chapter4 trans

8/6/2019 Chapter4 Trans

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Chapter 4.Transformer


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Transformer- Introduction Two winding transformers Construction and principles

Equivalent circuit Determination of equivalent circuit

parameters Voltage regulation

Efficiency Auto transformer 3 phase transformer


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Transformer- Introduction

Varieties of transformers


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Transformer is a device that makes use of themagnetically coupled coils to transfer energy

It is typically consists of one primary windingcoil and one or more secondary windings

The primary winding and its circuit is called

the Primary Side of the transformer The secondary winding and its circuit is called

the Secondary Side of the transformer


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If one of those winding, the primary, isconnected to an alternating voltagesource, an alternating flux will beproduced. The mutual flux will link theother winding, the secondary, and will

induced a voltage in it.


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Transformer- Introduction Transformers are adapted to numerous engineering

applications and may be classified in many ways:

Power level (from fraction of a volt-ampere (VA) to over athousand MVA),

Application (power supply, impedance matching, circuitisolation),

Frequency range (power, audio, radio frequency (RF)) Voltage class (a few volts to about750 kilovolts) Cooling type (air cooled, oil filled, fan cooled, water

cooled, etc.) Purpose (distribution, rectifier, arc furnace, amplifier

output, etc.).


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Power transmission


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Power transmission


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Transformer

4.1 Construction


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Transformer- construction

Basic components of single phase transformer


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Single phase transformer construction

A) re ty e B ) S ell ty e


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Transformer- constructionPrimar

Winding

econdarWinding

Multi-la er

Laminated

Iron Core

X1

X2H

1 H2

WindingTerminals


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4.2 Ideal Transformer

Transformer


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Transformer

i

e1

v1 v2

e2

The emf which induced in

transformer primary winding is

known as self induction emf asthe emf is induced due to to flux

which produced by the winding

itself.

While the emf which induced in

transformer secondary winding

is known as mutual induction

emf as the emf is induced due

to to flux which produced by the

other winding.


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Transformer

i

e1

v1 v2

e2

Acording to Faradays Law,theemf which induced in

the primary winding is,

e1 =dt

dN

J1

Since theflux is analternating flux,

tmak [sin!e1 = dt

tdN ak

)sin(1

[J

tmak [[ cos1!


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Transformer

i

e1

v1 v2

e2

e1

where,

tfmak [T cos21!

tE [cosmax1

!

max1E fN TJ 2max1=

2

max1

1

EErms ! fmax144.4 J!


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Transformer

i

e1

v1 v2

e2

e2 =

Similarly it can be shownthat,

dt

dN

J2

E2 rms fN max244.4 J

kf

f

E

E

1

2

max1

max2

1

2

44.4

44.4

J

Jk is transformation ratio


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Transformer

The voltage ratio ofinduced voltages on the secondary toprimary windings is equal to the turn ratio of the winding turn

number of the secondary winding to the winding turn numberof the primary winding. Therefore the transformers can beused to step up or step down voltage levels by choosingappropriate number their winding turns. In power system itsnecessary to step up the outputvoltage of a generator which

less than 30kV to up 500kV for long distance transmission.High voltage for long distance power transmission can reducecurrentflow in the transmission lines, thus line losses andvoltage drop can be reduced.


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2

NE m11

[*!

2

m22

[*!

Current, voltages and flux in an unloaded ideal transformer

Transformer- Ideal Transformer Winding resistances are zero, no leakage

inductance and iron loss

Magnetization current generates a flux thatinduces voltage in both windings


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2222

Transformer

i

e1

v1 v2

e2

Transformer on no load.


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Transformer- Ideal Transformer Loaded transformer

i

V1

E2E1

V2Z

L

I2

2

N1 N

2

1

W

he

n aload is conn

ect

ed to th

es

econdary output t

er

mina

ls o

fa transformer as shown in Figure 4.5, a current I2flows into the

load and into transformer secondary winding N2. The current I2which flowing in N2produces flux 2which opposite by Lenzslaw- to themain magnetic flux in the transformer core. Thiswill weaken or slightly reduce themain flux to .


