Plant Bulk electric power is produced by special plants 2.18 Comparison of the Various Power known as generating stations or power plants. Plants A generating station essentially employs a 9 CONTENTS CONTENTS 10 Principles of Power System prime mover coupled to an alternator for the production of electric power. The prime mover (e.g., steam turbine, water turbine etc.) converts energy from some other form into mechanical energy. The alternator converts mechanical energy of the prime mover into electrical energy. The electrical en- ergy produced by the generating station is transmitted and distributed with the help of conductors to various consumers. It may be emphasised here that apart from prime mover-alternator combination, a modern generating station employs several auxiliary equipment and instruments to ensure cheap, reliable and continuous service. Depending upon the form of energy converted into electrical energy, the generating stations are classified as under : (i) Steam power stations (ii) Hydroelectric power stations (iii) Diesel power stations (iv) Nuclear power stations 2.2 Steam Power Station (Ther mal Station) (Thermal A generating station which converts heat energy of coal combustion into electrical energy is known as a steam power station. A steam power station basically works on the Rankine cycle. Steam is produced in the boiler by utilising the heat of coal combustion. The steam is then expanded in the prime mover (i.e., steam turbine) and is condensed in a condenser to be fed into the boiler again. The steam turbine drives the alternator which converts mechanical energy of the turbine into electrical energy. This type of power station is suitable where coal and water are available in abundance and a large amount of electric power is to be generated. Advantages (i) The fuel (i.e., coal) used is quite cheap. (ii) Less initial cost as compared to other generating stations. (iii) It can be installed at any place irrespective of the existence of coal. The coal can be trans- ported to the site of the plant by rail or road. (iv) It requires less space as compared to the hydroelectric power station. (v) The cost of generation is lesser than that of the diesel power station. Disadvantages (i) It pollutes the atmosphere due to the production of large amount of smoke and fumes. (ii) It is costlier in running cost as compared to hydroelectric plant. 2.3 Schematic Arrangement of Steam Power Station Although steam power station simply involves the conversion of heat of coal combustion into electri- cal energy, yet it embraces many arrangements for proper working and efficiency. The schematic arrangement of a modern steam power station is shown in Fig. 2.1. The whole arrangement can be divided into the following stages for the sake of simplicity : 1. Coal and ash handling arrangement 2. Steam generating plant 3. Steam turbine 4. Alternator 5. Feed water 6. Cooling arrangement 1. Coal and ash handling plant. The coal is transported to the power station by road or rail and is stored in the coal storage plant. Storage of coal is primarily a matter of protection against coal strikes, failure of transportation system and general coal shortages. From the coal storage plant, coal is delivered to the coal handling plant where it is pulverised (i.e., crushed into small pieces) in order to increase its surface exposure, thus promoting rapid combustion without using large quantity of Generating Stations 11 excess air. The pulverised coal is fed to the boiler by belt conveyors. The coal is burnt in the boiler and the ash produced after the complete combustion of coal is removed to the ash handling plant and then delivered to the ash storage plant for disposal. The removal of the ash from the boiler furnace is necessary for proper burning of coal. It is worthwhile to give a passing reference to the amount of coal burnt and ash produced in a modern thermal power station. A 100 MW station operating at 50% load factor may burn about 20,000 tons of coal per month and ash produced may be to the tune of 10% to 15% of coal fired i.e., 2,000 to 3,000 tons. In fact, in a thermal station, about 50% to 60% of the total operating cost consists of fuel purchasing and its handling. 12 Principles of Power System 2. Steam generating plant. The steam generating plant consists of a boiler for the production of steam and other auxiliary equipment for the utilisation of flue gases. (i) Boiler. The heat of combustion of coal in the boiler is utilised to convert water into steam at high temperature and pressure. The flue gases from the boiler make their journey through super- heater, economiser, air pre-heater and are finally exhausted to atmosphere through the chimney. (ii) Superheater. The steam produced in the boiler is wet and is passed through a superheater where it is dried and superheated (i.e., steam temperature increased above that of boiling point of water) by the flue gases on their way to chimney. Superheating provides two principal benefits. Firstly, the overall efficiency is increased. Secondly, too much condensation in the last stages of turbine (which would cause blade corrosion) is avoided. The superheated steam from the superheater is fed to steam turbine through the main valve. (iii) Economiser. An economiser is essentially a feed water heater and derives heat from the flue gases for this purpose. The feed water is fed to the economiser before supplying to the boiler. The economiser extracts a part of heat of flue gases to increase the feed water temperature. (iv) Air preheater. An air preheater increases the temperature of the air supplied for coal burn- ing by deriving heat from flue gases. Air is drawn from the atmosphere by a forced draught fan and is passed through air preheater before supplying to the boiler furnace. The air preheater extracts heat from flue gases and increases the temperature of air used for coal combustion. The principal benefits of preheating the air are : increased thermal efficiency and increased steam capacity per square metre of boiler surface. 3. Steam turbine. The dry and superheated steam from the superheater is fed to the steam turbine through main valve. The heat energy of steam when passing over the blades of turbine is converted into mechanical energy. After giving heat energy to the turbine, the steam is exhausted to the condenser which condenses the exhausted steam by means of cold water circulation. 4. Alternator. The steam turbine is coupled to an alternator. The alternator converts mechanical energy of turbine into electrical energy. The electrical output from the alternator is delivered to the bus bars through transformer, circuit breakers and isolators. 5. Feed water. The condensate from the condenser is used as feed water to the boiler. Some water may be lost in the cycle which is suitably made up from external source. The feed water on its way to the boiler is heated by water heaters and economiser. This helps in raising the overall effi- ciency of the plant. 