MODULE I INTRODUCTION TO ENERGY RESOURCES Lesson 1 Lesson 2 Lesson 3 Types of Power Plant as to its Sources Intro to Renewable & Non-Renewable Energy Resources Power Plant Operation & Basic Problem Solving Module I 2 MODULE I INTRODUCTION TO ENERGY RESOURCES INTRODUCTION This module presents Familiarization of the different Power plant as to their flow of operations, and will discuss the power plants, as to its purpose, as to its source, as to its use and reserves. POWER PLANT ENGINEERING AND DESIGN- It is the art of selecting and placing the necessary power-generating equipment so that the maximum of return will result from the minimum expenditure over the working life of the plant; and the operation of the completed plant in manner to provide a cheap, reliable, and continuous service. OBJECTIVES After studying the module, you should be able to: 1. Evaluate the importance of the different power plants 2. Explain the different operations of a power plant. 3. Solve & analyze problems relative to the different power plant discussed. 4. Evaluate & give synthesis the different existing power plants in the Philippines. DIRECTIONS/ MODULE ORGANIZER There are three lessons in the module. Read each lesson carefully then answer the exercises/activities to find out how much you have benefited from it. Work on these exercises carefully and submit your output to your tutor or to the COE office. In case you encounter difficulty, discuss this with your tutor during the face-to-face meeting. If not contact your tutor at the COE office. Good luck and happy reading!!! -mev EE158 module -mev 3 Lesson 1 Types of Power Plants as to Sources ELECTRICAL POWER GENERATING STATION (POWER PLANT) A station in which are located prime movers, electric generators, and auxiliary equipment for converting mechanical, chemical, and/or nuclear energy into electrical energy. TYPES OF POWER PLANTS AS TO SOURCE 1. THERMAL PLANT – an electric generating station using heat as a source of energy. 1.1 Oil-fired Steam Plant – a thermal plant which makes use of heavy fuel oil (HFO), Light fuel oil (LFO), or Bunker C Oil as fuel for the production of electric energy. Example, Bataan 1 & 2 Thermal Plants. 1.2 Coal-fired Steam Plant – a thermal plant which makes use of pulverized coal as fuel in producing electric energy. Example, Batangas Coal-fired Thermal Power Plant. 1.3 Dendro-Thermal Power Plant – a thermal plant which makes use of wood in producing electric energy. 1.4 Geothermal Steam Plant – a thermal plant which makes use of generated heat from the earth’s magma as a fuel in producing electric energy. Example, Mak-Ban Geothermal Plant 1.5 Nuclear Steam Plant – a thermal plant which makes use of the steam generated in a reactor by heat from the fissioning of nuclear fuel as source of producing electric energy. Example- Bataan Nuclear Power Plant (NonOperational, and was commissioned) 1.6 Diesel Plant – a thermal plant wherein the prime mover is an internal combustion engine (ICE) and which makes use of diesel oil as the source of electric energy.(1590 Energy Corp.-Bauang) 1.7 Gas-Turbine Plant – a thermal plant in wherein the prime mover is a gas turbine engine and which makes use of combustible gas as fuel in producing electric energy. EE158 module -mev 4 1.8 Solar Thermal Plant - Solar thermal energy (STE) is a form of energy and a technology for harnessing solar energy to generate thermal energy or electrical energy for use in industry, and in the residential and commercial sectors. 1.9 Biomass Power Plant- Biomass, as a renewable energy sources, is a biological material from living, or recently living organisms. As an energy source, biomass can either be used directly or converted into other energy products such as biofuel. biomass is a plant matter used to generate electricity with steam turbines and gasifies or produced heat usually by direct combustion. Example (Phil-Bio Biomass Plant) Thermal Plant Basic Block Diagram Boiler – the pressure part of the steam generator wherein water is pumped into it, vaporizes, and delivers to the steam turbine. Steam Turbine – a versatile prime mover for the electric generator. Steam strikes the blade of the turbine which makes the shaft to rotate hence rotating the generator moving part to generate more power. EE158 module -mev 5 In which, Combined heat Pressures from the turbine are categorized into three (3) major parts a.) Low pressure turbine, b.) Intermediate pressure turbine, c.) High pressure turbine. Condenser – it is here where the steam is reduced to liquid and ready to be pumped again in the broiler. EE158 module -mev 6 Electric Generator – the mechanical power from the turbine converted into electrical power through this equipment. Exciter – provide excitation to the electric generator to generate emf and hence electric power. Transformer and Transmission Lines – power is distributed at higher voltage by means of these equipment and devices. EE158 module -mev 7 2.Hydro Electric Power Plant HYDRO PLANT/HYDRO ELECTRIC POWER PLANT – an electric generating station in which the prime mover is a turbine driven by falling water. 2.1 Run-of-River Plant – a hydroelectric generating station utilizing limited pondage or the stream flow as it occurs. 2.2 Plant with Storage Capacity - a hydroelectric generating station associated with a water storage reservoir. 2.3 Pumped-Storage Plant - a hydroelectric generating station wherein electric energy during periods of relatively high system demand by utilizing water which has been pumped into a storage reservoir usually during periods of relatively low system demand. Types of hydraulic turbines or waterwheels: Impulse type – suitable for very high heads, for net heads 1,300 ft. & above Reaction type/Francis type – suitable for medium heads, Propeller type – suitable for very low heads, for net heads up to 70ft. 3. Other types of Power Plant 3.1 Windmill Power Plant • - Wind power or wind energy is the use of wind to provide the mechanical power through wind turbines to turn electric generators and traditionally to do other work, like milling or pumping. • - Wind power is a sustainable and renewable energy, and has a much smaller impact on the environment compared to burning fossil fuels. Wind farms consist of many individual wind turbines, which are connected to the electric power transmission network. 3.2 Sea Wave Power Plant • -Tidal power or tidal energy is the form of hydropower that converts the energy obtained from tides into useful forms of power, although not yet widely used, tidal energy has potential for future electricity generation. 3.3 Solar Power 3.3.1 - Off-Grid Solar Power - SMALL SCALE OR for MICRO GRID PURPOSES EE158 module -mev 8 is a characteristic of facility and a lifestyle designed in an independent manner without reliance on one or more public utilities and autonomous in nature. The term "off-the-grid" traditionally refers to not being connected to the “Grid” (ex. Luzon Grid). Microgrids can serve hundreds to thousands of customers through a community and support the penetration of local energy. In microgrid, some houses may have some renewable sources that can supply their demand as well as that of their neighbors within the same community. The community microgrid may also be categorized as centralized source of energy & power. 3.3.2 HYBRID POWER are hybrid power systems that combine solar power from a photovoltaic system with another power generating energy source..(ex. Distribution Utilities, solar-wind, solar-D.U.,solar-diesel. Etc.) TYPE OF POWER PLANT AS TO USE: 1. Peaking Plant – a generating plant which is normally operated to provide power only during peak load periods. 2. Regulating Plant – a generating plant capable of carrying load for the time interval either during off-peak or peak periods, and usually responds to changes in system frequency. 3. Base Load Plant – a generating plant that can carry a minimal load over a given period of time and a plant that runs near its full rating continuously, day and night, all year long. TYPE OF PLANT RESERVE: a) Cold Reserve- portion of the installed reserve kept in operable condition and available for service but not for immediate loading. b) Operating Reserve- refers to the capacity in actual service in excess of peak load. c) Hot Reserve- refers to units available, maintained at operating temperature and ready for service although not in actual operation. d) Spinning Reserve- generating capacity connected to the bus ready to take load. EE158 module -mev 9 DIFFERENT TYPES OF PRIME MOVERS: 1. STEAM -Reciprocating Engine -Turbines 2. HYDRAULIC -Impulse wheel -Reaction wheel -Propeller wheel 3. INTERNAL COMBUSTION -Oil, Diesel, Gas THINK! On your own point of view, what are the existing Power plants available in our country today? THINK! Case: On our existing power plant on our Generation Mix, which type of power plant has the biggest part and why do you think that this has the biggest share? EE158 module -mev 10 Lesson 2 Intro to Renewable & Non-Renewable Energy Our country today is in need and focused on the usage of renewable energy as part of its sustainability to be partially independent from oil, diesel and coal energy generation plants and to further lessen the greenhouse effect. From Luzon Generation Mix, as of November 2013, 55.67% coal usage and 3.88% Diesel usage from generation plants, in which our country is highly dependent from these plants. In December 2008, our country enacted RA 9513, also known as the renewable Energy Act of 2008, which conforms to internationally accepted standards which enable our country to be part of a clean and green power generation allies. EE158 module -mev 11 Renewable energy is derived from natural processes that are replenished constantly. In its various forms, it derives directly from the sun, or from heat generated deep within the earth. Included in the definition is electricity and heat generated from solar, wind, ocean, hydropower, biomass, geothermal resources, and biofuels and hydrogen derived from renewable resources. Non-Renewable Energy A non-renewable resource (also called a finite resource) is a natural resource that cannot be readily replaced by natural means at a quick enough pace to keep up with consumption. An example is carbon-based fossil fuel. The original organic matter, with the aid of heat and pressure, becomes a fuel such as oil or gas. Earth minerals and metal ores, fossil fuels (coal, petroleum, natural gas) and groundwater in certain aquifers are all considered non-renewable resources, though individual elements are always conserved (except in nuclear reactions). EE158 module -mev 12 EE158 module -mev 13 THINK! Which is Cheaper in the Production of Energy? THINK! Write an essay why coal fired power plant dominates the generation matrix in the production of Electrical Energy? EE158 module -mev 14 Lesson 3 Power Plant Operation & Problem Solving COAL FIRED POWER PLANT Coal is the most abundant fossil fuel on the planet. It is the relatively cheap fuel, with some of the largest deposits in region that stable politically. Solid coal cannot directly replace natural gas or petroleum in most applications, is mostly used for transportation and natural gas not used for electricity generation is used for space water and industrial heating. Coal can be converted to gas or liquid fuel, but the efficiencies and economics of such processes can make them feasible. Vehicles or heaters may require modification to use coal derived fuels. Coal can produce more pollution than petroleum or natural gas. A fossil fuel power plant station is a type of power station that burns fossil fuels such as coal, natural gas or petroleum (oil) to produce electricity. Central station fossil power plant is designed on a large scale for continuous operation. In many centuries, such plants provide most of the electrical energy used. Fossil fuel power plant station have rotating machinery to convert heat energy of combustion into mechanical, which then operates an electrical generator. The prime mover may be steam turbine, a gas turbine or in a small plant, reciprocating internal combustion engine. All plants used the energy extracted from expanding gas-steam or combustion gasses. A very few MHD generators have been built which directly convert the energy of moving hot gas into electricity. By products of thermal power plant operation must considered in their design and operation. Waste heat energy, which remains due to the finite efficiency of the Carnot, Rankine, or diesel power cycle is heat energy released directly to the atmosphere, directly to the river or lake water or directly to the atmosphere. Using a cooling tower with river or lake water as EE158 module -mev 15 a cooling medium. The fuel gas from the combustion of the fossil fuel is discharged to the air. The gas contains carbon dioxide, water vapor as well as substances such as elemental nitrogen oxide, mercury traces of other metals and for coal fired plants fly ashes. Solid waste ash from coal fired boilers must also be removed. Some ash can be recycled for building materials. Fossil fueled power stations are major emitters of carbon dioxide a greenhouse gas (GHG) which according to consensus opinion of scientific organizations is a contributor of global warming as it has been observed over the last years. Brown coal emits about three times as much as carbon dioxide as natural gas, black coal emits about twice as much as carbon dioxide per unit of electric energy. Carbon capture and storage of emission is not expected to be available on a commercial economically viable basis until government supported legislation enacted. FUEL PROCESSING Coal is prepared for use by crushing the rough coal to pieces less than two inches or 5 centimeters in sizes. The coal is then transported from the storage yard to plant storage silos by rubberized conveyor belts at rates u at 250 tons per hour under full, load to 4000 tons per hour. In plants that burn pulverized coal, silos feed coal pulverizes (coal mills) they take larger two inch or fifty-one millimeter pieces, grind them to the consistency of face powder, sort them and mix them with primary combustion air which transport the coal to the boiler furnace and preheats the coal to the boiler in order to drive off the excess moisture content. A 500 MWE plant may have six such pulverizes, five of which can supply coal to the furnace at 250 tons per hour under full load. In plants that do not burn pulverized coal, the larger 2 inch or 51 millimeter pieces may be directly fed into the silos which then feed either mechanically distributors that drop the coal on a travelling grate or the cyclone burners a specific kind of combustor that can efficiently burn larger pieces of fuel. EE158 module -mev 16 Basic Conversion of Units, that is usable in solving in thermal power plants especially, coal fired power plant. Mass & weight (the weight of coal for every plant consumption) 1 Ton = 2,000 pounds (lbs.) 1Kg. = 2.2 lbs. 1 Long Ton = 2240 lbs. 1KgF = 9.80665 Newtons 1 Metric Ton = 1000 Kg. or 2,200 lbs. 1 Newton = 0.1019 Kg. = 0.0459 lbs. 1 KIP = 1,000 lbs. 1 Pound (lb.) = 0.4536 Kg. = 0.13825 N 1 slug = 14.59 Kg, = 32.17 lbs. 1 0z. = 28.35 grams Work & Energy (thermal heat Efficiency, capacity) 1 BTU (British thermal Unit) – 778-ft-lb - 252 cal. - 1055 joules - 778 ft-lb - 0.2929 W-Hr 1Quad = 1x1015 BTU 1 Therm =1x105 BTU 1 Ton = 12,000 BTU/Hr. 1 Kcal (Kilo Calories) – 4.187 KJ or 1cal. = 4.187 joules EE158 module -mev 17 1 ERG – 1 dyne-cm or 1x10-7 joule 1 Joule – 1 N-m Power 1 Horsepower – 550 ft-lb/sec, or 33,000 ft-lb/min - 2545 BTU/hr. or 42.4 BTU/min. -0.746 KW or 746 watts 1KW – 3414 BTU/Hr. or 860 Kcal/Hr. 0r in Electrical Energy, thus: 1KWHR = 3414 BTU = 860 Kcal = 3,600KJ 1 Watt – 1 Joule/sec, 1 N-m/sec, or 1x10-7 ergs/sec. 1 Boiler HP = 33,480 BTU/Hr. or 35,322 KJ/Hr. Overall Plant Thermal Efficiency o Pout HeatEnergyPower x PlantCapacity Pin Mass of coal x CalorificValue of coal to be used by the plant As such, o = Boiler Efficiency x Turbine Efficiency x Alt./Gen. Efficiency Or, o = Thermal Efficiency x Electrical Efficiency EE158 module -mev 18 Example 1: 1. To produce one KWHR, A coal fired power plant burns 0.9 lb. of coal with a heating value of 13,000 BTU/lb. What is the heat rate of the plant? Sol. (By Unit analysis) 13,000 BTU/ lb. x 0.9lbs./1KWHR = 11,700 BTU/KWHR /ans. Example 2: A certain coal fired power plant has a heat rate of 2.88x10 6 calories per KWHR production, Coal cost’s P2,500 per ton. How much is the fuel cost component of producing 1 KWHR? Sol. Kcal x (1KWHR ) KWHR P2500 x x Fuel Cost= BTU 860 Kcal 2.2lbs 100Kg Ton 13,000 x x x lb 3414BTU 1Kg 1Ton 2.88 x10 3 = P0.997 ≈ P1 per KWHR Example 3: A Coal fired power plant burns its imported fuel at a heating value of 14,000 BTU/lb. The plant has a turbine efficiency of 70% and an alternator efficiency of 95%. Determine the fuel cost component in producing 1KWHR in Philippine currency if coal cost 2 USD per pound. (Note: exchange rate for the moment is 1USD = P52.45) Sol. o = turbine x alternator = 0.70 x 0.90 = 0.63 x 100% = 63% BTU x ( KWHR ) KWHR BTU (0.63) x 14,000 lb 3414 m= m = 0.387 lbs. since, 1 lb. cost 2USD EE158 module -mev 19 therefore, 2USD P52.45 x 1lb. 1USD cost /KWHR = P 40.60 m = 0.387 lbs. x Example 4: A power plant consumes 21,542 Kg of coal per hour. The heating value of coal is 6,652,800 calories per Kg and the overall plant capacity is 30%. What is the MW output of the plant? Sol. 1 KWHR cal. Kg . x 21,542 x Kg . Hr . 