UNIT - II

TRANSFORMERS

Chetan Upadhyay

Introduction• A transformer is a static machines.• The word ‘transformer’ comes form the word ‘transform’.• Transformer is not an energy conversion device, but is a device that

changes AC electrical power at one voltage level into AC electrical power at another voltage level through the action of magnetic field, without a change in frequency.

• It can be either to step-up or step down.

Generation Station

TX1 TX1

Distributions TX1

TX1

Transmission System

33/13.5kV

13.5/6.6kV

6.6kV/415V

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Transformer Construction

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

• Core - type construction

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Transformer Construction

• Shell - type construction

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Ideal Transformer• An ideal transformer is a transformer which has no loses,

i.e. it’s 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

V1 – Primary VoltageV2 – Secondary VoltageE1 – Primary induced VoltageE2 – secondary induced VoltageN1:N2 – Transformer ratio

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Transformer Equation

• Faraday’s Law states that,– If the flux passes through a coil of wire, a voltage will be

induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time.

If we have N turns of wire,

dt

tdEmfV indind

)(

dt

tdNEmfV indind

)(

Lenz’s Law

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Transformer Equation

• For an ac sources,– Let V(t) = Vm sint

i(t) = im sint

Since the flux is a sinusoidal function;

Then:

Therefore:

Thus:

tt m sin)(

tNdt

tdNEmfV

m

mindind

cos

sin

mmindind fNNEmfV 2(max)

mmm

rmsind fNfNN

Emf

44.42

2

2)(

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Transformer Equation• For an ideal transformer

• In the equilibrium condition, both the input power will be equaled to the output power, 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

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Transformer Equation

aI

I

N

N

E

ETherefore

1

2

2

1

2

1,

Where, ‘a’ is the Voltage Transformation Ratio; which will determine whether the transformer is going to be step-up or step-down

E1 > E2For a >1

For a <1 E1 < E2

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Transformer Rating

• Transformer rating is normally written in terms of Apparent Power.

• Apparent power is actually the product of its rated current and rated voltage.

2211 IVIVVA Where,

I1 and I2 = rated current on primary and secondary winding.

V1 and V2 = rated voltage on primary and secondary winding.

Rated currents are actually the full load currents in transformer

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Example

1. 1.5kVA single phase transformer has rated voltage of 144/240 V. Finds its full load current.Solution

AI

AI

FL

FL

6240

1500

45.10144

1500

2

1

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Practical Transformer (Equivalent Circuit)

V1 = primary supply voltage

V2 = 2nd terminal (load) voltage

E1 = primary winding voltage

E2 = 2nd winding voltage

I1 = primary supply current

I2 = 2nd winding current

I1’ = primary winding current

Io = no load current

V1 = primary supply voltage

V2 = 2nd terminal (load) voltage

E1 = primary winding voltage

E2 = 2nd winding voltage

I1 = primary supply current

I2 = 2nd winding current

I1’ = primary winding current

Io = no load current

V1

I1 R1X1

RC

Ic

Xm

Im

Io

E1 E2

V2

I1’

N1: N2R2

X2

Load

I2

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Single Phase Transformer (Referred to Primary)

Actual Method

22

22

2

2

12 '' XaXORX

N

NX

22

22

2

2

12 '' RaRORR

N

NR

V1

I1 R1 X1

RC

Ic

Xm

Im

Io

E1 E2 V2

I2’ N1: N2R2

’ X2’

Load

I2

a

II

aVVORVN

NVE

22

2222

1'21

'

'

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Single Phase Transformer (Referred to Primary )

Approximate Method

22

22

2

2

12 '' RaRORR

N

NR

22

22

2

2

12 '' XaXORX

N

NX

V1

I1 R1X1

RC

Ic

Xm

Im

Io

E1 E2 V2

I2’ N1: N2R2

’ X2’

Load

I2

a

II

aVVORVN

NVE

22

2222

1'21

'

'

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Example Problem

1. A 10 kVA single phase transformer 2000/440V has primary resistance and reactance of 5.5 and 12 respectively, while the resistance and reactance of secondary winding is 0.2 and 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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Example 1 (Cont)

Solution

R1=5.5 , X1=j12 R2=0.2 , X2=j0.45 i) Refer to H.V side (primary)

R2’=(4.55)2 (0.2) = 4.14, X2’=j(4.55)20.45 = j9.32

Therefore,

R01=R1+R2’=5.5 + 4.13 = 9.64 X01=X1+X2’=j12 + j9. 32 = j21.32

V1 aV2

R01 X01

21.329.64

I1

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Example 1 (Cont)

AV

VAIFL 5

2000

1010 3

1

Solution

ii) Secondary voltage

p.f = 0.8

Cos = 0.8

=36.87o

Full load,

From eqn

o

oo

oo

V

Vj

aVIjXRV

8.06.422

)55.4()87.365)(32.2164.9(02000

))((0

2

2

2101011

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Transformer Losses

• 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)(

022

2012

1

22

212

1

)()(,

)()(

RIRIPreferredifor

PRIRIPP

cu

circuitshortcucopper

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Transformer Efficiency• 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

PPIV

IV

PP

P

PowerInput

PowerOutputEfficiency

%100cos

cos

%100cos

cos

2)(

)(

cucnload

cucloadfull

PnPnVA

nVA

PPVA

VA

Where, if ½ load, hence n = ½ , ¼ load, n= ¼ , 90% of full load, n =0.9

Where Pcu = Psc Pc = Poc

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Voltage Regulation

• 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 Defination b) Short – circuit Test c) Equivalent Circuit

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Voltage Regulation (Basic Definition)

• In this method, all parameter are being referred to primary or secondary side.

• Can be represented in either Down – voltage Regulation

%100.

NL

FLNL

V

VVRV

Up – Voltage Regulation

%100.

FL

FLNL

V

VVRV

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Tap Changer

A transformer tap is a connection point along a transformer winding that allows a certain number of turns to be selected.

By this means, a transformer with a variable turns ratio is produced, enabling voltage regulation of the output. The tap selection is made via a tap changer mechanism.

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Three Phase Transformers

3 single phase transformers connected together 1.Star/Delta winding arrangements 2. Easy to replace failed units Common core device 1. Lighter and cheaper than 3 individual units 2. 6 rather than 12 external connections 3. Whole transformer must be replaced if single

winding fails .

For both cases analysis procedure identical!

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