6. Cooling arrangement. In order to improve the efficiency of the plant, the steam exhausted from the turbine is condensed* by means of a condenser. Water is drawn from a natural source of supply such as a river, canal or lake and is circulated through the condenser. The circulating water takes up the heat of the exhausted steam and itself becomes hot. This hot water coming out from the condenser is discharged at a suitable location down the river. In case the availability of water from the source of supply is not assured throughout the year, cooling towers are used. During the scarcity of water in the river, hot water from the condenser is passed on to the cooling towers where it is cooled. The cold water from the cooling tower is reused in the condenser. 2.4 Choice of Site for Steam Power Stations In order to achieve overall economy, the following points should be considered while selecting a site for a steam power station : (i) Supply of fuel. The steam power station should be located near the coal mines so that transportation cost of fuel is minimum. However, if such a plant is to be installed at a place * Efficiency of the plant is increased by reducing turbine exhaust pressure. Low pressure at the exhaust can be achieved by condensing the steam at the turbine exhaust. Generating Stations 13 where coal is not available, then care should be taken that adequate facilities exist for the transportation of coal. (ii) Availability of water. As huge amount of water is required for the condenser, therefore, such a plant should be located at the bank of a river or near a canal to ensure the continuous supply of water. (iii) Transportation facilities. A modern steam power station often requires the transportation of material and machinery. Therefore, adequate transportation facilities must exist i.e., the plant should be well connected to other parts of the country by rail, road. etc. (iv) Cost and type of land. The steam power station should be located at a place where land is cheap and further extension, if necessary, is possible. Moreover, the bearing capacity of the ground should be adequate so that heavy equipment could be installed. (v) Nearness to load centres. In order to reduce the transmission cost, the plant should be located near the centre of the load. This is particularly important if d.c. supply system is adopted. However, if a.c. supply system is adopted, this factor becomes relatively less important. It is because a.c. power can be transmitted at high voltages with consequent reduced transmission cost. Therefore, it is possible to install the plant away from the load centres, provided other conditions are favourable. (vi) Distance from populated area. As huge amount of coal is burnt in a steam power station, therefore, smoke and fumes pollute the surrounding area. This necessitates that the plant should be located at a considerable distance from the populated areas. Conclusion. It is clear that all the above factors cannot be favourable at one place. However, keeping in view the fact that now-a-days the supply system is a.c. and more importance is being given to generation than transmission, a site away from the towns may be selected. In particular, a site by river side where sufficient water is available, no pollution of atmosphere occurs and fuel can be transported economically, may perhaps be an ideal choice. 2.5 Ef ficiency of Steam Power Station Efficiency The overall efficiency of a steam power station is quite low (about 29%) due mainly to two reasons. Firstly, a huge amount of heat is lost in the condenser and secondly heat losses occur at various stages of the plant. The heat lost in the condenser cannot be avoided. It is because heat energy cannot be converted into mechanical energy without temperature difference. The greater the temperature dif- ference, the greater is the heat energy converted* into mechanical energy. This necessitates to keep the steam in the condenser at the lowest temperature. But we know that greater the temperature difference, greater is the amount of heat lost. This explains for the low efficiency of such plants. (i) Thermal efficiency. The ratio of heat equivalent of mechanical energy transmitted to the turbine shaft to the heat of combustion of coal is known as thermal efficiency of steam power station. Heat equivalent of mech. energy transmitted to turbine shaft Thermal efficiency, ηthermal = Heat of coal combustion The thermal efficiency of a modern steam power station is about 30%. It means that if 100 calories of heat is supplied by coal combustion, then mechanical energy equivalent of 30 calories will be available at the turbine shaft and rest is lost. It may be important to note that more than 50% of total heat of combustion is lost in the condenser. The other heat losses occur in flue gases, radia- tion, ash etc. (ii) Overall efficiency. The ratio of heat equivalent of electrical output to the heat of combus- tion of coal is known as overall efficiency of steam power station i.e. * Thermodynamic laws. 14 Principles of Power System Heat equivalent of electrical ouput Overall efficiency, ηoverall = Heat of combustion of coal The overall efficiency of a steam power station is about 29%. It may be seen that overall effi- ciency is less than the thermal efficiency. This is expected since some losses (about 1%) occur in the alternator. The following relation exists among the various efficiencies. Overall efficiency = Thermal efficiency × Electrical efficiency 2.6 Equipment of Steam Power Station A modern steam power station is highly complex and has numerous equipment and auxiliaries. How- ever, the most important constituents of a steam power station are : 1. Steam generating equipment 2. Condenser 3. Prime mover 4. Water treatment plant 5. Electrical equipment. 