electrical machines ee 250 laboratory manual_hajvery university lahore pakistan

8/14/2019 Electrical Machines EE 250 Laboratory Manual_Hajvery University Lahore Pakistan

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The Hajvery University LahoreFaculty of Engineering

Bachelor of Electronic Engineering

EE-250 Electrical Machines

Laboratory Work Book


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CONTENTS

Lab.

No.

List of ExperimentsPage

No.

01 To operate the transformer on off load, loaded and short circuitconfigurations.

02To measure the losses and efficiency of a single phase transformer.

03

To examine transformation ratio of an Auto transformer usingmeasuring techniques, to measure current flow in various part of

auto transformer and thus develop the relationship between currentflow

04

To study the properties of separately excited DC Shunt generator

under no load and full conditions and to obtain saturation curve ofthe generator

05To study the properties of self excited DC Shunt generator under noload and full conditions

06To study the properties of series DC generators

07To measure the per phase winding resistance of the ACgenerator

08 To examine method of controlling voltage and frequencyfrom the A.C generator

09To observe the voltage line to line, generator at opencircuit

10To plot the characteristics curve of the AC generator usingshort circuit test

11To observe the effect of varied excitation (If) on thegenerator voltage at a constant speed

12To see the effect of varied speed on the generated voltagewith a constant field excitation.

13

To test the generator with resistive load (lamp load) andetc, to find out the voltage regulation.


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LAB SESSION 01

OBJECTIVE

1. TO operate a single phase transformer off- load, with a load applied and

With a short circuit applied.

2. To deduce the relationship b/w the secondary voltage and the loading andGive reasons for the deductions.

3. To measure the off-load voltage.

4. To define the terms, short circuit voltage, continuous short circuit currentAnd transient short circuit current.

5. To measures the short circuit voltage and, from this, calculate the

Continuous short circuit current.

APPARATUS

1 variable / isolating transformer1 single phase test transformer1 exercise panel2 resisters, 47 ohm / 11w2 moving coil multi meter

SCIRCUIT DIAGRAM

Fig. 2-1


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Fig.2-2

THEORY

The single phase- exercise transformer used in this exercise has atransformation ratio K of 1:1, i.e. the windings in the primary and secondarycoils have the same number of turns, N1 and N2. Therefore when no load isapplied the primary voltage V1 is the same as the voltage in the secondary,V2.

K= N2/N1 = V2/V1 = 1/1N1 = N2V1 = V2

When loaded, the winding resistance of the transformer has a similar effect asthe internal resistance of the voltage source, across which part of sourcevoltage is dropped. The transformer is design so that at the nominal current inthe secondary of I2N = 1A, the voltage available at the secondary is thenominal voltage V2N of 22 V. When off load, no voltage is drop internally, sothe off load voltage, V20 is larger than the nominal voltage, V2N.

The short circuit voltage Vsc is the voltage which must be applied primary,when the secondary is short circuited, so that the nominal current IN, flows inthe primary winding. It is usually quoted as percentage of the nominal voltage.

Vsc = V1SC / V1N . 100%

The magnitude of the short circuit voltage Vsc is a measure of a change involtage which occurs in the secondary and thus the internal resistance of thetransformer. The smaller the short circuit voltage, the less dependent is theoutput voltage on load.

The continuous short-circuit current, ICSC, is the current which flows in thesecondary circuit when it is permanently short-circuited . the value of ISC islarger for a smaller internal resistance of the transformer and thus the shortcircuit voltage, Vsc, is also smaller.


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Icsc = I 2N/Vsc . 100%

The current which flows , momentarily , immediately a short-circuit occurs,the transient short- circuit current ITSC, can be more than double the valueIcsc.

PROCEDURE

1.1 Construct the circuit as shown in the diagram, in section 2-1 slowlyincrease the Voltage from the variable transformer, until the secondaryvoltage of the test. Transformer is at the nominal value of V2N = 22V.

1.2 Measure the voltage at secondary of test transformer_________________

1.3 RL which you have attached to the secondary of test transformer_______

1.4 Measure the load current I2____________________

1.5 Check the measured value by calculation = V2N/RL=________

1.6 Give reasons for any deviation of the measured from theCalculated value, Write the answer as

By calculation, at the nominal secondary voltage V2N = 22 V,The secondary current______________________________

________________________________________________The value of current is ______because_________________

________________________________________________________________________________________________

2.1 Measure the load current I2 at the nominal secondary voltage, V2N of 22VAnd enter the value into the following table.

