Notes on Transformer

1

Chapter 4.

Transformer

2

Transformer- Introduction

Two winding transformers Construction and principles Equivalent circuit Determination of equivalent circuit

parameters Voltage regulation Efficiency Auto transformer 3 phase transformer

3


Varieties of transformers

4


5


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

It is typically consists of one primary winding coil 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

6


If one of those winding, the primary, is connected to an alternating voltage source, an alternating flux will be produced. The mutual flux will link the other winding, the secondary, and will induced a voltage in it.

7


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 a thousand MVA),

Application (power supply, impedance matching, circuit isolation),

Frequency range (power, audio, radio frequency (RF)) Voltage class (a few volts to about 750 kilovolts) Cooling type (air cooled, oil filled, fan cooled, water

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

output, etc.).

8


Power transmission

9


Power transmission

10

Transformer

4.1 Construction

11

Transformer- construction

Basic components of single phase transformer

N1 N

2Supply Load

Primary winding Secondary winding

Laminated iron core

12


Single phase transformer construction

A) Core type B) Shell type

13


14


Primary

Winding

Secondary

Winding

Multi-layer

Laminated

Iron Core

X1

X2H1 H

2

WindingTerminals

15

4.2 Ideal Transformer

Transformer

16

Transformer

i

e1

Φ

v1 v2

e2

The emf which induced in

transformer primary winding is

known as self induction emf as

the 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.

17

Transformer

i

e1

Φ

v1 v2

e2

Acording to Faraday’s Law, the emf which induced in the primary winding is,

e1 =dt

dN

1

Since the flux is an alternating flux,

tmak sin e1 = dt

tdN mak )sin(

1

tN mak cos1

18

Transformer

i

e1

Φ

v1 v2

e2

e1

where,

tfN mak cos21

tE cosmax1

max1E fN 2max1=

2

max1

1

EE rms fN max144.4

19

Transformer

i

e1

Φ

v1 v2

e2

e2 =

Similarly it can be shown that,

dt

dN

2

E2 rmsfN max244.4

kN

N

fN

fN

E

E

1

2

max1

max2

1

2

44.4

44.4

k is transformation ratio

20

Transformer

The voltage ratio of induced voltages on the secondary to primary windings is equal to the turn ratio of the winding turn number of the secondary winding to the winding turn number of the primary winding. Therefore the transformers can be used to step up or step down voltage levels by choosing appropriate number their winding turns. In power system it’s necessary to step up the output voltage of a generator which less than 30kV to up 500kV for long distance transmission. High voltage for long distance power transmission can reduce current flow in the transmission lines, thus line losses and voltage drop can be reduced.

21

2

NE m1

1

2

NE m2

2

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 that induces voltage in both windings

N1 N

2

mI

m

V1

E1

E2 = V

2

2222

Transformer

i

e1

Φ

v1 v2

e2

Transformer on no load.

23


Loaded transformer

i

Φ

V1E2E1

V2 ZL

I2

Φ2

N1 N2

Φ1

When a load is connected to the secondary output terminals of a transformer as shown in Figure 4.5, a current I2 flows into the load and into transformer secondary winding N2. The current I2

which flowing in N2 produces flux Φ2 which opposite –by Lenz’s law- to the main magnetic flux Φ in the transformer core. This will weaken or slightly reduce the main flux Φ to Φ’.

24


Loaded transformer

i

Φ

V1E2E1

V2 ZL

I2

Φ2

N1 N2

Φ1

The reduction of main flux Φ –by Faraday’s law- could also reduce the induced voltage in primary winding E1. Consequently E1 is now smaller than the supply voltage V1, then the primary current would be increased due to that potential differences. Therefore on loaded transformer, the primary current has an additional current of I1’.

25


Loaded transformer

i

Φ

V1E2E1

V2 ZL

I2

Φ2

N1 N2

Φ1

The extra current I1’ which flowing in the primary winding N1

produces flux Φ1 which naturally react according to Lenz’s law, demagnetize the flux Φ2. Therefore the net magnetic flux in the core is always maintained at original value, it is the main flux Φ (the flux which produced by the magnetizing current).

26


Loaded transformer

i

Φ

V1E2E1

V2 ZL

I2

Φ2

N1 N2

Φ1

The magneto motive force (mmf) source N2I2 at the secondary winding produces flux Φ2, while the mmf N1I1‘ produces flux Φ1. Since the magnitude of Φ1 equal to magnitude of Φ2 and the reluctance seen by these two mmf sources are equal, thus

N1I1‘ = N2I2

27

Currents and fluxes in a loaded ideal transformer


Loaded transformer

E2

Load

V2

I2

2

1

m

Im

+ I1

V1

E1

28


Turn ratio

If the primary winding has N1 turns and secondary winding has N2 turns, then:

The input and output complex powers are equal

1

2

2

1

2

1

I

I

E

E

N

Na

**IESSIE 222111

29


Functional description of a transformer:

When a = 1 Isolation Transformer

When | a | < 1 Step-Up Transformer Voltage is increased from Primary side to secondary side

When | a | > 1 Step-Down Transformer Voltage is decreased from Primary side to secondary side

30


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 Larger ones (Power Transformers) are given in kVA (apparent power)

