Unit 1ELECTROMAGNETISM

( Keelektromagnetan )

General Objective

To understand the basic principles of electromagnetism.

1.0 INTRODUCTION

The pattern of magnetic field of bar a magnet.

Arah uratdaya magnet

Attraction and Repulsion( tarikan dan tolakan )

Tarikan

Tolakan

1.1 CURRENT-CARRYING CONDUCTOR AND ELECTROMAGNETISM

( keelektromagnetan ke atas pengaliryang membawa arus )

• A flow of current through a wire produces a magnetic field in a circular path around the wire.The direction of magnetic line of flux around the wire is best remembered by the screw rule or the grip rule.

The field pattern of current flowing in the wire

Arus keluar

Arus masuk

( a) flow in the same direction ( b) opposite direction

tolakantarikan

If two closed current-carrying conductors flow in the same direction, magnetic flux

around that conductor will combine to create attraction between them. If closed

current-carrying conductors flow in opposite direction, these two conductors

will repulse each other

1.2. MAGNETIC FIELD STRENGTH, H (MAGNETISING FORCE)

• Magnetic field strength is defined as magnetomotive force, Fm

lNI

lF

H m == ampere turn / metre

N = bilangan lilitan pengalirI = arus yang mengalirl = panjang bahan magnet

1.3. MAGNETIC QUANTITY AND THEIR RELEVANT FORMULAE

• 1.3.1 Magnetic Flux and Flux density

-Magnetic flux,Φ is the amount of magnetic filed produced by a magnetic source.

- Flux density,B is the amount of fluxpassing through a defined area

unit for flux is the weber, wb

-AΦB =Flux density, Tesla

• Example 1.1• A current of 500mA is passed through a

600 turn coil wound of a iron of mean diameter 10cm. Calculate the magnetic field strength.

• An iron ring has a cross-sectional area of 400 mm2 and a mean diameter of 25 cm. it is wound with 500 turns. If the value of relative permeability is 250, find the totalflux set up in the ring. The coil resistance is 474 Ω and the supply voltage is 20 V.

Example 1.2

Fig. 1.1.

1.3.2 Permeability ( ketelapan )

the ratio of magnetic flux density to magnetic field strength is constant

HB

= a constant

( Kebolehan sesuatu bahan magnet untuk menghasilkan uratdaya magnet )

= 4π x 10-7 H/m 0μ=HB

rHB μμ0=

For air, free space and any other non-magnetic medium, the ratio

For all media other than free pace,

Cast iron μr = 100 – 250Mild steel μr = 200 – 800Cast steel μr = 300 – 900μr for a vacuum is 1

μ - absolute permeabilityμr - relative permeabilityμo - air permeability

where μ = μoμr

• 1.3.3 Reluctance ( Engganan )Reluctance, S is the magnetic resistance of a magnetic circuit

Al

rAl

BH

BAHlFS m

μμο

1===

Φ=

unit for reluctance is H-1

Perbadingan di antara Litar ElekrikDengan Litar magnet

Litar Elektrik Litar Magnet1. Arus2. Dge3. Rintangan

1. Uratdaya ( Fluks )2. Dgm3. Engganan

• Example 1.3A magnetic pole face has rectangular section having dimensions 200mm by 100mm. If the total flux emerging from the the pole is 150μWb, calculate the flux density.Example 1.4A flux density of 1.2 T is produced in a piece of cast steel by a magnetizing force of 1250 A/m. Find the relative permeability of the steel under these conditions.

• Example 1.5Determine the reluctance of a piece of metal of length 150mm, and cross sectional is 100mm2

when the relative permeability is 4000. Find also the absolute permeability of the metal.Exersice 1The maximum working flux density of a lifting electromagnet is 1.8 T and the effective area of a pole face is circular in cross-section. If the total magnetic flux produced is 353 mWb, determine the radius of the pole.

1.4 ELECTROMAGNETIC INDUCTION

When a conductor is moved across a magnetic field so as to cut through the flux, an electromagnetic force (e.m.f.) is produced in the conductor. This effect is known as electromagnetic induction. The effect of electromagnetic induction will cause induced current.

