Electrical Engineering2012-2016

POWER DISTRIBUTION & UTILIZATIONLAB MANUAL

Name: Rana Shawayz Tariq Roll No: 12EE05Electrical EngineeringSwedish College of Engineering & TechnologyRahim Yar KhanContentsSR.NOExperiment

01To learn the simple rules of safety.

02To measure the power by three voltmeter method.

03To measure the power by three ammeter method.

04To understand the voltage drop in short transmission line.

05To understand voltage regulation due to resistive, inductive and capacitive load.

06To understand the Ferranti effect of pi-model transmission line

07Voltage control in transmission line using shunt reactor.

08Understand construction of the line insulator

09Underground power cables

10Understand the Power factor correction with static capacitor

11Simulation of Power factor improvement using series capacitors

12MATLAB basics and Introduction to MATLAB Simulink

13Calculation of sag using MATLAB program

14MATLAB program for the analysis of short, medium and long transmission lines.

15 Designing of line parameter(inductance) using MATLAB

16 Simulation of Transmission Lines on MATLAB Simulink

LAB NO 1OBJECTIVE:Upon successful completion of this experiment, the students will be able to: To learn the simple rules of safety. To learn how to use the AC/DC power supply.Discussion: We know the location of the first AID supply in your shop or lab. Insist that every cut or bruise receives immediate attention. Regardless of how minor it seems to be. Notify your instructor about every accident. He will know what to do. If students follow the instruction with a degree of accuracy. These are no serious hazards of dangers in electro mechanical system of learning. Many people receive fatal shocks every year from the ordinary 220 volts electricity found in the home. A thorough safety program is a must for any one working with electricity. Electricity can be and even final to those who do not understand and practice the simple rules of SAFETY .the first rule of personal safety is always.THINK FIRST This rule applies to all industrial work as well as electrical workers. Develop good habits of workmanship. Learn to use tools correctly and safely. Always study the job at hand and think through yourprocedures. Your methods and the applications of tools, instruments and machines before acting.Never permit yourself to be distracted from your work and never distract another worker engaged in hazardous work. Dont be clown! Jokes are fun as so is horsing around but never moving machinery or electricity. There are generally three kinds of accidents which appear all too frequently among electrical students and technicians. Youre knowing and studying about them and observing simple rules will make you a safe person to work with. You could personally be saved from painful and expensive experiences you might be saved to live a rewarding retirement age.

ELETRIC SHOCK What about electric shocks? Are they fatal? The physiological effects of electric currents can generally be predicted by the chart shown in fig.

Fig.1Notes that it is the currents that do the damages. Currents about 100 mill amperes or only one tenth of an ampere or fatal. Work man who has contacted currents above 200 mill amperesmay live to see another day if given rapid treatment. Currents bellow 100 mill amperes can be serious and painful. A safe rule: dont place yourself in a position to get any kind of shocks.

What about VOLTAGE?Current depends upon voltage and resistance. Lets measure your resistance. Using your ohmmeter, measure your body resistance between these points: From left to right hand ohms (resistance). From hand to footohms (resistance) .Now wet your fingers and repeat the measurements: From right to left hand ..ohms (resistance). From hand to foot.ohms (resistance).The actual resistance varies of course, depending upon the points of contact and, as you have discovered, the conditions of your skin, the contact area. Notice how your resistance varies as you squeeze the probes more or less tightly. Skin resistance may vary between 250 ohms for wet skin and large contact area, to 500.000 ohms for dry skin. Considering the resistance of your body previously measured 100 mill amperes as a fatal current. What voltage might prove fatal for you to contact.Use the formula: volts =0.1*ohmsContact between two hands (dry).voltsContacts between one hand and foot (wet).volts

DO NOT ATTEMPT TO PROVE THIS!Nine rules for safe practice and to avoid electric shocks:1. Be sure of the conditions of the equipment and the dangers present BEFORE working on a piece of equipment. Many sportsmen are killed by supposedly unloaded guns; many technicians are killed by supposedly dead. Circuits,2. NEVER rely on safety devices such as fuses, relays and interlock systems to protect you. They may not be working and may fail to protect when most needed.3. NEVER remove the grounding prong of a three wire input plug. This eliminates the grounding feature of the equipment making it a potential shock hazard. 4. Disorganized mess of connecting leads, component and tools only leads to short circuits, shocks and accidents.5. Do not work on wet floor. Work on a rubber mat or an insulate floor.6. Donot Work Alone.It is good to have someone around to shot off the power give artificial respiration and to call a doctor.7. Work with one hand behind are in your pocket. A current in between a hands crosses your heart and can be more let than a current from hand to foot .a wise technician always work with one hand .watch your service man.8. Never talk to anyone while working .do not let yourself distracted. Also do not talk to anyone, if he is working on dangerous equipment. Do not be the cause of an accident.9. Always move slowly when work around electrical circuits .violent and rapid movements leads to accidental shock and shot circuits.BURNSAccidents caused by burns although usual not fatal, can be painfully serious .the dispatchment of energy produces heat.Four rules for safe practice and to avoids burns1. Resistors get very hot especially those who carry high current. Watch those five and ten work resistors .they will burn the skin of your finger. stay away from them until they cool off.2. Be on ground for all capacitor which retain a charge. Not only can you get a dangerous and fatal shot .you may also burn from electrical discharge. if the rate voltage of electrolytic capacitor is exceeded their polarities reversed they may get very hot may actually burst.3. Watch that hot soldering iron organ .do not place it on the bench where your arm might accidently hit it .never store it away while still .some innocent unsuspecting student may pick up.HOT SOLDER:HOT SOLDER can be particularly uncomfortable contact with your skin. Wait for soldered joints to cool. When disordering joints, dont take hot solder of so that you or your neighbor might get hit in the eyes on his clothes or body.

MECHNICAL INJURIES :This third class of safety rolls applies to all students whose work with tools and machinery. It is major concern of technician and safety lesson or found in correct of the use of tools. Five rolls for safe practice and to avoid mechanical injury. Metal corners and sharp edges on chassis and panels can cut and scratch. Improper selection of tool for the job can result in equipment damage and personal injury. Use proper eye protection when grinding chipping or working with hot metals. Protect your hand and clothes while working with battery acids and finish fluids. If u dont know ask your instructor.THE POWER SUPPLY : The power supply modules EMS-8829 provides both of necessary ac or dc power. Both fixed and variable single phase and three phase to perform all of the laboratory experiment the module must be connected to a three phase 220/380 V, 4 wire system. The power supply furnishes the following outputs Fixed 220/380 V, 3 phase power is brought out to 4 terminals labeled 1,2,3, and n. The current rating of the supply is 10 A per phase. Variable 220/380 V 3 phase power is brought out to four terminals labeled 4,5,6, and n. Variable three phase 0_380 V may be brought out from terminal 4,5,6 current rating of the supply is 3 A per phase. Fixed 220 V dc is brought out 2 terminals labeled 8 and n. Current rating is 1 A. Variable 0_220 V dc is brought out 2 terminals labeled 7 and n. Current rating is 5 A.The full current rating of various outputs cannot be used simultaneously. All power is removed from the output when the on off, breaker is in the off position.CAUTION : Power is still available behind the module phase with the breaker of never remove the power supply from the console without first remove the input power cable. The variable ac and dc outputs are controlled by the single control knob on the front of the module. The supply is fully protected against over and short circuit. Beside the 10 A three phase on off circuit breaker all of the output have their own circuit breaker. The rated current output may be exceeded considerably for short period of time without the harming the supply or tripping the breakers. All of the power sources may be used simultaneously provided that the total current drawn must not exceed the 10 A per Phase. Your supply if handled properly, will provide years of reliable operation and will present not danger to you.

