Siemens AC Motors

7/30/2019 Siemens AC Motors

1/106

1

Table of Contents

Introduction ..............................................................................2

AC Motors ................................................................................4

Force and Motion .....................................................................6

Energy ................................................................................... 11

Electrical Energy ................................................................... 13

AC Motor Construct ion ......................................................... 17

Magnetism ............................................................................. 23

Electromagnetism ................................................................. 25Developing a Rotating Magnetic Field .................................. 31

Rotor Rotation ....................................................................... 37

Motor Specifications............................................................. 42

NEMA Motor Characteristics ................................................ 46

Derating Factors .................................................................... 54

AC Motors and AC Drives .................................................... 56

Matching AC Motors to the Load ......................................... 61

Enclosures ............................................................................. 67

Mounting ............................................................................... 71

Siemens Motors ................................................................... 78

Above NEMA Motors ........................................................... 83

Review Answers ................................................................... 88

Final Exam ............................................................................. 89


2/106

2

Introduction

Welcome to another course in the STEP 2000 series, SiemensTechnical Education Program, designed to prepare ourdistributors to sell Siemens Energy & Automation productsmore effectively. This course covers AC Mot ors and relatedproducts.

Upon completion of AC Mot ors you should be able to:

Explain the concepts of force, inertia, speed, and torque

Explain the difference between work and power

Describe the construct ion of a squirrel cage AC motor

Describe the operation of a rotating magnetic field

Calculate synchronous speed, slip, and rotor speed

Plot starting torque, accelerating torque, breakdowntorque, and full-load torque on a NEMA torque curve

Apply derating factors as required by an application

Describe the relationship between V/Hz, torque, andhorsepower

Match an AC motor to an application and its load

Identify NEMA enclosures and mounting configurations

Describe Siemens Medallion, PE-21 Plus, verticalpump, and IEC motors

Describe torque characteristics and enclosures ofSiemens above NEMA motors


3/106

3

This knowledge will help you better understand customerapplications. In addition, you will be better able to describeproducts to customers and determine important differencesbetween products. You should complete Basics of Electricitybefore attempting AC Motors. An understanding of m any ofthe concepts covered in Basics of Electricity is required forAC Mot ors. You may also want to complete Basics of Cont rol

Components which discusses the application of controldevices for start, stop, and thermal protection of AC motors.

If you are an employee of a Siemens Energy & Automationauthorized distributor, fill out the final exam tear-out card andmail in the card. We will mail you a certificate of completion ifyou score a passing grade. Good luck with your efforts.

Medallion and PE-21 Plus are trademarks of Siemens Energy& Automation, Inc.

National Electrical Code and NEC are registered trademarksof the National Fire Protection Association, Quincy, MA 02269.Portions of the National Electrical Code are reprinted withpermission from NFPA 70-1999, National Electrical CodeCopyright, 1998, National Fire Protection Association, QuincyMA 02269. This reprinted material is not the complete andofficial position of the National Fire Protection Association onthe referenced subject which is represented by the standardin its entirety.

Underwriters Laboratories Inc. is a registered trademark ofUnderwriters Laboratories Inc., Northbrook, IL 60062. Theabbreviation UL shall be understood to mean UnderwritersLaboratories Inc.

National Electrical Manufacturers Association is located at2101 L. Street, N.W., Washington, D.C. 20037. Theabbreviation NEMA is understood to mean NationalElectrical Manufacturers Association.


4/106

4

AC Motors

AC motors are used worldwide in many residential,commercial, industrial, and ut ility applications. Motorstransform electrical energy into mechanical energy. An ACmotor may be part of a pump or fan, or connected to someother form of mechanical equipment such as a winder,conveyor, or mixer. AC motors are found on a variety ofapplications from those that require a single motor toapplications requiring several motors. Siemens manufacturesa wide variety of motors for various applications. The materialpresented in this course will help in selection of a motor for a

specific application.

Winder

Pump

Conveyor


5/106

5

NEMA Throughout this course reference is made to the NationalElectrical Manufacturers Association (NEMA). NEMA setsstandards for a wide range of electrical products, includingmotors. NEMA is primarily associated with motors used inNorth America. The standards developed represent generalindustry practices and are supported by manufacturers ofelectrical equipment. These standards can be found in NEMA

Standard Publication No. MG 1. Some large AC motors maynot fall under NEMA standards. These motors are built tomeet the requirements of a specific application. These arereferred to as above NEMA motors.

IEC The International Electrotechnical Commission (IEC) isanother organization responsible for motor standards. IEC area group of recommended electrical practices developed bycommittees from participating IEC countries. These standardsare different than NEMA standards. IEC standards areassociated w ith motors used in many countries, includingmotors used in North America. These standards can be foundin IEC 34-1-16. Motors which meet or exceed these standards

are referred to as IEC motors.


6/106

6

Force and Motion

Before discussing AC motors it is necessary to understandsome of the basic terminology associated with motoroperation. Many of these terms are familiar to us in someother context. Later in the course we will see how these termsapply to AC motors.

Force In simple terms, a force is a push or a pull. Force may becaused by electromagnetism, gravity, or a combination ofphysical means.

Net force Net force is the vector sum of all forces that act on an object,including friction and gravity. When forces are applied in thesame direction they are added. For example, if two 10 poundforces were applied in the same direction the net force wouldbe 20 pounds.

Object

10 LB

10 LB

Net Force = 20 LB

If 10 pounds of force were applied in one direction and 20pounds of force applied in the opposite direction, the netforce would be 10 pounds and the object would move in thedirection of the greater force.

Object10 LB 20 LB

Net Force = 10 LB


7/106

7

If 10 pounds of force were applied equally in both directions,the net force would be zero and the object would not move.

Object10 LB 10 LB

Net Force = 0 LB

Torque Torque is a twisting or turning force that causes an object torotate. For example, a force applied to the end of a levercauses a turning effect or torque at the pivot point.

Radius(Lever Distance)

Force

Pivot Point

Torque

Torque (T) is the product of force and radius (lever distance)and is typically measured in lb-ft (pound-feet).

Torque = Force x Radius

It can be seen that increasing force or increasing the radiusincreases torque. For example, if 10 pounds of force wereapplied to a lever 1 foot long there would be 10 lb-ft oftorque. Increasing the force to 20 pounds, or the lever to two

feet would increase the torque to 20 lb-ft. Inversely, a torqueof 10 lb-ft with a two feet radius would yield 5 pounds offorce.

Force10 Lb

Radius 1 Foot

Torque = 10 lb-ft


8/106

8

Inertia Mechanical systems are subject to the law of inertia. The lawof inertia states that an object will tend to remain in its currentstate of rest or motion unless acted upon by an external force.A soccer ball, for example, remains at rest until a playerapplies a force by kicking the ball. The ball will remain inmotion until another force, such as friction or the goal net,stops it.

Friction Because friction removes energy from a mechanical system,a continual force must be applied to keep an object in motion.The law of inertia is still valid, however, since the forceapplied is needed only to compensate for the energy lost. Inthe following illustration, a motor runs a conveyor. A largeamount of force is applied to overcome inertia and start the

system. Once the system is in motion only the energyrequired to compensate for various losses need be applied tokeep the conveyor in motion. These losses include:

Friction within motor and conveyor bearings Wind losses in the motor and conveyor Friction between conveyor belt and rollers

Belt

Rollers

BearingsPackage

Motor


9/106

9

Speed An object in motion travels a distance in a given time. Speedis the ratio of the distance traveled and the time it takes totravel the distance.

Speed =Distance

Time

A car, for example, may travel 60 miles in one hour. Thespeed of the car is 60 miles per hour (MPH).

60 MPH =60 Miles

1 Hour

Speed of a rotating object Speed also applies to a rotating object, such as the tire of acar or the shaft of a motor. The speed of a rotating object is ameasurement of how long it takes a given point on the

rotating object to make one complete revolution from itsstarting point. Speed of a rotating object is generally given inrevolut ions per minute (RPM). An object that makes tencomplete revolutions in one minute has a speed of 10 RPM.

Pivot Point

Starting Point

Acceleration An object can change speed. This change in speed is calledacceleration. Acceleration only occurs when there is a changein the net force acting upon the object, which causes a changein velocity. A car increases speed from 30 MPH to 60 MPH.

There has been a change in speed of 30 MPH. An object canalso change from a higher to a lower speed. This is known asdeceleration (negative acceleration).

30 MPH 60 MPH


10/106

10

Acceleration and deceleration also apply to rotating objects. Arotating object, for example, can accelerate from 10 RPM to20 RPM, or decelerate from 20 RPM to 10 RPM.

