single-phase transformers and ac machines-00

Electric Power / Controls

Single-Phase Transformersand AC Machines

Student Manual25988-00

A

ELECTRIC POWER / CONTROLS

SINGLE-PHASE TRANSFORMERSAND AC MACHINES

byTheodore Wildi

andthe Staff of Lab-Volt Ltd.

Copyright © 2001 Lab-Volt Ltd.

All rights reserved. No part of this publication may be reproduced,in any form or by any means, without the prior written permissionof Lab-Volt Ltd.

Legal Deposit – Fourth Trimester 2001

ISBN 978-2-89289-559-9 (2nd Edition)ISBN 2-89289-133-7 (1st Edition)

SECOND EDITION, DECEMBER 2001

Printed in CanadaApril 2010

III

Foreword

Electricity has been used since more than a century and the number of applicationsrequiring electricity is increasing constantly. As a result, the electrical power demandhas been rising since the early use of electricity. Many reasons explain whyelectricity is so popular.

One reason is the great versatility of electricity. We use it every day for cooling,heating, lighting, driving (through electric motors) etc. Furthermore, manyapparatuses that are part of our everyday life, such as telephones, televisions,personal computers, etc., require electrical power.

Another reason that explains the constantly rising demand for electricity lies in thefact that it is a highly reliable source of energy.

The Lab-Volt 0.2-kW Electromechanical Training System and related coursewareoffer a comprehensive program in the field of electrical power technology. It is anideal tool for preparing the students to the realities of the contemporary job market.

The program was developed by educators to satisfy educational requirements thatinclude industrial applications of electric power technology. The design objective wasto develop a low-power educational system with equipment that operates likeindustrial equipment.

The student manuals explain electrical principles as well as specific industrialapplications of the phenomenon discussed in each exercise. Hands-on exercisescarried out with the training system reinforce the student's knowledge of the theorybeing studied.

The method of presentation is unique in its modular concept and places emphasisupon electrical laboratory procedures performed by the individual student.

V

alternating current acAmerican wire gauge AWGampere Aampere-turn Atapplied voltage VA

average avgBritish thermal unit BTUcapacitance Ccapacitive reactance XC

clockwise cwcosine coscoulomb Ccounterclockwise ccwcounter electromotive force CEMFcurrent Icycles per second Hzdecibel dBdegree Celsius ECdegree Fahrenheit EFdegree (plane angle) ...Edirect current dcdivide ÷,/effective value (ac) rmselectromotive force EMFfarad Ffoot ft, N

frequency fgreater than >ground gndhenry Hhertz Hzhorsepower hphour himpedance Zinch in, Oinductance Linductance - capacitance LCkilohertz kHzkilohm kΩkilovar kvarkilovolt kVkilovolt-ampere kVAkilowatt kWkilowatthour kWhless than <load (resistance) RL

logarithm logmagnetomotive force MMFmaximum max.megahertz MHzmegavolt MVmegawatt MW

Symbols and Abbreviations

The user of this Student Manual may find some unfamiliar symbols andabbreviations. In general, Lab-Volt Educational System has adopted the "LetterSymbols for Units" IEEE Standard Number 260/USA Standard Number Y10.19,dated October 18, 1967.

The abbreviations have been adopted by Lab-Volt following a thorough study ofavailable abbreviations and guidelines published by the Institute of Electrical andElectronic Engineers (IEEE) and are consistent in nearly all respects with therecommendations of the International Organization for Standardization (ISO) andwith the current work of the International Electrotechnical Commission (IEC).

The symbols and abbreviations used in this manual are listed below. Each symbolderived from a proper name has an initial capital letter. Singular and plural forms areidentical.

Symbols and Abbreviations (cont'd)

VI

megohm MΩmicroampere µAmicrofarad µFmicrohenry µHmicrosecond µsmicrowatt µWmile mimilliampere mAmillifarad mFmillihenry mHmilliohm mΩmillisecond msmillivolt mVmilliwatt mWminimum min.minute (time) minminus !negative neg, !ohm Ωpeak pkphase φpicofarad pFpositive pos, +potential Epound-force lbfpound-force inch lbf@inpound-force foot lbf@ftpower (active) P

power (apparent) Ppower (instantaneous) Spower (reactive) Qpower factor PFreactance Xreactance (capacitance) XC

reactance (inductance) XL

reactive power varresistance Rresistance-capacitance RCresistance-inductance RLrevolutions per minute r/minrevolutions per second r/sroot-mean square rmssecond (time) ssine sinsource (current) IS

source (voltage) ES

tangent tantemperature Ttime ttotal current ITtotal power PT

volt Vvoltage (applied) VA

volt-ampere VAwatt Wwatthour Wh

VII

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . IX

Experiment 0 Safety and the Power Supply . . . . . . . . . . . . . . . . . . . . . . . 0-1

To learn the simple rules of safety. To learn how to use the AC/DCpower supply.

Experiment 1 The Single-Phase Transformer . . . . . . . . . . . . . . . . . . . . . 1-1

To study the voltage and current ratios of a transformer. To learnabout transformer exciting currents, volt-ampere capacity andshort-circuit currents.

Experiment 2 Transformer Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

To determine the polarity of transformer windings. To learn how toconnect transformer windings in series aiding. To learn how toconnect transformer windings in series opposing.

Experiment 3 Transformer Regulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1

To study the voltage regulation of the transformer with varyingloads. To study transformer regulation with inductive andcapacitive loading.

Experiment 4 The Autotransformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1

To study the voltage and current relationship of anautotransformer. To learn how to connect a standard transformeras an autotransformer.

Experiment 5 Transformers in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1

To learn how to connect transformers in parallel. To determine theefficiency of parallel connected transformers.

Experiment 6 The Distribution Transformer . . . . . . . . . . . . . . . . . . . . . . . 6-1

To understand the standard distribution transformer with a 120/220V secondary winding.

Experiment 7 Prime Mover and Torque Measurement . . . . . . . . . . . . . . 7-1

To learn how to connect a split-phase induction motor. To learnhow to connect the electrodynamometer. To learn how to use theProny brake.

Experiment 8 The Split-Phase Inductor Motor – Part I . . . . . . . . . . . . . . 8-1

To examine the construction of a split-phase motor. To measurethe resistance of its windings.

Table of Contents (cont'd)

VIII

Experiment 9 The Split-Phase Inductor Motor – Part II . . . . . . . . . . . . . . 9-1

To learn the basic motor wiring connections. To observe thestarting and running operation of the split-phase motor.

Experiment 10 The Split-Phase Inductor Motor – Part III . . . . . . . . . . . . 10-1

To measure the starting and operating characteristics of thesplit-phase motor under load and no-load conditions. To study thepower factor and efficiency of the split-phase motor.

Experiment 11 The Capacitor-Start Motor . . . . . . . . . . . . . . . . . . . . . . . . 11-1

To measure the starting and operating characteristics of thecapacitor-start motor. To compare its starting and runningperformance with the split-phase motor.

Experiment 12 The Capacitor-Run Motor . . . . . . . . . . . . . . . . . . . . . . . . . 12-1

To examine the construction of the capacitor-run motor. Todetermine its running and starting characteristics. To compare itsrunning and starting performance with the split-phase andcapacitor-start motors.

Experiment 13 The Universal Motor – Part I . . . . . . . . . . . . . . . . . . . . . . . 13-1

To examine the construction of the universal motor. To determineits no-load and full-load characteristics while operating onalternating current. To determine its no-load and full-loadcharacteristics while operating on direct current.

Experiment 14 The Universal Motor – Part II . . . . . . . . . . . . . . . . . . . . . . 14-1

To compare the starting torque on both AC and DC. To observethe effects of removing the compensating winding. To provide themotor with inductive compensation.

Appendices A Equipment Utilization ChartB Impedance Table for the Load Modules C Performing the Electrical Power Technology Courseware

Using the Lab-Volt Data Acquisition and Management System

IX

Introduction

The subject matter in this manual, Single-Phase Transformers and AC Machines,covers single-phase transformer. Polarity, regulation, and performance asautotransformers or when connected in parallel is explained in details. The subjectmatter of the second part covers the characteristics of the split-phase inductionmotor, capacitor-start motor and universal motor.

The exercises in this manual provide a systematic and realistic means of learning thesubject matter. Each exercise contains:

• an OBJECTIVE that clearly defines the objectives of the exercise;• a DISCUSSION of the theory involved;• a detailed step-by-step laboratory PROCEDURE in which the student observes

and measures important phenomena. Schematic diagrams facilitate connectingthe components;

• some REVIEW QUESTIONS to verify that the material has been wellassimilated.

The exercises can be carried out using either conventional instruments (AC/DCvoltmeters and ammeters, power meters, oscilloscope, etc.), or the Lab-Volt DataAcquisition and Management (LVDAM) System. Appendix C of this manual providesuseful guidelines to perform the exercises using the LVDAM system.

As a reference manual, we suggest to consult Electrical Machines, Drives, andPower Systems written by Theodore Wildi and published by Prentice Hall.

Note that the highlighted text in the manual only applies to the Imperial system ofunits.

0-1

Experiment 0

Safety and the Power Supply

OBJECTIVE

• To learn the necessary safety rules when working with electricity.

• To learn how to use adequately the AC/DC power supply.

DISCUSSION

TO ALL STUDENTS AND TEACHERS

Be sure to know the location of the first-aid kit in your class or lab at all times. Ensurethat all cuts and burns receive immediate care, no matter how minor they may seemto be. Notify your instructor about every accident. He will know what to do.

If students follow the instructions adequately, no serious accident will occur whileusing the Electro Mechanical Systems. There are many fatal shocks every yearcaused by ordinary electrical power found at home.

A thorough safety program is a necessity for anyone working with electricity.Electricity can be dangerous and even fatal to those who do not understand andpractice the simple rules of safety associated with it. There are many fatal electricalaccidents caused by well trained technicians who, either through over-confidence orcarelessness, disregard the basic rules of personal safety. The first rule of personalsafety is always:

“THINK FIRST”

This rule applies to all industrial work as well as to electrical workers. Develop goodwork habits. Learn to use your tools correctly and safely. Always study experimentsbeforehand and carefully think through all of the required procedures and methods.Make sure you know how to use all of your tools, instruments and machines beforeproceeding with an experiment. Never let yourself be distracted from your work andnever distract others around you. Do not joke around near moving machinery andelectricity.

There are generally three kinds of accidents which happen frequently to electricalstudents and technicians: electrical shocks, burns and mechanical injuries. Knowinghow to avoid them by observing simple rules will make you a safe person to workwith. Observing these precautions could save you from painful experiences (it couldeven save your life) and prevent expensive damage to the equipment.


0-2

Electric shock

Are electric shocks fatal? The physiological effects of electric currents can generallybe predicted by the chart shown in Figure 0-1.

Figure 0-1.

As you can see, it is the current that determines the intensity of an electrical shock.Currents above 100 mA are considered fatal. A person who has received a shockof over 2000 mA is in grave danger and needs immediate medical attention. Currentsbelow 100 mA are still serious and painful. As a safety rule: Do not put yourself in aposition where you could receive any kind of shock, no matter how low the currentis.


0-3

What about VOLTAGE?

Current depends upon voltage and resistance. We will now measure the resistanceof your body. Using your ohmmeter, measure your body resistance between thesepoints:

From right to left hand Ω

From hand to foot Ω

Now wet your fingers and repeat the measurements:

From right to left hand Ω

From hand to foot Ω

As you can see, the actual resistance of your body varies greatly depending on thepoints of contact, the condition of your skin, and the contact area. Notice how yourresistance varies as you squeeze the probes more or less tightly. Skin resistancemay vary between 250 Ω for wet skin and large contact areas, to 500 000 Ω for dryskin. Using the measured resistance of your body and considering 100 mA as a fatalcurrent, calculate what voltages could be fatal to you:

Use the formula: Volts = .1 x ohms.

Contact between two hands (dry): V

Contact between one hand and one foot (dry): V

Contact between two hands (wet): V

Contact between one hand and one foot (wet): V

DO NOT ATTEMPT TO PROVE THIS!

Here are nine rules to avoid electric shocks:

1. Be sure to check the condition of your equipment and the possible dangers beforeworking on a piece of equipment. The same way someone can be accidentallykilled by a supposedly unloaded gun, an electrical technician can receive shocksfrom a circuit that is supposedly turned off.

2. Never rely on safety devices such as fuses, relays and interlock systems toprotect you. They may not be working correctly and may fail to protect you whenmost needed.

3. Never remove the grounding prong of a three wire input plug. Without ground,electrical equipments become much more dangerous shock hazards.

4. Do not work on a cluttered desk. A disorganized working area full of connectingleads, components and tools leads to careless thinking, short circuits, shocks andaccidents. Develop systemized and organized work habits.


0-4

5. Do not work on a wet floor. This would greatly reduce your contact resistance tothe ground. Work on a rubber mat or an insulated floor.

6. Do not work alone. It’s an additional safety measure to have someone around toshut off the power, give artificial respiration or make an emergency call in caseof need.

7. Work with one hand behind you or in your pocket. A current between one handto the other would pass through your heart and is thus more lethal than a currentfrom hand to foot.

