Basic Modules for the Course of Intro to Electric Power Systems

7/29/2019 Basic Modules for the Course of Intro to Electric Power Systems

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Chapter 1.Introduction to Electrical Energy System

(Gii Thiu Chung V H Thng in Nng)

Electrical energy is produced through an energy conversion process. The electric powersystem is a network of interconnected components which generate electricity byconverting different forms of energy, (potential energy, kinetic energy, or chemicalenergy are the most common forms of energy converted) to electrical energy; andtransmit the electrical energy to load centers to be used by the consumer. The productionand transmission of electricity is relatively efficient and inexpensive, although unlikeother forms of energy, electricity is not easily stored and thus must generally be used as itis being produced.

The electric power system consists of three main subsystems: the generation subsystem,the transmission subsystem, and the distribution subsystem. Electricity is generated atthe generating station by converting a primary source of energy to electrical energy. Thevoltage output of the generators is then stepped-up to appropriate transmission levelsusing a step-up transformer. The transmission subsystem then transmits the power closeto the load centers. The power is then stepped-down to appropriate levels. Thedistribution subsystem then transmits the power close to the customer where it is stepped-down to appropriate levels for use by a residential, industrial, or commercial customer.In this chapter, a brief description of the common methods of converting energy toelectric power, and each power subsystem will be discussed.

1.1 Sources of EnergyElectricity is produced by converting energy from one form to electricity. The processused may be a direct conversion process, where the energy source is converted directly toelectricity. An example of this is solar photovoltaic cells, which converts the energyfound in solar radiation directly to electricity. An indirect conversion process consists ofconverting energy from one form, to an intermediate form, to electricity. Coal-firedgenerating plants are an example of this process, as the chemical energy released as heatby burning the coal is changed to rotating kinetic energy by the steam turbine, and thenthe rotating kinetic energy is converted to electricity. The majority of the electricityproduced today is produced through an indirect energy conversion process. Major

sources of energy for the production of electricity are fossil fuels, hydro energy, solarradiation, and nuclear energy.

Fossil fuels are coal, petroleum, and natural gas. Fossil fuels are a finite, non-renewableresource, and are the primary source for the production of electricity. These fuels areburned to release their chemical energy, which produces heat to power steam turbines.The steam turbines power rotating electric generators, which turn kinetic energy intoelectricity. No energy conversion process converts all the energy present in one form


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completely into the new form. Since the production of electricity from fossil fuelsinvolves several energy conversion steps, the overall efficiency of a fossil fuel powerplant is quite low, somewhere in the range of 40%.

The environmental effects of electricity generation from fossil fuels are also a concern.Fly-ash is the physical matter left after coal is burned. Since fly-ash is harmful tohumans, it must be disposed in an environmentally safe matter. Combustion of fossilfuels also produces carbon monoxide, carbon dioxide, sulfur dioxide, and nitrous oxidesgases. These are the "greenhouse gases" that contribute to acid rain and global warming.

Nuclear energy, just like fossil fuels, is a finite, non-renewable, energy source that usesan indirect conversion process to produce electricity. In the nuclear power plants, thefission reaction is used to produce electricity. The fission reaction involves the splittingof the nuclei of a heavy element. The heat output from this reaction is used to power asteam turbine, which is used to drive a rotating electric generator, just as in a fossil fuelpower plant. The advantage of nuclear fission compared to fossil fuels is the energycontent of fissionable materials. The energy content of uranium is approximately 1010Btu/kg, which is about one million times the energy content of fossil fuels. Thedisadvantage of nuclear fission is the environmental cost. After fission, the nuclear fuel,the reactor vessel where fission occurs, and the steam pipes are highly radioactive. Also,plant failures can lead to the release of radioactive steam into the atmosphere.

Solar radiation includes energy used directly as intercepted solar radiation, or indirectlyas wind and hydropower. Solar radiation is a renewable energy source. The averageincident power at the earth's surface is 182 W/m2, which corresponds to a daily averageenergy of 4.4 kWh/m2. Direct use of solar power includes active types involvingphotovoltaic cells, and passive types using radiation to heat solar collectors. Photovoltaiccells directly convert sunlight into electricity. Solar collectors are normally incorporatedinto a solar thermal system, converting sunlight into heat for various forms of use,including space heating, water heating, industrial process steam, and electricityproduction. At present, large-scale utilization of solar energy is limited by severalfactors, including the cost of solar cells and solar collector-heat exchanger systems, andby the requirement of an adequate energy storage system to smooth out the dailyvariation. Since solar cells generate direct current, inverting equipment is needed toobtain the desired alternating current for most large-scale operations. Since sunlight isavailable everywhere, use of solar radiation for energy production is not site specific.The best photovoltaic cells have efficiency in the 14% to 17% range. Photovoltaics areattractive from an environmental aspect, since there is no gases or wastes produced fromthe energy conversion process, and the fuel source does not have to be extracted from the

ground. However, the most efficient solar cells use gallium arsenide, which is a toxicmaterial. Solar cells are too new for an understanding of the disposal requirements andcosts involved for worn out cells.

Wind energy is a form of indirect use of solar radiation. Solar radiation produces windby heating the air. During the day, the air over land is heated much faster than air overwater bodies because the land absorbs much less sunlight, and the evaporation is less.The heated air over land expands, becomes lighter, and rises. The cooler, heavier air over


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large water bodies moves in to replace the lighter and warmer air, creating a horizontalmotion of air. During the night, since the land cools faster than water, the cool air movesseaward to replace the warm air that rises from the surface of the water. Wind energy toelectricity is an indirect energy conversion process, because turbine type wind generatorstransform the kinetic energy of the wind into rotary-shaft motion and, in turn, to electrical

energy. Since most wind turbine generators require a sustained wind speed of 20km/hour, the location of the wind turbine generator is important, as a consistent adequatewind velocity must be present. This means that bulk wind generators are limited tospecific sites, generally in coastal areas or mountain passes. Wind turbines have amaximum possible efficiency of 59.3%, with a more common efficiency of around 40%.The significant environmental problems associated with wind turbines are noise,aesthetics, and interaction with birds.

Hydropower is also an indirect means of using solar power to produce electricity, sincehydropower uses the stream-flow part of the hydrological cycle. In a hydropower plant,the potential energy of a mass of water in a reservoir a distance above the stream bed is

converted to kinetic energy by flowing through a hydraulic turbine. The resulting kineticenergy of the turbine drives an electric generator. Hydropower is available where ever asuitable site exists having enough stream flow, potential drop, and area. Industrializednations contain about 30% of all hydropower potential, and are responsible for about 80%of all electricity produced from hydro. Asia accounts for 30% of hydropower potential,and produces only 7% of electricity such produced. Africa accounts for 20% ofhydropower potential, yet produces only 2% of electricity such produced. Hydropower isvery attractive because it is a non-polluting renewable resource, but it can be verydisruptive environmentally. The dam and reservoir effect the normal ecology of thestream and the surrounding habitat by altering water use, changing natural water flowcycles of the stream, and taking up land area for the reservoir. New dams normally

requires the relocation of people and buildings.

Tidal energy uses the tidal flow of oceans to run a hydropower plant to produceelectricity. Basically, a dam encloses a tidal pool. The tidal pool fills during periods ofhigh tide, then empties during periods of low tide. The water flow into and out of thepool drives a reversible hydraulic turbine. Since the turbine is reversible, the flow ofwater into and out of the bay may be used to produce electricity. Therefore, tidal poweris available twice during each 12h 25min tidal period. The ideal sites for tidal energyhave a large difference in tides.

1.2 Electromechanical Energy Conversion

The most common method for bulk power generation is by rotary generators located inelectric power stations. These generators are electromechanical energy converters, alsoknown as electric machines. In practice, a mechanical prime mover coupled to thegenerator rotates the electrical conductors constituting the generator windings in amagnetic field, inducing a voltage in the generator windings. These windings supplyelectrical load on the generator. Conversely, if a current carrying conductor is placed in amagnetic field, the conductor experiences a force according to Ampere's law. In general,


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electric machines are reversible, and capable of operating both as generators and motors.There are three major types of rotating electric machines: dc commutator, induction, andsynchronous machines.

Electric

Energy

Mechanical

EnergyEnergy(fossil, hydro,wind, nuclear,

etc.)

PrimeMover

ElectricGenerator

Figure 1-1 Prime Mover Driving an Electric Generator

The source of mechanical energy for a rotating electric generator is known as the primemover. The prime mover is directly coupled to the generator. Energy sources for primemovers are thermal, hydro, and wind. The prime movers normally are turbines, but somethermal units use internal-combustion engines. A turbine is mechanical device that isforced to rotate by the pressure of a gas (such as steam for thermal units or air for windunits) or fluid (such as water for hydro units).