860,000 cal. Pin = 166,643.05 KW = 166.64 MW Pin = 6,652,800 But, o Pout Pin Therefore, Pout = Pin o Pout = (166.64 MW) (0.30) Pout = 50 MW Example 5: A coal fired thermal station has an overall efficiency of 15% and 0.75 Kg of coal is burnt per KWHR by the station. Determine the calorific value of coal in Kilocalories per Kilogram. (in this portion of problem, I will show you how to solve in different methods but will end up in the same answer...) Solution 1.(Using 1KWHR=3.6x106 J) 1 KWHR 4.187 J 0.75Kg Kg . x x 8.723x10 7 6 KWHR 3.6 x10 J 1cal. cal 1,146,405.541 EE158 module -mev cal 1Kcal Kcal x 1,146.405 Kg. 1,000cal. Kg Take the inverse of the value 20 1,146.405 Calorific Value = Kcal Kg 0.15 = 7,642.7 Kcal /ans. Kg Sol.2 (Using 1KWHR = 860 KWHR) 0.75Kg 1 KWHR Kg x 8.721 x10 4 KWHR 860Kcal Kcal = 1,146.667 Take the inverse Kcal Kg 1,146.667 Calorific Value = Kcal Kg 0.15 = 7,644.444 Kcal /ans. Kg Sol.3 (Using thermal Efficiency method) Given: o Pout HeatEnergyPower x PlantCapacity Pin Mass of coal x CalorificValue of coal to be used by the plant o = 15% Using, 1KWHR= 860 Kcal Kcal KWHR 15% = Kg 0.75 xq KWHR 860 Therefore,q (calorific value) = 7,644.444 Kcal/Kg..../Ans. EE158 module -mev 21 REMINDER! In dealing with problems that involve heat-rate, amount of fuel needed, cost of production of energy,use the mathematical technique normally called “unit analysis method & cancellation of units”..MEV HYDROELECTRIC POWER PLANT Hydroelectricity is a term referring to electricity generated by hydro power. The production of electrical power through the use of the gravitational force of failing or flowing or flowing water. It is the most widely used form of renewable energy accounting for 16% of global electricity consumption and 3427 terawatts-hour of electricity production in 2010 which continues the rapid rate of increase experienced up to the present situation. Hydro is also a flexible source of electricity since plants be ramped up and down very quickly to adapt to changing energy demand. However, damming interrupts, the flow of rivers and can harm local ecosystems and building large. Dams and reservoirs often involves displacing people and wildlife. Once a hydroelectric complex is constructed, the project produces no direct waste and has a considerably lower output level of the greenhouse gas carbon dioxide than fossil fuel powered energy plants. CONVENTIONAL (DAMS) Most hydroelectric power comes from the potential energy of dammed water driving a water turbine and a generator. The power extracted from the water depends on the volume and on the difference in height between the sourced and water outflow. This height difference is called head. The amount of potential energy in water is proportional to the head a large pipe “PENSTOCKS” called delivers water to turbine. EE158 module -mev 22 PUMPED STORAGE Is a type of hydroelectric power generation used by some power plants for load balancing, this method stores energy in the form of potential energy of water, pumped from lower elevation reservoir to a higher elevation? Low cost off peak electric power is used to run pumps. During periods of high electrical demands, the stored water is stored/ water released through the turbine to produce electric power. Although the losses of the pumping process make the plant a net consumer of energy overall, the system increases revenues by selling more electricity during periods of peak demand, when electricity prices highest. Pumped storage schemes currently provide the most commercially important means of large scale grid energy storage and improve the daily capacity factor of the generation system. RUN-OF-THE-RIVER Is a type of hydroelectric generation whereby little or no water storage is provided? Run of the river plants may either have no storage is provided. Run of the river plants may either have no storage at all, or a limited amount or storage, in which case the storage reservoir is referred to as a poundage. A plant without poundage has no storage and is, therefore, subject to a seasonal river flows and serves as a peaking power while a plant with poundage can regulate water flow and serve either as a peaking or base load power plant. ADVANTAGES OF HYDROELECTRIC POWER PLANT FLEXIBILITY Hydro is a flexible source of electricity since plants can be ramped up and down very quickly to adapt to changing energy demands. EE158 module -mev 23 LOW POWER COSTS The major advantage of hydroelectric is elimination of the cost of fuel. The cost of operating a hydroelectric is nearly immune to increases in the cost of fossils such as oil, natural, gas or coal and no imports are needed. The average cost of electricity from a hydro plant larger than 10 watts is 3 to 5% cents per kilowatt hour. Hydroelectric plants still in after 50-100 years. Operating labor is also usually low as plants are automated and have few personnel on site during normal operations. REDUCED CO2 EMISSIONS Since hydroelectric dams do not fossils, they do not directly produced carbon dioxide. While some carbon dioxide is produce during manufacturing and construction of the products, this is a tiny fraction of the operating emissions of equivalent fossil fuel electricity generations. OTHER USES OF THE RESERVOIR Reservoirs created by hydroelectric schemes often provide facilities for water sports and become tourist attraction themselves some countries aquaculture is common in reservoirs. Multi-use dams installed for irrigation support agriculture with a relatively constant water supply. Large hydro dams can control floods, which affect otherwise the people living downstream of the project. DISADVANTAGES ECOSYSTEM DAMAGE AND LOSS AF LAND Large reservoirs required for the operations of hydroelectric power stations result submersion of extensive area upstream of the dams, destroying biologically rich and productive lowland and river line, valley forest, marshland and grasslands. The loss of land is often exacerbated by habitat fragmentation of surrounding areas. EE158 module -mev 24 SILTATION AND FLOW SHORTAGE When water flows it has the ability to transport particles heavier than itself downstream, this has a negative effect on dams and subsequently their power stations particularly those on rivers or within catchment areas with siltation. Siltation’s can fill a reservoir and reduce its capacity to control floods along with causing additional horizontal pressure on the upstream portion of dam. Eventually, some reservoirs can be full of sediments and useless or over top during a flood and fail. Changes in the amount of live storage in a reservoir therefore reducing the amount of water that can be used for hydroelectricity. Types & Process of Hydraulic Turbine Used In Hydro Electric Plant a) Reaction Turbine – The water under the pressure is partly converted into velocity before it enters the turbine runner. Francis Type – The water enters the spiral case from the penstock, passes through the stay ring guided by the stationary stay ring vanes, then through the movable wicket gates through the runner and into the draft tube through which it flow into a tailrace or tail water reservoir. Propeller Type – It is the same as Francis type but it has an enshrouded * blades. Axial Flow Turbine – is a propeller type runner with either fix or adjustable blades. b) Impulse Turbine – The water under pressure is partly converted into velocity before it enters the turbine runner. It consists of one or more free jets of water discharging into an aerated space and impinging on a set of buckets attached around the periphery of a disk or wheel. *a covering or enclosure with or if with a shroud EE158 module -mev 25 How a Hydroelectric Power System Works? Flowing water is directed at a turbine (remember turbines are just advanced waterwheels). The flowing water causes the turbine to rotate, converting the water’s kinetic energy into mechanical energy. The mechanical energy produced by the turbine is converted into electric energy using a turbine generator. Inside the generator, the shaft of the turbine spins a magnet inside coils of copper wire. It is a fact of nature that moving a magnet near a conductor causes an electric current. Impulse Turbines Uses the velocity of the water to move the runner and discharges to atmospheric pressure. The water stream hits each bucket on the runner. No suction downside, water flows out through turbine housing after hitting. High head, low flow applications. Types : Pelton wheel, Cross Flow EE158 module -mev 26 Pelton Wheels Nozzles direct forceful streams of water against a series of spoonshaped buckets mounted around the edge of a wheel. Each bucket reverses the flow of water and this impulse spins the turbine. Suited for high head, low flow sites. The largest units can be up to 200 MW. Can operate with heads as small as 15 meters and as high as 1,800 meters. EE158 module -mev 27 drum-shaped elongated, rectangular-section nozzle directed against curved vanes on a cylindrically shaped runner “squirrel cage” blower water flows through the blades twice First pass : water flows from the outside of the blades to the inside Second pass : from the inside back out Larger water flows and lower heads than the Pelton. Reaction Turbines Combined action of pressure and moving water. Runner placed directly in the water stream flowing over the blades rather than striking each individually. lower head and higher flows than compared with the impulse turbines. Propeller Hydropower Turbine Runner with three to six blades. Water contacts all of the blades constantly. EE158 module -mev 28 Through the pipe, the pressure is constant Pitch of the blades - fixed or adjustable Scroll case, wicket gates, and a draft tube Types: Bulb turbine, Straflo, Tube turbine, Kaplan Bulb Turbine The turbine and generator are a sealed unit placed directly in the water stream. Francis Turbines The inlet is spiral shaped. Guide vanes direct the water tangentially to the runner. This radial flow acts on the runner vanes, causing the runner to spin. The guide vanes (or wicket gate) may be adjustable to allow efficient turbine operation for a range of water flow conditions. EE158 module -mev 29 EE158 module -mev 30 INSIDE A HYDROPOWER PLANT Reservoir stores the water coming from the upper river or waterfalls. Head Water is the water in the reservoir. Spillway is a weir in the reservoir which discharges excess water so that the head of the plant will be maintained. Dam is the concrete structure that encloses the reservoir. Silt Sluice is a chamber which collects the mud and through which the mud is discharged. Valve is a device that opens or closes the entrance of the water into the penstock. Trash rack is a screen which prevents the leaves branches and other water contaminants to enter into the penstock. Penstock is the channel that leads the water from the reservoir to the turbine. Surge Chamber is a standpipe connected to the atmosphere and attached to the penstock so that the water will be at atmospheric pressure. Tail Race is channel which leads the water from the turbine to the tail water. Tail Water is the water that is discharged from the turbine. Draft Tube is a device that connects the turbine outlet to the tail water so that the turbine can be set above the tail water level. Intake - Gates on the dam open and gravity pulls the water through the penstock, a pipeline that leads to the turbine. Water builds up pressure as it flows through this pipe. EE158 module -mev 31 Hydraulic Turbine- is a device that converts the energy of water into the mechanical energy. The water strikes and turns the large blades of a turbine, which is attached to a generator above it by way of a shaft. The most common type of turbine for hydropower plants is the Francis Turbine, which looks like a big disc with curved blades. A turbine can weigh as much as 172 tons and turn at a rate of 90 revolutions per minute (rpm), according to the Foundation for Water & Energy Education(FWEE). Generator is a device that converts the mechanical energy of the turbine into electrical energy. Transformer - The transformer inside the powerhouse takes the AC and converts it to higher-voltage current. Power lines - Out of every power plant come four wires: the three phases of power being produced simultaneously plus a neutral or ground common to all three. (Read How Power Distribution Grids Work to learn more about power line transmission.) Outflow - Used water is carried through pipelines, called tailraces, and reenters the river downstream. The water in the reservoir is considered stored energy. When the gates open, the water flowing through the penstock becomes kinetic energy because it's in motion. The amount of electricity that is generated is determined by several factors. Two of those factors are the volume of water flow and the amount of hydraulic head. The head refers to the distance between the water surface and the turbines. As the head and flow increase, so does the electricity generated. The head is usually dependent upon the amount of water in the reservoir. EE158 module -mev 32 CONVENTIONAL (DAMS) Pumped storage EE158 module -mev 33 Run of River AMBUKLAO HYDRO-POWER PLANT EE158 module -mev 34 SAN ROQUE HYDRO-POWER PLANT Theoretical Power Output of a Hydraulic Turbine P = Qwh where: P = theoretical power of the turbine (watt) Q = volume discharge of water (m3/sec) w = specific weight of water (9810 N/m3) h = head of water (m) P o out Pin P= Qwh k where: P = theoretical power of the turbine (horsepower) K = constant for conversion factor EE158 module -mev 35 Discharge of water ft3/sec ft3/min m3/sec m3/min Specific weight (water) 62.4 lbs/ft3 62.4 lbs/ft3 9810 N/m3 9810 N/m3 Head or Elevation feet feet meter meter Constant for conversion (k) 550 33000 746 44760 The Law of Proportionality of turbines of varying sizes (the variation of power speed and discharge with runner size and head): For constant runner For variable diameter and For constant head diameter head P ∞ h3/2 P ∞ d2 P ∞ h3/2d2 N ∞ h1/2 N ∞ 1/d N ∞ h1/2/d Q ∞ h1/2 Q ∞ d2 Q ∞ h1/2/d2 where: d = nominal diameter of the turbine runner Basic Conversions: Volume: 1L= 0.03531ft2 = 0.2642 gal. = 0.22 British Gallon = 0.2642 US Gallon 1ft3 = 7.481 gal. = 28.3 liters 1 acre-ft. = 43,560 ft3 1 Gallon = 4 quarts = 3.79 liters 1 Quart = 2 pints = 32 fluid ounces 1 Quart = 0.95 liter 1 Pint = 0.47 liter 1 British gallon = 1.2 US gallons 1 Pound = 16 Ounces 1 cu. Meter = 1,000 liters = 35.5 cu.feet Area: 1 ha = 104 m2 = 2.47 acres 1 acre = 4047 m2 = 43,560 ft2 = 0.405 ha 1 are = 100 sq.m 1 centare = 1 sq m. Length (height for the head) 1 Km = 3281 ft. = 0.54 naut. mile = 0.6214 statute mile 1 naut.mile = 6080 ft. 1 statute mile = 5280 feet 1 mile = 5280 ft. 1 yd =3 ft. EE158 module -mev 36 1m = 3.281 ft. 1cm = 0.3937 in 1 cable length = 720 ft. 1 fathom = 6 ft. 1 span = 9 inches 1 vara = 33 1/3 inches 1 furlong = 201.1684 meters Velocity (water flow rate to be used) Q 1mph = 1.467 ft/s = 0.8684 knot 1 knot = 1.688 ft/s 1km/h = 0.6214 mph Specific Speed – is the speed of a hydraulic turbine to generate one horsepower under a head of one foot. N= Nsh5/4 √BHP where: Ns = specific speed of the turbine (rpm) N = actual speed of the turbine (rpm) h = head of water available (foot) BHP = brake horsepower output of the turbine a. Gross Head, hg Gross Head is the difference between head water and tail water elevation. hg =hhw - htw where: hg= gross head hhw = head water elevation htw = tail water elevation b. Friction Head Loss, hf Friction Head Loss is the head lost by the flow in a stream or conduit due to frictional disturbances set up by the moving fluid and its containing conduit and by intermolecular friction. a. Using Darcy’s Equation: hf = EE158 module -mev b. Using Morse Equation hf = 37 where: hf = friction head (in m) f = coefficient of friction L = total length in m g = 9.81 m/s2 D = inside diameter ( in m) Note: Friction head loss is usually expressed as a percentage of the gross head. c. Net head or Effective head, h: Net Head or Effective head is the difference between the gross head and the friction head loss. h = hg -hf d. Penstock efficiency,ep: Penstock efficiency is the ratio of the net head to the gross head. ep = e. Volume flow rate of water, Q: The volume flow rate of water is the product of the velocity and the crosssectional area. Q = AV f. Water Power, Pw Water power is the power generated from an elevated water supply by the use of hydraulic turbines. Pw= γQh Where: γ = specific weight