8. Never talk to anyone while working. Do not let yourself be distracted by yoursurroundings. Similarly, do not talk to someone working on dangerous equipment.

9. Always move slowly when working around electrical circuits. Fast and carelessmovements lead to accidental shocks and short circuits.

Burns

Accidents resulting in burns, although usually not fatal, can be painful and serious.They are generally caused by the production of heat due to electrical energydissipation. Here are four rules to avoid electrical burns:

1. Resistors may get very hot, especially those that carry high currents. Watch outespecially for five and ten watt resistors. They can severely burn your skin. Do nottouch them until they cool off.

2. Be wary of capacitors: they may still retain a charge. Not only would you receivea dangerous and sometimes fatal shock, you could also be burned from theelectrical discharge. If the rated voltage of electrolytic capacitors is exceeded ortheir polarities reversed, they may get very hot and actually burst.

3. Watch out for hot soldering irons or guns. Do not leave one where your arm mightaccidentally touch it. Never store a soldering iron away while it is still hot.Unknowing students could be burned by picking it up.

4. Hot solder can be painful when put in contact with naked skin. Wait for solderedjoints to cool off. When desoldering joints, do not shake hot solder off. You or aneighbor could receive hot solder in the eye or on his body.

Mechanical injuries

This third class of safety rules concerns all students who work with tools andmachinery. Here are five rules to avoid mechanical injury:

1. Metal corners and sharp edges on chassis and panels can cut and scratch. Filethem until they are smooth.

2. Improper tool selection for a job can result in damage to the equipment andpersonal injuries.

3. Use proper eye protection when grinding, chipping or working with hot metalswhich might splatter.


0-5

4. Protect your hands and clothes when working with battery acids, etchants, andfinishing fluids. These liquids can cause severe burns.

5. If you are unsure about something, ask your instructor.

The Power Supply

The Power Supply provides all of the necessary AC/DC power, both fixed andvariable, single phase and three-phase, to perform all of the experiments presentedin this manual.

The module must be connected to a three-phase, 240 /415 V, four wire (with fifthwire ground) system. Power is brought in through a five prong, twist-lock connectorlocated at the rear of the module. An input power cable with mating connector isprovided for this purpose.

The power supply provides the following outputs:

1. Fixed 24 V ac is made available for the use of accessory equipments such asmeters.

2. Fixed 120 /208 V, 3φ power is brought out to four terminals, labeled 1, 2, 3 andN. Fixed 208 V 3φ may be obtained from terminals 1, 2 and 3. Fixed 208 V acmay be obtained between terminals 1 and 2, 2 and 3 or 1 and 3. Fixed 120 V acmay be obtained between any one of the 1, 2 or 3 terminals and the N terminal.The current rating of this supply is 15 A per phase.

3. Variable 120 /208 V, 3φ power is brought out to four terminals, labeled 4, 5, 6 andN. Variable 3φ 0-208 V may be obtained from terminals 4, 5 and 6. Variable0-208 V ac may be obtained between terminals 4 ans 5, 5 and 6 or 4 and6.Variable 0-120 V ac may be obtained between any one of the 4, 5 or 6terminals and the N terminal. The current rating of this supply is 5 A per phase.

4. Fixed 120 V dc is brought out to terminals labeled 8 and N. The current rating ofthis supply is 2 A.

5. Variable 0-120 V dc is brought out to terminals labeled 7 and N. The currentrating of this supply is 8 A.

The full current rating of the various outputs cannot be used simultaneously. If morethan one output is used at a time, reduced current must be drawn. The neutral Nterminals are all connected together and joined to the neutral wire of the AC powerline. All power is removed from the outputs when the on-off breaker is in the offposition (breaker handle down).

CAUTION!

Power is still available behind the module face with the breaker off!Never remove the power supply from the console without firstremoving the input power cable from the rear of the module.

The variable AC and DC outputs are controlled by the single control knob on thefront of the module. The built-in voltmeter will indicate all the variable AC and thevariable and fixed DC output voltages according to the position of the voltmeter


0-6

selector switch. The power supply is fully protected against overload or short circuit.Besides the main 15 A 3φ on-off circuit breaker on the front panel, all of the outputshave their own circuit breakers. They can be reset by a common button located onthe front panel.

The rated current output may be exceeded considerably for short periods of timewithout harming the supply or tripping the breakers. This feature is particularly usefulin the study of DC motors under overload or starting conditions where high currentsmay be drawn.

All of the power sources can be used simultaneously, providing that the total currentdrawn does not exceed the 15 A per phase input breaker rating. Your power supply,if handled properly, will provide years of reliable operation and will present no dangerto you.

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of this manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

CAUTION!

High voltages are used in this Experiment! Do not make or modifyany banana jack connections with the power on unless otherwisespecified!

G 1. Examine the construction of the Power Supply. On the front panel of themodule, identify the following elements:

a. The three-pole circuit breaker on-off switch.

b. The three lamps indicating the operation of each phase.

c. The AC/DC voltmeter.

d. The AC/DC voltmeter selector switch.

e. The variable output control knob.

f. The fixed 24 V ac receptacle.

g. The fixed 120 /208 V output terminals (labeled 1, 2, 3 and N).

h. The variable 0-120 /208 V output terminals (labeled 4, 5, 6 and N).

i. The fixed DC output terminals (labeled 8 and N).

j. The variable DC output terminals (labeled 7 and N).

k. The common reset button.


0-7

G 2. State the AC or DC voltage and the rated current available from each of thefollowing terminals:

a. Terminals 1 and N = V A

b. Terminals 2 and N = V A

c. Terminals 3 and N = V A

d. Terminals 4 and N = V A

e. Terminals 5 and N = V A

f. Terminals 6 and N = V A

g. Terminals 7 and N = V A

h. Terminals 8 and N = V A

i. Terminals 1, 2 & 3 = V A

j. Terminals 4, 5 & 6 = V A

k. The low power connector = V A

G 3. Examine the interior construction of the module. Identify the following items:

a. The 3φ variable autotransformer.b. The filter capacitors.c. The thermal-magnetic circuit breakers.d. The solid state rectifier diodes.e. The five prong twist lock connector.

G 4. Insert the Power Supply into the console. Make sure that the on-off switchis in the off position and that the output control knob is turned fullycounterclockwise for minimum output. Insert the power cable through theclearance hole in the rear of the console and into the twist-lock moduleconnector. Connect the other end of the power cable into a source of 3φ120 /208 V.

G 5. a. Set the voltmeter selector switch to its 7-N position and turn the powersupply on by placing the on-off breaker switch in its “up” position.

b. Turn the control knob of the 3φ autotransformer and note that the DCvoltage increases. Measure and record the minimum and maximum DCoutput voltage as indicated by the built-in voltmeter.

V dcminimum = V dcmaximum =


0-8

c. Return the voltage to zero by turning the control knob to its full ccwposition.

G 6. a. Place the voltmeter selector switch into its 4-N position.

b. Turn the control knob and note that the AC voltage increases. Measureand record the minimum and maximum AC output voltage as indicatedby the built- voltmeter.

V acminimum = V acmaximum =

c. Return the voltage to zero and turn off the power supply by placing theon-off breaker switch in its “down” position.

G 7. What other AC voltages are affected by turning the control knob?

Terminals and = V ac

Terminals and = V ac

Terminals and and = V ac

G 8. For each of the following conditions:

a. Connect the 250 V ac meter across the specified terminals.b. Turn on the power supply.c. Measure and record the voltage.d. Turn off the power supply.

Terminals 1 and 2 = V ac



Terminals 1 and N = V ac



e. Are any of these voltages affected by turning the control knob?

G Yes G No

G 9. a. Set the voltmeter selector switch to its 8-N position.

b. Turn on the power supply.

c. Measure and record the voltage

Terminals 8 and N = V dc


0-9

d. Is this voltage affected by turning the control knob?

G Yes G No

e. Turn off the power supply.

G 10. For each of the following positions of the voltmeter selector switch:

a. Turn on the power supply and rotate the control knob to its full cwposition.

b. Measure and record the voltage.c. Return the voltage to zero and turn off the power supply.







1-1

Experiment 1

The Single-Phase Transformer

OBJECTIVE

• To study the voltage and current ratios of a transformer.

• To learn about transformer-exciting currents, volt-ampere capacity and short-circuit currents.

DISCUSSION

Transformers are probably the most universally-used pieces of equipment in theelectrical industry. They range in size from miniature units in transistor radios to hugeunits, weighing tons, used in central power distributing stations. However, alltransformers have the same basic properties which you are about to examine.

When mutual induction exists between two coils or windings, a change in currentthrough one induces a voltage in the other. Every transformer has a primary windingand one or more secondary windings. The primary winding receives electrical energyfrom a power source and couples this energy to the secondary winding by means ofa changing magnetic field. The energy appears as an electromotive force across thesecondary winding, and if a load is connected to the secondary, the energy istransferred to the load. Thus, electrical energy can be transferred from one circuit toanother, with no physical connection between the two. Transformers areindispensable in AC power distribution, since they can convert electrical power at agiven current and voltage into an equivalent power at some other current andvoltage.

When a transformer is in operation, AC currents flow in its windings and analternating magnetic field is set-up in the iron core. As a result, copper and ironlosses are produced which represents active power (watts) and causes thetransformer to heat up. Establishing a magnetic field requires reactive power (vas)which is drawn from the power line. For these reasons the total power delivered tothe primary winding is always slightly larger than the total power delivered by thesecondary winding. However, we can say, to a good approximation, that in mosttransformers:

a) Primary power (watts) = Secondary power (watts)

b) Primary volt-amperes (VA) = Secondary volt-amperes (VA)

c) Primary vars = Secondary vars.

When the primary voltage is raised beyond its rated value, the iron core (laminations)begins to saturate, and the magnetizing (exciting) current increases rapidly.


1-2

Transformers are subject to accidental short-circuits caused by natural and man-made disasters. The short-circuit currents can be very large and, unless interrupted,will quickly burn out a transformer. It is the purpose of this Experiment to show thesemajor points.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!

High voltages are present in this Experiment! Do not make anyconnections with the power on! The power should be turned offafter completing each individual measurement!

G 1. Examine the construction of the Single-Phase Transformer paying particularattention to the transformer, connection terminals and the wiring.

a. The transformer core is made up of thin sheets (laminations) of steel.Identify it.

b. Note that the transformer windings are brought out to terminals mountedon the transformer coil.

c. Note that these windings are then wired to the connection terminalsmounted on the module face.

G 2. Identify the three separate transformer windings marked on the moduleface:

a. List the rated voltage for each of the three windings:

terminals 1 to 2 = V ac




1-3

b. List the rated voltage between the following connection terminals:








c. List the rated current for each of the following connections:

terminals 1 to 2 = A ac





G 3. Using the lowest range of your ohmmeter, measure and record the DCresistance of each winding:

terminals 1 to 2 = Ω








G 4. You will now measure the unloaded secondary voltages with 120 V acapplied to the primary winding.


1-4

Figure 1-1.

a. Connect the circuit shown in Figure 1-1.

b. Turn on the power supply and adjust for 120 V ac as indicated by thevoltmeter across the power supply terminals 4 and N.

c. Measure and record the output voltage E2.

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

e. Repeat (b, c and d) measuring the output voltage E2 for each of thelisted windings.

f. winding 1 to 2 = V ac

winding 3 to 4 = V ac








1-5

G 5. a. Do your measured voltages correspond well with the rated values?Explain.

G Yes G No

b. Could you measure the value of magnetizing (exciting) current? Explain.

G Yes G No

G 6. Windings 1 to 2 and 5 to 6 each have 500 turns of wire. Winding 3 to 4 has865 turns. Calculate the following turn ratios:

a.

b.

G 7. a. Connect the circuit shown in Figure 1-2. Notice that current meter I2short-circuits winding 5 to 6.

Figure 1-2.


1-6

CAUTION!

Before turning on the power supply, verify that the voltagecontrol knob is set to zero.

b. Turn on the power supply and gradually increase the voltage until theshort-circuit current I2 is 0.4 A ac.

c. Measure and record E1, I1, and I2.

I1 = A ac

E1 = V ac

I2 = A ac


e. Calculate the current ratio:

I1 / I2 =

f. Is this current ratio equal to the turn ratio? Explain.

G Yes G No

Figure 1-3.

G 8. a. Connect the circuit shown in Figure 1-3. Notice that winding 3 to 4 isshort-circuited by the current meter I3.

CAUTION!

Before turning on the power supply, verify that the voltagecontrol knob is set to zero.


1-7

b. Turn on the power supply and gradually increase the voltage until thecurrent through the primary winding I1 is 0.4 A ac.

c. Measure and record I3 and E1.

I3 = A ac

E1 = V ac


e. Calculate the current ratio:

I1 / I3 =

f. Is this current ratio equal to the inverse of the turn ratio? Explain.

G Yes G No

Figure 1-4.

G 9. You will now determine the effect of core saturation upon the exciting-current of a transformer.

a. Connect the circuit shown in Figure 1-4. Notice that power supplyterminals 4 and 5 are now being used. These terminals will furnishvariable 0 to 208 V ac.

b. Turn on the power supply and adjust for 25 V ac as indicated by thevoltmeter across power supply terminals 4 and 5.


