JJ102 Electrical Technology CHAPTER_5 Tansfomers [Compatibility Mode]

Prepared By:

Sheilani bt Shaari

Department of Electrical Engineering, PKB

Course Learning Outcome (CLO)

Department of Electrical Engineering PKB

Introduction

Transformer

Construction


Construction

Operating Principle

Transformer Ratio

Types of Transformer

GENERAL OBJECTIVE

Understand the principles of transformer.

SPECIFIC OBJECTIVES

At the end of the unit you will be able to :



1. Explain the operating principles of

transformer

2. Describe the relationship between primary

winding and secondary winding.

3. Name types of transformer.

4. Explain step up and step down transformer

A transformer is a device which uses the

phenomenon of mutual induction to

change the values of alternating voltages

and currents.


One of the main advantages of alternating

current transmission and distribution


an alternating voltage an alternating voltage

can be increased or

decreased by

transformers.

Basically the transformer :

LOSSES are generally LOW

EFFICIENCY is HIGH


EFFICIENCY is HIGH

Have a LONG LIFE and VERY STABLE

Transformers have various range in size from:

the miniature units used in electronic applications


the large power transformers used in power stations.

applications

A transformer consists of 2 windings

connected by a magnetic core.

1 winding is connected to a power supply and

the other to a load.


the other to a load.

N1 N2AC supplyLoad

Flux

Secondary windingPrimary winding Core

Transformer ratio is the comparison between

primary and secondary elements

Flux produced by primary winding induces secondary winding via

corePrimary Winding


Primary Turns, Np

Secondary Turns,

NsPrimary Voltage, Vp

Secondary Voltage,Vs

coreSecondary winding

Primary Winding

Conductors CUT the flux flux CUT the Conductors

Current Flow through Primary

Winding


Secondary winding

Open Circuit

Produce Magnetic Flux

Apply AC Voltage (V1)at Primary winding

E1

Induce emf, E1

E2

Produce emf, E2

Current Flow through Secondary

Winding

The induced e.m.f., E in a coil of N turns is

given by

E = -N volts

dtd


where

is the rate of change of flux.

dt

dtd

In the ideal transformer, losses are neglected

and a transformer is considered to be 100%

efficient.

the rate of change of flux is the same for


the rate of change of flux is the same for

both primary and secondary

=

The induced e.m.f per turn is constant

1

1

2

2

Assuming no losses,

E1 = V1 and E2 = V2

V V


=

=

1

1V 2

2V

2

1

VV

2

1N

The input powers = output power

V1I1 = V2I2


Or the primary and secondary volt-amperes are equal.

Thus

=

2

1

VV

1

2

II

Combining equations gives

1N1V 2I


= =

2

1N2

1

VV

1

2

II

Transformer ratio is the comparison

between primary and secondary

elements.


elements.

Flux which is produced by primary

winding completely inducing to

secondary winding.

Therefore, e.m.f induced in all the

winding are similar for primary and

secondary winding.


secondary winding.

==

/ windinge.m.f x Np / windinge.m.f x Ns

ndingprimary wiin induced e.m.f of Totalwindingsecondary in induced e.m.f of Total

sNNp

Constant K is known as transformer ratio:

K = s2s

VV

EE

NN

==


Where

E2 = secondary e.m.f induced

E1 = primary e.m.f induced

Vs = secondary terminal voltage

Vp = primary terminal voltage

p1p VEN

If Ns < Np then K < 1

this is step-down transformer

If Ns > Np then K > 1


Np(3000

)

Ns

(2000)

If Ns > Np then K > 1

this is step-up transformer

If Ns = Np then K = 1

this is coupling transformer

Np(100)

Ns

(2000)

Ns

(100)Np

(100)

The number of windings for the three transformers are

Np = 100, Ns = 2000

Np = 3000, Ns = 2000

Np = 100, Ns =100


Np = 100, Ns =100

Calculate the value of K for each transformer then determine the type of transformer and draw the symbol of transformer to differentiate the number of windings

A transformer is to be used to provide

a 60 V output from a 240 V A.C supply.

If the secondary is wound with 500

turns. Calculate

Vs or V2 Vp or V1


turns. Calculate

(a) the turns of ratio required and

(b) the number of primary turns

N2

A 2000/200V, 20kVA transformer has 66 turns

in the secondary. Calculate

(i) primary turns

(ii) primary and secondary full-load currents

Vp or V1 Vs or V2

N2N1

I1 & I2

P


Neglect the losses.

There 2 types of single-phase double-wound

transformer construction

core type and shell type


the winding surround considerable part of the core

the core surround is considerable portion of the winding.

Single phase core type transformer

consists of two windings and one

core

Core type provides better


Core type provides better

insulation between the primary

and secondary because of its

inherent two winding.

The main application for a core

type transformer is to increase

voltage from a standard 240v

supply up to 415v, instead of using

three phase supply.

shell type consists of one

winding and two cores.

Shell types provide a higher

degree of mechanical


degree of mechanical

protection to the winding

because it is surrounded by

core.

Shell type transformer is

used in transmission of

multiple high and low

voltages.

The advantages of each type are:

Core type is the most commonly

used method of construction, the


used method of construction, the

smaller core means less weight and

expense.

Shell type is used for larger

transformer because they can be

made with a reduce height.

(Q1 Q7)


(Q1 Q7)

GENERAL OBJECTIVE

Understand the principles of auto transformer.