31


Example 4.1

Determine the turns-ratio of a 5 kVA 2400V/120V Power Transformer

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

This means it is a Step-Down transformer

32


Example 4.2

A 480/2400 V (r.m.s) step-up ideal transformer delivers 50 kW to a resistive load. Calculate:

(a) the turns ratio, (0.2)(b) the primary current, (104.17A)(c) the secondary current. (20.83A)

33


Nameplate of transformer

34


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

35


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

loada ZZ2

1

Vac

Zload

T

V1

V2

I1

I2

36

Vac a2ZloadV1

I1

Vac

/k Zload

V2

I2

a) Equivalent circuit when

secondary impedance is

transferred to primary side and

ideal transformer eliminated

b) Equivalent circuit when

primary source is transferred to

secondary side and ideal

transformer eliminated

Thévenin equivalents of transformer circuit


37

Transformer- practical transformer

Practical Transformer

38

4.3 Equivalent Circuits

Transformer

39

Transformer- equivalent circuit

mI

1

V1

V2

l1

l2

R1

I2

R2

N1

N2

40


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).

41

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.

Transformer- practical equivalent circuit

42

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

43


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.

44


The power before transferred = The power after transferred.

I22R2 = I1’

2R2’

Therefore R2’= (I2/ I1’ ) 2 R2

= a2R2

45


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.

46

Transformer- Approximate 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

47


Example 4.3

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

48

4.4 Determination of

Equivalent Circuit

Parameter

Transformer

49

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

2. Short-circuit test.

Transformer- o/c-s/c tests

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:

50

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


51

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

52

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

53

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.

54


Short circuit test

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

55


Short circuit test

V

A W

Xe1Re1

I sc

Vsc

P sc

Simplified circuit for calculation of series impedance

2

2

11 RaRRe

2

2

11 XaXXe

Primary and secondary impedances are combined

56


Short circuit test evaluation

2

1

2

11

121

eee

sc

sce

sc

sce

RZX

I

VZ

I

PR

57


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.

58


Example 4.4

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, Re1=0.21ohm, Xe1=0.31 ohm)

59

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.

60

ZV2

I2’ I2

V12’=V20

R1’ R2

Ze2

X1’

Rc’ Xm’

X2

Ze2 = R1’ + R2 + jX1’ + jX2

= Re2 + jXe2


61

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 )

62

I2Xe2

I2Re2

I2

V2

OA

θ2


63

I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

A

B

θ2


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

64

I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2


Voltage drop = AM = OM – OA

= AD + DN + NM

65

I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2


AD = I2 Re2 cosθ2

DN=BL= I2 Xe2 sinθ2

66

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)

67

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

68

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 Re2 cosθ2 + I2 Xe2 sin θ2 +


69

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 isvery small compared with V2. Thus the votage regulationformula can be reduced to:

70

I2Xe2

I2Re2

I2

V2

O

V20

I2Re2

I2Xe2

C

MNDA

B

θ2

L

θ2

θ2


Voltage regulation =

20

222222 sincos

V

XIRI ee

71

Transformer- voltage regulation

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)

72


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)

73


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

74


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

75


Example 4.5

Based on Example 4.3 calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of

(i) 0.8 lagging (0.0336pu,425V)

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

76

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

77


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

78


Efficiency on full load

where S is the apparent power (in volt amperes)

scocFLFL

FLFL

PPS

S

cos

cos

79


Efficiency for any load equal to n x full load

where corresponding total loss =

scocFLFL

FLFL

PnPSn

Sn

2cos

cos

scoc PnP 2

80


Example 4.6

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%)

81


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

82

2

2

2222

222

cos

cos

ec RIPVI

VI

22

2

22

22

cos

cos

ec RI

I

PV

V

From transformer efficiencyformula,

Efficiency =

= (4.11)

Transformer-Maximum Efficiency

83

If V2 is approximately constant, hence for a load of givenpower factor, the efficiency is a maximum when thedenominator of Equation 4.11 is a minimum, it is when

0cos 22

2

22

2

RI

I

PV

dI

d c

022

2

e

c RI

P

Or I22Re2 = Pc (4.12)

Hence the efficiency is a maximum when the variable lossI22Re2 (copper loss) is equal to the constant core loss.

Transformer-Maximum Efficiency

84


85

Example 4.7

From Example 4.6: [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 maximum efficiency for 0.7 power factor. Also find the load at maximum efficiency.

86

Assignment 2 SEE2053

Question:

Explain in detail the occurrence of hysteresis loss and eddy current loss in a transformer core and how these losses can be reduced.

Please submit this assignment before or on 24th October 2010.

87

Transformer- Auto transformer

It is a transformer whose primary and secondary coils are in a single winding

Autotransformer

88


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

89


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

90


A Step Down Autotransformer:

and

91


A Step Up Autotransformer:

and

92


Example 4.7

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.

93


Solution

94

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

95


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

96



IA

IB

IC

N

IAN

ICN

IBN

Iab

Ibc

Ica

Ib

Ia

Ic

97



VCA

VAB

N

VA N

VC N

VBN

Vbc

Vbc

Vab

Vca

Vab

A

C

B

a

c

b

VB C Vbc

98

Three phase transformer


99


Transformer

100


Transformer

101


Transformer

102


Transformer

103

Transformer Three phase 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.

104


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

105



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

106



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

107



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

108



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

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Notes on Transformer

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