1-8

c. Measure and record the exciting-current I1 and the output voltage E2 foreach of the input voltages listed in Table 1-1.


E1V ac

I1mA ac

E2V ac

25

50

75

100

125

150

175

200

Table 1-1.

G 10. a. Plot your recorded current values on the graph of Figure 1-5. Draw asmooth curve through your plotted points.

b. Note that the magnetizing current increases rapidly after a certain inputvoltage has been reached.

c. Was the voltage ratio, between the two windings, affected by the coresaturation? Explain.

G Yes G No


1-9

Figure 1-5.

REVIEW QUESTIONS

1. If the short-circuit current through secondary winding 9 to 6 were 1 A ac whatwould be the current through the primary winding 1 to 2?

I1-2= A ac

2. If the secondary winding 7 to 8 is short-circuited and the primary winding 5 to 6is drawing 0.5 A ac:

a) Calculate the short-circuit current through winding 7 to 8.

I7-8= A ac


1-10

b) Why would these tests be performed as quickly as possible?

3. If 120 V ac were applied to winding 3 to 4, what would be the voltages across:

a) winding 1 to 2 = V ac

b) winding 5 to 9 = V ac

c) winding 7 to 8 = V ac

d) winding 5 to 6 = V ac

4. Which of the two windings in procedure 7 dissipates the most heat? Explain.

5. If 120 V ac were applied to winding 1 to 2 with winding 5 to 6 short-circuited:

a) What would be the current in each winding?

b) How much larger is this current than the normal value?

c) How much more heat than normal is generated in the winding under theseconditions?

2-1

Experiment 2

Transformer Polarity

OBJECTIVE

• To determine the polarity of transformer windings.

• To learn how to connect transformer windings in series aiding.

• To learn how to connect transformer windings in series opposing.

DISCUSSION

When the primary winding of a transformer is energized by an AC source, analternating magnetic flux is established in the transformer core. This alternating fluxlinks the turns of each winding on the transformer, thereby inducing AC voltages inthem. Consider the circuit shown in Figure 2-1.

Figure 2-1.

By definition an AC voltage is continually changing its value and its polarity,therefore, the voltage across the primary winding (terminals 1 and 2) keeps changingthe polarity of terminal 1 with respect to terminal 2. Terminals 1 and 2 can neverhave the same polarity. Terminal 1 must always be positive or negative with respectto terminal 2. Thus, the alternating magnetic flux induces voltages in all of the otherwindings causing an AC voltage to be produced across each pair of terminals. Theterminals of each winding also change polarity with respect to each other.

When we speak of “the polarity” of transformer windings we are identifying all of theterminals that are the same polarity (positive or negative) at any instant of time.Polarity marks are employed to identify these terminals. These marks may be blackdots, crosses, numerals, letters or any other convenient means of showing which


2-2

terminals are of the same polarity. For example, in Figure 2-1, we have used blackdots. These black dots, “polarity marks”, indicate for a given instant in time

when 1 is positive with respect to 2,3 is positive with respect to 4,6 is positive with respect to 5,7 is positive with respect to 8,

and 10 is positive with respect to 9.

It should be noted that a terminal cannot be positive by itself. It can only be positivewith respect to some other terminal. Therefore, at any given instant of time, terminals1, 3, 6, 7 and 10 are all positive with respect to terminals 2, 4, 5, 8 and 9.

When batteries (or cells) are connected in series, to obtain a higher output voltage,the positive terminal of one battery must be connected to the negative terminal of theother battery. Connected in this manner, their individual voltages will add. Similarly,if transformer windings are to be connected in series, so that their individual voltagesadd, the “polarity mark” terminal of one winding must connect to the “unmarked”terminal of the other winding.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. a. Connect the variable DC voltage output of the Power Supply(terminals 7 and N) to terminals 1 and 2 of the transformer (primarywinding) through a DC ammeter.

b. Turn on the Power Supply and set the voltage control knob so that thecurrent is approximately equal to the nominal current of the transformerprimary winding.

c. Turn off the Power Supply without modifying the setting of the voltagecontrol knob.

d. Connect a DC voltmeter across terminals 3 and 4 of the transformer(secondary winding) as indicated in Figure 2-2.


2-3

Figure 2-2.

e. Note the deflection of the DC voltmeter at the moment you close thepower supply switch. If the voltmeter pointer momentarily deflects to theright, then terminals 1 and 3 have the same polarity mark. (Terminal 1is connected to the positive side of the DC supply and terminal 3 isconnected to the positive side of the voltmeter).

f. Which two terminals are positive in windings 1 to 2 and 3 to 4?

g. Disconnect the DC voltmeter from winding 3 to 4 and connect it acrosswinding 5 to 6. Repeat (e).

h. Which two terminals are positive in windings 1 to 2 and 5 to 6?

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

G 2. In this procedure you will see the effect of connecting two windings of atransformer in series and the importance of polarity.

a. Using your AC Voltmeter, connect the circuit shown in Figure 2-3. Notethat terminals 1 and 5 are connected together.


2-4

Figure 2-3.

b. Turn on the power supply and adjust for exactly 104 V ac (one half ofthe rated voltage of winding 3 to 4).

c. Measure and record the voltages across the following terminals:

E1 to 2 = V ac

E5 to 6 = V ac

E2 to 6 = V ac


e. Remove the connection between terminals 1 and 5. Connect terminals1 and 6 together and connect the voltmeter across terminals 2 and 5 asshown in Figure 2-4.


2-5

Figure 2-4.

f. Turn on the power supply and adjust for exactly 104 V ac.

g. Measure and record the voltage across the following terminals:

E1 to 2 = V ac

E5 to 6 = V ac

E2 to 6 = V ac

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

i. Explain why the voltage with the two windings in series is approximatelyzero in one case and nearly 120 V ac in the other.

j. Which terminals have the same polarity?

G 3. a. Consider the circuit shown in Figure 2-5. Note that winding 3 to 4 isconnected to a 104 V ac power source. Do not connect the circuit at thistime!


2-6

Figure 2-5.

b. What would be the induced voltage across winding 1 to 2?

E1 to 2 V ac

c. If winding 1 to 2 is connected in series with winding 3 to 4, what threepossible output voltages can be obtained?

E = V ac and V ac and V ac.

d. Connect the circuit shown in Figure 2-5 and place the windings inseries, joining terminals 1 and 3.

e. Turn on the power supply and adjust for 104 V ac. Measure and recordthe voltage between terminals 2 and 4.

E2 to 4 = V ac

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

g. Remove the connection between terminals 1 and 3 and join terminals1 to 4.

h. Turn on the power supply and adjust for 104 V ac. Measure and recordthe voltage between terminals 2 to 3 ans 1 to 2.

E2 to 3 = V ac E1 to 2 = V ac


j. Do the results of (e) and (h) check with your prediction in (c)? Explain.

G Yes G No


2-7

k. Which terminals have the same polarity?

REVIEW QUESTIONS

1. Assuming you have a 120 V ac power source and that all of the windings onyour transformer module develop their rated voltage, show in the spacesprovided, how you would connect the windings to obtain the following voltages.

a) 240 V:

b) 88 V:

c) 180 V:

d) 92 V:

3-1

Experiment 3

Transformer Regulation

OBJECTIVE

• To study the voltage regulation of the transformer with varying loads.

• To study transformer regulation with inductive and capacitive loading.

DISCUSSION

The load on a large power transformer in a sub-station will vary from a very smallvalue in the early hours of the morning to a very high value during the heavy peaksof maximum industrial and commercial activity. The transformer secondary voltagewill vary somewhat with the load and, because motors and incandescent lamps andheating devices are all quite sensitive to voltage changes, transformer regulation isof considerable importance. The secondary voltage is also dependent upon whetherthe power factor of the load is leading, lagging or unity. Therefore, it should beknown how the transformer will behave when it is loaded with a capacitive, aninductive or a resistive load.

If a transformer were perfect (ideal) its windings would have no resistance.Furthermore, it would require no reactive power (vars) to set up the magnetic fieldwithin it. Such a transformer would have perfect regulation under all load conditionsand the secondary voltage would remain absolutely constant. But, practicaltransformers do have winding resistance and they do require reactive power toproduce their magnetic fields. The primary and secondary windings possess,therefore, an overall resistance R and an overall reactance X. The equivalent circuitof a power transformer having a turn ratio of 1 to 1, can be approximated by thecircuit shown in Figure 3-1. The actual transformer terminals are P1 P2 on the primaryside and S1 S2 on the secondary.

In between these terminals we have shown the transformer as being composed ofa perfect (ideal) transformer in series with an impedance consisting of R and X,which represents its imperfections. It is clear that if the primary voltage is heldconstant, then the secondary voltage will vary with loading because of R and X.

An interesting feature arises with a capacitive load, because partial resonance is setup between the capacitance and the reactance X so that the secondary voltage E2may actually tend to rise as the capacitive load value increases.


3-2

Figure 3-1.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


Figure 3-2.

G 1. Using your Single-Phase Transformer, Power Supply, Resistive Load, ACAmmeter and AC Voltmeter, connect the circuit shown in Figure 3-2.


3-3

G 2. a. Place all of the Resistive Load switches in their open position for zeroload current.

b. Turn on the power supply and adjust for exactly 120 V ac as indicatedby voltmeter E1.

c. Measure and record in Table 3-1 the input current I1, the output currentI2 and the output voltage E2.

d. Adjust the load resistance ZL to 1200 Ω. Make sure that the inputvoltage remains at exactly 120 V ac. Measure and record I1, I2 and E2.

e. Repeat (d) for each of the listed values in Table 3-1.


ZL(ohms)

I2(mA ac)

E2(V ac)

I1(mA ac)

4

1200

600

400

300

240

Table 3-1.

G 3. a. Calculate the transformer regulation using the no-load and full-loadoutput voltages from Table 3-1.

= %

b. Does the primary winding VA equal the secondary winding VA for everyvalue of load resistance in the Table? Explain.

G Yes G No

G 4. a. Repeat procedure 2 using the Inductive Load in place of the resistanceload.

b. Record your measurements in Table 3-2.


3-4

ZL(ohms)

I2(mA ac)

E2(V ac)

I1(mA ac)

4

1200

600

400

300

240

Table 3-2.

G 5. a. Repeat procedure 2 using the Capacitive Load in place of the resistanceload.

b. Record your measurements in Table 3-3.

ZL(ohms)

I2(mA ac)

E2(V ac)

I1(mA ac)

4

1200

600

400

300

240

Table 3-3.

G 6. You will now construct an output voltage E2 vs output current I2 regulationcurve for each type of transformer load.

a. Plot your recorded values of E2 (at each value of I2 listed in Table 3-1)on the graph of Figure 3-3.

b. Draw a smooth curve through your plotted points. Label this curve“resistive load”.

c. Repeat (a) for the inductive (Table 3-2) and capacitive (Table 3-3)loads. Label these curves “inductive load” and “capacitive load”.


3-5

Figure 3-3.

REVIEW QUESTIONS

1. Explain why the output voltage increases when capacitance loading is used.

2. A transformer has a very low impedance (small R and X):

a) What effect does this have on the regulation?

b) What effect does this have on short-circuit current?


3-6

3. Very large transformers are sometimes designed not to have optimum regulationproperties in order for the associated circuit breakers to be within reasonablesize. Explain.

4. Will transformer heating be approximately the same for resistive, inductive orcapacitive loads of the same VA rating? Explain.

G Yes G No

4-1

Experiment 4

The Autotransformer

OBJECTIVE

• To study the voltage and current relationship of an autotransformer.

• To learn how to connect a standard transformer as an autotransformer.

DISCUSSION

There is a special type of transformer which physically has only one winding.Functionally, though, the one winding serves as both the primary and secondary.This type of transformer is called an autotransformer. When an autotransformer isused to step up the voltage, part of the single winding acts as the primary, and theentire winding acts as the secondary. When an autotransformer is used to step downthe voltage, the entire winding acts as the primary, and part of the winding acts asthe secondary.

Figure 4-1 (a) and Figure 4-1 (b) show autotransformers connected for both step-upand step-down operation.

The action of the autotransformer is basically the same as the standard two-windingtransformer. Power is transferred from the primary to the secondary by the changingmagnetic field, and the secondary in turn, regulates the current in the primary to setup the required condition of equal primary and secondary power. The amount ofstep-up or step-down in voltage depends on the turns ratio between the primary andsecondary, with each winding considered as separate, even though some turns arecommon to both the primary and secondary.

Voltages and currents in the various windings can be found by two simple rules:

a) Primary apparent power (VA) equals Secondary apparent power (VA).

(VA)P = (VA)S (1)

EP IP = ES IS (2)

b) The primary (source) voltage and the secondary (load) voltage are directlyproportional to the number of turns N.

EP / ES = NP / NS (3)

Thus, in Figure 4-1 (a):

The Autotransformer

4-2

and in Figure 4-1 (b):

Figure 4-1.

These equations depend upon one important fact, that voltage EA to B and EB to C addin the same direction and do not oppose each other. We have assumed that thevoltages are in phase.

