EEEB413 tansformers v2

Electro-Mechanical Systems

EEEB413

TRANSFORMERS

Introduction

INTRODUCTION

• Applications, Types, and Construction of

Transformers

• Applications: Transfers Electric Energy, changing

the voltage level (or current level), through a

magnetic field (In our study)

• Other applications: e.g., voltage & current sampling

and measurement, impedance transformation

• It has two or more coils wrapped around a common

electromagnetic core

• Generally, flux in the core is common among the

coils

INTRODUCTION• One winding is connected to source of ac power, the 2nd (& 3rd )

supplies power to loads

• Winding connected to source named “Primary”

• Winding connected to load named “Secondary”

• If there is another one is called “Tertiary”

• Importance of Transformers:

• Main: to transfer electrical energy over long distances (from power plants to load centers)

• In modern power system electric energy is generated at voltages between 12 to 25 kV, Transformers step up voltage between 110 kV to 1200 kV for transmission over long distances with very small losses

• in Malaysia 500kV, 275kV and 132kV for transmission (frequency: 50 Hz)

• Then Transformers step down to 33 kV, 22 kV, 11kV, 6.6kV for local distribution & finally supply safely homes, offices & factories at voltages as low as 240 V (single phase)

Power Loss in Transmission Lines

RIPLoss

2

Transformers play a key role in the transmission of electric power.

Power Transmission

INTRODUCTION

• Transformers are classified, based on types

of core structure, into: (both use thin

laminations)

1- Core form,transformer windings

wrapped on two legs as shown

2- Shell form, transformer windings

Wrapped only on center leg as shown:

(leakage flux is minimized)

Single phase transformer construction

A) Core type B) Shell type

INTRODUCTION

INTRODUCTION• The primary and secondary windings are wrapped one on

top of the other to:

Reduce the leakage flux

• And the low-voltage winding innermost to :

Simplify insulating of the high-voltage winding

from the coreTypes of transformers :

• Step up/Unit transformers – located at output of a generator to step up the voltage level to transmit the power

• Step down/Substation transformers – Located at main distribution or secondary level transmission substations to lower the voltage levels for distribution 1st level purposes

• Distribution Transformers – located at small distribution substation. It lowers the voltage levels for 2nd level distribution purposes.

• Special Purpose Transformers - E.g. Potential Transformer (PT) , Current Transformer (CT)

Oil immersed Distribution

Transformers

Dry Type Distribution Transformers

Nameplate of Transformer

TWO WINDING TRANSFORMER

CONNECTION

• SINGLE PHASE

TRANSFORMER

CONSTRUCTION

Basic components of single phase transformer

N1 N

2Supply Load

Primary winding Secondary winding

Laminated iron core

Ideal Transformer

IDEAL TRANSFORMER

(SINGLE PHASE)

• a lossless transformer with an input winding

and an output winding in which magnetic core

has an infinite permeability

• Figure below shows: ideal transformer and

schematic symbols of a transformer

IDEAL TRANSFORMER

• Np : turns of wire on its primary side

• Ns : turns of wire on its secondary sides

• The relationship between the primary and secondary

voltage is as follows: (a: is the turns ratio)

• The relationship between primary and secondary

current is Np ip(t) = Ns is(t)

•

a

N

N

tv

tv

s

p

s

p

ati

ti

s

p 1

IDEAL TRANSFORMER

• In terms of Phasor quantities:

• Vp/Vs=a , Ip / Is=1/a

• while:

1- phase angles of Vp and Vs are the same

2- phase angles of Ip and Is are the same

• ideal transformer turn ratio affects the magnitude of

voltages & currents not their angles

• Now: given primary circuit voltage is positive at

specific end of coil, what would be the polarity of

secondary circuit’s voltage?

IDEAL TRANSFORMER

• It is possible to specify the secondary’s polarity

only if transformers were opened & it windings

examined

• To avoid requirement of this examination,

transformers employ a dot convention:• If the primary voltage is +ve at the dotted end of the winding wrt the

undotted end, then the secondary voltage will be positive at the dotted

end also. Voltage polarities are the same wrt the dots on each side of

the core.