An electromechanical energy converter converts mechanical energy into electricalenergy, and vice versa. A generator converts energy from mechanical to electrical form,and modulates in response to an electric signal. A motor converts energy from electricalto mechanical form, and modulates in response to an electrical signal derived frommechanical speed. Rotating machines, if lossless, operate on the principle ofelectromechanical power equivalence as given by

Pmech=Mwm = vi=Pelectrical

Where: M is mechanical torque (N-m), wm is mechanical angular velocity (rad/s), v is

instantaneous electrical volts (volts), and i is instantaneous electrical current (amperes).Electric generators are governed by Faraday's law of electromagnetic induction: anelectromotive force (emf) is induced in a conductor "cutting" magnetic lines of flux.Specifically if a conductor of length l(m) moves with a velocity u in a uniform magneticfieldB (tesla), such that l, u, andB are mutually perpendicular, then the induced emf inthe conductor is given byE=Blu.

Example Problem1.1.An ideal energy converter develops 500 N-m of torque while running at 3000 rpm. If

the input voltage is 1000 volts, determine the input current for this generator.

amps

x

v

Mi

viM

m

m

08.1571000

60

23000500

=

==

=

1.2. Calculate the power output of the ideal energy converter.


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kWx

Power

PowerM m

08.15760

23000500 =

=

=

1.3 Power System Load

An electric load (or demand) is the power requirement of any device or equipment thatconverts electric energy into light, heat, or mechanical energy. The power system load isthe total of all such loads connected to the system. As such, the load is never constant,varying daily, weekly, monthly, and yearly as loads are added or subtracted from thepower system. The minimum system load for a given period is called the base load. Themaximum system load for a given period is known as the peak load or peak demand. Thepeak demand is usually quite short in duration. The operation of generation plants mustbe closely coordinated with the load demands to ensure that enough generation capacityis on line. On weekdays, the base load generally begins increasing at about 5:00 a.m.,and hits peak load around 7:00 p.m. Maximum yearly peak loads generally occur during

the summer.

As more people and businesses connect to an electric system, the amount of load on thesystem will increase. Load forecasting is performed to ensure that power systemgenerating capacity will be adequate to meet these future load demands. Power stationstake years to build, so it is necessary to plan well in advance. One important part of loadforecasting is the idea of the load growth rate. This is the estimated rate at which load onthe power system will increase. It is generally based on historical data. The growth rateof the system loadL is mathematically represented by

aLdt

dL=

where a is the constant of proportionality, also known as the per-unit growth rate. Thesolution to this equation is written as

ateLL 0=

whereL0 is the value ofL at t= 0. At any two values of time, t1 andt2, the ratio of thecorrespondingL1 andL2 is

)(

1

2 12 ttaeL

L =

This equation may be used to determine the time tksuch thatL2 = kL1 andt2-t1 = tk, givenby

a

ktk

ln=

When talking about the growth rate of a quantity, the term "doubling time" is often used.The doubling time is the period necessary to double the initial value of loadL, given aconstant value ofa.

aat

693.02ln2 ==

Doubling time is used to describe how long it will take, at a constant growth rate, to usetwice what is currently used. For example, assuming a present peak energy demand of


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10000 MW, and a 10 percent growth rate in the peak demand, the doubling time forenergy demand is 6.93 years. This means that peak energy demand in about 7 years willequal 20 GW, or twice the current peak demand. Obviously, setting a steady growth ratefor the use of any quantity is unrealistic, as the growth rate depends on many factors.

1.4 Environmental Impact of Electricity Generation andTransmission

All energy conversion methods used to produce electricity have some environmentalimpact. The impact may have an active effect like the emission of airborne pollutants, ormay have a passive effect like aesthetics or habitat modification. Even methodsconsidered environmentally friendly, like wind, solar, and hydro, have some impact onthe environment. Not only does the production of electricity have an environmentalimpact, but the transmission of electricity, with concerns over electromagnetic fields,aesthetics, and land use, has an impact as well.

The whole cycle of electricity generation must be considered when looking at theenvironmental impact. This includes the production and transportation of fuel for theconversion process. This is especially true of fossil fuel and nuclear power plants, whichuse large quantities of fuel taken from the earth.

Fossil fuel power plants generally have the most widespread effect on the environment,as the combustion process produces airborne pollutants that spread over a wide area.Nuclear power plants have the most potentially dangerous effect. An operating accidentat a nuclear station could allow a large release of radioactive particles to occur. Solar,hydro, and wind power plants generally have a small effect on the environment.

Fossil Fuel Power PlantsFossil fuel power plants produce environmental problems including land and water use,air emissions, thermal releases, climatic and visual impacts from cooling towers, solidwaste disposal, ash disposal (for coal), and noise. Due to the need for large amounts ofsteam, plants can have a great effect on water use. For example, a typical 500 MW coalfired power plant uses 25 x 109 l/GW-year of water, which must be taken from a watersource, and then cooled to return to the water source with as little environmental effect aspossible. The biggest effect fossil fuel plants have overall is the emission of airpollutants, particularly SOX, NOX, CO, CO2, and hydrocarbons. CO, CO2 andhydrocarbons are the greenhouse gases, possibly responsible for global warming. SOXand NOX can produce acid when released into the atmosphere, leading to the production

of acid rain. Table 1.1 list approximate amounts of airborne pollutants produced.Generally, air emissions are controlled by the use of scrubbers and precipitators located atthe plant.

Plant Type CO NOX SO2 CO2

Coal 0.11 3.54 9.26 1090


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Oil 0.19 2.02 5.08 781

Gas 0.20 2.32 0.004 490

Table 1.1 Power Plant Emissions (g/kWh)

HydroThe use of hydropower to produce electricity can have both positive and negative effectson the environment. At some sites, a dam may help with flood control, flow regulation,or the reservoir may provide recreational opportunities. At other sites, the dam may haveadverse effects on the hydrological cycle, water quality of the stream, stream ecology,fish migration, and cause the destruction of landscapes and ecosystems. Building newhigh-head dams requires the displacement and compensation of populations. Low-headdams generally have a benign effect on the environment. Dam failures can lead tocatastrophic floods.

Nuclear Power Plants

Nuclear power plants have one environmental issue no other form of electrical powerplant does. An accident at a nuclear power plant may release large amounts ofradioactive particles, possibly resulting in a direct loss of life, and rendering a large landarea immediately around the plant unlivable. The largest regular environmental impact isthe disposal of the high level nuclear waste contained in spent fuel rods, as this wastemust be stored safely for thousands of years. A long term issue is the decommissioningof nuclear power plants. Decommissioning is shutting down a nuclear plant after itsoperational life is over. At this point the entire reactor vessel becomes a high levelradioactive waste that must be disposed. The current methods of decommissioning a

plant are to completely remove and dispose of all radioactive components, to entomb thereactor in concrete, or simply to shut the plant down and restrict access until theradioactivity dies out.

Transmission of bulk electricity from the generating station to the load uses wiressuspended on large towers, known as transmission lines. Traditionally these lines havebeen viewed only as an aesthetic nuisance that could cause communications interferenceand be a hazard to low flying aircraft. Today, there are other issues considered about theeffect of transmission lines on the environment. Greater concern is placed on the effectof the lines on the natural habitat. The major new issue is the effect of electromagneticfields (EMFs) on human health.

1.5 The Generation Subsystem

The electric power system consists of three main subsystems: the generation subsystem,the transmission subsystem, and the distribution subsystem. Electricity is generated atthe generating station by converting a primary source of energy to electrical energy.


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Alternating current synchronous generators are the main source of electrical energy.These generators convert energy from a primary form to the electrical form. Energy isinput to the generator in the form of mechanical torque from a prime mover or turbine.The turbine in turn is powered by a moving fluid either steam or water. The currentsources of nearly all the electrical energy distributed come from: the conversion of

chemical energy of fossil fuels, kinetic energy of water, and the nuclear fission energy.Some renewable sources of energy like wind, solar, and geothermal are also used. Thesynchronous generators typically range in size from 50 kW to over 1300 MW, andoperate at voltage levels from 480 V to 25 kV. The output delivered by the generators isa balanced set of three phase ac voltages.

