Power Electronics and Drives U4

7/26/2019 Power Electronics and Drives U4

1/58

iUNIT 4Electrical drive system

TEL 202/05

Power Electronics and Drives

Electrical DriveSystem

Unit 4


2/58

ii WAWASAN OPEN UNIVERSITYTEL 202/05 Power Electronics and Drives

COURSE TEAM

Course Team Coordinator: Dr. Magdalene Goh Wan Ching

Content Writer: Associate Professor Lim Soo King

Instructional Designer: Ms. Jeanne ChowAcademic Member: Mr. Tan Yee Chyan

COURSE COORDINATOR

Dr. Magdalene Goh Wan Ching

EXTERNAL COURSE ASSESSOR

Associate Professor Cheong Kuan Yew, Universiti Sains Malaysia

PRODUCTION

In-house Editor: Ms. Jeanne Chow

Graphic Designer: Ms. Valerie Ooi

Wawasan Open University is Malaysias first private not-for-profit tertiary institution dedicated to

adult learners. It is funded by the Wawasan Education Foundation, a tax-exempt entity established

by the Malaysian Peoples Movement Party (Gerakan) and supported by the Yeap Chor Ee Charitable

and Endowment Trusts, other charities, corporations, members of the public and occasional grants

from the Government of Malaysia.

The course material development of the university is funded by Yeap Chor Ee Charitable and

Endowment Trusts.

2014 Wawasan Open University

All rights re served. No part of this publication may be reproduced, stored in a retrieval system or

transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or

otherwise, without prior written permission from WOU.

Wawasan Open University(KPT/JPT/DFT/US/P01)Wholly owned by Wawasan Open University Sdn. Bhd. (700364-W)

54, Jalan Sultan Ahmad Shah, 10050 Penang.Tel: (604) 2180333 Fax: (604) 2279214Email: [email protected]

Website: www.wou.edu.my


3/58

iiiUNIT 4Electrical drive system

Contents

Unit 4 Electrical Drive System

Unit overview

Unit objectives

4.1 Introduction to motor drives

Objectives

Introduction

Criteria for selecting drive components Matching the motor and the load Thermal consideration for motor Matching the motor with the power electronic

converter Switching frequency and motor inductance Selection of speed and position

Servo drive control and current limiting Current limiting in adjustable speed drives

Suggested answers to activities

4.2 dc motor drives

Objectives

Introduction

Equivalent circuit of dc motor

Permanent-magnet dc motor

dc motor with a separately excited field winding

Effect of armature current

Form factor

Torque pulsation

Servo drives


1

1

3

3

3

5

5

8

11

12

12

1313

15

17

17

17

17

19

21

23

23

24

25

29


4/58

iv WAWASAN OPEN UNIVERSITYTEL 202/05 Power Electronics and Drives

4.3 ac motor drives

Objectives

Introduction

Induction motor drives

Synchronous motor drives


Summary of Unit 4

Suggested answers to self-tests

References

Glossary

31

31

31

31

39

44

45

47

51

53


5/58

1UNIT 4Electrical drive system

Unit Overview

In this Unit 4 on Electrical drive system, you will learn three main topics, whichare introduction to motor drives, dc motor drives, and ac motor drives. In eachtopic, you will study and achieve what have been specified in the learning objectivesvia tutorials, activities and self-tests.

In section one on dc motor drives, you will learn and achieve the ability andconfidence to explain the general control of motor drives; explain the concept ofdesigning servo motor drives; explain the design concept of variable speed motordrives; and select the components and electric motor for motor drives.

In section two on dc motor drives, you will learn and achieve the ability andconfidence to examine the equivalent circuit of a dc motor; analyse the characteristicsof permanent magnetic dc motor; apply the concept on how to increase the powerof a dc motor by designing with excited wind stator; interpret the effect of armaturecurrent on the performance of dc motor; and discriminate the difference betweena normal dc motor and a dc servo motor.

In section three on ac motor drives, you will learn and achieve the ability andconfidence to describe the basic principles of an inductor motor; describe perphase representation of the induction motor; describe the equivalent circuit of theinductor motor; design ac synchronous motor drives, describe and analyse per phaserepresentation of the synchronous motor; and describe and analyse the equivalentcircuit of the synchronous motor.

Unit Objectives

By the end of Unit 4, you should be able to:

1. Select the components including electric motor for motor drives.

2. Distinguish between a dc motor and an ac motor.

3. Design and implement drive circuit for a dc motor.

4. Design and implement drive circuit for an ac motor.

5. Distinguish between an induction ac motor and a synchronous ac motor.


6/58

2 WAWASAN OPEN UNIVERSITYTEL 202/05 Power Electronics and Drives


7/58


4.1 Introduction to Motor Drives

Objectives

By the end of this section, you should be able to:

1. Explain the concept of general control of motor drives.

2. Explain the concept of designing servo motor drives.

3. Explain the design concept of variable speed motor drives.

4. Select the components and electric motor for motor drives.

Introduction

Converting power from one form to another form and using it to control the motionof a motor are the two majoring applications of power electronics. The operationmust be as efficient as possible because large amount of power is being controlledand if there is a small percentage loss, it would mean that the circuit may overheatand fail. Accurate control of power is required to assure that the load is properly

serviced and the source is not overloaded. Rapid switching of large current andvoltage generates considerable amount of electromagnetic interference that mayshorten the life span of the electronic drive circuit or even damage it permanently.Thus, careful design of drive circuit and implementation are required to avoid thisnoise interference affecting this circuit.

A general block diagram for the control of motor drives is shown in Figure 4.1.The system basically consists of an electric motor, a power electronic converter, anda process requirement and feedback to the process control computer. The processcontrol computer which is based on feedback, determines the requirements of theparameters of motor such as time of response, accuracy etc., which are controlled

within the operating requirements to the motor drives. Take for example; a servo-drive for servo motor drive is needed for accurately control the robot and adjustablespeed drive is needed for air conditioner, which is based on requirements like coldnessof the room, outside temperature etc. As you have learnt earlier, power electronicconverter converts the required power needed for driving the motor.


8/58


Power source

ControllerPower

ElectronicConverter

Motor Process

ProcessControl

Computer

Figure 4.1 Block diagram showing control of motor drives

In servo applications of motor drives, the response time and accuracy with whichthe motor follows the speed and position commands are extremely important. Thus,the servo motor drives require speed or position feedback for a precise control likethe one shown in Figure 4.2. The speed and/or position sensor is part of the drivesystem because the sensor will feedback signal to the controller telling the speedand position of the motor.

Input power

Inputcommand

ControllerPower

ElectronicsConverter

Motor LoadSpeed and/or Position

Sensor

Figure 4.2 Block diagram of servo motor drives

For large number of applications, the accuracy and response time of the motor tofollow the speed command is not important. Thus, one can see that a feedback loopto control the process is outside the motor drive. Unlike the servo motor drives, thesensor is mounted in the motor to detect the position and motion of the motor shaft.

An example of such motor drives is shown in Figure 4.3. The figure basically showsthe block diagram for an adjustable speed drive for an air conditioner. The speed ofthe drive is depending on the desired temperature, measured outside temperatureand humidity, and indoor temperature and humidity.