• If the primary current of the transformer flows into the dotted end of the

primary winding, the secondary current will flow out of the dotted end of

the secondary winding

IDEAL TRANSFORMER

• Power in Ideal Transformer

-Power supplied to Transformer:Pin= Vp Ip cosθp

-Power supplied to loads : Pout=Vs Is cosθs

• Since V & I angles unaffected by ideal transformer θp = θs=θ

• Using the turn ratio; Vp/Vs=a , Ip / Is=1/a

• Pout = Vp / a (a Ip) cosθ = Pin

• similiar relation for reactive power Q & S

• Qin= Vp Ip sinθp= Vs Is sinθs = Qout

• Sin= Vp Ip = Vs Is = Sout

• Equivalent circuit

I2

I1 = I

2 /T

E2 = V

2

V1 = E

1 = T E

2

V1

E1

T

Equivalent circuit of an ideal transformer

IDEAL TRANSFORMER

IDEAL TRANSFORMER

Impedance Transformation

• Load impedance ZL = Vs/Is and apparent impedance of primary circuit: Z’L=Vp/Ip

• Vp = aVs

• Is = a Ip

• Z’L=Vp/Ip= aVs / Is /a= a² ZL

• With a transformer,

it is possible to match

magnitude of a load

impedance with source

impedance by picking proper turn ratio

IDEAL TRANSFORMER

• Analysis of CCT.s containing Ideal Transformer

In equivalent cct.

a) Voltages & impedances replaced by scaled values,

b) polarities reversed if the dots on one side of transformer windings are reversed compared to dots on the other side of transformer windings

• Example 1: A single phase power system consists of a 480 V, 50 Hz generator supplying a load Zload=4+j3 Ωthrough a transmission line of impedance: Zline=0.18+j0.24 Ω

• a) what is the voltage at load? What is the transmission losses?

IDEAL TRANSFORMER• b) a 1:10 step-up transformer placed at the

generator end of transmission line & a step

down transformer placed at load end of line.

What is load voltage ? What is transmission

losses?

IDEAL TRANSFORMER

• a) IG=Iline=Iload

• Iline = V / (Zline + Zload)=480 / [(0.18b+ j 0.24)+(4+j3)]

= 480 / (4.18 + j 3.24) = 480 / 5.29 =

90.8 A

• Load voltage : Vload= Iline Zload=(90.8 )(4+j3)=

454 V

• And the losses are :

• Pline=(Iline)² Rline=(90.8)²(0.18)=1484 W

00 0 8.37

8.378.37

9.0

IDEAL TRANSFORMER

b) need to convert the system to a common voltage

• Need two steps to be followed:1- eliminate T2 referring to load to Transmission line’s voltage

2-eliminate T1 by referring transmission line’s elements & equivalent load to

source side

step 1: Z’load=a² Zload = (10/1)² (4+j3)=400+j300Ω

Zeq=Zline+Z’load=400.18+j300.24Ω=500.3 Ω88.36

IDEAL TRANSFORMER

• Step 2: total impedance reflected cross T1 to source

side

• Z’eq=a² Zeq =a² (Zline+Z’load)

• =(1/10)²(0.18+j0.24+400+j300)=

• = 0.0018+j0.0024+4+j3=5.003 Ω88.36

IDEAL TRANSFORMER

• The generator’ current is :

• IG=480/ [5.003 ] =95.94

• Now it can be worked back to find Iline & Iload through T1

• Np1IG=NS1IlineIline = Np1/Ns1 IG =(1/10)(9.594 )

• Working back through T2:

• Np2IG=NS2Iline

• Iload = Np2/Ns2 Iline =(10/1)(95.94 )=95.94

• The load voltage:

• Vload =Iload Zload=(95.94 )(5 )=479.7

Volts

88.3688.36

88.36

88.3688.36

88.36 87.36 01.0

IDEAL TRANSFORMER

• The line losses are given:

Ploss= Iline² Rline = 9.594 ² 0.18 = 16.7 W

• Note: rising transmission voltage of power

system reduced transmission losses

by a factor of 90

Also voltage at load dropped much

less

Real/Practical

Transformer

REAL SINGLE PHASE

TRANSFORMER

• Operation of a real Transformer

• primary connected to ac source, secondary

open circuited

REAL SINGLE PHASE

TRANSFORMER

• The transformer’s hysteresis curve is shown

• Based on Faraday’s law:

• eind= dλ /dt

Where λ = ∑ φi (on N turn)