Acceleration Deceleration

10 RPM 20 RPM 20 RPM 10 RPM

Review 11. A ____________ is a push or a pull.

2. An object has 20 pounds of force applied in onedirection and 5 pounds of force applied in theopposite direction. The net force is ____________pounds.

3. A twisting or turning force that causes an object torotate is known as ____________ .

4. If 40 pounds of force were applied to a lever 2 feetlong, the torque would be ____________ lb-ft .

5. The law of ____________ states that an object w ill tendto remain in its current state of rest or motion unlessacted upon by an external force.

6. ____________ is the ratio of distance traveled and time.

7. The speed of a rotating object is generally given in____________ per ____________ .

8. ____________ only occurs when there is a change in an

objects state of motion.


11/106

11

Energy

Work Whenever a force of any kind causes motion, work isaccomplished. Work is generally expressed in foot-poundsand is defined by the product of the net force (F) applied andthe distance (d) moved. If twice the force is applied, twice thework is done. If an object moves twice the distance, twice thework is done.

W = F x d

Pow er Power is the rate of doing w ork, or work divided by time.

Power =Work

Time

Power =Force x Distance

Time


12/106

12

Horsepower Power can be expressed in foot-pounds per second, but isoften expressed in horsepower (HP). This unit w as defined inthe 18th century by James Watt. Watt sold steam engines andwas asked how many horses one steam engine wouldreplace. He had horses walk around a wheel that would lift aweight. He found that each horse would average about 550foot-pounds of work per second. One horsepower is

equivalent to 550 foot-pounds per second or 33,000foot-pounds per minute.

The following formula can be used to calculate horsepowerwhen torque (in lb-feet) and speed are known. An increase oftorque, speed, or both will cause an increase in horsepower.

HP =T x RPM

5250


13/106

13

Electrical Energy

In an electrical circuit, voltage applied to a conductor willcause electrons to flow . Voltage is the force and electron flowis the motion.

BatteryVoltage = Force

Electron Flow = Motion

The rate at which work is done is called power and isrepresented by the symbol P. Power is measured in wattsrepresented by the symbol W. The watt is defined as therate work is done in a circuit when 1 amp flows with 1 volt

applied.

Pow er consumed Power consumed in a resistor depends on the amount ofin a resistor current that passes through the resistor for a given voltage.

This is expressed as voltage (E) times current (I).

P = E x I

or

P = EI


14/106

14

In the following simple circuit power consumed in the resistorcan be calculated.

+

_ R=6

I=2 amps

12 volts

P = EI

P = 12 volts x 2 amps

P = 24 watts

Pow er in an AC circuit Resistance is not the only circuit property that affects powerin an AC circuit. Capacitance and inductance also affect power.Power consumed by a resistor is dissipated in heat and notreturned to the source. This power is used to do useful workand is called true power. True power is the rate at whichenergy is used and is measured in watts (W). Current in an ACcircuit r ises to peak values and diminishes to zero many timesa second. The energy stored in the magnetic field of aninductor or plates of a capacitor is returned to the sourcewhen current changes direction. This power is not consumedand is called reactive power. Reactive power is measured involt-amps reactive (VAR). Power in an AC circuit is the vector

sum of true power and reactive power. This is called apparentpower. Apparent power is measured in volt-amps (VA).


15/106

15

The formula for apparent power is:

P = EI

True power is calculated from a trigonometric function, thecosine of the phase angle (cos as shown on the previouspage). The formula for true power is:

P = EI cos

Pow er factor Power factor is the ratio of true power to apparent power inan AC circuit. Power factor is equal to the cosine .

PF = cos

Horsepow er and kilow atts AC motors manufactured in the United States are generallyrated in horsepower (HP). Equipment manufactured in Europe

is generally rated in kilowatts (KW). Horsepower can beconverted to kilowatts with the following formula:

KW = .746 x HP

For example, a 25 HP motor is equivalent to 18.65 KW.

18.65 KW = .746 x 25 HP

Kilowatts can be converted to horsepower with the followingformula:

HP = 1.341 x KW

The power formula for a single-phase system is:

KW =V x I x PF

1000

The power formula for three-phase power is:

KW =V x I x PF x 1.732

1000

Note that voltage, current, and power factor are provided bythe motor manufacturer.


16/106


17/106

17

AC Motor Construction

AC induction motors are commonly used in industrialapplications. The following motor discussion will centeraround three-phase, 460 VAC, asynchronous, inductionmotors. An asynchronous motor is a type of motor where thespeed of the rotor is other than the speed of the rotatingmagnetic field. This type of motor is illustrated below. Thethree basic parts of an AC motor are the rotor, stator, andenclosure.

Enclosure

Stator

Rotor


18/106

18

Stator construction The stator and the rotor are electrical circuits that perform aselectromagnets. The stator is the stationary electrical part ofthe motor. The stator core of a NEMA motor is made up ofseveral hundred thin laminations.

Stator windings Stator laminations are stacked together forming a hollowcylinder. Coils of insulated wire are inserted into slots of thestator core.


19/106

19

Each grouping of coils, together with the steel core itsurrounds, form an electromagnet. Electromagnetism is theprinciple behind motor operation. The stator w indings areconnected directly to the power source.

Rotor construction The rotor is the rotating part of the electromagnetic circuit.The most common type of rotor is the squirrel cage rotor.Other types of rotor construction will be mentioned later in

the course. The construction of the squirrel cage rotor isreminiscent of rotating exercise wheels found in cages of petrodents.


20/106

20

The rotor consists of a stack of steel laminations with evenlyspaced conductor bars around the circumference.

The laminations are stacked together to form a rotor core.Aluminum is die cast in the slots of the rotor core to form aseries of conductors around the perimeter of the rotor.Current flow through the conductors form the electromagnet.The conductor bars are mechanically and electricallyconnected with end rings. The rotor core mounts on a steelshaft to form a rotor assembly.

Steel Laminations

Conductor Bars

End Ring

Shaft


21/106

21

Enclosure The enclosure consists of a frame (or yoke) and two endbrackets (or bearing housings). The stator is mounted insidethe frame. The rotor fi ts inside the stator w ith a slight air gapseparating it from the stator. There is no direct physicalconnection between the rotor and the stator.

The enclosure also protects the electrical and operating partsof the motor from harmful effects of the environment in whichthe motor operates. Bearings, mounted on the shaft, supportthe rotor and allow it to turn. A fan, also mounted on the shaft,is used on the motor shown below for cooling.


22/106

22

Review 31. Identify the following components from the

illustration:

A. ____________B.____________C._____________

C

B

A

2. The ____________ and the ____________ are two partsof an electrical circuit that form an electromagnet.

3. The ____________ is the stationary electrical part of anAC motor.

4. The ____________ is the rotating electrical part of anAC motor.

5. The ____________ ____________ rotor is the mostcommon type of rotor used in AC motors.


23/106


24/106

24

By drawing lines the way the iron filings have arrangedthemselves, the following illustration is obtained. Broken linesindicate the paths of magnetic flux l ines. Field lines existoutside and inside the magnet. The magnetic lines of fluxalways form closed loops, leaving the north pole and enteringthe south pole. They return to the north pole through themagnet.

N S

Unlike poles attract The polarity of the magnetic field affects the interactionbetween separate magnets. For example, when the oppositepoles of two magnets are brought within range of each otherthe lines of flux combine and tend to pull or attract themagnets.

N S SN

Like poles repel When poles of like polarity of two magnets are brought w ithinrange of each other the lines of flux produce a force that tendsto push or repel the magnets. For this reason it is said thatunlike poles attract and like poles repel. The attracting andrepelling action of the magnetic fields is important in theoperation of AC motors.

NSN S


25/106

25

Electromagnetism

When current flows through a conductor a magnetic field isproduced around the conductor. The magnetic field is madeup of lines of flux, just like a natural magnet. The size andstrength of the magnetic field will increase and decrease asthe current flow strength increases and decreases.

Current Flow

Increased Current Flow


26/106

26

Left-hand rule for A definite relationship exists between the direction of currentconductors flow and the direction of the magnetic field. The left-hand rule

for conductors demonstrates this relationship. If a current-carrying conductor is grasped with the left hand with thethumb pointing in the direction of electron flow, the fingerswill point in the direction of the magnetic lines of flux.

In the following illustration it can be seen that when theelectron flow is away from the viewer (indicated by the plussign) the lines of flux flow in a counterclockwise directionaround the conductor. When the electron flow reverses andcurrent flow is towards the viewer (indicated by the dot) thelines of flux reverse direction and flow in a clockwisedirection.