LAB NO 2To measure the power by three voltmeter method

Equipment: Three voltmeters Power Supply Connecting wires Wattmeter Ammeter

Circuit Diagram:

Theory:On analyzing the given circuit, it is observed that R-L load and R load are connected in series and same current passes through both loads. The voltage drop in the inductor leads the current.

Graphically it can be represented as:

Resolve V into components & consider ABC

(V1)2 = (V2+V3cos )2+(V3sin)2 (V1)2 = (V2)2 +2(V2)( V3)cos +( V3)2(cos ) 2+( V3)2(sin ) 2(V1)2-( V2)2 = 2(V2)( V3)cos + +( V3)2(cos2 +sin2 )

cos = ((V1)2-( V2)2-( V3)2)/2(V2)( V3)------------------------------------(1)2(V2)( V3)cos = (V1)2-( V1)2-( V3)2V3cos = ((V1)2-( V2)2-( V3)2)/2(V2)I* V3cos = I*(( V1)2-( V2)2-( V3)2)/2(V2)

P = I*(( V1)2-( V2)2-( V3)2)/2(V2)---------------------------------------------(2)

Precautions: Make sure that connections are tight. Before switching ON power supply show connections to instructor. Make sure current does not exceed 5A.

Observations & Calculations:

Sr.No

I(Amp)V1(Volts)V2(Volts)V3(Volts)Pw(wattmeterreading)WattPc(calculated)Watt%Error

1

2

3

4

5

%error = [(Pw-Pc)/Pw]*100

Questions:1-Why there is difference b/w calculated and measured values?2-Calculate the formulas of real power and power factor of above circuit by connecting R-C load instead of R-L load?

LAB NO 3To measure the power by three ammeter method

Equipment: Three Ammeters Power Supply Connecting wires Wattmeter

Circuit Diagram:

Theory:On analyzing the given circuit, the total current I1 is divided into I3 and I2. I2 current is passing through resistor therefore it is in phase wih applied voltage, while I3 is passing through inductor and resistor therefore it lags by angle with respect to applied voltage.

Graphically it can be represented as:

Resolve I into components & consider ABC (I1)2 = (I2+I3cos )2+(I3sin)2 (I1)2 = (I2)2 +2(I2)(I3)cos +(I3)2(cos ) 2+(I3)2(sin ) 2(I1)2-(I2)2 = 2(I2)(I3)cos + (I3)2(cos2 +sin2 )

cos = ((I1)2-(I2)2-(I3)2)/2(I2)(I3)------------------------------------(1)2(I2)(I3)cos = (I1)2-(I2)2-(I3)2I3cos = ((I1)2-(I2)2-(I3)2)/2(I2)V*I3cos = V*((I1)2-(I2)2-(I3)2)/2(I2)

P = V*((I1)2-(I2)2-(I3)2)/2(I2)---------------------------------------------(2)

Observations & Calculations:

S.No

V(volts) I1(Amp) I2(Amp) I3(Amp) Pw(wattmeter reading) Pc(calculated) %Error

1

2

3

4

5

%error = [(Pw-Pc)/Pw]*100

Questions:1-Why there is difference b/w calculated and measured values?2-Calculate the formulas of real power and power factor of above circuit by connecting R-C load instead of R-L load?

LAB NO4OBJECTIVE:Upon successful completion of this experiment, the students will be able to: To understand the voltage drop in short transmission line.Apparatus:3- variable supply IT-6000Ammeters IT-6036Voltmeters IT-6038Transmission line module IT-60023- Resistive load IT-60043- Capacitive load IT-60063- Inductive load IT-6005Cables.Energy meter IT-6052

Theory:Transmission line possesses resistance R, inductance L, leakage conductance G and capacitance. All low voltages overhead lines having length upto 80 km are categorize as short line. In a short line , the shunt capacitance C and shunt conductance G are neglected. The series resistance R and series inductance for the total length of the line is considered. A single phase supply line is short in length and operate at low voltage. It has two conductors. Each conductor has resistance R1 and inductance L1. The inductance is affected equivalent to the inductive reactance X1=2pifL1. A balance three phase circuit consisting of three separate identical single phase circuit therefore the calculation for balance three phase line are carried out in similar manner as singe phase line, the difference being that per phase base is adopted. In this line all the given voltages are line to line values that all the current are line current.

Short transmission line: In short transmission line the parameters are resistance and inductive reactance. The equivalent circuit and vector diagram of a short transmission line are shown in the figure given below. In the equivalent circuit short transmission line is represented by the lumped parameters R and L. R is the resistance (per phase) L is the inductance (per phase) of the entire transmission line. As said earlier the effect of shunt capacitance and conductance is not considered in the equivalent circuit.The line is shown to have two ends : sending end (designated by the subscript S) at the generator, and the receiving end (designated R) at the load.Connections:

Procedure:-Step 01Make the connections as they are shown in figure.Step 02Switch on the power and adjust the voltage to 200.Step 03Now by keeping the capacitive load zero and changing the values of resistive and inductive load, obtain the value of voltage at output terminal.Step 04Then by keeping the line capacitor constant and changing the values of resistive and inductive load , get the value at output terminal.Step 05Now take the values of output by keeping the inductive and resistive load constant and varying the value of capacitive load.Step 06Now by disconnecting all load, taking the value of output.Table (a)Case 1: changing resistive and inductive load.InputOutputV.D

Table (b)Case 2: resistive and inductive load varying with line capacitance.InputOutputV.D

Table (c) Case 3: inductive and resistive load constant and varying capacitive load.InputOutputV.D

Table (d) Case 4: by disconnecting load.InputOutputV.D

Review Questions:-Q1: what isthe range of distance for short transmission line?Q2: what are the parameters in short transmission line?

Q3: what is the voltage range in short transmission line?

Q4: what is the effect of capacitance and conductance in short transmission line?Q5: what happened when all the loads are disconnected at the output terminal?

