CH 1 Transformer

8/12/2019 CH 1 Transformer

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ELECTRICAL MACHINE

BMT 3013CHAPTER 1: SINGLE AND THREE PHASE

TRANSFORMER


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LEARNINGOUTCOMES

At the end of the chapter, students should be able to:

Understand the fundamental laws in the dynamic

magnetic systemsand their relation to the electrical

machines.

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INTRODUCTIONTOELECTRICALMACHINE

An electrical machine is a device which converts

electrical power (voltages and current) into mechanical

power (torque and rotational speed) and/or vice versa.

Electromechanical energy conversion the process

which an electrical machine deals with the energytransfer either from mechanical to electrical form or from

electrical to mechanical form.

Electrical motor an electrical machine which converts

an electrical energy into the mechanical energy.

Electric generator an electrical machine which

converts mechanical energy into an electrical energy.

3

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INTRODUCTIONTOELECTRICALMACHINE

(CONT..)

Many electric machines are capable of performing

both as motors and generators;

The capability of a machine performing as one or

the other is often through the action of a magnetic

field, to perform such conversions.

To understand how an electrical machines works,

the key is to understand how the electromagnet

works.

The principles of magnetismplay an important role

in the operation of an electrical machines.

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REVISIONOFMAGNETISM

Magnetismproperty by virtue of which a piece of

solid body (natural magnet) attracts iron piece and

piece of some other metals.

Polestwo ends of a magnet

The end of magnets which adjusting in the direction

of North is called N pole while other is S pole

Behavior of two magnets which are brought near to

each other is governed by Laws of Magnetism

5

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Thus,

=12

2

Where

1and 2are the pole strength

d is the distance between the two poles

K is constant which depends on the nature of thesurrounding

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REVISIONOFMAGNETISM(CONT..)


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MAGNETICFIELD

Magnetic Fieldthe region around magnet within

which influence of the magnet can be experienced.

The presence of magnetic field is represented by

imaginary lines around magnet which called

magnetic lines of force.

The total number of lines of force existing in a

particular magnetic fieldmagnetic flux

Line of flux have a fixed direction

1.ExternalN-pole to S-pole

2.InternalS-pole to N-pole

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MAGNETICFIELD(CONT..)

Unlike electric fields (which start on +q and end onq),magnetic field encircle their current source.

The field weakens as you move away from the wire Amperescircuital law - the integration path length

is longeridH

.

A circular magnetic field

develops around the wirefollows r ight-hand rules

field is perpendicular to

the wire and that the

field's direction dependson which direction the

current is flowing in the

wire


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MAGNETICFIELD(CONT..)

o Magnetic field due to circularconductor

Solenoid an arrangement inwhich a long conductor is woundwith number of turns on a core,close together to for a coil.

The right hand thumb rule : holdthe solenoid in the right handsuch that curled fingers point inthe direction of the currentthrough the curled conductor,then the outstretched thumbalong the axis of the solenoidpoints to the North pole of

solenoid or points in the directionof flux lines inside the core.

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REVIEWOFELECTROMAGNETISM

The basic idea behind an electromagnet is

extremely simple: a magnetic field around the

conductor can be produced when current flows

through a conductor.

In other word, the magnetic field only exists whenelectric current is flowing

By using this simple principle, you can create all

sorts of things, including motors, solenoids,

read/write heads for hard disks and tape drives,speakers, and so on

ELECTRIC

ALMACHINE(DEI202

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http://electronics.howstuffworks.com/motor.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/speaker.htmhttp://electronics.howstuffworks.com/speaker.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/cassette.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/hard-disk.htmhttp://electronics.howstuffworks.com/motor.htm


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ELECTRICAL MACHINE (DEI 2023) 13

EXAMPLEOFELECTROMAGNETIC

An electromagnet can be made by winding theconductor into a coil and applying a DC voltage.