The load current, of course, cannot exceed the current carrying capacity of thewinding. Once this is known it is relatively easy to calculate the VA load which aparticular autotransformer can supply.

A disadvantage of the autotransformer is the lack of isolation between the primaryand secondary circuits, because the primary and secondary both use some of thesame turns.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


The Autotransformer

4-3

G 1. Using your Single-Phase Transformer, Power Supply, Resistive Load, ACAmmeter and AC Voltmeter, connect the circuit shown in Figure 4-2. Notethat winding 5 to 6 is connected as the primary winding across the 120 V acsource. The center tap of the winding, terminal 9, is connected to one sideof the load and the 6 to 9 portion of the primary winding is connected as thesecondary winding.

G 2. a. Place all of the Resistive Load switches in their open positions for zeroload current.

b. Turn on the power supply and adjust for exactly 120 V ac as indicatedby voltmeter E1. (This is the rated voltage for winding 5 to 6).

c. Adjust the load resistance RL to 120 Ω.

d. Measure and record currents I1, I2 and the output voltage E2.

I1 = A ac

I2 = A ac

E2 = V ac

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

Figure 4-2.

G 3. a. Calculate the apparent power in the primary and secondary circuits.

E1 x I1 = (VA)P

E2 x I2 = (VA)S

The Autotransformer

4-4

b. Are these two apparent powers approximately equal? Explain.

G Yes G No

c. Is this a step-up or step-down autotransformer?

G 4. Connect the circuit shown in Figure 4-3. Notice that winding 6 to 9 is nowconnected as the primary winding across the 60 V ac source. The 5 to 6winding is now connected as the secondary winding.

Figure 4-3.

G 5. a. Place all of the Resistive Load switches in their open positions for zeroload current.

b. Turn on the power supply and adjust for exactly 60 V ac as indicated byvoltmeter E1. (This is the rated voltage for winding 6 to 9).

c. Adjust the load resistance RL to 600 Ω.

d. Measure and record currents I1, I2 and the output voltage E2.

I1 = A ac

I2 = A ac

E2 = V ac

The Autotransformer

4-5


G 6. a. Calculate the apparent power in the primary and secondary circuits.

E1 x I1 = (VA)P

E2 x I2 = (VA)S

b. Are these two apparent powers approximately equal? Explain.

G Yes G No

c. Is this a step-up or step-down autotransformer?

REVIEW QUESTIONS

1. A standard transformer has a rating of 60 kVA. The primary and secondaryvoltages are rated respectively at 600 V and 120 V.

a) What is the rated current for each winding?

b) If the primary winding is connected to 600 V ac, what kVA load can beconnected to the secondary winding?

2. If the same transformer in Question 1 is connected as an autotransformer to600 V ac:

a) What would be the available output voltages using different connections?

The Autotransformer

4-6

b) Calculate the kVA load that the transformer can supply for each of theoutput voltages.

c) Calculate the winding currents for each of the output voltages and statewhether they exceed the rated values.

3. Using the Single-Phase Transformer and a fixed source of 120 V ac, whatwindings would you use as the primary and secondary for an output voltage of:

a) 148 V ac

b) 328 V ac

c) 224 V ac

d) 300 V ac

5-1

Experiment 5

Transformers in Parallel

OBJECTIVE

• To learn how to connect transformers in parallel.

• To determine the efficiency of parallel-connected transformers.

DISCUSSION

Transformers may be connected in parallel to furnish load currents greater than therated current of each transformer. There are two precautions to be observed whenconnecting transformers in parallel.

1) The windings to be paralleled must have identical output voltage ratings.

2) The windings to be paralleled must have identical polarities.

Very large short-circuit currents can be developed if these rules are not followed. Infact, transformers, circuit breakers and associated circuitry can be severelydamaged, or may even explode, if these short-circuit currents are large enough.

The efficiency of any machine or electrical device is given by the ratio of outputpower to input power. (Apparent power and reactive power are not used incalculating transformer efficiency). The equation for percent efficiency is:

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. Using your Single-Phase Transformer, Power Supply, Single-PhaseWattmeter, Resistive Load, AC Ammeter, and AC Voltmeter, connect the


5-2

circuit shown in Figure 5-1. Note that the two transformers are connectedin parallel. The primary windings (1 to 2) are connected together to the120 V ac power source. The wattmeter will indicate the input power. Eachsecondary winding (3 to 4) is connected in parallel to the load RL. Ammetersare inserted to measure load current IL and transformer secondary currentsI1 and I2.

Connect two resistance sections of the Variable resistance module in seriesto implement the resistive load RL shown in Figure 5-1. This is required todissipate the large amount of power involved in this exercise.

Figure 5-1.

G 2. Place all the resistance switches in their open positions for zero loadcurrent. Note that the windings are connected for voltage step-up operation(120 V primary to 208 V secondary).

G 3. Have your circuit wiring approved by the instructor before proceeding.

G 4. a. Turn on the power supply and slowly advance the voltage output controlknob while noting the transformer secondary current meters I1. If thewindings are properly phased, no load or secondary currents should beflowing.

b. Adjust the power supply voltage to 120 V ac as indicated by thevoltmeter connected across the wattmeter.


5-3

G 5. a. Gradually increase the load RL until the load current IL equals500 mA ac. Check to see that the input voltage is exactly 120 V ac.

b. Measure and record the load voltage, load current, transformersecondary currents and the input power.

EL = V ac

IL = A ac

I1 = A ac

I2 = A ac

Pin = W

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

G 6. a. Calculate the load power.

El x IL = W

b. Calculate the circuit efficiency.

Pout / Pin x 100 = %

c. Calculate the transformer losses.

Pin ! Pout = W

d. Calculate the power delivered by transformer 1.

I1 x EL = W

e. Calculate the power delivered by transformer 2.

I2 x EL = W

G 7. Is the load reasonably distributed between the two transformers? Explain.

G Yes G No


5-4

REVIEW QUESTIONS

1. Show how you would parallel connect the transformers to the source and theload in Figure 5-2. Windings 1 to 2 and 3 to 4 are rated for 2.4 kV ac andwindings 5 to 6 and 7 to 8 are rated for 400 V ac.

Figure 5-2.

2. The efficiency of a transformer which supplies a pure capacitive load is zero.Explain.

3. Name the losses which cause a transformer to heat up.

4. How does the efficiency of your Single-Phase Transformer compare to theefficiency of DC Motor/Generator? Explain.

6-1

Experiment 6

The Distribution Transformer

OBJECTIVE

• To understand the standard distribution transformer with a 120/240 V secondarywinding.

DISCUSSION

The majority of distribution transformers which supply homes and stores with powerhave one primary high voltage winding. The secondary winding furnishes 120 V forlighting and the operation of small appliances and also 240 V for electric stoves,water heaters, clothes dryers, etc. The secondary may be a center-tapped single-winding or two separate windings connected in series.

This Experiment will show how such a transformer reacts under various loadconditions.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. Using your Single-Phase Transformer , Resistive Load, Power Supply, ACAmmeter and AC Voltmeter, connect the circuit shown in Figure 6-1. Notethat the primary winding (3 to 4) is connected to the 0-208 V ac output of thepower supply, terminals 4 and 5. The transformer secondary windings 1 to2 and 5 to 6 are connected in series to obtain 240 V ac between points Aand B. Use single independent resistance sections for R1 and R2.


6-2

Figure 6-1.

G 2. a. Place all of the resistance switches in their open positions.

b. Turn on the power supply and adjust for 208 V ac as indicated by thepower supply voltmeter.

c. Measure and record in Table 6-1 the transformer total output voltage ET,the voltages across each of the loads E1 and E2, the line currents I1 andI2, and the current in the neutral line IN.

G 3. a. Place 300 Ω in each load circuit by closing the appropriate switches.

b. Measure and record all quantities in Table 6-1.

c. Why is the neutral line current zero?

PROCEDURENUMBER

R1Ω

R2Ω

I1mA

I2mA

INmA

E1V

E2V

ETV

2 (c) 4 4

3 (b) 300 300

4 (b) 300 1200

5 (c) 300 1200

6 (f) 400 400

Table 6-1.


6-3

G 4. a. Place 1200 Ω in the R2 load while leaving 300 Ω in load R1.

b. Measure and record all quantities.


d. Is the neutral line current equal to the difference of the line currents?

G 5. a. Disconnect the neutral line from the transformer by removing theconnection between the transformer and the neutral current meter IN.

b. Turn on the power supply and adjust for 208 V ac as indicated by thepower supply voltmeter.

c. Measure and record all quantities.


e. If the load R1 and R2 were incandescent lamps in a home, what wouldbe noticeable?

G 6. a. Reconnect the neutral line from the transformer to the neutral currentmeter IN.

b. Replace the load R2 with the Inductive Load.

c. Adjust R1 for a resistance of 400 Ω.

d. Adjust R2 for an inductive reactance XL of 400 Ω.

e. Turn on the power supply and adjust for 208 V ac.

f. Measure and record all quantities.

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


6-4

h. Is the current in the neutral line equal to the arithmetic difference of theline currents? Explain.

G Yes G No

REVIEW QUESTIONS

1. A home is equipped with a 120/240 V ac power system and the followingelectrical loads are in operation:

Line 1 to Neutral7 60 W lamps1 100 W lamp1 (5 A ac) motor

Line 2 to Neutral1 200 W television1 1200 W toaster4 40 W lamps

Line 1 to Line 21 2 kW dryer1 1 kW stove

a) Calculate the currents in Line 1, Line 2 and the Neutral (assume a 100%power factor for all appliances).

Line 1 = A ac

Line 2 = A ac

Neutral = A ac


6-5

b) If the neutral wire opened, which lamps would get brighter and which onesdimmer?

2. A 2400 V to 120/240 V distribution transformer has a capacity of 60 kVA:

a) What is the rated secondary (240 V) line current?

b) If the load is all placed on one side (line to neutral, 120 V) what is themaximum load that the transformer will carry without overheating?

7-1

Experiment 7

Prime Mover and Torque Measurement

OBJECTIVE

• To learn how to connect a split-phase induction motor.

• To learn how to connect the electrodynamometer.

• To learn how to use the Prony brake.

DISCUSSION

The split-phase induction motor is the simplest alternating current single-phasemotor; it is the least expensive type and for this reason, is the most frequently used.Its speed varies very little from no load to full load.

The split-phase motor does not constitute the main object of this Experiment;however, you will only learn how to connect it as a driver for the electrodynamometerand the Prony brake. The complete study of its characteristics will be taken up inlater Experiments.

You will note that the motor is started with two windings and that one of them, theauxiliary (starting) winding, is disconnected from the supply by a centrifugal switchwhen the motor has reached approximately 75% of its normal running speed. Youwill also note that when a split-phase induction motor is severely overloaded, itsspeed decreases rapidly until the centrifugal switch closes again and reenergizes theauxiliary winding. When such a condition occurs, the motor should be disconnectedimmediately from the power source in order to prevent it from burning.

The load imposed on a motor can be measured by two different means; these twotorque measuring devices are the Prony brake and the electrodynamometer. TheProny brake is an entirely passive device (no electrical power is required) while theelectrodynamometer requires external power.

The Prony brake is a friction brake which is used to act as a load for any type ofmotor and to measure the torque developed by these motors. This brake is entirelymechanical, and consists of a spring balance mounted in a standard full-size module.It is built to accurately measure the torque developed by any rotating machine placedon its left-hand side.

A self-cooling friction wheel is slipped over the shaft of the machine being tested,and attached to its output pulley by means of two screws. The friction belt of theProny brake is then removed from inside the module and slipped over the frictionwheel. The braking torque applied to the machine can be varied by turning theknurled wheel (LOAD) in the upper left corner of the module. A second knurled wheel(TORQUE PRESET) in the upper center allows to bring the spring balance back intoequilibrium by aligning the red line in the right-hand window (ZERO) with the black


7-2

line; the torque can then be read directly on the 0-3.4 NAm [0-30 lbfAin], 360E circularscale, in steps of 0.02 NAm [0.2 lbfAin].

The accuracy is better than 2% and the torque is continuously adjustable over thefull range from no load to locked rotor. When one wants to apply a known torque tothe machine, the TORQUE PRESET wheel must first be set so that the calibratedcircular scale reads exactly the desired torque value and the LOAD wheel must thenbe turned in such a way that the red line in the ZERO window is aligned with theblack line.

The electrodynamometer is a device used to accurately measure the torquedeveloped by motors of all kinds. It is actually an electrical brake in which the brakingforce can be varied electrically rather than by mechanical friction. Theelectrodynamometer is a more stable, easier to adjust, device than the mechanicalfriction brakes.

The electrodynamometer consists of a stator and a squirrel-cage rotor. The stator,unlike other electromechanical devices, is free to turn, but its motion is restricted bya helical spring.

In normal operation, DC current is applied to the stator winding. This sets up amagnetic field which passes through both the stator and the rotor. As the rotor turns(being belt-coupled to the driving motor), a voltage is induced in the rotor bars, andthe resulting eddy currents react with the magnetic field causing the stator to turn inthe same direction as the rotor.