R/ Ohm I2 /A V2/V

23. 5 22

47

3.1 Evaluate the measured result from 2.1

When the loading on the transformer is reduced, the load current I 2 ____________________________. Thus the voltage losses in the transformer ___________and the secondary voltage V2 ______________________ When off- load (R= ),the voltage at the output terminals of the Transformer is ______________

___________________________________________________________


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4.1 Construct the circuit as shown in the diagram 2-2 (short-circuit operation).Slowly increase the voltage from the variable transformer until theNominal current of 1 A flow in the primary circuit.

Measure the voltage applied at the primary, V1sc = ________________________

Calculate the value of Vsc in % of the nominal voltageVSC = (V1SC/V1N ) * 100 % = _________________

Calculate the value of continuous short - circuit current, Icsc.Icsc = (I2N / VSC) * 100 % = ____________________


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LAB SESSION 02

BJECTIVEO

1. To measure the delivered power of a transformer when off-load, or when a

Short circuit is applied.

2. To determine by measurements, the losses which occur in the iron coreAnd the windings of a transformer.

3. Determine the total losses of a transformer.

4. Calculate the efficiency of a transformer from the values obtained byMeasurement.

APPARATUS

1 Variable / isolating transformer1 Single phase test transformer1 Exercise Panel2 Resistors, 47 Ohm / 11 W2 Moving coil multi meter1 Universal power meter

SIRCUIT DIAGRAMC

Fig. 2-1


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Fig.2-2

THEORY

The power losses, PL in a transformer consist of iron losses PFE and copperlosses in the windings, PCU.

PL= PFE + PCU

Iron losses are dependent on loading of transformer and can be determined inan off load test. The copper losses which occur here can be ignored since theoff load current is very small.

Copper losses changes corresponding to the square of the current strengthand are measured in short circuit test at the nominal current. The iron lossesare ignored here because the magnetic flux in short-circuit operation is verysmall.

The efficiency of a transformer is given by the ratio of output power to theinput power.

= POUT/PIN

PROCEDURE

1.1 Construct the circuit as shown in diagram 2-1.

1.2 Slowly increase the voltage from the variable transformer, until theSecondary voltage of the test transformer is at the nominal value of V2N=22 V.

1.3 Measure the power consumed by the test transformer, PIN.PIN = __________________________ W

1.4 Calculate the delivered power at the load resistor, Pout.Pout = (V2N)^2 /R = ______________ W

1.5 Determine the power losses PL and efficiency of the transformer.PL = PIN POUT = ________________ W


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1.6 = POUT/PIN = ____________________ %

1.7 Remove both load resistors and power consumed by the test transformer,When off-load PIN.PIN =__________________________ W (Off load)

1.8 PIN when off-Load is equivalent to the _____________ losses of theTransformer

2.1 Construct the circuit as shown in diagram 2-2

2.2 Slowly increase the voltage from the variable transformer until the nominalCurrent of 1 A flow in the primary circuit

2.3 Measure the power consumed by the test transformer, when output isShort circuited, PIN.PIN= _________________ W (Short Circuit)

2.4 The power consumed when operated with short circuit is equivalent to the_______________ Losses of a transformer

2.5 Total power losses PL = PFE + PCU = ______________ W

2.6 Explain why there is the difference from the value determined in 1.1 Step?

The total power loss PL equal to the ________ of the iron and copper lossIs _________ than the value calculated in step 1.1, because in off loadTest some _________ occurs and in the short circuit test, some ________Occurs


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LAB SESSION 03

OBJECTIVE

1. Describe the construction of an auto transformer

2. Examine using measurement techniques, the transformation ratio of anAuto transformer

3. Measure the current flow in the various parts of an auto transformer andThus develops the relationship between the current flows

4. State, and give reasons for the advantage of an auto transformer compareTo other types of transformer

APPARATUS

1 Variable / isolating transformer

1 Single phase test transformer

1 Exercise panel2 resistors 47 ohm / 11 W

2 moving - coil multi meter

SCIRCUIT DIAGRAM

Fig 2-1


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Fig.2-2

Fig.2-3

THEORY

An auto transformer contains only one winding which is tapped at the requiredvoltage transformation points thus an increase in voltage is possible (step up)as well as decrease in voltage (step down)