+

Current Flow

Away From ViewCurrent Flow

Toward View


27/106

27

Electromagnet An electromagnet can be made by w inding the conductor intoa coil and applying a DC voltage. The lines of flux, formed bycurrent flow through the conductor, combine to produce alarger and stronger magnetic field. The center of the coil isknown as the core. In this simple electromagnet the core is air.

Air Core

DC Voltage

Adding an iron core Iron is a better conductor of flux than air. The air core of anelectromagnet can be replaced by a piece of soft iron. When apiece of iron is placed in the center of the coil more lines offlux can flow and the magnetic field is strengthened.

Iron Core

DC Voltage

Number of turns The strength of the magnetic field in the DC electromagnetcan be increased by increasing the number of turns in the coil.The greater the number of turns the stronger the magneticfield will be.

5 Turns 10 Turns

DC Voltage DC Voltage


28/106

28

Changing polarity The magnetic field of an electromagnet has the samecharacteristics as a natural magnet, including a north and southpole. However, when the direction of current flow through theelectromagnet changes, the polarity of the electromagnetchanges. The polarity of an electromagnet connected to an ACsource will change at the same frequency as the frequency ofthe AC source. This can be demonstrated in the following

illustration. At Time 1 current flow is at zero. There is nomagnetic field produced around the electromagnet. At Time 2current is flowing in a positive direction. A magnetic fieldbuilds up around the electromagnet. The electromagnetassumes a polarity with the south pole on the top and thenorth pole on the bottom. At Time 3 current flow is at its peakpositive value. The strength of the electromagnetic field is atits greatest value. At Time 4 current flow decreases and themagnetic field begins to collapse, until Time 5 when currentflow and magnetic field are at zero. Current immediatelybegins to increase in the opposite direction. At Time 6 current

is increasing in a negative direction. The polarity of theelectromagnetic field has changed. The north pole is now ontop and the south pole is on the bottom. The negative half ofthe cycle continues through Times 7 and 8, returning to zeroat Time 9. This process will repeat 60 times a second with a60 Hz AC power supply.

S

N

S

N

S

N

N

S

N

S

N

1

6 8

9

7

2 4 5

3

S


29/106

29

Induced voltage A conductor moving through a magnetic field will have avoltage induced into it. This electrical principle is used in theoperation of AC induction motors. In the following illustrationan electromagnet is connected to an AC power source.Another electromagnet is placed above it. The secondelectromagnet is in a separate circuit. There is no physicalconnection between the two circuits. Voltage and current arezero in both circuits at Time 1. At Time 2 voltage and current

are increasing in the bottom circuit. A magnetic field builds upin the bottom electromagnet. Lines of flux from the magneticfield building up in the bottom electromagnet cut across thetop electromagnet. A voltage is induced in the topelectromagnet and current flows through it. At Time 3 currentflow has reached its peak. Maximum current is flowing in bothcircuits. The magnetic field around the coil continues to buildup and collapse as the alternating current continues toincrease and decrease. As the magnetic field moves throughspace, moving out from the coil as it builds up and backtowards the coil as it collapses, lines of flux cut across the topcoil. As current flows in the top electromagnet it creates itsown magnetic field.

0

Ammeter

0

Ammeter

S

N

N

S

SN

0

Ammeter

Time 1 Time 2 Time 3


30/106

30

Electromagnetic at t ract ion The polarity of the magnetic field induced in the topelectromagnet is opposite the polarity of the magnetic field inthe bottom electromagnet. Since opposite poles attract, thetop electromagnet will follow the bottom electromagnetwhen it is moved.

0 0

Ammeter Ammeter

S S

N N

N N

S S

Old Position New Position

Review 41. Magnetic lines of flux leave the ____________ pole of a

magnet and enter the ____________ pole.

2. Identify which magnets will attract each other andwhich magnets will repel each other in the follow ing

illustration.

N

N

S

N

S

S

N

S

S

N

S

S

N

S

N

N

A.

B.

C.

D.

____________

____________

____________

____________

3. When current flows through a conductor a____________ ____________ is produced around theconductor.

4. Which of the following will not increase the strength ofthe magnetic field?

A. Increase the current flowB. Increase the number of turns in a coilC. Add an iron core to a coilD. Increase the frequency of the AC power

supply


31/106

31

Developing a Rotat ing M agnetic Field

The principles of electromagnetism explain the shaft rotationof an AC motor. Recall that the stator of an AC motor is ahollow cylinder in which coils of insulated wire are inserted.

Stator coil arrangement The following schematic illustrates the relationship of thecoils. In this example six coils are used, two coils for each ofthe three phases. The coils operate in pairs. The coils are

wrapped around the soft iron core material of the stator.These coils are referred to as motor w indings. Each motorwinding becomes a separate electromagnet. The coils arewound in such a way that when current flows in them one coilis a north pole and its pair is a south pole. For example, if A1were a north pole then A2 would be a south pole. Whencurrent reverses direction the polarity of the poles would alsoreverse.

A1

C1

C2

A2

B1

B2

Motor Windings

Iron Core

(N)

(S)


32/106

32

Pow er supply The stator is connected to a 3-phase AC power supply. In thefollowing illustration phase A is connected to phase A of thepower supply. Phase B and C would also be connected tophases B and C of the power supply respectively.

A1

C1

C2

A2

B1

B2 (N)

(S)

+

0

A CB

_

To Phase A

To Phase A

Phase windings (A, B, and C) are placed 120apart. In thisexample, a second set of three-phase windings is installed.The number of poles is determined by how many times aphase winding appears. In this example, each phase windingappears two times. This is a two-pole stator. If each phasewinding appeared four t imes it would be a four-pole stator.

B1

B2

A2

A1

C1

C2

2-Pole Stator Winding


33/106

33

When AC voltage is applied to the stator, current flowsthrough the windings. The magnetic field developed in aphase winding depends on the direction of current flowthrough that winding. The following chart is used here forexplanation only. It wi ll be used in the next few illustrations todemonstrate how a rotating magnetic field is developed. Itassumes that a positive current flow in the A1, B1 and C1

windings result in a north pole.

A1A2B1B2C1C2

PositiveNorthSouthNorthSouthNorthSouth

NegativeSouthNorthSouthNorthSouthNorth

Winding Current Flow Direction

Start It is easier to visualize a magnetic field if a start t ime is pickedwhen no current is flowing through one phase. In thefollowing illustration, for example, a start time has beenselected during which phase A has no current flow , phase Bhas current flow in a negative direction and phase C hascurrent flow in a positive direction. Based on the above chart,B1 and C2 are south poles and B2 and C1 are north poles.Magnetic lines of flux leave the B2 north pole and enter thenearest south pole, C2. Magnetic lines of f lux also leave the

C1 north pole and enter the nearest south pole, B1. Amagnetic field results, as indicated by the arrow.

B1

B2

A2

A1

C1

C2

Start

C

A

B

Resultant Magnetic Field

Current Flow In A Negative Direction

Current Flow In A Positive Direction

Current Flow At Zero

Magnetic Lines Of Flux


34/106

34

Time 1 If the field is evaluated at 60intervals from the starting point,at Time 1, it can be seen that the field will rotate 60. At Time 1phase C has no current flow, phase A has current flow in apositive direction and phase B has current flow in a negativedirection. Following the same logic as used for the startingpoint, w indings A1 and B2 are north poles and windings A2and B1 are south poles.

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

Start 1

60

C

A

B




60


35/106

35

Time 2 At Time 2 the magnetic field has rotated 60. Phase B has nocurrent flow. Although current is decreasing in phase A it isstill flowing in a positive direction. Phase C is now flowing in anegative direction. At start it was flowing in a positivedirection. Current flow has changed directions in the phase Cwindings and the magnetic poles have reversed polarity.

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

Start 1 2

60

120

C

A

B




60

360rotation At the end of six such time intervals the magnetic field will

have rotated one full revolution or 360. This process willrepeat 60 times a second on a 60 Hz power supply.

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

B1

B2

A2

A1

C1

C2

Start 1 2 3 4 5 6

60

360

C

A

B


36/106

36

Synchronous speed The speed of the rotating magnetic field is referred to assynchronous speed (NS). Synchronous speed is equal to 120times the frequency (F), divided by the num ber of poles (P).

N =S120F

P

If the frequency of the applied power supply for the two-polestator used in the previous example is 60 Hz, synchronousspeed is 3600 RPM.

N =S120 x 60

2

N = 3600 RPMS

The synchronous speed decreases as the number of polesincrease. The following table shows the synchronous speedat 60 Hz for the corresponding number of poles.