of water = 9.81 kN/m3 g. Turbine efficiency, et: Turbine efficiency is the ratio of the turbine power output to the water power output. et = et = or Pt = γQhet h. Electrical or Generator Efficiency, egen: Electrical or Generator Efficiency is the ratio of the generator output to the turbine power output. eG = egen = EE158 module -mev or Pgen = Pt egen = γQhetegen 38 i.Generator Speed, N N= Where: N = angular frequency, rpm f = frequency (usually 60 hertz) P = no. of poles (even number) j. Hydraulic Efficiency, eh: Hydraulic efficiency is the ratio of the utilized head to the net head. eh = Where: hw = utilized head h = net head k. Head of Impulse Turbine (Pelton) Impulse Turbine is a power-generation prime mover in which fluid under pressure enters a stationary nozzle where its pressure (potential) energy is converted to velocity (kinetic) energy and absorbed by the rotor. EE158 module -mev 39 l. Head of Reaction Turbine (Francis and Kaplan) Reaction Turbine is a power – generation prime mover utilizing the steady flow principle of fluid acceleration where nozzles are mounted on the moving element. m. Peripheral Coefficient, Φ Peripheral coefficient is the ratio of the peripheral velocity (V p) to the velocity of the jet (Vj). Φ= Φ= Where: D = diameter of runner N = angular speed H = net head n. Specific Speed of Hydraulic Turbine, Ns Specific speed is a number used to predict the performance of the hydraulic turbines. a. In English units: where: Ns = N = angular speed, rpm h = net head, ft b. In SI units: where: Ns = N = angular speed, rpm h = net head, m EE158 module -mev 40 o. Total efficiency, etotal: etotal = ehemev where: eh = hydraulic efficiency em = mechanical efficiency ev = volumetric efficiency p. Turbine type recommendation based on head Net Head Up tp 70 ft 70 ft to 110 ft 110 ft to 800 ft 800 ft to 1300 ft 1300 ft and above Type of Turbine Propeller Type Propeller or Francis Type Francis Type Francis Type or Impulse Type Impulse Type Example Problems: Problem 1: The volumetric flow of a river is 20 m 3/s. A hydraulic turbine is proposed to be installed in the power plant site with an available head of 24 m. If the specific speed of the turbine is 68 rpm and with an efficiency of 88%, determine the operating speed of the turbine. Solution: Qwh (20)(9810)(24) = = 6312.06 hp 746 746 Pout = ŋPin = 0.88(6312.06) = 5554.6 hp 5/4 24 m 3.281 (68) ( ) 5/4 x Nsh 1m N= = √BHP √5554.6 N = 214 rpm Pin = EE158 module -mev 41 Problem 2: For a given head, the diameter of the runner of a hydraulic turbine is 10 m while developing a BHP of 200. If the diameter of the runner is increased to 15 m, what will be its new brake horsepower rating? Solution: Using the law of proportionality: P ∞ d2 for constant head P1 d2 P2 = P1 = ( ) P2 d1 2 15 P2 = (200) ( ) 10 P2 = 450 BHP ( d2 d1 ) 2 Problem 3: A hydroelectric plant generates 100 MW at an available head of 200 m and at overall efficiency of 75%, what quantity of water in cubic meter per second is required? Solution: Pout Pin 100 Pin = 0.75 Qwh Pin = Pin = ŋ= Pout ŋ = 133.33 MW Q= Pin wh 133.33 x 106 (9810)(200) Q = 67.956 m3 per sec Q= Problem 4: A power plant gets water from dam from a height of 122.45 m at the rate of 1000 cubic meters per minute. If the output of the plant is 15,000 kW, what is the plant efficiency? Solution: Pin = Qwh 44760 Pout = Pin ŋ = 74.92% ŋ= EE158 module -mev (1000)(9810)(122.45) = 26,837.23 hp 44760 15,000 X 100% 26837.23 (0.746) = 42 Characteristics of cascaded units ŋ overall = Poutput(GEN) Pinput(PM) ŋoverall = (ŋPM)( ŋGEN) where: Poutput(GEN) = output power of the generator (watt) Pinput(GEN) = input power of the generator (watt) Poutput(PM) = output power of the prime mover (watt) Pinput(GEN) = input power of the prime mover (watt) ŋPM = prime mover efficiency ŋGEN = generator efficiency ŋoverall = total or overall efficiency of the cascaded units Problem 5: A waterfall is 60 meters high. It discharges at a constant rate of 1.0 cubic meter per second. A mini – hydroelectric plant is to be constructed below the waterfalls. The turbine efficiency is 80% and the generator efficiency is 95%. Calculate the kW output of the generator. Solution: Pin = Qwh = (1)(9810)(60) = 588.6 kW Poutput = Pinput [(ŋturbine)(ŋgenerator)] Poutput = 588.6(0.8)(0.95) Poutput = 447.336 kW Relationship between pulley diameter and speed N1 N2 = d2 d1 Note: The speed of a pulley is inversely proportional to the diameter where: N1 = speed of the driving pulley N2 = speed of the driven pulley d1 = diameter of the driving pulley EE158 module -mev 43 d2 = diameter of the driven pulley Problem 6: A generator is rated 600 kVA, 240 V, 60 Hz, 3-phase, 6-poles and wye connected. What will be the speed of the driving pulley if the driven and driving pulleys are 1 ft and 2 ft in diameter respectively? 120f N2 = P 120(60) N2 = 6 N2 = 1200 rpm N1 N2 = d2 d1 d2 ) d1 N1 = 600 rpm N1 = N2 synchronous speed ( = 1200 ( 1 ) 2 Power developed (horsepower) on the driven pulley of a generator P = 2πNFr 33,000 where: N = speed of the driven pulley (rpm) r = radius of the driven pulley (foot) F = force exerted by the driving belt (pound) P = power developed on the driven pulley (horsepower) In SI units, P = 2πNFr 44,760 where: N = speed of the driven pulley (rpm) r = radius of the driven pulley (meter) F = force exerted by the driving belt (newton) P = power developed on the driven pulley (horsepower) EE158 module -mev 44 Problem 7: The pulley of an old gen-set has a diameter of 20 inches. The belt exerts a pull of 353 pounds on the pulley. The gen-set runs at 900 rpm. What is the approximate kW rating of the gen-set? 1 ft 2π(900)(353) (10 in x ) 2πNFr 12 ft P = = 33,000 33,000 P = 50.41 hp 0.746 kW P = 50.41 hp x 1 hp P = 37.605 kW REMINDER! In dealing with hydro problems, take note on the consistency of units from SI to English units…MEV EE158 module -mev 45 DIESEL POWER PLANT EE158 module -mev 46 Diesel Engine – is an excellent prime mover for electric power generation in capacities of 101 hp to 5070 hp. These are also widely used in hotels, utility companies, municipalities and private industries. The design of diesel electric power plant includes the following elements; the stationary diesel engine, fuel system, lubricating system, cooling system, intake and exhaust system, starting system, and the governing system. Applications of Internal Combustion Engines (ICE) 1. Portable generating units in which may be moved from site to site where electrical power is required temporarily. 2. Standby units, normally idle, which can be activated when there is a failure of central station power where interruption would mean financial losses or danger. 3. Engine-generator installed in power plants where they are normal primary source of electrical power generated for public, industrial, or institutional consumption. Types of Internal Combustion Engines (ICE) 1. Gasoline Engine 2. Gas Engine 3. Hesselman Engine 4. Injection Gasoline Engine 5. Injection Gas Engine 6. Vaporizing Oil Engine 7. Diesel Engine 8. Gas-Diesel Engine 9. Dual-Fuel Engine EE158 module -mev 47 Advantages of the diesel engines: 1. 2. 3. 4. 5. 6. The cost of diesel fuel is cheaper than other fuels. It needs no long warming up. It has no standby losses. It has uniformly high efficiency of all size. It has a simple plant lay out. It needs no large water supply. PERFORMANCE OF DIESEL POWER PLANT 1. Heat supplied by fuel, Q,: Q s = m fQ h Where: mf = mass flow rate of fuel Qh = heating value of fuel Sample problem: (Calculating the heat supplied by Fuel) What is the heat that can be supplied by 39 kg/hr diesel fuel whose heating value is 43,912 kJ/kg? Solution: Q s = m fQ h Qs = (39/3600)43,912 kJ/s or kW Answer: Qs = 475.71 kW 2. Air – Fuel Ratio, A/F: = Where: ma = mass of air mf = mass of fuel Sample Problem: ( Calculating the Air-Fuel Ratio) The density of air entering the engine is 1.19 kg/m3 whose volume flow rate is 0.15 m3/s. If the mass flow rate of fuel is 121.38 kg/hr, what is the air-fuel ratio? Solution: = EE158 module -mev 48 Where: mf = 121.38 kg/hr = 0.034 kg/s Solving for the mass of air; ma: ma = ρaVa = 1.19(0.51) = 0.6069 kg/s then; = Answer: = 18 kg air/kg fuel 3. Piston Displacement, VD: Piston displacement is the volume displaced by the piston as it moves from top dead center to bottom dead center. VD = ( ) LN ncnp Where: D = bore or diameter of the piston L = length of stroke or stroke N = engine speed nc = no. of cylinders np = no. of piston actions NOTE: for 4-stroke engine divide the engine speed “N” by 2. Sample problem: (Calculating the Piston displacement) Determine the piston displacement of 35 cm x 45 cm, 4 –stroke, 1200 rpm, 8-cylinder diesel engine? Solution: Vd = Vd =( )LNncnp )(0.45)[ ](8) Answer: Vd = 3.46 m3/s 4. Piston Speed: Piston speed is the total distance a piston travels in a given time. Piston speed = 2 LN EE158 module -mev 49 Where: 2L = distance travelled by piston in one revolution N = angular speed in rpm or rps Sample problem: (calculating the piston speed of a Diesel Engine) Solution: The piston speed (V): Piston speed = 2LN = 2(0.35)(1100/60) Answer: Piston speed = 12.83 m/s 5. Indicated Power, Pind: Indicated power is the power developed by the action of piston within a cylinder, so named because it is measured by use of an indicator. Pind = PmiVD Where: Pmi = indicated mean effective pressure @ Calculating the indicated mean effective pressure using data provided by the planimeter. Planimeter measures the area of actual P – V diagram traced by engine indicator Pmi = Where: Ac = area of indicator card diagram Sc = spring scale Lc = length of indicator card diagram @ If working cylinder and crankcase are to be considered: Pmi = ( )wc – ( )cc NOTE: Crankcase compression is Used for scavenging. Where: wc = working cylinder cc = crankcase EE158 module -mev 50 Sample problem: (Calculating the indicated power of Diesel Engine) The cylinder diameters of an eight – cylinder, single acting, four-stroke diesel engine are 750 mm and the stroke is 1125 mm. The indicated mean effective pressure in the cylinder is 586 kPa, when the engine is running at 110 rpm. Calculate the indicated power. Soulution: Pind = Pmi L A N nc np = 586(1.25)[ (0.75)2][ Answer ](8)(1) = 2,135.82 kW 6. Brake Power, Pb Brake power is the delivered power to a shaft. Brake power is always less than the indicated power for a given engine, because some of the work developed by the cylinders is used cylinders is used to overcome the friction of running the engine. The often term of brake is shaft power. Pb = and T=Fr NOTE:Brake power is calculated Where: T = brake torque N = engine rotative speed in rpm F = brake force or brake load R = brake arm or torque arm using either prony brake or dynamometer. Sample problem: (Calculating the Brake power of a Diesel Engine) The flywheel of a rope brake is 1.22 m diameter and the rope is 24mm diameter. When the engine is running at 250rev/min the load on the brake is 480 N on one end of the rope and 84 N on the other end. Calculate the brake power. Solution: Pb = The load (F) on the brake (F): F = 480 – 84 = 395=6 N The radius (r): r = (1220 + 24) = 622 mm EE158 module -mev 51 r = 0.622 m The torque (T): T = 396 (0.622) = 246.31 N-m The Brake power (Pb): Pb = = 6,448.43 W Answer Pb = 6.448kW @ Brake power in terms of brake mean effective pressure and piston displacement: Pb = PmbVD Where: Pmb = brake mean effective pressure Pb = brake power VD = piston displacement Sample problem: (Calculating the brake mean effective pressure) A single acting, 8 cylinder, 4 stroke cycle diesel engine with a bore to stroke of 142.1-mm x 210.45-mm operates at 1200 rpm. The load on the brake arm at 1 m is 150 kg. What is the brake mean effective pressure in kPa? Solution: Pb = Pmb VD The volume displacement (VD): VD = [ (0.1412)2](0.21045)[ ](8)(1) = 0.267 m3/s The brakepower (Pb): Pb = Where; the torque(T): T = [150(0.00981)](1) T = 1.4715 kN-m The brakepower (Pb): Pb = Pb = 184.91kW The mean effective pressure(Pmb): 184.91 = Pmb (0.267) Answer Pmb = 692.55 kPa EE158 module -mev 52 7. Friction Power, Pf: Friction power is the power dissipated in an engine through friction. Friction Power = Indicated Power – Brake Power Pf = Pind Pb @ Calculating friction power using Morse test method applied to multicylinder engines: Consider a six – cylinder engine The indicated power if all six cylinders are firing Pind(6) = Pb(6) – Pf(6) If one cylinder is cut, or five cylinder are firing Pind(5) = Pb(5) – Pf(5) Derived from two equations above, equating the friction power: Pind(6) – Pind(5) = Pind(1) = Pb(6) – Pb(5) Thus, the total indicated power for six cylinder engine is, Pind(6) = 6(Pb(6) – Pb(5)) Note: no matter how many cylinders are firing; the friction power is constant. (Pf(6) = Pf(5)) Sample problem: (Calculating the indicated power) A six cylinder, four stroke diesel engine with 76 mm bore x 89 mm stroke was run in the laboratory at 200rpm, when it was found that the engine torque was 153.5 N-m with all cylinders firing but 123 N-m when one cylinder was out. The engine consumed 12.2 kg of fuel per hour with a heating value of 54,120 kJ/kg of air at 15.60C. Determine the indicated power. Solution: Pind(6) = 6(Pb(6) –Pb(5)) Pind(6) =6[ - ] Answer Pind(6) = 38.83 kW 8. Mechanical Efficiency, em Mechanical Efficiency is the ratio of the brake power to the indicated power. em = EE158 module -mev or em = 53 where: Pb = brake power Pind = indicated power Pmi = indicated mean effective pressure Pmb = indicated mean brake power Sample problem: (Calculating the indicated power) What is the mechanical efficiency of a 0.5 MW diesel engine if the friction power is 70 kW. Solution: The mechanical efficiency: em = The indicated power: Pind = Pb + Pf Pind = 500 + 70 kW Pind = 570 kW Substitute: em = Answer: em = 0.8772 or 87.72 % 9. Electrical or Generator Efficiency, egen Electrical or Generator Efficiency is the ratio of the generator power to the brake power. egen = where: Pgen = Generator Power Pb = Brake Power Sample problem: (Calculating the generator efficiency given the brakepower) A 16-cylinder V – type diesel engine is directly coupled to ¾ MW AC generator. Calculate the generator efficiency of the engine if the brake power is 833.33 kW. Solution: egen = = Answer: egen = 0.9 or 90% EE158 module -mev 54 10. Thermal efficiencies,et Thermal Efficiencies is the ratio of the work done by heat engine to the heat supplied by the fuel. a. Indicated thermal efficiency, eti Indicated thermal efficiency is the ratio indicated power to the heat supplied by the fuel. eti = Sample problem: (Calculating the indicated thermal efficiency) A 3.5 MW diesel power plant uses 3500 gallons in 24 hours period. What is the indicated thermal efficiency of the engine if the generator and mechanical efficiencies are 90% and 92% respectively? Oil is 280 API. Solution: eti = Solving for Qh: Qh = 41,130 + 139.6 (28) = 45038.80 kJ/kg Solving for mfuel: S.G.15.6 oC = = 0.887 ρfuel = 0.887(1000) = 887 kg/m3 mfuel = ρfuel Vfuel = 887[3500( )] = 11750.53 kg Mass of fuel per second: mfuel = = 0.136 kg/s Then; eti = Answer: eti = 0.5714 04 57.14% b. Brake thermal efficiency,eth Brake thermal efficiency is the ratio of the brake power to the heat supplied by the fuel. eth = EE158 module -mev 55 Sample problem: (Calculating the brake thermal efficiency) A supercharged six cylinder, four stroke cycle diesel engine of 10.48 cm bore and 12.7 cm stroke has a compression ratio of 15. When it tested on a dynamometer with a 53.34 cm arm at 2500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45822.20 kJ/kg heating value are burned during a 6 min. test and air metered to the cylinders at the rate of 0.182 kg/s. Find the brake thermal efficiency. Solution: etb = Solving for Pb: T = 81.65(0.00981)(0.5334) T = 0.42725 N-m Pb = = = 111. 