TransmissionSubstation

Sub-TransmissionSubstation

Extra-High-VoltageTransmission Substation

TransmissionLines

Turbine

Generator

Step-up

Transformer

500 kV

220 kV

DistributionTransformer

110 kV

0.4 kV

Distribution Subsystem (35, 22,10, 6 kV)

T

Figure 1.2 Example of a Power System in VietnamIn fossil fuel plants, the source of heat energy is due to the combustion of the fossil fuel.When coal is used, crushed coal is conveyed to a pulverizer where the fuel is ground to aconsistency of face powder. The pulverized coal is then fed to burners where itundergoes combustion. The thermal energy produced is then used to heat water in aboiler to produce steam, and stored in the form of internal energy of the steam. Thesteam flows into the turbine where this internal energy is released as mechanical energy.


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Alternatively, natural gas or petroleum can also be combusted to produce the source ofheat energy.

The turbine is a device used to convert the stored energy of high-pressure and high-temperature steam into rotational energy. The steam fed to the turbine is passed through

a series of stages, each of which consist of stationary blades and moving blades attachedto a rotor. The stationary blades accelerate the steam to high velocity, and this kineticenergy is converted into shaft rotation by the moving blades. The turbine then providesthe rotational energy to the synchronous generator.

Combustion turbines are an important part of the utility system generation mix. Thecombustion turbine has characteristics that make it attractive both for generation to meetpeak loads and for base-load operation. Combined cycle plants using combustionturbines, heat recovery steam generators, and a steam turbine have the advantage ofsuperior heat rates.

In a simple cycle combustion turbine, air at atmospheric pressure and temperature iscompressed to raise it temperature and pressure. The pressurized air is partly combustedwith fuel and passed to the turbine. The fuel used could be gaseous, or liquid fuels, orcombinations of different fuels.

The gases exhausting from the combustion turbine are quite hot. In a combined cycleplant, these gases are used to generate steam for a steam turbine. This systemincorporates some of the features of a conventional fossil-fired unit.

With the advent of energy consciousness, and the drive to improve energy efficiency,cogeneration has become an important source of energy production. Electricity is

produced sequentially with thermal energy. Two different schemes are normally used inco-generation. In the topping system electricity is first produced ("on top"). Waste heatis used in some industrial process or to provide heating. In the bottoming system, theheat energy is first used in the process. Waste heat from the process is used to produceelectricity ("on bottom").

In a hydroelectric plant, the electrical energy is derived from falling water. Water istypically stored by a dam. It is then delivered using pen-stocks or pipes to the hydraulicturbine, which in turn supplies mechanical energy to the generator. Hydraulic turbinesare of two basic types, impulse and reaction. The impulse turbine is also known as thePelton wheel. It is used in high-head plants, where the head (fall of water) is 305 metersor more. The impulse turbine usually has a horizontal shaft. In the Pelton wheel, kineticenergy is derived by converting the fall of water into one or more high-velocity jetslocated around the periphery of the wheel and directed into spoon-shaped buckets.Reaction type turbines are of two general types, Francis and propeller. In both types thewater passages are completely filled with water. The energy which drives the wheel is inboth kinetic and "pressure head" form.


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The output of the generators is fed to a unit transformer which steps-up the voltage to theappropriate transmission level.

1.6 The Transmission Subsystem

The electrical power produced at generating stations is transported or moved to areasclose to load locations by the transmission subsystem. Transmission lines are a majorcomponent of this subsystem.

Transmission lines could be underground or overhead. The latter kind are more common.Transmission line voltages range from 110 kV to 765 kV. Overhead transmission linesspan long distances and use bare conductors. As a result, they must be connected byinsulated mechanical connections to towers or poles which support them. A transmissionline consists of conductors, insulators, mechanical supports, and usually shield wires.Aluminum conductor, steel reinforced (ACSR), conductors are most commonly used for

high-voltage transmission lines. Compared to other conductor material, they have lowcost and high strength-to-weight ratio. Suspension type insulators are used. The materialmost commonly used for insulators is porcelain. New polymer materials are now beingused for making insulators. Shield wires are used to protect the energized conductorsfrom lightning.

Underground transmission is mainly used in urban areas. Underground lines cost on anaverage, about 8-15 times more than overhead lines. Underground cables also have thedisadvantage of poor accessibility.

Transformers are an important part of the transmission subsystem. The transformer is a

static device which transfers electrical energy from one circuit magnetically coupled withanother. In doing so, they can transform voltage levels in the magnetically coupledcircuits. Different types of transformers are used in the transmission subsystem. Powertransformers are used to step-up, or step-down voltages. On load tap changingtransformers are used to regulate voltages. Auto-transformers are used for most tiesbetween moderate-voltage and high voltage transmission, high-voltage and extra-highvoltage circuits. Phase shifting transformers are used to regulate the flow of real power.

Switchgear constitutes an essential component of the transmission subsystem. Undernormal operating conditions, it provides the means to perform routine switchingoperations, e.g., disconnecting and isolating various equipment for maintenance,

inspection, or replacement, transferring load, isolating regulators, etc. Under abnormalconditions, switchgear provides the means for automatically isolating parts of the systemin trouble to prevent damage and to localize the problem. The main components of theswitchgear include, circuit breakers, disconnecting switches, fuse, instrumenttransformers, buses and connections, supporting insulators, protective and control relays,and control switches.


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A substation houses all equipment involved in the switching or regulating of electricity.Substations can be large or small. Control can be automatic or manual.

1.7 The Distribution Subsystem

The sub-transmission designates the circuits, which deliver energy from thetransmission subsystem to the distribution subsystem. Usually the transmissionsubstations supply the sub-transmission system, but it is still referred to as the sub-transmission. Many sub-transmission systems were previously transmission lines. Loadgrowth and demand for more power resulted in the transmission voltage being too low.As a result, in Vietnam, voltages from110 kV down to 35, 22, 10, 6 kV are found in sub-transmission systems.

The distribution subsystem includes the primary circuits and the distribution substationsthat supply them, the distribution transformers, the secondary circuits, including thecomponents all the way up to the entrance of the customer's premises, and the protectiveand control devices. The primary circuit is three-phase and is operated in the 6-35 kVrange. The secondary circuits serve most of the customers at levels of 220 volts single-phase, or 380 volts three-phase.

The various other components used in the distribution subsystem include: distributiontransformers, automatic circuit reclosers, cut-outs (a combination fuse and knife switchused on power poles), surge arresters, separable connectors, capacitors and voltageregulators, and metering.

1.8 Power Industry Structure

The traditional structure of the electric power system was a single company that ownedand operated all of the components of the system. This included the generation,transmission and distribution subsystems along with all communications and controlsystems. This structure is changing dramatically since the deregulation of the industrywas implemented in some industrialised countries in the early 1990s. Since this time,each company has been reorganized to fit the expected structure of the industry within thevery near future. This structure assumes that the companies are segmented intogeneration (GENCO), transmission (TRANSCO), distribution (DISTCOS) andindependent system operator (ISO) companies or independent contract administration(ICA) business units. The GENCOs are responsible for providing (selling) electric

generation according to contracts along with any auxiliary services as required for thecustomer or the transmission grid. The TRANSCOs are responsible for the maintenanceof the transmission equipment to enable the transmission grid to transport electricityaccording to schedules with auxiliary services as required. The DISTCOs are responsiblefor providing (purchasing) electricity along with any auxiliary services to the ultimatecustomer(s). The ICA is responsible for matching the schedules (transmission gridsupport, schedule control and accounting, control and communication facilities andpersonnel) such that the transmission grid can transport the electricity and such that


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auxiliary services are available as required. The ISO is responsible for coordination ofthe maintenance schedules subject to the contracts and operating restrictions of thetransmission system.

Chapter 2.Introduction to Polyphase Networks Analysis

(Phn Tch Mch in Nhiu Pha)

Systems with more than one phase are generally called polyphase. Today, the large-scale generation, transmission, and distribution of electric power is by means of the 3phase ac system; that is, three individual single-phase voltages and currents having a

120 phase relationship to each other and intermingled on three wires (excluding aneutral). The three-phase system has been adopted because it provides for a constantrather than pulsating power flow to motors, and because it is an efficient system as far asthe amount of copper required per kilowatt transmitted. Networks with different numberof phases are used for some specialized applications.

This chapter starts with an elementary discussion of alternating voltages and currents andwith associated energy flows. It ends with an analysis with balanced three-phase circuits.

2.1 Sinusoidal Steady State AnalysisAt the nodes of a power system, the voltage waveform can be assumed to be purelysinusoidal and of constant frequency. In these notes, we deal with the phasorrepresentation of sinusoidal voltages and currents, and use the boldface quantity V andIto indicate these phasors. | V| and |I| designate the magnitudes of the phasors.