1. Steam generating equipment. This is an important part of steam power station. It is con- cerned with the generation of superheated steam and includes such items as boiler, boiler furnace, superheater, economiser, air pre-heater and other heat reclaiming devices. (i) Boiler. A boiler is closed vessel in which water is converted into steam by utilising the heat of coal combustion. Steam boilers are broadly classified into the following two types : (a) Water tube boilers (b) Fire tube boilers In a water tube boiler, water flows through the tubes and the hot gases of combustion flow over these tubes. On the other hand, in a fire tube boiler, the hot products of combustion pass through the tubes surrounded by water. Water tube boilers have a number of advantages over fire tube boilers viz., require less space, smaller size of tubes and drum, high working pressure due to small drum, less liable to explosion etc. Therefore, the use of water tube boilers has become universal in large capac- ity steam power stations. (ii) Boiler furnace. A boiler furnace is a chamber in which fuel is burnt to liberate the heat energy. In addition, it provides support and enclosure for the combustion equipment i.e., burners. The boiler furnace walls are made of refractory materials such as fire clay, silica, kaolin etc. These materials have the property to resist change of shape, weight or physical properties at high tempera- tures. There are following three types of construction of furnace walls : (a) Plain refractory walls (b) Hollow refractory walls with an arrangement for air cooling (c) Water walls. The plain refractory walls are suitable for small plants where the furnace temperature may not be high. However, in large plants, the furnace temperature is quite high* and consequently, the refrac- tory material may get damaged. In such cases, refractory walls are made hollow and air is circulated through hollow space to keep the temperature of the furnace walls low. The recent development is to use water walls. These consist of plain tubes arranged side by side and on the inner face of the refractory walls. The tubes are connected to the upper and lower headers of the boiler. The boiler water is made to circulate through these tubes. The water walls absorb the radiant heat in the furnace which would otherwise heat up the furnace walls. (iii) Superheater. A superheater is a device which superheats the steam i.e., it raises the tempera- ture of steam above boiling point of water. This increases the overall efficiency of the plant. A superheater consists of a group of tubes made of special alloy steels such as chromium-molybdenum. These tubes are heated by the heat of flue gases during their journey from the furnace to the chimney. * The size of furnace has to be limited due to space, cost and other considerations. This means that furnace of a large plant should develop more kilocalories per square metre of furnace which implies high furnace temperature. Generating Stations 15 The steam produced in the boiler is led through the superheater where it is superheated by the heat of flue gases. Superheaters are mainly classified into two types according to the system of heat transfer from flue gases to steam viz. (a) Radiant superheater (b) Convection superheater The radiant superheater is placed in the furnace between the water walls and receives heat from the burning fuel through radiation process. It has two main disadvantages. Firstly, due to high furnace temperature, it may get overheated and, therefore, requires a careful design. Secondly, the temperature of superheater falls with increase in steam output. Due to these limitations, radiant superheater is not finding favour these days. On the other hand, a convection superheater is placed in the boiler tube bank and receives heat from flue gases entirely through the convection process. It has the advantage that temperature of superheater increases with the increase in steam output. For this reason, this type of superheater is commonly used these days. (iv) Economiser. It is a device which heats the feed water on its way to boiler by deriving heat from the flue gases. This results in raising boiler efficiency, saving in fuel and reduced stresses in the boiler due to higher temperature of feed water. An economiser consists of a large number of closely spaced parallel steel tubes connected by headers of drums. The feed water flows through these tubes and the flue gases flow outside. A part of the heat of flue gases is transferred to feed water, thus raising the temperature of the latter. (v) Air Pre-heater. Superheaters and economisers generally cannot fully extract the heat from flue gases. Therefore, pre-heaters are employed which recover some of the heat in the escaping gases. The function of an air pre-heater is to extract heat from the flue gases and give it to the air being supplied to furnace for coal combustion. This raises the furnace temperature and increases the thermal efficiency of the plant. Depending upon the method of transfer of heat from flue gases to air, air pre-heaters are divided into the following two classes : (a) Recuperative type (b) Regenerative type The recuperative type air-heater consists of a group of steel tubes. The flue gases are passed through the tubes while the air flows externally to the tubes. Thus heat of flue gases is transferred to air. The regenerative type air pre-heater consists of slowly moving drum made of corrugated metal plates. The flue gases flow continuously on one side of the drum and air on the other side. This action permits the transference of heat of flue gases to the air being supplied to the furnace for coal combustion. 2. Condensers. A condenser is a device which condenses the steam at the exhaust of turbine. It serves two important functions. Firstly, it creates a very low *pressure at the exhaust of turbine, thus permitting expansion of the steam in the prime mover to a very low pressure. This helps in converting heat energy of steam into mechanical energy in the prime mover. Secondly, the condensed steam can be used as feed water to the boiler. There are two types of condensers, namely : (i) Jet condenser (ii) Surface condenser In a jet condenser, cooling water and exhausted steam are mixed together. Therefore, the tem- perature of cooling water and condensate is the same when leaving the condenser. Advantages of this type of condenser are : low initial cost, less floor area required, less cooling water required and low maintenance charges. However, its disadvantages are : condensate is wasted and high power is re- quired for pumping water. In a surface condenser, there is no direct contact between cooling water and exhausted steam. It consists of a bank of horizontal tubes enclosed in a cast iron shell. The cooling water flows through the tubes and exhausted steam over the surface of the tubes. The steam gives up its heat to water and is itself condensed. Advantages of this type of condenser are : condensate can be used as feed water, less pumping power required and creation of better vacuum at the turbine exhaust. However, disad- * By liquidating steam at the exhaust of turbine, a region of emptiness is created. This results in a very low pressure at the exhaust of turbine. 16 Principles of Power System vantages of this type of condenser are : high initial cost, requires large floor area and high mainte- nance charges. 