No. of Poles

2468

10

Synchronous Speed

360018001200900

720


37/106

37

Rotor Rotation

Permanent magnet To see how a rotor w orks, a magnet mounted on a shaft canbe substituted for the squirrel cage rotor. When the statorwindings are energized a rotating magnetic field isestablished. The magnet has its own magnetic field thatinteracts with the rotating magnetic field of the stator. Thenorth pole of the rotating magnetic field attracts the southpole of the magnet, and the south pole of the rotatingmagnetic field attracts the north pole of the magnet. As therotating magnetic field rotates, it pulls the magnet alongcausing it to rotate. This design, used on some motors, is

referred to as a permanent magnet synchronous motor.

B1

B2

A2

A1

C1

C2S

Rotating Magnetic Field

Magnet

Induced voltage The squirrel cage rotor acts essentially the same as theelectromagnet magnet. When power is applied to the stator, current flows

through the winding, causing an expanding electromagneticfield which cuts across the rotor bars.

A1

C1

C2

A2

B1

B2

Magnetic Field Of Coil A1

RotorConductor Bars

Stator

Rotor


38/106

38

When a conductor, such as a rotor bar, passes through amagnetic field a voltage (emf) is induced in the conductor. Theinduced voltage causes a current flow in the conductor.Current flows through the rotor bars and around the end ring.The current flow in the conductor bars produces magneticfields around each rotor bar. Recall that in an AC circuit currentcontinuously changes direction and amplitude. The resultant

magnetic field of the stator and rotor continuously change.The squirrel cage rotor becomes an electromagnet withalternating north and south poles.

The following drawing illustrates one instant in t ime duringwhich current flow through winding A1 produces a north pole.The expanding field cuts across an adjacent rotor bar,inducing a voltage. The resultant magnetic field in the rotortooth produces a south pole. As the stator magnetic fieldrotates the rotor follows.

A1

C1

C2

A2

B1

B2 S

N


39/106

39

Slip There must be a relative difference in speed between therotor and the rotating magnetic field. If the rotor and therotating magnetic field were turning at the same speed norelative motion would exist between the two, therefore nolines of flux w ould be cut, and no voltage would be induced inthe rotor. The difference in speed is called slip. Slip isnecessary to produce torque. Slip is dependent on load. An

increase in load will cause the rotor to slow down or increaseslip. A decrease in load wil l cause the rotor to speed up ordecrease slip. Slip is expressed as a percentage and can bedetermined with the following formula.

% Slip =N - N

N

S R

Sx 100

For example, a four-pole motor operated at 60 Hz has asynchronous speed (NS) of 1800 RPM. If the rotor speed atfull load is 1765 RPM (NR), then slip is 1.9%.

% Slip =1800 - 1765

1800x 100

% Slip = 1.9%

Wound rotor motor The discussion to this point has been centered on the morecommon squirrel cage rotor. Another type is the wound rotor.A major difference between the wound rotor motor and thesquirrel cage rotor is the conductors of the wound rotorconsist of wound coils instead of bars. These coils areconnected through slip rings and brushes to external variableresistors. The rotating m agnetic field induces a voltage in therotor windings. Increasing the resistance of the rotorwindings causes less current flow in the rotor w indings,decreasing speed. Decreasing the resistance allows morecurrent flow, speeding the motor up.

Slip Ring

Brush

Wound Rotor

External Rotor Resistance


40/106

40

Synchronous motor Another type of AC motor is the synchronous motor. Thesynchronous motor is not an induction motor. One type ofsynchronous motor is constructed somewhat like a squirrelcage rotor. In addition to rotor bars coil w indings are added.The coil windings are connected to an external DC powersupply by slip rings and brushes. On start AC is applied to thestator and the synchronous motor starts like a squirrel cage

rotor. DC is applied to the rotor coils after the motor reachesmaximum speed. This produces a strong constant magneticfield in the rotor which locks in step with the rotating magneticfield. The rotor turns at the same speed as synchronousspeed (speed of the rotating magnetic field). There is no slip.Variations of synchronous motors include a permanentmagnet rotor. The rotor is a permanent magnet and anexternal DC source is not required. These are found on smallhorsepower synchronous motors.

Slip Ring

Brush

Rotor Bar

External DC

Power SupplyCoil


41/106

41

Review 51. The following illustration represents a ____________

pole motor. If winding A1 is a south pole then windingA2 is a ____________ pole.

A1

C1

C2

A2

B1

B2

2. The speed of the rotating magnetic field is referred toas ____________ speed.

3. The synchronous speed of a 60 Hz 4-pole motor is____________ RPM.

4. The difference in speed between the rotor andsynchronous speed is ____________ .

5. A 2-pole motor is operating on a 60 Hz power supply.The rotor is turning at 3450 RPM. Slip is _________ %.


42/106

42

Motor Specifications

Nameplate The nameplate of a motor provides important informationnecessary for selection and application. The followingdrawing illustrates the nameplate of a sample 30 horsepowerAC motor. Specifications are given for the load and operatingconditions as well as motor protection and efficiency.

Voltage and amps AC motors are designed to operate at standard voltages andfrequencies. This motor is designed for use on 460 VACsystems. Full-load current for this motor is 34.9 amps.


43/106

43

RPM Base speed is the nameplate speed, given in RPM, where themotor develops rated horsepower at rated voltage andfrequency. It is an indication of how fast the output shaft willturn the connected equipment when fully loaded with propervoltage and frequency applied.

The base speed of this motor is 1765 RPM at 60 Hz. It isknown that the synchronous speed of a 4-pole motor is 1800RPM. When fully loaded there will be 1.9% slip. If theconnected equipment is operating at less than full load, theoutput speed (RPM) will be slightly greater than nameplate.

% Slip =1800 - 1765

1800x 100

% Slip = 1.9%

Service factor A motor designed to operate at its nameplate horsepowerrating has a service factor of 1.0. This means the motor canoperate at 100% of its rated horsepower. Some applicationsmay require a motor to exceed the rated horsepower. Inthese cases a motor w ith a service factor of 1.15 can bespecified. The service factor is a multiplier that may beapplied to the rated power. A 1.15 service factor motor canbe operated 15% higher than the motors nameplatehorsepower. The 30 HP motor with a 1.15 service factor, forexample can be operated at 34.5 HP. It should be noted thatany motor operating continuously at a service factor greaterthan 1 will have a reduced life expectancy compared tooperating it at it s rated horsepower. In addition, performancecharacteristics, such as full load RPM and full load current, willbe affected.


44/106

44

Class insulation The National Electrical Manufacturers Association (NEMA) hasestablished insulation classes to meet motor temperaturerequirements found in different operating environments. Thefour insulation classes are A, B, F, and H. Class F is commonlyused. Class A is seldom used. Before a motor is started, itswindings are at the temperature of the surrounding air. This isknown as ambient temperature. NEMA has standardized on

an ambient temperature of 40C, or 104F within a definedaltitude range for all motor classes.

Temperature wil l rise in the motor as soon as it is started.Each insulation class has a specified allowable temperaturerise. The combination of ambient temperature and allowedtemperature rise equals the maximum winding temperature

in a motor. A motor with Class F insulation, for example, has amaximum temperature rise of 105C when operated at a 1.0service factor. The maximum w inding temperature is 145C(40ambient plus 105rise). A margin is allowed to providefor a point at the center of the motors windings where thetemperature is higher. This is referred to as the motors hotspot.

180160140

120100

80604020

0

180160140

120100

80604020

0

180160140

120100

80604020

0

180160140

120100

80604020

0

8060

5

10

10

15

105125

Class B80 C Rise

10 C Hot Spot

Class A60 C Rise

5 C Hot Spot

Class F105 C Rise

10 C Hot Spot

Class H125 C Rise

15 C Hot Spot

The operating temperature of a motor is important to efficient

operation and long life. Operating a motor above the limits ofthe insulation class reduces the motors life expectancy. A10C increase in the operating temperature can decrease themotors insulation life expectancy as much as 50%.


45/106

45

Mot or design The National Electrical M anufacturers Association (NEMA) hasestablished standards for motor construction andperformance. NEMA design B motors are most comm onlyused.

Efficiency AC motor efficiency is expressed as a percentage. It is anindication of how much input electrical energy is converted tooutput mechanical energy. The nominal efficiency of thismotor is 93.6%. The higher the percentage the moreefficiently the motor converts the incoming electrical powerto mechanical horsepower. A 30 HP motor w ith a 93.6%efficiency would consume less energy than a 30 HP motorwith an efficiency rating of 83%. This can mean a significantsavings in energy cost. Lower operating temperature, longerlife, and lower noise levels are typical benefits of high

efficiency motors.