9/58


Inputpower

Adjustablespeed drives

Temperatureand humidity

Desiredtemperature System

Controller

PowerElectronicsConverter

MotorAir

Conditioner

IndoorSensors

BuildingCooling

Load

Indoortemperatureand humidity

Measuredoutdoor

temperatureand humidity

Figure 4.3 An adjustable speed drive for an air conditioner

Criteria for selecting drive components

For a motor drive system to perform to optimum condition, the matching ofmechanical load and drive components are important. We shall discuss five criteriafor selecting components for motor drives. They are matching the motor and theload; thermal consideration for motor; matching the motor and the power electronicconverter; switching frequency and the motor inductance; selection of speed andposition sensors; and servo drive; and current limiting in adjustable speed drives.

Matching the motor and the load

Before selecting the drive components, the load parameters and requirements suchas the load inertia, maximum speed, speed range and direction must be known. Themotion profile like the speed (L) and position (L) are a function of time such aswhat has been shown in Figure 4.4 (a)and Figure 4.4 (b), which are important andneeded to be specified.


10/58


Speed L

0 t

(a) Speed profile

PositionL

0 t

(b) Position profile

Load torqueTL

0 t

(c) Load-torque profile

Figure 4.4 Load profile

By means of mechanical modelling, one can obtain a load-torque profile. If theprimarily inertial load with negligible damping is known, the load-torque (TL) profilecorresponding to the speed (L) profile as shown in Figure 4.4 (a)can be known andis shown in Figure 4.4 (c). From the profile, the peak torque and maximum speedrequirement for the motor are established.

One way to drive a rotating load is to couple it directly to the motor. In such adirect coupling, the problem and losses associated with a gearing can be avoided.

Another way is using gear mechanism like what is shown in Figure 4.5. Assuming

the energy efficiency of the gear to be 100%, then the torque on two sides of thegear are related as


11/58


Tm = L =

L =M = a (4.1)

TL M M L

where =d

is the angular frequency and is the number of teeth, and ais the dt

coupling ratio. The subscript M and L denote motor and load respectively.

nm

Lm

TLTm

Jm Bm

Motor

TWLn

L

JL, B

L,X

L

Load

Figure 4.5 Coupling mechanism using gear

In the feed-screw driven mechanism as shown in Figure 4.6, the torque and forceare related with equation (4.2).

Tm =VL =

xL =S

= a (4.2)TL M M 2

where the linear velocity is VL=dXL, S is the pitch of the feed screw in m/turn,dt

XLis the displacement of the load, and a is the coupling ratio.

The electromagnetic torque (Tem) required for the gear couple motor as shown inFigure 4.6can be calculated on the basis of energy consideration in terms of theinertias, required load acceleration, coupling ratio (a), and working torque (TWL)

or force (FWL) andd L is the load acceleration. Therefore,

dt

Tem=1

d L [JM+ a

2JL] + aTWL+ L(BM+ a

2BL) (4.3) a dt a

where BM

and BL

are respectively the damping factors for motor and load. Theequivalent total inertia isJeq=Jm+ a

2JL. Similarly, the equivalent total damping Beq= Bm+ a

2BL, and the equivalent work torque of the load is TWeq= aTWL. Thus, theelectromagnetic torque (Tem) can be written in terms of all these equivalences andis equal to


12/58


Tem=JeqdM + BeqM+ TWeq (4.4)

dt

Similarly for feed-screw coupling type motor as shown in Figure 4.6, theelectromagnetic torque (Tem) is defined by equation (4.5).

Tem=1

dVL [Jeq+Js+ a

2(MT+ MW) + aFWL (4.5) a dt

wheredVL is the linear acceleration of the load.

dt

Milling good

Work piece

Table

Motor

Pitch s (m/turn) J

m= Motor inertia

JS= Feed-screw inertia

m

, Tm

FWL

MTM

W

FL,x

L

Figure 4.6 Coupling mechanism using feed-screw

Activity 4.1

State the criteria for selecting motor drive component.

Thermal consideration for motor

Besides the parameter criteria selection mentioned in the previous subsection suchas the load-torque, electromagnetic torque (Tem) produced by the motor, a functionof time for the motor, which is proportional to motor current can be known. It istrue provided that the flux in the air gap of the motor is kept constant. The typicalillustration of electromagnetic torque (Tem) of motor and its corresponding motorcurrent are shown in Figure 4.7.


13/58


Tem

T2

T3

T1

t1

t2

t3

t4

t5

t6

T6

t

T5

0

T4

tperiod

(a) Electromagnetic torque

I1

0

I2

I3

I4

I5

I6

Motor

Current

t

(b) Motor current

Figure 4.7

The motor current as shown in Figure 4.7 (b)is the dc current during various timeinterval for a dc motor. For an ac motor, the motor current shown is the approximatedrmsac current drawn at various time interval.

The power loss (PR) in the winding resistance (RM) of the motor due to the motorcurrent is a large part of the total motor losses, which is eventually converted into heatthat will heat up the motor. The resistive loss is proportional to the current square(Irms) and hence it is also proportional to electromagnetic torque (Tem) square. For arepetitive motor current waveform like the one shown in Figure 4.7 (b)with period(tperiod) comparable with motor thermal time constant, the heating and maximumtemperature rise of the motor can be calculated based on the resistive power loss(PR) averaged over the time period (tperiod). Thus, the power loss (PR) is defined as

PR= RMI2rms (4.6)

where the square of rmscurrent (I2rms) is defined as

m

I k2t k

I2

rms=k = 1

(4.7) t period

where I2kis the instantaneous rmscurrent square for a specific time interval tk.


14/58


Thus, the average power loss (PR) due to resistive winding (RM) of the motor is

m

I k2t k

PR= RMk = 1

(4.8) t period

m is equal to 6 for the example shown in Figure 4.7 (b).

Since the motor current is proportional to the motor torque (T em), the rmsmotortorque square (T2em, rms) over a period (tperiod) is defined as

m I k

2t k

T2em, rms= k1k = 1 (4.9)

t period

Comparing equations (4.7) and (4.9), it yields that rms motor torque square (T2em,rms) is equal to equation (4.10).

T2em, rms= k1I2rms (4.10)

where k1is the constant.

From equations (4.6) and (4.10), it yields that the average power loss (P R) is equalto expression (4.11).

PR= k2T2em, rms (4.11)

where k2is constant, which is k2=RM.

k1

In addition to power loss (PR) due to winding of the motor, there are other powerlosses, which are friction and windage (PFW), PEHdue to eddy current and hysteresiswithin the motor lamination, and Psdue to switching frequency ripple in motor

current, and stray power loss (Pstay) etc. To sum the total power loss (P loss), equation(4.12) accounts for them.

Ploss= PR+ PFW+ PEH+ Ps+ Pstray (4.12)

Under steady state condition, the motor temperature rise () is in degree centigradeand it is equal to be

= PlossRTH (4.13)

where RTHis the thermal resistance of the motor in centigrade per watt. In general,the power loss like (PR) is lower with the increase speed of motor with constantthermal resistance (RTH). If the motor is installed with self-cool fan, then the thermalresistance (RTH) of the motor would be reduced, pushing the motor to operate far


15/58


in the safe operation area preventing it from heat damage. From the torque profile,the motor should be chosen such that its rms electromagnetic torque (Tem) of themotor falls in the safe operating area (SOA) of the power-temperature graph.

Activity 4.2

The current profile of a motor is shown below. Find the rms currentand the power loss if the resistance of the motor winding is 5.

