Energy Conversion

Magnetism and Electromagnetism

Centuries ago, it was discovered that certain types of mineral rock possessed unusual properties of attraction to the metal iron. One particular mineral, called lodestone, or magnetite, is found mentioned in very old historical records (about 2500 years ago in Europe, and much earlier in the Far East) as a subject of curiosity. Later, it was employed in the aid of navigation, as it was found that a piece of this unusual rock would tend to orient itself in a north-south direction if left free to rotate (suspended on a string or on a float in water). A scientific study undertaken in 1269 by Peter Peregrinus revealed that steel could be similarly "charged" with this unusual property after being rubbed against one of the "poles" of a piece of lodestone.

Unlike electric charges (such as those observed when amber is rubbed against cloth), magnetic objects possessed two poles of opposite effect, denoted "north" and "south" after their self-orientation to the earth.

Magnetomotive Force or MMFMagnetomotive force is a force that sets up or tends to set

up magnetic flux in a circuit by passing an electric current through a number of turns of a wire.

Where:F – mmf(ampere-turn)N – number of turnsI – current carried(ampere)

In cgs unit,

Example• A solenoid has 250 turns. What is the

magnetomotive force when the current is 0.12 A?

• The coil in a magnetic contactor requires a 0.5 A to provide a magnetizing force of 500AT. How many turns are necessary?

Magnetic Flux(Φ)Magnetic flux is the number of magnetic

lines of forces in a magnetic field.

maxwell – unit of magnetic flux equal to one line of force.weber – SI unit of magnetic flux equal to lines or maxwell

Flux Density (β) It is given by the flux passing per unit area

through a plane at right angles to the flux. It is usually designated by β and is measured in .

Tesla – SI unit of magnetic flux density equal to webers per square meter.

Gauss – cgs unit of magnetic flux density equal to maxwells per square centimeter.

ExampleHow many magnetic lines of force will pass

a 5 sq. cm perpendicular area on a magnetic field having a flux density of 2000 gauss?

The phenomena of magnetism and electromagnetism are dependent upon a certain property of the medium called its permeability. Every medium is supposed to possess two permeabilities: (i)absolute permeability (μ) and (ii) relative permeability (μr). For measuring relative permeability, vacuum or free space is chosen as the reference medium. It is allotted an absolute permeability of μ0 = 4π × 10−7 henry/metre. Obviously, relative permeability of vacuum with reference to itself is unity. Hence, for free space, absolute permeability μ0 = 4π × 10−7 H/m relative permeability μr = 1.

Permeabilitythe ability of a material to conduct

magnetic flux through it.

Magnetic Field Strength (H)Magnetic field strength at any point within a

magnetic field is numerically equally to the force experienced by a N-pole of one weber placed at that point. Hence, unit of H is N/Wb.

In cgs, Oersted – gilbert per centimeter

Infinitely long Straight Wire

ExampleIf a current 5A flows through a long wire of

radius 0.004 meter, what is the magnetic field intensity produced 0.02 meter away from the surface?

Reluctance- is the property of a material that opposes

flux flow. It is equal to the ratio of the magnetomotive force in a magnetic circuit to the magnetic flux through any cross section of the magnetic circuit.

Magnetic Circuit- a closed path in which magnetic inductin

of flux flows

Reluctance in Series

Reluctance in Series

Electromagentism(page 279 theraja)

DC Generators

Introduction

• Electrical GeneratorA machine that converts mechanical energy into

electrical energy. The energy conversion is based on the principle of the of Faraday’s first law of Electromagnetic Introduction, which states that whenever a conductor cuts magnetic lines of flux, an emf is developed in the conductor

Operating Principle

• Generating an AC Voltage

• The difference between AC and DC generators:– AC generators use slip rings– DC generators use commutators– Otherwise, the machine constructions are essentially the

same.