2.1.1 Phasor RepresentationThe phasor is a complex number that contains the amplitude and phase angle informationof a sinusoidal function. Using Euler's identity, which relates the exponential function tothe trigonometric function, the phasor concept can be developed.

e =cos j sin (2.1)

Equation (2.1) provides us an alternative way of expressing the cosine and sine function.The cosine function can be represented as the real part of the exponential function, andthe sine function can be represented as the imaginary part of the exponential function asfollows


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cos = Re{e} (2.2)

andsin =Im{e

} (2.3)

Where represents the real part ofand Im represents the imaginary part of.Re

A sinusoidal voltage function (we have chosen to use the cosine function in analyzing thesinusoidal steady state), can be written in the form suggested by Eq. (2.2)

}{Re

}{Re

)cos()(

jtj

m

tj

m

m

eeV

eV

tVv

=

=

+=+ (2.4)

We can move the coefficient Vm inside the argument, and also reverse the order of the

two exponential functions inside the argument without altering the result.

v =Re{Vmee

t} (2.5)

In Eq. (2.5) the coefficient of the exponential et

is a complex number that carries theamplitude and phase angle of the given sinusoidal function. This complex number is bydefinition the phasor representation or phasor transform, of the given sinusoidalfunction. Thus

V = Vme

=P{Vm cos( t + ) } (2.6)

Where the notation P{Vm cos( t + ) } depicts "the phasor transform of )cos( +tVm ."

Hence, the phasor transform P transforms the sinusoidal function from the time domain tothe complex-number domain.

Equation (2.6) is the polar representation of a phasor, we can also obtain the rectangularrepresentation of the phasor as

V = Vm cos +jVmsin (2.7)

2.1.2 Power Calculation in Single-Phase AC CircuitsOur aim is to determine the average power that is either delivered to or absorbed from apair of terminals by a sinusoidal voltage and current. Figure 2.1 depicts the problem.Here, v andi are steady-state sinusoidal signals. With the use of passive sign convention;the power at any instant is

p = vi (2.8)


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_

+

Ieff

CircuitVeff

Figure 2.1 The Basic Calculation to Determine Average Power

The power is measured in watts when the voltage is in volts and the current is in amperes.First we write expressions forv andi

v =Vm cos( t + v) (2.9)

andi = Im cos( t + i) (2.10)

Here, v is the voltage phase angle and i is the current phase angle. Using the instant

when current is passing through a positive maximum as the reference for zero time, andexpressing v andi with respect to this reference we have

v =Vm cos( t + v i)

=Vm cos(t+ )(2.11)

Where = v i

i =Im cos t (2.12)

The angle in these equations is positive for current laggingthe voltage and negative forleadingcurrent. A positive value ofp shows that energy is absorbedat the terminals.Substituting Eqs. (2.11) and (2.12) into Eq. (2.8), the instantaneous power is given by

)cos()cos( ttIVp mm += (2.13)

The average power associated with sinusoidal signals is given by the average of theinstantaneous power over one period

+

=Tt

t

avg pdt

T

p0

0

1(2.14)

where T is the period of the sinusoidal function. Substituting Eq. (2.13) into Eq. (2.14),the average power can be determined. More information, however can be obtained byexpanding Eq. (2.13) using the trigonometric identity

[ ])cos()cos(2

1)cos()cos( BABABA ++=


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Letting A = t+ , andB = t, we obtain from Eq. (2.13)

p =VmIm

2cos +

VmIm

2cos(2t+ ) (2.15)

Expanding the second term on the right-hand side of Eq. (2.15) using the trigonometricidentity cos( A +B) =cos Acos B sin Asin B, we get

p =VmIm

2cos +

VmIm

2cos()cos 2t

VmIm

2sinsin 2t

(2.16)

The average value ofp is given by the first term on the right hand side of Eq. (2.16)because the integral of eithercos2 t orsin 2 tover one time period is zero. Hence, the

average power is

P=VmIm

2cos (2.17)

Pis also referred to as the realoractivepower. The third term on the right hand side ofEq. (2.16), the term containing sin, is alternatively positive and negative and has anaverage value of zero. This component of instantaneous power p is called theinstantaneous reactive power. The maximum value of this pulsating power is designatedas Q, and is calledreactive power. Hence,

Q =VmIm

2sin (2.18)

PandQ carry the same dimension. However, in order to distinguish between real andreactive power, we use the term VArs (volt-ampere reactive) for reactive power. Sincewe have used current as the reference, Q is positive for inductors ( = 90) and negativefor capacitors ( = -90). The angle is referred to as the power factor angle. Thecosine of this angle is called thepower factor. Lagging power factor implies that currentlags voltage. Leading power factor implies that the current leads the voltage.

The average power given by Eq. (2.17) and the reactive power given by Eq. (2.18) can bewritten in terms of effective or rms values

cos

cos22

cos2

effeff

mm

mm

IV

IV

IVP

=

=

=

(2.19)

and, by similar manipulation,


16/67

Q =VeffIeffsin (2.20)

Complex poweris the complex sum of average real power and reactive power, or

S=P+jQ (2.21)

Complex power has the same dimensions as real or reactive power. However, in order todistinguish complex power from real and reactive power, we use the term volt amps.Thus we use volt amps for complex power, watts for average real power, and vars forreactive power. We can think ofP, Q, and |S | as the sides of a right-angled triangle asshown in Figure 2.2.

|S|

Q

P

Figure 2.2 The Power Triangle

Combining Eqs. (2.17), (2.18), and (2.21) we get

S=VmIm

2

cos +jVmIm

2

sin

=VmIm

2[cos +j sin]

=VmIm

2e

j=

1

2VmIm

(2.22)

Using effective values of the sinusoidal voltage and current, Eq. (2.22) becomes

S= VeffIeff (2.23)

Equations (2.22) and (2.23) show that if the phasor current and voltage are known at apair of terminals, the complex power associated with that pair of terminals is either onehalf the product of the voltage and conjugate of the current or the product of the rmsphasor voltage and the conjugate of the rms phasor current.

jQPS +== *2

1VI (2.24)

or


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(2.25)jQPS ffff +== ee IV*

Equations (2.24) and (2.25) have several useful variations. In order to demonstrate thesevariations, we first replace the circuit in the box in Figure 2.1 by equivalent impedance Zas shown in Figure 2.3.

_

+

Ieff

Veff Z

Figure 2.3 General Circuit Replaced with Equivalent Impedance

Expressing the voltage as the product of the current times the impedance, we obtain

ffff Z ee IV = (2.26)

Substituting Eq. (2.26) into Eq. (2.25) yields

jQPXjR

jXR

Z

ZS

ffeff

ff

ff

ffff

+=+=+=

=

=

22

2

2

)(

e

e

e

ee

III

I

II

(2.27)

from which,

RRP mff22

2

1IIe == (2.28)

and

XXQ meff22

2

1II == (2.29)

In Eq. (2.29), X is the reactance of either the equivalent inductance or the equivalentcapacitance of the circuit; it is positive for inductive circuits and negative for capacitivecircuits.

Another useful variation of Eq. (2.25) is obtained by replacing the current with thevoltage divided by the impedance:


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jQPZZ

Sffff

ff+==

=

2

eee

VVV (2.30)

If theZis a pure resistance element,

RP

ff

2

eV

= (2.31)and ifZ is a pure reactive element

XQ

ff

2

eV= (2.32)

Xis positive for an inductor and negative for a capacitor in Eq. (2.32).

2.2 Balanced Three-Phase Circuits

Electric power is supplied by three-phase generators. It is then transformedappropriately using transformers and transmitted in the form of three-phase power exceptat the lowest voltage levels of the distribution system where single phase power is used.

There are two main reasons for using three-phase power. First, the instantaneous powersupplied to motors is constant torque and therefore motors run much smoother. Second,three phase power requires less conductor-cost than single-phase power for the samedelivered power.

Figure 2.5 represents a three-phase (3) circuit. The circuit is said to be a balanced 3

circuit if the impedances are equal and the three voltage source phasors are equal inmagnitude and are out of phase with each other by exactly 120. In discussing three-phase circuits, it is standard practice to refer to the three phases as a, b, and c.Furthermore, the a-phase is almost always used as the reference phase. The threevoltages that comprise the three-phase set are referred to as the a-phase voltage, the b-phase voltage, and the c-phase voltage.