The lines of flux, formed by current flow through the

conductor, combine to produce a larger and stronger

magnetic field. The center of the coil is known as the core. In this

simple electromagnet the core is air.


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ADDINGANIRONCORE

Iron is a better conductor of fluxthan air. The air coreof an electromagnet can be replaced by a piece of soft

iron.

When a piece of iron is placed in the center of the coil

more lines of flux can flow and the magnetic field isstrengthened.


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STRENGTHOFMAGNETICFIELD(CONT)

Because the magnetic field around a wire iscircular and perpendicular to the wire, an easyway to amplify the wire's magnetic field is to coilthe wire

The strength of the magnetic field in the DC

electromagnet can be increased by increasing the

number of turns in the coil. The greater the

number of turns the stronger the magnetic

field will be.


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16

FARADAYSLAWANDLENZSLAW

FaradaysLaw :If a magnetic flux, , in a coil is changing intime (n turns),hence a voltage, V

ab

is induced

Lenzs Law : if the loop is closed, a connected to b, thecurrent would flow in the direction to produce the flux insidethe coil opposing the original flux change. (in other words,LenzsLaw will determine the polarity of the induced voltage)

ab

tNV

V = induced voltage

N = no of turns in coil

= change of flux in coil

t = time interval

Vt

If no turns :

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17

FARADAYSLAW

The effect of magnetic field:

Induced Voltage from a Time Changing

Magnetic Field

Production of Induced Force on a Wire Induced Voltage on a Conductor Moving

in a Magnetic Field

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VOLTAGEINDUCEDFROMATIME

CHANGINGMAGNETICFIELD


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VOLTAGEINDUCEDINACONDUCTOR

MOVINGINAMAGNETICFIELD

Faradays Law for moving conductors : For coils in which wire(conductor) is moving thru the magnetic flux, an alternateapproach is to separate the voltage induced by time-varying fluxfrom the voltage induced in a moving conductor.

This situation is indicates the presence of an electromagnetic fieldin a wire (conductor). This voltage described by Faradays Law is

called as the flux cutting or Electromotive force, or emf.

The value of the induced voltage is given by

E =Blvwhere

E= induced voltage (V)

B= flux density (T)

l= active length of the conductor in the magnetic field (m)

v= relative speedof the conductor (m/s)

The polarity of induced

voltage is given by theright-hand rule.

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20

INDUCEDFORCE

The electrical circuit consists of battery, resistor, two stationary rails, and movablebar that can roll or slide along the rails with electrical contact.

When switch is closed:

Current will not start immediately as inductance of the circuit.

(However time constant L/R is very small). Hence, current quickly

reach V/R.

Force is exerted on the bardue to interaction between current and

magnetic fluxto the right and made the bar move with certain velocity.

The mechanical power out of the bar.

F= ilB

Force induced on

the conductor:

Unit: (N)

The direction of

force is given by

the right-hand

rule.

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21

INDUCEDFORCE(CONT)

The motion of the bar produces an

electromagnetic force. The polarity of

the emfis+ve where the current enters

the moving bars. The moving bargenerates a back emfthat opposes the

current.

The instantaneous electrical power into the bar =

mechanical output power

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22

PRODUCTIONOFAMAGNETICFIELD

The production of a magnetic field by a current is

determine byAmpereslaw:

netIdlH

H = magnetic field intensity

dl = differential element of

length along the path of

integration

cl

NiH lc = mean path length

Magnetic field intensity:

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23


HB u = magnetic permeability ofmaterial

r0 u0= permeability of free spaceur= relative permeability of

material

m/H104 70

The strength of the magnetic field flux produced in the

core also depends on the material of the core.