LAB NO 5Objective:To understand voltage regulationdue to resistive, inductive and capacitive load.Apparatus: 3- variable supplyIT-6000 3- Resistive loadIT-6004 3- Capacitive loadIT-6006 3- Inductive loadIT-6005 Transmission line moduleIT-6002 Three voltmeters IT-6038Theory:In electrical engineering, particularly power engineering, voltage regulation is the ability of a system to provide near constant voltage over a wide range of load conditions.In electrical power systems it is a dimensionless quantity defined at the receiving end of a transmission line as:

where Vnl is voltage at no load and VRated is voltage at full load. A smaller value of VR is usually beneficial.The Voltage Regulation formula could be visualized with the following; "Consider power being delivered to a load such that the voltage at the load is the load's rated voltage VRated, if then the load disappears, the voltage at the point of the load will rise to Vnl."Why Voltage Regulation is Required?Ideally, the output of most power supplies should be a constant voltage. Unfortunately, this is difficult to achieve. There are two factors that can cause the output voltage to change. First, the ac line voltage is not constant. The so-called 115 volts ac can vary from about 105 volts ac to 125 volts ac. This means that the peak ac voltage to which the rectifier responds can vary from about 148 volts to 177 volts. The ac line voltage alone can be responsible for nearly a 20 percent change in the dc output voltage. The second factor that can change the dc output voltage is a change in the load resistance. In complex electronic equipment, the load can change as circuits are switched in and out. In a television receiver, the load on a particular power supply may depend on the brightness of the screen, the control settings, or even the channel selected. Voltage regulator:A voltage regulator is an electricity HYPERLINK "http://en.wikipedia.org/wiki/Regulator_%28automatic_control%29"regulator designed to automatically maintain a constant voltage level. A voltage regulator may be a simple "feed-forward" design or may include negative feedback HYPERLINK "http://en.wikipedia.org/wiki/Control_theory"control loops. It may use an electromechanical mechanism, or electronic components. Depending on the design, it may be used to regulate one or more AC or DC voltages.Electronic voltage regulators are found in devices such as computer power supplies where they stabilize the DC voltages used by the processor and other elements. In automobile alternators and central power station generator plants, voltage regulators control the output of the plant. In an electric power distribution system, voltage regulators may be installed at a substation or along distribution lines so that all customers receive steady voltage independent of how much power is drawn from the line.Circuit diagram:

Procedure:Step 1Connect the 3 phase supply with transmission line module with the help of connecting probes.(Do not turn on supply)Step2Now connect the transmission line module with over voltage and phase failure relay.Step3Now turn on the 3 phase power supply and note reading on energy and voltmeter mounted on relay.this procedure will give us no load voltage reading.Step4Now connect the inductive,capacitive and resistive load to relay. This will give reading under load conditions.Step5 Increase the resistive and inductive load keeping the capacitive load constant and note reading on voltmeter Increase the capacitive load keeping the inductive and capacitive load constant. Increase the more capacitive load and still keep the inductive and resistive load constant.Step6Construct a table.

Table:S.NOVoltage at no load(Vnl)(volts)Voltage at full load(vfl)(volts)VR=(Vnl-Vfl/Vfl)*100

1

2

3

4

5

Review Questions:-Q1. What is Voltage Regulation?------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q2.What are the main causes of voltage failure?------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q3. What is the effect of variation of resistance ?------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.4 What is a voltage regulator?----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.5 Write the names of some voltage regulators.------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Lab No 6Objective:To understand the Ferranti effect of pi-model transmissionline.EQUIPMENT REQUIRED:Apparatus required for the experiment are given below Three phase supply IT-6000 AmmeterIT-6036 VoltmeterIT-6038 3 power supply IT-6017 Under-over current relay IT-6028 Line model IT-6002 1-3 inductive load IT-6005 1-3 resistive load IT-6004 1-3 capacitive loadIT-6006 Cables ammeter IT-6036 VoltmeterIT-6038

THEORY OF FERRANTI EFFECT:Alongtransmissionline drawsa substantial quantity of charging current. If such a line is open circuited or very lightly loaded at the receiving end, the voltage at receiving end may become greater than voltage at sending end due to capacitive reactance. This is known as Ferranti effect.This effect is due to the voltage drop across the line inductance being in phase with the sending end voltages. Therefore inductance is responsible for producing this phenomenon. The Ferranti effect will be more pronounced the longer the line and the higher the voltage applied.The relative voltage rise is proportional to the square of the line length.Due to high capacitance, the Ferranti effect is much more pronounced in underground cables, even in short lengths.It was first observed during the installation of underground cables in Sebastian Ziani de Ferranti's 10,000 volt distribution The figure shown below is representing a transmission line by an equivalent pi()-model. The voltage rise is proportional to the square of the line length.

The Line capacitance is assumed to be concentrated at the receiving end. In the phasor diagram shown above OM = receiving end voltage VrOC = Charging current drawn by capacitance = IcMN = Resistive drop NP = Inductive reactance drop Therefore; OP = Sending end voltage at no load and is less than receiving end voltage (Vr) Since, resistance is small compared to reactance, resistance can be neglected in calculating Ferranti effect. From -model, Vs=Vr-Impedance drop Under open circuit condition Ir=0 and hence, Vs=Vr-IcR-jwL*Ic i.e. receiving end voltage is greater than sending end voltage and this effect is called Ferranti Effect. It is valid for open circuit condition of long line. When load current is increased of R-L loads the resultant current is not remains leading, because of the inductive drop. Hence, receiving end voltage (Vr) is lesser than sending end voltage (Vs) under full load conditions.

CIRCUIT DIAGRAM:

Fig.1PROCEDURE:First we arrange the required apparatus to perform the experiment. Then we make the connection according to the connection diagram given. We supply the 3 electrical power to the model. model consists of a module to which a bank of inductive load, capacitive and resistive load is connected. After making the connection we disconnect the capacitive load and on the power supply. We note the sending end voltage and receiving end voltage and we notice that sending end voltage is higher than the receiving end voltage as it would be in normal condition. After that we connect capacitive load to the transmission line and then supply the 3 power again to the system and note the voltage at sending and receiving end. Now we observed that receiving end voltage is greater than the sending end voltage. This effect is increases as capacitive load increases means receiving end voltage becomes greater than the sending end voltage with the increased capacitive load. This effect is known as Ferranti effect.