Bb

Nn

Van

N

BC

Vcn Vbn

c

Aa


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Figure 2.4 Three-Phase Balanced Circuit

Since the phase voltages are out of phase by 120, two possible phase relationships canexist between the a-phase voltage and the b- and c- phase voltages. One possibility is

that the b-phase voltage lags the a-phase voltage by 120, in which case the c-phasevoltage must lead the a-phase voltage by 120. This phase relationship is known as theabc, orpositivephase sequence. The other possibility is for the b-phase voltage to leadthe a-phase voltage by 120, in which case the c-phase voltage must lag the a-phasevoltage by 120. This phase sequence is known as the acb, ornegativephase sequence.In phasor notation, the two possible sets of balanced phase voltages are

Va

Vc

VbVb

Va

Vc

Positive Sequence Negative Sequence

Figure 2.5 Phasor Diagram: Three Phase Voltages

o

o

o

120

120

0

+=

=

=

m

m

m

V

V

V

c

b

a

V

V

V

(2.33)

and

(2.34)o

o

o

120

120

0

=

+=

=

m

m

m

V

V

V

c

b

a

V

V

V

The phase sequence of the voltages given by Eq. (2.33) is the abc, or positive

sequence. The phase sequence of the voltages given by Eq. (2.34) is the acb, or negative

sequence. Another important characteristic of a set of balanced three-phase voltages isthat the sum of the voltages is zero. Thus, from either Eq. (2.33) or Eq. (2.34)

0=++ cba VVV (2.35)

This relationship holds for any set of balanced three-phase variables.Components of balanced three-phase circuits can be connected either in a Y-connection


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or a - connection. The phase quantities and the line quantities for these connections arerelated as follows.

2.2.1 Delta Connection

Figure 2.6 illustrates a - connected balanced three-phase load. The relationshipsdeveloped however, can be applied to any component, e.g., generator, transformer, etc.

A

B C

ICA

IAB

IBC

a

b

c

IaA+

+

++

+

VBC

VCAVAB

Z

_

_Z

_

Z

IaA

IbB

Vab

Vbc

_

_

Vca

_

+

Figure 2.6 Delta Connection

In the - circuit, the line-to-line voltage Vab is equal to the phase voltage V . To

demonstrate the relationship between the phase currents and line currents, we assume a

positive phase sequence and let

AB =V

I represent the magnitude of the phase current. Thenselecting IAB arbitrarily as a reference phasor we have

(2.36)o0= IBAI

(2.37)o120= ICBI

and

(2.38)o120+= ICAI

We can express the line currents in terms of the phase currents by direct application of

Kirchoff's current law:

o

oo

303

1200

=

==

I

IICAABA IIIa(2.39)


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o

oo

1503

0120

=

==

I

IIABBCbB III(2.40)

o

oo

903

120120

=

==

I

IIBCCAcC III

(2.41)

Comparing Eqs. (2.39)-(2.41) with Eqs. (2.36)-(2.38) we see that the magnitude of the

line currents is 3 times the magnitude of the phase currents and that the set of linecurrents lags (leads) the set of phase currents by 30 for positive (negative) sequences.

2.2.2 Wye Connection

Figure 2.7 illustrates a Y- connected balanced three-phase load. The relationshipsdeveloped however, can be applied to any component, e.g., generator, transformer, etc.

Vab

+

_

A

IaA_

C B

IAN

ZY

N

+

_VAN

ICNZY

IBNZY+

VCN_

VBN+

_

b

a

c _IcC

IbB

Vca

Vbc

+

+

Figure 2.7 Wye Connection

In the Y-circuit, the line current IaA is equal to the phase current . To

demonstrate the relationship between the line-to-line voltages and the line-to-neutralvoltages, we assume a positive, or abc, sequence and let

IAN =I

Vbe the magnitude of the line

to neutral or phase voltage. We arbitrarily choose the line-to-neutral voltage of a-phase

as the reference. We then have(2.42)o0= VANV

(2.43)o120= VBNV

(2.44)o120+= VCNV


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We can express the line-to-line voltages in terms of the line-to-neutral voltages by directapplication of Kirchoff's voltage law:

oo 303120 === VVVBNANAB VVV (2.45)ooo 903120120 === VVVCNBNBC VVV (2.46)

ooo 15030120 === VVVANCNCA VVV (2.47)

Equations (2.45) - (2.47) reveal that the magnitude of the line-to-line voltage is

3 times the magnitude of the line-to-neutral or phase voltage, and the set of line-to-linevoltages leads (lags) the set of line-to-neutral voltages by 30 for positive (negative)sequences.

2.2.3 Power Calculations in Balanced Three-Phase Circuits

The total power in a three-phase balanced circuit, i.e., power delivered by a three-phasegenerator or absorbed by three-phase load is determined by adding the power in each ofthe three phases. In a balanced circuit this is the same as multiplying the power in anyone phase by 3 since the power is the same in all phases.

If the magnitude V of the voltages to neutral for a Wye-connected circuit is

cnbnanV VVV === (2.48)

and if the magnitude I of the phase current for a Wye-connected circuit is

cnbnanI III === (2.49)

the total three-phase power is cos3 IVP= (2.50)

where is the phase angle difference between the phase current , and the phase

voltage V . IfV

I

L andIL are the magnitudes of line-to line voltage VL and line current

IL, respectively,

LL IIandV

V == 3 (2.51)

Substituting Eq. (2.51) into Eq. (2.50) yields

cos3 LLIVP=(2.52)

The total vars are


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sin3 IVQ =(2.53)

sin3 LLIVQ =2.54)

and the voltamperes of the load are

LLIVQPS

jQPS

322 =+=

+=

(2.55)

Equations (2.50), (2.54), and (2.55) are used for calculating P,Q, andS inbalanced three-phase networks since the quantities usually known are line-to-line voltage,line current, and the power factor,

cos=pf

When we refer to a three-phase system, balanced conditions are assumed unlessotherwise specified, and the terms voltage, current, and power, unless otherwisespecified, are understood to mean line-to-line voltage, line current, and total three-phasepower, respectively.

If the circuit is -connected, the voltage across each phase is the line-to-line voltage, andthe magnitude of the current through each phase is the magnitude of the line current

divided by 3 (see Section 2.1.1), or

3

LL

IIandVV == (2.56)

The total three-phase power is

cos3 IVP=(2.57)

Substituting Eq. (2.56) in Eq. (2.57) we obtain

cos3 LLIVP=(2.58)

which is identical to Eq. (2.52). It follows that Eqs. (2.54) and (2.55) are also validregardless of whether a particular circuit is connected or Y.

2.2.4 Per-Phase Analysis

From the analysis in Sections 2.2.1, and 2.2.2, we observe that in balanced three-phasecircuits the currents and voltages in each phase are equal in magnitude and displaced

from each other by 120. This characteristic results in a simplified procedure to analyzebalanced three-phase circuits. In this procedure it is necessary only to compute results in


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one phase and subsequently predict results in the other phases by using the relationshipthat exists among quantities in the other phases.

Chapter 3.Transformers and Per-Unit Systems

(My Bin p v Hn V Tng i)

A primary function of the transformer is to convert electrical energy at one voltage levelto voltages at another level. The transformation may be to increase voltages or todecrease voltages, depending on the application. Transformers are the essential parts of

most power systems since they are untilised to interconnect different parts oftransmission and distribution power grids that operate at different voltage levels.Transformers also are found in almost all power supplies for small power electronicequipment such as TV sets, laptops, chargers A key application of power transformersis to reduce the current before transmitting electrical energy over long distances throughwires. Most wires have resistance and so dissipate electrical energy at a rate proportionalto the square of the current through the wire. By transforming electrical power to a high-voltage, and therefore low-current form for transmission and back again afterwards,transformers enable the economic transmission of power over long distances.Consequently, transformers have shaped the electricity supply industry, permittinggeneration to be located remotely from points of demand. All but a fraction of the

world's electrical powerhas passed through a series of transformers by the time it reachesthe consumer. In this chapter, we will deal with some simplified models of transformerswithout discussing about electromagnetic elements in details.

3.1 The Ideal Transformer

For analysis of transformers, it is convenient to start by using the ideal transformerrelations. Suppose we consider two coupled coils on a steel core of high magneticpermeability, a simplified model for this transformer can be shown in Figure 3.1.

i2i1

N1 : N2

v1

_

+

_

+

v2

Figure 3.1 Ideal Transformer
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If the flux varies sinusoidally, there will be a sinusoidal voltage generated in each turn ofeach coil according to Faraday. This quantity is called volt-per-turn and occupies animportant role in transformer design. If windings 1 and 2 have N1 and N2 turns,respectively, then:

2

2

1

1

N

v

N

v= (3.1)

One of the ideal transformer relations is that there can be no energy absorbed, stored orlost in the device. Whatever complex power enters one winding must leave the other.Therefore, we have

*22

*11 iviv = (3.2)

and

1

2

2

1

N

N

i

i= (3.3)

The transformer also tends to transform impedances. In Figure 3.3, some impedance is

connected to one side of the ideal transformer. We can find an equivalent impedance 'Z viewed from the other side of the transformer.