Two laws of electromagnetic inductioni. Faraday’s law

Conductor cuts flux

Flux cuts conductor

This induced electromagnetic field is given by θ°

Where ,

B = flux density, T= length of the conductor in

the magnetic field, mv = conductor velocity, m/s

ii. Lenz’z Law

Bar magnet move in and move out from a solenoid

Example 1.6A conductor 300mm long moves at a uniform speed of 4m/s at right-angles to a uniform magnetic field of flux density 1.25T. Determine the current flowing in the conductor when i. its ends are open-circuitedii. its ends are connected to a load of 20 Ωresistance.

Exersice 2

A conductor of length 0.5 m situated in and at right angles to a uniform magnetic field of flux density 1 wb/m2 moves with a velocity of 40 m/s. Calculate the e.m.f induced in the conductor. What will be the e.m.f induced if the conductor moves at an angle 60º to the field.

Solution to Example 1.3Magnetic flux, Φ = 150 μWb = 150 x 10-6 WbCross sectional area, A = 200mm x 100mm

= 20 000 x 10-6m2

Flux density,

= 7.5 mT

6

6

102000010150

AΦB −

−

××

==

Solution to Example 1.4

= 764

HB rμμ 0=

)1250)(104(2.12

0−×

==πμ

μH

Br

Solution to Example 1.5Reluctance,

=

= H-1

AlS

rμμ 0

=

)10100)(4000)(104(10150

67

3

−−

−

×××

π

rμμμ 0=

)4000)(104( 7−×π

Absolute permeability,

= 5.027 x 10-3 H/m

Solution To Example 1.6

i. If the ends of the conductor are open circuited, no current will flow even though1.5 V has been induced.

ii. From Ohm’s lawmA

REI 75

205.1

==

Unit 2

GENERATOR

• OBJECTIVES

To apply the basic principle of DC generator, construction

principle and types of DC generator.

2.0 Introduction

• A generator is a machine that convertsmechanical energy into electrical energy

by using the principle of magnetic induction.

Penjana Arus Terus

Rajah Blok

Sumber Tenaga TenagaTenaga Mekanikal Elektrikal

Rajah Blok Penjana

Pengalir

The Princple of the DC Generator( base on the Faraday’s Law )

Medan Magnet

The Princple of the DC Generator

The Princple of the AC Generator

• Whenever a conductor is moved within a magnetic field in such a way that the

conductor cuts across magnetic lines of flux, voltage is generated in the conductor.

- Faraday’s Law

• The POLARITY of the voltage depends on the direction of the magnetic lines of flux

and the direction of movement of the conductor. To determine the direction of

current in a given situation, the RIHGT-HAND RULE FOR GENERATORS is used.

Right-Hand Rule

• The AMOUNT of voltage generated depends on :i. the strength of the magnetic field,ii. the angle at which the conductor cuts

the magnetic field, iii. the speed at which the conductor is

moved, and iv. the length of the conductor within the

magnetic field.

Model DC Generator

The Princple of the Generating AC Voltage

The Princple of the Generating DC Voltage

2.1 THE PATHS OF THE DC GENERATOR

1. Armature ( angker )

2. Stator ( penetap )

Cross sectional of the DC Generator

Steam Turbine Generator

Hidro Electric Station

Alternator

Nuclear Power Generator

Wind Power Generator

Small Generator

2.2. Types of DC Generator

Stator ( penetap )

Gelung Medan

Armature ( Angker )

(Gelung angker)

Stator and Armature

Bahagian Angker( Gelung Angker )

Bahagian Stator(Gelung Medan)

Types of DC Generator

• Separately-excited generators• Self-excited generators

i. Shunt-wound generatorii. Series-wound generatoriii. Compound-wound generator

a. Short compound generator b. Long compound generator

DC PowerSupply

Penjana Ujaan Berasingan

MedanAngker

1. Penjana Ujaan Berasingan

1. Series-wound generator

2. Penjana Ujaan Diri

2. Shunt-wound generator

3. Compound-wound generator

Example 2.1

A shunt generator supplies a 20 kWload at 200 V. If the field windingresistance, Rf = 50Ω and thearmature resistance Ra = 40 mΩ,determine