The stator rotation is limited by the helical spring and the amount that it turns ismarked off on a scale attached to the external stator housing.

The electrodynamometer is calibrated from !0.3 to 3 NAm [!3 to 27 pound-force-inches (lbfAin)] which is more than adequate for the testing of 0.2 kW [¼ hp] motorseven when they are tested at overload conditions.

The power output of a motor depends upon its speed and the torque it develops.This relationship is given by the following equation:

(1)

where: Pout = Mechanical Output Power in watts (W)N = Speed in revolutions per minute (r/min)T = Torque in NewtonAmeter (NAm)

(1)

where: Pout = Mechanical Output Power in horse power (hp)N = Speed in revolutions per minute (r/min)T = Torque in pound-force-inches (lbfAin)


7-3

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


Split-Phase Induction Motor

G 1. Examine the front face of the Capacitor-Start Motor. (The entire Capacitor-Start Motor will be fully described in a later Experiment.)

a. Note the two separate windings. The main winding is rated 5 A while theauxiliary winding is marked “intermittent”. A circuit breaker protecting theauxiliary winding will trip if the winding is left connected to the input line(120 V) for longer than a few seconds.

b. Identify the centrifugal switch connected between terminals 6 and 7.The centrifugal switch should always be connected in series with theauxiliary winding.

c. Note also the capacitor connected between terminals 4 and 5. Thecapacitor is not used in the split-phase motor wiring; it will be used inthe capacitor start motor wiring.

G 2. Using the Power Supply, Capacitor-Start Motor, and AC Ammeter, connectthe circuit shown in Figure 7-1.

Terminals 1 and N on the power supply provide fixed single-phase 120 Vpower to the main winding in parallel with the series connection of thecentrifugal switch and the auxiliary winding. The capacitor connectedbetween terminals 4 and 5 is not used when this module is operated as asplit-phase motor.


7-4

Figure 7-1.

a. Have your instructor check your completed circuit.

b. Turn on the power supply. The motor should start running immediatelyand you should hear the click of the centrifugal switch as it opens.

c. Note the indication of the current meter.

I = A ac

d. Note whether the direction of rotation is clockwise or counterclockwise.

Rotation =

e. Measure with your hand tachometer the speed of the motor withoutload.

Speed without load = r/min

f. Turn off the power supply. You should hear the closing of the centrifugalswitch as the motor slows down and comes to a rest.

g. Disconnect the motor.

Electrodynamometer

Note: If your list of equipment does not include anelectrodynamometer, proceed to procedure 12.

G 3. a. Examine the construction of the Electrodynamometer.

b. Note the cradle construction for the electrodynamometer housing. (Thisis also called trunnion mounting.)


7-5

c. Note the helical spring at the rear of the machine. This spring has beenaccurately calibrated against the graduations marked on the front of thehousing.

d. Note the mechanical stops that limit the rotational travel of the statorhousing.

e. Identify the stator winding that is attached to the inside of the housing.(This winding carries DC current.)

f. Identify the two wire leads that carry DC current to the stator winding.(They enter the housing through the center of the helical spring.)

g. Identify the bridge rectifier located at the rear of the module. (This bridgefurnishes DC power for the stator magnetic field.)

h. Identify the variable autotransformer mounted on the front face of themodule. The braking effect of the electrodynamometer is controlled bythe strength of the stator magnetic field, which is proportional to the DCoutput of the bridge rectifier, which is varied by the variableautotransformer.

i. Identify the two AC connections terminals mounted on the module face.

G 4. a. Connect the electrodynamometer to the fixed AC output of the powersource by connecting the two input terminals of the electrodynamometerto terminals 1 and N of the power source.

DO NOT APPLY POWER AT THIS TIME!

b. Set the electrodynamometer variable transformer control knob to itsmid-position.

c. Lower the front face of the module so that you may turn the pulley byhand.

G 5. a. Turn on the power supply.

b. Keeping one hand in your pocket, for safety reasons, carefully reach inand try to turn the pulley. Caution is advised because there are severallive terminals exposed when the front panel is dropped.

Do you feel a drag when you turn the pulley?

G Yes G No

Does the stator housing tend to turn in the same direction as the pulley?

G Yes G No


7-6

G 6. a. Remove your hand from the inside of the module and advance thecontrol knob, thereby increasing the magnetic stator field.

b. Carefully reach in and turn the pulley. Did the drag increase?

G Yes G No

c. Repeat (a) but this time reduce the stator magnetic field.

d. Carefully reach in and turn the pulley. Did the drag increase?

G Yes G No

e. Turn off the power supply.

G 7. a. Couple the capacitor-start motor to the electrodynamometer with thetiming belt.

b. Connect the motor as shown in Figure 7-1.

c. Connect the electrodynamometer to terminals 1 and N of the powersupply. (There should now be two connection leads at these terminalsof the power supply, one to the capacitor-start motor and one to theelectrodynamometer.)

d. Set the electrodynamometer control knob at its full ccw position (toprovide minimum starting load for the motor).

G 8. Apply power and note if the motor revolves in a cw direction. If not, reverseits rotation by interchanging the connections to one of the motor windings.(The electrodynamometer torque can only be measured for cw rotation).

G 9. Increase the load on the motor (the electrodynamometer braking action) byvarying the control knob on the electrodynamometer until the scale markedon the stator housing indicates 1 NAm [9 lbfAin]. (The numeral 1 [9] should bedirectly beneath the red vertical line on the window beneath the pulley.)

G 10. a. Measure and record the motor current with a 1 NAm [9 lbfAin] load on themotor.

I = A ac

b. Measure and record the motor speed with a 1 NAm [9 lbfAin] load.

Speed with load = r/min


7-7

G 11. Gradually increase the load on the motor, by advancing the variableautotransformer control knob on the electrodynamometer, until the motorstalls. Quickly note the maximum value measured by theelectrodynamometer and immediately turn off the power supply.

“Breakdown” torque = NAm [lbfAin]

Prony Brake

Note: If your list of equipment does not include a Prony brake,proceed to REVIEW QUESTIONS.

CAUTION!

The friction wheel may become very hot during thisexperiment.

G 12. a. Examine the construction of the Prony Brake.

b. Remove the friction wheel from inside the module and note the holes inthe web of the pulley; they are for cooling purposes. Slip the frictionwheel over the synchronous motor shaft and secure it firmly to the motorpulley by tightening the two screws in the grooves of the motor pulley.

c. Note the floating plate inside the module. In operation, this plate ispulled in one direction by the friction belt and in the other direction by aspring which is connected to the fixed frame by means of a rack gear.A pinion gear, driven by the rack, turns the front plastic disc which hasa red line to read the torque. The TORQUE PRESET wheel effectivelyincreases the spring tension and consequently, the force which tendsto turn the floating plate.

d. The LOAD wheel effectively increases the tension on the friction belt,and tends to turn the floating plate in the opposite direction.

e. Note the friction belt attached at one end to the floating plate and at theother end to the LOAD wheel screw.

f. The complete system constitutes a spring balance.

G 13. a. Turn the LOAD wheel downwards to release the tension on the frictionbelt, and slip the belt over the friction pulley mounted on the motor.Leave the belt loose.

b. Connect the motor as shown in Figure 7-1.

c. Turn on the power supply and note if the motor revolves in a cwdirection. If not, reverse its rotation by interchanging the connections toone of the motor windings. (The Prony brake can only measure torquefor cw rotation).


7-8

d. Measure the motor current.

I = A ac

e. Vary the TORQUE PRESET wheel until the circular scale indicates9 lbfAin. This does not impose any torque on the motor and the currentshould not change. You have just preset the balance to the indicatedtorque.

f. Turn slowly the LOAD wheel upwards to tighten the belt over the frictionpulley. Note the gradual increase in the motor current. Keep turning theLOAD wheel until the hairlines in the right-hand window are aligned.The balance is now in equilibrium and is imposing a torque of 1 NAm[9 lbfAin] on the motor.

g. Measure and record the motor speed with a 1 NAm [9 lbfAin] load.

Speed with load = r/min

G 14. a. Gradually increase the load on the motor by turning the LOAD wheel.Then bring the balance into equilibrium. Repeat successively these lasttwo steps until the motor stalls. Immediately turn off the power supply.

b. Record the value of the “breakdown” torque as indicated by thecalibrated dial.

“Breakdown” torque = NAm [lbfAin]

REVIEW QUESTIONS

1. How can you reverse the direction of rotation of a split-phase induction motor?

2. Calculate the developed motor power (P) [horsepower] in procedure 10 (b) or13 (g).

P = W [hp]

3. Which torque measuring device is easier to use, the electrodynamometer or theProny brake?


7-9

4. Where is the power (heat) dissipated in the electrodynamometer?

5. Where is the power (heat) dissipated in the Prony brake?

8-1

Experiment 8

The Split-Phase Inductor Motor – Part I

OBJECTIVE

• To examine the construction of a split-phase motor.

• To measure the resistance of its windings.

DISCUSSION

Some means must be provided for getting two phases from the standard single-phase power supplied to homes if it is to be used to start and run an AC motor. Theprocess of deriving two phases from one is known as phase-splitting and is usuallybuilt into the stator circuit of the AC motor. Two-phase power creates the rotatingmagnetic field.

One method is a special auxiliary winding built into the stator called the start(auxiliary) winding to differentiate it from the actual run (main) winding of the stator.In split-phase AC motors, the start winding is used only for starting the motor andhas a high resistance and low inductive reactance. The run winding has lowresistance and high reactance. When power is first applied, both windings areenergized. Because of their different inductive reactances, the run winding currentlags the start winding current, creating a phase difference between the two. Ideally,the phase difference should be 90E; but in practical motors, it is much less.Nevertheless, the windings develop fields that are out of phase, which creates arotating magnetic field in the stator. This applies torque to the rotor, starting themotor.

When the motor gets up to operating speed, the rotor is able to follow thealternations of the magnetic field created by the run winding without the field of thestart winding. The start winding is then switched out of the circuit by a mechanicaldevice called a centrifugal switch, because it is operated by the centrifugal forcecreated by the rotor revolutions. The direction of a split-phase rotating field can bereversed by reversing the connections to the start winding. This changes thedirection of the initial phase shift, creating a magnetic field rotating in the oppositedirection.

The motor speed depends essentially upon the AC power line frequency and thenumber of poles on the stator.

The split-phase motor, like a single-phase induction motors, vibrates mechanicallyat twice the power line frequency.


8-2

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. Examine the construction of the Capacitor-Start Motor, paying particularattention to the motor, centrifugal switch, connection terminals and thewiring.

The capacitor, mounted in the rear of the module, is used only when themodule is connected as a capacitor-start motor.

G 2. Viewing the motor from the front of the module:

a. The main stator winding is composed of many turns of large diameterwire. Identify the main winding.

b. The auxiliary stator winding, wound inside the main stator winding, iscomposed of fewer turns of smaller diameter wire. Identify the auxiliarywinding.

c. Does the auxiliary winding exactly straddle the main winding? Explain.

G Yes G No

d. How many main stator poles are there?

e. How many auxiliary poles are there?

f. This is a pole motor.

g. Note that there are a number of slots distributed in each pole.

h. Note the construction of the rotor.

i. Note the rotor aluminum end ring.

j. Note that the fan is integrally cast with the end ring.


8-3

k. Note the air gap separating the rotor and the stator.

l. Estimate the air gap distance in mm [inch].

G 3. Viewing the motor from the rear of the module:

a. Identify the centrifugal switch mechanism mounted on the shaft.

b. Pull outward on the centrifugal weights and note the action of theinsulated sleeve.

c. Note that the stationary electrical contacts open when the weights arepulled out.

d. If the coiled springs on the centrifugal switch were stiffer, would theelectrical contacts open at a lower or higher shaft speed?

G 4. Viewing the front face of the module:

a. The main winding (many turns of heavy wire) is connected to terminals and .

b. The auxiliary winding (fewer turns of finer wire) is connected toterminals and .

c. The centrifugal switch contacts are connected to terminals and .

d. The capacitor (not used in the split-phase motor wiring) is connected toterminals and .

e. Note that the current rating for the main winding is marked 5 A while theauxiliary winding is marked “intermittent”.

Note: The circuit breaker protecting the auxiliary winding will tripif the winding is left across the input line 120 V) for longer than afew seconds.

CAUTION!

Always connect the centrifugal switch in series with theauxiliary winding and the input line, unless you areinstructed not to do so.


8-4

G 5. Using your ohmmeter measure and record the resistance of the:

main winding = Ω

aux winding = Ω

G 6. Using your Power Supply, DC Voltmeter/Ammeter and Capacitor-StartMotor, connect the circuit shown in Figure 8-1.

Figure 8-1

G 7. Turn on the power supply and adjust for exactly 5 V dc as indicated by thevoltmeter across the main winding (terminals 1 and 2).

Imain winding = A dc

Rmain winding = E/I = Ω

G 8. Return the voltage to zero and turn off the power supply. Connect the circuitshown in Figure 8-2.

G 9. Turn on the power supply and adjust for exactly 5 V dc as indicated by thevoltmeter across the auxiliary winding (terminals 3 and 4).