Magnetic flux density:

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24


cl

NiHB

BA

cl

NiA

Total flux:

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25

MAGNETICCIRCUIT

iRV

Ni

Electric circuit equation:

Magnetic circuit equation:

Analogy: Electric circuit & Magnetic circuit

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26

MAGNETICSATURATION& HYSTERESIS

INACMAGNETICFIELD

unmagnetized Material

Iron becomes

magnetically

saturated

Magnetism increase as

magnetic field magnetized

unmagnetized iron

a

b

c

d

Applied field is reduced; the magnetism

reduced thru diff. curve since iron tends to

retains magnetized state - hence produced

permanent magnet, Residual Flux, res

AC increased in negative direction,

magnetic field reversed , the iron

reversely magnetized until saturated

againIf continue apply ac current, curve

continue to follow S-shaped curve(hysteresis curve)

The area enclosed by hysteresis curve is energy loss per unit volume per cycleheats the iron and is one

reason why electric machines become hot

Therefore, it is required to select magnetic materials that have a narrow hysteresis loop

Hm

Magnetic field density

Bm

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27

HYSTERESISLOSS

During a cycle of variation of i (hence H),

there is a net energy flow from the source to

the coil-core assembly and return to the

source.Energy flowing is greater than energy

returned.

This energy loss goes to heat the core.

The loss of power in the core due to thehysteresis effect is called hysteresis loss.

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EDDYCURRENTLOSS

Voltage will be induced in the path ofmagnetic core because of time variation of flux

enclosed by the path.

A current, known as an eddy current will flowaround the path.

Because core has resistance, a power loss

will be cause by the eddy current and appearas heat in the core.

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EDDYCURRENTLOSS

Eddy current can be reduced in 2 ways:

1. Adding a few percent of silicon to iron to

increase the resistivity.

2. Laminate core with thin laminations and

insulated from each other.

Hysteresis loss + eddy current loss = Core loss

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LEARNINGOUTCOMES

At the end of the lecture, student should to:

Understand the principle and the nature of

static machines of transformer.

Perform an analysis on transformers which

their principles are basic to the

understanding of electrical machines.

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INTRODUCTION

A transformer is a static machines. The word transformercomes form the word transform.

Transformer is not an energy conversion device, but is adevice that changes AC electrical power at one voltagelevel into AC electrical power at another voltage levelthrough the action of magnetic field, without a change infrequency.

It can be either to step-upor step down.

GenerationStation

TX1 TX1

Distributions

TX1

TX1

Transmission

System

33/13.5kV 13.5/6.6kV

6.6kV/415V

Consumer

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TRANSFORMERCONSTRUCTION

Two types of iron-core construction:a) Core - type construction

b) Shell - type construction

Core - type construction

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TRANSFORMERCONSTRUCTION

Shell - type construction



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IDEALTRANSFORMER

An ideal transformer is a transformer which has no loses,

i.e. its winding has no ohmic resistance, no magnetic

leakage, and therefore no I2 R and core loses.

However, it is impossible to realize such a transformer in

practice.

Yet, the approximate characteristic of ideal transformer will

be used in characterized the practical transformer.

V1 V2

N1 : N2

E1 E2

I1 I2

V1Primary Voltage

V2Secondary Voltage

E1Primary induced Voltage

E2secondary induced Voltage

N1:N2Transformer ratio



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TRANSFORMEREQUATION

Faradays Law states that,

If the flux passes through a coil of wire, a voltage will be

inducedin the turns of wire. This voltage is directly

proportional to the rate of change in the flux with respect of

time.

If we have Nturns of wire,

dt

tdEmfV indind

)(

dt

tdNEmfV indind

)(

Lenzs Law

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TRANSFORMEREQUATION

For an ac sources,

Let V(t) = Vmsint

i(t) = imsint

Since the flux is a sinusoidal function;

Then:

Therefore:

Thus:

tt m sin)(

tN

dt

tdNEmfV

m

mindind

cos

sin

mmindind fNNEmfV 2(max)

mmm

rmsind fNfNN

Emf

44.42

2

2)(

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TRANSFORMEREQUATION

For an ideal transformer

In the equilibrium condition, both the input power will be equaled to the outputpower, and this condition is said to ideal condition of a transformer.