46/106

46

NEMA Motor Characteristics

Standard motor designs Motors are designed with certain speed-torque characteristicsto match speed-torque requirements of various loads. Thefour standard NEMA designs are NEMA A, NEMA B, NEMA C,and NEMA D. NEMA A is not used very often. NEMA B ismost commonly used. NEMA C and NEMA D are used forspecialized applications. A motor must be able to developenough torque to start, accelerate and operate a load at ratedspeed. Using the sample 30 HP, 1765 RPM motor discussedpreviously, torque can be calculated by transposing theformula for horsepower.

HP =

T = T = T = 89.2 Lb-Ft

T x RPM

5250

HP x 5250

RPM

30 x 5250

1765

Speed-torque curve for A graph, like the one shown below, shows the relationshipNEMA B mot or between speed and torque the motor produces from the

moment of start until the motor reaches full-load torque atrated speed. This graph represents a NEMA B motor.

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100


47/106

47

Starting torque Starting torque (point A on the graph) is also referred to aslocked rotor torque. This torque is developed when the rotoris held at rest with rated voltage and frequency applied. Thiscondit ion occurs each time a motor is started. When ratedvoltage and frequency are applied to the stator there is a briefamount of time before the rotor turns. At this instant a NEMAB motor develops approximately 150% of its full-load torque.

A 30 HP, 1765 RPM motor, for example, will developapproximately 133.8 Lb-Ft of torque.

300

275

250

225

200

175

150125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

A

Full-Load Torque89.2 Lb-Ft

133.8 Lb-Ft

Starting Torque


48/106

48

Accelerat ing and The magnetic attraction of the rotating magnetic field willbreakdown torque cause the rotor to accelerate. As the motor picks up speed

torque decreases slightly until it reaches point B on the graph.As speed continues to increase from point B to point C torqueincreases until it reaches its maximum at approximately200%. This torque is referred to as accelerating or pull uptorque. Point C is the maximum torque a motor can produce.

At this point a 30 HP motor w ill develop approximately 178.4Lb-Ft of torque. If the motor were overloaded beyond themotors torque capability, it would stall or abruptly slowdown at this point. This is referred to as breakdown or pullouttorque.

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100


133.8 Lb-Ft

178.4 Lb-FtStarting Torque

AB

C

Breakdown

(Pullout) Torque

Accelerating Torque


49/106

49

Full-load t orque Torque decreases rapidly as speed increases beyondbreakdown torque (point C), until it reaches full-load torque ata speed slightly less than 100% synchronous speed. Full-loadtorque is the torque developed when the motor is operatingwith rated voltage, frequency and load. The speed at whichfull-load torque is produced is the slip speed or rated speedof the motor. Recall that slip is required to produce torque. If

the synchronous speed of the motor is 1800 RPM and theamount of slip is 1.9%, the full-load rated speed of the motoris 1765 RPM. The full-load torque of the 1765 RPM 30 HPmotor is 89.2 Lb-Ft. NEMA design B motors are generalpurpose single speed motors suited for applications thatrequire normal starting and running torque such asconveyors, fans, centrifugal pumps, and machine tools.

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100


133.8 Lb-Ft

178.4 Lb-FtStarting Torque

Accelerating Torque

Breakdown

(Pullout) Torque

Slip 1.9%

AB

C

D


50/106

50

Starting current and Starting current is also referred to as locked rotor current,full-load current and is measured from the supply line at rated voltage and

frequency with the rotor at rest. Full-load current is the currentmeasured from the supply line at rated voltage, frequencyand load with the rotor up to speed. Starting current istypically 600-650% of full-load current on a NEMA B motor.Starting current decreases to rated full-load current as the

rotor comes up to speed.

100

200

300

400

500

600

700

Time

Mo

tor

Full-Load

Amps

(%) Starting Current

Full-Load Current

NEMA A motor NEMA sets limits of starting (locked rotor) current for NEMAdesign B motors. When special load torque or load inertiarequirements result in special electrical designs that will yieldhigher locked rotor current (LRA), NEMA design A may result.This designation also cautions the selection of motor control

components to avoid tripping protective devices duringlonger acceleration times or higher than normal startingcurrent.


51/106

51

NEMA C motor Starting torque of a NEMA design C motor is approximately225%. A NEMA C, 1765 RPM, 30 HP motor will developapproximately 202.5 Lb-Ft of starting torque. Hard to startapplications such as plunger pumps, heavily loadedconveyors, and compressors require this higher startingtorque. Slip and full-load torque are about the same as aNEMA B motor. NEMA C applies to single speed motors from

approximately 5 HP to 200 HP.

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

202.5 Lb-FtA

B

Breakdown Torque


52/106

52

NEMA D motor The starting torque of a NEMA design D motor isapproximately 280% of the motors full-load torque. A NEMAD, with a full-load rated speed of 1765 RPM, 30 HP motor willdevelop approximately 252 Lb-Ft of starting torque. Veryhard to start applications, such as punch presses, cranes,hoists, and oil well pumps require this high starting torque.NEMA D motors have no true breakdown torque. After initial

starting torque is reached torque decreases until full-loadtorque is reached. NEMA D motors typically are designedwith 5 to 8% slip or 8 to 13% slip.

300

275

250

225

200

175150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

252 Lb-Ft

Slip 8%

A

Multispeed and ASD These specialized motor designs are uniquely designed or(adjustable speed drive) selected to specific load requirements. NEMA design

classifications are not appl icable to these specialized motors.

Soft starts Various special configurations of motor controls are selectedwhen starting/accelerating torques must be more accuratelycontrolled, or when starting current must be limited. In thecases of part winding start or wye-delta start, the motor

windings must be designed with unique connections for thespecial controls. In cases such as reduced voltageautotransformer or electronic soft starts, relatively standardmotors may be approved for these special applications.


53/106

53

Review 61. A 30 HP motor with a 1.15 service factor can be

operated at ____________ HP.

2. A motor with Class F insulation has a maximum____________ temperature rise.

3. The starting torque of a NEMA B motor isapproximately ____________ % of full-load torque.

4. ____________ torque refers to point on a torque curvewhere a motor is overloaded beyond the motorstorque capability, causing the motor to stall orabruptly slow down.


54/106

54

Derating Factors

Several factors can effect the operation and performance ofan AC motor. These need to be considered when applying amotor.

Voltage variat ion AC motors are designed to operate on standardized voltagesand frequencies. The following table reflects NEMAstandards.

60 Hz

115 VAC

200 VAC

230 VAC

460 VAC

575 VAC

50 Hz

380 VAC

400 VAC

415 VAC

220/380 VAC

A small variation in supply voltage can have a dramatic affecton motor performance. In the following chart, for example,when voltage is 10% below the rated voltage of the motor,the motor has 20% less starting torque. This reduced voltagemay prevent the motor from getting its load started or

keeping it running at rated speed. A 10% increase in supplyvoltage, on the other hand, increases the starting torque by20%. This increased torque may cause damage duringstartup. A conveyor, for example, may lurch forward atstartup. A voltage variation will cause similar changes in themotors starting amps, full-load amps, and temperature rise.

+20

+15

+10+5

0

-5

-10

-15

-20

0

0

+5 +10 +15-5-10-15

Below Above

% Voltage Variation

%Change

InM

otor

Per

formance


55/106

55

Frequency A variation in the frequency at which the motor operatescauses changes primarily in speed and torque characteristics.A 5% increase in frequency, for example, causes a 5%increase in full-load speed and a 10% decrease in torque.

Frequency

Variation

+5%-5%

% Change

Full-Load Speed

+5%-5%

% Change

Starting Torque

-10%+11%

Altitude Standard motors are designed to operate below 3300 feet.Air is thinner and heat is not dissipated as quickly above 3300feet. Most motors must be derated for altitude. The followingchart gives typical horsepower derating factors, but thederating factor should be checked for each motor. A 50 HPmotor operated at 6000 feet, for example, would be deratedto 47 HP, providing the 40C ambient rating is still required.

Altitude

3300 - 5000

5001 - 6600

6601 - 8300

8301 - 9900

9901 - 11,500

Derating Factor

0.97

0.94

0.90

0.86

0.82

50 HP X 0.94 = 47 HP

The ambient temperature may also have to be considered.The ambient temperature may be reduced from 40C to 30Cat 6600 feet on many motors. A motor with a higher insulationclass may not require derating in these conditions.

Ambient

Temperature (C)

40

30

20

Maximum

Altitude (Feet)

3300

6600

9900


56/106

56

AC M otors and AC Drives

Many applications require the speed of an AC motor to vary.The easiest way to vary the speed of an AC induction motor isto use an AC drive to vary the applied frequency. Operating amotor at other than the rated frequency and voltage has aneffect on motor current and torque.