(a) the terminal voltage(b) the e.m.f. generated in the armature

2.3. E.m.f generated ( Voltan janaan, dge )

cZn2p Φ

generated e.m.f, Eg =

Where ; Z = number of armature conductors,Φ = useful flux per pole in WebersΡ = number of pairs of polesn = armature speed in rev/s

( c=2 for a wave winding and c= 2p for a lap winding )

Example 2.2.

An 8-pole generator, wave winding connected armature has 600 conductor and is driven 625 rev/min. If the flux per pole is 20mWb, determine the generated e.m.f.

Solution

Z = 600, c = 2 for a wave windingP = 4 pairs, n = 625/60 rev/min, Φ = 30 × 10-3 Wb

cZn2p Φ

Dge, Eg =

2

)60

625)(10 2(4)(20 3-×

Example 2.3.

A 4-pole generator has a lap winding armature, with 50 slots and 16 conductors per slot. The useful flux per pole is 30mWb. Determine the speed at which the machine must be driven to generate an e.m.f. of 240 volts.

E = 240 V, Z = 50 x 16 = 800c = 2p (for a lap winding), Φ = 30 × 10-3 Wb

Ans : ( 10 rev/s )

2.4 Power Losses and Efficiency

For any type of machine, output power is different from input power. The difference is caused by power losses that had happened whenever one type of energy is converted

or delivered to the other type.

The principal losses of machine are:• Copper loss ( I2R )• Iron losses, due to hysteresis and eddy

current • Friction and windage losses• Brush and contact losses ( vB )

2.4.1. Efficiency of DC generatorThe efficiency of an electrical

machine is the ratio of the outputpower and input. The greek letter ‘η’(eta) is used to signify efficiency, the

efficiency has no units.

efficiency, η = %100)powerinput poweroutput ( ×

%100)IIV

IV( 2aLL

LL ×+++ CVIR ffa

η =

%100)( ×+

=DVVo

Voη

Example 2.4

A shunt generator supplies 96 A at a terminal voltage of 200 volts. The armature and shunt field resistances are 0.1Ω and 50Ω respectively. The iron and frictional losses are 2500 W. Find :

(i) e.m.f generated.(ii) copper losses(iii) efficiency

Example 2.5A 75 kW shunt generator is operated at 230 V. The stray losses are 1810 W and shunt field circuit draws 5.35 A. The armature circuit has a resistance of 0.035 Ω and brush drop is 2.2 V. Calculate :

1. total losses2. input of prime mover3. efficiency at rated load.

Unit 3

DC Motor

DC motors are very useful in many

applications of our everyday life, for

example controlling, such as crane, tape driver, lift

system and others.

OBJECTIVES

• General ObjectiveTo apply the basic principles of DC motor

operation, types of DC motor and their application

• Specific Objectives

• Explain the principle operation of DC motor

• List the types of DC motor• State the left-handed rule for motors• List the advantages and disadvantages of

the different types of DC motors.• State typical applications of DC motors

3.1 INTRODUCTION

DC Motor is a machine that convertselectrical energy into mechanical

energy.

Electrical Mechanical Load

Energy Energy

Blok Diagram

Electrical DC Motor

armature( rotor )

Stator

3. 2 THE PARTS OF DC MOTOR

Stator RotorFan

Carbon BrushCommutator

shaft

Cross sectional of the Stator

field coil

Armature

Armature

Armature coil

Commutator

shaft

Armature

TEST YOUR UNDERSTANDING

1 . What is a DC motor?

2. State the uses of DC motors.

3. What are the parts of DC motors?

3. 3 PRINCIPLE OF OPERATION

Fleming′s Left Hand Rule

When a wire carrying current sits into a magnetic field, a force is created on the wire causing it to move

perpendicular ( tegak lurus ) to the magnetic field. The greater the current in the wire, or the greater the magnetic

field, the faster the wire moves because of the greater force created. If the wire sits parallel with the magnetic

field, there will be no force on the wire.