Motorcurrent

(A)2.5

2.0

1.5

1.0

0

2.5

1s 2s 4s 5s

6s

I5

t

I4

I3

I2

I1

Matching the motor with the power electronic converter

Power electronic design and its control depend on the type of motor to be driven.In general, power electronic converter provides a controlled voltage to the motor inorder to control the current flowed in the motor, hence the electromagnetic torqueproduced by the motor. We shall discuss a few aspects to be considered for motorin power electronic.

The power electronic needs to be able to supply sufficiently enough current for themotor since large amount of peak current is required. A large current shall meanlarge junction temperature of the electronic device. This would result in powerloss within the semiconductor device. Large current shall also mean that the heatgenerated can be large, affecting the temperature of the motor. If the duration ofthe peak current is small as compared with the thermal time constant of the motor,then it is considered safe for the motor.

Motor irrespective of dc or ac type, it produces back electro motive force emf (e) that

opposes the voltage (V) applied to it. The back emf can be viewed from a simplifiedcircuit of a motor drive as in Figure 4.8.


16/58


50HzInput

PowerElectronicConverter

Motor

Li

eV

++

Figure 4.8 A simplified circuit to motor drive showing back emf

Hence, the control of torque shall be the rate of current (di ), which is defined as dt

di=

V e(4.14)

dt L

From equation (4.14), one can see that the ability to control the motor current isto ensure that the output voltage (V) is greater than the back emf(e).

Switching frequency and motor inductance

In servo motor drive, the motor current should be able to respond quickly to the loadrequirement. Thus, it requires low inductance value as you can see from equation(4.14). In the steady state, the ripple current should be small so that it can preventpower loss (Ps) due to switching. But in order to achieve this, it requires largeinductance value. Thus, there is a contradiction whereby a compromised inductanceshould be selected for motor and switching frequency.

Selection of speed and position

In selection of speed and position sensors, direct or indirect coupling, sensor inertia,

possibility and avoidance of torsional resonance, and maximum sensor speed are to beconsidered. To control the instantaneous speed within a specified range, the ripple inthe speed sensor should be small. This is needed to prevent error in the incrementalposition encoder, which is often used to measure the speed and position. If suchsensor is used at low speed, the number of pulse output per revolution should belarge to provide instantaneous speed measurement with sufficient accuracy. Similarly,accurate position information will require an incremental position encoder withlarge number of pulse output per revolution.


17/58


Servo drive control and current limiting

In most practical applications of servo drive control, a very fast response to a sudden

change in position or speed is needed, in which it requires a large peak current andtorque. This may be prohibitive in terms of cost of converter. Therefore, convertercurrent is limited by the controller. The design usually contains inner current loopwhereby the actual current is measured and compared with reference current. Theerror between them is then used to control the converter output current by using acurrent-regulated modulation. A block diagram illustrating such control is shownin Figure 4.9.

A tachometer registers the speed of the motor and feedbacks to the proportionalintegrated (PI) amplifier. If the actual speed is higher than the reference speed, thenthe PI amplifier will adjust the speed and send the torque reference to the torque-

to-computer so that a command can be issued to slow down the motor. Likewise,if the speed is too low, the torque-to-computer will issue command to increase thespeed of the motor.

Figure 4.9 Block diagram of control of servo drives with inner current loop

Current limiting in adjustable speed drives

In adjustable speed drives, the current is kept from exceeding its limit by meansof limiting the rate of change of control voltage with time. A block diagram of themotor drives illustrating the ramping limiter to limit the motor current is shownin Figure 4.10. It mainly consist of a ramp limiter that has the function to controlthe voltage supply to power electronic converter.


18/58


Adjustable-speed drive

Control

voltageRampLimiterSystemRegulatorReference

Actual

+

Motor ProcessPower

ElectronicsConverter

Figure 4.10 A block of motor drives showing the ramping limiter

Summary

In this section, you have achieved the learning objectives as specifiedin the objective section. You learnt the general knowledge of howto control the motor drive, explain the designing concept of servomotor drives and variable speed motor drives, and the selectioncriteria for choosing the components and motors for motor drives.

Self-test 4.1

The load speed profile is shown below. The gear ratio L is 2. The

M

load inertia is JL= 10 kgm2and motor inertia is JM= 2.5 kgm

2.Ignoring the damping factor, draw the torque profile of the motor.

Speed L

1000

01s 2s 3s 4s 5s 6s 7s

t

Self-test 4.2

Based on the parameters stated in Self-test 4.1 and using the resultof the angular speed (M) of the motor obtained from Self-test4.1, plot the electromagnetic torque (Tem) profile of the motor fora cycle of operation.


19/58



Feedback

Activity 4.1

The parameter criteria are the load inertia, maximum speed, speedrange, and direction of motion.

Activity 4.2

The rmscurrent square is equal tom

I k2t k

I2rms=k = 1 =

1.02 1 + 22 1 + 1.52 2 + 2.52 1 + (2.5)2

t period 6

=22

= 3.67A. Thus, the rms current is 1.91A.

6

The power loss is equal tom

I k2t k

PR= RMk = 1 = 3.67 5 = 18.3W

t period


20/58



21/58


4.2 dc Motor Drives

Objectives


1. Examine the equivalent circuit of a dc motor.

2. Analyse the characteristics of permanent magnetic dc motor.

3. Apply the design concept to increase the power of a dc motor with excited wind stator.

4. Interpret the effect of armature current on the performance of a dc motor.

5. Discriminate between a normal dc motor and a dc servo motor.

Introduction

Traditionally, dc motor drives have been used for speed and position controlapplications. Nowadays, the use of ac servo drives application is increasing. In spite

of that, in application where an extremely low maintenance is required, dc drivescontinue to be used because of their low initial cost and excellent drive performance.

Equivalent circuit of dc motor

In a dc motor, the magnetic flux (f) is established by the stator either by meansof permanent magnet like what is shown in Figure 4.11 (a), whereby the magneticflux stays constant or by means of field winding as shown in Figure 4.11 (b). If themagnetic saturation in the flux path can be neglected, then the magnetic field (f)can be expressed in terms of field current (If) by equation (4.15).

f= kfIf (4.15)

where kfis the constant of proportionality.


22/58


f= constant

If

Permanentmagnets

NSNSN S

f= k

fIf

(a) Permanent magnet motor (b) dc motor with field winding

Figure 4.11 A dc motor

The rotor located in the centre slot of the motor is called armature winding. Ithandles the electric power. This is in contrast to most ac motor, where the power inhandling winding is on the stator for the ease of handling large amount of power.

The electromagnetic torque of a dc motor is produced by the interaction of the field(f) and the armature current (ia), which follows equation (4.16).

Tem= ktfia (4.16)

where ktis the torque constant of the motor. In the armature circuit, a back emf(ea)

is produced by the rotation or armature conductor at speed (M) in the presenceof a field flux (f). Thus,

ea= kefM (4.17)

where keis the voltage constant of the motor.

The electrical power (Pe) is equation to eaia, which is

Pe= eaia= kefMia (4.18)and the mechanical power (Pm) is equal to

Pm= MTem= ktfMia (4.19)

In practice, a controllable voltage (Vt) is applied to the armature terminals toestablish the current ia. Therefore, the armature current (ia) in the armature circuitis determined by the source voltage (Vt) and the back emf(ea), the resistance (Ra) ofthe armature winding and inductance (La) of the armature. Thus, the supply voltage

(Vt) can expressed as

Vt= ea+ Raia+ Ldia (4.20)

dt


23/58


The equivalent circuit illustrating equation (4.18) is shown in Figure 4.12. Theinteraction of electromagnetic torque (Tem) with the load torque (TL) as shown inequation (4.3), determines how the angular speed (M) of the motor is being built

up as shown in equation (4.21).