854 kW Solving for mf: mf = = 0.00794 kg/s thus; etb = = 0.307 c. Combined or Over – all thermal efficiency, etc Combined or over – all thermal efficiency is the ratio of the electrical or generator power to heat supplied by the fuel. etc = Sample problem: (calculating the combined thermal efficiency) A 16 – cylinder V –type diesel engine is directly coupled to 1 MW AC generator. Calculate the combined thermal efficiency if the heat supplied by the fuel is 2.5 MW. Solution: etc = = = 0.40 or 40% EE158 module -mev 56 11. Engine efficiencies, et a. Indicated engine efficiency, eei Indicated engine efficiency is the ratio of the indicated thermal efficiency to the ideal thermal efficiency. eei = Sample problem: (Calculating the cycle efficiency) What is the indicated engine efficiency of diesel engine if the indicated thermal efficiency is 35% and the cycle efficiency is 45%. Solution: Answer: eei = eei = 77.78% b. Brake engine efficiency: eeb Brake engine efficiency is the ratio of the brake thermal efficiency to the ideal thermal efficiency. eeb = Sample problem: (Calculating engine efficiency) A 500 kW diesel has a rate of 12,000 kJ/kW-hr. The compression ratio is 16:1 cut of ratio of 2.3. Assume k = 1.32. Calculate the engine efficiency based on the output of 500kW. Solution: Let: eeb = brake engine efficiency eeb = The brake engine efficiency,etb: etb = = 0.30 or 30% Solving for cycle efficiency, e e = 1- [ Answer e = 0.5195 or 51.95% EE158 module -mev ] 57 The engine efficiency, eeb: eeb = = .5777 eeb = 57.77% c. Combined oe over – all engine efficiency, eec Combined or over –all engine efficiency is the ratio of the combined or over- all thermal efficiency to the ideal thermal efficiency. eec = Sample problem: If the over-all thermal efficiency of a diesel engine is 33% and the diesel cycle efficiency 49%, what is the combined engine efficiency? Solution: eec = 0.33/0.49 Answer: eec = 67.35% 12. Volumetric efficiency, ev Volumetric efficiency is the ratio of the volume of air drawn into a cylinder to the piston displacement. ev = where: Va = VD = piston displacement Sample problem: Determine the volumetric efficiency of 35 cm x 45 cm, 4 stroke 1200 rpm, 8 cylinder diesel engine if the air drawn in the engine is 3 m 3/s? Solution: The volume displacement, Vd: Vd = ( =( )LNn )(0.45)[ Vd = 3.46 m3/s EE158 module -mev ](8) 58 The volumetric efficiency, ev ev = Answer ev = = 0.87 or 87% 13. Specific Fuel Consumption, m Specific fuel consumption is the mass flow rate of fuel consumed per unit power developed. It is also known as specific propellant consumption. a. Indicated specific fuel consumption, mi mi = Sample problem: (Calculating the indicated specific fuel consumption) What is the indicated specific fuel consumption of a six cylinder, four stroke diesel engine with 76 mm bore x 89 mm stroke and indicated power of 38.83 kW if the engine consumed 12.2 kg/hr of fuel? Solution: mi = b. Brake Specific fuel consumption, mb mb = Sample problem: (Calculating the brake specific fuel consumption) A four-stroke, 8-cylinder engine with bore and stroke of 9 in and 12 in respectively and speed of 950 rpm has a brake mean effective pressure of 164 psi. What is the brake specific fuel consumption in lb per hp-hr if the engine consumed 468.55 lbs of fuel per hour? Solution: The brake mean effective pressure consumption, m b mb = EE158 module -mev 59 Solving for Pb: Pb = Where: A= ( )2 = 0.44 ft2 Pb = = 1,201.40 Hp Then: mb = Answer: mb = 0.39 lb/ hp-hr c. Combined or over – all Specific consumption ,mc mc = Sample problem: (Calculating the combined specific fuel consumption) A 16-cylinder V - type diesel engine is directly coupled to 1 MW AC generator. Calculate the combined specific fuel consumption if the fuel consumed is 1 kg/s. Solution: mc = Answer: mc = 3.6 kg/kW-hr 14. Heat Rate, HR Heat rate is the rate of energy charge per unit of power. To calculate the heat rate is to multiply the specific fuel consumption by the heating value of the fuel. A. Indicated Heat Rate, HRi: HRi = Sample problem: (Calculating the indicated heat rate) A four cylinder, 4 stroke cycle, 20 cm x 25 cm x 55 rpm diesel engine has a mean effective pressure of 150 psi. If the heat supplied by the fuel is 50 kW, what is the indicated heat rate? EE158 module -mev 60 Solution: HRi = Solving for Indicated power, Pind: Pind = Pm L A N n Pind = [150( )](0.25)[ ][ ](4) Pind = 148.875 kW then; HRi = Answer: HRi = 1209.07 B. Brake Heat Rate, HRb: HRb = Sample problem: (Calculating the brake heat rate) A 305 mm 475 mm four stroke single acting diesel engine is rated at 150 kW at 2600 rpm. Fuel consumption at rated load is 0.26 kg/kW-hr with a heating value of 43,912 kJ/kg. Calculate the brake heat rate. Solution: HRb = HRb = mcQh HRb = (0.26 kg/kW-hr)(43,912 kJ/kg) Answer: HRb = 11,417.12 kJ/kW-hr C. Combined or over –all Heat Rate, HRc: HRc = Sample problem: (Calculating the combined heat rate) The kilowatt output of a generator coupled to a diesel engine is 1.5 MW . If the mass of fuel with heating value 45,000 kJ/kg consumed by the engine is 0.04 kg/s, what is the combined heat rate? EE158 module -mev 61 Solution: HRc = Answer: HRc = 4,320 15. Generator Speed , N N= Where: N = speed in rpm f = frequency = 60 Hz(if not given) P = no. of poles Sample problem: (Calculating the generator speed) An eight-cylinder , two cycle, single acting diesel engine rated at 1250 Hp at standard condition is to be directly coupled to a 24-pole alternator, 3 phase, 60 cycles. What is the generator speed? Solution: N= = Answer: N = 300 rpm 16. Energy Stream ITEM 1. Useful Output (brake output) 2. Cooling Loss 3. Exhaust Loss 4. Radiation, Friction and Unaccounted Losses Total EE158 module -mev HEATLOSS (kJ/hr) 3600 Pb HEAT INPUT (kJ/kg) mfQ f mwCwΔTw M fQ h mgCρgΔTg M fQ h By difference M fQ h (%) COOLING LOSS 100 – ∑ 1 to 3 100% 62 17. Diesel Engine with closed cooling system By Energy Balance Qwj = Qw mwjCpwjΔTwj = mwCpwΔTw mwjΔTwj = mwΔTw 18. Waste Heat Recovery Boiler By energy balance Qsteam = Qgas Ms(hs-hf) = mgCpgΔtg Ms = Boiler efficiency, eboiler: eboiler = 19. = Engine operated at high altitudes P = Ps ( P = Ps ( ) SI units ) English units Where: Ps = standard power or power at sea level Pact = new pressure or actual barometric pressure in in.Hg T = new temperature or actual absolute temperature in 0R Note: 1. The decrease is pressure is approximately 1 in.Hg per 1000ft elevation. Pact = 29/92 - in.Hg 2. The decrease in temperature is approximately 3.60R per 1000ft elevation. T = 520 – 0 R Sample problem: (Calculating the maximum power delivered at a given elevation) What maximum power can be delivered by 1500 kW engine at 2800 ft elevation considering the pressure effect alone. EE158 module -mev 63 Solution: P = Ps ( ) Solving for Pact Pactual = 29.92 - = 29.92 - = 27.12 in.Hg T = 5200R (for ther’s no effect in temperature) Answer: P = Ps ( ) = 1,359.63 kW. GEOTHERMAL POWER PLANT Geothermal Power Plant – is the facility in which the electrical energy is produced from hot spring, steam vent or geyser. Geothermal Energy – is heat energy naturally occurring with the earth. It comes from two words “geo” meaning earth and “thermal” meaning concerning heat. EE158 module -mev 64 Magma – a molten metal within the earth which basically nickel-iron in composition whose stored energy heats the surrounding water thereby producing steam or hot water. Its temperature reaches as high as 12000C. EE158 module -mev 65 Well-bore product the effluent coming out from the geothermal well produced after drilling. This can be purely steam or hot water, or a mixture of both. Steam-dominated geothermal field refers to a geothermal plant with its well producing all steam as the well-bore product. Liquid-dominated geothermal field refers to a geothermal plant in which the well-bore product is practically all hot water (Pressurized). Fumarole is a crack in the earth through which geothermal substance passes. TYPES OF GEOTHERMAL PLANT 1. Dry or Superheated Geothermal Plant: 2. Separated Steam or “Single Flashed Geothermal Plant”: EE158 module -mev 66 3. Separated Steam/Hot-Water-Flash or “Double Flash” Geothermal Plant 4. Single Flashed Plant with Pumped Wells: 5. Binary Geothermal Plant EE158 module -mev 67 Performance and Design of Geothermal Power Plant 1. Mass flow rate of steam, ms : m s = x mg where: mg =mass of ground water x = quality after throlling (Process ① → ②) Solving for x: h1 = h2 = hf2 + xhfg2 2. Turbine Output: Wt = ms(h3 – h4) et Where: et = turbine efficiency 3. Generator Output: Wgen = Turbine Output x Generation Eff. Wgen = ms (h3 - h4) etegen 4. Heat Rejected in the Condenser: QR = ms (h4 – h5) 5. Over-all Plant Efficiency: ec = ec = SOURCES OF GEOTHERMAL ENERGY 1. Hydrothermal Fluids – basically made up of hot water, steam and minerals. It is the only form of energy currently being tapped for significant commercial heat and electric energy supply. 2. Geopressurized brines – represent a special subset of hydrothermal fluids typically found at depths exceeding at 3km and is characterized as hot water existing at pressure above the normal hydraulic gradient and containing dissolved methane. 3. Hot dry rock – is a water free, impermeable rock at high temperature and practically drilling depth to extract energy, high pressure water maybe injected through one or more wells to create new to enhance existing natural fracture system with limited access to ground water flow. 4. Magma – is characterized by or practically molten rock with temperatures reaching as high as 12000C. EE158 module -mev 68 THE GEOTHERMAL POWER PLANT IN THE PHILIPPINES 1. Tiwi Geothermal Power Plant, 330 MW (Albay) 2. Makiling – Banahaw (Mak-Ban) Geothermal Power Plant, 330 MW (Los Banos , Laguna) 3. Tongonan Geothermal Power Plant, 112.5 MW (Leyte) 4. Palimpinon Dauin Geothermal Power Plant, 112.5 MW (Negros Oriental) EE158 module -mev 69 NUCLEAR POWER PLANT NUCLEAR POWER PLANT is a power plant in which nuclear energy is converted into heat to be used in producing steam for turbines, which in turn drive generators that produce electric power. TYPICAL NUCLEAR POWER PLANT EE158 module -mev 70 Fuel Core is a radioactive material, U235 and U238 which is source of energy. Moderator is a device that slows down the neutrons to thermal energy, made of Carbon and Beryllium. Control Rods are boron coated steel rods used to control the reactor. Reflector is a device made of lead or carbon which surrounds the core to bounds back any leakage of neutrons. Thermal Shield is a device that prevents escape of radiation from reactor vessel. Reactor Drum is a device that encloses the fuel core and components. Biological Shield is a concrete or lead which absorbs any leakage of radiation and protects operators from exposure to radioactivity. Control Cubicle is a device that contains the meters that show the operating quantities in the reactor. EE158 module -mev 71 Containment Vessel is a concrete the prevent the spread of radiation in case of major explosion. Coolant is a substance that absorbs the heat from the fuel core then releases the heat to the water in the steam generator. Coolant Pump is a device (pump) that circulates the coolant. Turbine generator is a device that generates the electric power. Condenser is a device that converts steam coming from the turbine into liquid. Feedwater Pump a device that delivers the feedwater to the steam generator. TYPES OF NUCLEAR POWER REACTORS 1. PRESSURIZED WATER NUCLEAR POWER REACTOR (PWR) A pressurized water reactor uses water under pressure as both the coolant and moderator in the reactor. The water pressure must be sufficiently high so that we can have water at 550 -6600F without boiling in the core. The fission of the fuel produces heat that is carried away by circulating water under pressure. 2. FAST-BREEDER NUCLEAR POWER REACTOR A fast - breeder reactor depends upon the fission of fast neutrons rather than thermal neutrons. Therefore, no moderator or moderating material can be used. Further the term “breeder” implies that the reactor produces more fuel due to the absorption of neutrons than is burned up. This is possible since on the average 2.5 neutrons are produced due to its fission of a uranium atom and only one of this is needed to keep the chain reaction going. 3. ADVANCE GAS – COOLED NUCLEAR POWER REACTOR (AGCR) Gas cooled nuclear power reactor uses carbon dioxide as coolant. Advantages of Gas Cooled Power Reactor EE158 module -mev 72 1. High temperature is possible at low pressure since boiling is not a problem. 2. Plant thermal efficiencies increasing cycle temperatures Steam can be produced in heat exchangers heated by gas. 3. Gases can be nonhazardous. Disadvantages of Gas Cooled Power Reactor 1. The power required to pump the gas compared to that for pumping liquid. 2. Large flow rates and high velocities are required. 3. Care must be taken not to use gases that are corrosive to moderating materials at the high temperatures employed. 4. LIQUID – COOLED NUCLEAR POWER REACTOR There are two types of liquid – cooled reactor: the water-cooled reactor and the liquid metal – cooled reactor. Advantages of water-cooled reactor 1. Low cost of coolant 2. Coolant is also a moderator 3. Pumping power is small and design of pumps is relatively simple 4. Low viscosity Disadvantages of water-cooled reactor 1. High-pressures are required to obtain high temperatures. 2. Detrimental to large conversion ratios because hydrogen in the water is a good absorber. 3. Problem of corrosion of materials in contact with the water. Advantages of liquid metal-cooled reactor 1. High thermal conductivity 2. Can operate atmospheric pressure 3. Good heat transfer characteristic 4. Stable at high temperatures Disadvantages of liquid metal-cooled reactor 1. High Cost EE158 module -mev 73 2. Danger of reaction with air or water makes for difficulty in handling. 5. BOILING WATER NUCLEAR POWER REACTOR (BWR) Boiling water nuclear power reactor is the simplest nuclear reactor. The feed water from the power turbine goes directly into the reactor and picks up the heat from the fuel core, thus the feedwater also serves as the coolant. Advantages of boiling water reactor 1. Moderator and coolant are the same (water). This is the same as in a pressurized-water reactor. 2. The steam formed goes directly to the turbine, thus eliminating the extra heat exchange present in the pressurized water plant. 3. The pumping power is much less than that for pressurized water plants because of lower operating pressures to produce the same temperature and elimination of the extra loop. 4. The stresses in the pressure vessel are lower because of lower operating pressures, thus decreasing costs. Disadvantages of boiling water reactor 1. Presence of the steam in the reactor will result in more neutron leakage and will require a more highly enriched fuel. (For example: enrichment of U235) 2. There will be some depositing radioactive elements in the pump, turbine, and condensing. This can become a hazard under normal maintenance procedures. 3. The leakage of the less dense steam is more apt to occur than in a pressurized water reactor Other Nuclear Reactors 6. Modulated Fuel Reactor (CANDU) 7. Heavy Water Reactor 8. Graphite Moderated Light Water EE158 module -mev 74 Alternative Methods of disposing Nuclear Wastes MODULE SUMMARY EE158 module -mev Sending nuclear waste into outer space. Burying them in polar ice caps Dumping them deep into the ocean Shooting them into the sun. Land Burial. 75 Pressurized Water Reactor EE158 module -mev 76 1 Gram of Uranium Pellet EE158 module -mev 77 REMINDER! In module I, you have learned about the introduction of the different power plant as to their sources, application and how they are very much available in our country. That is, Philippines is one of the countries who has the most expensive energy and that’s for a fact that we as future Electrical Engineers must be aware of the availability of the different plants that is and will be discovered and may be able to be available in the near future to be able to shift to plants that is more reliable and can lessen the cost of energy in every household. There are three lessons in module I. Lesson 1 consists of the different types of power plants as to its sources, most particularly to thermal plant that has most essential part in our energy grid, and most important is as to its source, components and operation. A part of it also, is the hydro energy that plays the major role in the production of energy and of its kind. The characteristics of hydro plant makes it very unique in nature because of its environmental part that its duality. Why duality? first it produces energy, second, the excess water can also be used for farmer’s irrigation purposes for their plants and harvests. Lesson 2 deals with the discussion of renewable & non-renewable power plant. Base load plants in our country, whether we like it or not are more of a non-renewable type. That is, they carry big load of energy as its big share, and mostly has a reciprocating effect in our environment, mostly coal fired and diesel plant has its big share in the energy industry. Lesson 3 deals now with the different power plant operation and how it operates and its components from a raw material to its finished product which is the energy output produced by this plants. The efficiencies of these plants must be measured and through its high energy input, its energy output must be compensated to achieve the highest possible energy efficiencies. Congratulations! You have just studied Module I. now you are ready to evaluate how much you have benefited from your reading by answering the summative test. Good Luck!!! -mev EE158 module -mev 78 SUMMATIVE TEST Essay: 1. Enumerate and discuss the different thermal power plants and how they are interlinked with each other. 2. From the different thermal plants discussed above, which of these plants exists but do not/did not operate due to what probable reasons. 3. From the hydro power energy, which of these are the best possible design to further develop and had an advantage over the other types of hydro power plants. 