• Φav.=λ / N

• And e ind= N d Φav/dt

REAL SINGLE PHASE

TRANSFORMER

• Voltage Ratio of realizing the leakage flux in a real

Transformer

• φp=φm+φLp

• φS= φm+φLS

• Since φm >>φLS , φm >>φLp

• φm can be employed to determine the induced voltage

in the windings and approximately : Vp(t)/Vs(t)=Np/NS=a

• As smaller the leakage fluxes, the better ideal

transformer turn ratio approximate the real transformer

turn ratio

REAL SINGLE PHASE

TRANSFORMER

• Magnetization Current in a Real Transformer

• ac source even when the secondary is open

circuited supply a current to produce flux in real

ferromagnetic core (as seen in chapter One)

• There are two components in the current:

(a) magnetization current iM, required to produce

flux

(b) core-loss current ih+e supplies hysteresis & eddy

current losses of core

REAL SINGLE PHASE

TRANSFORMER

• Magnetization curve of a typical transformer

core can be considered as a saturation curve

REAL SINGLE PHASE

TRANSFORMER

• Knowing the flux in the core magnitude of magnetization current can be found from curve

• Ignoring the leakage flux in the core:

• φav = 1/Np∫ vp(t) dt

• If vp(t) = Vm cos ωt φav= 1/Np∫ Vm cos ωt dt =

= Vm/(ω Np) sin ωt

• If current required to produce a given flux determined at different times from the magnetization curve (above), the magnetization current can be found

REAL SINGLE PHASE

TRANSFORMER

• Finding magnetization current

REAL SINGLE PHASE

TRANSFORMER

• Magnetizing current (another example)

REAL SINGLE PHASE

TRANSFORMER

• Note (Magnetization Current):

1 – magnetization current is nonsinusoidal

2 - once peak flux reaches the saturation point, a small increase in peak flux results in a very large increase in magnetization current

3 - fundamental component of magnetization current lags the voltage applied by 90◦

4 - higher harmonics (odd one) are present in the magnetization current and may have relatively large amount compared to the fundamental &

as core driven further into saturation, larger the harmonic components become

REAL SINGLE PHASE

TRANSFORMER

• Other components of no-load current of

transformer

• is required to supply the hysteresis and eddy

current losses in the core

• assuming sinusoidal flux in the core , eddy

current loss in core proportional to dφ/dt and is

largest when flux pass 0

• Eddy and hysteresis loss shown in Fig 1 and

the total current required to produce flux in the

core shown in Fig 2

REAL SINGLE PHASE

TRANSFORMER• Exciting Current (components: e+h & m)

Fig 1 Fig 2

REAL SINGLE PHASE

TRANSFORMER

• Current Ratio & Dot Convention

• A current flowing into dotted

end of winding produces

a pos. mmf, while current

flowing to undotted end of winding

proguces neg. mmf

• Two current flowing into dotted ends of their respective windings produce mmfs that add

• If one current flows into a dotted end of a winding and one flows out of dotted end, then mmfs will subtract each other

REAL SINGLE PHASE

TRANSFORMER

• In this situation as shown in last figure

• ζP =NPIP , ζS=-NSIS

• ζnet= NPIP-NSIS

• The net mmf produce net flux in core

• ζnet= NPIP-NSIS = φ R

• Where R; reluctance of transformer core

• Since R of well designed is very small until core saturate ζnet= NPIP-NSIS ≈ 0

• Therefore until core unsaturated NPIP ≈ NSIS

• IP/IS ≈ NP /NS =1/a

REAL SINGLE PHASE

TRANSFORMER

• To convert a real transformer to an ideal transformer following assumptions are required:

1- core must have no hysteresis or eddy current

2- magnetization curve must have shape shown (infinite permeabilty before satuartion)ζnet=0 and

NPIP=NSIS

3- leakage flux in core must be zero, implying all flux in core couples both windings

4- resistance of transformer windings must be zero

REAL SINGLE PHASE

TRANSFORMER

Practical Transformer

Equivalent Circuit

mI

1

V1

V2

l1

l2

R1

I2

R2

N1

N2

EQUIVALENT CIRCUIT

I1

R1

V1

X1

I2

R2

V2

X2

N1:N

2

Development of the transformer equivalent circuits

The effects of winding resistance and leakage flux are respectively accounted for by

resistance R and leakage reactance X (2πfL).

EQUIVALENT CIRCUIT

• In a practical magnetic core having finite

permeability, a magnetizing current Im is

required to establish a flux in the core.

• This effect can be represented by a magnetizing

inductance Lm. The core loss can be

represented by a resistance Rc.