Tem=JdM + BM+ TWL(t) (4.21)

dt

whereJand Bare the total equivalent inertia and damping respectively of the motorload system and TWLis the equivalent work torque of the load.

TWL

JL

BL

Tem

ea

La

M

f

Ra

ia

Vt

+

+

Figure 4.12 A dc motor equivalent circuit

Permanent-magnet dc motor

A dc motor with permanent magnet stator as shown in Figure 4.11 (a)produces aconstant field flux (f). In steady state, the equations pertaining its electromagnetictorque (Tem), armature voltage (Ea), and supply voltage (Vt) can be established basedon the equivalent circuit shown in Figure 4.13. They are

Tem= kTIa (4.22)

Ea= kEM (4.23)

Vt= Ea+ RaIa (4.24)

where kT= ktfand kE= kef.


24/58


Vt

+

+

Ea= k

E

M

Tem

= kTIa

M

Ra

Ia=

Vt Ea

Ra

Figure 4.13 Equivalent circuit of a permanent magnet dc motor

From the above equations, the steady state angular speed (M) of the motor asthe function of electromagnetic torque (Tem) can be established for a given supplyvoltage (Vt) and finds to be

M=1 (Vt RaTem) (4.25)

kE kT

The plot of equation (4.25) as shown in Figure 4.14 shows that the torques areincreased, the torque-angular speed characteristics for a given supply voltage (Vt) isessentially vertical except for a droop due to voltage drop across (IaRa) the windingof the armature.

Tem

Rated

Rated

Noted: Vt1

< Vt2

< Vt3

< Vt4

< Vt5

0 M

Vt5

Vt4

Vt3

Vt2

Vt1

Figure 4.14 Torque-angular speed characteristic of a permanent magnetic dc motorfor various supply voltages


25/58


In a continuous steady state, the current of the armature (Ia) should not be exceedingits rate value and therefore the torque should not exceed the rated torque as shownin Figure 4.14. This is the illustration of the limitation of a permanent magnet

dc motor, whereby the maximum angular speed of the motor is limit by the ratedangular speed of the motor. Figure 4.15shows the steady state operating limits ofthe torque and current, and it also shows the terminal voltage required as a functionof speed and the corresponding voltage across the armature (Ea).

Per unitquantities

1.0

1.00

Tem

, Ia

Vt

Ea

M

(per unit)

Figure 4.15 Continuous torque-angular speed capability of a permanent magnetic

dc motor

Activity 4.3

Using the graph shown in Figure 4.14, derive the gradient of thegraph and shows that it is large for a small armature resistance.

dc motor with a separately excited field winding

Permanent magnetic dc motor has limitation in terms of power delivery and angularspeed. The limitations can be overcome if the field (f) is produced by field wind ofstator, whereby the stator is supplied with a dc current (If) as shown in Figure 4.11(b). This way offers the most flexible in controlling the dc motor, whereby the fieldwinding is excited by a separately controlled dc voltage (Vf) as shown in Figure 4.16.


26/58


ia

Ra

La

if(If= Vf)

Rf

Vt

ea

+

+

+

Lf

Rf

Vf

Figure 4.16 The equivalent circuit of separate excited dc motor

The equation of angular frequency of the motor shown in equation (4.25) can bewritten to include the magnetic field generated by the dc voltage. Thus,

M=1 (Vt Ra Tem) (4.26) kef ktf

where kT= ktfand kE= kef . kfand keare torque constant and voltage constantrespectively. Equation (4.26) shows that both field flux (f) and supply voltage (Vt)can be controlled to yield the desired torque and angular speed. With field flux (f)at its rated value, equations (4.22) to (4.24) for electromagnetic torque (Tem), voltageacross the armature (Ea), and supply voltage (Vt) are equally applicable for this typeof dc motor. This is also true for its torque-angular speed characteristic followingthe torque-speed characteristic of a permanent magnetic dc motor for various supplyvoltages as shown in Figure 4.14.

With field (f) keeping constant and equals to its rated value, the motors torque-

angular speed capability is shown in the left portion of the graph shown in Figure4.17. The region of constant field is also called constant torque region.


27/58


Ea

Tem

,f, I

f

Tem,Ia, f, If Vt,Ia

Vt

Ea

Per unitquantities

Constant torque region

(f= rated)

Field weakening or

constant power region(fis decreased)

1.0

0 1.0

M

(per unit)

Figure 4.17Continuous torque-angular speed capability of a separate excited dc motor

At constant magnetic field (f), the angular speed (M) of motor increases linearlywith the supply voltage (Vt). At the right hand side of the graph, it shows at constantpower region, the magnetic field is decreasing. This is the region whereby the angularspeed is kept increasingly operating beyond its maximum rating. The magnetic field(f) has to be decreased while keeping input supply voltage (Vt) constant to maintain

a constant voltage (Ea) across the armature. It is in the constant power region becausenot only the supply voltage (Vt) is kept constant, the armature current is (I a) alsokept constant. Note that in the magnetic field weak region, the angular speed of themotor may exceed 50 100% of its rated speed.

Effect of armature current

In sc motor drives, the output voltage of the power electronic converter containsan ac ripple voltage superimposed on the dc voltage. Ripple voltage in the terminal

voltage can lead to ripple in armature current (Ia) with the listed effects here. Theyare form factor and torque pulsation.

Form factor

The form factor of the dc motor armature current is defined as:

Form factor =Ia(rms) (4.27)

Ia(average)


28/58


Based on equation (4.27), the form factor will be unity provided the ripple currentripple (ia) is purely dc. The more ripple armature current (ia) derivates from pure dc,the higher will be the form factor. The higher the form factor will result in higher

power loss because power depends on the rmssquare of the ripple current and theresistance of armature winding. The end result is higher power loss, which shallmean high heating causing loss in motor efficiency.

If the form factor is much higher than one, this implies that the peak armaturecurrent is much larger than its average armature current, which can cause excessivearcing in the commutator and brushes. To avoid this serious damage to the motorthat is caused by peak armature current, the motor has to derate. This shall meanthat the maximum power or torque has to keep below the rating in order to keepthe temperature below its specified limit and also to protect the commentator andbrushes.

Activity 4.4

If the rms current of the armature is 2.0 A and the average currentis 1.9 A, calculate the form factor of the armature current of thedc motor.

Torque pulsation

Since the instantaneous electromagnetic torque (Tem) developed by the motor isproportional to the instantaneous armature current (Ia), a ripple current (ia) resultsin a ripple in the torque and also the angular speed if the inertia is not large. A highfrequency torque ripple will generally have lesser angular speed fluctuation thanthe low frequency ripple. This is because the filter circuit at the output of powerelectronic is a low pass filter, whereby high frequency ripple voltage will be attenuated.

Activity 4.5

If the ripple current of armature is too high, what will be the effecton the angular frequency of the motor? Being an engineer, what isyour suggestion to solve this problem?


29/58


Servo drives

As mentioned earlier, in servo applications, the speed and accuracy of response

are important. In spite of the increasing popularity of ac servo motor, dc servodrives are still widely used today. If it is not because of the disadvantages of havingcommutator and brushes, the dc motor would be ideal for servo application. Thisis because the instantaneous electromagnetic torque (Tem) can be linearly controlledby the armature current (Ia).