From the ideal transformer circuit, note that,

Hence, substitute in (i)

m

m

fNE

fNE

22

11

44.4

44.4

1

2

2

1

2211 coscos

I

I

V

V

IVIV

poweroutputpowerInput

(i)

2211 VEandVE 37


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TRANSFORMEREQUATION

aI

I

N

N

E

ETherefore

1

2

2

1

2

1,

Where, ais the Voltage Transformation Ratio; which will

determine whether the transformer is going to be step-upor step-down

E1> E2For a >1

For a


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ELECTRICALMACHINE(DEI2023

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TRANSFORMERRATING

Transformer rating is normally writtenin terms ofApparent

Power.

Apparent power is actually the product of its rated current

and rated voltage.

2211 IVIVVA

Where,

I1and I2= rated current on primary and secondary winding.

V1and V2= rated voltage on primary and secondary winding.

Rated currents are actually the full load currentsin

transformer39


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)

EXAMPLE

1. 1.5kVA single phase transformer has rated voltage of144/240 V. Finds its full load current.

Solution

AI

AI

FL

FL

250.6240

1500

417.10144

1500

2

1

40


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EXAMPLE

2. A single phase transformer has 400 primary and

1000 secondary turns. The net cross-sectional

area of the core is 60m2

. If the primary winding isconnected to a 50Hz supply at 520V, calculate:

a) The induced voltage in the secondary winding

b) The peak value of flux density in the core

Solution

N1=400 V1=520V A=60m2

N2=1000 V2=?

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EXAMPLE

3. A 25kVA transformer has 500 turns on the primary and 50

turns on the secondary winding. The primary is connected

to 3000V, 50Hz supply. Find:

a) Full load primary current

b) The induced voltage in the secondary winding

c) The maximum flux in the core

Solution

VA = 25kVA

N1=500 V1=3000VN2=50 V2=?

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)

EXAMPLE3 (CONT)

a) Know that,

b) Induced voltage,

c) Max flux

VIIEE

AI

I

I

N

N

a

3003.8333.83000

3.8350

33.8500

2

112

2

1

2

2

1

AV

VAI

IVVA

FL 33.83000

1025 3

1

1

mWb

fNE

27

)50)(50(44.4300

44.4

44


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PRACTICALTRANSFORMER

(EQUIVALENTCIRCUIT)

V1 = primary supply voltage

V2 = 2ndterminal (load) voltage

E1 = primary winding voltageE2 = 2

ndwinding voltage

I1 = primary supply current

I2 = 2ndwinding current

I1 = primary winding current

Io = no load current

Ic = core current

Im = magnetism current

R1= primary winding resistanceR2= 2

ndwinding resistance

X1= primary winding leakage reactance

X2= 2ndwinding leakage reactance

Rc= core resistance

Xm= magnetism reactance

V1

I1 R1X1

RC

Ic

Xm

Im

Io

E1 E2

V2

I1

N1: N2R2

X2

Load

I2

45

SINGLE PHASE TRANSFORMER (REFERRED


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SINGLEPHASETRANSFORMER(REFERRED

TOPRIMARY)

1) Transferred secondary parameter to primary winding

2

2

22

2

2

12 '' RaRORR

NNR

2

2

22

2

2

12 '' XaXORX

N

NX

a

II

aVVORVN

NVE

22

222

2

1'

21

'

'

V1

I1 R1 X1

RC

Ic

Xm

Im

Io

E1 E2 V2

I2 N1: N2

R2

X2

Load

I2

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TOPRIMARY)

2) Transferred core resistance and magnetism reatance

2

2

22

2

2

12 '' RaRORR

N

NR

V1

I1 R1X1

RC

Ic

Xm

Im

Io

E1 E2 V2

I2 N1: N2R2

X2

Load

I2

2

2

22

2

2

12 '' XaXORX

N

NX

a

II

aVVORVN

NVE

22

222

2

1'

21

'

'

47



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TOPRIMARY)

3. Equivalent resistance and reactance

2

2

22

2

2

12 '' RaRORR

N

NR

V1

I1

R01 X01

aV2

2

2

22

2

2

12 '' XaXORX

N

NX

'

'

2101

2101

XXX

RRR

222

2

1'

2 ' aVVORVN

NV

In some application, the excitation

branch has a small current compared

to load current, thus it may be

neglected without causing seriouserror.