4. From the different hydro power plants discussed above, which of these plants do not exist, and what could be the probable cause? Problem Solving: solve the requirement/s of the said unknown parameters: (show you solution in an orderly manner). 1. A sixteen-cylinder, two cycle, single acting diesel engine rated at 1500 Hp at standard condition is to be directly coupled to a generator running @ 3 phase for 13,200 volts output, what would be the possible number of poles necessary to run the diesel engine at 150 rpm? 2. A coal fired thermal power plant has the following data: Coal consumption per day: 68.54 slugs Units of energy per day: 4,000 KWhr Calorific value of fuel used: 12,000 Kcal/Kg Alternator Efficiency: 95% Turbine Efficiency: 93% Solve a.) specific heat combustion of fuel b.) overall efficiency 3.A hydroelectric station has a turbine efficiency of 90% & generator efficiency of 89%. The effective head of water is 150m. Calculate the amount of volume of water used in delivering a load of 50 MW for 8 hours of operation. EE158 module -mev 79 MODULE II POWER PLANT ECONOMICS Lesson 1 Lesson 2 Lesson 3 EE158 module -mev Other Sources of Energy Introduction to Power Plant Economics Load Determination & Load Graphs 80 MODULE II POWER PLANT ECONOMICS INTRODUCTION This module presents Familiarization of other Power plants sources as to their flow of operations, and will discuss the power plants economics as to its use, load determination, graphs, interpretations and reserves. POWER PLANT ENGINEERING AND DESIGN- It is the art of selecting and placing the necessary power-generating equipment so that the maximum of return will result from the minimum expenditure over the working life of the plant; and the operation of the completed plant in manner to provide a cheap, reliable, and continuous service. OBJECTIVES After studying the module, you should be able to: 1.Evaluate the importance of the different power plants economics. 2.Explain the different operations of other sources of a power plant. 3.Solve & analyze problems relative to power plant economics & other sources of power plants. 4.Give synthesis the different existing power plants in the Philippines as to their economics. DIRECTIONS/ MODULE ORGANIZER There are three lessons in the module. Read each lesson carefully then answer the exercises/activities to find out how much you have benefited from it. Work on these exercises carefully and submit your output to your tutor or to the COE office. In case you encounter difficulty, discuss this with your tutor during the face-to-face meeting. If not contact your tutor at the COE office. Good luck and happy reading!!! -mev EE158 module -mev 81 Lesson 1 Other Sources of Energy Wind power Plant (Windmill Power Plant) Windmills are any various mechanisms, such as mill, pump or electric generator, operated by the force of wind against vanes or sails radiating about a horizontal shafts. (old type design of a windmill typically used for water distribution on land fields. EE158 module -mev 82 EE158 module -mev 83 PERFORMANCE OF WIND POWER PLANT 1. Pump Power, Wp Wp = Where: γ = specific weight of water γ = 9.81 kN/m3 Q = volume flow rate H = net head ep = pump efficiency 2. Kinetic Energy, KEair KEair = maVa2 Where: ma = mass flow rate of air Va = average velocity of air 3. Volume of air, Va Va = If wind velocity is given: Volume = Area x Velocity Va = x velocity 4. Aerodynamic Efficiency, ea: ea = Calculating the Power Output of a wind: Power k Cp1 / 2 AV 3 Where: P = Power output, kilowatts Cp = Maximum power coefficient, ranging from 0.25 to 0.45, dimension less (theoretical maximum = 0.59) ρ = Air density, lb/ft3 EE158 module -mev 84 A = Rotor swept area, ft2 or π D2/4 (D is the rotor diameter in ft, π = 3.1416) V = Wind speed, mph k = 0.000133 A constant to yield power in kilowatts. (Multiplying the above kilowatt answer by 1.340 converts it to horse- power [i.e., 1 kW = 1.340 horsepower]). The rotor swept area, A, is important because the rotor is the part of the turbine that captures the wind energy. So, the larger the rotor, the more energy it can capture. Note: The air density, ρ, changes slightly with air temperature and with elevation. The ratings for wind turbines are based on standard conditions of 59° F (15° C) at sea level. A density correction should be made for higher elevations as shown in the Air Density Change with Elevation graph. A correction for temperature is typically not needed for predicting the long-term performance of a wind turbine. Although the calculation of wind power illustrates important features about wind turbines, the best measure of wind turbine performance is annual energy output. The difference between power and energy is that power (kilowatts [kW]) is the rate at which electricity is consumed, while energy (kilowatt-hours [kWh]) is the quantity consumed. An estimate of the annual energy output from your wind turbine, kWh/year, is the best way to determine whether a particular wind turbine and tower will produce enough electricity. To get a preliminary estimate of the performance of a particular wind turbine, use the formula below. AEO k D 2 V 3 Where: k = 0.01328 AEO = Annual energy output, kWh/year D = Rotor diameter, feet V = Annual average wind speed, mph EE158 module -mev 85 OTHER NON-CONVENTIONAL POWER SOURCES ① Tidal Power ② Thermoionic Converter ③ Fuel Cell ④ Low Thermal Head Plant ⑤ Magneto Hydrodynamic Plant (Non-Conventional Power Plant) – Renewable Energy SOLAR POWER PLANT Solar Power Plant is the conversion of the energy of the sun’s radiation to useful work. TYPES OF SOLAR COLLECTORS ① Flat plate ② Concentrating ③ Focusing SOLAR ENERGY RECEIVED AT THE EARTH’S SURFACE Es = qS ( 1 – I ) A Where: Qs = solar energy without atmospheric interference i = atmospheric interference A = surface area of solar collector Qsun = Qw + PE + Qloss EE158 module -mev 86 Other types of Solar Plant • 1.OFF-GRID POWER PLANT • -SMALL SCALE OR MICRO GRID PURPOSES • 2. HYBRID POWER • 3. SOLAR THERMAL PLANT (Solar Thermal Plant) EE158 module -mev 87 (Hybrid Power Systems) EE158 module -mev 88 EE158 module -mev 89 OFF – GRID SOLAR POWER SYSTEM (Small scale to Micro Grid system) An off-grid solar system is designed for the power needs of mid- to large-size homes upto communities of independent power. Unlike gridtied solar systems, off-grid systems have no connection to the utility grid, and must make all the electricity necessary to power your home and community. Off-grid solar systems operate from the stored energy in a battery bank. Advantages: Disadvantages: 1. Solar Materials are High in cost beacause of importation. 2. Solar deep cycle batteries have short life especially on an under designed off-grid power.But for computed and properly designed off-grid power can last up to 3-6 years. EE158 module -mev 90 The light from the Sun, made up of packets of energy called photons, falls onto a solar panel and creates an electric current through a process called the photovoltaic effect. Each panel produces a relatively small amount of energy, but can be linked together with other panels to produce higher amounts of energy as a solar array. The electricity produced from a solar panel (or array) is in the form of direct current (DC). EE158 module -mev 91 Charge Controllers Pulse-Width Modulation (PWM) • The PWM charge controller is a good low cost solution for small systems only, when solar cell temperature is moderate to high (between 45°C and 75°C). EE158 module -mev 92 Maximum Power Point Tracking (MPPT) • Maximum Power Point Tracking The MPPT controller will harvest more power from the solar array. The performance advantage is substantial (10% to 40%) when the solar cell temperature is low (below 45°C), or very high (above 75°C), or when irradiance is very low. Irradiance • irradiance is the radiant flux (power) received by a surface per unit area. The SI unit of irradiance is the watt per square meter (W.m⁻²). The CGS unit erg per square centimeter per second (erg·cm⁻²·s⁻¹) BASIC STEPS HOW TO DESIGN AN OFF-GRID SOLAR POWER SYSTEM 1. Figure out the input power to the inverter when the load is on to decide how many inverters are needed, especially the rating needed for the set-up of the system. - inverters must be “pure sine wave” in nature There are inverters in the market which they claim to be pure sine wave, but actually they are modified sine wave which will eventually damage your load system especially your sensitive electronic appliances. 2. Figure out the amperage the charge controller provides to the inverter to decide how many charge controllers needed for your design. 3. Calculate how much energy the load consumes per day basis. - note: your energy is expressed in KWHR For example: An electric Fan of 50 watts will operate at 12-hour basis = 0.6 KWHR 4. Calculate how many solar panels you will need in your specific location. Note: solar panels are dependent on your specific location such as, too cloudy, rainy areas, too hot season, areas where cold season like Baguio or tagaytay. Mono crystalline – used for cold, rainy seasons like in Baguio because of slightly higher efficiency than poly. Poly crystalline – are used for hot seasons and dependable to sustain voltage regulation and wattage in your solar panel. EE158 module -mev 93 Note: Do not mix any different kind of solar panel, check the characteristics and specifications, because different brands different in its specifications. -doing this, some installers connect a series of 12v to obtain a 24v solar panel requirement, doing this may damage the solar charge controller and may lessen its life and even voids the warranty. 5. Calculate how many batteries you will need for the system to work properly and not stressed out, to further prolong its natural life span. - over and under charging your batteries could further destroy and shorten your batteries life span and can void the warranty. - Without sunlight for certain number of days. Efficiency of the inverter: ηinverter = output power/ input power such that; Input power = output power/ ηinverter Example: (An assumption of 50% efficiency of inverter at its full stressed load. Input power = output power/ ηinverter = 100 w/ 50% = 200 watts Since, the input power to the inverter (200w) is less than its rated power (450w), we only need one inverter (but still dependent on the inverter model and capability. Number of charge controller -we know the power coming out of the charge controller to the inverter is 200w Such that; P= V.I And I= P/V , therefore 200/12 = 16.67Amps -note (12v system is used for this example, it could either be 12, 24, 48 volts system where the charge controller could automatically detect. -since the charge controller capacity is 10amps, and we are drawing 16.67Amps, therefore; we need 2 charge controllers, or 20 Amps rating of charge controller @ 80% safety factor. LOAD ENERGY USAGE Example: -we know the lightbulb consumes 100w for 2hours per day, we need to convert that to KWHR per day: 100w x 1kw/1000w x 2hrs./day = 0.2 KWHR per day EE158 module -mev 94 That means to say, that the energy coming from the charge controller is: ηinverter = output energy input energy input energy = output energy inverter = 0.2kwhr = 50% 0.4kwhr day DE (direct sunlight energy) receives ; 2.88 Kwhr/m2 per day, this unit is sometimes called sun-hours, also having units of hours/day of full sunlight. Since you will use the solar panel that produces 20w full sunlight, you can use it to calculate the kwh/day that the solar panel produces, likewise, 0.0576kwhr day 1kw 2.88hours 20w = x x = 1 solar panel day 1 solar panel 1,000w - You know you have to provide a 0.4 kwhr/day, so; to calculate how many solar panels you need, that is: 0.4 kwhr day = 6.94 solar panels 0.0576 kwhr day 1 solar panel that is approximately = 7 solar panels@ 20w each Month January Feb March April May June July Aug Sep Oct. Nov Dec year EE158 module -mev RESULTS Solar AC Energy Radiation(irradiance) (KWH) 2 (KWH/m /day) 3.48 344 4.46 394 4.48 450 5.13 454 5.34 460 5.73 473 5.57 468 5.70 485 5.02 420 4.71 425 3.50 317 2.88 270 4.69 4960 Energy Value 30.27 34.67 39.60 39.95 40.48 41.62 41.18 42.68 36.96 37.40 27.90 23.76 436.48 95 An example how irradiance is used in an example given as it is measured as its lowest in the month of December. Note: -this is just a basis approximation based on an experiment done HOW MANY BATTERIES NEEDED? - - To start, decide on number of day/s (hours if more specific) that the batteries will need to operate w/o sunlight. For example, I choose 3days in this exercise. You know that the load uses 0.4 kwhr/day and over three days, this is approximately equal to 1.2 kwhr, what would be the battery size in ampere-hours & how many pieces?, if 12V-5AH is only available? To convert 1.2 kwhr in amp-hours, you do the following: 1.2 kwhr x 1,000 ampere hours 1 0.1 kilo ampere hours x 1 kilo ampere hours 12 volts = 100 ampere-hours - Since each battery is 5Amp-hrs only, you will now need 20 batteries to get 100 amp-hour Note: - This Method is more likely applicable to light load only like light bulbs, LED lightings. But if with light inductive loads, specifically with motor, electric funs, refrigerators, A/C..etc.. the formula for battery loading is necessary: Battery Bank (AH) ( Load in watts)(hrs. of operation)(DoD)(Batt. disc Loss Factor) ( BattAging factor) ( System voltage) (Comp..Efficiency) ( Inverter Efficiency) Where: DoD = Depth of discharge @ 50% = 2 Batt.Discharge Factor = 1.3 Batt. Aging Factor = 1.25 Sys.Voltage = either 12,24,48 Volts system Comp. Efficiency = 0.7, 70 Inverter Efficiency = 0.85 or 85% EE158 module -mev 96 Load one 100w AC light bulb For 2 hours/day all year long (computed req’ment) Battery (ies) 5 amp-hr each (needed 20 pcs) 100w 0.2 kwhr/day Inverter ηinverter = 50% @450w maximum (needed one) 200w 200w 0.4 kwhr/day Note: in this design, everything is 12 volts except the output of the inverter which is 220 volts EE158 module -mev Charge Controller 100% @ 10Amp max. (Needed 2) 0.4 kwhr/day Solar Photovoltaic Module(s) @ 20w each (Needed 7) 97 Lesson 2 INTRODUCTION TO POWER PLANT ECONOMICS GENERAL DEFINITIONS: Before the economics of the power plant is studied in detail, the following terms must be defined: 1. Load Factor: It is defined as the ratio of the average load to the peak load during a certain prescribed period of time. The load factor of a power plant should be high so that the total capacity of the plant is utilized for the maximum period that will result in lower cost of the electricity being generated. It is always less than unity. 2. Utility Factor: It is the ratio of the units of electricity generated per year to the capacity of the plant installed in the station. It can also be defined as the ratio of maximum demand of a plant to the rated capacity of the plant. Supposing the rated capacity of a plant is 200 MW. The maximum load on the plant is 100 MW at load factor of 80 percent, then the utility will be 100 x 0.8 = X 100 = 40 percent 200 3. Plant Operating Factor: It is the ratio of the duration during which the plant is in actual service, to the total duration of the period of time considered. 4. Capacity Factor: It is the ratio of the average load on a machine or equipment to the rating of the machine or equipment, for a certain period of time considered. 5. Demand Factor: The actual maximum demand of a consumer to always less than his connected load since all the appliances in his resident will not be in operation at the same time or to their fullest extent. This ratio of the maximum demand of a system to its connected load is termed as demand factor. It is always less than unity. 6. Diversity Factor: Supposing there is a group of consumers. It is known from experience that the maximum demands of the individual EE158 module -mev 98 consumers will not occur at one time. The ratio of the sum of the individual maximum demands to the maximum demands of the total group is known as diversity factor. It is always greater than unity. 7. Load Curve: It is curve showing the variation of power with time. It shows the value of a specific load for each unit of the period covered. The unit of time considered may be hour, days, weeks, months or years. 8. Load Duration Curve: It is the curve for a plant showing the total time within a specified period, during which the load equaled or exceeded the values shown. 9. Dump Power: This term is used in hydro-plants and it shows the power in excess of the load requirements and it is made available by surplus water. 10. Firm Power: It is power which should always be available even under emergency conditions. 11. Prime Power: It is power, may be mechanical, hydraulic or thermal that is always available for conversion into electric power. Recall on the first discussion as to plant as to its reserve: 12. Cold Reserve: It is that reserve generating capacity which is not in operation but can be made available for service. 13. Hot Reserve: It is that reserve generating capacity which is in operation but in service. 