3. Prime movers. The prime mover converts steam energy into mechanical energy. There are two types of steam prime movers viz., steam engines and steam turbines. A steam turbine has several advantages over a steam engine as a prime mover viz., high efficiency, simple construction, higher speed, less floor area requirement and low maintenance cost. Therefore, all modern steam power stations employ steam turbines as prime movers. Steam turbines are generally classified into two types according to the action of steam on moving blades viz. (i) Impulse turbines (ii) Reactions turbines In an impulse turbine, the steam expands completely in the stationary nozzles (or fixed blades), the pressure over the moving blades remaining constant. In doing so, the steam attains a high velocity and impinges against the moving blades. This results in the impulsive force on the moving blades which sets the rotor rotating. In a reaction turbine, the steam is partially expanded in the stationary nozzles, the remaining expansion takes place during its flow over the moving blades. The result is that the momentum of the steam causes a reaction force on the moving blades which sets the rotor in motion. 4. Water treatment plant. Boilers require clean and soft water for longer life and better effi- ciency. However, the source of boiler feed water is generally a river or lake which may contain suspended and dissolved impurities, dissolved gases etc. Therefore, it is very important that water is first purified and softened by chemical treatment and then delivered to the boiler. The water from the source of supply is stored in storage tanks. The suspended impurities are removed through sedimentation, coagulation and filtration. Dissolved gases are removed by aeration and degasification. The water is then ‘softened’ by removing temporary and permanent hardness through different chemical processes. The pure and soft water thus available is fed to the boiler for steam generation. 5. Electrical equipment. A modern power station contains numerous electrical equipment. However, the most important items are : (i) Alternators. Each alternator is coupled to a steam turbine and converts mechanical energy of the turbine into electrical energy. The alternator may be hydrogen or air cooled. The necessary excitation is provided by means of main and pilot exciters directly coupled to the alternator shaft. (ii) Transformers. A generating station has different types of transformers, viz., (a) main step-up transformers which step-up the generation voltage for transmission of power. (b) station transformers which are used for general service (e.g., lighting) in the power station. (c) auxiliary transformers which supply to individual unit-auxiliaries. (iii) Switchgear. It houses such equipment which locates the fault on the system and isolate the faulty part from the healthy section. It contains circuit breakers, relays, switches and other control devices. Example 2.1. A steam power station has an overall efficiency of 20% and 0·6 kg of coal is burnt per kWh of electrical energy generated. Calculate the calorific value of fuel. Generating Stations 17 Solution. Let x kcal/kg be the calorific value of fuel. Heat produced by 0·6 kg of coal = 0·6 x kcal Heat equivalent of 1 kWh = 860 kcal Electrical output in heat units Now, ηoverall = Heat of combustion 860 or 0·2 = 0 ⋅ 6x 860 ∴ x = = 7166·67 kcal/kg 0 ⋅6 × 0⋅2 Example 2.2. A thermal station has the following data : Max. demand = 20,000 kW ; Load factor = 40% Boiler efficiency = 85% ; Turbine efficiency = 90% Coal consumption = 0·9 kg/kWh ; Cost of 1 ton of coal = Rs. 300 Determine (i) thermal efficiency and (ii) coal bill per annum. Solution. (i) Thermal efficiency = ηboiler × ηturbine = 0·85 × ·9 = 0·765 or 76·5 % (ii) Units generated/annum = Max. demand × L.F. × Hours in a year = 20,000 × 0·4 × 8760 = 7008 × 10 kWh 4 Coal consumption/annum = a0 ⋅ 9f e7008 × 10 j = 63,072 tons 4 1000 ∴ Annual coal bill = Rs 300 × 63072 = Rs 1,89,21,600 Example 2.3. A steam power station spends Rs. 30 lakhs per annum for coal used in the station. The coal has a calorific value of 5000 kcal/kg and costs Rs. 300 per ton. If the station has thermal efficiency of 33% and electrical efficiency of 90%, find the average load on the station. Solution. Overall efficiency, ηoverall = 0·33 × 0·9 = 0·297 Coal used/annum = 30 × 10 /300 = 10 tons = 10 kg 5 4 7 Heat of combustion = Coal used/annum × Calorific value = 10 × 5000 = 5 × 10 kcal 7 10 Heat output = ηoverall × Heat of combustion = (0·297) × (5 × 10 ) = 1485 × 10 kcal 10 7 Units generated/annum = 1485 × 10 /860 kWh 7 Units generated / annum 1485 × 10 7 ∴ Average load on station = = = 1971 kW Hours in a year 860 × 8760 Example 2.4. The relation between water evaporated (W kg), coal consumption (C kg) and kWh generated per 8-hour shift for a steam generating station is as follows : W = 13500 + 7·5 kWh....(i) C = 5000 + 2·9 kWh....(ii) (i) To what limiting value does the water evaporating per kg of coal consumed approach as the station output increases ? (ii) How much coal per hour would be required to keep the station run- ning on no load ? Solution. (i) For an 8-hour shift, weight of water evaporated per kg of coal consumed is 18 Principles of Power System W 13500 + 7 ⋅ 5 kWh = C 5000 + 2 ⋅ 9 kWh As the station output (i.e., kWh) increases towards infinity, the limiting value of W/C approaches 7·5/2·9 = 2·6. Therefore, the weight of water evaporated per kg of coal consumed approaches a limiting value of 2·6 kg as the kWh output increases. (ii) At no load, the station output is zero i.e., kWh = 0. Therefore, from expression (ii), we get, coal consumption at no load = 5000 + 2·9 × 0 = 5000 kg ∴ Coal consumption/hour = 5000/8 = 625 kg Example 2.5. A 100 MW steam station uses coal of calorific value 6400 kcal/kg. Thermal efficiency of the station is 30% and electrical efficiency is 92%. Calculate the coal consumption per hour when the station is delivering its full rated output. Solution. Overall efficiency of the power station is ηoverall = ηthermal × ηelect = 0·30 × 0·92 = 0·276 Units generated/hour = (100 × 10 ) × 1 = 10 kWh 3 5 Electrical output in heat units Heat produced/hour, H = ηoverall 105 × 860 = = 311 ⋅ 6 × 106 kcal (∵ 1 kWh = 860 kcal) 0 ⋅ 276 H 311⋅ 6 × 106 ∴ Coal consumption/hour = = = 48687 kg Calorific value 6400 TUTORIAL PROBLEMS 1. A generating station has an overall efficiency of 15% and 0·75 kg of coal is burnt per kWh by the station. Determine the calorific value of coal in kilocalories per kilogram. [7644 kcal/kg] 2. A 75 MW steam power station uses coal of calorific value of 6400 kcal/kg. Thermal efficiency of the station is 30% while electrical efficiency is 80%. Calculate the coal consumption per hour when the station is delivering its full output. [42 tons] 3. A 65,000 kW steam power station uses coal of calorific value 15,000 kcal per kg. If the coal consump- tion per kWh is 0·5 kg and the load factor of the station is 40%, calculate (i) the overall efficiency (ii) coal consumption per day. [(i) 28·7% (ii) 312 tons] 4. A 60 MW steam power station has a thermal efficiency of 30%. If the coal burnt has a calorific value of 6950 kcal/kg, calculate : (i) the coal consumption per kWh, (ii)the coal consumption per day. [(i) 0·413 kg (ii) 238 tons] 5. A 25 MVA turbo-alternator is working on full load at a power factor of 0·8 and efficiency of 