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TOSECONDARY)

2

111

2

1

21 ''

a

RRORR

N

NR

a

VVORV

N

NV 111

1

21 ''

I1 R1X1

RC

Ic

Xm

Im

IoI2 R2

X2

V2

2

111

2

1

21 ''

a

XXORX

N

NX

aV1

49

1) Transferred primary parameter to secondary winding



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TOSECONDARY)

2

111

2

1

21 ''

a

XXORX

N

NX

2102

2102

'

'

XXX

RRR

I1R02 X02

V2

a

V1

2

111

2

1

21 ''

a

RRORR

N

NR

a

V

VORVN

N

V

1

111

2

1 ''

11 ' aII

Neglect the excitation branch

50

2. Equivalent resistance and reactance


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EXAMPLE

4. For the parameters obtained from the test of 20kVA

2600/245 V single phase transformer, refer all the

parameters to the high voltage side if all the

parameters are obtained at lower voltage side.

Rc = 3.3, Xm=j1.5, R2= 7.5, X2= j12.4

Solution

Given

Rc

= 3.3, Xm

=j1.5,

R2= 7.5, X2= j12.4

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)

EXAMPLE4 (CONT)

i) Refer to H.V side (primary)

R2=(10.61)2(7.5) = 844.65,

X2=j(10.61)2(12.4) = 1.396k

Rc=(10.61)2(3.3) = 371.6,

Xm=j(10.61)2(1.5) = j168.9

61.10245

2600

2

1

2

1 V

V

E

Ea

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POWERFACTOR

Power factor = angle between Current and voltage,cos

V

I

= -ve

V

I

= +ve

VI

= 1

Lagging Leading unity

Power factor always lagging for real

transformer.

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EXAMPLE

5. A 10 kVA single phase transformer 2000/440V has

primary resistance and reactance of 5.5and 12

respectively, while the resistance and reactance of

secondary winding is 0.2and 0.45 respectively.

Calculate:i. The parameter referred to high voltage side and draw the

equivalent circuit

ii. The approximate value of secondary voltage at full load of 0.8

lagging power factor, when primary supply is 2000V.

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EXAMPLE5 (CONT)

SolutionR1=5.5 , X1=j12

R2=0.2 , X2=j0.45

i) Refer to H.V side (primary)

R2=(4.55)2(0.2) = 4.141,

X2=j(4.55)2

0.45 = j9.316

Therefore,

R01=R1+R2=5.5 + 4.141 = 9.641

X01=X1+X2=j12 + j9. 32 = j21.316

55.4440

2000

2

1

2

1 V

V

E

Ea

V1 aV2

R01 X01

21.329.64

I1

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EXAMPLE5 (CONT)

Solutionii) Secondary voltage

p.f = 0.8

Cos = 0.8

=36.87o

Full load,

From eqn.

AV

VAIFL 5

2000

1010 3

1

o

oo

oo

V

Vj

aVIjXRV

7.1214.417

)55.4()87.365)(316.21641.9(02000

))((0

2

2

2101011

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TRANSFORMERLOSSES

Generally, there are two types of losses;i. Iron losses:- occur in core parameters

ii. Copper losses:- occur in winding resistance

i. Iron Losses

ii. Copper Losses

circuitopenccciron PRIPP 2)(

02

2

201

2

1

2

2

21

2

1

)()(,

)()(

RIRIPreferredifor

PRIRIPP

cu

circuitshortcucopper

Pocand Pscwill be discusses later in transformer test

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TRANSFORMEREFFICIENCY

To check the performance of the device, by comparing the

output with respect to the input.