SPECIFIC OBJECTIVES




1. Explain the operating principles of auto

transformer

2. Describe the differences between auto

transformer and application

3. Explain ideal transformer and its

relationship with power losses

Introduction

Auto Transformer

Losses


Losses

Efficiency

+ m

Magnetic Flux, (Wb)flux has to change from +m

to -m in the half cycle


Time,T (s)

- m

f

Average rate of change of flux

= 2 m =4 f m webers /second

f21


Average e.m.f induced per turn is

E1/N1= 4 f m volts

A sinusoidal wave the r.m.s or

effective value is 1.11 times the

average value


RMS value of e.m.f induced per turn

E1/N1= 1.11 4 f m

Hence r.m.s value of e.m.f induced for

primary:

E1 = 4.44 N1 f m volts



secondary:


A 250 kVA, 1100 V / 400 V, 50 Hz

single-phase transformer has 80 turns

on a secondary. Calculate :

the primary and secondary currents.

P Vp or V1

N2

Vs or V2

I & I


the primary and secondary currents.

the number of primary turns.

the maximum values of flux.

I1 & I2

N1

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

PNp

Vp or V1

Ns


i. primary and secondary currents on

full-load

ii. secondary e.m.f. and

iii. the maximum core flux

I1 & I2

V2

m

An autotransformer has only a single

winding which is tapped at some

point along the winding.

The primary and secondary windings


The primary and secondary windings

are not electrically separate, hence

if an open-circuit occurs in the

secondary winding the full primary

voltage appears across the

secondary

Primary no. of turns

Primary Current

Secondary Current

Ai1

Secondary

Laminated Ferromagnetic Core


of turns

Secondary no. of turns

B

C Load

N1

N2

i2

V1

V2

Primary Voltage

Secondary Voltage

Connected to the end of the end of

the winding


IPIS

AC voltage is applied across a portion of the winding.

VP NP

IS - IP

NS VS

IS

VP NP

IS - IP

IP

NS VS230 V 230 V

115 V

115 V

LOWER voltage is produce across another portion of the same winding for STEP DOWN

HIGHER voltage is produce across another portion of the same winding for STEP UP

Auto Transformer ratio (n) is the ratio of the smaller voltage to the larger voltage

(Neglecting the losses, the leakage reactance and the magnetizing current)

n = = = V I N


n = = =

The nearer the ratio of transformation to unity, the greater the economy of conductor material.

1

2

VV

2

1

II

1

2

NN

For the same current density in the

windings and the same peak values

of the flux and flux density

the I2R loss in the auto transformer is


the I2R loss in the auto transformer is

lower and

the efficiency higher than in the two

winding transformer.

Generally, an auto transformer

should not be used for

interconnecting high voltage and low

voltage systems.


voltage systems.

The most important thing is the

common connection to the earth ,

otherwise there is a risk of serious

shock.

Mainly used for :

interconnecting systems that are

operating at roughly the same


operating at roughly the same

voltage and

starting cage-type induction motors.

saving in a cost since less copper is

needed.

less volume, hence less weight.


less volume, hence less weight.

higher efficiency, resulting from lower I2R

losses

continuously variable output voltage is

achievable if a sliding contact is used.

smaller percentage voltage regulation

There is a direct metallic connection

between the input and the output, where

the coupling in a double wound

transformer is magnetic only when giving


transformer is magnetic only when giving

electrical isolation of the two winding.

In the event of an open circuit fault in the

common part of the winding, the input

voltage of a step down autotransformer

would appear on the output terminals.

An ideal transformer is one which

has no losses,

its winding have no ohmic resistance,


its winding have no ohmic resistance,

no magnetic leakage and

no I2R and core losses.

An ideal transformer consists of two

purely inductive coils wound on the

loss free core.

2 cases consider for transformer

with losses but no magnetic leakage.

No Load


is assumed as an ideal No Load

Loaded/ Full load

is assumed as an ideal transformer which there is no losses.

The losses only can occur in a loaded transformer. There are two types of losses :iron loss in the core and copper loss in the winding.

Copper losses

can result from power loss in the form of heating of the conductors.


If R1 and R2 are the primary and secondary

winding resistances than

total copper loss =I12R1 + I2

2R2

Core losses

(i) Hysteresis loss

the heating of the core so the internal

molecular structure reversals which occur as


molecular structure reversals which occur as

the magnetic flux alternates.

Core losses

(i) Hysteresis loss

The loss is proportional to the area of

the hysterisis loop and thus low loss nickel


the hysterisis loop and thus low loss nickel

iron alloys are used for the core since

their hysteresis loop have small areas.

Core losses

(ii)Eddy current loss

the heating of the core due to emfs being

induced not only in the transformer windings


induced not only in the transformer windings

but also in the core.

Core losses

(ii)Eddy current loss

These induced e.m.f.s set up circulating

currents called eddy currents. Owing to the low


currents called eddy currents. Owing to the low

resistance of the core, eddy currents can be

quite considerable and can cause a large power

loss and excessive heating of the core.

total losses in transformer

= Pc + I 2R + I 2R


= Pc + I12R1 + I22R2

total core loss

=powerinput poweroutput

lossespower input poweroutput

+


=

powerinput lossespower input +

22

212

122

22

p.f.p.f.

RIRIPVIVI

c +++

=powerinput poweroutput

power input losses -powerinput


=

= 1 -

powerinput power input

powerinput losses

The primary and secondary windings of a 500

kVA transformer have resistances of 0.42 and 0.0019 respectively. The primary and secondary voltages are 11 000


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 :

full load

half load

In a 50 kVA transformer, the iron loss

is 500 W and full-load copper loss is

800W. Find the efficiency at :


1. full-load and

2. half-load

at 0.8 p.f. lagging.


Transformer Drum

(Q8 Q18)


(Q8 Q18)


AC ELECTRICAL MECHINES

Documents

JJ102 Electrical Technology CHAPTER_5 Tansfomers [Compatibility Mode]