EQUIVALENT CIRCUIT

Rc :core loss component,

Xm : magnetization component,

R1 and X1 are resistance and reactance of the primary winding

R2 and X2 are resistance and reactance of the secondary winding

I1

R1

V1

X1

I2

R2

V2

X2

N1:N

2

I’1

Rc

Xm

Ic

I0

Im

EQUIVALENT CIRCUIT

• The impedances of secondary side such

as R2, X2 and Z2 can be moved to primary

side and also the impedances of primary

side can be moved to the secondary side,

base on the principle of:

The power before transferred

=

The power after transferred

EQUIVALENT CIRCUIT

I22R2 = I1’

2R2’

Therefore R2’= (I2/ I1’ ) 2 R2

= a2R2

EQUIVALENT CIRCUIT

I1

R1

V1

X1 I

2R’2

V2

X’2

N1:N

2

I’1

Rc

Xm

Ic

I0

Im

V2’

V2' = a V2 , I1' = I2/a

X2' = a2 X2 , R2' = a2 R2a = N1/N2

The turns can be moved to the right or left by referring all quantities to the

primary or secondary side.

The equivalent circuit with secondary side moved to the primary.

EQUIVALENT CIRCUIT

• For convenience, the turns is usually not shown

and the equivalent circuit is drawn with all

quantities (voltages, currents, and impedances)

referred to one side.

I1

R1

V1

X1 R’2X’

2

N

V2’

Z2’

I0

Rc

Xm

Ic Im

I’1

EQUIVALENT CIRCUIT

• Example 2

A 100kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and 0.035 ohm respectively. The supply voltage is 2200V. Calculate:

(a) the equivalent impedance referred to the primary circuit (2.05 ohm)

(b) the equivalent impedance referred to the secondary circuit

EQUIVALENT CIRCUIT

Determination of

Equivalent Circuit

Parameters

• The equivalent circuit model for the actual transformer

can be used to predict the behavior of the transformer.

• The parameters R1, X1, Rc, Xm, R2, X2 and N1/N2 must be

known so that the equivalent circuit model can be used.

• These parameters can be directly and more easily

determined by performing tests:

1. No-load test (or open-circuit test).

2. Short-circuit test.

Transformer- o/c-s/c tests

• No load/Open circuit test

– Provides magnetizing reactance (Xm) and core

loss resistance (RC)

– Obtain components are connected in parallel

• Short circuit test

– Provides combined leakage reactance and

winding resistance

– Obtain components are connected in series


Transformer- open circuit test

• No load/Open circuit test

V

A

X1 R1

X m R c

X2 R2

W

V oc

I oc

P oc

Equivalent circuit for open circuit test, measurement at the primary side.

Simplified equivalent

circuitV

A

Xm Rc

W

Voc

Ioc

Poc

Transformer- open circuit test

• Open circuit test evaluation

Q

VX

P

VR

IVQIV

P

ocm

oc

occ

ococ

ococ

oc

22

0

1

0 sincos

Transformer- short circuit test

• Short circuit test– Secondary (normally the LV winding) is shorted, that

means there is no voltage across secondary terminals; but a large current flows in the secondary.

– Test is done at reduced voltage (about 5% of rated voltage) with full-load current in the secondary. So, the ammeter reads the full-load current; the wattmeter reads the winding losses, and the voltmeter reads the applied primary voltage.



R2

I scV

A W

X1R1 X2P sc

Vsc

Equivalent circuit for short circuit test, measurement at the primary side

V

A W

a2R2

X1R1 a2X2

I sc

P sc

Vsc

Simplified equivalent circuit for short circuit test



V

A W

Xe1Re1

I sc

Vsc

P sc

Simplified circuit for calculation of series impedance

22

11 RaRReq

22

11 XaXX eq

Primary and secondary

impedances are

combined


• Short circuit test evaluation

21

211

121

eeeq

sc

sceq

sc

sceq

RZX

I

VZ

I

PR


• Equivalent circuit obtained by measurement

Xm Rc

X e1 R e1

Equivalent circuit for a real transformer resulting from the open and

short circuit tests.


• Example 3

Obtain the equivalent circuit of a 200/400V, 50Hz

1-phase transformer from the following test data:-

O/C test : 200V, 0.7A, 70W - on L.V. side

S/C test : 15V, 10A, 85W - on H.V. side

(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm,

Xe=0.31 ohm)

Voltage Regulation

Transformer – voltage regulation

• Voltage Regulation

– Most loads connected to the secondary of a transformer are designed to operate at essentially constant voltage. However, as the current is drawn through the transformer, the load terminal voltage changes because of voltage drop in the internal impedance.