The higher the efficiency, the better the system.

%100cos

cos

%100

%100,

22

22

cuc

lossesout

out

PPIVIV

PP

P

PowerInput

PowerOutputEff iciency

%100cos

cos

%100cos

cos

2)(

)(

cuc

nload

cuc

loadfull

PnPnVA

nVA

PPVA

VA

Where, if load, hence n = ,

load, n= ,

90% of full load, n =0.9Where Pcu = PscPc = Poc

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VOLTAGEREGULATION

The purpose of voltage regulation is basically to

determine the percentage of voltage drop

between no load and full load.

Voltage Regulation can be determine based on 3

methods:a) Basic Definition

b) Shortcircuit Test

c) Equivalent Circuit

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VOLTAGEREGULATION(BASICDEFINITION)

In this method, all parameter are being referred to primary

or secondary side.

Can be represented in either

Downvoltage Regulation

UpVoltage Regulation

%100.

NL

FLNL

V

VVRV

%100.

FL

FLNL

V

VVRV

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VOLTAGEREGULATION(SHORTCIRCUIT


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ELECTRICA

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(

TEST)

In this method, direct formula can be used.

%100cos

.1

. V

VRV

fpscsc If referred to primary side

%100cos

.2

. V

VRV

fpscsc If referred to secondary side

Note that:

is for Lagging power factor

+ is for Leading power factor

Iscmust equal to IFL

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VOLTAGEREGULATION(EQUIVALENT


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ELECTRICA

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(

CIRCUIT)

In this method, the parameters must be referred to primary orsecondary

%100sincos

.1

.01.011

V

XRIRV

fpfp If referred toprimary side

If referred to

secondary side

Note that:

+ is for Lagging power factor

is for Leading power factor

j terms ~0

%100sincos

.2

.02.022

V

XRIRV

fpfp

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EXAMPLE

6. In example 5, determine the Voltage regulation by usingdownvoltage regulation and equivalent circuit.

Solution

Downvoltage Regulation

Know that, V2FL=422.6V

V2NL=440V

Therefore,

%95.3

%100440

6.422440

%100.

NL

FLNL

V

VVRV

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ELECTRICA

LMACHINE(DEI2023)

EXAMPLE6 (CONT)

Equivalent Circuit

I1=5A R01=9.641 X01= 21.316 V1=2000V, 0.8 lagging p.f

%13.5

%1002000

)6.0(32.21)8.0(64.95

%100

sincos

. 1

.01.011

V

XRI

RV

fpfp

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ELECTRICA

LMACHINE(DEI2023)

EXAMPLE

7. A short circuit test was performed at the secondary side of 10kVA,

240/100V transformer. Determine the voltage regulation at 0.8lagging power factor if

Vsc=18V

Isc=100

Psc=240W

Solution

Check:

Hence, we can use short-circuit method

,

100100

10000

2

2

scFL

FL

II

AV

VAI

%100cos

.

2

.

V

VRV

fpscsc 65


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ELECTRICA

LMACHINE(DEI2023)

EXAMPLE7 (CONT)

o

scsc

scsc

scscscsc

o

fp

fpscsc

IV

P

IVPthatKnow

Hence

fpGiven

VVRV

34.82)100)(18(

240cos

cos

cos,

87.368.0cos,

8.0.

%100cos.

1

1

1

.

2

.

%62.12

%100100

87.3634.82cos18.

oo

RV 66


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EXAMPLE8 (CONT)

.732.7861.346.63643.8

643.833.8

72

46.63)33.8)(72(

268cos

cos

cos,

87.368.0cos,

8.0.

%100cos

.

0101

1

1

1

.