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i

V1

E2E1

V2Z

L

I2

2

N1 N

2

1

The reduction ofmain flux by Faradays law- could alsoreduce the induced voltage in primary winding E1.Consequently E1is now smaller than the supply voltage V1,then the primary current would be increased due to thatpotential differences. Therefore on loaded transformer, theprimary current has an additional current of I1.


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i

V1

E2E1

V2Z

L

I2

2

N1 N

2

1

Theextra current I1which flowing in the primary winding N1produces flux 1which naturally react according to Lenzslaw, demagnetize theflux 2. Therefore the net magneticflux in the core is always maintained at original value, it isthemain flux (theflux which produced by themagnetizingcurrent).


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i

V1

E2E1

V2Z

L

I2

2

N1 N

2

1

Themagneto motiveforce (mmf) source N2I

2at the

secondary winding produces flux 2, while themmf N1I1produces flux 1. Since themagnitude of 1equal tomagnitude of2and the reluctance seen by these two mmfsources areequal, thus

N1I1 = N2I2


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Currents and fluxes in a loaded ideal transformer



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Transformer- Ideal Transformer Turn ratio

Ifthe primary winding has N1 turns and

secondary winding has N2 turns, then:

The inputand outputcomplex powers are equal

1

2

2

1

2

1

I

I

E

E

N

Na !!!

**IESSIE 222111 !!!


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Transformer- Ideal Transformer Functional description of a transformer:

When a = 1 Isolation Transformer

When | a | < 1 Step-Up Transformer Voltageis increased from Primary

side to secondary side

When | a | > 1 Step-Down TransformerVoltage is decreased fromPrimary side to secondary

side


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Transformer- Ideal Transformer Transformer Rating

Practical transformers are usually rated

based on:Voltage Ratio (V1/V2) which gives us the

turns-ratio

Power Rating, small transformers are

given in Watts (real power) and Largerones (Power Transformers) are given inkVA (apparentpower)


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Transformer- Ideal Transformer

Example 4.1

Determine the turns-ratio of a 5 kVA2400V/120V Power Transformer

Turns-Ratio = a = V1/V2 = 2400/120 =20/1 = 20

This means itis a Step-Down transformer


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Transformer- Ideal Transformer

Example 4.2

A 480/2400 V (r.m.s) step-up idealtransformer delivers 50 kW to a resistiveload. Calculate:

(a) the turns ratio, (0.2)

(b) the primary current, (104.17A)(c) the secondary current. (20.83A)


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Transformer- IdealTransformer Nameplate oftransformer


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Transformer- IdealTransformer Equivalentcircuit

Equivalent circuit of an ideal transformer


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Transformer- IdealTransformer Transferring impedances through a transformer

2

22

2

2

1

11

I

V

I

V

I

VZ a

a

a!

!!

Equivalent circuit of an ideal transformer

load

a2

1

!


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a) Equivalent circuit when

secondary impedance istransferred to primary side and

ideal transformer eliminated

b) Equivalent circuit when

primary source is transferred tosecondary side and ideal

transformer eliminated

Thvenin equivalents of transformer circuit

Transformer- IdealTransformer


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Transformer- practical transformer PracticalTransformer


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4.3 Equivalent Circuits

Transformer


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Transformer- equivalent circuit


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Transformer- equivalent circuitDevelopment of the transformer equivalent circuits

The effects of winding resistance and leakage flux are

respectively accounted for by resistance R and leakagereactance X (2fL).


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In a practical magneticcore having finitepermeability, a magnetizing currentIm is

requiredto establish a flux in the core. This effectcan be represented by a

magnetizing inductance Lm. The core loss canbe represented by a resistance Rc.

Transformer- practical equivalentcircuit


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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

Transformer- practical equivalent circuit


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Transformer- practical equivalent circuit The impedances of secondary side such

as R2, X2 and Z2 can be moved to

primary side and also the impedancesof primary side can be moved to thesecondary side, base on the principle

of:The power before transferred = The

power after transferred.


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Transformer- practical equivalent circuitThe power before transferred = The power

after transferred.

I22R2 = I1

2R2

Therefore R2= (I2/ I1) 2 R2

= a2R2


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Transformer- practical equivalent circuitThe 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.


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Transformer- Approximate equivalentcircuit

For convenience, the turns is usually notshown and the equivalent circuit is drawn

with allquantities (voltages, currents, andimpedances) referred to one side.