Essential Parts

• Magnetic frame or Yoke• Pole-cores and Pole-shoes• Pole coils or Field coils• Armature core• Armature windings or Conductors• Commutator• Brushes and Bearings

• YokeThe outer frame or yoke serves double purpose:

— It provides mechanical support for the poles and acts as a protecting cover for the whole machine

— It carries the magnetic flux produced by the poles

• Pole cores and Pole shoesThe pole shoes serve two purposes:

— they spread out the flux in the air gap and reduces the reluctance of the magnetic path

— support the exciting coils

• Pole coilsThe field coils or pole coils are former wound for the

correct dimension. Then the former is removed and wound coil is put into place over the core.

When current is passed through these coils, they electromagnetise the poles which produce the necessary flux that is cut by revolving armature conductors.

• Armature CoreIt houses the armature conductors or coils and causes

them to rotate and hence cut the magnetic flux of the field magnets. Its most important function is to provide a path of very low reluctance to the flux through the armature from a N-pole to a S-pole.

• Armature WindingsThe armature windings are usually former wound.

These are first wound in the form of flat rectangular coils and are then pulled into their proper shape in a coil puller. The conductors are placed in the armature slots whish are lined

with tough insulating material.

• CommutatorThe function of the commutator is to facilitate

collection of current from the armature conductors.

• Brushes and BearingsThe brushes whose function is to collect current from

commutator, are usually made of carbon or graphite and are in the shape of a rectangular block.

The Armature Winding

• Lap windingForms a loop as it expands around the armature core.

windingiveretrogress

windingeprogressiv

on so andduplex for 2 simplex,for factor(1ty multiplici m

pitchfront Y

pitchback Y

where

2

f

b

mYY fb

• Wave windingForms a wave as it expands around the armature core.

windingiveretrogress

windingeprogressiv

on so andduplex for 2 simplex,for factor(1ty multiplici m

pitchfront Y

pitchback Y

poles ofnumber P

elements windingofnumber total Z

pitch average Y

where

2

2

f

b

fb YY

YP

mZY

• Total number of elements or conductors

• Number of armature current paths(a)

• Coil pitch

poleperslots

slotsinspancoilY

mamPa

slotsofnumbertotalslotelementsZ

s

wavelap

2

) )(/(

• ExampleIn a lap winding the front pitch is 17 and the back is

19. What is the average pitch?

Solution

182

17192

ave

ave

fb

ave

Y

Y

YYY

Generated EMF of a DC Generator

constantality proportion k

pathscurrent armature ofnumber a

)pole(weberper flux

conductors ofnumber total Z

pm)rotation(r core armature of speed N

poles ofnumber P

emf(volt) generated E

60

where

kNEa

PNZE

• ExampleA 4 pole dc generator with simplex wave winding has

72 slots. The fluxper pole is 2.88 x 106. What is the speed of the prime mover when the open circuit voltage of the generator is 120 volts?

Solution

rpmN

N

)s per slot conductor(assume a

PNZE

868

10)1)(2(60

)1088.2)(722)((4120

2

1060

86

8

Types of DC Generator According to Excitation

• Separately ExcitedThe field windings of the generator is excited from a

separate source usually a battery

ag

aaL

La

L

LL

EIP

RIVE

II

V

PI

• ExampleA DC generator has no-load output voltage of 120

volts. Its armature resistance is 0.95 ohm and its field coils are separately energized. If the load is rated 2000 W at 115 V. Neglecting the effect of armature reaction, what power could be delivered to the load?

Solution

WP

P

RIP

AI

I

RR

EI

R

R

P

VR

L

L

LLL

L

L

La

L

L

L

L

LL

29.1667

)6.6(894.15

894.156.695.0

120

6.62000

115

2

2

2

2

• Self – ExcitedThe field windings of the generator is supplied or

excited from its own generated emf.– Shunt Generator– Series Generator– Compound Generator• Long Shunt• Short Shunt

• Shunt generator

aaL

ag

shLa

L

LL

sh

Lsh

RIVE

EIP

III

V

PI

R

VI

• ExampleA shunt generator delivers 450A at 230V and the

resistance of the shunt field and armature are 50Ω and 0.03Ω respectively. Calculate the generated emf.