Iaux winding = A dc

Raux winding = E/I = Ω


8-5

Figure 8-2.

G 10. a. Return the voltage to zero and turn off the power supply.

b. Compare the results of procedure 5 with the results of procedure 7and 9.

c. Note that although the main winding has many more turns of wire thanthe auxiliary winding, its resistance is lower. Explain why.

REVIEW QUESTIONS

1. If a split-phase motor has two poles on the main winding, how many poles areneeded for the auxiliary winding?

2. How many poles are there respectively on the running and the starting windingof an 8 pole split-phase motor?

Running Winding = poles.

Starting Winding = poles.


8-6

3. Why is an auxiliary winding necessary?

4. Why must the auxiliary winding be different from the main winding in a split-phase motor?

5. What would happen if the starting and running windings were identical?

9-1

Experiment 9

The Split-Phase Inductor Motor – Part II

OBJECTIVE

• To learn the basic motor wiring connections.

• To observe the starting and running operation of the split-phase motor.

DISCUSSION

When power is applied to a split-phase induction motor, both the running (main) andthe starting (auxiliary) windings draw from 4 to 5 times their normal full load current.This means that the heat loss in these windings is from 16 to 25 times higher thannormal. As a result, the starting period must be kept short to prevent overheating ofthe windings.

The high starting currents also produce a proportionally high current in the squirrel-cage rotor, so that the entire motor heats up very quickly during start-up.

The smaller diameter wire employed in the auxiliary winding of split-phase motorsis particularly susceptible to overheating and will burn out if it is not disconnectedfrom the power line within 4 to 6 seconds.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. Your power supply must be adjusted for an output of 100 V ac to performthe procedures in this Experiment.

a. Connect an AC voltmeter across power supply terminals 4 and N.

b. Turn on the power supply and adjust for an output of 100 V ac asindicated by the voltmeter. Do not touch the voltage output control forthe remainder of this Experiment unless told to do so.


9-2

c. Turn off the power supply.

G 2. a. Connect the main winding of the capacitor-start motor, terminals 1 and2, to the pre-adjusted 100 V ac output of the power supply, terminals 4and N.

b. Close the power supply switch for no longer than 3 seconds.

c. Did the motor growl?

G Yes G No

d. Did the motor turn?

G Yes G No

G 3. a. Lower the front face of the module. Carefully reach in behind the frontface of the module so that you may give the motor shaft a quick turn byhand at the moment of power supply switch closure.


c. Did the motor turn?

G Yes G No

d. What determined the direction of rotation of the motor?

e. Return the front face of the module to its normal position.

G 4. a. Disconnect the main winding, terminals 1 and 2, from the power supply.

b. Connect the auxiliary winding, terminals 3 and 4, to the pre-adjusted100 V ac output of the power supply terminals 4 and n.

c. Close the power supply switch for no longer than 3 seconds.

d. Did the motor make a growling sound?

G Yes G No

e. Did the motor turn?

G Yes G No


9-3

G 5. a. Connect the main winding, terminals 1 and 2, in parallel with theauxiliary winding, terminals 3 and 4.

b. Connect the parallel windings to the pre-adjusted 100 V ac output of thepower supply.

c. Close the power supply switch for no longer than 3 seconds.

d. Did the motor start?

G Yes G No

e. Was the motor noisy?

G Yes G No

f. Note the direction of rotation.

G 6. a. Interchange the leads connecting the two windings in parallel.


c. Note the direction of rotation.

d. Give a rule for reversing the rotation of a split-phase motor.

G 7. Connect the circuit shown in Figure 9-1. The centrifugal switch is connectedin series with the auxiliary winding and both windings are connected inparallel across the 100 V ac power source terminals 4 and N. Note that thecapacitor, connected between terminals 4 and 5, is not used when themodule is operated as a split-phase motor.


9-4

Figure 9-1.

G 8. a. Close the power supply switch. The output voltage control shouldremain at its 100 V setting.

b. Did the motor start?

G Yes G No

c. Did the centrifugal switch operate?

G Yes G No

d. Estimate the starting time.

T = s

e. Using your hand tachometer, measure the running speed.

Speed = r/min

f. Reduce the input voltage to 80 V ac as indicated by the voltmeter andmeasure the running speed.

Speed = r/min

g. Return the voltage to 100 V ac and turn off the power supply.

G 9. Connect the circuit shown in Figure 9-2. Note that both windings areconnected in parallel and that the centrifugal switch is in series with theparallel connected motor windings and the 100 V ac power supply terminals4 and N.


9-5

Figure 9-2.

G 10. Before applying power to the motor answer the following questions:

a. Will current flow through both windings?

G Yes G No

b. Will a starting torque be developed?

G Yes G No

c. Will the motor start to turn?

G Yes G No

d. What will eventually happen?

G 11. a. Close the power supply switch and note what happens.

b. Observe the operation of the centrifugal switch.

c. At approximately what speed does the centrifugal switch close?

Speed = r/min



9-6

REVIEW QUESTIONS

1. Will a single-phase induction motor start if only the running (main) or the starting(auxiliary) winding is excited?

2. Will such a motor run on one winding once it has been started?

G Yes G No

3. How could your reverse the rotation of the motor?

4. What will happen to your motor when power is applied if springs twice as stiff areused on the centrifugal switch?

5. Explain in detail the behavior of your motor in procedure 11.

6. If the running winding and the auxiliary winding were connected in series, wouldthe motor turn? Explain.

G Yes G No


9-7

7. Does the speed of a split-phase motor change appreciably with a change in theapplied voltage?

G Yes G No

10-1

Experiment 10

The Split-Phase Inductor Motor – Part III

OBJECTIVE

• To measure the starting and operating characteristics of the split-phase motorunder load and no-load conditions.

• To study the power factor and efficiency of the split-phase motor.

DISCUSSION

The starting current of a split-phase motor is usually four to five times normal full-load current. This produces two effects: 1) the motor heats very rapidly duringstart-up; and 2) the high starting current can cause a large line voltage drop so thatthe starting torque may be seriously reduced.

The no-load current is usually 60% to 80% of the full-load current, which is highcompared to three-phase motors. Most of the no-load current is used to produce themagnetic field in the motor, and only a small portion is used to overcome mechanicalfriction and the copper and iron losses. Because of the large magnetizing current,the power factor of these motors is rarely more than 60%, even at full-load.

Split-phase motors tend to be much noisier than their three-phase counterparts,because of the inherent 120 cycle mechanical vibration. This vibration can bereduced by using resilient rubber mounting supports.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


Starting Currents

G 1. Using your Capacitor-Start Motor, Power Supply, and AC Ammeter, connectthe circuit shown in Figure 10-1.


10-2

Figure 10-1.

G 2. Close the power supply switch and measure the current through the mainwinding as quickly as possible - within 3 seconds.

Imain winding = A ac

G 3. a. Disconnect your leads from the main winding and connect them to theauxiliary winding, terminals 3 and 4 as shown in Figure 10-2.

Figure 10-2.

b. Repeat procedure 2. Remember to take your measurement as quicklyas possible.

Iauxiliary winding = A ac

G 4. a. Connect both windings in parallel, terminals 1 to 3 and 2 to 4 as shownin Figure 10-3.


10-3

Figure 10-3.

b. Couple the electrodynamometer to the capacitor-start motor with thetiming belt.

c. Connect the electrodynamometer to the fixed 120 V ac output of thepower supply, terminals 1 and N.

d. Set the electrodynamometer control knob at its full cw position (toprovide a maximum starting load for the split-phase motor).

e. Close the power supply switch and measure the starting current asquickly as possible - within 3 seconds.

Istarting = A ac

No-Load Operation

G 5. Using your Single-Phase Wattmeter, AC Ammeter and AC Voltmeter,connect the circuit shown in Figure 10-4.

Note that the module is wired as a standard split-phase motor.

Figure 10-4.


10-4

G 6. a. Turn on the power supply and adjust for 120 V ac as indicated by thevoltmeter across the motor.

b. Measure and record in Table 10-1 the line current, the power and motorspeed. Note and record the relative motor vibration.

c. Repeat (b) for each of the input voltages listed in the Table 10-1.


E(volts)

I(amps)

P(watts)

SPEED(r/min) VIBRATION

120

90

60

30

Table 10-1.

Full-Load Operation

G 7. a. Couple the electrodynamometer to the capacitor-start motor with thetiming belt.

b. Connect the input terminals of the electrodynamometer to the fixed120 V ac output of the power supply, terminals 1 and N.

c. Set the electrodynamometer control knob at its full ccw position (toprovide a minimum starting load for the split-phase motor).

G 8. a. Turn on the power supply and adjust for 120 V ac.

b. Measure and record in Table 10-2 the line current, the power and motorspeed.

c. Repeat (b) for each of the torques listed in the Table 10-2, maintainingthe input voltage at 120 V ac.



10-5

TORQUE(NAm)

I(amps) VA Pin

(watts)SPEED(r/min)

Pout(watts)

0

0.3

0.6

0.9

1.2

Table 10-2.

TORQUE(lbfAin)

I(amps) VA P

(watts)SPEED(r/min)

Pout(hp)

0

3

6

9

12

Table 10-2.

G 9. a. Calculate and record in the Table 10-2, the apparent power (in VA)delivered to the motor for each of the listed torques.

b. Calculate and record in the Table 10-2, the developed mechanicaloutput power for each of the listed torques. Use the formula:

where Pout = Mechanical Output Power in watts (W)N = Speed in revolution per minute (r/min)T = Torque in NewtonAmeter (NAm)

where Pout = Mechanical Output Power in horse power (hp)N = Speed in revolution per minute (r/min)T = Torque in pound-forceAinches (lbfAin)


10-6

G 10. You will now determine the maximum starting torque developed by thecapacitor-start motor.

a. Disconnect the wattmeter and metering modules from your circuit.

b. Connect the input of your capacitor-start motor to terminals 2 and N ofthe power supply (fixed 120 V ac).

c. Set the electrodynamometer control knob at its full cw position (formaximum loading).

d. Close the power supply switch and quickly measure the developedtorque on the electrodynamometer scale. Open the power supplyswitch.

Starting Torque = NAm [lbfAin]

REVIEW QUESTIONS

1. From Table 10-2 state the no-load (0 NAm [lbfAin] torque):

a) apparent power = VA

b) active power = W

c) reactive power = var

d) power factor =

2. From Table 10-2 state the full-load (1.2 NAm [9 lbfAin] torque):


b) active power = W


d) power factor =

e) power delivered = W [hp]

electrical equivalent = W

f) efficiency of the motor = %

g) motor losses = W

3. What is the approximate full-load current of your capacitor-start motor?

I = A ac


10-7

4. How much larger is the starting current than the full-load operating current?

5. Based on procedures 1, 2 and 3, explain why the starting (auxiliary) windingheats much faster than the main winding.

6. Does the no-load speed of a split-phase motor change greatly with changes inthe applied voltage?

G Yes G No

7. How many times greater is the starting torque than the normal full-load torque?

11-1

Experiment 11

The Capacitor-Start Motor

OBJECTIVE

• To measure the starting and operating characteristics of the capacitor-start motor.

• To compare its starting and running performance with the split-phase motor.

DISCUSSION

When the split-phase rotating field was described, it was stated that the differentresistance-reactance ratio of the two windings was designed to give the differencein time phase of the currents in the windings necessary to produce a rotatingmagnetic field.

In two-phase machines, where the windings are identical but displaced in space by90E, the ideal time phase displacement of the winding currents is 90E.

For both two-phase and split-phase motors the torque developed at starting can becalculated using the relationship:

T = k I1I2 sin α

where k is a machine constant, I1 and I2 are the currents in the windings, and α is theangle between the currents.

Because of the small magnitude of α in the split-phase machine the developedtorque is relatively low. It is possible to increase α by adding capacitance in serieswith the auxiliary winding. If too much capacitance is added, the impedance of thewinding is increased to the point that there is an unacceptable reduction in thecurrent which more than offsets the benefit gained from increasing α.

The optimum value of C is that where the product of the sine of α and the auxiliarywinding current is a maximum.

The capacitor and the start winding are disconnected by a centrifugal switch, just asin the case of the standard split-phase motor. Reversing the direction of rotation ofa capacitor start motor is the same as in the case of the split-phase motor, that is,reverse the connections to the start or to the running winding leads.

EQUIPMENT REQUIRED



11-2

PROCEDURE

CAUTION!


G 1. Using your Capacitor-Start Motor, Power Supply, and AC Ammeter, connectthe circuit shown in Figure 11-1. Note that the fixed 120 V ac output of thepower supply, terminals 1 and N are being used.

Figure 11-1.

G 2. Close the power supply switch and measure the current through the mainwinding as quickly as possible - within 3 seconds.

Imain winding = A ac

G 3. a. Disconnect the leads from the main winding and connect them to theauxiliary winding and capacitor, as shown in Figure 11-2.

Figure 11-2.

b. Repeat procedure 2.

Note: Remember to take your measurement as quickly aspossible.

Iauxiliary winding = A ac


11-3

G 4. a. Connect both windings in parallel, terminals 1 to 3 and 2 to 5, as shownin Figure 11-3.