Volts per hertz A ratio exists between voltage and frequency. This ratio isreferred to as volts per hertz (V/Hz). A typical AC motormanufactured for use in the United States is rated for 460 VACand 60 Hz. The ratio is 7.67 volts per hertz. Not every motor

has a 7.67 V/Hz ratio. A 230 Volt, 60 Hz motor, for example,has a 3.8 V/Hz ratio .

460

60

230

60= 7.67 V/Hz = 3.8 V/Hz

Flux (), magnetizing current (IM), and torque are alldependent on this ratio. Increasing frequency (F) withoutincreasing voltage (E), for example, wil l cause acorresponding increase in speed. Flux, how ever, willdecrease causing motor torque to decrease. It can be seenthat torque (T = kIW) is directly affected by flux (). Torqueis also affected by the current resulting from the applied load,represented here by IW. Magnetizing current (IM) will alsodecrease. A decrease in magnetizing current will cause acorresponding decrease in stator or line (IS) current. Thesedecreases are all related and greatly affect the motors abilityto handle a given load.

E

F

T = k I W

I =ME

2 FL M


57/106

57

Constant torque AC motors running on an AC line operate with a constantflux () because voltage and frequency are constant. Motorsoperated with constant flux are said to have constant torque.Actual torque produced, however, is determined by thedemand of the load.

T = k I W

An AC drive is capable of operating a motor with constant flux() from approximately zero (0) to the motors ratednameplate frequency (typically 60 Hz). This is the constanttorque range. As long as a constant volts per hertz ratio ismaintained the motor will have constant torquecharacteristics. AC drives change frequency to vary the speedof a motor and changes voltage proport ionately to maintainconstant flux. The following graphs illustrate the volts perhertz ratio of a 460 volt , 60 Hz motor and a 230 volt, 60 Hzmotor. To operate the 460 volt motor at 50% speed with the

correct ratio, the applied voltage and frequency would be 230volts, 30 Hz. To operate the 230 volt motor at 50% speed withthe correct ratio, the applied voltage and frequency would be115 volts, 30 Hz. The voltage and frequency ratio can bemaintained for any speed up to 60 Hz. This usually defines theupper limits of the constant torque range.

230

230

115

0 0

460

0 030 3060 60

230

30= 7.67 V/Hz

115

30= 3.8 V/Hz

460

60= 7.67 V/Hz

230

60= 3.8 V/Hz


58/106

58

Constant horsepower Some applications require the motor to be operated abovebase speed. The nature of these applications requires lesstorque at higher speeds. Voltage, however, cannot be higherthan the rated nameplate voltage. This can be illustrated usinga 460 volt, 60 Hz motor. Voltage will remain at 460 volts forany speed above 60 Hz. A motor operated above its ratedfrequency is operating in a region known as a constant

horsepower. Constant volts per hertz and torque ismaintained up to 60 Hz. Above 60 Hz the volts per hertz ratiodecreases, with a corresponding decrease in torque.

Frequency V/Hz

30 Hz 7.6760 Hz 7.6770 Hz 6.690 Hz 5.1

Flux () and torque (T) decrease:

E

F T = k I W

230

0

460

0 30 60 90

Constant

Horsepower

Constant

Torque

Decreasing V/Hz

Horsepower remains constant as speed (N) increases andtorque decreases in proportion. The following formulaapplies to speed in revolutions per minute (RPM).

HP (remains constant)=5250

T (decreases) N (increases)x


59/106

59

Reduced voltage and A NEMA B motor that is started by connecting it to the powerfrequency start ing supply at full voltage and full frequency will develop

approximately 150% starting torque and 600%starting current. AC drives start at reduced voltage andfrequency. The motor will start with approximately 150%torque and 150% current at reduced frequency and voltage.The torque/speed curve shifts to the right as frequency and

voltage are increased. The dotted lines on the torque/speedcurve illustrated below represent the portion of the curve notused by the drive. The drive starts and accelerates the motorsmoothly as frequency and voltage are gradually increased tothe desired speed. An AC drive, properly sized to a motor, iscapable of delivering 150% torque at any speed up to speedcorresponding to the incoming line voltage. The onlylimitations on starting torque are peak drive current and peakmotor torque, whichever is less.

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90100

603010 20 40 50

Frequency - Hz

2

Some applications require higher than 150% starting torque.A conveyor, for example, may require 200% rated torque forstarting. If a motor is capable of 200% torque at 200% current,and the drive is capable of 200% current, then 200% motortorque is possible. Typically dr ives are capable of producing150% of drive nameplate rated current for one (1) minute. If

the load requires more starting torque than a drive candeliver, a drive with a higher current rating would be required.It is appropriate to supply a drive with a higher continuoushorsepower rating than the motor when high peak torque isrequired.


60/106

60

Select ing a motor AC drives often have more capability than the motor. Drivescan run at higher frequencies than may be suitable for anapplication. Above 60 Hz the V/Hz ratio decreases and themotor cannot develop 100% torque. In addition, drives canrun at low speeds, however, self-cooled motors may notdevelop enough air flow for cooling at reduced speeds andfull load. Each motor must be evaluated according to its own

capability before selecting it for use on an AC drive.

Harmonics, voltage spikes, and voltage rise times of ACdrives are not identical. Some AC drives have moresophisticated filters and other components designed tominim ize undesireable heating and insulation damage to themotor. This must be considered when selecting an AC drive/motor combination. Motor manufacturers will generallyclassify certain recommended motor selections based onexperience, required speed range, type of load torque, andtemperature limits.

Distance between Distance from the drive to the motor must also be taken intodrive and motor Consideration. All motor cables have line-to-line and line-to-

ground capacitance. The longer the cable, the greater thecapacitance. Some types of cables, such as shielded cable orcables in metal conduit, have greater capacitance. Spikesoccur on the output of AC drives because of the chargingcurrent in the cable capacitance. Higher voltage (460 VAC) andhigher capacitance (long cables) result in higher currentspikes. Voltage spikes caused by long cable lengths canpotentially shorten the life of the AC drive and motor. Whenconsidering an application where distance may be a problem,contact your local Siemens representative.

Service factor on AC drives A high efficiency motor w ith a 1.15 service factor isrecommended when used on an AC drive. Due to heatassociated w ith harmonics of an AC drive, the 1.15 servicefactor is reduced to 1.0.


61/106

61

M atching AC M otors to the Load

One way to evaluate whether the torque capabilities of amotor meet the torque requirements of the load is tocompare the motors speed-torque curve with the speed-torque requirements of the load.

300

275

250

225

200

175

150

125

100

75

50

25

0

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed % Synchronous Speed

0 010 1020 2030 3040 4050 5060 6070 7080 8090 90100 100

Motor Load

Load characterist ics tables To find the torque characteristics a table, similar to the partialone shown below, can be used. NEMA publication MG 1 isone source of typical torque characteristics.

Load Description

Load Torque as % Full-

Load Drive Torque

Break-

away

Accel-

erating

Peak

Running

200150

100

150

30

40

40

75

150110

100

100

50

110

40

110

125100

100

100

40

100

100

100

Actuators:

Screw-down (rolling mills)Positioning

Agitators

Liquid

Slurry

Blowers, centrifugal:

Valve closed

Valve open

Blowers, positive displacement,

rotary, bypassed

Calenders, textile or paper


62/106

62

Calculating load torque The most accurate way to obtain torque characteristics of agiven load is to obtain them from the equipmentmanufacturer. A simple experiment can be set up to showhow the torque of a given load can be calculated. In thefollow ing illustration a pulley is fastened to the shaft of a loadthat a motor is to drive. A cord is wrapped around the pulleywith one end connected to a spring scale. The torque can be

calculated by pulling on the scale until the shaft turns andnoting the reading on the scale. The force required to turn theshaft, indicated by the scale, times the radius of the pulleyequals the torque value. It must be remembered that theradius is measured from the center of the shaft. If the radius ofthe pulley and shaft were 1 foot, for example, and the forcerequired to turn the shaft were 10 pounds, the torquerequirement is 10 Lb-Ft. The amount of torque required toturn the connected load can vary at different speeds.

Torque = Force (F) x Radius (R)

Force (F)

Radius (R)

Scale


63/106

63

Centrifugal pump When a motor accelerates a load from zero to full-load speedthe amount of torque it can produce changes. At any pointduring acceleration and while the motor is operating at full-load speed, the amount of torque produced by the motormust always exceed the torque required by the load. In thefollow ing example a centrifugal pump has a full-load torque of600 Lb-Ft. This is equivalent to 200 HP. The centrifugal pump

only requires approximately 20% of full-load torque to start.The torque dips slightly after it is started and then increases tofull-load torque as the pump comes up to speed. This istypically defined as a variable torque load.