Solution

AI

I

III

AI

AI

I

R

VI

a

a

shLa

L

sh

sh

sh

Lsh

6.454

6.4450

450

6.450

230

VE

E

RIVE

g

g

aag

64.243

)03.0)(6.454(230

• Series Generator

ag

seaaL

La

L

LL

EIP

RRIVE

II

V

PI

)(

• ExampleA DC series generator is supplying a current of 10A to a

load through a feeder of total resistance, 1.5Ω. The generated voltage is 550 volts. The armature and series field resistances are respectively. Determine the voltage:a. at the terminals of the generatorb. at the feeder

Solution

VV

V

RIVV

VV

V

RRIEV

L

L

feederaTL

T

T

seaaT

524

)5.1(10539

539

)6.05.0(10550

)(

• Long Shunt Generator

)( seaaL

ag

shLa

L

LL

sh

Lsh

RRIVE

EIP

III

V

PI

R

VI

• ExampleA long shunt comound generator delivers a load

current of 50A at 500V and has armature, series field and shunt field resistances of 0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the armature current. Allow 1V per brush for contact drop.

Solution

AI

I

III

AI

AI

I

R

VI

a

a

shLa

L

sh

sh

sh

Lsh

52

250

50

2250

500

VE

E

dropbrushRRIVE

g

g

seaag

16.506

2)05.003.0)(52(500

)(

• Short Shunt Generator

seLLsh

seLaaL

ag

shLa

L

LL

sh

Lsh

RIVV

RIRIVE

EIP

III

V

PI

R

VI

• ExampleA short shunt compound generator delivers a load

current of 30A at 220V, and has armature, series-field and shunt-field resistances of 0.05Ω, 0.30Ω and 200Ω respectively. Calculate the induced emf and the armature current. Allow 1.0V per brush for contact drop.

Solution

AI

I

III

AI

I

R

VI

VV

V

RIVV

a

a

shLa

sh

sh

sh

shsh

sh

sh

seLLsh

145.31

145.130

145.1200

229

229

)3.0(30220

VE

E

dropbrushRIRIVE

g

g

seLaag

56.232

2)3.0)(30()05.0)(145.31(220

• Losses and Efficiency– Armature I2R loss

I2R loss in the armature windings

– Shunt field I2R loss I2R loss in the shunt field windings

shshsh

aaa

RIP

RIP

2

2

– Series field I2R loss I2R loss in the series field windings

– Friction and Windage loss The mechanical power required to drive the unexcited machine at normal speed

– Core loss or Iron loss Loss due to the magnetic field at no-load rated voltage corrected for effect of IR drop

a. Eddy current lossb. Hysteresis loss

sesese RIP 2

Power Stages

• Mechanical efficiency

motordrivingofoutput

IE ag

m

• Electrical efficiency

• Overall or Commercial Efficiency

ag

e IE

VI

generatedwattstotal

circuitloadinavailablewatts

suppliedpowermechanical

circuitloadinavailablewattsc

• ExampleThe field circuit of a 200kW, 230V shunt generator is

8A when running full load at rated terminal voltage. If the combined brush and armature resistance is 0.03ohm, solve for the electrical efficiency of the generator.

Solution

kWP

P

RIP

I

I

III

AI

I

V

PI

a

a

aaa

a

a

shLa

L

L

L

LL

1.23

)03.0(56.877

56.877

856.869

56.869230

200000

2

2

%91.88

%10084.11.23200

200

%100

84.1

)230(8

shaL

L

sh

sh

shshsh

PPP

P

kWP

P

VIP

• Voltage RegulationPercent voltage regulation is the percentage rise in the

terminal voltage of a generator when its load is removed.

voltageterminalloadfullV

voltageterminalloadnoV

V

VVVR

FL

NL

FL

FLNL

where

%100%

Parallel Operation of DC Generators

In order to increase the capacity of a system serving different loads, a number of generators are connected in parallel

Basic requirements1. The same external characteristics2. Terminal polarity must be the same3. Terminal voltage must be equal in magnitude

• ExampleTwo identical 600kW, 230V dc generators are

operating in parallel and take equal shares of an 800kW, 250V busload. The voltage regulation of each machine is 5%. If one of the generators is accidentally tripped off from the line, what is the voltage of the remaining busload?