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Transformer- equivalent circuit Example 4.3

A 100

kVA transformer has400

turns on the primaryand80 turns on the secondary. The primary andsecondary resistance are 0.3 ohm and0.01 ohmrespectively and the corresponding leakagereactances are 1.1 ohm and0.035 ohm respectively.The supply voltage is 2200V. Calculate:

(a) the equivalent impedance referred to the primarycircuit(2.05 ohm)(b) the equivalent impedance referred to thesecondary circuit


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4.4D

etermination ofEquivalent Circuit

Parameter

Transformer


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

2. Short-circuit test.

Transformer- o/c-s/c tests The equivalentcircuit model for the actual

transformer can be usedto predictthe behavior ofthe transformer.

The parameters R1, X1, Rc, Xm, R2, X2 and N1/N2mustbe known so thatthe equivalentcircuit modelcan be used.

These parameters can be directly and more easily

determined by performing tests:


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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- o/c-s/c tests


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Transformer- open circuit test No load/Open circuit test

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

Simplified equivalent

circuit


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Transformer- open circuit test Open circuit test evaluation

Q

V

XP

V

R

IVQIV

P

oc

m

oc

oc

c

ococ

ococ

oc

22

0

1

0 sincos

!!

!

! UU


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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 thesecondary.

Test is done at reduced voltage (about5% ofrated voltage) with full-load current in the

secondary. So, the ammeter reads the full-loadcurrent; the wattmeter reads the winding losses,and the voltmeter reads the applied primaryvoltage.


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Transformer- short circuit test Short circuit test

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

Simplified equivalent circuit for short circuit test


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Transformer- short circuit test Short circuit test evaluation

21211

121

eee

sc

sce

sc

sce

RZX

I

VZ

I

PR

!

!!


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Transformer- o/c-s/c tests Equivalentcircuit obtained by measurement

Equivalent circuit for a real transformer resulting from the

open and short circuit tests.


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Transformer- o/c-s/c tests Example 4.4

Obtain the equivalentcircuit 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)


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Transformer voltage regulation Voltage Regulation

Mostloads 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 loadterminalvoltage changes because of voltage drop in theinternal impedance.

To reduce the magnitude of the voltage change, thetransformer should be designed for a low value of theinternal impedance Zeq

The voltage regulation is defined as the change inmagnitude of the secondary voltage as the loadcurrentchanges from the no-load to the loadedcondition.


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ZV2

I2 I2

V12=V20

R1 R2

Ze2

X1

Rc Xm

X2

Ze2 = R1 + R2 + jX1 + jX2

= Re2 + jXe2

Transformer voltage regulation


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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)


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I2Xe2

I2Re2

I2

V2

OA

2



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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

A

B

2


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


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


Voltage drop = AM = OM OA

= AD + DN + NM


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


AD = I2 Re2 cos2

DN=BL= I2 Xe2 sin2


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


Applying Phytogrus theorem to OCN triangle.

(NC)2 = (OC)2 (ON)2

= (OC + ON)(OC - ON) 2(OC)(NM)


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


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

NC = LC LN = LC BD

= I2

Xe2

cos2

- I2

Re2

sin2


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2

20

2

222222

2

sincos

V

RIXIee

UU

20

2

222222

2

sincos

V

RIXIee

UU

NM =

AM = AD + DN + NM

= I2Recos

2 + I2Xe2sin

2 +



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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


thus,votage regulation = (AM)/V20 per unit

In actual practice the term NM is negligible since its value isvery small compared with V 2. Thus the votage regulation

formula can be reduced to:


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I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

2

L

2

2


Voltage regulation =

20

222222 sincos

V

XIRIee

UU s


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Transformer- voltage regulation The voltage regulation is expressed as

follows:

NL

LNL

V

VVregulationVoltage

2

22 !

V2 L= secondar voltage (no-load condition)

V2L = secondar voltage (full-load condition)


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Transformer- voltage regulation For the equivalentcircuit referred to the

primary:

1

21

V

VVregulationVoltage

'!

V1 = no-load voltage

V2 = secondar voltage referred to the primar (full-load condition)


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Transformer- voltage regulation Consider the equivalentcircuit referred to the

secondary,

Re2

Xe2

NL

ee

V

sinXIcosIregulationVoltage

2

222222 UsU!

(-) : power factor leading

(+) : power factor lagging


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Transformer- voltage regulation Consider the equivalentcircuit referred to the primary,

1

211211

V

sinXIcosIregulationVoltage ee

UsU!