Figure 11-3.

b. Couple the electrodynamometer to the capacitor-start motor with thetiming belt.

c. Connect the input terminals of the electrodynamometer to the fixed120 V ac output of the power supply, terminals 1 and N.

d. Set the electrodynamometer control knob at its full cw position toprovide a maximum starting load for the capacitor-start motor.

e. Close the power supply switch and measure the starting current asquickly as possible - within 3 seconds.

Istarting = A ac

G 5. Compare your results from procedures 2, 3 and 4 with the results fromprocedures 2, 3 and 4 of Experiment 10.

a. What conclusions can you make about the main winding currents?

b. What conclusions can you make about the auxiliary winding currents?


11-4

c. What conclusions can you make about the starting current for each typeof motor?

G 6. Using your Single-Phase Wattmeter, Electrodynamometer, AC Ammeterand AC Voltmeter, connect the circuit shown in Figure 11-4.

Note that the module is wired as a standard capacitor-start motor.

Figure 11-4.

G 7. Set the electrodynamometer control knob at its full ccw position to provideminimum starting torque for the capacitor-start motor.


b. Measure and record in Table 11-1, the line current, the power and motorspeed.

c. Repeat (b) for each of the torques listed in the Table.



11-5

TORQUE(NAm)

I(amps) VA Pin

(watts)SPEED(r/min)

Pout(watts)

0

0.3

0.6

0.9

1.2

Table 11-1.

TORQUE(lbfAin)

I(amps) VA P

(watts)SPEED(r/min)

Pout(hp)

0

3

6

9

12

Table 11-1.

G 9. a. Calculate and record in the Table 11-1, the apparent power delivered tothe motor for each of the listed torques.

b. Calculate and record in the Table 11-1, the developed power (Pout)[horsepower] for each of the listed torques.

G 10. You will now determine the starting torque developed by the capacitor-startmotor. Because the current is very high when the motor starts, you willmeasure the starting torque when the supply voltage is half the rated valueand then, you will determine the starting torque for the rated value bycalculations.

a. Disconnect the wattmeter and metering modules from your circuit.

b. Set the electrodynamometer control knob to its full CW position (formaximum loading).

c. Turn on the power supply and set the voltage applied to the motor tohalf the rated voltage value shown on the motor nameplate. Measurethe developed torque on the electrodynamometer scale. Turn off thepower supply.

Starting torque (half the rated voltage value) = N·m [lbfAin]


11-6

d. Calculate the starting toque developed by the motor when the supplyvoltage corresponds to the rated value. The starting torque is nearlyproportional to the square of the applied voltage; thus the starting torqueobtained at rated value would be four times greater than at half therated value.

Starting torque (rated voltage value) = N·m [lbfAin]

REVIEW QUESTIONS



b) active power = W


d) power factor =



b) active power = W


d) power factor =




g) motor losses = W

3. What is the approximate full-load current of your capacitor-start motor?

I = A ac



11-7

5. Compare these results with those found for the split-phase motor(Experiment 10).

12-1

Experiment 12

The Capacitor-Run Motor

OBJECTIVE

• To examine the construction of the capacitor-run motor.

• To determine its running and starting characteristics.

• To compare its running and starting performance with the split-phase andcapacitor-start motors.

DISCUSSION

Single-phase motors are all rather noisy because they vibrate at 120 Hz whenoperated on a 60 Hz power line. Various attempts to reduce this noise, such asresilient rubber mounting, are never totally effective in eliminating this vibration,particularly when the motor is directly coupled to a large resonant-prone fan.

The capacitor-run motor is very useful in this type of application, because the motorcan be designed to have low vibration under full-load. The capacitor serves to shiftthe phase on one of the windings so that the voltage across the winding is at 90Efrom the other winding, thus making the capacitor-run motor a truly two-phasemachine at its rated load. Because the capacitor remains in the circuit at all times nocentrifugal switch is required.

When running at no-load the motor is always noisier than at full-load, because onlyunder full-load does it run as a true two-phase machine. If the proper value ofcapacitance is chosen, the currents through each of the two equal stator windings,under full-load, can be made such that the power factor is close to 100%. However,the starting torque is quite low and the capacitor-run motor is not recommended forsevere starting conditions.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!



12-2

G 1. Examine the construction of the Capacitor-Run Motor, paying particularattention to the motor, capacitor, connection terminals and the wiring.


a. Both stator windings are composed of many turns of wire. Identify thestator windings.

b. Do the stator windings appear to be identical?

G Yes G No

c. Do the windings exactly straddle each other? Explain.

G Yes G No

d. How many poles are there? Explain.

e. This is a pole motor.

f. Note that there are a number of slots distributed in each pole.

g. Note the construction of the rotor.

h. Note the rotor aluminum end ring.

i. Note that the fan is integrally cast with the end ring.

G 3. Viewing the motor from rear of the module:

a. Note the capacitor and its rating.

b. Is this capacitor electrolytic? Explain.

G Yes G No


a. Note the two stator windings.

b. One winding is connected to terminals and .


12-3

c. The other winding is connected to terminals and .

d. Note that the voltage and current ratings for each winding are identical.

e. The capacitor is connected to terminals and .

G 5. Using your Capacitor-Run Motor, Power Supply, Single-Phase Wattmeter,Electrodynamometer, AC Ammeter and AC Voltmeter, connect the circuitshown in Figure 12-1.

Figure 12-1.

G 6. a. Couple the electrodynamometer to the capacitor-run motor with thetiming belt.


c. Set the electrodynamometer control knob at its full ccw position (toprovide a minimum starting load for the capacitor-run motor).


b. Measure and record in Table 12-1 the line current, the power and motorspeed.


12-4

TORQUE(NAm)

I(amps) VA Pin

(watts)SPEED(r/min)

Pout(watts)

0

0.3

0.6

0.9

1.2

Table 12-1.

TORQUE(lbfAin)

I(amps) VA Pin

(watts)SPEED(r/min)

Pout(hp)

0

3

6

9

12

Table 12-1.


d. Was there a noticeable difference in the level of motor vibrationbetween no-load and full-load?

G Yes G No


G 8. a. Calculate and record in the Table 12-1, the apparent power delivered tothe motor for each of the listed torques.

b. Calculate and record in the Table 12-1 the developed mechanical outputpower for each of the listed torques. Use the formula:

where Pout = Mechanical Output Power in watts (W)N = Speed in revolution per minute (r/min)T = Torque in NewtonAmeter (NAm)


12-5

where Pout = Mechanical Output Power in horse power (hp)N = Speed in revolution per minute (r/min)T = Torque in pound-forceAinches (lbfAin)

c. Set the electrodynamometer control knob to its full cw position (formaximum loading).

d. Close the power supply switch and quickly measure the developedtorque on the electrodynamometer scale and the starting current. Openthe power supply switch.

Starting Torque = NAm [lbfAin]

Starting Current = A ac

REVIEW QUESTIONS



b) active power = W


d) power factor =



b) active power = W


d) power factor =




g) motor losses = W

3. What is the approximate full-load current of your capacitor-run motor?

I = A ac


12-6


5. Compare these results with those found for the split-phase and capacitor-startmotors (Experiments 10 and 11).

6. How can you change the direction of rotation of a capacitor-run motor?

7. Can you explain why oil-filled capacitors must be used for capacitor-run motorsinstead of the more economical AC electrolytic types?

13-1

Experiment 13

The Universal Motor – Part I

OBJECTIVE

• To examine the construction of the universal motor.

• To determine its no-load and full-load characteristics while operating onalternating current.

• To determine its no-load and full-load characteristics while operating on directcurrent.

DISCUSSION

The AC/DC universal motor is found in portable tools such as electric drills, saws,sanders, etc., and in home appliances such as vacuum cleaners, electric mixers,blenders, etc., where high speed, power and small size are an advantage.

It is closer in concept to the DC motor than to the AC motor and has some inherentdisadvantages, which can be avoided in purely AC induction motor. The maindisadvantage is the need for commutation and brushes.

The universal motor is basically a series DC motor which is specially designed tooperate on AC as well as on DC. A standard DC series motor has very poorcharacteristics when operated on AC, mainly due to two reasons:

a) The high reactance of both the armature and field windings limits AC currentto a much lower value than DC current (for the same line voltage).

b) If solid steel is used for the stator frame, AC flux will produce large eddycurrents in the frame with consequent heating.

The reactance of the armature winding can be lowered by placing a compensatingwinding on the stator so that the fluxes oppose or “cancel” each other. This samecompensating winding can be connected in series with the armature winding. In thiscase, the motor is said to be conductively compensated. Under these conditions, theuniversal motor will have similar operating characteristics whether on AC or DCpower.

The compensating winding may be simply shorted upon itself, so that it behaves likea short-circuited secondary of a transformer (the armature winding acting as theprimary). The induced AC current in the compensating winding again opposes or“bucks” the armature current and the motor is said to be inductively compensated.The reactance of the field winding can be kept low by limiting the number of turns.


13-2

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


G 1. Examine the construction of the Universal Motor, paying particular attentionto the motor, brushes, connection terminals and wiring.

G 2. Viewing the motor from the rear of the module:

a. Identify the armature winding.

b. Identify the stator.

c. Identify the main series winding.

d. Identify the compensating winding.


a. Identify the commutator.

b. Identify the brushes.

c. The neutral position of the brushes is indicated by a red line marked onthe motor housing. Identify it.

d. The brushes can be positioned on the commutator by moving the leverto the right or left of the red line. Move the lever both ways and thenreturn it to its neutral position.


a. The main series winding is connected to terminals and .

b. The compensating winding is connected to terminals and .

c. The brushes (commutator and armature winding) are connected toterminals and .


13-3

Finding the Neutral

G 5. You will now determine the neutral brush position for your motor by usingalternating current. Using your Power Supply, AC Voltmeter and UniversalMotor, connect the circuit shown in Figure 13-1. Terminals 4 and N on thepower supply will furnish variable 0-120 V ac as the voltage output controlis advanced.

DO NOT APPLY POWER AT THIS TIME!

Figure 13-1.

G 6. Unlock the Universal Motor and move it forward approximately 10 cm [4 in].Reach in behind the front face of the module and move the brushpositioning lever to its maximum cw position. Do not slide the module backin place (you will later move the brushes again).

G 7. Turn on the power supply and adjust the output control until approximately80 V ac is applied to the armature. The AC voltage that appears across thecompensating winding is induced by the AC current through the armature.

G 8. a. Carefully reach in behind the front face of the module (preferablykeeping one hand in your pocket) and move the brushes from oneextreme position to another. You will notice that the induced voltageacross the compensating winding increases and then drops again asyou approach the other extreme position.

b. Leave the brushes at the position where the induced voltage ismaximum. This is the neutral point of your Universal Motor. Each timeyou use the Universal Motor the brushes should be set at the neutralposition.

c. Return the voltage to zero and turn off the power supply. Slide yourUniversal Motor back in place.

G 9. Connect the armature and compensating winding in series, across the0-120 V ac output of the power supply as shown in Figure 13-2.


13-4

Figure 13-2.


b. If the line current is less than 1 A ac with 30 V ac applied, thecompensating winding is producing a flux in the same direction as thearmature thereby increasing the inductance (and reactance). If thisoccurs, interchange the leads to the armature or to the compensatingwinding.

c. Measure and record the line current.

I = A ac

Note: If the armature revolves, the brushes are not exactly at theneutral position.


G 11. Using your Single-Phase Wattmeter and Electrodynamometer, connect thecircuit shown in Figure 13-3. (Remember to keep the armature and thecompensating winding compensating winding connections as inprocedure 10).


13-5

Figure 13-3.

G 12. a. Couple the electrodynamometer to the universal motor with the timingbelt.


c. Set the electrodynamometer control knob at its full ccw position (toprovide a minimum starting load for the universal motor).


b. Measure and record in Table 13-1, the line current, the power and motorspeed.

Note that there is very little sparking at the brushes.



TORQUE(NAm)

I(amps) VA Pin

(watts)SPEED(r/min)

Pout(watts)

0

0.3

0.6

0.9

1.2

Table 13-1.


13-6

TORQUE(lbfAin)

I(amps) VA P

(watts)SPEED(r/min)

Pout(hp)

0

3

6

9

Table 13-1.

G 14. a. Calculate and record in the Table 13-1 the apparent power delivered tothe motor for each of the listed torques.

b. Calculate and record in the Table 13-1 the developed power (Pout)[horsepower] for each of the listed torques.

G 15. Replace the AC ammeter and voltmeter with DC meters, remove the Single-Phase Wattmeter and connect the input to the variable DC output, terminals7 and N, of the power supply as shown in Figure 13-4.

Figure 13-4.

G 16. Repeat procedures 13 and 14 using DC power instead of AC power andcomplete Table 13-2. To obtain P, multiply V by I.


13-7

TORQUE(NAm)

I(amps)

Pin(watts)

SPEED(r/min)

Pout(watts)

0

0.3

0.6

0.9

1.2

Table 13-2.

TORQUE(lbfAin)

I(amps)

P(watts)

SPEED(r/min)

Pout(hp)

0

3

6

9

Table 13-2.