Figure 4.18shows a servo dc motor operating in a closed-loop to deliver controlledspeed and position. The position and speed transducer is used to measure theposition and speed of the load. The information is used to feedback to the positionand speed controller. It is then compared with the position and speed reference fortaking appropriate action. This circuit also contains the armature current monitoringmeasurement that is used to feedback to the current controller for ensuring that themotor is not operating beyond its rated value.

Controller

i*a

+

Position/Speed

Controller

Position/speed

reference

PowerElectronicsConverter

Position/Speed

TransducerMotor

Speed

Position

LoadCurrentController

Controlvoltage

ia

t

Figure 4.18 A closed-loop position and speed controlled servo dc motor

For analysing small signal performance of the motor-load system around a steadystate operating point, the following sets of equations are applicable.

Vt= ea+ Raia+ Lad

(ia) (4.28)

dtea= keM (4.29)

Tem= kTia (4.30)

Tem= TWL+ BM+Jd(M) (4.31)

dt

If one takes Laplace transform of these equations, where the Laplace variablesrepresent only the small signal values, then equations (4.28) to (4.31) will be

Vt(s) = Ea(s) + (Ra+ sLa)Ia(s) (4.32)

Ea(s) + kEa(s) (4.33)


30/58


Tem(s) = kTIa(s) (4.34)

Tem(s) = TWL(s) + (B+ sJ)M(s) (4.35)

m(s) = sM(s) (4.36)

These equations from (4.32) to (4.36) which are meant for motor-load systemcan be represented by the transfer function block as shown in Figure 4.19. Theinputs to the motor-load system are the armature terminal voltage Vt(s) and theload torque [TWL(s)]. Applying superposition principle by applying one input at atime by setting the other input to be zero, it yields equation (4.37) the angularfrequency of the motor.

M(s) = kT Vt(s) Ra+ sLa TWL (4.37) (Ra+ sLa)(s J+ B) + kTkE (Ra+ sLa)(s J+ B) + kTkE

TWL

(s )

Tem

(s )Ia(s )

Ea

(s )

Vt(s )

++

M

(s )k

T

kE

m(s ) 1

B+ sj

1

Ra+ sLa1

s

Figure 4.19 Block diagram representing the motor-load system without feedback

Equation (4.37) results in two closed-loop transfer functions G1(s) and G2(s), whichare

G1(s) =M(s) | = kT (4.38)

Vt(s) TWL(s) = 0

(Ra+ sLa)(s J+ B) + kT kE

For TWL(s) = 0 and

and

G1(s) =M(s) | = Ra+ sLa (4.39)

TWL(s) Vt(s) = 0 (Ra+ sLa)(s J+ B) + kT kE

for Vt(s) = 0.


31/58


From equation (4.38), for a system with small damping and without load, then theequation is equal to

1G1(s)|Vt(s) = 0=kT = (4.40)

s Jm(Ra+ sLa) + kT kE

a m a mkE(s2

L J

+s

R J

+1)

kTkE

kTkE

where m=RaJm and e=

La are defined as mechanical time constant and electricalkTkE Ra

time constant respective. Thus, equation (4.40) becomes

G1(s)|Vt(s) = 0=1

(4.41) kE(s

2me+ sm+ 1)

In general, the mechanical time constant is much larger than electrical time constant.Thus, equation (4.41) is approximately equal to equation (4.42) after replacing smby s(m+e).

G1(s)|Vt(s)=1

(4.42) kE(sm+ 1)(se+ 1)

The electrical time constant (e) determines how quickly the armature current isbuilt up as shown in Figure 4.20in response to a step change in input voltage (Vt)in the terminal voltage where the rotor speed is assumed to be constant. The graphindicates that the change of armature current (ia) of the motor is exponentialramping up, which is proportional to (1 et/te).

Figure 4.20 Electrical time constant with angular speed of motor is assumed constant


32/58


The mechanical time constant determines how quickly the speed of the motor is builtup in response to the change in input voltage (Vt) in the terminal voltage providedthe electrical time constant (e) is assumed to negligible. Neglecting electrical time

constant (e), the change in speed from a steady state condition can be obtainedfrom equation (4.43).

1

mM(s) =

Vt(s) =Vt =

Vt kE(se+ 1) kEs(sm+ 1) kE

s

(

s+ 1

)

m

Notice that from equation (4.43), vt(s) =Vt, then from equation (4.43), it yields

s

equation (4.44).

M(t) =Vt(1 et/m) (4.44)

kE

The plot of equation (4.44) is illustrated in Figure 4.21.

Figure 4.21 Mechanical time constant with load torque assumed to be constant


33/58


Summary

In this section, you have learnt and understood how to examinethe equivalent circuit of a dc motor; analysed the characteristics ofpermanent magnetic dc motor on how to apply the design conceptto increase the power of a dc motor with excited wind stator; howto interpret the effect of armature current on the performance of dcmotor; and discriminate the difference between a normal dc motorand a dc servo motor.

Self-test 4.3

A permanent magnetic dc servo motor has the following given

parameters; kT =0.5 Nm

, kE =53 V

, Ra = 0.4 , andA 1000rpmm= 12.0 ms.

1. Calculate the terminal voltage (Vt) of the motor if it is required

to deliver torque (Tem) of 5.0 Nm with angular speed (

M) of 1,500 rpm.

2. Calculate the change of angular speed of the motor at time = 2.0 min if the change of terminal voltage (Vt) is 20 V.


Feedback

Activity 4.3

Use equation M=1

(VtRaTem) and re-arrange it, it yieldskE kT

Tem= kEkt

M Vt. Thus, the gradient of the graph is kEkT.

Ra Ra

The armature resistance (Ra) is in denominator of the equation ofgradient. Thus, for a small Ra, it yields a large gradient.


34/58


Activity 4.4

The form factor of the armature current follows equation (4.27),

which is Form factor = Ia(rms) = 2.0 = 1.05. Ia(average) 1.9

Activity 4.5

If the ripple current armature is too high, it will torque pulsationsince the torque is dependent on armature current, which is

Tem= kTIa. Torque of motor is also equal to Tem= J dM with dt

assumption of no damping. This will cause variation in angularfrequency (M) of the motor.

As an engineer, a method to overcome this problem is to designlow-pass LC filter circuit at the output of power electronic voltageconverter to have higher critical frequency than what it has presentlyin the circuit.


35/58


4.3 ac Motor Drives

Objectives


1. Describe the basic principle of an inductor motor.

2. Describe per phase representation of an induction motor.

3. Describe the equivalent circuit of an inductor motor.

4. Design ac synchronous motor drives.

5. Describe and analyse per phase representation of a synchronous motor.

6. Describe and analyse the equivalent circuit of a synchronous motor.

Introduction

In this section, you will learn two types of ac motor drives, which are inductor motor

drive and synchronous motor drives.ac induction motor, which is an asynchronous motor, is the workhorse of industrybecause of its low cost and rugged construction. When operated directly with 50 Hzac utility input power, it can operate at a nearly constant angular speed. However,by means of power electronic converter, it is possible to vary the angular speed ofthe motor. The inductor motor drives can be classified into two broad categoriesdepending on their applications, which are adjustable speed drives and servo drives.

ac synchronous motor is used as servo drives in the applications such as computerperipheral equipment, robotic, and adjustable speed drives in a variety of applications

such as load proportional capacity-modulated heat pump, large fan, and compressor.