In contrast to a transformer with separate windings power can be produce at

the output by a current transformer from the input to a output sides so thatonly part of the power has to be transformed via the magnetic flux of the ironcore thus copper for the windings and iron core material can be saved in theconstruction of an auto transformer


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The transformer used in this exercise is operated as an auto transformerwhere the secondary winding remains open circuit and the centre tap of theprimary winding is used for tapping off the secondary voltage V2

Since input and output conductivity connected, it is not permitted to use anauto transformer as a protecting isolation transformer

PROCEDURE

1.1 Construct the circuit as shown in the diagram 2.1

1.2 Slowly increase the voltage from the variable transformer until the primaryVoltage V1 is at normal value of 22 V

1.3 Measure the secondary voltage V2 and state the transformation ratio K

V1= _________ VV2= _________V

K=__________

2.1 Construct the circuit as shown in diagram 2.2

2.2 Slowly increase the voltage from the variable transformer until the primaryvoltage V1 of the test transformer is at 11V

2.3 Measure the secondary voltage V2 and state the transformer ratio

V2= _________V V1= _________ V

K=__________

2.4 Evaluate the result from 1.3 and 2.3

By changing the connections the primary voltage U1 can be _____ or__________ Which is equivalent to a __________ of the transformation

Ratio.

3.1 Construct the circuit as shown in diagram 2.3

3.2 Slowly increase the voltage from the variable transformer until theSecondary voltage V2 of the test transformer is at 11V

3.3 Measure the currents I1, I1, I2U2 = 11VI1= __________A

I2= __________A

I1= __________A


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3.4 From the currents measured, deduce the relationship which exitsBetween value of currents

I1= I1- I2= _________= __________A

3.5 Describe the flow of current in an auto- transformerIn the lower part of the winding the current which flows in equalTo______________ primary current I1 and secondary current I2

In the upper part of the winding the current which flows is________________


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LAB SESSION 04

PRE LAB STUDY

)dSeparately Excite(t GeneratorDC ShunAn excitation power is needed to produce the stator magnetic field. If thispower is supplied by an external voltage source, we call the methodindependent or Separately Excited Generator. If the generator feed its ownmagnetic field, the process is one of Self Excitation. Problem can occur duringrunning up, for an excitation current (IF) flows only when the voltage (V) isapplied to the field winding. Due to the residual magnetism of the voltage andthe resultant excitation current influence one another in such a way that thefinal values reached very quickly we must ensure that the residual magnetismis not cancelled out by a wrong polarity in the excitation winding.

The DC Separately Excited Shunt Generator

The DC Shunt generators construction is identical to that of the DC Shuntmotor. In operation, however, its rotor is driven by the mechanical power,and DC voltage taken from the commutator brushes on the rotor shaft. Inorder to generate electricity, the rotor must rotate within a magnetic field. ADC voltage is, therefore, applied to the stator to create the necessarymagnetic field. This can be accomplished by connecting a battery or someother separate source of DC voltage to the stator terminals. When thestator is thus separately excited, the generator is called the SeparatelyExcited Generator. By varying the field excitation, a fairly wide range ofoutput voltages may be obtained. However, it is expensive and, in manycases, impractical or inconvenient to provide a separate source of DC

excitation.

Procedural NotesThe driving motor must rotate. According to the direction of the outputvoltage, if the meter shows reverse voltage, change the connection leadsof the meter.

DiscussionA DC machine can run either as a motor or as a generator. A motor convertselectrical power into mechanical while a generator converts mechanical powerinto electrical power. A generator must therefore, be mechanically driven inorder that it may produce electricity.

Since the field winding is an electromagnet, current must flow through it toproduce a magnetic field. This current is called the Excitation Currentand can be supplied to the field winding in one of the two ways; it cancome from a separate, external DC Source, in which case the generator iscalled a separately excited generator; or it can come from the generatorsown output, in which case the generator is called a self-excited generator.Assume that a DC current excites the shunt field, thereby setting up a

magnetic flux in the generator. If the rotor (or more correctly, the armature)is rotated by applying mechanical effort to the shaft, the armature coils willcut the magnetic flux. And a voltage will induce in them. This is AC and inorder to get DC out of the generator, a rectifier must be employed. Thecommutator and the brushed carried out this role.