Z

I2I1N1 : N2

V1

_

+

Figure 3.2 Impedance Coupling

Noting that

221

2

12 ZIVandI

N

NI ==

The ratio between input voltage and current is:

1

2

2

1

22

1

1

'

1 ZVN

N

VN

N

IZV

===

We may derive the expression of the equivalent impedance viewed from the primary sideof the transformer:

ZN

NZ

2

2

1'

= (3.4)


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From the ideal transformer relations we see that the voltamperes into one winding of atwo-winding transformer must equal the voltamperes out of a second winding. Thevoltampere rating of a two winding transformer is then given as the voltampere rating ofeither winding since the two are equal. In large power transformers the nameplate gives avoltamperes (or kVA or MVA) rating for the device as well as the voltage ratings of the

two windings. The current ratings then follow from these data since S = VI. Smalltransformers, for example, those used in electronic power supplies, are often rated bygiving the voltage and current ratings of each winding, from which the voltampere ratingwould follow if desired.

No mention was made ofpower in the above statements. In a practical transformer therelative phase angle of voltage and current has almost no effect on the voltage and currentcapabilities of the windings, and hence the magnitude ofS is the important factor andhow S is divided intoPandQ is immaterial to the rating.

3.2 Three-Phase TransformerA three-phase transformer is simply three single phase transformers. There are anumber of ways of winding them, and a number of ways of interconnecting them. Oneither side of a transformer connection (i.e. the high voltage and low voltage sides), it ispossible to connect transformers windings either line to neutral (wye), or line to line(delta). Thus we will have four connecting combinations: wye-wye, delta-delta, wye-delta, delta-wye.

Ignoring all the imperfections, connection of transformers in either wye-wye or delta-delta is reasonably easy to understand. On the other hand, the interconnections of wye-delta or delta-wye transformer are a little more complex. Figure 3.3 shows a delta-wyeconnection, in what might be called wiring diagram form. A more schematic (andmore common) form of the same picture is shown in Figure 3.4.

ab YaYbYcc

Figure 3.3 Delta-Wye Transformer Connection

Assume that N and NY are numbers of turns. If the three individual transformers areconsidered to be ideal, the following voltage and current constraints exist:


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( )

= baY

aY vvN

Nv

( )

= cbY

bY vvN

Nv

(

= acYcY vvNNv ) (3.5)

( )cYaYY

a iiN

Ni =

( )aYbYY

b iiN

Ni =

( )bYcYY

c iiN

Ni =

where each of the voltages are line-neutral and the currents are in the lines at thetransformer terminals.

Now, consider what happens if a -Y transformer is connected to a balanced three-phasevoltage source, so that:

( )

=

=

=

+

3

2

3

2

Re

Re

Re

tj

c

tj

b

tj

a

eVv

eVv

eVv

Where: Re denotes the real part; V is the line-neutral (phase) voltage amplitude, anunderline beneath the variable means it is a vector. Then the complex amplitudes on thewye side are:

63

2

31

jY

jY

aY eVN

Ne

N

NV

=

=

23

2

3

2

3

jY

jjY

bY eVN

Nee

N

NV

=

=

6

5

3

2

31

j

Yj

YcY eVN

N

eN

N

V =

=

Two observations should be made here: The ratio of voltages (that is, the ration of either line-line or line-neutral) is

different from the turns ratio by a factor of 3

All wye side voltages are shifted in phase by 30o with respect to the delta sidevoltage.


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It can be proved that impedances transform across transformers by the square of thevoltage ratio, no matter what connection is used.

As an example of some of the things said above, suppose that we read from the

nameplate of a large transformer at a hydroelectric generating station the following ratingdata: 40 MVA, 115/24 kV. These data now tell us other things by using the idealtransformer relations, for example:

79.424/115/ 21 ==NN

since the voltage ratio and the turns ratio are the same under rating standards of largetransformers. Also the rated current of the high voltage winding, which we callI1 is givenby

currentratedamperes82.200

101153/1040 361

=

=I

and

windingvoltagelowfor thecurrentratedamperes25.96210243/1040

362

==I

It will be noted that this latter figure forIcould also have been obtained by using theideal transformer relationI1/I2 = N1/N2if more convenient.

3.3 An Actual Transformer

The ideal transformer relations give very good answers to many transformer problems,as in the examples preceding this section. For some problems, however, we must takeaccount of the departures from perfection to get an adequate answer to a transformer

problem. The first imperfection we will discuss is that of the core. The core is notinfinitely permeable, it does require ampere turns to establish the flux, and in addition,there are internal energy losses in the core when the flux varies with time.

Figure 3.4 shows the hysteresis properties of a transformer core. Each time the magneticfield is reversed, a small amount of energy is lost due to hysteresis within the core. For agiven core material, the loss is proportional to the frequency, and is a function of the peakflux density (Bm) to which it is subjected.
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Figure 3.4 (a) Hysteresis Loops of Steel; (b) The Normal Magnetization Curve

Transformer losses arising from the magnetic circuit, sometimes callediron loss. Theselosses are independent with load current, and may furthermore be expressed as "no-load"loss. Iron losses are caused mostly by hysteresis and eddy current effects in the core, andtend to be proportional to the square of the core flux for operation at a given frequency.

The hysteresis loss occurs as an inherent property of the magnetic material. The internalstructure of a ferromagnetic material is organized into domains and these domains arereoriented as the magnetic flux density - B vector goes through a cyclic change inmagnitude or direction. An internal energy loss appearing is the result. The energy lossmay be minimized by suitable alloying and heat treatment of the metal. The treatment

processes may also affect the mechanical properties, however, so compromises must bemade.

If the frequency of the applied voltage were reduced but the range ofB in the coremaintained (by applying lower voltage), a similar loop would be observed but withsmaller area than that originally observed. The reason for the larger area with higherfrequency is the effect of eddy currents in the steel. The steel is a conductor and, as theflux in the steel varies, voltages are induced within the closed contours in the material.Currents flow as a result of the voltage and an I2R loss occurs known as eddy currentloss. The loss is reduced by building the core from sheets of steel called laminations andby increasing the resistivity of the material by alloying.

Since the core flux is proportional to the applied voltage, the iron loss can be representedby a resistanceRc (or a conductance Gc=1/Rc) in parallel with the ideal transformer. Acore with finite permeability requires a magnetizing current IM to maintain the mutualflux in the core. The magnetizing current is in phase with the flux; saturation effectscause the relationship between the two to be non-linear, but for simplicity this effecttends to be ignored in most circuit equivalents. With a sinusoidal supply, the core fluxlags the induced EMF by 90 and this effect can be modeled as a magnetising reactance
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Xc ( or a susceptance BBc=1/Xc) in with the core loss component. Rparallel c andXc aresometimes together termed the magnetising branch of the model. If the secondarywinding is made open-circuit, the current I0 taken by the magnetising branch representsthe transformer's . Figure 3.5 shows the use of a fictitious circuit added tothe ideal transformer to account for exciting current.

no-load current

Iex

Gc Bc

Figure 3.5 Transformer Equivalent Circuit Accounted For Exciting Current

The equivalent circuit of Figure 3.5 is an approximation, but it is valid for most purposes.The advantages of using a linear model far outweigh the slight error introduced by alinear model. In any case the exciting current of a modern transformer of any size at allis only a very small percentage of the full load current, so if there is a slight error in thelinear representation, it amounts to very little in terms of the total current passed by thetransformer. The model to use for a certain study is of course a matter of experience andengineering judgment. For heavy load or short circuit studies the exciting current branchis normally omitted entirely. For light load or no load the exciting current branch may beincluded. For studies involving the wave form distortion of the exciting current the linearmodel is completely inadequate.

Evaluation ofGcandBBcof Figure 3.5 involves an approximation to best model the actualtransformer in some sense of a most useful model. The usual method of evaluating theparameters is to choose GcandBcB sothat the exciting current has the same rms value asthe actual exciting current and the power loss in Gc is the same as the actual core loss.For example, suppose that a certain transformer is tested by applying rated voltage to a10.5-kV winding with no load on the other winding, and it is observed that a current of10 amperes flows and a power of 10000 watts is drawn. We solve forGc andBBc as

follows: ;P G Ec=2 ( )21050010000

cG= ; ; ;

; ; ;

siemen107.90 6=cG I EYex c=

cY1050010 = siemen1038.9526=cY Y G Bc c

2 2= + c2

siemen1005.948 622 == ccc GYB .