97%. Find the quantity of cooling air required per minute at full load, assuming that 90% of the total losses are dissipated by the internally circulating air. The inlet air temperature is 20º C and the temperature rise is 3 3 30º C. Given that specific heat of air is 0·24 and that 1 kg of air occupies 0·8 m. [890 m /minute] 6. A thermal station has an efficiency of 15% and 1·0 kg of coal burnt for every kWh generated. Determine the calorific value of coal. [5733 kcal/kg] 2.7 Hydr o-electr Hydro-electr ic P o-electric ower Sta Po tion Station A generating station which utilises the potential energy of water at a high level for the generation of electrical energy is known as a hydro-electric power station. Generating Stations 19 Hydro-electric power stations are generally located in hilly areas where dams can be built conve- niently and large water reservoirs can be obtained. In a hydro-electric power station, water head is created by constructing a dam across a river or lake. From the dam, water is led to a water turbine. The water turbine captures the energy in the falling water and changes the hydraulic energy (i.e., product of head and flow of water) into mechanical energy at the turbine shaft. The turbine drives the alternator which converts mechanical energy into electrical energy. Hydro-electric power stations are becoming very popular because the reserves of fuels (i.e., coal and oil) are depleting day by day. They have the added importance for flood control, storage of water for irrigation and water for drink- ing purposes. Advantages (i) It requires no fuel as water is used for the generation of electrical energy. (ii) It is quite neat and clean as no smoke or ash is produced. (iii) It requires very small running charges because water is the source of energy which is avail- able free of cost. (iv) It is comparatively simple in construction and requires less maintenance. (v) It does not require a long starting time like a steam power station. In fact, such plants can be put into service instantly. (vi) It is robust and has a longer life. (vii) Such plants serve many purposes. In addition to the generation of electrical energy, they also help in irrigation and controlling floods. (viii) Although such plants require the attention of highly skilled persons at the time of construc- tion, yet for operation, a few experienced persons may do the job well. Disadvantages (i) It involves high capital cost due to construction of dam. (ii) There is uncertainty about the availability of huge amount of water due to dependence on weather conditions. (iii) Skilled and experienced hands are required to build the plant. (iv) It requires high cost of transmission lines as the plant is located in hilly areas which are quite away from the consumers. 2.8 Schematic Arrangement of Hydr Schematic o-electr Hydro-electr ic P o-electric ower Sta Po tion Station Although a hydro-electric power station simply involves the conversion of hydraulic energy into electrical energy, yet it embraces many arrangements for proper working and efficiency. The sche- matic arrangement of a modern hydro-electric plant is shown in Fig. 2.2. The dam is constructed across a river or lake and water from the catchment area collects at the back of the dam to form a reservoir. A pressure tunnel is taken off from the reservoir and water brought to the valve house at the start of the penstock. The valve house contains main sluice valves and automatic isolating valves. The former controls the water flow to the power house and the latter cuts off supply of water when the penstock bursts. From the valve house, water is taken to water turbine through a huge steel pipe known as penstock. The water turbine converts hydraulic energy into mechanical energy. The turbine drives the alternator which converts mechanical energy into electrical energy. A surge tank (open from top) is built just before the valve house and protects the penstock from bursting in case the turbine gates suddenly close* due to electrical load being thrown off. When the * The governor opens or closes the turbine gates in accordance with the changes in electrical load. If the electrical load increases, the governor opens the turbine gates to allow more water and vice-versa. 20 Principles of Power System gates close, there is a sudden stopping of water at the lower end of the penstock and consequently the penstock can burst like a paper log. The surge tank absorbs this pressure swing by increase in its level of water. 2.9 Choice of Site for Hydr o-electric Power Stations Hydro-electric The following points should be taken into account while selecting the site for a hydro-electric power station : (i) Availability of water. Since the primary requirement of a hydro-electric power station is the availability of huge quantity of water, such plants should be built at a place (e.g., river, canal) where adequate water is available at a good head. (ii) Storage of water. There are wide variations in water supply from a river or canal during the year. This makes it necessary to store water by constructing a dam in order to ensure the generation of power throughout the year. The storage helps in equalising the flow of water so that any excess quantity of water at a certain period of the year can be made available during times of very low flow in the river. This leads to the conclusion that site selected for a hydro-electric plant should provide adequate facilities for erecting a dam and storage of water. (iii) Cost and type of land. The land for the construction of the plant should be available at a reasonable price. Further, the bearing capacity of the ground should be adequate to with- stand the weight of heavy equipment to be installed. (iv) Transportation facilities. The site selected for a hydro-electric plant should be accessible by rail and road so that necessary equipment and machinery could be easily transported. It is clear from the above mentioned factors that ideal choice of site for such a plant is near a river in hilly areas where dam can be conveniently built and large reservoirs can be obtained. 