Induction motor drives

Lets begin the study of inductor motor drives by understanding the behaviour ofthis motor type and how it is used to control its angular speed (M). Two simpleexamples of an inductor motor driving a centrifugal pump are shown in Figure4.22. Showing in the figures are the constant speed drive type and adjustable-speeddrive type.


36/58


(a) Constant speed drive type

(b) Adjustable-speed drive type

Figure 4.22 Centrifugal pump

For the constant speed pump shown in Figure 4.22 (a), it would cause energy

loss across the throttling valve if the flow rate is to be reduced by partially closingthe throttling valve. However, for the adjustable-speed design type shown inFigure 4.22 (b), the desired flow rate can be controlled by eliminating the energy lossacross the throttling valve by adjusting the motor speed. For this motor-load systemtype, the input power decreased significantly as the angular speed is also decreasedto reduce the flow rate. The decrease in power can be calculated by looking at thetorque of centrifugal pump, where the torque is

Torque k1(speed)2 (4.45)

Thus, the power requirement by the pump from the motor is defined by

Power k2(speed)2 (4.46)

where k1and k2are constant of proportionality.

Majority applications of induction motor involve three-phase operation. Thus,the stator of an induction motor is designed to consist of three phase windingsdistributed in the stator slots. These windings are displaced by 120 in space withrespect to each other. If a balanced set of three phase sinusoidal signals of frequency

(f) are applied to the stator, it results in balanced sets of current, which establishesa flux density distribution (Bag) in the air gap with a constant amplitude and ratoteswith a constant angular speed in which it is called synchronous angular speed (s)radian per second. The synchronous angular speed (s) in ap-pole motor suppliedby frequency (f) can be obtained from equation (4.47).


37/58


s=2 /(p /2)

=2

(2f) =2 (4.47)

1/f p p

The synchronous angular frequency (s) should be synchronised with the suppliedfrequency (f) of the applied voltage and current to the windings of stator. In termsof revolution per minute (rpm), the synchronous speed (ns) should be

ns=60 s =

120f (4.48)

2 p

The air gap flux (ag) caused by the flux density distribution (Bag) rotates at a

synchronous speed relative to the stationary stator winding. As the result, counteremf called air gap voltage (Eag) is induced in each of the stator phase frequency (f).This can be illustrated by means of the equivalent circuit for per-phase voltage shownin Figure 4.23. Vsis per phase voltage, which is line voltage divided by 3, Eagisthe air gap voltage, Rsis the resistance of the stator winding, and Llsis the leakageinductance of the stator winding. The magnetised current component (Im) of thestator current (Is) establishes the air gap flux. From the magnetic circuit analysis,it can be seen that

Nsag= Lmim (4.49)

where Nsis the equivalent number of turn per phase of stator winding and Lmis themagnetised inductance as shown in Figure 4.23.

From Faradays law, the air gap voltage (eag), which is the counter emf, is equal to

eag= Nsd ag (4.50)

dt

With the air gap flux (ag) links to the stator phase winding, the air gap flux is

ag(t) = agsint and the air gap voltage (eag) is equal to

eag= Nsagcost (4.51)

after differentiating ag(t) = as sint to getd ag and then substituting it into

dt

equation (4.50).

In terms of rmsvalue, the air gap voltage (eag) has an rmsvalue of

Eag= k3fag (4.52)

where k3is a constant depending on the number of turns of the stator winding andEagdenotes the rmsvalue of air gap voltage.


38/58


Figure 4.23 The equivalent circuit per-phase representation of an induction motor

If the motor is rotating at the synchronous speed (s), then there will not be anyrelative motion between the air gap flux and the rotor. Thus, there will not haveany induced rotor voltage, rotor current, and rotor torque. At any other angularspeed (r) of the rotor in the same direction of the air gap flux rotation, the motoris slipping with respect to the air gap at a relative speed called the slip angularspeed (sl). A relationship between the synchronous angular speed (s) and betweenother angular speed and slip angular speed is given by

sl= s r (4.53)

If the slip angular speed (sl) is normalised with synchronous speed is called slip(s), it is defined as

Slip (s) per unit =Slip angular speed

= s r (4.54)

Synchronous speed s

Therefore, from equations (4.53) and (4.54), the slip angular speed (sl) of air gapflux with respect to the motor is calculated from equation (4.55).

Slip angular speed (sl) = s r= ss (4.55)

From Faradays law, the induced voltage in the rotor circuit is at the slip frequency(fsl) and it is proportional to the slip speed (sl). Thus,

fsl= slf= sf (4.56)

s

The induced emf(Er) obtained in the rotor conductor can be obtained by replacingfin equation (4.52) by the slip frequency (fsl). If one assumes that squirrel-cage

motor is replaced by a three-phase short-circuited winding with the same equivalentnumber of turns (Ns) per phase, one gets

Er= k3fslag (4.57)


39/58


Since the rotor squirrel-cage winding is short circuited by the end rings, the inducedvoltages at the slip frequency result in a rotor current (Ir) at the slip frequency (fsl),which is equal to

Er= Rs Ir +j2fslLlsIr (4.58)

where Rr and Llr are the resistance and the leakage inductance of the per-phaseequivalent rotor winding. Since sl + r = s, the slip frequency rotor currentproduces a field that rotates at the slip speed with respect to the rotor, and atthe synchronous speed with respect to the stator. The interaction of agand fieldproduced by the rotor current results in an electromagnetic torque. Losses in therotor winding resistance are

Pr= 3RrI2r (4.59)

Take equation (4.58) and multiply it byf

and using equation (4.52) and equation fsl

(4.57), it yields equation (4.60).

Eag=f

Er=fRr Ir+j2fLlrIr (4.60)

fsl fsl

As it has been indicated in Figure 4.23,fRr represents the sum of Rr andfsl

Rr(ffsl) . From equation (4.60), all rotor quantities are referred to N s, which is fsl

the stator number of turns. By multiplying both sides of equation (4.60) by I *randtaking the real part Re[ErI

*r], the power crossing the air gap, which is air gap power

(Pag) is

Pag= 3f

RrI2r (4.61)

fsl

From equation (4.61) and equation (4.59), the electromechanical power (Pem) isequal to

Pem= Pag= 3RrffslI2r (4.62)

fsl

and the electromagnetic torque is equal to

Tem=Pem (4.63)

r


40/58


This equation is also equal to equation (4.64) from equations (4.53), (4.62) and(4.63).

Tem=

Pag

(4.64) s

Thus, as it is shown in Figure 4.23, the equivalent circuit of per-phase representationof an induction motor, the loss in rotor resistance and the per-phase electromechanical

power are shown by splitting the resistancef(Rr)in equation (4.60) into Rfand fslRr(f fsl). fsl

The total current (Is) drawn by the stator is the sum of the magnetising current (Im)

and the equivalent rotor current (Ir), which is

Is= Im+ Ir (4.65)

The phasor diagram for the stator current and voltage is shown in Figure 4.24,whereby the magnetising current (Im) is produced by air gap field (ag), which lagsthe air gap voltage (Eag) by 90. The rotor current (Ir) which is responsible to producethe electromagnetic torque (Tem) lags the air gap voltage (Eag) by the power factorangle (r) of the rotor circuit, whereby the power factor angle is defined as

2fL lrr= tan

12fslLlr = tan1

(4.66) Rr

rR f

fl s

Figure 4.24 The phase diagram of per-phase representation of an equivalent circuitfor an induction motor

From electromagnetic theory, the torque (Tem) produced is

Tem= k4agIrsin (4.67)


41/58


where = 90 + r and k4 is emf constant or torque constant that has unitVs(wb rad)1. It is the angle between the magnetising current (Im), which producesair gap flux (ag) and rotor current (Ir), which represents the rotor field.