CALCULATIONS:1.without line capacitance on R&L,C loadS.No.V1V2Vd(V1-V2)

1

2

3

2. with line capacitance on R,LloadS.No.V1V2Vd

1

2

3

3. with line capacitance on R,L loadS.No.V1V2Vd

1

2

3

4. with line capacitance without any loadS.No.V1V2Vd

1

2

3

LAB NO 7OBJECTIVE:Upon successful completion of this experiment, the students will be able to: Voltage control in transmission line using shunt reactor.Apparatus: 3- variable supplyIT-6000 3- Resistive loadIT-6004 3- Capacitive loadIT-60O6 3- Inductive loadIT-6005 Transmission line moduleIT-6002 Energy measuring unitIT-6052Voltmeter & AmmeterTheory: Shunt Reactorsare used inhigh voltageenergy HYPERLINK "http://en.wikipedia.org/wiki/Energy_transmission"transmissionsystems to stabilize the voltage during load variations. Shunt reactor are generally employed at the end of long transmission line. During charging of line due to Ferranti effect receiving end voltage increases dangerously to control this shunt reactors are used.A traditional shunt reactor has a fixed rating and is either connected to thepower lineall the time or switched in and out depending on the load. The fixed or mechanically switched reactors may be used for the absorption or generation of reactive power, the amount of reactive power produced is fixed and the response time is slow.

Circuit Diagram:

Procedure:1-Connect all circuit elements as shown in diagram in schematic form.2-Keep all loads at zero 1st then switch ON the power supply & note the reading at sending end.3-Just connect capacitive load at the receiving end of transmission line & note the vr.4-Then gradually exceed the inductive load & seek its effects at power factor & voltage at receiving end side. Write down the reading.5-Take the difference between receiving & sending end voltage.

Serial#Sending end V (Vs) in voltsReceiving end V (Vr) in volts Vs - Vr

1.

2.

3.

REVIEW QUESTION:Q 1: Explain voltage regulation?----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q 2: What type of load are causing +ive voltage regulation and why?--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q 3: What are the importance of shunt reactors?--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

LAB NO 8OBJECTIVE:Upon successful completion of this experiment, the student will be able to: Understand construction of the line insulator Understand why voltage across the insulators is not equal. Understand different types of line insulator and their characteristicsEQUIPMIENT: PIN TYPE INSULATOR SUSPENSION TYPE INSULATOR STRAIN OR TENSION TYPE INSULATORDISCUSSION:Overhead line insulator is used to separate line conductor form each other and from supporting structures electrically.Insulator must have the following characteristics The insulator should have high permittivity to with stand high electrical stresses. The insulator should be able to with stand overvoltages due to lightning, switching, or other causes under severe weather conditions in addition to the normal working voltages. It should posses high mechanical strength to bear the conductor load under worst loading condition It needs to have high resistance to temperature changes to reduce damages from power flashover Leakage current to earth should be minimum to keep the corona loss and radio interference with in reasonable limits Insulator material should not be porous and should be imperious to gases in atmosphere and should be free from impurity and cracks which may lower the permittivity.Failure of insulator occur either by puncture or flashover .in case of a puncture the arc passes through the body of the insulator. Flashover is caused by an arc discharge between the conductor and earth through air surrounding the insulator. it is either due to line surges or due to the formation of wet conducting layer over the insulator surface.Thickness of material is provided to prevent the puncture conditions .increasing the resistance to leakage resistance to prevent a flashover. The length of the leakage path is made large by constructing several layers called petticoats or rainsheds.they keep the inner surface relatively dry in wet weather and provide sufficient leakage resistance to prevent a flashover.For satisfactory operation, the flashover should occur before puncture .the ratio of puncture voltage is called the factor of safety is kept as high as possible. The rain sheds should have shapes of equipotential surface and they should be constructed along the lines of electrostatic field around the pin. Figure show the different types of insulatorsTypes of insulator:There are three main types of insulators used for overhead lines:1. Pin type insulator 2. Suspension type insulator3. Strain or tension type insulatorPin type insulator:Pin type insulator is supported on a forged steel or bolt which is secured to the cross arm of the supporting structure. single piece type pin insulator are used for lower voltages but for higher voltages two or more pieces are cemented together to provide sufficient thickness of porcelain and adequate leakage path or more creep age path. Figure shows a single piece insulator. Tow piece and tree piece insulators are shown in figures respectively.Suspension type insulators:It consist number of separate insulator connected with each other by metal links to form a flexible chain or a string. The insulator string is suspended from the cross arm of the sport. Suspension type insulators offer following advantages: Each unit design for operating voltage of about 11 kv .so that connecting several units to suit the service voltage con assemble a string. Additional unit can be introducing to the same unit to cop with future higher voltage. In case of damage to one of the units only the damaged insulator is replaced. The string is free to saving in a direction .the tension in the successive span is balanced .the line can therefore be designed for longer spans and higher mechanical loading. There is decrease in ability to lightning disturbances if the string is suspended from a metallic supporting structure, which works as alighting shield for conductor. There are two types of suspension insulators: Cap-and pin type. Hewlett or inter link type.Cap and pin type is more common. a galvanized cast iron is forged steel cap and a galvanized forged steel pin are connected to porcelain in the cap and pin type construction. the units are joined together either by ball and socked or clevis pin connections .Hewlett or interlink type units employs porcelain having two curved channels with planes at right angles to each other. Interlink type insulator is mechanically stronger than cap and pin type. the disadvantage of Hewlett type insulator is that the porcelain between the links is highly stressed electrically and therefore its puncture strength is lesser as compared to other types. figure shows the interlink type insulator.Strain or tension insulators:These types of insulators are designed for handling mechanical stresses at angle position where there is a change in the direction of the line or at termination of the line. Shackle and pin insulator serve the purpose for low voltage lines. for high voltage lines having longer spans and greater mechanical loading, suspension insulator strings are arranged in a horizontal position.

10 kV ceramic insulators, Showing sheds

Egg shaped strain insulator

Disc type insulator Pin Insulator

Disc Insulator internal structure

Review questions:Q.1 What is the difference between flashover and puncture of insulation?----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.2 Where do strain or tension insulators are used?---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Q.3 Write few cases of flashover?---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

LAB NO 9Objective:Upon successful completion of this experiment, the students will be able to: UNDERGROUND POWER CABLESPerformance objective : Understand the constructional parts of the power cables Understand the dielectric loss of power cable Learn the skin effect, proximity effect and current rating of cableDiscussion:Residential load today have a trend towards their growing density. This requires rugged construction, greater service reliability safety, and better appearance. The interface front external disturbances like storms, lightening, ice, tress, etc, should be reduced to a minimum. These difficulties are easily overcome by the use of underground cables and a trouble free service is achieved under a variety of environment conditions.Underground cables eared to be used In areas where it is almost impractical to use overhead lines, i.e transmission lines through sea, cross-over or terminal connections in substations or air-field crossing. Increased working voltages of the overhead lines require, the cable to be insulated for such voltages in order to meet the requirement of the overhead lines. The design of power cables is, therefore, governed by the requirements of the overhead lines.Supply interruptions due to external influence will be reduced wit the underground cables, but if a fault occurs due to any reason it will not be easily located. For long distance transmission, cables cannot be used to their large charging current.Cables construction:A power cable consists of three main components Conductor Insulation Sheath