2.10 Constituents of Hydr o-electric Plant Hydro-electric The constituents of a hydro-electric plant are (1) hydraulic structures (2) water turbines and (3) electrical equipment. We shall discuss these items in turn. 1. Hydraulic structures. Hydraulic structures in a hydro-electric power station include dam, spillways, headworks, surge tank, penstock and accessory works. (i) Dam. A dam is a barrier which stores water and creates water head. Dams are built of concrete or stone masonary, earth or rock fill. The type and arrangement depends upon the Generating Stations 21 topography of the site. A masonary dam may be built in a narrow canyon. An earth dam may be best suited for a wide valley. The type of dam also depends upon the foundation conditions, local materials and transportation available, occurrence of earthquakes and other hazards. At most of sites, more than one type of dam may be suitable and the one which is most economical is chosen. (ii) Spillways. There are times when the river flow exceeds the storage capacity of the reservoir. Such a situation arises during heavy rainfall in the catchment area. In order to discharge the surplus water from the storage reservoir into the river on the down-stream side of the dam, spillways are used. Spillways are constructed of concrete piers on the top of the dam. Gates are provided between these piers and surplus water is discharged over the crest of the dam by opening these gates. (iii) Headworks. The headworks consists of the diversion structures at the head of an intake. They generally include booms and racks for diverting floating debris, sluices for by-passing debris and sediments and valves for controlling the flow of water to the turbine. The flow of water into and through headworks should be as smooth as possible to avoid head loss and cavitation. For this purpose, it is necessary to avoid sharp corners and abrupt contractions or enlargements. (iv) Surge tank. Open conduits leading water to the turbine require no* protection. However, when closed con- duits are used, protection becomes necessary to limit the abnormal pressure in the conduit. For this reason, closed conduits are always provided with a surge tank. A surge tank is a small res- ervoir or tank (open at the top) in which water level rises or falls to reduce the pressure swings in the con- duit. A surge tank is located near the beginning of the conduit. When the turbine is running at a steady load, there are no surges in the flow of water through the conduit i.e., the quantity of water flowing in the conduit is just sufficient to meet the turbine requirements. However, when the load on the turbine decreases, the governor closes the gates of turbine, reducing water supply to the turbine. The excess water at the lower end of the conduit rushes back to the surge tank and increases its water level. Thus the conduit is prevented from bursting. On the other hand, when load on the turbine increases, addi- tional water is drawn from the surge tank to meet the increased load requirement. Hence, a surge tank overcomes the abnormal pressure in the conduit when load on the turbine falls and acts as a reservoir during increase of load on the turbine. (v) Penstocks. Penstocks are open or closed conduits which carry water to the turbines. They are generally made of reinforced concrete or steel. Concrete penstocks are suitable for low * Because in case of open conduits, regulating gates control the inflow at the headworks and the spillway discharges the surplus water. 22 Principles of Power System heads (< 30 m) as greater pressure causes rapid deterioration of concrete. The steel pen- stocks can be designed for any head; the thickness of the penstock increases with the head or working pressure. Various devices such as automatic butterfly valve, air valve and surge tank (See Fig. 2.3) are provided for the protection of penstocks. Automatic butterfly valve shuts off water flow through the penstock promptly if it ruptures. Air valve maintains the air pressure inside the penstock equal to outside atmospheric pressure. When water runs out of a penstock faster than it enters, a vacuum is created which may cause the penstock to collapse. Under such situations, air valve opens and admits air in the penstock to maintain inside air pressure equal to the outside air pressure. 2. Water turbines. Water turbines are used to convert the energy of falling water into mechani- cal energy. The principal types of water turbines are : (i) Impulse turbines (ii) Reaction turbines (i) Impulse turbines. Such turbines are used for high heads. In an impulse turbine, the entire pressure of water is converted into kinetic energy in a nozzle and the velocity of the jet drives the wheel. The example of this type of turbine is the Pelton wheel (See Fig. 2.4). It consists of a wheel fitted with elliptical buckets along its periphery. The force of water jet strik- ing the buckets on the wheel drives the turbine. The quantity of water jet falling on the turbine is controlled by means of a needle or spear (not shown in the fig- ure) placed in the tip of the nozzle. The movement of the needle is controlled by the governor. If the load on the turbine decreases, the governor pushes the needle into the nozzle, thereby reducing the quantity of water striking the buckets. Reverse action takes place if the load on the turbine increases. (ii) Reaction turbines. Reaction turbines are used for low and medium heads. In a reaction turbine, water enters the runner partly with pressure energy and partly with velocity head. The impor- tant types of reaction turbines are : (a) Francis turbines (b) Kaplan turbines A Francis turbine is used for low to medium heads. It consists of an outer ring of stationary guide blades fixed to the turbine casing and an inner ring of rotating blades forming the runner. The guide blades control the flow of water to the turbine. Water flows radially inwards and changes to a downward direction while passing through the runner. As the water passes over the “rotating blades” of the runner, both pressure and velocity of water are reduced. This causes a reaction force which drives the turbine. A Kaplan turbine is used for low heads and large quantities of water. It is similar to Francis tur- bine except that the runner of Kaplan turbine receives water axi- ally. Water flows radially inwards Bhakra Dam Generating Stations 23 through regulating gates all around the sides, changing direction in the runner to axial flow. This causes a reaction force which drives the turbine. 3. Electrical equipment. The electrical equipment of a hydro-electric power station includes alternators, transformers, circuit breakers and other switching and protective devices. Example 2.6. A hydro-electric generating station is supplied from a reservoir of capacity 5 × 10 cubic metres at a head of 200 metres. Find the total energy available in kWh if the overall 6 efficiency is 75%. Solution. Weight of water available is W = Volume of water × density = (5 × 10 ) × (1000) 6 3 (3mass of 1m of water is 1000 kg) = 5 × 10 kg = 5 × 10 × 9·81 N 9 9 Electrical energy available = W × H × ηoverall = (5 × 10 × 9·81) × (200) × (0·75) watt sec 9 (5 × 109 × 9 ⋅ 81) × (200) × (0 ⋅ 75) = kWh = 2.044 × 106 kWh 3600 × 1000 Example 2.7. It has been estimated that a minimum run off of approximately 94 m 3/sec will be available at a hydraulic project with a head of 39 m. Determine (i) firm capacity (ii) yearly gross output. Assume the efficiency of the plant to be 80%. Solution. Weight of water available, W = 94 × 1000 = 94000 kg/sec Water head, H = 39 m Work done/sec = W × H = 94000 × 9·81 × 39 watts = 35, 963 × 10 W = 35, 963 kW 3 This is gross plant