– To reduce the magnitude of the voltage change, the transformer should be designed for a low value of the internal impedance Zeq

– The voltage regulation is defined as the change in magnitude of the secondary voltage as the load current changes from the no-load to the loaded condition.

ZV2

I2’ I2

V12’=V20

R1’ R2

Ze2

X1’

Rc’ Xm’

X2

Ze2 = R1’ + R2 + jX1’ + jX2

= Re2 + jXe2


ZV2

I2’ I2

V12’=V20

R1’ R2

Ze2

X1’

Rc’ Xm’

X2

Applying KVL, V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2

Or V2 = V20 - I2 (Ze2 )


I2Xe2

I2Re2

I2

V2

OA

θ2


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

A

B

θ2

V20 = I2 (Ze2 ) + V2 = I2 (Re2 + jXe2 ) + V2


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

Voltage drop = AM = OM – OA

= AD + DN + NM


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

AD = I2 Re2 cosθ2

DN=BL= I2 Xe2 sinθ2


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

Applying Phytogrus theorem to OCN triangle.

(NC)2 = (OC)2 – (ON)2

= (OC + ON)(OC - ON) ≈ 2(OC)(NM)


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

Therefore NM = (NC)2/2(OC)

NC = LC – LN = LC – BD

= I2 Xe2 cosθ2 - I2 Re2 sinθ2


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

20

2

222222

2

sincos

V

RIXI ee

20

2

222222

2

sincos

V

RIXI ee

NM =

AM = AD + DN + NM

= I2 Recosθ2 + I2 Xe2 sin θ2 +


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

thus,

votage regulation = (AM)/V20 per unit

In actual practice the term NM is negligible since its value is very small

compared with V2. Thus the votage regulation formula can be reduced

to:


I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2

Voltage regulation =

20

222222 sincos

V

XIRI ee


• The voltage regulation is expressed as

follows:

NL

LNL

V

VVregulationVoltage

2

22

V2NL= secondary voltage (no-load condition)

V2L = secondary voltage (full-load condition)


Transformer- voltage regulation

• For the equivalent circuit referred to the

primary:

1

21

V

VVregulationVoltage

'

V1 = no-load voltage

V2’ = secondary voltage referred to the primary (full-load condition)


• Consider the equivalent circuit referred to the

secondary,

I2' R

1'

V2NL

X1' R

2X

2

V2 Z

2

I2

Re2

Xe2

NL

ee

V

sinXIcosRIregulationVoltage

2

222222

(-) : power factor leading

(+) : power factor lagging


• Consider the equivalent circuit referred to the primary,

1

211211

V

sinXIcosRIregulationVoltage ee

I1

R1

V1

X1 R

2'X

2'

Z’2

I1'

V2

'

Re1

Xe1

(-) : power factor leading

(+) : power factor lagging


• Example 4

Based on Example 3 calculate the voltage

regulation and the secondary terminal

voltage for full load having a power factor

of

(i) 0.8 lagging (0.0336pu,14.8V)

(ii) 0.8 leading (-0.0154pu,447V)

Transformer’s Efficiency

Transformer- Efficiency

• Losses in a transformer

– Copper losses in primary and secondary

windings

– Core losses due to hysteresis and eddy

current. It depends on maximum value of flux

density, supply frequency and core

dimension. It is assumed to be constant for all

loads


• As always, efficiency is defined as power output to

power input ratio

• The losses in the transformer are the core loss

(Pc) and copper loss (Pcu).

lossesP

P

)P(powerinput

)P(poweroutput

out

out

in

out

2

2

2222

222

ec RIPcosIV

cosIV


• Efficiency on full load

where S is the apparent power (in volt amperes)

scocFLFL

FLFL

PPS

S

cos

cos


• Efficiency for any load equal to n x full load

where corresponding total loss =

scocFLFL

FLFL

PnPSn

Sn

2cos

cos

scoc PnP 2


• Example 5

The following results were obtained on a 50 kVA

transformer: open circuit test – primary voltage,

3300 V; secondary voltage, 400 V; primary power,

430W.Short circuit test – primary voltage,

124V;primary current, 15.3 A; primary power,

525W; secondary current, full load value.