In addition to the core loss and the requirement for exciting current, there are other waysin which an actual transformer differs from the ideal transformer model. For one thing,some of the applied voltage is absorbed in IR drop in the winding resistance. We canmodify the ideal transformer model to account for winding resistance by adding twoseries resistors R1 andR2 in either side of the windings. Now, each winding has a
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resistance which, while not zero, is kept low in order to minimize losses and increaseefficiency.

Flux leakage results in a fraction of the applied voltage dropped without contributing tothe mutual coupling, and thus can be modeled as self-inductancesXl1 andXl2 in series

with the perfectly-coupled region. These series reactances (leakage reactance equivalent)play a significant factor in transformer performance.

N2N1Xl1

Iex

Gc Bc

R1 Xl2 R2

Figure 3.6 The Equivalent Circuit of An Actual TransformerThe secondary impedanceR2 andXl2 is frequently moved (or "referred") to the primary

side after multiplying the components by the impedance scaling factor 2

2

1 )(N

N

2

2

2

1 RN

N

2

2

2

1lX

N

N

N2N1Xl1

Iex

Gc Bc

R1

Figure 3.7 Alternative Equivalent Circuit for the TransformerThe net series resistance and reactance are known simply as the impedance of thetransformer. This is an item of data that is available from the manufacturer of thetransformer. We designate the net resistance and reactance by the symbolsReqandXeqinthe figure where:

( )R R N N Req = +1 1 22

2/

and

( )X X N N Xeq l l= +1 1 22

2/
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Open and short-circuit tests:The parameters of the equivalent circuits of Figure 3.8 may be determined by thedesigner from the physical dimensions and material properties of the transformer. On theother hand, an actual transformer may be tested electrically to determine these values.For the test, we apply rated voltage to the left side of the transformer of Figure 3.8(a)with the right side open-circuited. Since the output current is zero, the current throughReqandXeqis zero from properties of an ideal transformer. We thus "see" only the shuntbranch and determine Gc andBBc from the instrument readings.

To determine the series impedances Req andXeq, we short-circuit one side, and it isconvenient to use the model of Figure 3.8(b) in this case. With a short circuit onN2, thevoltage is zero on this winding and also is zero acrossN1according to the properties of anideal transformer. As a result we can ignore the shunt exciting current branch and we"see" onlyReqandXeq.

Xeq

Iex

Gc Bc

Req

(a)

Xeq

Iex

Gc Bc

Req

(b)

Figure 3.8 Simple Equivalent Circuit for the Transformer3.4 Introduction to Per-Unit SystemsPer-unit systems are nothing more than normalizations of voltage, current, impedancepower, reactive power, and apparent power (volt-ampere). These normalizations ofsystem parameters because they provide simplifications in many network calculations.


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As we will discover, the transmission system and several portions of the distributionsystem are operated at voltages in the kV range. This results in large amounts of powerbeing transmitted in the range of kilowatts to megawatts, and kilovoltamperes tomegavoltamperes. As a result, in analysis, it is useful to scale, or normalize quantitieswith large physical values. This is commonly done in power system analysis and is

referred to as the per-unit system. This helps in understanding how certain types ofsystem behave. The numerical per-unit value of any quantity is its ratio to the chosenbase quantity of the same dimensions. Thus a per-unit quantity is a normalizedquantitywith respect to a chosen base value.

Normalization of Voltage and Current:

The basis for the per-unit system of notation is the expression of voltage and current asfractions of base levels. Thus the first step in setting up a per-unit normalization is topick up base voltage and current.

Consider the simple situation in Figure 3.9. For this network the complex amplitudes ofvoltage and current are: IZV = (an underline beneath the variable means it is a vector).

We start by defining two base quantities, Vbase for voltage and Ibase for current. In manycases, these will be chosen to be nominal or rated values. For generating plants, forexample, it is common to use the rated voltage and rated current of the generator as basequalities. In other situations, such as system stability studies, it is common to use astandard, system wide base system.

_

+v1

i1

_

+V Z

Figure 3.9 Example

The per-unit voltage and current are then simply:

basebase I

Ii

V

Vv == ,

With IZV = , we have

izV

IZ

I

I

V

IZv

base

base

basebase

===

Where the per-unit impedance is:


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basebase

base

Z

Z

V

IZz ==

This leads to a definition for a base impedance for the system:

base

basebase

I

V

Z =

And there is also a base power, with for a single phase system is:

basebasebase IVP =

As long as Vbase and Ibase are expressed in RMS (Root Mean Square). It is interestingto note that, as long as normalization is carried out in a consistent way, there is noambiguity in per-unit notation. That is, peak quantities normalized to peak base will bethe same, in per-unit, as RMS quantities normalized to RMS bases. This advantage iseven more striking in polyphase systems.

Three Phase Systems:

In power system calculations the nominal voltage of lines and equipment is almostalways known, so the voltage is a convenient base value to choose. The apparent power(volt-ampere - S) is usually chosen as a second base. In equipment this quantity isusually known and makes a convenient base. The choice of these two base quantities willautomatically fix the base of current, impedance, and admittance. In a system study, thevolt-ampere base can be selected to be any convenient value such as 100 MVA, 1000MVA, etc.

The same volt-ampere base is used in all parts of the system. One base voltage in a

certain part of the system is selected arbitrarily. All other base voltages must be relatedto the arbitrarily selected one by the turns ratio of the connecting transformers.

For single-phase systems or three-phase systems where the term current refers to linecurrent, where the term voltage refers to line to neutral voltage, and where the term volt-amperes refers to volt-amperes per phase, the following formulae relate the variousquantities:

)(

)(

NLbase

base

baseV

SI

=

( ) ( )

)(

2

)(

)(

2

)()(

base

NLbase

NLbasebase

NLbase

base

NLbase

base

S

V

VI

V

I

VZ

===

In performing per-phase analysis, the bases for the quantities in the circuit representationare volt-amperes per-phase or kilo-volt-amperes per phase, and volts or kilovolts fromline to neutral. System specification is usually given in terms of total three-phase volt-amperes or kilo-volt-amperes or mega-volt-amperes and line-to-line volts or kilovolts.This may result in some confusion regarding the relation between the per-unit value ofline-to-line voltage and the per-unit value of phase voltage (line to neutral voltage). In a


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per-phase circuit, the voltage required for the solution is the line to neutral voltage eventhough a line-to-line voltage may be specified as a base. The base value of the line toneutral voltage is the base value of the line-to line voltage divided by 3 . Since this isalso the relation between line-to-line and line to neutral voltages of a balanced three-phase system, the per-unit value of a line to neutral voltage on the line to neutral voltage

base is equal to the per-unit value of the line-to-line voltage at the same point on the line-to -line voltage base if the system is balanced. Similarly, the three-phase volt-amperes isthree times the volt-amperes per-phase, and the base value of the three-phase volt-amperes is three times the base value of the per-phase volt-amperes. Therefore, the per-unit value of the three-phase volt-amperes on the three-phase volt-ampere base isidentical to the per-unit value of the volt-amperes per-phase on the volt-ampere per-

phase base.

In a three-phase system, normally, a given value of base voltage is a line-to-line voltage,and a given value of base kilo-volt-amperes or base mega-volt-amperes is the total three-phase base.

The values of base impedance and base current can be computed from base values ofvoltage and volt-amperes as shown earlier in the section. If the base values of volt-amperes and voltage are specified as the volt-amperes for the total three phases andvoltage from line-to-line in a balanced three-phase system respectively, we have

)(

)3(

3 LLbase

base

baseV

SI

=

( ) ( )

)3(

2

)(

)(

2

)()(

33 base

LLbase

NLbasebase

LLbase

base

LLbase

baseS

V

VI

V

I

VZ

===

Networks With Transformers:

One of the most important advantages of the use of per-unit systems arises in the analysisof networks with transformers. Properly applied, a per-unit normalization will causenearly all ideal transformers to disappear from the per-unit network, thus greatlysimplifying analysis.

+

_

V2

I2I11 : N

V1

_

+

Figure 3.10 An Ideal Transformer


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To show how this comes about, consider the ideal transformer as shown in Figure 3.10.The ideal transformer imposes the constraints that:

1212

1, I

NIVNV ==

Note that an underline beneath the variable means it is a vector. Normalized to base

quantities on the two sides of the transformer, the per-unit voltage and current are:

2

22

2

22

1

11

1

11

,

,

basebase

basebase

I

Ii

V

Vv

I

Ii

V

Vv

==

==

Note that if the base quantities are related to each other as if they had been processed bythe transformer:

1212

1, IINVV basebasebase ==

then 21 vv = and 21 ii = , as if the ideal transformer were not there.