2

.

sideprimarytoconnectedbecausejXRjZ

I

VZ

IV

P

IVPthatKnow

Hence

fpGiven

V

VRV

o

sc

sc

scsc

o

scsc

scsc

scscscsc

o

fp

fpscsc

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EXAMPLE8 (CONT)

%68.2%100

2400

)6.0(72.7)8.0(86.32400

20000

%100

sincos

.,.2

%68.2%100

2400

87.364.63cos72.

%100cos

.,.1

1

.01.011

1

.

V

XRI

RVcircuitEquivalent

RV

V

VRVmethodCircuitShort

fpfp

oo

fpscsc

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EXAMPLE8 (CONT)

%68.2

%100240

58.233240

%100.

79.058.233

240

24004.6364.887.362400

2000002400

,.3

2

2

20111

NL

FLNL

o

ooo

VVVRV

VV

V

aVZIV

Def inationBasic

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EXAMPLE8 (CONT)

%12.97%100)268()5.0(170)8.0)(20000)(5.0(

)8.0)(20000)(5.0(

%34.97%100)268()1(170)8.0)(20000)(1(

)8.0)(20000)(1(

2)(

2)(

loadhalf

loadfull

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MEASUREMENTONTRANSFORMER

There are two test conducted on transformer.

i. Open Circuit Test

ii. Short Circuit test

The test is conducted to determine the parameter of

the transformer.

Open circuit test is conducted to determine

magnetism parameter, Rcand Xm.

Short circuit test is conducted to determine the

copper parameter depending where the test is

performed. If performed at primary, hence the

parameters are R01and X01and vice-versa.

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OPEN-CIRCUITTEST

Measurement are at high voltage side

From a given test parameters,

Rc XmVoc

Ic Im

Voc

Ioccosoc

Ioc

Voc

Ic

Im

Ioc

sinoc

oc

Note:

If the question asked parameters referred to

Low voltage side, the parameters (Rcand Xm) obtained

need to be referred to low voltage sidem

ocm

c

occ

mc

ococm

ococc

ococ

ococ

ococococ

I

VX

I

VR

XandRThen

II

II

HenceIV

P

IVP

,

,,

sin

cos

,

cos

cos

1

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SHORT-CIRCUITTEST

Measurement are at low voltage side

If the given test parameters are taken on primary

side, R01 and X01 will be obtained. Or else, vice-

versa.

X01R01

For a case referred to

Primary side

010101

01

1

,

cos

cos

jXRZ

I

VZ

Hence

IV

P

IVP

sc

sc

sc

scsc

scsc

scscscsc

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EXAMPLE9 (CONT)

A

II

A

II

IVP

o

ococm

o

ococc

o

oc

ococococ

2.49

6.69sin5.52

sin

26.18

6.69cos5.52

cos

6.69

)208)(5.52(

3800cos

cos

1

Ioccosoc

Ioc

Voc

Ic

Im

Iocsinoc

oc

From Open Circuit Test,

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ELECTRICA

LMACHINE(DEI2023)

EXAMPLE9 (CONT)

AV

VAIFL 4.217

2300

10500 3

11

o

sc

scscscsc IVP

53.72)4.217)(95(

6200cos

cos

1

From Short Circuit Test,First, check the Isc

Since IFL1=Isc , all reading are actually taken on the primary side

42.013.0

53.7244.053.724.217

95

01

j

I

VZ

oo

sc

sc

sc

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EXAMPLE9 (CONT)

Equivalent circuit referred to high voltage side,

V2=aV2V1Rc

1392

Xm

517.21

R010.13

X010.42

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ELECTRICA

LMACHINE(DEI2023)

EXAMPLE9 (CONT)

Efficiency,

%59.97

%1003800)5.0)(6200()866.0)(10500)(5.0(

)866.0)(10500)(5.0(

%100cos

cos

%74.97

%10038006200)866.0)(10500(

)866.0)(10500(

%100cos

cos

23

3

22

1

3

3

ocscL

ocsc

FL

PPnnVA

nVA

PPVA

VA

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CH 1 Transformer

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Transcript of CH 1 Transformer