Solution

AII

II

AI

I

V

PI

AII

II

V

PII

LL

LL

LOAD

LOAD

L

LOADLOAD

FLFL

FLFL

FL

FLFLFL

16002

3200

3200250

800000

69.2608230

600000

21

21

21

21

21

machine. trippeditsby offgiven load the todue is therefore

busload in the change The line. thefrom trippedhas 2gen Assume :NOTE

1

1

11

84.226

)230(05.0

)69.2608(

)(%

)(

)(%

VI

VI

VVR

IVI

I

VVR

I

V

FL

FL

FL

L

VV

V

VVV

VV

V

II

LNEW

LNEW

LOLDLNEW

BUS

95.242

05.7250

05.7

160084.2261

DC Motors

Introduction

• Electric motorA machine that converts electrical energy into

mechanical energy

Direct current motors are seldom used in ordinary industrial applications because all electric utilities supply AC. However, for special applications such as in steel mills, mines, and electric trains, it is sometimes advantageous to transform the AC into DC in order to use DC motors. The reason is that the torque-speed characteristics of DC motors can be varied over a wide range while retaining high efficiency.

Speed Characteristics of a DC Motor

)(

Z

a

)(

60

voltemfcounteroremfbackE

poleperflux

conductorsofnumber

polesofnumberP

pathscurrentarmatureofnumber

rpmspeedN

wherePZ

EaN

b

b

Motor Torque

By the term torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts.

T=F x r

• Torque in the Armature

)(

)(

)(

)(

283.6

weberpoleperflux

conductorsofnumberZ

polesofnumberP

pathsarmatureofnumbera

rpmrotationarmatreofspeedN

amperecurrentarmatureI

rnewtonmetedevelopedtorqueT

wherea

IPZT

a

a

• Shaft torqueThe torque which is available for doing useful

work is known as shaft torque Tsh. It is so called because it is available at the shaft.

N-mN

output.=Tsh

559

• Example

A DC motor takes an armature current of 110A at 480V. The armature circuit resistance is 0.2Ω. The machine has 6 poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05Wb. Calculate (i)the speed and (ii)the gross torque developed by the armature.

Solution

rpmN

N

PZ

EaN

VE

E

RIVE

b

b

b

aasb

636

)05.0)(864)(6(

)458)(6(60

)(60

458

)2.0(110480

mNT

T

a

IPZT

a

a

aa

3.7566

)110)(05.0)(864)(6(159.0

159.0

Speed of a Motor

• For series motor

• For shunt motor

2

1

1

2

1

2

Φ

Φ

E

E

N

N

b

b

1

2

1

2

b

b

E

E

N

N

Types of DC Motor

• Shunt MotorThe armature and the field coils are connected in

parallel

sh

ssh

sham

aasb

R

VI

III

RIVE

• ExampleA shunt motor is taking 72A at 120V while developing

an output of 10bhp. Armature resistance is 0.05ohm. Shunt field resistance is 60 ohms. Determine the counter emf.

Solution

AI

I

III

AI

I

R

VI

a

a

shma

sh

sh

sh

ssh

70

272

260

120

VE

E

RIVE

b

b

aasb

5.116

)05.0(70120

• Series Motor

)( seaasb

am

RRIVE

II

• Long Shunt Compound Motor

)( seaasb

sh

ssh

sham

RRIVE

R

VI

III

• ExampleA long shunt compound motor draws a line current of

42A from a 230V dc source. The armature resistance is 0.1 ohm while the series and shunt field resistances are 0.2 ohm and 50 ohm respectively. If the iron and friction losses amount to 500W, determine the overall efficiency of the machine.