Conductors :Copper and aluminum are used as conductor materials in cables. Aluminum occupies a greater space than copper for a given conductance. Solid or number of bare wires made o either copper or aluminum are used to make a power cable. For a conductor having more than three wires are arranged round a central wire such that there are 6 in the first layer, 12 in the second, 18 in the third and so on. In this way number of wires in conductor are7, 9,37,62,91 etc. The size of the conductor represented by 7/9, 19,b, 37/c, etc. in which first figure represents the number of strands and the second figure A, B, C, etc represents the diameter in cm or mm of the individual wires making the conductor.Stranded conductors having more than one layer of wires are made such that the directions of lay wires in adjacent layers are opposite to each other. Fig 7.1 and 7.2 shows the standard wire called segmental conductor and shaped conductors respectively.Insulation:The dielectric compound, as insulators for power cables should possess the following main properties:a. High insulation resistanceb. High dielectric strengthc. Good mechanical propertiesd. Capable of being operated at high temperature e. Low thermal resistancef. Low power factorThe most commonly used dielectric in power cables is impregnated paper, but rubber, polyvinyl chlorides(PVC), polythene, cross-linked polthylene (XPLE). Polyvinyl chloride(PVC) cables is being used today for distributions purpose as an alternate to paper insulated cable because of its several advantages over insulated cables.Polythene insulated power cable are non hygroscopic, used in cables for submarine and damp soil.Sheath:Metal sheath is required to protect the cable from moisture, which would affect the insulation. Following materials are being used as sheathing Lad Lead alloys AluminumCorrugated seamless aluminum (CSA) sheath is used these days. It has better bending, reduces thickness and loser weigh. It is mainly used in high voltage oil-filled cables and telephone cables.Protective covering:To protect cable from mechanical damage, corrosion and electrolytic action protective coverings are applied to sheath. Bitumen and bituminous material (paper, hessian etc) or polyvinyl chloride is used against corrosion and electrolytic action. Layers of fibrous material permeated with waterproof compound applied to the exterior of the cabke are called serving.Armoring:One or two layers of galvanized steel wires or two layers or metal tape Armoring is applied over hessian or jute bedding to protect the sheath from mechanical demage.Double wire armor is used for cables requiring increased tensile strength.Presence of magnetic material within the alternating magnetic field of single-core cable produce excessive losses, therefore cables are either left unarmored or, if necessary, they are armored with non-magnetic materials like tin-bronze tapes or wires.In case of multi core cables the reluctant alternating magnetic field is zero. Aluminum is preferred as an armor material due to its non-magnetic properties, high conductivity and mechanical strength. It is particularly useful for single core cables working on ac.Fig. 7.4 shows the mass-impregnated paper insulating submarine cables. Table 7.1 shows the constructional parts of mass impregnated paper insulating submarine cables.Table 7.11Conductor

2Conductor shielding

3Insulation

4Insulation shielding

5Lead sheath

6Plastic jacket

7Tape armor

8Optical armor

9Steel wire armor

10Serving

Review Questions:Q1. Explain skin effect__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q2. What is proximity effect?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q3. What is three main cause of dielectric loss in the power cables? _____________________ _____________________ _____________________Q4. Draw the pencil sketch of circuit oil-filled cable and write its main components?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

LAB NO 10Objective: Understand the Power factor correction with static capacitor.Apparatus: 3- variable supplyIT-6000 3- Resistive loadIT-6004 3- Capacitive loadIT-6006 3- Inductive loadIT-6005 Transmission line moduleIT-6002 Energy measuring unitIT-6052 Voltmeter & AmmeterTheory : The electrical energy is the almost exclusively generated, transmitted and distributed in the form of alternating current and 70% of the loads are inductive in nature, hence have low lagging power factor. The low power factor causes an increase in current, resulting in additional losses of active power (P)in all the elements of power system from generation down to utilization devices. In order to ensure supply system economical, it is important to have power factor close to unity.Power factor: The cosine of angle between voltage and current in an AC circuit is known as power factor (Pf). In an AC circuit, there is generally a phase difference between voltage and current. The term Cos is called Pf. If the circuit is inductive, the current lags behind the voltage and Pf is called lagging. If the circuit is capacitive, the current leads the voltage and Pf is called leading. But if the load is resistive, current and voltage will be in phase and Pf is called unity.The power factor of a circuit can be defined in one of the following three ways: Power factor = Cos = Cosine of angle between V and I Power factor = R/Z = Resistance / Impedance Power factor =VI Cos / VI= Active power / Apparent power

Note: The reactive power is neither consumed in the circuit nor does it do any useful work. It merely flows back and forth both directions in the circuit. A var meter measures the reactive power.Disadvantage of low Power factor:The power factor plays an important role in an a.c circuits since power consumed depends upon this factor.P = VL IL Cos (for single phase supply)IL = P / (VL Cos)(for single phase supply)P = 3 VL IL Cos (for 3 phase supply)IL = P / (3 VL Cos)(for 3 phase phase supply)

It is clear from above that for fixed power and voltage, the load current is inversely proportional to the Pf. Lower the Pf, higher is the load current and vice-versa. A power factor less than unity results in the following disadvantages: Larger KVA rating of equipment Greater conductor size Large copper size Poor voltage regulation Reduced handling capacity of systemCauses of low Power factor: Most of the AC motors are of induction type (1 and 3 induction motors). These motors work at extremely low Pf on small load (0.2 to 0.3) and rise to 0.8 or 0.9 at full load. Arc lamps, electric discharge lamps and industrial heating furnace operate at low lagging power factor. The load on the power system is varying, high during morning and evening and low at other times. During low period, supply voltage is increased which increase the magnetization current. This result in the decreased power factor Pf.

Power factor improvement:Normally the power factor of the whole load on a large generating station is in the range of 0.8 to 0.9 lagging. In case of low Pf, it is important to take special steps to improve it. This can be achieved by the following equipment. Static capacitors Synchronous condenser Phase advancersStatic capacitors:Capacitors (generally known as static capacitors) draw leading current and partly or completely neutralize the lagging reactive component of load current. This raises the Pf of the load. For 3 loads the capacitor can be connected in delta or star as shown in the figure 1.

Procedure:Step 01Connect to the indicator and load resistor as in figure-2.Step 02Switch on power and adjust supply voltage 220 ( L-L ) V.Step 03With the help of variable resistive and inductive loads set the load current 2A at 0.5 lagging power factor.Step 04Switch on the capacitive load: increase it in steps as shown in table and complete the table (a).Table (a)S. NoV1I1Cos I2V2V.D=V1-V2

1

2

3

4

5

6

Review Questions:Part 01Draw the characteristic curve between power factor and line current (Pf Vs A1)?

Part 02Q1.What are the causes of low power factor?

Q2.Write methods of improving power factor?