capacity. (i) Firm capacity = Plant efficiency × Gross plant capacity = 0·80 × 35,963 = 28,770 kW (ii) Yearly gross output = Firm capacity × Hours in a year = 28,770 × 8760 = 252 × 10 kWh 6 Example 2.8. Water for a hydro-electric station is obtained from a reservoir with a head of 100 metres. Calculate the electrical energy generated per hour per cubic metre of water if the hydraulic efficiency be 0·86 and electrical efficiency 0·92. Solution. H = 100 m ; discharge, Q = 1 m /sec ; ηoverall = 0·86 × 0·92 = 0·79 3 Water head, Wt. of water available/sec, W = Q × 1000 × 9·81 = 9810 N Power produced = W × H × ηoverall = 9810 × 100 × 0·79 watts = 775 × 10 watts = 775 kW 3 ∴ Energy generated/hour = 775 × 1 = 775 kWh Example 2.9. Calculate the average power in kW that can be generated in a hydro-electric project from the following data Catchment area = 5 × 109 m2 ; Mean head, H = 30 m Annual rainfall, F = 1·25 m ; Yield factor, K = 80 % Overall efficiency, ηoveall = 70 % If the load factor is 40% , what is the rating of generators installed ? 24 Principles of Power System Solution. Volume of water which can be utilised per annum = Catchment area × Annual rainfall × *yield factor = (5 × 10 ) × (1·25) × (0·8) = 5 × 10 m 9 9 3 Weight of water available per annum is W = 5 × 10 × 9·81 × 1000 = 49·05 × 10 N 9 12 Electrical energy available per annum = W × H × ηoverall = (49·05 × 10 ) × (30) × (0·7) watt-sec 12 ( 49 ⋅ 05 × 10 ) × (30) × (0 ⋅ 7) 12 kWh = 2 ⋅ 86 × 10 kWh 8 = 1000 × 3600 ∴ Average power = 2·86 × 10 /8760 = 32648 kW 8 Average demand 32648 Max. demand = = = 81620 kW Load factor 0⋅4 Therefore, the maximum capacity of the generators should be 81620 kW. Example 2.10. A hydro-electric power station has a reservoir of area 2·4 square kilometres and capacity 5 × 10 m. The effective head of water is 100 metres. The penstock, turbine and generation 6 3 efficiencies are respectively 95%,90% and 85%. (i) Calculate the total electrical energy that can be generated from the power station. (ii) If a load of 15,000 kW has been supplied for 3 hours, find the fall in reservoir level. Solution. (i) Wt. of water available, W = Volume of reservoir × wt. of 1m of water 3 = (5 × 10 ) × (1000) kg = 5 × 10 × 9·81 N 6 9 Overall efficiency, ηoverall = 0·95 × 0·9 × 0·85 = 0·726 Electrical energy that can be generated = W × H × ηoverall = (5 × 109 × 9·81) × (100) × (0·726) watt-sec. (5 × 109 × 9 ⋅ 81) × (100) × (0 ⋅ 726) = kWh = 9,89,175 kWh 1000 × 3600 (ii) Let x metres be the fall in reservoir level in 3 hours. Area of reservoir × x 2 ⋅ 4 × 10 × x 6 = = 222 ⋅ 2 x m 3 Average discharge/sec = 3 × 3600 3 × 3600 Wt. of water available/sec, W = 222·2x × 1000 × 9·81 = 21·8x × 10 N 5 Average power produced = W × H × ηoverall = (21·8x × 105) × (100) × (0·726) watts = 15·84x × 10 watts = 15·84x × 10 kW 7 4 But kW produced = 15,000 (given) ∴ 15·84x × 104 = 15,000 15,000 or x = 4 = 0 ⋅ 0947 m = 9.47 cm 15 ⋅ 84 × 10 Therefore, the level of reservoir will fall by 9·47 cm. * The total rainfall cannot be utilised as a part of it is lost by evaporation or absorption by ground. Yield factor indicates the percentage of rainfall available for utilisation. Thus 80% yield factor means that only 80% of total rainfall can be utilised. Generating Stations 25 Alternative method Vol. of reservoir = 5 × 10 = ⋅ 6 Level of reservoir = 2 083 m Area of reservoir 2 ⋅ 4 × 106 kWh generated in 3 hrs = 15000 × 3 = 45,000 kWh If kWh generated are 9,89,175 kWh, fall in reservoir level = 2·083 m If kWh generated are 45,000 kWh, fall in reservoir level = 2 ⋅ 083 × 45,000 = 0 ⋅ 0947 m = 9.47 cm 9,89175 , Example 2.11. A factory is located near a water fall where the usable head for power genera- tion is 25 m. The factory requires continuous power of 400 kW throughout the year. The river flow in a year is (a) 10 m3/sec for 4 months, (b) 6 m3/sec for 2 months and (c) 1·5 m3/sec for 6 months. (i) If the site is developed as a run-of-river type of plant, without storage, determine the standby capacity to be provided. Assume that overall efficiency of the plant is 80%. (ii) If a reservoir is arranged upstream, will any standby unit be necessary ? What will be the excess power available ? Solution. (i) Run of river Plant. In this type of plant, the whole water of stream is allowed to pass through the turbine for power generation. The plant utilises the water as and when available. Conse- quently, more power can be generated in a rainy season than in dry season. 3 (a) When discharge = 10 m /sec Wt. of water available/sec, w = 10 × 1000 kg = 104 × 9·81 N = w × H × ηoverall = (10 × 9·81) × (25) × (0·8) watts 4 Power developed = 1962 × 10 watts = 1962 kW 3 3 (b) When discharge = 6 m /sec Power developed = 1962 × 6*/10 = 1177·2 kW 3 (c) When discharge = 1·5 m /sec Power developed = 1962 × 1·5/10 = 294 kW 3 3 It is clear that when discharge is 10 m /sec or 6 m /sec, power developed by the plant is more 3 than 400 kW required by the factory. However, when the discharge is 1·5 m /sec, power developed falls short and consequently standby unit is required during this period. ∴ Capacity of standby unit = 400 − 294 = 106 kW (ii) With reservoir. When reservoir is arranged upstream, we can store water. This permits regulated supply of water to the turbine so that power output is constant throughout the year. (10 × 4) + (2 × 6) + (1 ⋅ 5 × 6) Average discharge = = 5 ⋅ 08 m 3 / sec. 12 ∴ Power developed = 1962 × 5·08/10 = 996·7 kW Since power developed is more than required by the factory, no standby unit is needed. ∴ Excess power available = 996·7 − 400 = 596·7 kW Example 2.12. A run-of-river hydro-electric plant with pondage has the following data : Installed capacity = 10 MW ; Water head, H = 20 m Overall efficiency, ηoverall = 80% ; Load factor = 40% * If discharge is 10 m3/sec, power devloped = 1962 kW If discharge is 1 m3/sec, power devloped = 1962/10 If discharge is 6 m3/sec, power devloped = 1962 × 6/10 26 Principles of Power System 3 (i) Determine the river discharge in m /sec required for the plant. 