Transforming Form One Base To Another:

In most instances, the per-unit impedance of a component is specified on the ratedcomponent base which is different from the base selected for the part of the system inwhich the component is located. When performing calculations, all impedances in anyone part of the system must be expressed on the same impedance base. As a result, it isnecessary to have a means of converting per-unit impedances from one base to another.The process of changing this per-unit value of impedance to per-unit on a new base can

be done as follows:

Note that impedance in Ohms (ordinary units) is given by:

.__ newbasenewoldbaseold ZzZzZ == Here, replace( )

base

basebase

S

VZ

2

= , we can write:

( ) ( )

newbase

newbase

new

oldbase

oldbase

oldS

Vz

S

Vz

_

2

_

_

2

_=

This yields a convenient rule for converting from the old base to the new one:

( )( ) oldnewbase

oldbase

oldbase

newbase

new zV

V

S

Sz

2

_

2_

_

_=

In other word: )()( 2old

new

new

oldoldnew

baseVA

baseVA

baseV

baseVZUnitPerZUnitPer =


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Chapter 4.Electric Power Generation and Transmission Systems

(H Thng Pht v Truyn Ti in)

The electrical power system is a network of interconnected components which generateelectricity by converting different forms of energy, (thermal, hydro, and nuclear are themost common forms of energy converted) to electrical energy; and transmit the electricalenergy to load centers to be used by the consumer. Electricity is generated at thegenerating station by converting a primary source of energy to electrical energy. Theoutput of the generators is then stepped-up to appropriate transmission levels using astep-up transformer. The transmission subsystem then transmits the power close to theload centers.

4.1 Electric Power GenerationAlmost all the energy conversion process take advantage of the synchronous ACgenerator coupled to a steam, gas, or hydro turbine such that the turbine converts steam,gas, or water flow into rotational energy, and the synchronous generator then converts therotational energy of the turbine into electrical energy. It is the turbine-generatorconversion process that is by far most economical and consequently most common in theindustry today. In this section, we will briefly look at this conversion process withparticular emphasis on the synchronous machine and the controls that are used to governits behavior. A basic turbine-generator can be illustrated in Figure 4.1.

Vt++

__

Steam Power at voltage VtTorque atspeed

Turbine

Governor

ref Vref

Generator

ExcitationSystem

Figure 4.1 Block Diagram for Turbine-Generator System

The governor and excitation systems are known asfeedback control systems because it isthe feedback loops which provide for good control of certain parameters. The governorand excitation systems are typical feedback controllers in that the quantities to becontrolled (speed and voltage, respectively) are also providing the feedback signal.


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The generator is classified as a synchronous machine because it is only at synchronousspeed that it can develop electromagnetic torque. If the nominal system frequency is ,

synchronous speed is computed as

f

( )emp

2

=

where fe 2= is the frequency in rad/sec and is the number of poles on the rotor of

the machine. The machine speed in RPM can be computed as

p

fp

Ns120

= .

The synchronous generator has two iron structures. The rotoris the revolving part of themachine, and is located inside the stator, which is the stationary part of the machine.Because hydro-turbines are relatively slow, the number of poles must be high in order toproduce 50 Hz voltages (60 Hz in North America). Salient pole construction is simplerand more economical when a large number of poles are required. Steam plants, on theother hand, have very high speeds (1500 and 3000 RPM steam-turbine-generators are

typical), and saliency would create significant mechanical stress at these speeds.Therefore,smooth orroundrotor construction is employed for these generators. The twotypes of rotor construction are illustrated in Figure 4.2.

(a) (b)

Figure 4.2 Salient Pole (a) and Round (b) Rotor Construction

A magnetic field is provided by the DC-current carryingfield winding, which induces thedesired AC voltage in the armature winding. The field winding is always located on therotor where it is connected to an external DC source via slip rings and brushes or to arevolving DC source via a special brushless configuration. The armature consists of

three windings, all of which are wound on the stator, physically displaced from eachother by 120 degrees. It is through these windings that the electrical energy is producedand distributed. A typical layout for a 2 pole, round rotor machine would appear as inFigure 4.3.


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C

C

A

A

B

B

Figure 4.3 Winding Layout for Two-Pole Round Rotor Synchronous MachineThe DC current in the revolving field windings on the rotor produces a revolvingmagnetic field. We denote the flux associated with this field that links the armaturewindings as f (the subscript f indicates field windings). By Faradays Law of

Induction, this rotating magnetic field will induce voltages in the three armaturewindings. Because these three windings are physically displaced by 120 degrees (for atwo-pole machine), the induced voltages will be phase displaced in time by 120 degrees.

If each of the three armature windings is connected across equal impedances, balancedthree phase currents will flow in them. These currents will in turn produce their ownmagnetic fields. We denote the flux associated with each field as a , b , and c . The

resultant field with associated flux obtained as the sum of the three component fluxes a ,

b , and c is the field ofarmature reaction. We designate the associated flux as ar .

Using electromagnetic field theory and a trigonometric identity, one can show thatar

revolves at the same velocity as the rotor. Therefore the two fields represented by f

and ar are stationary with respect to each other. The armature field is effectively

locked in with the rotor field and the two fields are said to be rotating in synchronism.The total resultant field is the sum of the field from the rotor windings and that associatedwith armature reaction: farr += .

A voltage is induced in each of the three armature windings according todt

dNv r

=

where is the number of winding turns. Because r is a sinusoidal function of time,

the negative sign captures the fact that the induced voltage will lag the flux by 90degrees. Letting , , and be the voltages induced in winding a by the fluxesrE arE fE r ,

ar , and f , respectively, we can represent the relationships in time between the various

quantities using the phasor diagram, illustrated in Figure 4.4.


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f

r

ar

Ef

Ear

IaEr

Figure 4.4 Phasor Diagram for Synchronous Generator

Regarding Figure 4.4, take note that All voltages lag their corresponding fluxes by 90 degrees. The current in winding a, denoted by , is in phase with the flux it producesaI ar If (no load conditions), thenIa = 0 0=ar , and in this case, Fr = , and .Fr EE = All resistances have been neglected.A Generator Equivalent Circuit can be illustrated in Figure 4.5.

Vt

+

__

Ia

Ef

jXa

Er

+jXl

jX

Load

Figure 4.5 Equivalent Circuit Model of Synchronous Generator

Defining as thesynchronous reactance, we have thatarls XXX +=

asft IjXEV =

The phasor diagram corresponding to the equivalent circuit, when the load is inductive, isshown in Figure 4.6.


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-jXSIa

Ef

Ia

Vt

Figure 4.6 Phasor Diagram for Equivalent Circuit Inductive Load

The phasor diagram corresponding to the equivalent circuit, when the load is capacitive,is shown in Figure 4.7.

-jXSIa

Ia

EfVt

Figure 4.7 Phasor Diagram for Equivalent Circuit Capacitive Load

When the load is inductive, the current lags the voltageIa Vt; the generator is said to be

operating lagging. When the load is capacitive, the current leads the voltage ; the

generator is said to be operating leading. The angle between and is

Ia tV

Ia tV i, i.e.,

I Ia a= | | I ifVt =| Vt | 0. This implies that

< 0i lagging, > 0i leading

The excitation voltage magnitude is higher in the lagging case. We sometimes refer

to the lagging case as overexcited operation; here we have that

|| fE

||cos|| tf VE > , where

is the angle between and . The leading case results in under-excited operation; in

this case we have

fE tV

tf VE


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s

f

s

tf

s

tff

s

tf

ajX

Ej

jX

VE

jX

VEjE

jX

VEI

sin||cos||sin||cos||0||+

=

+=

=

=

s

tf

s

f

aX

VEj

X

EI

cos||sin||

Equating real and imaginary parts of the above equation, we haves

f

aX

EI

sincos =

ands

tf

aX

VEI

=

cossin . Multiplying both sides of the previous equations by

yields

tV3

s

ft

atoutX

EV

IVP

sin3

cos3 ==

s

t

s

ft

atoutX

V

X

EVIVQ

23cos3

sin3 ==

We can note that, reactive power out of the machine is positive when the machine isoperated overexcited, i.e., when it is lagging implying 0


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outmech PP = . (In reality, in steady-state operation so that ).

Consider what happens to this lossless machine operating at

0>lossP lossoutmech PPP +=

maxPPout = ( = 90 ) when

the steam valve opening is increased so that becomes slightly larger. In this case,

the power anglemechP

increases beyond , and the electrical power begins to decrease.

However, the mechanical power is only dependent on the steam valve opening, i.e., it isunaffected by the decrease in . This can only mean that . The difference

caus

Basic Modules for the Course of Intro to Electric Power Systems

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