14. Spinning Reserve: It is that reserve generating capacity which is connected to the bus and is ready to take the load. COST ANALYSIS The following cost analysis refers to the thermal stations. The cost analysis for hydro-plants will be taken in the next article. The annual cost of supplying electric energy from a thermal station is split into the following items: 1. Fixed cost. 2. Operating or running cost. 3. Profit on the investment. EE158 module -mev 99 Fixed Cost The fixed cost depends upon the capital outlay or the total investment of the plant. The total investment includes: cost of land, cost of buildings and equipment, cost of installation, engineering fees, cost of erection of the transmission lines and sub-stations. The annual fixed cost consists of the interest on the total investment, taxes and insurance and the annual contribution to the fund to pay off the capital expended after the useful life of the plant. This remains fixed throughout the year or throughout the useful life of the plant. It varies directly with the installed capacity of the plant and it does not depend upon the factor whether the plant supplies any electric power or not. Cost of the land and building will depend upon the location of the plant. If the plant is situated near the cities, the land will be costlier than the case if it is located away from the cities or the residential areas. But it is always better and economical to locate the plant near the load center to reduce the transmission and distribution charges. The cost of the equipment is carried over in the books from year to year and there are two methods to do so; original cost, or replacement cost. In the first method, the cost of the equipment is established at the time of its erection and installation. This cost remains in the books as long as the equipment is in operation with suitable allowances for the depreciation of the equipment. In the second method, the cost of the equipment is estimated periodically and the cost will depend upon the price level at the time of the cost estimation. This is a very costly procedure. Interest is included to represent the income which would have been derived from the investment if it would have been used otherwise or loaned in the market. To avoid accidents from fire and explosions, the equipment and the buildings must be insured and regularly inspected. Depreciation is the most important item in the fixed costs. It represents the decrease in the value of the property due to continuous wear and tear and also due to obsolescence. The first factor can be reduced by proper maintenance of the equipment and the buildings, but the second factor is quite unpredictable. To account for the depreciation, a certain fixed amount of money is set aside every year so that the total amount accumulated at the end of the useful life of the equipment is the original EE158 module -mev 100 cost of the equipment less the expected salvage value. There are two methods to accumulate the amount: straight line method and the sinking fund method. In the first method, the same fixed amount is set aside every year throughout the useful life of the equipment. The interest earned on this amount is not added to the depreciation reserve but is considered as the income of the firm or the company. In the sinking fund method, the interest earned is compounded periodically and it forms a part of the depreciation reserve. The total amount accumulated by this method is the sinking fund plus the interest earned and the sum should be equal to the original cost of the equipment minus the salvage value. Operating or Running Costs These costs are based on the energy output as measured in kWhr and these are directly proportional to the station load. These include: (a) Cost of fuel including its handling and ash handling in the case of steam plant. (b) Cost of labor i.e. the salaries and wages. (c) Cost of water for: (1) Boiler feed, condensers, cooling and house service for steam power plants. (2) Jacket cooling water, and water to be used in cooling towers, for i.e. engine power plants. (3) Intercoolers and gas coolers in gas turbine. (d) Repairs and maintenance. (e) Oil, waste, stores and other supplies. (f) Transmission and distribution costs. (g) Customer’s expenses. In the case of steam power plant, the cost of the coal is made up of two parts: a small fixed component and a large running component. The first component is that which is needed to keep the station in readiness to meet the demand for power and it is independent of whether any energy is sent out or not. The running component is the cost of coal equivalent of the heat converted into electric power plus the heat rejected in the condensers. It is directly proportional to the energy sent out in kWh. The amount of the fuel required per kWh can be determined by dividing the heat rate by the heating value of the coal. The amount of coal and its cost will depend upon the plant capacity, its efficiency, load factors, transportation charges and the cost of storage etc. Labor is needed both for the operation and the maintenance of the plant and its requirement is proportional to the capacity of the plant and EE158 module -mev 101 the quality of the labor employed. Skilled labor will cost more than unskilled labor but the plant will operate more economically and the maintenance charges will be reduced. The consumer’s expenses include the cost of meter reading, billing, collecting of revenue, promotion of electrical appliances etc. Investor’s Profit If the power plant is the public property, as this case in India, then the customers will be the taxpayers to share the burden of the government. For this purpose, there is an item in the rates to cover taxes in place of the investor’s profit. These taxes will be paid by the consumers in the form of electric consumption bills. This amount is collected in twelve instalments per year or six instalments per year. COST OF HYDRO-PLANTS The cost analysis of a hydro-plant is different from those of the other plants in this respect that the fixed cost is the major item of the total cost and the operating cost is relatively much smaller whereas in the steam and other plants, the operating cost is a large part of the total cost. The total annual cost of a hydro-electric scheme can be broken into two components as usual: 1. Fixed Costs (a) Interest on the capital. (b) Amortization of the capital. 2. Running Costs (a) Operating costs including salaries and wages. (b) Maintenance and repairs. (c) Rates and taxes. (d) Stores, oil and other supplies. It is a general practice to include the rates, taxes and the insurance in the fixed charges. The fixed charges of a hydro-plant are about 60 to 70 percent of the total cost of power and these do not depend upon the station output. The running charges depend upon the station output but not so much as in the thermal power plant. EE158 module -mev 102 The total capital outlay or the investment on a hydro-plant includes the following items: 1. Preliminary surveys and investigations of the topography and geology of the proposed plant site. 2. Purchase of water rights and the land needed to provide for sufficient storage or poundage. 3. Compensation to the oustees. 4. Cost of preparation of detailed designs and specifications. 5. Costs of testing the materials of construction. 6. Costs of carrying out experimental work and model tests on designs for hydraulic structures. 7. The actual costs of construction and the purchase and installation of the equipment. 8. Interest on the capital during construction. 9. Working capital during the period of load development. In addition to the above items, new roads, railway lines, residential houses and even new towns will have to be constructed. The total cost of construction of a hydro-plant is invariably higher than that of a thermal plant of equal capacity. Therefore, the annual charges for interest and depreciation are comparatively high. The cost per kW for a hydro-plant is higher, the smaller the quantity of water stored. A typical cost analysis of a hydro-plant is: Reservoir, dam and water ways : 55% Power plant and equipment : 20% Land : 15% Structures : 10% Another important item is called the transmission liability which refers to the transmission charges for conveying the electricity from the plant site to the load center. Figure (17.1) shown a typical sub-division of the various items forming the capital cost for the various plants. The areas of the circles represent the comparative unit costs. EE158 module -mev 103 1. 2. 3. 4. 5. 6. 7. 8. 9. Turbo-generators and condensers Land building and foundation Misc. Switch yard Switching and wiring Piping Fuel handling Turbine and generator Cooling, fuel system and other auxiliaries. ECONOMICS OF COMBINED-HYDRO AND STEAM POWER PLANTS It has been already mentioned that if country or a region is neither rich in fuel reserves nor in hydro resources then to generate the electricity at the low economical cost, a combined operation of the hydro and steam power plants will give the best results. The following benefits can be derived from the combined operation: 1. Flexibility of Operation: The major drawback of steam power plant is that appreciable time is required to start up and synchronize the plant and then to increase the load. Therefore, it is necessary that the capacity of the steam plant connected to the system should be such as to cope with the rapid rate of load increase and also it should possess adequate margin of safety against breakdown. Moreover, to reduce the stresses and strains on the boiler joints due to temperature variation, it is necessary to keep sufficient boiler plant hot throughout the day and night to meet the peak load requirements. This represents a continuous loss of energy which cannot be avoided. EE158 module -mev 104 On the other hand, a hydro-plant can be run up and synchronized quickly and the load can be raised from no load to full load very rapidly. This plant can also meet the sudden variations of load demand easily and so it is particularly suitable for peak load operation. Therefore, if the steam and hydro-plant are combined in a system, its operation will become flexible. The running of the steam plants will become easier as they can operate within the flatter top portion of the daily load curve and the violent fluctuations of load will be met by the hydro-plant. 2. Security of Supply: As far as the reliability of operation and security of electric supply is concerned, there is not much to choose between a steam plant and a hydro-plant. The reliability of steam plant depends upon the coal supplies and that of the hydro-plant on the steam flow. Due to greater diversity, a combined steam and hydro-plant system would be more reliable than the individual steam or hydro-steam. 3. Improved Utilization of Hydro-power: As already pointed out, during the periods of heavy flow, the hydro-plant can generate the base load and the steam plant may not generate as much energy and this will result in a saving of fuel. During drought periods, the steam plant can generate more energy and the water can be held back in reservoirs to make the capacity of the hydroplant effective. 4. Spare Plant: In any electric system, a spare plant is needed to provide for any unforeseen excess demand, for covering any temporary loss due to breakdown and due to maintenance and repairs of the main plant. A hydro-plant is simple and more reliable than a steam plant and for this plant only few maintenance and repairs are needed and if the combined system adopted, the steam plant can be taken out of commission during the periods of lower electric demand and a spare plant need not be provided. But it is always judicious to carry a spare plant in a combined system, it is possible to achieve an economy in the overall amount of the spare plant. EE158 module -mev 105 Economic Factors The combined system of steam and hydro-plants can be grouped into three categories: 1. Bulk of hydro power resources and fuel has to be imported. 2. Bulk of coal reserves with hydro resources deficient. 3. Considerable resources of both hydro and thermal power. In the first case, there might be various plant sites and the comparison will have to be made of the cost of producing equivalent electric power and its transmission to the load center. Some sites will be suitable for base load plants and others might be suitable for peak load service. The cost of electric generating at various sites will have to be studied for various load factors. The load factor to be adopted will be that which will give the lowest overall cost of production. There is one drawback of the above system. The maximum run-off may not occur during the periods of maximum demand and for the hydroplant to supply the full demand throughout the year, a storage will have to be built to cover the seasonal variations of run-off. This storage may be either impracticable or very costly to provide. For this, a sufficient thermal plant will have to be provided on the hydro system to make up its deficiencies and also for supplying the peak loads. The load factor of the thermal plant will be high only for a small period and its annual load factor will be low. In the second case it will be economical to develop the hydro-plants for peak load service. The reason for this is that in hydro-plants, the cost of storage is the major item in the total cost and this item does not depend much on the annual load factor, therefore, the capital cost per kW will be less for low load factor than for high load factor. It is found that the incremental costs resulting from reduced load factor i.e. increased amount of installed capacity is less than the capital cost of raising the installed capacity of a steam plant. An advantage of increased capacity of the hydroplant is that the spoilage can be decreased at times of high run-off by operating the plant for longer periods and thereby saving the cost of fuel in the thermal stations. For the third case, the hydro-plants must be planned to generate enough electric power during adverse periods so that they can meet the requirements of that portion of the load curve to which they have been assigned. EE158 module -mev 106 For any particular scheme, there must be some ratio of the hydroplant to the total demand which gives the lowest annual cost of electric generation. In areas where both cheap fuel and favorable hydro sites are abundant, this ratio will be much higher. But in areas where fuel is cheap and the cost of development of hydro-plant is high, the ratio maybe quite low. For any combined system, the economic balance between the capacity of the hydro station and thermal station will depend upon the following factors: 1. Shape of load curve. 2. Cost of fuel. 3. Availability of condensing water. 4. Availability of suitable hydro sites and their distance from the load center. 5. Run-off and its seasonal variation. 6. Cost of storage. Combined Operation The operation of all the stations in the integrated system should be carefully coordinated to achieve the maximum overall economy. For this purpose, there should be a centralized supervision station to supervise all the operations including generation and switching. But a big centralized control will be too unwieldy. So, the whole region is divided geographically in smaller parts. Each part will be supervised from a group control center. These control centers will have telephone connections with one another, with the power stations under their respective control and also with the system central control room. PLANT SELECTION After the capacity of the plant has been established, the question of selection of the type of plant and its location arises. As already mentioned, the major source of electric power in India are coal and water. The state wise position was given in Chapter 1. The diesel engine or gas turbine or a nuclear power plant can also be considered depending upon the availability of oil fuel, natural gas and nuclear fuel. That type of plant is to be selected, which will give the lowest cost of electric generation. If the power plant is to be set up for an industrial complex, then the problem of its location is automatically solved. If this site is near a lake, river or sea then the condensing steam power plant may be economical. If EE158 module -mev 107 water supply is scarce then the possibility of setting up either a steam plant or a diesel plant using spray ponds or cooling towers should be investigated. If the site is near a river with favorable hydro site, then the hydro-plant can be set up. Anyhow before deciding the type of plant, the total annual cost for each must be calculated. If the power plant is to be set up for a regional system, then the factors to be analyzed are more complex. There may be more than one suitable site for the location of a steam plant or a hydro-plant. In studying alternate plant sites, the final selection will depend upon the total annual cost for each case. For a steam plant, the location will be governed by (1) availability and cost of fuel and its transportation (2) water supply (3) cost and nature of land (4) problems of ash handling, noise, smoke and heat if the location near residential houses. For the hydro-plants, the factors to be considered will be topography, geology and hydrology of the site. All the above factors have been discussed in previous chapters. Equipment Selection Once the type of the power plant is selected, the size of the equipment will depend upon many factors. With all the plants, higher the load factor and larger the size of the plant, higher the working efficiency. Moreover, the cost per unit installed is decreased if the plant size increases, but a large plant will require bigger investment and it will work at low load factor. The following considerations must be taken into account while selecting the equipment for a steam power plant: 1. Operational consideration. 