Left-hand rule for DC motors

conductor

( field )

field diretion

power supply

DC motor rotation

3.4 TYPES OF DC MOTOR

• The series DC motor• The shunt DC motor• The compound DC motor

Series DC motor

Shunt DC motor

Compound DC motor

Question

1. What are three major types of DC motor?

2. Draw the schematic diagram of series, shunt and compound of DC motors.

3.5 BACK ELECTROMOTIVE FORCE

• Φ = useful flux per pole in webers• Nr = the speed in revolution per minute• P = the number of pairs of poles

602 pNE rbΦ

=

• Torque ( Daya kilas )

nIET ab

a Π=

260

Example

A 350 V shunt motor runs at its normal speed of 12rev/s when the armature current is 90 A. The resistance of the armature is 0.3 Ω.

Find the speed when the armature current is 45 A and a resistance of 0.4 Ω is connected in series with the armature, the shunt field remaining constant.

3.6 FACTORS THAT INFLUENCE SPEEDCONTROL OF DC MOTOR

The speed of a dc motor is changed by changing the current in the field or by changing the current in the armature.

Controlling motor speed.

3.7 REVERSE DIRECTION METHOD

• The direction of rotation of a dc motor depends on the direction of the magnetic field and the direction of current flow in the armature.

3.8 EFFICIENCY AND POWER LOSSES

Kecekapan

%100powerinput poweroutput ,efficiency ×=η

%100)(2

×−−−

=VI

CVIRIVI faaη

Example

A 320 V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 Ω and the armature resistance is 0.2 Ω, determine the overall efficiency of the motor.

Unit 4 AC ELECTRIC MACHINES

OBJECTIVES

To analyze the basic principles of operationof an AC generator and the differences

between DC generator and AC generatorby using commutator and slip ring.

4.1 INTRODUCTION

An electric generator is a device used to convert mechanical energy into electrical energy.The generator is based on the principle of "electromagnetic induction" discovered by Michael Faraday Law’s.

The simple of Electric Generator

AC Generator

COMMUTATOR

DC generator

Output voltage

(AC waveform)

(DC waveform)

The amount of voltage generated depends on the following:

• The strength of the magnetic field.• The angle at which the conductor cuts the

magnetic field.• The speed at which the conductor is moved• The length of the conductor within the

magnetic field.

4.2 THE DIFFERENCES BETWEEN ACGENERATOR AND DC GENERATOR

The difference between AC and DC generator is that the DC generator results when you replace the slip rings of an elementary generator with

commutator, changing the output from AC to pulsating DC.

AC generator is also called Alternator.

4.3 E.M.F. Equation of an AlternatorE rms / phase = 2.22 KdKp f Φ Z volts

where, Z = No. of conductors or coilΦ = Flux per pole in webersP = Number of rotor polesN = Rotor speed in r.p.mKd = Distribution factorKp = Pitch factor

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pfN 120

Star-connected Delta-connected

Example 4.1 A 3-phase, 50 Hz star-connected alternator has 180 conductors per phase and flux per pole is 0.0543 wb. Find:-a) e.m.f. generated per phaseb) e.m.f. between line terminals.

Assume the winding to be full pitchedand distribution factor to be 0.96.

Exersice 1Find the number of armature conductors in series per phase required for the armature of a 3-phase, 50Hz, 10-pole alternator. The winding is star-connected to give a line voltage of 11000. The flux per pole is 0.16 wb. Assume Kp = 1 and Kd = 0.96.

4.4 AC motor

Stator for an AC motor.

Rotor

Rotor

Differences between AC Motor and DC motor

In general, AC motors cost less than DC motors. Some types of AC motors do not use brush carbon and commutators.

What is the advantage of AC motor over DC motor ?