The applied per-phase stator voltage (Vs) is given by

Vs= Eag+ (Rs+j2fL ls)Is (4.68)

For an induction motor of normal design, the following condition stated in equation(4.69) is true in the rotor circuit at low value of fslcorresponding to normal operation.

2fslL lr


42/58


The ratio of power loss (Pr) in the rotor and the electromechanical output power(Pem) is defined as

Pr% =

Pr

=

fsl

(4.76) Pem ffsl

After going through so much theory and mathematical equation governing the basicprinciple for operating induction motor, lets summarise the important equation fora frequency controlled induction motor in Figure 4.25.

Synchronous angular frequency s= k

7f

Slips =

s

r

s

Slip frequency fsl= sf

Ratio of power loss in the rotor and the electromechanicaloutput power

Pr% =

Pr

=f

sl

Pem

ffsl

Applied per-phase stator voltage Vsk

3

agf

Total current (Is) I

s= I2

m+ I2

r

Electromagnetic torque Tem

k4k

52

agf

sl

Rotor current Irk

5

agf

sl

Figure 4.25 Important equations for a frequency controlled induction motor

Activity 4.6

A four-pole, three-phase induction motor has per-phase equivalentcircuit shown below. Determine its synchronous speed (s) persecond, and rotor speed (r) for 200 V per phase 50 Hz input andhaving slip of 2.0%.


43/58


Synchronous motor drives

The structure of a synchronous motor is shown in Figure 4.26. Figure 4.26 (a)

shows a two-pole permanent magnetic rotor type, while Figure 4.26 (b)shows thetwo-pole salient-pole wound rotor type.

(a) A two pole type permanent-magnet rotor

(b) A two pole type salient-polewound rotor

Figure 4.26 Structure of synchronous motor

The field winding on the motor produces flux (f) in the air gap. The flux rotatesat a synchronous angular speed srad/s, which is the same as the rotor speed. Theflux (fa) linking one of the stator phase windings with an example phase avariessinusoidally with time.

fa(t) = fsint (4.77)

where

= 2f=p

s (4.78) 2

pis the number of poles in the motor. If one assumes that Nsis an equivalent numberof turns in each stator phase winding, the induced emf in phase ais equal to

efa(t) = Nsdfa = Nsfcost (4.79)

dt


44/58


The induced voltage in stator winding is called excitation voltage whereby its rmsvalue is equal to

Efa=

Nsf=

kENs

(4.80) 2 2

where kEis ratio of the peak phase voltage and speed.

In the normal convention, the amplitudes of voltage and current phasor arerepresented by their rmsvalue; the amplitudes of flux phasors are represented by theirpeak values. Being sinusoidal with time, efaand facan be represented as phasor att = 0, where Efa= Efais reference phasor in Figure 4.27and from equation (4.77),

fa= jfa (4.81)

From equations (4.79) to (4.81) and Figure 4.27, voltage across the phase ais

Efa=jNs

fa= Efa2

Reference

Esa

= jLaIa

Eag, a

Efa

sa

fa

ag, a

( 90)

90

Ia

p

Figure 4.27 Phasor diagram of per phase representation of synchronous motor

In synchronous motor drives, the stator is supplied with a set of balanced three-phasecurrent, whose frequency is controlled to bef, which is equation (4.78).

f=p s (4.82)

4

The fundamental frequency components of these stator motor currents produce aconstant amplitude flux (s) in the air gap, which rotates at synchronous speed (s).The amplitude of sis proportional to the amplitudes of the fundamental frequencycomponents in the stator.


45/58


In three-phase motor, the flux links with phase adue to sproduced by all thesestator currents is sa(t). sa(t) is proportional to the phase a current ia(t) i.e.,

sa(t) =

Laia(t)

(4.83) Ns

where the armature inductance (La) is 3/2 times the self-inductance of phase a.Therefore, from equation (4.83), the synchronous back emfof the armature (esa)is equal to

esa(t) = Nsd sa = La

dia (4.84) dt dt

Assuming the fundamental component at the supplied current to the stator phase

ato be

ia(t) = 2Iasin(t + ) (4.85)

Differentiation of ia(t) in equation (4.85) with respect to time (t) and rewrite equation(4.84), it yields equation (4.86)

esa(t) = 2IaLacos(t + ) (4.86)

where

is the torque angle. Here iaand eascan be represented as phasor, which iswhen the angular frequency is equal to zero i.e.,

Ia= Iaej( )/2 (4.87)

As shown in Figure 4.27, phasor for stator phase voltage is

Esa=jLaIa= LaIaej (4.88)

The resultant air gap flux [ag, a(t)] linking the stator phase a is the sum of fa(t) andsa(t), which is defined by equation (4.89).

ag, a= fa(t) + sa(t) (4.89)

In which, it can be represented by phasor i.e.,

ag, a= fa+ sa (4.90)


46/58


The air gap voltage eag, a(t) due to resultant air gap flux (ag) linking phase ais

ag, a(t) = Nsd ag, a = efa(t) + esa(t) (4.91)

dt

Using equations (4.89), (4.79) and (4.84), equations (4.81) and (4.88) combinedwith equation (4.91) would yield

Eag, a= Efa+ Esa= Efa+jLaIa (4.92)

Based on equation (4.92) and phasor diagram, a per phase equivalent circuit of asynchronous motor is shown in Figure 4.28, where Rsand Llsare the stator windingresistance and leakage inductance respectively. If one includes the voltage drop across

Rsand Lls, then the per phase terminal voltage in phase ais

Va= Eag, a= Efa+ (Ra+jLs)Ia (4.93)

Synchronous inductance Ls

Lls

Eag, a E

fa

Esa

+

+

+ +

Rs

Ia

Va

La

Figure 4.28 Per phase equivalent circuit of a synchronous motor

The phasor representation of equation (4.93) is shown in Figure 4.29, where ais

the angle between the current and terminal voltage phasor.

Reference

Eag, a

Va

Ia

(Rs+ jL

ls)I

a

ap

Figure 4.29 Phasor representation of the terminal voltage for per phase of asynchronous motor


47/58


From the phasor diagrams shown in Figure 4.27and Figure 4.29, the electromagnetictorque (Tem) can be obtained from the electrical power that has been converted intomechanical power (Pem). The mechanical power (Pem) is

Pem= 3EfaIacos(1 ) (4.94)

2

and the electromagnetic torque (Tem) is equal to

Tem=Pem (4.95)

s

From equations (4.94), (4.95) and (4.80), the electromagnetic torque is equal to

Tem= ktfIasin (4.96)

Activity 4.7

State two of the conditions for an ac motor to be claimed assynchronous motor.

Summary

In this section on ac motor drives, you have learnt and understoodthe basic working principle of an inductor motor includingequivalent circuit and per phase phasor diagram of the voltageand current of the induction motor, and the working principleand design of an ac synchronous motor, which also includes theequivalent circuit of the motor and the per-phase phasor diagram

of the motor.

Self-test 4.4

For the same induction motor shown in Activity 4.6, determinethe Ircurrent and torque for 100 V 25 Hz per phase input if theslip is 1%.