300

275

250

225

200

175

150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

600 Lb-Ft

120 Lb-Ft


64/106

64

A motor has to be selected that can start and accelerate thecentrifugal pump. By comparing a 200 HP NEMA B motorcurve to the load curve, it can be seen that the motor w illeasily start and accelerate the load.

300

275

250

225

200

175

150

125

100

75

5025

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

600 Lb-Ft

120 Lb-Ft

Screw down actuator In the following example a screw down actuator is used. Thestarting torque of a screw down actuator is approximately200% of full-load torque. Comparing the loads requirement(200%) with the NEMA design B motor of equivalent

horsepower, it can be seen that the loads starting torquerequirement is greater than the motors capability (150%).The motor, therefore, will not start and accelerate the load.

300

275

250

225

200

175150

125

100

75

50

25

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

Load Motor


65/106

65

One solution would be to use a higher horsepower NEMA Bmotor. A less expensive solution might be to use a NEMA Dmotor of the same horsepower requirements as the load. ANEMA D motor would easily start and accelerate the load.

300

275

250

225

200

175

150

125

100

75

5025

0

% Synchronous Speed

0 10 20 30 40 50 60 70 80 90 100

Load Motor

The motor selected to dr ive the load must have sufficienttorque to start, accelerate, and run the load. If, at any point, themotor cannot produce the required torque the motor will stall

or run in an overloaded condition. This will cause the motorto generate excess heat and typically exceed current limitscausing protective devices to remove the motor from thepower source. If the overload condition is not corrected, orthe proper motor installed, the existing motor will eventuallyfail.


66/106

66

Review 71. A motor rated for 460 VAC operating on an a supply

of 437 VAC (-5%) will have a ____________ % changein motor performance.

2. Using the altitude derating table the Derating Factorssection, a 200 HP motor operated at 5500 feet would

be derated to ____________ HP.

3. The volts per hertz ratio of a 460 Volt 60 Hz motor is____________ V/Hz.

4. When applying an AC motor to an AC drive a motorwith a ____________ service factor is recommended.

5. If the radius of a pulley and shaft were 2 feet, and theforce required to turn the shaft were 20 pounds, theamount of torque required to turn the load is

____________ Lb-Ft.


67/106

67

Enclosures

Recall that the enclosure provides protection fromcontaminants in the environment in which the motor isoperating. In addition, the type of enclosure affects thecooling of the motor. There are two categories of enclosures:open and totally enclosed.

Open drip proof (ODP) Open enclosures permit cooling air to flow through the motor.The rotor has fan blades that assist in moving the air throughthe motor. One type of open enclosure is the drip proofenclosure. The vent openings on this type of enclosure

prevent liquids and solids falling from above at angles up to15from vertical from entering the interior of the motor anddamaging the operating components. When the motor is notin the horizontal position, such as mounted on a wall, a specialcover may be necessary to protect it. This type of enclosurecan be specified when the environment is free fromcontaminates.

Vents

Vents


68/106

68

Totally enclosed In some cases air surrounding the motor contains corrosivenon-ventilated (TENV) or harmful elements which can damage the internal parts of a

motor. A totally enclosed motor enclosure restricts the freeexchange of air between the inside of the motor and theoutside. The enclosure is not airtight, however, and a seal atthe point where the shaft passes through the housing keepsout water, dust, and other foreign matter that could enter the

motor along the shaft. The absence of ventilating openingsmeans all heat dissipates through the enclosure by means ofconduct ion. Most TENV motors are fractional horsepower.TENV motors are used, however, for larger horsepowerspecial applications. For larger horsepower applications theframe is heavily ribbed to help dissipate heat more quickly.TENV motors can be used indoors and outdoors.


69/106

69

Totally enclosed The totally enclosed fan-cooled motor is similar to the TENVfan cooled (TEFC) except an external fan is mounted opposite the drive end of

the motor. The fan provides additional cooling by blowing airover the exterior of the motor to dissipate heat more quickly.A shroud covers the fan to prevent anyone from touching it.With this arrangement no outside air enters the interior of themotor. TEFC motors can be used in dirty, moist, or mildly

corrosive operating conditions. TEFC motors are morewidely used for integral HP applications.

Fan

Explosion proof (XP) The explosion proof motor enclosure is similar in appearanceto the TEFC, however, most XP enclosures are cast iron. The

application of motors used in hazardous locations is subjectto regulations and standards set by regulatory agencies suchas the National Electrical Code and UnderwritersLaboratories for XP motors used in the United States.


70/106

70

Hazardous environments Although you should never specify or suggest the type oflocation, it is important to understand regulations that apply tohazardous locations. It is the users responsibility to contactlocal regulatory agencies to define the location as Division I orII and to comply w ith all applicable codes. There are twodivisions.

Division I Hazardous materials are normally present in the atmosphere.A division I location requires an explosion proof motor.

Division II Atmosphere may become hazardous as result of abnormalconditions. This may occur if, for example, a pipe breaks thatis the conduit for a hazardous chemical.

Classes and groups Once the location is defined as hazardous the location isfurther defined by the class and group of hazard. Class I,Groups A through D are chemical gases or liquids such asgasoline, acetone, and hydrogen. Class II, Groups E, F, and G

include flammable dust, such as coke or grain dust. Class III isnot divided into groups. It includes all ignitable fibers and lintssuch as clothing fiber in textile mills.

Class I

Groups A - D

Gases and Liquids

Groups E - G

Flammable Dust Ignitible Fibers

Gasoline

Acetone

Hydrogen

Ethyl

Coke Dust

Grain Dust

Metallic Dust

Rayon

Jute

Class II Class III

In some cases it may be necessary for the user to define thelowest possible ignition temperature of the hazardousmaterial to assure the motor complies with all applicablecodes and requirements.


71/106

71

Mounting

NEMA dimensions NEMA has standardized frame size motor dimensions.Standardized dimensions include bolt hole size, mountingbase dimensions, shaft height, shaft diameter, and shaftlength. Existing motors can be replaced without reworking themounting arrangement. New installations are easier to designbecause the dimensions are known. Letters are used toindicate where a dimension is taken. For example, the letterC indicates the overall length of the motor. The letter Erepresents the distance from the center of the shaft to thecenter of the mounting holes in the feet. The actual

dimensions are found by referring to a table in the motor datasheet and referencing the letter to find the desired dimension.


72/106

72

NEMA divides standard frame sizes into two categories:fractional and integral. Fractional frame sizes are designated48 and 56 and include pr imarily horsepower ratings of lessthan one horsepower. Integral or medium horsepowermotors are designated by frame sizes ranging from 143T to445T. A T in the motor frame size designation of integralhorsepower motors indicates the motor is built to current

NEMA frame standards. Motors built prior to 1966 have a Uin the motor frame size designation, indicated they are built toprevious NEMA Standards.

143 =

326 =

T

U

Current NEMA Standards

Previous NEMA Standards

The frame size designation is a code to help identify key

frame dimensions. The first two digits, for example, are usedto determine the shaft height. The shaft height is the distancefrom the center of the shaft to the mounting surface. Tocalculate the shaft height divide the first two digits of theframe size by 4. In the follow ing example a 143T frame sizemotor has a shaft height of 3 inches (14 4).

Shaft

Height

143T

14 4 = 3"

Frame


73/106

73

The third digit in the integral T frame size number is theNEMA code for the distance between the centerlines of themounting bolt holes in the feet of the motor.

14 T3Distance

Between

Bolt Holes

Frame

The dimension is determined by matching the third digit in

the frame number with a table in NEMA publication MG-1. Itcan be seen that the distance between the centerlines of themounting bolt holes in the feet of a 143T frame is 4.00 inches.

Frame

Number

Series

140

160

180

200

210

220

250

280

320

4.00

4.50

5.00

5.25

5.50

6.25

7.00

8.00

3.50

4.00

4.50

4.50

5.00

5.50

6.25

7.00

4.00

4.50

5.00

5.00

5.50

6.25

7.00

8.00

4.00

4.50

5.00

5.50

5.50

6.25

7.00

8.00

9.00

4.50

5.00

5.50

6.50

6.25

6.75

8.25

9.50

10.50

4.50

5.00

5.50

6.50

6.25

6.75

8.25

9.50

10.50

D 1 2 3 4 5

Third/Fourth Digit In Frame Number


74/106

74

IEC dimensions IEC also has standardized dimensions which differ fromNEMA. Many motors are manufactured using IECdimensions. IEC dimensions are shown in the followingdrawing.