4.5 Types of AC Motor

1. Series AC Motor

2. Synchoronous Motors

3. Induction Motors

Types of starting induction motor

1. Capasitor-Start

2. Resistance-Start.

• SlipThe actual mechanical speed (nr) of the rotor is often expressed as a fraction of the synchronous speed (ns) as related by slip (s), defined as

s

rs

nnn −

S =

where ns =

Pf120

s

rs

nnn −

× 100%Percent slip, %s =

and fr = sf

fr = frequency rotor

Example 4.2

Determine the synchronous speed of the six pole motor operating from a 220V, 50Hz source.

Example 4.3 The stator of a 3-phase, 4 pole induction motor is connected to a 50 Hz supply. The rotor runs at 1455 rev/min at full load. Determine:

a) the synchronous speed b) the slip

Example 4.4

The frequency of the supply to the stator of an 8-pole induction motor is 50 Hz and the rotor frequency is 3 Hz. Determinei. the slipii. the rotor speed

Example 4.5

A 4-pole, 3 phase, 50 Hz induction motor runs at 1440 rev/min at full load. Calculate a) the synchronous speedb) the percent of slipc) the frequency of the rotor.

Unit 5TRANSFORMER

OBJECTIVES

To understand the basic principles of a transformer, construction

principle, transformer ratio, current and core, type of transformer and

uses.

At the end of the unit you will be able to :• explain the operating principles of a

transformer• describe transformer construction• explain transformer ratio for voltage,

current and winding coil.• calculating of the efficiency.• describe auto transformer

Secondary winding

Primarywinding

transformer construction

CoreCoil /

Winding

5.1 Introduction

The basic transformer is an electrical device that transfers alternating-current energy from one circuit to another circuit

by magnetic flux of the primary and secondary windings of the transformer.

A transformer circuit

A transformer circuit

Primary winding

Secondary winding

Simbol of the transformer

SecondaryPrimary

The uses of the transformerA transformer is a device which used to change the values of alternating voltages or currents to step-up or step-down.

High-voltage transformer

Sub-station transformer

5.2 Transformer Ratio, K

• Ep = 4.44 Np f Φm volts• Es = 4.44 Ns f Φm volts

• If K < 1 i.e. Ns < Np : step-down• If K > 1 i.e. Ns > Np : step-up• If K = 1 i.e. Ns = Np : coupling

equations of ideal transformer

Transformer rating :The rating of the input power of the transformer.

example : 25 kVA ( kV x Arus )

sVVp

sΝpN

pII

s= =

Example 5.1

A 2000/200V, 20kVA transformer has 66 turns in the secondary. Calculate

(i) primary turns

(ii) primary and secondary

Example 5.2A 250 kVA, 1100 V / 400 V, 50 Hz single-phase transformer has 80 turns on a secondary. Calculate :

a) the values of the primary and secondary currents.

b) the number of primary turns.c) the maximum values of flux.

Example 5.3

An ideal 25 kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Calculate(i) primary and secondary currents(ii) secondary e.m.f. and(iii) the maximum core flux

Michael-Faraday

5.3 Auto-transformer

An auto-transformer is a transformer having a part of its winding common to the primary and secondary circuits

auto-transformer

Advantages and disadvantages of auto transformers

- a saving in a cost- less volume, hence less weight.- higher efficiency- a continuously variable output - a smaller percentage voltage regulation.

5.4 Efficiency and losses of a transformer

The losses which occur in the transformer :-i. copper losses, I2Rii. Core losses , Pc

- hysterises- eddy current

Efficiency = powerinput poweroutput

lossespower output poweroutput +

=

ssppcss

ss

RIRIPVIVI

22p.f.p.f.

+++××

=

Example 5.4

The primary and secondary windings of a 500 kVA transformer have resistances of 0.42 Ω and 0.0019 Ω respectively. The primary and secondary voltages are 11 000 V and 400 V respectively and the core loss is 2.9 kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on :

i. full loadii. half load

‘Manusia tidak pernah merancanguntuk gagal, tetapi gagal untuk

merancang’

“Selamat maju jaya”