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The voltage induced in the coils (and, therefore, the DC voltage at thebrushes) depends only upon two things- the speed of rotation and thestrength of magnetic field. If the speed is doubled, the voltage becomestwice. If the field strength is increased by 20%, the voltage also increasedby 20%.

Although separate excitation requires a separate DC power source, it isuseful in case where generator must respond quickly and precisely to anexternal control source, or when the output voltage must be varied over awide range.

With no electrical load connected to the generator, no current flows andonly a voltage appears at the output. But is a resistance load is connectedacross the output, current will flow and the generator will begin to deliverelectric power to the load.

The machine which drives the generator must then furnish additional

mechanical power to the generator. This is often accompanied by thegenerator, together with a drop in speed.


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LAB SESSION 04

OBJECTIVE

1. To study the properties of the separately excited DC Shunt generator underNo-load and full load conditions

2. To obtain the saturation curve of the generator.3. To obtain the armature voltage Vs armature current load curve of the

Generator

APPARATUS

Driving Motor (DC PM Motor) controlled with DC Drive, 3000 RPM, 180 Volts,2.5 Amp.DC Excitation Supply: 190 Volts, 0.6 Amps.DC Volt Meter: (0-300 Volts)

DC Amp Meter: (0-10 Amps)RPM MeterFuses: 8 Amps & 6 AmpsEmergency Stop SwitchDC Shunt Generator: 190 Volts, 1.6 Amps

SCIRCUIT DIAGRAM

Fig.1-1


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PROCEDURE

1.1 Connect the circuit shown in the fig1-1:

1.2The relationship between the no-load voltage V0 and the speed of therotation n is measured at the rated excited current IF.

1.3The speed is constant and the excitation current IF is varied, No loadvoltage V0 is measured.

WHEN EXCITATION CURRENT IS CONSTANTIF = 0.6 AmpsBy varying the speed of driven motor

n(RPM)

0 500 1000 1500 2000 2500 2800

V0 inVA out

HEN SPEED IS CONSTANTWn=2800 RPMBy varying the field excitation current

IFin

0 0.1 0.2 0.3 0.4 0.5 0.6

V0

= VAOutput V

Return the power supply voltage to zero and turn off the power supply.

LOAD CHARACTERISTICS2.1Connect the circuit diagram in the previous fig, and connected the load

after the proper output voltage and measure the values.

2.2 Operating behavior of an independently excited shunt generator.

2.3 The generator is driven at the rated speed of n=2800 RPM.

2.4 An excitation current IF = 0.6 Amps.

2.5 Keeping the speed if the generator constant, the terminal voltage isMeasured

:RESULTS

I load

VA out


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LAB SESSION 05

PRE LAB STUDY

EXCITED SHUNT GENERATOR-THE DC SELF

The self-excited DC shunt generator provides its own field excitation. Theshunt field is connected to the armature through the commutator so that direct

current flows through the shunt field. This arrangement leaves the self-excitedgenerator independent of any separate sources of DC excitation.

However, regulation is poorer than that of the separately-excited generatorbecause the field current is dependent on armature voltage. If the generatorload increases, output voltage decreases due to armature reaction and the IRdrop across the armature winding resistance. Since excitation voltage comesfrom the generator output, a decrease in output voltage. Output voltage canbe varied to compensate for load changes by manually adjusting the fieldrheostat. However, automatic adjustment is more convenient and more rapidlyresponsive to load changes. Such devices as the Terrill regulator, the diactorand the saturable reactor or magnetic amplifier provide automatic means ofvarying field flux inversely with changes in output voltage.

In laboratory experiments previously small output voltage was noted evenwhen no field excitation current was flowing in the shunt coil. This was due toresidual magnetism, which in a separately excited is little more than aninteresting phenomenon. In a self excited generator, however, the residualvoltage which, in turn causes a small current to flow in the field coil. theresulting magneto motive for either aids or opposes the residual magnetismdepending on the manner in which the shunt terminals are connected to the

armature. If the residual flux is aided. The generated voltage will rise or "buildup" this rise will in turn cause an increase in field current, a further rise in flux.And a still larger output voltage build up continuous the field current cause bythe generator output voltage is sufficient to develop the required excitation tosustain the output voltage at the desired level.