Q3.What is the importance of power factor in the supply system?

Q4.Why line current reduces as load capacitor increased when RL load was same in the experiment?

LAB NO11Objective:` improvement using series capacitorsApparatus: Single phase ac supplyIT-6017 AC ammeterIT-6036 TWO ac voltmeterIT-6038 WattmeterIT-6048 Variable resistive load IT-6004 Variable Inductive load IT-6005 Variable capacitive loadIT-6006 Connecting leadsTheory: The power factor of a circuit implies that how efficiently power is being consumed or utilized in the circuit .The greater the power factor of a circuit,greater is the ability of the circuit to utilize apparent power.Thus if the power is 0.5,it means that50%of the power is being utilized. However,it is desired that power factor of a circuit to be as close to unity as possible.The cosine of the angle between voltage and current in a circuit is also known as power factor cosThe power factor of an alternating circuit is defined as ratio of active power to the apparent power.Mathematically, Power factor=cos=KW/KVAWhereKW=active power delivered or absorbed by the circuit.KVA=Apparent power of the circuit.The ratio of resistance to the impedance is also known as power Factor.Alternatively, Power Factor=R\ZWhereR=Resistance of the circuit.Z=Impedance of the circuit.Disadvantages of low power factorA power factor less than unity results in the following disadvantages: Large KVA rating of equipment. Greater conductor size. Large copper losses Poor voltage regulation. Reduced handling capacity of system.Improving the power factor Power factor improvement using capacitor is done using following connections: Series Shunt

Series capacitors:Series Capacitors are connected in series with lines but they are hardly used in the distributed system because there is a requirement for a large amount of complex Engineering investigation. Figure shows that how series capacitor compensates forInductive reactance. A series capacitor is a capacitor (negative) reactance in series with theCircuits inductive (positive) reactance with the effect of compensating for part or all of it.Therefore, the primary effect of the series capacitor is to minimize the voltage drop caused By the inductive reactance in the circuit.A series capacitor can even be considered as a voltage regulator that provides voltage rise which increased automatically and instantaneously as the load increase .Also, a seriesCapacitor produced more net voltage rise than a shunt capacitor at lower factor, which creates more voltage drop. However, a series capacitor improves the system power factor much less than a shunt capacitor and has a little effect on the source current

Consider a feeder circuit and its voltage-phasor diagram as shown in Fig a and c. The voltagedrop in the feeder can be expressed as

=IRcos+IsinWhere R=resistance of feeder circuit=inductive reactance of the circuit Cos=receiving end power factorSin=sine of the receiving end power factor angleAs it can be seen from the phasor diagram the magnitude of the second term in the above equation for voltage drop is much larger than the first. The difference gets to be much larger when the power factor is smaller and the the ratio of R/ is small.However, when a series capacitor is applied, as shown in Fig b and d. the resultant lower voltage Drop can be calculated as cos+I()sinWhere R=resistance of feeder circuit

Procedure: connect the circuit as shown below.

Record V, I and Wvaluesin the table. Calculate p.f, Q and S by using relations given below. Connect capacitor in series of R,L load as shown below.

Record values of V,I, W and Vc in the table. Calculate p.f,Q, C and S using relation given below. Change the value of capacitor and repeat the step 5,6.

Observations and calculation:Cos=W/(VI) = S=VI Q=VIsinXc =Vc/IC=1/( 2 XX 50 X Xc) TableS.NoV(Volts)I(Amps)S(VA)Vc(Volts)W(Watts)cosQ(vars)Xc(Ohms)C(Farads)

123036.183031626624.40.8049804.48.00072

223536.98671.516671970.834836.54.490.00070

324037.690241707670.40.8547524.520.000701

424538.59432.51738300.60.8844794.500.000707

525039.2980017789180.914062.44.510.000705

Review Questions1.What are the advantages and disadvantages of using series capacitor?-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2. How much power factor improvement has been achieved in this experiment?----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3. What other methods can be employed for p.f correction?---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

LAB NO11

MATLAB BASICS

Engineering Problem Solving

Engineering often involves applying a consistent, structured approach to the solving of problems.A general problem-solving approach and method can be defined, although variations will be required for specific problems. Problems must be approached methodically, applying an algorithm, or step-by-step procedure by which one arrives at a solution.

Problem-Solving Process

The problem-solving process for a computational problem can be outlined as follows:

1. Define the problem.2. Create a mathematical model.3. Develop a computational method for solving the problem.4. Implement the computational method.5. Test and assess the solution.The boundaries between these steps can be blurred and for specific problems one or two of the steps may be more important than others. Nonetheless, having this approach and strategy in mind will help to focus our efforts as we solve problems

MATALB the program Solver Matlab is an interactive system for doing numerical computations. A numerical analyst called Cleve Moler wrote the first version of Matlab in the 1970s. It has since evolved into a successful commercial software package. Matlab relieves you of a lot of the mundane tasks associated with solving problems numerically. This allows you to spend more time thinking, and encourages you to experiment. Matlab makes use of highly respected algorithms and hence you can be confident about your results. Powerful operations can be performed using just one or two commands. You can build up your own set of functions for a particular application. Excellent graphics facilities are available, and the pictures can be inserted into LATEX and Word documents.

MATLAB is a high level computer language used for the programming of complex engineering problems. MATLAB stands for Matrix Laboratory. In MATLAB we have many tool boxes each containing functions related to specific operations, just as we have libraries in JAVA or C language. Communications, Control System, Image processing and Image acquisition are some of the mostcommonly used tool boxes. MATLAB has extensive capabilities of two and three dimensional plotting. We will use these plotting facilities of MATLAB to aid in the physical interpretation of equations, particularly those related to fields and potentials.In MATLAB, there are two modes of executing a program; the first one is command line execution and the second one is M-File or DOT M (.m) file execution. For command line execution commands are entered in the command window where (.m) file, complete program is first written in a file and saved with DOT M (.m) extension. The complete program is then RUN like any other programming language. We will use DOT M files throughout this lab.