2. Economic considerations. 3. Thermodynamic considerations. Operational Considerations: The equipment selected should be reliable i.e. it should be of improved performance and design. The machine should be simple in construction and operation so that unskilled labor can be employed which will be cheaper. The machine should be easy to be dismantled and reassembled and there should be accessibility for the inspection purpose. Operation controls both for when the machine is working and for emergency, should be provided with the machine. Economic Considerations: Economic considerations are studied on the basis of theoretical efficiency based on thermodynamic consideration and depends upon kJ/kW; and practical efficiency based on cost considerations i.e. cost of machine, depreciation, fixed charges and interest, working efficiency at various loads; cost of coal including the cost EE158 module -mev 108 of transportation etc. Cost of machine will depend upon the size of the machine, condition of temperature and pressure and the year of manufacture. The cost per kW, floor space per kW and volume of building per kW decreases as the size of the machine is increased. Also the working efficiency increases with the increase in size. However, there will be more fixed charges with bigger size of the equipment. The capital cost per kW is reduced and the thermal efficiency is increased with the increase in the unit size. If the plant is small, then there is no need to go in for higher temperature and pressure, condition plant as it would prove uneconomical. It is safe practice not to purchase the latest design of the machine. The type of the steam turbine will depend upon its use. For industries, back pressure or non-condensing and extraction or condensing machine may be used but for power undertaking it is always better to use a condensing unit. Thermodynamic Considerations: Following are the thermodynamic considerations for the plant design: 1. Superheating of steam. 2. Use of high temperature and pressure. 3. Heat cycles. Superheating is invariably practiced in modern power plants for the following reasons: (a) more work can be obtained due to greater specific volume and heat content of the steam (b) smaller plants (c) cost saving due to decreased rate of steam. But the limitation of the superheating is the decreased strength of steel at high temperature which makes it necessary to use special alloy steels for super heater tubing when temperature of steam exceeds about 425oC which makes the plant costlier. The decrease in the rate of steam as a result of superheating does not mean only a saving in the cost of fuel but it also means smaller size of the unit, saving in space, lesser repairs and maintenance, less power consumption by the auxiliaries and lesser breakdown. High pressures and temperatures are desirable from the thermodynamic considerations but as the pressure increases, the thickness of metal increases with a resulting increase in cost. As the temperature increases, better and the better materials are required resulting in the increase in plant cost. Moreover, high temperatures introduce problems such as thermal expansion, change in structural properties of materials. The effect of heat cycles on the performance of a steam plant has already been discussed in Chapter 6. The economy and the theoretical EE158 module -mev 109 thermal efficiency increase by incorporating heaters and by the reheat cycle. In case of i.e., engines, the cycles become more efficient and the efficiency increases as the comprehension ratio is raised. But higher compression ratios increase the maximum cylinder pressure requiring heavier cylinder walls, pistons, crankshafts etc. and the plant becomes expensive. Regarding the choice between two stroke cycle and four stroke cycle, the former is generally lighter in weight and less expensive and the latter has higher thermal efficiency. Modern i.e. engines incorporate the supercharger to increase the power per unit weight of the engine and at a higher thermal efficiency. But the super charging will be justified only if there is enough increase in power output which will balance the additional cost of the supercharger. The type of fuel available will determine the type of cycle to be adopted. Otto cycle is adopted if a plentiful and cheap supply of gasoline is available and if cheap fuel oil is available then the diesel engine will have favored. The latter case is usually adopted in practice since fuel oil is cheaper than gasoline. A diesel engine usually operates at higher thermal efficiencies as compared to a steam plant of the same capacity. But the maximum unit size of the diesel engine is very much limited as compared to a steam turbine and as the coal is cheaper than fuel oil, the diesel engine plants have not found favor with large regional power systems. They are usually suitable for municipal, commercial, institutional and industrial purposes where steam is not required and water is scarce. The thermal efficiency of a gas turbine increases as the simple cycle is modified by incorporating reheaters, regeneration and intercooling but tis additional expense will increase the cost of the plant. Due to limited unit size, these plants are not suitable for base load power plants and are usually adopted to supply peak loads. COMBINATION OF WIND POWER WITH OTHER SOURCES OF POWER No firm power can be obtained from wind unless there is some form of storage. However, it may be used in combination with steam and hydropower. Combination of Wind and Steam Power: For a power system predominantly supplied by steam power plants, the value of wind power will be only in the fuel saved by its operation in combination with steam plants. For its economic justification the cost of wind energy may be compared with the cost of generating additional units from the existing steam plants. Wind EE158 module -mev 110 power would not be economical beyond the total incremental cost of generating in the steam station. If the steam station is less efficient and operates at low load factors, it might be economical to replace the energy generated by the steam plant by the electric energy generated by wind power. Combination of Wind and Water Power: The firm capacity of an electric system can be increased by operating the wind power in combination with storage hydro-power stations, particularly if there is diversity between the occurrence of rain and wind. During windy periods, the wind energy can be fed into the electric system. During this period, the output of the hydro-stations can be reduced. This will result in the conservation of stored water which can be more beneficially used afterwards. SINKING FUND METHOD OF DEPRECIATION In this method a fixed amount is set aside and put into the reserve fund each year, but the interest earned on this amount collected in the sinking fund plus the interest earned throughout the life of the equipment should be equal to the original value of the equipment minus its salvage value. In this method, the annual amount to be set aside will be less than that calculated by straight line method. Let A = Annual deposit in the reserve fund, rupees S = Sum to be provided at the end of useful life, rupees n = expected useful life, years I = Annual interest earned by the investment of A. Accumulation of first year’s investment will be = A (1 + I) n – 1 Accumulation of second year’s investment will be = A (1 + I) n – 2 Similarly, the accumulation of (n – 2) year’s investment will be = A (1 + I) 2 Accumulation of first year’s investment will be = A (1 + I) n-1 Accumulation of the nth year’s investment will be =A Total accumulation S = A [1 + (1 + I) + (1 + 1)2 + ….. + (1 + I) EE158 module -mev n–2 + (1 + I) n – 1] 111 This is geometrical progression with a common ratio of r = (1 + I) A [(1+I)n – 1] = [(1+I)n – 1] S= A 1+I–1 I I A= S (1+I)n -1 A i.e., the ratio of the annual depreciation to the S capital invested is termed as the depreciation rate. The ratio Example.1 If the expected life of an equipment is 15 years and the interest earned on the capital invested is 5 percent, find the depreciation rate. Solution: Depreciation rate, n = 15 years I = 5% A I = S (1 + I)n – 1 0.05 = (1 + 0.05)15 – 1 = 0.0463 Example.2 The original value of an equipment is Rs. 250,000 and its salvage value at the end of its useful life of 20 years is Rs 25,000. Find the value of the equipment at the end of 10 years of its use by the following methods: (a) Straight line depreciation. EE158 module -mev 112 (b) Sinking fund depreciation, when it is compounded annually at 8%. Solution: (a) Straight line method Original value = Rs. 250,000 Salvage value = Rs. 25,000 Total Depreciation = Rs. 250,000 – Rs. 25,000 = Rs. 225,000 225,000 Depreciation per year = = Rs. 11,250 20 Depreciation at the end of 10 year = Rs. 11,250 x 10 = Rs. 112,500 Value of the equipment at the end of years = Rs. 250,000 – Rs. 112,500 = Rs. 137,500 Ans. (b) Sinking Fund Method Total amount to be provided in the sinking fund end of 20 year = Rs. 250,000 – Rs. 25,000 S = Rs. 11,250 Now I = 8% I 0.08 Annual deposit, A = Rs. S [ ] = Rs. 225,000 [ n (1 + I) – 1 (1 + 1.08)20 – 1 Rs. 225,000 ( 0.08 225,000 x 0.08 = ) = Rs. 4.66 – 1 3.66 Total amount collected at the end of 10 years will be [(1 + I)n – 1] [(1 + 1.08)10 – 1] = Rs. A = Rs. 4,920 I 0.08 Rs. 4,920 [ 2.16 – 1 = ] = Rs. 71,250 0.08 Value of equipment at the end of 10 years = Rs. 250,000 – Rs. 71,250 = Rs. 178,750 EE158 module -mev at the ] = Rs. 4920 113 Lesson 3 Load Determination & Load Graphs The load demand on a power system is governed by the consumers and for a system supplying industrial and domestic consumers, it varies within wide limits. This variation of load can be considered as daily, weekly, monthly or yearly. Typical load curves for a large power system are shown in Fig. 17.3 these curves are for a day and for a year and these show the load demanded by the consumers at any particular time. Such load curves are termed as “Chronological load Curves”. If the ordinates of the chronological load curves are arranged in the descending order of magnitude with the highest ordinates on left, a new type of load known as “load duration curve” is obtained. Fig. 17.4 shows such a curve. If any point is taken on this curve then the abscissa of this point will show the number of hours per year during which the load exceeds the value denoted by its ordinate. Another type of curve is known as “energy load curve” or the “integrated duration curve”. This curve is plotted between the load in kW or MW and the total energy generated in kWh. If any point is taken on this curve, abscissa of this point show the total energy in kWh generated at or below the load given by the ordinate of this point. Such a curve is shown in Fig. 17.5. In Fig. 17.4, the lower part of the curve consisting of the loads which are to be supplied for almost the whole number of hours in a year, represents the “Base Load”, while the upper part, comprising loads which are required for relatively few hours per year, represents “Peak Load”. EE158 module -mev 114 Ideal and Realized Load Curves: From the stand-point of equipment needed and operating routine, the ideal load on a power plant would be one of constant magnitude and steady duration. However, the shape of the actual load curve (more frequently realized) departs far from this ideal, Fig. 17.3. The cost to produce one unit of electric power in the former case would be from ½ to ¾ of that for the latter case, when the load does not remain constant or steady but varies with time. This is because of the lower first cost of the equipment due to simplified control and the elimination of various auxiliaries and regulating devices. Also, the ideal load curve will result in the improved operating conditions with the various plant machines (for example turbine and generators etc.) operating at their best efficiency. The reason behind the shape of the actual realized load curve is that the various users of electric power (industrial, domestic etc.) impose highly variable demands upon the capacity of the plant. Effect of variable Load on Power Plant Design: The characteristics and method of use of a power plant equipment is largely influenced by the extent of variable load on the plant. Supposing the load on the plant increases. This will reduce the rotational speed of the turbo-generator. The governor will come into action operating a steam valve and admitting more steam and increasing the turbine speed to its normal value. This increased amount of steam will have to be supplied by the seam generation. The governor response from load to turbine is quite prompt, but after this point, the governing response will be quite slower. EE158 module -mev 115 The reason is explained as given below: In most automatic combustion control systems, steam pressure variation is the primary signal used. The steam generates with imbalance between heat transfer and steam demand long enough to suffer a slight but definite decrease in steam pressure. The automatic combustion controller must then increase fuel, air and water flow in the proper amount. This will affect the operation of practically every component of auxiliary equipment in the plant. Thus, there is a certain time lag element present in combustion control. Due to this, the combustion control components should be of most efficient design so that they are quick to cope with the variable load demand. Variable load results in fluctuating steam demand. Due to this it becomes very difficult to secure good combustion since efficient combustion requires the coordination of so many various services. Efficient combustion is readily attained under steady steaming conditions. In diesel and hydropower plants, the total governing response is prompt since control is needed only for the prime mover. The variable load requirements also modify the operating characteristics built into equipment. Due to non-steady load on the plant, the equipment cannot operate at the designed load points. Hence for the equipment, a flat topped load efficiency curve is more desirable than the peaked one. Regarding the plant units, if their number and sizes have been selected to fit a known or correctly predicted load curve, then, it may be possible to operate them at or near the pint of maximum efficiency. However, to follow the variable load curve very closely, the total plant capacity has usually to be sub divided into several power units of different sizes. Sometimes, the total plant capacity would be more nearly coincide with the variable load curve, if more units of smaller unit size are employed than a few units of bigger unit size. Also, it will be possible to load the smaller units somewhere near their most efficient operating points. However, it must be kept in mind that as the unit size decreases, the initial cost per kW of capacity increases. Again, duplicate units may not fit the load curve as closely as units of unequal capacities. However, if identical units are installed, there is a saving in the first cost because of the duplication of sizes, dimensions of pipes, foundations, wires insulations etc. and also because spare parts requires are less. EE158 module -mev 116 Effect of Variable Load on Power Plant Operation: In addition to the effect of variable load on power plant design, the variable load conditions impose operation problems also, when the power plant is commissioned. Even though the availability for service of the modern central power plant is very high, usually more than 95%, the public utility plants commonly remain on the “readiness-to-service” bases. Due to this, they must keep certain of their reserve capacity in “readiness-to-service”. This capacity is called “spinning reserve” and represents the equipment standby at normal operating conditions of pressure, speed etc. normally, the spinning reserve should be at least equal to the least unit actively carrying load. This will increase the cost of electric generation per unit (kWh). In a steam power plant, the variable load on electric generation ultimately gets reflected on the variable steam demand on the steam generator and on various other equipment. The operation characteristics of such equipment are not linear with load, so, their operation becomes quite complicated. As the load on electrical supply systems grow, a number of power plants are interconnected to meet the load. The load is divided among various power plants to achieve the utmost economy in the whole system. When the system consist of one base load plant and one or more peak load plants, the load in excess of base load plant capacity is dispatched to the best peak system, all of which are nearly equally efficient, the best load distribution needs thorough study and full knowledge of the system. Coordinate Base Load and Peak Load Power Plants: If the load represented by Fig. 17.4 is to be supplied from one power plant only, then the installed capacity of the plant should be equal to the peak load. Such a plant will be uneconomical since the peak load occurs only for a short period in a year and therefore the capacity equal to the difference of peak load and base load will remain idle for the major part of the year. Hence such a power would not be supplied from a single power station. There would be some station supplying the base load and others, possible of different type, supplying the peak load. One method of meeting this varying load demand is to coordinate the operation of hydro and steam stations. The steam plant capacity and the available water power energy are fitted into load curve. Peak load demand can be conveniently met by hydro-stations, the base load being supplied by the steam power plants. A hydro-station can be started up quickly at any time to meet a sudden emergency. Also the load on hydro-station can be EE158 module -mev 117 reduced more quickly than is possible with steam plant. There are two methods for utilizing the hydro-electric power for supplying the peak load. 1. By storing the natural run-off from a catchment area during hours of light load and employing the water to operate the station at full capacity during periods of peak demand. 