The self excited generator provides its own field flux from the generatedoutput voltage. If the output voltage varies, the field flux also varies, so achange in output voltage is magnified by a reduction in field flux, whichcauses less voltage to be induced in the armature winding and a furtherreduction in generated output voltage.


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LAB SESSION 05

AUTIONC

Do not make any.High voltages are present in this laboratory experimentThe power should be turned off after.connection with the power on

.dividual measurementcompleting each in

OBJECTIVE

1. To study the properties of the self excited DC shunt generator under noLoad and full load condition.

2. To learn how to connect the self excited generator.

3. To obtain the armature voltage vs. armature current load curve of theGenerator

APPARATUS

Driving motor (DC PM Motor) controlled with DC Drive, 3000 RPM, 180 V,2.5 AmpereDC Volt meter (0-300Volts)

DC Amp Meter (0-10Amp)RPM MeterFuses.8Amp & 6 AmpsEmergency stop switchDC Shunt Generator (190Volts, 1.6Amp)Connecting leads


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SCIRCUIT DIAGRAM

Fig.2-2

THEORYThe separately excited generator has many applications. However it doeshave the disadvantage that a separate DC current power source is needed toexcite the shunt field. This is costly and some time inconvenient; and the self-excited DC generator is often more suitable.

In a self excited generator, the field winding is connected to the generatoroutput. It may be connected across the output, in series with the output, or acombination of the two. The way in which the is connected (shunt, series orcompound) determine many of the generator's characteristics.

All of the above generators can have identical construction. Self excitation ispossible because of the residual magnetism in the stator pole pieces. As thearmature rotates a small voltage is induced across its windings. When thefield winding is connected in parallel (shunt) with the armature a small fieldcurrent will flow. If this small field current is flowing in the proper direction, thisfurther increases the armature voltage and thus, a rapid voltage buildupoccurs.

If the field current flows in the wrong direction, the residual magnetism willbe reduced and voltage build up cannot occur. In this case, interchangingthe shunt field leads will correct the situation. It is a purpose of thislaboratory. Experiment to show these major points


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PROCEDURE

Do not make any.High voltages are present in this laboratory experiment

The power should be turned off after.ection with the power onconn.completing each individual measurement

1.1 Connect the circuit shown in the fig 2-2

1.2 Turn on the power supply of the driven motor.

1.4 Measured the voltage at terminal points E1 & E2 of the generator, outputIs build up _________

1.5 If not turn the connection supply off and interchange the shunt field leadsFor excitation

1.6 Measure the open circuit armature voltage VA=____________

1.7 Vary the field rheostat and notice if the armature voltage VAChanges_________

1.8Adjust the speed of the generator by driven motor=2800 r.p.m

1.9 Fixed the rated speed of the generator by driven motor and measured

Different load value characteristic

Fixed speed = 2800 r.p.m

RLLoad

IAOutput(A)

VAVoltage

Output


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LAB SESSION 06

AUTIONC


.each individual measurementcompleting

OBJECTIVE

1.1 To study the properties of series DC generator

1.2 To learn how to connect the series generator

1.3 To obtain the armature voltage vs. armature current load curve of theGenerator

APPARATUS

Driving Motor (DC PM Motor) controlled with DC Drive, 3000 RPM 180 V, 2.5AmpereDC Volt meter (0-300Volts)DC Amp Meter (0-10Amp)

RPM MeterFuses.8 Amp & 6 AmpsEmergency stop switchDC Shunt Generator (190Volts, 1.6Amp)Connecting leads


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SUIT DIAGRAMCIRC


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THEORY

When field winding is connected in series with armature winding, thegenerator is called a series generator. The exciting current through a fieldwinding of a series generator is the same current the generator delivers to

the load.

If the load has high resistance only a minimum output voltage can begenerated because of the minimum field current. On an open circuit, thegenerator will have a minimum output voltage due to its residualmagnetism. If the load draws current, the excitation current increases, themagnetic field become stronger and the generator delivers an outputvoltage.

You can see then that in a series generator, change in load current greatlyaffect the generator output voltage. A series generator ha s very poor

voltage regulation and is not recommended for use as a power source,Series generators have been used on a DC distribution system as linevoltage booster.