REVIEW QUESTIONS

1. From Table 13-1 state the AC operating no-load (0 NAm [lbfAin] torque):


b) active power = W


d) power factor =

e) motor speed = r/min

2. From Table 13-1 state the AC operating full-load (1.2 NAm [9 lbfAin] torque):


b) active power = W


d) power factor =

e) motor speed = r/min


13-8

f) power delivered = W [hp]


g) efficiency of the motor = %

h) motor losses = W

3. From Table 13-2 state the DC operating no-load (0 NAm [lbfAin] torque):

a) power = W

b) motor speed = r/min

4. From Table 13-2 state the DC operating full-load (1.2 NAm [9 lbfAin] torque):

a) power = W

b) motor speed = r/min

c) power delivered = W [hp]


d) efficiency of the motor = %

e) motor losses = W

5. Compare the universal motor operating characteristics on AC and DC and listthe major differences.

14-1

Experiment 14

The Universal Motor – Part II

OBJECTIVE

• To compare the starting torque on both AC and DC.

• To observe the effects of removing the compensating winding.

• To provide the motor with inductive compensation.

DISCUSSION

The starting torque of a universal motor is determined by the current that flowsthrough the armature and field windings. Due to the inductive reactance of thesewindings the AC starting current (with equal supply voltages). Consequently, thestarting torque on AC power will be lower than the starting torque on DC power.

The compensating winding has the important role of reducing the overall reactanceof the motor. However, it also has an equally important part in opposing armaturereaction, thereby improving commutation. An uncompensated universal motor willlose most of its power. Sparking at the brushes will also be markedly worse.

EQUIPMENT REQUIRED


PROCEDURE

CAUTION!


Starting Torque

G 1. Using your Universal Motor, Electrodynamometer, Power Supply and DCVoltmeter/Ammeter, connect the circuit shown in Figure 14-1.

Remember to keep the armature and compensating winding connections asin the previous Experiment (procedure 10).


14-2

Figure 14-1.

G 2. a. Couple the electrodynamometer to the universal motor with the timingbelt.


c. Set the electrodynamometer control knob at its full cw position (toprovide a maximum starting load for the universal motor).

d. Make sure the brushes on the universal motor are set at their neutralposition.

G 3. a. Turn on the power supply and adjust for 30 V dc as indicated by thevoltmeter across the motor windings.

b. Measure and record the motor current and the torque developed.

I = A dc

Torque = NAm [lbfAin]


G 4. Reconnect your circuit for AC operation as shown in Figure 14-2.


14-3

Figure 14-2.

G 5. a. Turn on the power supply and adjust for 30 V ac as indicated by thevoltmeter across the motor windings.

b. Measure and record the motor current and the torque developed.

I = A ac

Torque = NAm [lbfAin]


G 6. Explain the results of procedures 3 and 5.

Uncompensated Operation

G 7. Eliminate the compensating coil winding by reconnecting your circuit asshown in Figure 14-3.


14-4

Figure 14-3.

G 8. a. Set the electrodynamometer control knob at its full ccw position (toprovide a minimum starting load).

b. Turn on the power supply and adjust for 120 V ac.

c. Carefully increase the electrodynamometer loading to 0.3 NAm [3 lbfAin]of torque.

d. Measure and record the motor current and speed.

I = A ac

AC speed = r/min

e. Note the sparking at the brushes.


G 9. a. Compare the above results with the results from Table 13-1 inExperiment 36.

b. Is the sparking at the brushes worse than in Experiment 36?

G Yes G No

G 10. Reconnect your circuit for DC operation as shown in Figure 14-4.


14-5

Figure 14-4.

G 11. Repeat procedure 8.

I = A dc

DC speed = r/min

G 12. a. Compare the above results with the results from Table 13-2 inExperiment 36.

b. Is the sparking at the brushes worse than in Experiment 36?

G Yes G No

Inductive Compensation

G 13. a. You will now observe the effect of using inductive compensation whilethe motor is operating on AC.

b. Reconnect your circuit for AC operation as shown in Figure 14-3.Change the ammeter scale to 0-8 A ac. Short out the compensating coilby connecting a lead directly across its terminals.

c. Turn on the power supply and adjust for 120 V ac.

d. Load the motor to 0.9 NAm [9 lbfAin] torque.


14-6

e. Measure and record the motor current, motor speed and developedtorque.

I = A ac

AC speed = r/min

AC torque = NAm [lbfAin]

f. Is the sparking at the brushes about the same as was observed whenthe motor was conductively compensated (Experiment 36)?

G Yes G No

g. While the motor is operating, remove the short across the compensatingcoil.

CAUTION!

Hold the shorting lead by the insulated connectors; do nottouch the exposed terminals!

h. Explain what happened.


G 14. a. Reconnect your circuit for DC operation as shown in Figure 14-3.

b. Repeat procedure 13.

I = A ac

DC speed = r/min

DC torque = NAm [lbfAin]

c. Was there any change when the short was removed?

G Yes G No

G 15. Does the inductively-compensated motor work as well on DC as on AC?Explain.

G Yes G No


14-7

REVIEW QUESTIONS

1. Explain the difference between an inductively-compensated and a conductively-compensated universal motor.

2. Of the two types of motors mentioned in Question 1, which one is the betteradapted on either AC or DC? Explain.

3. Explain why a compensating winding is necessary in an AC series motor.

4. Would a universal motor operate better of worse on a 25 Hz power sourcecompared to a 60 Hz power source? Explain.


14-8

5. Name some tools and appliances (other than those previously mentioned) whichcontain universal motors.

A-1

Appendix A

Equipment Utilization Chart

MODEL EQUIPMENT1

ELECTRICAL POWER TECHNOLOGYEXPERIMENT

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

8110 Mobile Workstation 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

8251 Capacitor-Start Motor 1 1 1 1 1

8253 Capacitor-Run Motor 1

8254 Universal Motor 1 1

8311 Resistive Load 1 1 1 1

8321 Inductive Load 1 1

8331 Capacitive Load 1

8341 Single-Phase Transformer 1 1 1 1 2 1

8412 DC Voltmeter/Ammeter 1 1 1 1

8425 AC Ammeter 1 1 1 1 1 1 1 1 1 1 1

8426 AC Voltmeter 1 1 1 1 1 1 1 1 1 1 1 1 1

8431 Single-Phase Wattmeter 1 1 1 1 1

8821 Power Supply 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

8911² Electrodynamometer 1 1 1 1 1 1

8920 Digital Tachometer 1 1 1 1 1 1 1

8942 Timing Belt 1 1 1 1 1 1

8946 Analog Multimeter 1 1

8951 Connections Leads 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 The module storage facilities Storage Cabinet have not been included in this chart2 The Electrodynamometer Module EMS 8911 may be replaced by a Prony Brake Module EMS 8913.

B-1

Appendix B

Impedance Table for the Load Modules

The following table gives impedance values which can be obtained using either theResistive Load, Model 8311, the Inductive Load, Model 8321, or the CapacitiveLoad, Model 8331. Figure B-1 shows the load elements and connections. Otherparallel combinations can be used to obtain the same impedance values listed.

IMPEDANCE (Ω) SWITCH POSITIONS FOR LOAD ELEMENTS

120 V60 Hz

220 V50 Hz

240 V50 Hz

1 2 3 4 5 6 7 8 9

1200 4400 4800 I

600 2200 2400 I

300 1100 1200 I

400 1467 1600 I I

240 880 960 I I

200 733 800 I I

171 629 686 I I I

150 550 600 I I I I

133 489 533 I I I I

120 440 480 I I I

109 400 436 I I I I

100 367 400 I I I I I

92 338 369 I I I I I

86 314 343 I I I I I I

80 293 320 I I I I I I I

75 275 300 I I I I I I I

71 259 282 I I I I I I

67 244 267 I I I I I I I

63 232 253 I I I I I I I I

60 220 240 I I I I I I I I

57 210 229 I I I I I I I I I

Table B-1. Impedance table for the load modules.

Impedance Table for the Load Modules (cont'd)

B-2

Figure B-1. Location of the load elements.

C-1

Appendix CPerforming the Electrical Power Technology

Courseware Using the Lab-Volt Data Acquisitionand Management System

The exercises in the Electric Power / Controls courseware have been designed tobe performed using conventional instruments (AC/DC voltmeters and ammeters,power meters, etc). All these exercises can also be carried out using the Lab-VoltData Acquisition and Management (LVDAM) System.

The LVDAM System consists of the Data Acquisition Interface (DAI) module, model9062, and the corresponding LVDAM software. The system includes a user manual(p.n. 30328-EX) designed to familiarize users with the operation of the LVDAMSystem.

The Electrodynamometer (model 8911) and Precision Hand Tachometer (model8920) are replaced in the LVDAM System by the Prime Mover / Dynamometermodule (model 8960). In some exercises, the Prime Mover / Dynamometer modulecan also replace the Synchronous Motor/Generator (model 8241) to drive rotatingmachines mechanically. Refer to the manual titled AC/DC Motors and Generators(p.n. 30329) to familiarize yourself with the operation of the Prime Mover /Dynamometer module.

When performing the Electric Power / Controls courseware with the LVDAM System,the following guidelines should be taken into account:

• The "AC and DC voltmeters" are implemented using the high-voltage inputs E1,E2, and E3 of the DAI module. The voltage values are displayed on meters E1,E2, and E3 in the Metering application of the LVDAM System.

• The "AC and DC ammeters" are implemented using the high-current inputs I1,I2, and I3 of the DAI module. The current values are displayed on meters I1, I2,and I3 in the Metering application of the LVDAM System.

• The "Single-Phase Wattmeter" (model 8431) is implemented using one high-voltage input combined with one high-current input. This can be done using theinputs E1 with I1, E2 with I2, and E3 with I3 of the DAI module. The powervalues are displayed on meters PQS1, PQS2, and PQS3 in the Meteringapplication of the LVDAM System.

Note that the voltage and current values used in the power measurements canbe displayed on the voltage and current meters in the Metering application of theLVDAM System. This is a useful feature which, in certain cases, can reduce thenumber of inputs required to measure the various parameters in a circuit. As anexample, in a circuit where the line-to-line voltage is measured using input E1,it would be a wise choice to use inputs E1 and I1 to measure single-phasepower in this circuit. This would prevent the line-to-line voltage from beingmeasured twice and using two voltage inputs.

• The "Three-Phase Wattmeter" (model 8441) is implemented using two"wattmeters" (each of them being implemented using one high-voltage inputcombined with one high-current input). This can be done using inputs E1 with I1to produce P1, E2 with I2 to produce P2, and E3 with I3 to produce P3. The

Performing the Electrical Power TechnologyCourseware Using the Lab-Volt Data Acquisition

and Management System (cont'd)

C-2

power values are displayed on meters PQS1, PQS2, or PQS3 in the Meteringapplication of the LVDAM System. As an example, using inputs E1 with I1 andE3 with I3, and selecting functions PQS1 and PQS3, allow one to measure W1and W2 present on the Three-Phase Wattmeter.

EXERCISES WHERE THE NUMBER OF AVAILABLE INPUTS IS EXCEEDED

In some exercises of the Electric Power / Controls courseware, four or five high-current inputs are required to perform all current measurements. Unfortunately, onlythree high-current inputs are available in the LVDAM system.

In the exercises where it is asked to measure the currents IF, IA, and the line currentsI1, I2, and I3, the number of available inputs is exceeded. Since the line currents areusually measured only to make sure that the motor operation is normal, you mayconsider measuring only one line current (or all but one at the same time).

In some exercises, the implementation of the Single-Phase Wattmeter, or Three-Phase Wattmeter, using high-voltage and high-current inputs also causes thenumber of available inputs to be exceeded.

The exercises where the number of available inputs is exceeded are listed below.A solution is also suggested for each case.

• The DC Separately-Excited Shunt Generator

In the circuits of Figures 1, 2 and 3, use input I1 to measure current IF, input I2to measure current IA, and input I3 to measure line current I3.

• The DC Self-Excited Shunt Generator

In the circuits of Figures 1 and 2, use input I1 to measure current IA, and inputsI2 and I3 to measure line currents I2 and I3.

• The DC Compound Generator


• The DC Series Generator


• Transformers in Parallel

In the circuit of Figure 1, the implementation of the Single-Phase Wattmeterusing one high-voltage input and one high-current input from the LVDAM Systemincreases the number of required high-current inputs to four.

Use a programmable meter programmed to calculate I1 + I2 to measure the loadcurrent IL (I1 + I2 = IL in this circuit).

Performing the Electrical Power TechnologyCourseware Using the Lab-Volt Data Acquisition

and Management System (cont'd)

C-3

• Frequency Conversion

In the circuits of Figures 1 and 2, the implementation of the Three-PhaseWattmeter using two high-voltage inputs and two high-current inputs from theLVDAM system increases the number of required high-current inputs to four.

Use inputs E1-I1 and E3-I3 to implement the Three-Phase Wattmeter. Instep 3d, use temporarily input I3 (currently used in the power measurementcircuit) to adjust the DC excitation of the synchronous motor. Once theadjustment is completed, return input I3 in the power measurement circuit.

single-phase transformers and ac machines-00

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