BASIC TERMINOLOGY:MATLAB has some basic terms and predefined variables you should get familiar with before going any further. One of these is the variable ans. When you type most commands to MATLAB, they will return values. You can assign these values to variables by typing in equations. For example, if you type>>x = 5MATLAB will printx = 5Review of commands

Display command>>Disp(I am the student Of SCET)InputA=input(Enetr The 1st No.)B=(Enter The 2nd no);Fprintf A=2Fprintf(The Value of a is % A )

Rational and Logical Operators (LOGICALS)MATLAB represents true and false by means of the integers 0 and 1.true = 1, false = 0If at some point in a calculation a scalar x, say, has been assigned a value, we may make certain logical tests on it:x == 2 is x equal to 2?x ~= 2 is x not equal to 2?x > 2 is x greater than 2?x < 2 is x less than 2?x >= 2 is x greater than or equal to 2?x > x = pix =3.1416>> x ~= 3, x ~= pians =1ans =0>> A=1:9;>> B=8-A;>> tf1=A> tf2=A>Btf2 = 0 0 0 0 1 1 1 1 1>> tf3=A==Btf3 = 0 0 0 1 0 0 0 0 0>> tf4=A-(A>2)tf4 = 1 2 2 3 4 5 6 7 8

Operators : OR, AND, NOT, AND, XOR clcA=[0 1 0]B=[0 1 0]XOR=xor(A,B)AND=and(A,B)OR=or(A,B)NOT=not(A)NOT=~(A)

ResultA =

0 1 0

B =

0 1 0

XOR =

0 0 0

AND =

0 1 0

OR =

0 1 0

NOT =

1 0 1

LAB NO11Objective:Introduction to MATLAB Simulink

EQUIPMENT REQUIRED: A computer running Matlab 7.0 or higher version

THEORETICAL BACKGROUND:-The MATLAB and Simulink environments are integrated into one entity, and thus we can analyze, simulate, and revise our models in either environment at any point. We invoke Simulink from within MATLAB. We begin with a few examples and we will discuss generalities in subsequent chapters. Throughout this text, a left justified horizontal bar will denote the beginning of an example, and a right justified horizontal bar will denote the end of the example. These bars will not be shown whenever an example begins at the top of a page or at the bottom of a page. Also, when one example follows immediately after a previous example, the right justified bar will be omitted.In the Command Window, we type:>>simulinkAlternately, we can click on the Simulink icon shown in Figure 1.3. It appears on the top bar onMATLABs Command Window.

Figure 3.01 The Simulink Library BrowserExercise with Simulink:Now we are to draw the sine wave and cosine wave on Simulink.

y_1= Sin x and y_2= cos x ..>>> (2.11)

Figure 3.02: - Dragging the Sine Wave block function into File Untitled

Figure 3.03:- File Equation with added Sine wave and Scope block

Scope Result

Figure 3.04 Sine waveFigure 3.05:-Block parameters Sine waveTo generate the cos wave we have to change the parameter as follows Phase (rad) = pi / 2

LAB NO13Objective: Calculation of sag using MATLAB program.Apparatus: MATLAB software Theory:Sag is provided in transmission lines in order to lessen the tension of the transmission lines. Sag literally means to be bend in shape. In overhead transmission lines , the difference in level between points of supports (towers or utility poles) and the lowest point on the conductor is called a sag.

Explanation: While erecting an overhead line, it is very important that the conductors are under safe tension. If the conductors are too much stressed between the supports ( towers, utility poles), then the stress on the conductors may reach to an unsafe level and the conductor may break due to excessive pressure ( i,e tension). in order to permit safe tension in the conductors, the conductors ( i.e the transmission lines) are not fully stretched but are allowed to have a dip or a sag.

The sag can be calculated by the parabolic and catenary method.

For the parabolic method it is calculated by using formula

Sigma=w*l^2/8*H

For the catenary method it is calculated by using formula

Sigma=H/w(coshw*l/(2H)-1)

But when the span is of unequal length then the sag is calculated using the formula

Sigma=w*l^2/(8*T)

Matlab program

Example

A transmission line conductor at a river crossing is supported from two towers at heights of 30m and 90m above water level. the horizontal distance between the towers is 270m.If the tensioning the conductor is 1800kgf and the conductor weights1kgf/meter. Find the clearance between conductor and the water at a point midway b/w the towers. Assume parabolic configuration.

MATLAB codel=270; %l is in meter(m) and it is a horizontal distance between towers.T=1800; % tension in the conductor ,in kg/fh1=30;h2=90;h=h2-h1; % h is the difference in level between two supports . w=1; % weight of conductor/meter.x=l/2-T*h/(w*l);%value of spanx1=-x+l/2;sigma1=w*x1^2/(2*T)sigma2=w*x^2/(2*T)sigma=w*(l-x)^2/(2*T)clearance=h1+(sigma1-sigma2) %clearance between conductor and water

Resultssigma1 = 44.4444sigma2 = 19.5069sigma = 79.5069clearance = 54.9375

Review QuestionsQ1- What Is Sag?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q2- What Is The Difference B/W Sag And Span?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q3- What Is the Benefit Of Providing Sag In A Transmission Line?_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

LAB NO14Objective;Matlab program for the analysis of short,medium and long transmission lines.Tool:MatlabTheory:Transmission lines are classified asshort, medium and long.Short Lines:When the length of the line isless than about 80Kmthe effect of shunt capacitance and conductance is neglected and the line is designated as ashort transmission line.For these lines the operating voltage isless than 20KV.Medium lines:Formedium transmission linesthe length of the line is in between80km - 240kmand the operating line voltage wil be in between21KV-100KV.In this case the shunt capacitance can be assumed to be lumped at the middle of the line or half of the shunt capacitance may be considered to be lumped each end of the line.The two representations of medium length lines are termed asnominal-T and nominal- respectively.Long Lines: Linesmore than 240Kmlong and line voltageabove 100KVrequire calculations in terms of distributed parameters.Such lines are known aslong transmission lines.This classification on the basis of length is more or less arbitrary and the real criterion is the degree of accuracy required.Matlab program;

Matlab program for the analysis of short,medium and long transmission lines

Matlab code

Q: The receiving end voltage of power system is 33KV. The receiving end power is 15000KVA. Power factor of the system is 0.85. If line resistance per km per phase is 0.29 ohm. The line reactance per km per phase is 0.65 ohm. Find the results if the line length is 8km.Result for 8 Km.STUDYING THE CHARECTERISTICS OF A TRANSMISSION LINE***************************************************Receiving end line voltage in volt =33000length of transmission line in km =8Rated power at receiving end load =15000000power factor=0.85Vr = 1.9053e+004Ir = 2.6243e+002 -1.6264e+002iresistance/km/phase=0.29inductive reactance/km/phase=0.65Z = 2.3200 + 5.2000i

SHORT TRANSMISSON LINE**********************A = 1B = 2.3200 + 5.2000iC = 0D = 1

Vs= 2.0507e+004 +9.8732e+002iIs =2.6243e+002 -1.6264e+002ireg =7.7592Vs_line =3.5519e+004 +1.7101e+003iPs =1.6627e+007eff =90.2152

Review QuestionsQ1- Differentiate between long , medium and short transmission line ?_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q2- what are the short ,medium and long line parameters?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Q3- What is the benefit of thisMATLAB program?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

LAB NO15Objective Designing of line parameter(inductance) usingMATLABTool: MATLABTheory:Parameters of Transmission Line Transmission line has four electrical parameters - resistance, inductance, capacitance and conductance. The inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. The shunt conductance characterizes the leakage current through insulators, which is very small and can be neglected. The parameters R, L and C are essential for the development of the transmission line models to be used in power system analysis both during planning and operation stages. While the resistance of the conductor is best determined from manufactures data, the inductances and capacitances can be evaluated using formula. The student is advised to read chapter 4 or any other text book before taking up the experiment. The transmission lines are represented by an equivalent circuit model with approximate circuit parameters on per phase basis. This model can be used to compute voltages, currents, power flows, efficiency and regulation etc. Normally the lines are classified into short, medium and long lines for the purpose of modeling.inductance of a symmetrical 3- phase system

A transmission line is said to be symmetrical when its conductors are suited at the corners of a equilateral triangle. Such an arrangement is also sometimes referred to as equilateral spacing and it is shown in the fig below.