2. Pumped storage system. In this water is pumped into a high level reservoir at off-peak periods, and is utilized to drive the turbines and generators at the time of peak demand. The system has already been discussed in Chapter 16. Peak load can also be supplied by diesel engine power plant and gas turbine power plants. Base load stations operate almost continuously i.e. at a load factor of about 80%. They are shut down only for a small periods for maintenance and overhaul. The load factor of peak load plants is very low, e.g. 5 to 15%, since they operate only for a small period in a day, week, month or a year. As already discussed the cost supplying the electric energy may be divided into two parts: (a) Fixed cost, which mainly consists of the interest on the capital cost and depreciation. It is independent of the amount of electric energy actually supplied. It is however, approximately proportional to the capacity of plant installed i.e. proportional to kW. (b) Running cost, which depends upon the actual energy generated, i.e. it is proportional to the kWh. Since the electric power plants are very expensive to install, therefore, it is desirable and even essential to generate as much energy as possible in order to spread the fixed cost over the highest possible number of units (kWh) supplied. Therefore, the plants should run at a high load factor and then the cost per unit will be minimum. If the plant is idle for most of the period, it will generate only a small number of units and hence the fixed charges will be shared by small number of units resulting in high cost per unit supplied. Therefore, the load factor has a very important effect on the cost of the electric energy supplied from a power plant. Hence, the base load are cheaper to be supplied whereas the peak load units are expensive to produce. A base load power station should have the highest possible efficiency. For peak load plants, since the units to be generated are small, efficiency is not of much importance of course, the capital cost should be minimum since it is to be distributed over the small number of units supplied or generated. Since a peak load plant may have to be started once or perhaps twice in a day and possibly for some unexpected EE158 module -mev 118 emergency condition, it should be capable of quick starting and quick load pick up. Significance of Various Factors (a) Load Factor: High load factor is a desirable quantity. Higher load factor means greater average load, resulting in greater number of power units generated for a given maximum demand. Thus, fixed cost, which is proportional to the maximum demand, can be distributed over a greater number of units (kWh) supplied. This will lower the overall cost of the supply of electric energy. (b) Diversity Factor: High diversity factor (which is always greater than unity) is also a desirable quantity. With a given number of consumers, higher the value of diversity factor, lower will be the maximum demand on the plant, since, Diversity Factor = Sum of the individual maximum demands Maximum demand of the total group So, the capacity of the plant will be smaller, resulting in lower fixed charges. (c) Plant Capacity Factor: Since the load and diversity factors are not involved with ‘reserve capacity’ of the power plant, a factor is needed which will measure the reserve, likewise the degree of utilization of the installed equipment. For this, the factor “Plant factor, capacity factor of Plant capacity factor” is defined as, Plant Capacity Factor = Actual Energy Produced Maximum possible energy that might have been produced during the same period Thus, the annual plant capacity factor will be, = Annual kWh produced Plant capacity (kW) x hours of the year The difference between load and capacity factors is an indication of reserve capacity. (d) Plant Use Factor: This is a modification of plant capacity factor in that only the actual number of hours that the plant was in operation are used. Thus, annual plant use factor is, EE158 module -mev 119 = Annual kWh produced Plant capacity (kW) x number of hours of plant operation Example.3. Explain a method of constructing a load duration curve using a load curve. The following data were collected from the daily load curves of a power system during a year. Load (kW) 15,000 12,000 and over 10,000 and over 8,000 and over 6,000 and over 4,000 and over 2,000 and over Duration (Hrs) 87 876 1752 2628 4380 7000 8760 Construct annual load duration curve and find load factor of system (Assume a smooth curve). Solution: The load duration curve is shown in Fig. 17.6 To get average demand find energy consumed or area under curve first. To find energy used: Energy consumed = Area under the curve = Area 1 + Area 2 + … + Area 12 = 500 x 114.8 MW hrs Energy consumed in one year Average demand = No. of hrs. in a year 500 x 114.8 = = 6.45 MW 365 x 24 Max. demand = 15 MW 6.54 Load factor of system = X 100 = 4.36% 15 To draw Load Duration Curve Using Load Curve: This can be best explained by considering an example. Now take the load curve shown in Fig. 17.7 and 17.8. EE158 module -mev 120 At various ordinates draw horizontal lines (show dotted) and note the length over which the curve cuts these lines. Then at corresponding ordinates on load duration side for the abscissa equal in length to the total no. of lines cut by load curve. 3 cases have been illustrated. After obtaining a no. of points like this draw a smooth curve through them. This gives the load duration curve. EE158 module -mev 121 Example.4. Draw the chronological daily load curve and the load duration curve from the following observation. If the capacity of the plant serving this load is 100 MW, find the load factor and the plant utilization factor. The changes in load are linear. Time 12 P.M. 2 A.M. 6 8 12 M Load, MW 20 10 10 50 50 Time 12:30 P.M. 1 5 6 12 Load, MW 40 50 50 70 20 Solution: The chronological load curve and the load duration curve are shown in Fig. 17.9. From the load duration curve, the average load can be estimated. Lave = Average load for the period = E/h where E = total energy in load curve for period h = total number of hours in period Lave comes out to be about 42.8 MW Lave Load factor = Lmax Now Lmax = 70 MW 42.8 Load factor = = 61.14% 70 L Utilization factor = max Cap where Cap = Rated capacity of the plant Utilization 70 = 70% factor = 100 EE158 module -mev 122 Example.5. The yearly duration curve of a certain plant can be considered as a straight line from 20 MW to 3 MW. To meet this load, three turbine-generator units, two rated at 10 MW each and one at 5 MW, are installed. Determine: (a) Installed capacity (b) Plant factor (c) Maximum demand (d) Load factor (e) Utilization factor. Solution: The load duration curve is shown below in Fig. 17.10. (a) Installed capacity = 2 x 10 + 5 = 25 MW (b) Plant factor, capacity factor or use factor Lave = Cap 8760 x 3 + ½ x 8760 E x 17 Now Lave = = = 11.5 MW h 8760 Plant 11.5 = 46% factor = 25 (c) Maximum demand = 20 MW (d) 11.5 Load factor Lave = = 57.5% = Lmax 20 (e) Utilization factor Lmax 20 = = 80% = Cap 25 Example.6. A residential consumer has 10 lamps of 40 W each connected his residence. His demand is Mid-night to 5 AM ……………….. 40 W EE158 module -mev 123 5 AM to 6 PM 6 PM to 7 PM 7 PM to 9 PM 9 PM to 12 mid-night ……………….. ……………….. ……………….. ……………….. no load 320 W 360 W 160 W (a) Plot the load curve (b) Find Average load (c) Max. load (d) Load Factor (e) Energy consumption during day. Solution: (a) The load curve: It is drawn in Fig. 17.11. (b) Average load = = Energy consumed in 24 hrs. = 24 hrs. 40 x 5 + 320 x 1 + 2 x 360 + 3 x 160 24 Area of load curve 24 = (c) Max. load = 360 W Average demand 71.7 x 100 Load factor = = 360 Maximum = demand (d) E Energy consumed during one day = Average load x 24 hrs = 71.7 x 24 watt hrs = 71.7 x 24 x 10-3 kWh = 1.72 kWh EE158 module -mev 1720 = 71.7 W 24 717 360 = 19.9% 124 Example 7. A power system has the following loads 1. Residential lighting load Maximum demand = 1000 kW L.F. = 20% Diversity between = 1.3 consumers 2. Commercial load Maximum demand = 2000 kW L.F. = 30% Diversity between consumers = 1.1 3. Industrial Load Maximum demand = 5000 kW L.F. = 0% Diversity between consumers = 1.2 Overall diversity factor may be taken as 1.4. Find: (a) Max. demand on system. (b) Daily energy consumption of each type of load and total energy consumption. (c) Overall load factor. (d) Connected loads of each type assuming that demand factor for each is 100%. Solution: (a) Group diversity factor Sum of individual max. demands = Actual max. demand of the group Maximum demand of the system 1000 + 2000+ 5000 8000 = = = 5628 kW 1.4 1.4 (b) Average demand Average demand = Maximum demand x L.F. = 1000 x 0.2 + 2000 x 0.3 + 5000 x 0.8 = 4800 kW Daily energy consumption = 4800 x 24 = 115,200 kWH (c) Overall L.F. = (d) EE158 module -mev Average Demand Maximum Demand 4800 = 5628 = 85.3% 125 Maximum demand Demand factor Now maximum demand = 1000 x 1.3 + 2000 x 1.1 + 5000 x 1.2 = 1300 + 2200 +600 = 9500 kW Connected load = 9500 x 1 = 9500 kW Connected load = Example.8. A consumer has following connected load: 10 lamps of 60 W each 2 heaters of 1000 W each Max. demand = 1500 W. on the average he uses 8 lamps, for 5 hrs a day and each heater for 3 hrs a day. Find his average demand, load factor and monthly energy consumption. Solution: (a) Average demand: Daily energy consumption = (60 x 8 x 5 + 2 x 1000 x 3) 10-3 x 1.2 = (2400 + 6000) 10-3 kW = 8.4 kW This is for 24 hrs. 8400 Average Demand = = 350 watts 24 (b) Load factor = Ave. Demand Max Demand (c) Monthly energy consumption 350 x 100 1500 = = 23.3% = Daily energy consumption x 30 = 8.4 x 30 = 252 kWh Example 9. The following data is given for a steam power plant: Maximum demand 25,000 kW Load factor 40% Coal consumption 0.86 kg per kWh Boiler efficiency 85% Turbine efficiency 90% Price of coal Rs. 55 per tonne Determine: 1. Thermal efficiency of the station. 2. Coal bill of the station for one year. Solution: Thermal efficiency of the plant = Boiler efficiency x Turbine efficiency = 0.85 x 0.90 = 76.6% Now average demand on the station EE158 module -mev = Maximum demand x load factor 126 = 25000 x 0.4 = 10000 kW Energy generated per year = 1000 x 8760 kWh Coal consumption = 1000 x 8760 x 0.86 N per year Coal bill per year = 10000 x 8760 x 0.86 x 55 1000 = Rs. 4,143,480 Example.10. The load shown by the load duration curve, Fig. 17.12, is to be carried by a base-load station having a capacity of 18000 kW and a standby station having a capacity of 20000 kW. Determine the load factors, use factors and capacity factors of the two station. Solution: The load shown by the shaded area is to be met by the standby station. The remaining load is to be met by the base-load station. From the curve, Annual standby output = 7,350,000 kWh Annual base load station output = 101,350,000 kWh Peak load on standby station = 30,000 – 18,000 = 12,000 kW Time during which standby station was in use = 2190 hours (from the figure) 1. Standby Station: Average load 7350000 Load factor = = = 0.2797 Maximum load 12000 x 2190 Units of electricity generated per year Use factor = Capacity of the plant (units generated during actual use) 7350000 = = 0.167 20000 x 2190 Average load or units generated per year Capacity factor = Capacity of plant (units) per year EE158 module -mev 127 = 7350000 20000 x 8760 = 0.043 2. Base Load Station: Load factor = 101350000 18000 x 8760 = 0.642 Since the plant works throughout the year, the use factor and the capacity factor will also be 0.642. Example 11. A 10,000 kW steam plant is erected at a cost of Rs. 2000 per kW. Assume that the bonds in the amount of the total cost were sold. They are to mature in 15 years which is also the estimated life of the plant. Salvage value is estimated at 5% of the first cost. Interest on bonds is 4%, on sinking fund deposit 3 ½ %. Determine the amount of annual payment on the investment and also the sinking fund accumulation after 5 years. Solution: In equation (17.1), the ratio A/S is frequently referred to as the “sinking fund factor” which will be I = (1 + I)n – 1 I = 3.5%, n = 15 years 0.035 = 0.052 (1.035)15 – 1 Sinking fund annual payment, A = S x 0.052 Sinking fund factor = S = 10000 x 2000 x 0.95 A = 10000 x 2000 x 0.95 x 0.052 = Rs. 988000 Annual interest payment = 0.4 x 10000 x 2000 = Rs. 800000 Total = Rs. 1788000 0.035 (1.035)15 – 1 A Accumulated amount after 5 years, S = 0.186 988000 = 0.186 Now five year accumulation factor = EE158 module -mev = 0.186 = Rs. 5311828 128 In module II, you have learned about the introduction to power plant’s load economics and analysis through graphs that shows emphasis on the relavance on its usage. There are three lessons in module II. Lesson 1 deals with the additional resources of energy that could be very useful to society and our environment, just like solar energy that is fastly growing in today’s scenarios of renewable energy. While Lesson 2 deals with the introduction to power plant economics that clearly illustrates the different power plants usage as to its, dependability, fuel consumption, number of hours of operation that will show its power plant demand, utilization factors, etc., Lesson 3 deals with load determination and load graphs, that will illustrate the learner the different usage of power plants as to its efficiency, demand and as to how it operates without any shutdown or interruptions that can cause its losses and dependability. Congratulations! You have just finished the course. now you are ready to evaluate how much you have benefited from your reading by answering the summative test. Good Luck!!! -mev SUMMATIVE TEST Solve for the following unknown parameters 1. The annual peak load on a 15,000-kW power plant is 10,500 kW. Two substations are supplied by this plant. Annual energy dispatched through substation A is 27,500,000 kW-hr with a peak load at 8,900 kW, 16,500,000 are sent through substation B with a peak load at 6,650 kW. Neglect line losses. Find the diversity factor between substations and capacity factor of the power plant. a. b. 1.48, 0.446 1.48, 0.335 c. 1.75, 0.335 d. 1.75, 0.446 2. What is the annual capacity factor of the plant if the annual energy produced in a 150 MW power plant is 500,000,000 kW-hrs? a. 38.05 % c. 56.785 % b. 44.04 % d. 34.44 % EE158 module -mev 129 3. Calculate the use factor of a power plant if the capacity factor is 35% and it operates 8000 hrs .during the year? a. b. 38.325 % 33.825 % c. 35.823 % d. 32.538 % 4. A 75 MW power plant has an average load of 35,000 kW and a load factor of 65%. Find the reserve over peak. a. b. 21.15 MW 23.41 MW c. 25.38 MW d. 18.75 MW II. Define the following according to its function and uses. 1.Reserve Capacity 2. Use Factor 3. Capacity Factor 4. Demand Factor 5. Annual Load Factor? EE158 module -mev