Calculate the efficiency at full load and half load

for 0.7 power factor.

(97.3%, 96.9%)


• For constant values of the terminal voltage V2 and load power factor angle θ2 , the maximum efficiencyoccurs when

• If this condition is applied, the condition for maximum efficiency is

that is, core loss = copper loss.

02

dI

d

2

2

2 ec RIP

Autotransformer

Transformer- Autotransformer

• It is a transformer whose primary and

secondary coils are in a single winding

Autotransformer


• Same operation as two windings transformer

• Physical connection from primary to secondary

• Sliding connection allows for variable voltage

• Higher kVA delivery than two windings connection


• Advantages:– A tap between primary and secondary sides which

may be adjustable to provide step-up/down capability

– Able to transfer larger S apparent power than the two winding transformer

– Smaller and lighter than an equivalent two-winding transformer

• Disadvantage:– Lacks electrical isolation


• A Step Down Autotransformer:

and


• A Step Up Autotransformer:

and


• Example 6

An autotransformer with a 40% tap is supplied by a 400-V, 60-Hz source and is used for step-down operation. A 5-kVA load operating at unity power factor is connected to the secondary terminals.

Find:

(a) the secondary voltage,

(b) the secondary current,

(c) the primary current.


• Solution

3 Phase Transformer

Three phase transformers

• The three-phase transformer can be built by:

– the interconnection of three single-phase transformers

– using an iron core with three limbs

• The usual connections for three-phase transformers are:

– wye / wye seldom used, unbalance and 3th harmonics problem

– wye / delta frequently used step down.(345 kV/69 kV)

– delta / delta used medium voltage (15 kV), one of the transformer can be removed (open delta)

– delta / wye step up transformer in a generation station

• For most cases the neutral point is grounded

Transformer -3 phase transformer


• Analyses of the grounded wye / delta transformer

Each leg has a

primary and a

secondary winding.

The voltages and

currents are in phase

in the windings

located on the same

leg.

The primary phase-to-

line voltage generates

the secondary line-to-

line voltage. These

voltages are in phase

A B C

VAN VB N VC N

VabVbc Vca

N

a b c



IA

IB

IC

N

IAN

ICN

IBN

Iab

Ibc

Ica

Ib

Ia

Ic



VCA

VAB

N

VA N

VC N

VBN

Vbc

Vbc

Vab

Vca

Vab

A

C

B

a

c

b

VB C Vbc

107

• Three phase transformer

Transformer

Transformer

Transformer


Transformer Construction

Iron Core

The iron core is made of thin

laminated silicon steel (2-3 %

silicon)

Pre-cut insulated sheets are

cut or pressed in form and

placed on the top of each

other .

The sheets are overlap each

others to avoid (reduce) air

gaps.

The core is pressed together

by insulated yokes.

Transformer


Transformer Construction Winding

The winding is made of copper or

aluminum conductor, insulated with

paper or synthetic insulating material

(kevlar, maylard).

The windings are manufactured in

several layers, and insulation is placed

between windings.

The primary and secondary windings

are placed on top of each others but

insulated by several layers of insulating

sheets.

The windings are dried in vacuum and

impregnated to eliminate moisture.

Small transformer winding

Transformer



Iron Cores

The three phase transformer iron

core has three legs.

A phase winding is placed in

each leg.

The high voltage and low voltage

windings are placed on top of

each other and insulated by

layers or tubes.

Larger transformer use layered

construction shown in the

previous slides.

A B C

Three phase transformer iron core

Transformer



The dried and treated

transformer is placed in a steel

tank.

The tank is filled, under vacuum,

with heated transformer oil.

The end of the windings are

connected to bushings.

The oil is circulated by pumps

and forced through the radiators.

Three phase oil transformer

Transformer



The transformer is equipped with

cooling radiators which are

cooled by forced ventilation.

Cooling fans are installed under

the radiators.

Large bushings connect the

windings to the electrical system.

The oil is circulated by pumps

and forced through the radiators.

The oil temperature, pressure

are monitored to predict

transformer performance.

Three phase oil transformer

Transformer



Dry type transformers are used

at medium and low voltage.

The winding is vacuumed and

dried before the molding.

The winding is insulated by

epoxy resin

The slide shows a three phase,

dry type transformer.

Dry type transformer

EEEB413 tansformers v2

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Transcript of EEEB413 tansformers v2