3 (ii) If on a particular day, the river flow is 20 m /sec, what load factor can the plant supply ? Solution. (i) Consider the duration to be of one week. Units generated/week = Max. demand × L.F. × Hours in a week = (10 × 10 ) × (0·4) × (24 × 7) kWh 3 = 67·2 × 10 kWh 4... (i) 3 Let Q m /sec be the river discharge required. Wt. of water available/sec, w = Q × 9·81 × 1000 = 9810 Q newton Average power produced = w × H × ηoverall = (9810 Q) × (20) × (0·8) W = 156960 Q watt = 156·96 Q kW Units generated/week = (156·96 Q) × 168 kWh = 26,369 Q kWh... (ii) Equating exps. (i) and (ii), we get, 26,369 Q = 67·2 × 10 4 67 ⋅ 2 × 10 4 ∴ Q = = 25.48m3/sec 26,369 3 (ii) If the river discharge on a certain day is 20 m /sec, then, Power developed = 156·96 × 20 = 3139·2 kW Units generated on that day = 3139·2 × 24 = 75,341 kWh 75,341 Load factor = 4 × 100 = 31.4% 10 × 24 Example 2.13. The weekly discharge of a typical hydroelectric plant is as under : Day Sun Mon Tues Wed Thurs Fri Sat 3 Discharge(m /sec) 500 520 850 800 875 900 546 The plant has an effective head of 15 m and an overall efficiency of 85%. If the plant operates on 40% load factor, estimate (i) the average daily discharge (ii) pondage required and (iii) in- stalled capacity of proposed plant. Generating Stations 27 Solution. Fig. 2.5 shows the plot of weekly discharge. In this graph, discharge is taken along Y-axis and days along X-axis. 500 + 520 + 850 + 800 + 875 + 900 + 546 (i) Average daily discharge = 7 4991 = 7 = 713 m3/sec 7 (ii) It is clear from graph that on three dyas (viz., Sun, Mon. and Sat.), the discharge is less than the average discharge. Volume of water actually available on these three days = (500 + 520 + 546) × 24 × 3600 m = 1566 × 24 × 3600 m 3 3 Volume of water required on these three days = 3 × 713 × 24 × 3600 m3 = 2139 × 24 × 3600 m3 = (2139 − 1566) × 24 × 3600 m = 495 × 10 m 3 5 3 Pondage required (iii) Wt. of water available/sec, w = 713 × 1000 × 9·81 N Average power produced = w × H × ηoverall = (713 × 1000 × 9·81) × (15) × (0·85) watts = 89180 × 10 watts = 89180 kW 3 Installed capacity of the plant Output power 89180 = = = 223 × 103 kW = 223 MW Load factor 0×4 TUTORIAL PROBLEMS 1. A hydro-electric station has an average available head of 100 metres and reservoir capacity of 50 million cubic metres. Calculate the total energy in kWh that can be generated, assuming hydraulic efficiency of [10·423 × 10 kWh] 6 85% and electrical efficiency of 90%. 2. Calculate the continuous power that will be available from hydroelectric plant having an available head of 300 meters, catchment area of 150 sq. km, annual rainfall 1·25 m and yield factor of 50%. Assume penstock, turbine and generator efficiencies to be 96%, 86% and 97% respectively. If the load factor is 40% what should be the rating of the generators installed ? [7065 kW, 17662 kW] 3. A hydroelectric plant has a reservoir of area 2 sq. kilometres and of capacity 5 million cubic meters. The net head of water at the turbine is 50 m. If the efficiencies of turbine and generator are 85% and 95% respectively, calculate the total energy in kWh that can be generated from this station. If a load of 15000 [5·5 × 10 kWh ; 27·8 cm] 5 kW has been supplied for 4 hours, find the fall in reservoir. 3 4. It has been estimated that a minimum run-off of approximately 94 m /sec will be available at a hydraulic project with a head of 39 m. Determine the firm capacity and yearly gross output. [3600 kW, 315·36 × 106 kWh] Hint. Wt. of water flowing/sec = a f 94 × 100 3 kg 1000 2 5. A hydroelectric power station is supplied from a reservoir having an area of 50 km and a head of 50 m. If overall efficiency of the plant is 60%, find the rate at which the water level will fall when the station is generating 30,000 kW. [7·337 mm/hour] 6. A hydro-electric plant has a catchment are of 120 square km. The available run off is 50% with annual rainfall of 100 cm. A head of 250 m is available on the average. Efficiency of the power plant is 70%. Find (i) average power produced (ii) capacity of the power plant. Assume a load factor of 0.6. [(i) 3266 kW (ii) 5443 kW] 28 Principles of Power System 2.11 Diesel Power Station A generating station in which diesel engine is used as the prime mover for the generation of electri- cal energy is known as diesel power station. In a diesel power station, diesel engine is used as the prime mover. The diesel burns inside the engine and the products of this combustion act as the “working fluid” to produce mechanical energy. The diesel engine drives the alternator which converts mechanical energy into electrical energy. As the generation cost is considerable due to high price of diesel, therefore, such power stations are only used to produce small power. Although steam power stations and hydro-electric plants are invariably used to generate bulk power at cheaper cost, yet diesel power stations are finding favour at places where demand of power is less, sufficient quantity of coal and water is not available and the transportation facilities are inad- equate. These plants are also used as standby sets for continuity of supply to important points such as hospitals, radio stations, cinema houses and telephone exchanges. Advantages (i) The design and layout of the plant are quite simple. (ii) It occupies less space as the number and size of the auxiliaries is small. (iii) It can be located at any place. (iv) It can be started quickly and can pick up load in a short time. (v) There are no standby losses. (vi) It requires less quantity of water for cooling. (vii) The overall cost is much less than that of steam power station of the same capacity. (viii) The thermal efficiency of the plant is higher than that of a steam power station. (ix) It requires less operating staff. Disadvantages (i) The plant has high running charges as the fuel (i.e., diesel) used is costly. (ii) The plant does not work satisfactorily under overload conditions for a longer period. (iii) The plant can only generate small power. (iv) The cost of lubrication is generally high. (v) The maintenance charges are generally high. 2.12 Schematic Arrangement of Diesel Power Station Fig. 2.6 shows the schematic arrangement of a typical diesel power station. Apart from the diesel- generator set, the plant has the following auxiliaries : (i) Fuel supply system. It consists of storage tank, strainers, fuel transfer pump and all day fuel tank. The fuel oil is supplied at the plant site by rail or road. This oil is stored in the storage tank. From the storage tank, oil is pumped to smaller all day tank at daily or short intervals. From this tank, fuel oil is passed through strainers to remove suspended impurities. The clean oil is injected into the engine by fuel injection pump. (ii) Air intake system. This system supplies necessary air to the engine for fuel combustion. It consists of pipes for the supply of fresh air to the engine manifold. Filters are provided to remove dust particles from air which may act as abrasive in the engine cylinder. (iii) Exhaust system. This system leads the engine exhaust gas outside the building and dis- charges it into atmosphere. A silencer is usually incorporated in the system to reduce the noise level. Generating Stations 29 (iv) Cooling system. The heat released by the burning of fuel in the engine cylinder is partially converted into work. The remainder part of the heat passes through the cylinder walls, piston, rings etc. and may cause damage to the system. In order to keep the temperature of the engine parts within the safe operating limits, cooling is provided. The cooling system consists of a water source, pump and cooling towers. The pump circulates water through cylinder and head jacket. The water takes away heat form the engine and itself becomes hot. The hot water is cool