48/58


Self-test 4.5

A four pole 460 V 10 ph motor is supplying its rated power to acentrifugal load at 50 Hz and it has rated speed equal to 1500 rpm.Calculate its speed, rated slip, and rated slip frequency at 50 Hz.

Self-test 4.6

A permanent magnet synchronous motor has the ratio of the peakphase voltage induced to the rotational speed equal to 25V/1000rpm

at p= 2, and n =10. Calculate the terminal frequency, per phasevoltage, and per phase resistance if the load draws 10 A rms perphase from the motor.


Feedback

Activity 4.6

The synchronous speed per minute is equal to ns=60 s =

120f,

2 p

which is equal to2

f revolution per second. This gives rise top

2 50 = 25 rev/s.

4

The rotor speed (r) can be calculated from s = s r. This

s

implies that r= 25(1 s) rev/s = 25(1 0.02) = 24.5 Rev/s, whichis 154 rad/s.

Activity 4.7

Two of the conditions are; the angular frequency of the motor isequal to the supplied line frequency and a balanced set of threephase sinusoidal signals of frequency (f) are applied to the statorthat results in balanced sets of current.


49/58


Summary of Unit 4

Summary

In this Unit 4 on Electrical drives system, you have learnt three maintopics, which are introduction to motor drives, dc motor drives,and ac motor drives. In each topic, you have studied and achievedwhat have been specified in the learning objectives via tutorials,learning activities, and self-tests.

In section one on dc motor drives, you have learnt and achieved

the ability and confidence to explain the general control of motordrives; explain the concept of designing servo motor drives; explainthe design concept of variable speed motor drives; and select thecomponents and electric motor for motor drives.

In section two on dc motor drives, you have learnt and achievedthe ability and confidence to examine the equivalent circuit of a dcmotor; analyse the characteristics of permanent magnetic dc motor;apply the concept on how to increase the power of a dc motor bydesigning with excited wind stator; interpret the effect of armaturecurrent on the performance of dc motor; and discriminate thedifference between a normal dc motor and a dc servo motor.

In section three on ac motor drives, you have learnt and achievedthe ability and confidence to describe the basic principle of aninductor motor; describe per phase representation of the inductionmotor; describe the equivalent circuit of the inductor motor;design ac synchronous motor drives, describe and analyse per phaserepresentation of the synchronous motor; and describe and analysethe equivalent circuit of the synchronous motor.


50/58



51/58


Suggested Answers to Self-tests

Feedback

Self-test 4.1

Since L= 2 then the coupling ratio (a) is equal to

1. Thus, the

m 2

ratio of Land Mis L =

1. This implies that the angular speed

m 2

of motor is twice the angular speed of load, in which its profile isshown below.

Between 0 and 1s, the torque of the motor is 2000 rad/s2 2.5kgm2= 5000Nm.

Between 1s to 3s, the torque of the motor is 0 2.5kgm2= 0Nm.

Between 3s to 4s, the torque of the motor is 2000 2.5 kgm2=5000Nm.

Between 4s to 5s, the torgque of the motor is 0 rad/s 2 2.5 kgm2

= 0Nm.

The torque profile of the motor is shown as follows.


52/58


Self-test 4.2

The electromagetic torque (Tem) is equal Tem= JeqdM since the

dt

damping can be ignored.

The equivalent inertia (Jeq) is equal toJeq=JM+ a2JL= 2.5 + 0.5

210 = 5.0 kgm2.

Thus, the electromagnetic torque (Tem) is Tem= 5 dMdt

From the result of Self-test 4.1, the angular speed (M) of the motoris as follows:

For time t = 0 to 1s, the electromagnetic torque (Tem) is Tem =5 2000 = 10,000Nm.

For time t = 1s to 3s, the electromagnetic torque (Tem) is Tem=5 0 = 0Nm.

For time t = 3s to 4s, the electromagnetic torque (Tem) is Tem= 5 (2000) = 10,000Nm

For time t = 4s to 5s, the electromagnetic torque (T em) is Tem= 5 0 = 0Nm.

The electromagnetic torque (Tem) profile is shown as follows:


53/58


Self-test 4.3

1. The terminal voltage can be calculated using equation

Vt= kEM RaTem.

kT

Thus, Vt= 53 1.5 + 0.4 5.0

= 99.5V. 0.5

2. The change angular speed (Vt) of motor at time = 2.0 min for

Vtof 3.0V is M(t) =Vt (1 et/m) = 3.0 (1 e2/12)

kE 53/1000

8.69rpm.

Self-test 4.4

The synchronous speed per minute is equal to ns=60s =

120f,

2 p

which is equal to2

f revolution per second. This gives rise top

2 25 = 12.5 rev/s.

4

The rotor speed (r) can be calculated from s =s r . This

s

implies that r= 12.5(1 s) rev/s = 12.0 Rev/s, which is 75.39 rad/s.

The Ircurrent can be found from the analysis of the equivalent circuit

using current divider rule, which is


54/58


Ir=100s

A. 0.075s + 0.0146 +j(0.272s 0.0012)

After substituting s = 0.01, the rotor current is equal to

Ir=1

=1

= 65.0A. 0.0153 +j(0.00152) 0.0153

The torque per phase is equal to

Tem=I2r 0.04/s =

I2r 0.04 = 215.2Nm. 2 12.5 25s

Since it is a three-phase motor, the total torque is equal to 215.2 3 = 645.5Nm

Self-test 4.5

At 50Hz, the ns=120

f=120

50 = 1,500 rev/min. p 4

The rated slip is s =ns nr =

1500 1455= 0.03

ns 1500

The rated slip frequency is equal tofsl= sf= 0.03 50 = 1.8 rev/s

Self-test 4.6

The frequency of stator is ns =60 s

=120

f . ns is equal to 2 p

10,000rpm and the number phasep= 2. This frequency (f) of the

supplyf=2

10,000 = 166.67Hz. 120

KEis given to be 24V/1000rpm. Thus, per phase a, the rms voltage

is Efa=kENs =

25 10= 176.7V.

2 2

The per phase resistance is 176.7V/10A = 17.67.


55/58


References

Bose, B (2006) Power Electronics and Motor Drives, New York: Academic Press.

Jacob, J M (2002) Power Electronics: Principles & Application, New York: DelmarThomson Learning.

Mohan, N and Undeland, T M and Robbins, W P (2003) Power electronics: converters,applications, and design, New York: John Wiley.


56/58



57/58


Coupling ratio

Equivalent damping factor

Damping factor of load

Damping factor of motor

Air gap back emf

Induced emffrom phase a

Back emfof stator per phase a

Working force

Armature ac current

Armature current

Field current

Magnetising current

Rotor current

Inertia of load

Inertia of motor

Voltage constant of motor

Current constant of motor

Armature inductance

Slip leakage inductance

Pole

Air gap power

Electrical power

Electromechanical power

Glossary

a

Beq

BL

BM

eag

efa

esa

FWL

ia

Ia

f

Im

Ir

JL

JM

ke

kT

La

Lls

p

Pag

Pe

Pem


58/58


Pm

Ra

Tem

TL

Tm

TWL

s

L

ag

ag,a(t)

f

fa

sa(t)

sl

s

M

Mechanical power

Armature resistance

Electromagnetic torque of motor

Torque of load

Torque of motor

Working torque

Motor position

Load position

Air gap flux

Air gap flux per phase a

Field flux

Field flux from phase a

Stator current

Slip frequency

Synchronous frequency

Angular speed of motor

Documents

Power Electronics and Drives U4