Fig. Symmetrical three phase line

The inductance of a symmetrical three phase line can calculated by using the formula which has been calculated by using above fig L=0.2lnFrom the equation we found that inductance per conductorof a three phase symmetrically spaced line s equal to the inductance per conductor of a single phase line of equal length and with equal spacing between conductors.Matlab code%program For the calculation of inductance of symmetrical 3-phase line>>D=input('Enter the spacing between the conductor A and B and C in meters =');>>d=input('Enter the diameter of three phase transmission line in milimeters =');>>f=input('Enter the value of frequency in HZ=');>>r=0.5*0.001*d>>req=0.7788*r>>L=2e-7*log(D/req)>>disp('inductance is in H/m')>>XL=2*pi*f*L>>disp('inductance reactance is in ohm/m')

ExampleA two conductor single phase line operates at 50 Hz, the diameter of each conductor is 20mm and the spacing between the conductors is 3m.calculate 1.inductance of conductor2.inductive reactanceSolution By using by above matlab program we have results asEnter the spacing between the conductor A and B and C in meters =3Enter the diameter of three phase transmission line in milimeters =>> 20Enter the value of frequency in HZ=>> 50r = 0.0100req =0.0078L = 1.1908e-006inductance is in H/mXL = 3.7409e-004inductance reactance is in ohm/minductance of unsymmetrical 3-phase lineFor an unsymmetrical spaced 3-phase line the inductance and the voltage drops will be different for all the phases even under balanced current conditions.

When a unsymmetrical spaced line is properly transposed, the average of flux 3positions in the transpositions cycle and dividing the sum by 3

The inductance of unsymmetrical 3-phase system can be found by using the formula L=2 ln( )1/3

Matlab code%program For the calculation of inductance of unsymmetrical 3-phase line>>Dab=input('Enter the spacing between the conductor A and B in meters =');>>Dbc=input('Enter the spacing between the conductor B and c in meters =');>>Dca=input('Enter the spacing between the conductor C and A in meters =');>>d=input('Enter the diameter of three phase transmission line in milimeters =');>>f=input('Enter the value of frequency in HZ=');>>Deq=(Dab*Dbc*Dca)^0.33>>r=0.5*1e-3*d>>req=0.7788*rL=2e-7*log(Deq/req)>>disp('inductor is in H/m')>>XL=2*pi*f*L>>disp('inductor is in ohm/m')

ExampleA 3-phase, 50 Hz line consist of 3 conductors each of diameter 21mm. The spacing between the conductors is as follow AB=3m, ,BC=5m, CA=3.6m Find 1.inductance of line 2.inductive reactanceSolutionBy using matlabEnter the spacing between the conductor A and B in meters = 3Enter the spacing between the conductor B and c in meters =>> 5Enter the spacing between the conductor C and A in meters =>> 3.6Enter the diameter of three phase transmission line in milimeters =>> 21Enter the value of frequency in HZ=>> 50Deq =

3.7298

r =

0.0105

req =

0.0082

L =

1.2245e-006

inductor is in H/m

XL =

3.8470e-004 inductor is in ohm/m

Review Question1 What are the parameters of a transmission line?__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

LAB NO16Objective:Simulation of Transmission Lines on MATLAB Simulink

This demonstration illustrates steady-state and transient simulation of Medium Transmission Line and use of the Powergui block

CASE STUDY:-

Nominal T Model Nominal Model

Fig 4 Nominal T Model Fig. 5 Nominal ModelThis circuit is a simplified model of a 132 kV three-phase power system. Only one phase of the transmission system is represented. The equivalent source is modeled by a voltage source (132 kV rms/sqrt(3) or 160 kV peak, 50 Hz) in series with its internal impedance (Rs + Ls) .The source feeds a RL load through a 150 km transmission line. The line distributed parameters (R L and C).The load (50 MW 0.87 Mvar per phase is modeled by a parallel RLC load block. Current and Voltage Measurement blocks provide signals for visualization purpose.

Click On the Powergui block and get the required results.

Powergui Graphical user interface for the analysis of circuits and systems Library:Powerlib ( in Simulink)

Demonstration

1. Simulation using a continuous solver Start the simulation and observe line voltage and load current transients during load switching and note that the simulation starts in steady-state.Use the zoom buttons of the oscilloscope to observe the transient voltage.2. Using the Powergui to obtain steady-state phasors and set initial statesOpen the Powergui block and select "Steady State Voltage and Currents" to measure the steady-state voltage and current phasors..

Using the Powergui select now Initial States Setting to obtain the initial state values (voltage across capacitors and current in inductances).Now, reset all the initial states to zero by clicking the to zero" button and then "Apply" to confirm changes. Restart the simulation and observe transientsat simulation starting. Using the same Powergui window, you can also set selected states to specific values.

Fig. 6 Powergui Block

3. Discretizing your circuit and simulating at fixed stepsThe Powergui block can also be used to discretize your circuit and simulate it at fixed steps.

Open the Powergui. Select Discretize electrical model" and specify a sample time of 50e-6 s. The state-space model will now be discretizedusing trapezoidal fixed step integration. The precision of results is now imposed by the sample time. Restart the simulation and compare simultion resultswith the continuous integration method. Vary the sample time of the discrete system and note the impact on precision of fast transients.4. Using the phasor simulation methodYou will now use a third simulation technique. The "phasor simulation" method consists to replace the circuit state-space model by a set of algebraicequations evaluated at a fixed frequency and to replace sinusoidal voltage and current sources by phasors (complex numbers). This method allows a fastcomputation of voltage and current phasors at a selected frequency, disregarding fast transients. It is particularly efficient to study electromechanicaltransients of generators and motors involving low frequency oscillation modes.

By Changing the Values of Comports in Block we can have a Complete AnalysisSteady State Voltage and Current Measurement by Graphical User Interface:

Fig. 7 Steady State ToolMODELING IN MATLAB/SIMPOWER SYSTEM

Fig. 8 Block Representation of Medium line A type-T

Fig. 9 Block Representation of Medium line B type-PI

SCOPE OUTPUT:Fig. 10 Scope Results (a) and (b) for both Models