Mounting positions The typical floor mounting positions are illustrated in thefollow ing drawing, and are referred to as F-1 and F-2mountings. The conduit box can be located on either side ofthe frame to match the mounting arrangement and position.The standard location of the conduit box is on the left-handside of the motor when viewed from the shaft end. This isreferred to as the F-1 mounting. The conduit opening can beplaced on any of the four sides of the box by rotating the boxin 90steps.

F-1 Position

(Standard)

F-2 Position

Shaft

Conduit Box


75/106

75

With modification the foot-mounted motor can be mountedon the wall and ceiling. Typical wall and ceiling mounts areshown in the following illustration. Wall mounting positionshave the prefix W and ceiling mounted positions have theprefix C.

Assembly W-1

Assembly W-5

Assembly C-1

Assembly W-2

Assembly W-6

Assembly C-2

Assembly W-3

Assembly W-7

Assembly W-4

Assembly W-8

Mounting faces It is sometimes necessary to connect the motor directly to the

equipment it drives. In the following example a motor isconnected directly to a gear box.

Gear Box

Motor


76/106

76

C-face The face, or the end, of a C-face motor has threaded boltholes. Bolts to mount the motor pass through mating holes inthe equipment and into the face of the motor.

ARNING

Threaded Bolt Holes

D-flange The bolts go through the holes in the flange of a D-flange

motor and into threaded mating holes of the equipment.

Through Bolt Holes


77/106

77

Review 81. A type of open enclosure that prevents liquids and

solids falling from above at angles up to 15fromvertical from entering the interior of the motor is an

____________ ____________ ____________ .

2. A type of enclosure that is closed and uses a fan

mounted on the shaft to supply cooling is referred as____________ ____________ ________________________ .

3. Gasoline is defined as a Class ____________ hazard.

4. The NEMA dimension from the center of the shaft tothe mounting surface is designated by the letter

____________ .

5. The letter ____________ in the motor frame size

designation indicates a motor is built to current NEMAstandards.

6. The shaft height can be determined by dividing thefirst two digits of an integral frame designation by

____________ .

7. A motor intended to be mounted on the wall with theconduit box facing up and the shaft facing left is anAssembly ____________ .

8. A ____________ motor has threaded bolt holes tomount a motor to another piece of equipment.


78/106

78

Siemens Motors

Siemens manufactures a wide range of AC motors. Thefollowing information provides only an introduction to thesemotors. Contact your local Siemens representative for moreinformation on any of the motors discussed or otherSiemens AC motors.

Medallion motors Medallion motors represent the newer family of Siemensenclosed motors. Medallion EPACT efficiency motors arehigh performance motors designed to meet therequirements of the U.S. Energy Policy Act of 1992 (EPAct).

EPACT efficiency motors are available from 1 to 200 HP inboth ODP and TEFC enclosures. Depending on the specificmotor, EPACT efficiency motors are wound for 900, 1200,1800, or 3600 RPM when used on a 230 or 460 volt powersupply. Premium efficiency Medallion motors are availablefrom 1 to 400 horsepower. Depending on the specific motor,premium efficiency motors are wound for 900, 1200, 1800,or 3600 RPM w hen used on a 230/460 (460 only above 20HP) volt power supply. EPACT and premium efficiencymotors are also available for use on other voltage sourcessuch as 575 volt systems. Contact your Siemensrepresentative for information and lead times.


79/106

79

Medallion motors are available with longer shafts for beltdriven applications and vertical mounting for applicationssuch as pumps. Various protective devices, such asthermocouples and thermistors, can be installed as an option.These devices are wired into the motor controller and shutdown the motor if temperature becomes excessive. Spaceheaters can also be used to keep the temperature of the

motor above the dew point in areas that are damp or humid.Space heaters are turned off when the motor is running andon when the motor is stopped.

Medallion motors are also available in two-speedconfigurations. A two-speed motor often has two sets ofwindings, each wound with a different number of poles. Thewindings are brought out to an external controller. The motorcan be run at either speed. Typical speed selections are 900or 1200 RPM at low speed, and 1800 RPM at high speed.

T1

T3T2

L1

L2

L3

T11

T12T13

1

2

3

11

12

13

Low Speed: Close 1, 2, 3Open 11, 12,13

High Speed: Close 11, 12, 13Open 1, 2, 3

Speed =120 x Frequency

Number of Poles- Slip

PE-21 Plus motors PE-21 Plus motors are premium efficiency motors availablefrom 1 to 500 HP. Premium efficiency motors typically costslightly more than standard efficiency motors, but payback isin energy savings.


80/106

80

The following example is used to show energy savingsavailable over the life of a premium efficiency motor. A 25 HPmotor (Motor 1) with an efficiency of 86.5% costing $590, anda PE-21 Plus with an efficiency of 93% costing $768 arecompared. Using a process life of 60,000 hours at $.08 akilowatt hour, the PE-21 Plus will save $7,055 in total operatingcost.


81/106

81

Vertical pump motors Vertical hollow shaft pump motors are designed for verticalpump applications. The motors are squirrel cage inductiontype w ith NEMA design B torque and current characteristics.Motors are rated from 25 to 250 HP and 1800 RPM. Verticalpump motors are designed for 460 volt, 3-phase, 60 Hzsystems. Thermostats and space heaters are optional.

These motors have NEMA standard P flange mounting shaftwith a hollow shaft which accommodates the driven shaft toextend through the rotor. The coupling for connecting themotor shaft to the driven shaft is located in the top of themotor.

In addit ion to hollow shaft, more conventional solid shaftmotors are supplied where the motor shaft is coupled to thedriven shaft below the P flange face. Vertical solid shaftmotors designed for in-line pump applications are availablefrom 3 - 100 HP at 3600 RPM, and 3 - 250 HP at 1200 and 1800RPM.


82/106

82

IEC motors IEC motors are manufactured to meet specifications of theInternational Electrotechnical Commission, IEC 34. Standardvoltages at 50 Hz are 220, 400, 500, or 660 volts. Standardvoltage at 60 Hz is 460 volts. Siemens IEC motors areavailable with 2, 4, 6, and 8 poles. IEC motors are alsoavailable for multispeed applications. Siemens IEC motorsare available from 0.12 KW (0.16 HP) to 630 KW (840 HP).

While mounting flange dimensions, shaft height, shaftextensions, and other performance standards clearly differwith comparable NEMA motors, a closer comparison willshow remarkably similar characteristics. Perhaps the greatestobstacle to working with IEC motors is familiarity w ith uniqueterminology and the ability to correlate with more familiarNEMA standards.


83/106

83

Above NEMA Motors

Motors that are larger than the NEMA frame sizes are referredto as above NEMA motors. These motors typically range insize from 200 to 10,000 HP. Some above NEMA motorsmanufactured by Siemens may also be consideredMedallion motors. There are no standardized frame sizes ordimensions because above NEMA motors are typicallyconstructed to meet the specific requirements of anapplication. Siemens offers large motors in seven basic framesizes: 30, 500, 580, 680, 708, 800, and 1120 frames. For eachframe size Siemens has standard frame dimensions similar to

NEMA dimensions. For specific application informationcontact your local Siemens representative.

C

W

XW

2F

X

B

Keyway Length

Yoke Width

AB

AC

XD

XL

AF

2E

E

D

HT

AA

Pipetap

N-W

N


84/106

84

The customer typically supplies specifications for startingtorque, breakdown torque, and full-load torque based onspeed-torque curves obtained from the driven equipmentmanufacturer. There are, however, some minimum torquesthat all large AC motors must be able to develop. These arespecified by NEMA.

Locked Rotor Torque

60% of Full-Load TorquePull-Up Torque 60% of Full-Load TorqueMaximum Torque 175% of Full-Load Torque

Above NEMA motors require the same adjustment foraltitude and ambient temperature as integral frame sizemotors. When the motor is operated above 3300 feet a higherclass insulation should be used or the motor should bederated. Above NEMA motors with class B insulation caneasily be modified for operation in an ambient temperaturebetween 40C and 50C. Above 50C requires special

modification at the factory.

Enclosures Environmental factors also affect large AC motors. Enclosuresused on above NEMA motors look differently than those onintegral frame size motors.

Open drip proof (ODP) The open drip proof enclosure provides the same amount ofprotection as the integral frame size open motor. Thisprovides the least amount of protection for the motorselectrical components. It is typically used in environmentsfree of contaminants.


85/106

85

Horizontal drip proof The weat

Siemens AC Motors

Documents

Transcript of Siemens AC Motors