Re1

Xe1

(-) : power factor leading

(+) : power factor lagging


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Transformer- voltage regulation Example 4.5

Based on Example 4.3 calculate thevoltage regulation and the secondaryterminalvoltage for fullload having apower factor of

(i) 0.8lagging (0.0336pu,14.8V)(ii) 0.8leading (-0.0154pu,447V)


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Transformer- Efficiency Losses in a transformer

Copper losses in primary and secondary

windings Core losses due to hysteresis and eddy

current. Itdepends on maximum value offlux density, supply frequencyand core

dimension. It is assumedto be constantforall loads


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Transformer- Efficiency As always, efficiency is defined as power outputto

power input ratio

The losses in the transformer are the core loss(Pc) and copper loss (Pcu).

lossesP

P

)P(powerinput)P(powe

routput

out

out

in

out

!

!L

222222

222

ec IPcosIV

cosIV

U

U!L


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Transformer- Efficiency Efficiencyon full load

where S is the apparentpower (in voltamperes)

scocFLFL

FLFL

PPSS

! U UL cos

cos


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Transformer- Efficiency Efficiency for any load equal to n x f ull load

where corresponding total loss =

scocFLFL

FLFL

PnPSn

Sn

vv

v!2cos

cosU

UL

scoc PnP v2


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Transformer- Efficiency Example 4.6

The following results were obtained on a 50 kVA

transformer: open circuittest primaryvoltage,3300 V; secondaryvoltage, 400 V; primarypower,430W.Shortcircuittest primaryvoltage,124V;primarycurrent, 15.3 A; primarypower,525W; secondarycurrent, full load value. Calculatethe efficiency at full load and half load for 0.7power factor.

(97.3%, 96.9%)


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Transformer- Efficiency For constantvalues ofthe terminal voltage V2 and

load power factor angle 2 , the maximum efficiencyoccurs when

Ifthis condition is applied, the condition for

maximum efficiency is

thatis, core loss = copper loss.

0

2

!LdI

d

222 ec IP !


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Transformer- Efficiency


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Transformer- Auto transformer Same operation as two windings

transformer

Physical connection from primary tosecondary

Sliding connection allows for variablevoltage

Higher kVA delivery than two windingsconnection


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Transformer- Auto transformer Advantages:

A tap between primary and secondary sides which

may be adjustable to provide step-up/downcapability

Able to transfer larger S apparent power than thetwo winding transformer

Smaller and lighter than an equivalent two-

winding transformer Disadvantage:

Lacks electrical isolation


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Transformer- Auto transformerA Step Down Autotransformer:

and


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Transformer- Auto transformerA Step Up Autotransformer:

and


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Transformer- Auto transformer Example 4.7

An autotransformer with a 40% tap is supplied by a

400-V, 60-Hz source andis used for step-downoperation. A5-kVA load operating at unity powerfactor is connected to the secondary terminals.

Find:

(a) the secondary voltage,(b) the secondary current,(c) the primary current.


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Transformer- Auto transformer Solution


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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 VBN VCN

Vab Vbc Vca

N

a b c


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IA

IB

IC

N

IAN

ICNIBN

Iab

Ibc

Ica

Ib

Ia

Ic


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VCA

VAB

N

VAN

VCN

VBN

Vbc

Vbc

Vab

Vca

Vab

A

C

B

a

c

b

VB C Vbc


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Three phase transformerTransformer -3 phase transformer


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Three phase transformerTransformer


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Transformer Three phase transformer

TransformerConstruction

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 eachother .

The sheets are overlap each

others to avoid (reduce) air

gaps.

The core is pressed together

by insulated yokes.


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TransformerConstruction Winding

The winding is made of copper or

aluminum conductor, insulated withpaper or synthetic insulating material

(kevlar, maylard).

The windings are manufactured in

several layers, and insulation is

placed between windings.

The primary and secondary windingsare 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


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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 ofeach 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


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The dried and treated

transformer is placed in a steeltank.

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


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The transformer is equipped with

cooling radiators which arecooled 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


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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

chapter4 trans

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