PROCEDURE


.completing each individual measurement

1.1 Connect the circuit according to the diagram. Connect the meter withthe output of the generator.

1.2 Start the driven motor, and measure the output voltage at the terminal

E1, E2 of the generator at open load VA = ____________________

nSpeedVA

1.3 Connect a ohm load in the circuit by closing switch note whether VAincrease. If not, turn off the power supply and interchange the seriesfields leads at terminal D1, D2.

1.4 Measure the VA = __________________ at IA = _________________


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1.5 Adjust the lead resistance to obtain each of the listed in the table.At speed 2800 r.p.m

Power calculateIAVARL


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LAB SESSION 07

OBJECTIVE

To measure the per phase winding resistance of the AC generator

SCIRCUIT DIAGRAM

Fig 1-1: Synchronous generator Y or Star connecter field winding

PROCEDURE

The three terminals U, V, W are connected in Y connected connection. Thisconnection is form a common (neutral) terminal as shown in fig 1-1

1.1 Measure the resistance between each pair of the line terminals andRecord it in table 1-1

R12 (L1.L3) R23 (L2.L3) R31 (L3.L1) RAVG Ra=0.5 RAVG

Table 1-1

1.2We will take the average of the three measurements to determine the per-phase winding resistance Ra.


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LAB SESSION 8

OBJECTIVETo examine method of controlling voltage and frequency from the A.Cgenerator

PROCEDURE

1.1 The Stator winding of generator is connected in Star or Y connections.

1.2 Set the field excitation switch to the manual excitation, connect theManual supply D.C excitation supply F1, F2, to the generator excitationTerminal F1(Red), F2(Black)

1.3 Prime mover speed knob at fully zero position (induction motor)

1.4 Switch on the main supply of the Prime mover (inventors supply), Main

Power switch and enable switch one direction forward or reverse

1.5 Use the speed control knob of the prime mover to increase the speed to1500 R.P.M

1.6 Use the volt meter and frequency meter to measure the output voltageAnd frequency of the generator. The terminal output of the generator L1,L2, L3 and DC voltmeter ampere meter for DC field excitation.

1.7 Increase the excitation control and note the change in generator terminalVoltage, dont exceed the excitation current up (0-30 V or 0-3 A)

Excitationvoltage

ExcitationCurrent

OutputVoltage

FrequencyR.P.M

Table 1-1

1.8 Measure the frequency out put voltage, excitation current and voltage inThe table 2-1 uses the manual excitation knob to maintain the out-putVoltage 380 V line to line

1.9 Increase the speed o prime mover at 1800 rev/min and again take theMeasurement in table 2-1

1.10 Reduce the speed to back 1500 rev/min and switch of the prime moverOFF position

Result Analysis1. Comment on the relationship between excitation and terminal voltage

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________


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2. given that that the machine has 4 poles, thin the link between theFrequency and the speed of the generator

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3. Comment on the relationship between frequency, voltage and field current.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________


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LAB SESSION 9

OBJECTIVETo observe the voltage line to line, generator at open circuit

SCIRCUIT DIAGRAM

Excitation Winding (Rotor)

AC Generator Y or Star ConnectedField winding (Stator) Connection for an open circuit

UREPROCED

!Please be careful.ory experimentHigh voltages are present in this laborat

1.1 Connect the circuit according to the diagram in fig.1-1

3.02.752.502.252.01.751.501.251.00.750.50.250.00MeasurementField CurrentMeasuredLine Voltage

ComputedPer phase

Voltage

1.2 Ensure the prime mover is on, speed is at 1500 rev/min (RPM) switchclosed the DC field excitation power supply (0-250 V), for generator use 0-


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30 V only, terminal F1, F2. Connect the AC voltmeter between line to linevoltages (L1, L2) or (L1, L3) or any two line of the Y connected generatorwinding to measure the open circuit line to line voltage.

Excitation

Even when the field excitation winding current is zero, there may be someinduced e.m.f in the field winding (stator) due to residual flux in the machinerecord this voltage as the reading in the table 3-1

Now field excitation winding supply switch is on, increase the field current if insmall increments and record the field current with line voltage for each valueof the field current.

If possible, take reading until the line voltage is nearly 150 % of its rated value(380 V). Calculate the per phase induced e.m.f. The is per phase no loadvoltage of the generator. Sketch the per phase no load voltage as a functionof the field excitation winding current with the field winding current along x-axis. Now plot a curve by obtained value, this is called the magnetizationcharacteristic of the three phase synchronous machine.