Solution

WP

P

RIP

AI

I

III

AI

I

R

VI

a

a

aaa

a

a

shma

sh

sh

sh

ssh

876.139

1.0)4.37(

4.37

6.442

6.450

230

2

2

WP

P

PPPPP

WP

P

VIP

WP

P

RIP

losses

losses

frictionironshsealosses

sh

sh

shshsh

se

se

sease

628.1977

5001058752.279876.139

1058

230)6.4(

752.279

2.0)4.37(

&

2

2

%53.799660

628.19779660

9660

)42(230

input

lossesinput

input

output

input

input

msinput

P

PP

P

P

WP

P

IVP

• Short Shunt Compound Motor

)semaasb

sh

semssh

sham

RIRIVE

R

RIVI

III

ExampleA 220V short shunt compound motor has an armature

resistance of 0.4Ω, a shunt field resistance of 110Ω and a series field resistance of 0.6Ω. If this motor draws an armature current of 50A at rated load, determine horsepower developed in the armature.

Solution

VE

E

RIRIVE

AI

II

III

II

II

R

RIVI

b

b

aasemsb

m

mm

sham

msh

msh

sh

semssh

97.168

)4.050()6.0718.51(220

718.51

)005454.02(50

005454.02110

)6.0(220

hpP

P

IEP

d

d

abd

325.11746

)50)(97.168(746

Speed Regulation

Speed regulation is the percentage rise in speed when the mechanical load of the motor is removed.

voltageterminalloadfullN

voltageterminalloadnoN

N

NNNR

FL

NL

FL

FLNL

where

%100%

Power Developed in the Armature

)(

)(

)(

voltemfcounteroremfbackE

amperecurrentarmatureI

wattarmaturetheindevelopedpowerP

where

IEP

b

a

d

abd

Losses and Efficiency

• Armature loss

• Shunt field loss

• Series field loss

aaa R=IP 2

shshsh R=IP 2

sesese R=IP 2

• Efficiency

input

lossesinput

input

ouput

P

PP

P

P

Power Stages

ExampleA 100 volt shunt motor is developing 6 hp while

operating at an overall efficiency of 86%. The armature and shunt field resistances are 0.06 and 50Ω respectively. Determine stray power losses.

Solution

AI

I

V

PI

WP

P

PP

WP

hp

WhpP

m

m

s

inm

in

in

outin

out

out

04.52100

65.5204

65.520486.0

4476

4476

)1

746(6

WP

RIP

AI

I

III

AI

I

R

VI

a

aaa

a

a

shma

sh

sh

sh

shsh

24.150

)06.0(04.50

04.50

204.52

250

100

22

WP

P

PPPPP

WP

IVP

stray

stray

strayshaoutin

sh

shssh

41.378

20024.150447665.5204

200

)2)(100(

Speed Control of Shunt Motors

• Variation of Flux or Flux Control MethodBy decreasing the flux, the speed can be increased and

vice versa. The flux of a d.c motor can be changed by changing Ish with help of a shunt field rheostat.

• Armature or Rheostatic Control Method

This method is used when speeds below the no load speed are required. As the supply voltage is normally constant, the voltage across the armature is varied by inserting a variable rheostat or resistance in series with the armature circuit.

• Voltage Control Method– Multiple Voltage Control

In this method, the shunt field of the motor is connected permanently to a fixed exciting voltage but the armature is supplied with different voltages by connecting it across one of the several different voltages by means of suitable switchgear. The armature speed will be approximately proportional to these different voltages. The intermediate speeds can be obtained by adjusting the shunt field regulator.

– Ward-Leonard System This system is used where an unusually wide and

very sensitive speed control is required as for colliery winders, electric excavators, elevators and the main drives in steel mills and blooming and paper mills.

Speed control of Series motors

• Flux control methodVariation in the flux of series motor can be brought

about in any one of the following ways.– Field Divertors

The series winding are shunted by a variable resistance known as field divertors.

– Armature Divertor A divertor across the armature can be used for

giving speeds lower than the normal speed.

– Trapped Field Control Field This method is often used in electric traction. The

number of series field turns in the circuit can be changed at will. With full field, the motor runs at its minimum speed which can be raised in steps by cutting out some of the series turns.

– Paralleling Field Coils In this method several speeds can be obtained by

regrouping the field coils

• Variable Resistance in Series with MotorBy increasing the resistance in series with the

armature the voltage applied across the armature terminals can be decreased.