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LAB SESSION 10

OBJECTIVETo plot the characteristics curve of the AC generator using short circuit test

SCIRCUIT DIAGRAM

Fig.1-1

PROCEDURE

1. Start the prime mover and set the speed 1500 rev/min.

2. Set the field excitation winding current to zero by setting the variable DCfield source voltage to zero

3. Place a short circuit the three lines of a field winding in (stator) as shown infig.1-1

Ampere meter AC am- meter is included in one of the lines to measure theline current

The three phase induction motor (Prime mover is running) or rotating.

MeasurementFieldexcitationcurrent (A)

MeasuredShort circuitcurrent (A)


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5. Now increase the field excitation current in a small increments and recordthe short circuit current in the field winding (Stator) in table 4-1, For eachsetting of the excitation winding current. Take quite a few data points but themaximum value of the line current should not exceed 150% of its rated valueas recorded in table.

6. Plot the per-phase as a function of the field excitation winding current.Note that the phase current in the Y or Star connected machine is same asthe line current. The field excitation winding current should be along x-axis,this plotter curve is called short circuit characteristics of the synchronousmachine.


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LAB SESSION 11

OBJECTIVETo observe the effect of varied excitation (If) on the generator voltage at aconstant speed

PROCEDURE1.1 Stator winding is connected as Star of Y connection

1.2 Set the field switch to manual excitation, set the potentiometer and theprime mover speed to zero(fully anti clock wise)

1.3 Switch on the main supply, increase the speed of prime mover to 1500rev/min

1.4 Increase the excitation current to change the out put voltage in steps fromMinimum to maximum of and then increments back to minimum. Record

Generated voltage and excitation current (if) at each increment.

Note. Take reading progressively (dont go back to take a reading) or yourresults will be wrong

Increase excitation Decrease ExcitationV VIf If

Repeat the procedure for constant speed of.a. 20% (1800 rev/min)

b. 40 % (2000 rev/min)

6. Reduce the speed back and switch off the primer mover.Plot curves of voltage against increasing and decreasing current for eachspeed, typical shape of the curve at one speedNote that your curve has the same shape as a magnetizing curve.Note that they show the effect of Hysteresis and that

V B (Flux density)


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LAB SESSION 12

OBJECTIVETo see the effect of varied speed on the generated voltage with a constantfield excitation.

PROCEDURE

1. Generator is connected in a STAR or Y connection.2. Set the manual excitation and adjust speed control of (prime mover) to

zero.3. Start the prime mover and adjust the speed to nominal 1599rev-min

-1.

Speed

Table 6-1

4. Keep a constant excitation current IF of 3A and vary the speed inincrements of 50 rev-min-1 to a max of 2000 rev-min -1. At eachincrement, record the output terminal voltage and speed in Table 6-1.

5. Repeat for constant excitation current 2.5A and 3.5A.

Result Analysis:

Plot your result as curve of Voltage against speed.

V N and IFSoV IF N

If u studied the electromagnetic theory, examine your curves and connect

their agreement with the fundamental equations


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LAB SESSION 13

OBJECTIVETo test the generator with resistive load (lamp load) and etc, to find out thevoltage regulation.

SCIRCUIT DIAGRAM

Fig.1-1

PROCEDURE

1.1 Manual Excitation: When a synchronous generator supplies a threePhase load, its terminal voltage may be higher or lower than its no loadVoltage. By drawing the phasor diagram, one can verify the terminalVoltage at any load is lower than the no-load voltage.

1.2 Set the excitation to manual and adjust control to minimum.

1.3 We use the set of three lamps as balance load Y connected or threeVariable resistors in Y connected in Fig 1-1.

1.4 Start the prime mover and set to normal speed (1500 rev/min)

1.5 Switch on the excitation output and increase the field current to adjust theOutput voltage at rated value (voltage).

1.6 Data table for voltage regulation.


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Full Load Voltage No Load Voltage Percent Voltage Regulation

Table 7-1

Now connect the load in order to measure the load voltage, record thevalue in the given table 7-1, and compute the percentage regulation

electrical machines ee 250 laboratory manual_hajvery university lahore pakistan

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