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1 Week © Vocational Training Council, Hong Kong. Basic Operating Principles Basic Operating Principles of Transformers of Transformers EEE3441 Electrical Machines Department of Electrical Engineering Lecture
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1Week © Vocational Training Council, Hong Kong.

Basic Operating Principles Basic Operating Principles of Transformersof Transformers

EEE3441 Electrical MachinesDepartment of Electrical Engineering

││││ Lecture ││││

2Week © Vocational Training Council, Hong Kong.

EEE3441 Electrical Machines

Basic operating principles of following transformers

are introduced

� Single-phase Transformers

� Three-phase transformers

� Auto-transformers

� Instrument transformers

In this Lecture In this Lecture …………

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• Transformer is a machine that has no moving parts

but is able to transform alternating voltages and

currents from high to low (step-up transformer )

and vice versa (step-down transformer).

• Transformers are used extensively in all branches of

electrical engineering from the large power

transformer employed in the T&D network to the

small transformer of an electronic amplifier.

Introduction

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45VA 1-Ø Transformer

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Single-phase transformer

• A simple single-phase transformer consists of

two coils wound on a closed iron core as shown:

Secondary winding

Primary winding

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Construction of Single-phase transformer

Basically, a transformer has two windings:

• primary winding

• Secondary winding

Each winding consists of many turns and are wound

on a laminated iron core. The iron core is insulated

with the windings. The core itself forms a closed

iron magnetic circuit. Consequently, the windings

encircle the core and the core encircles the

windings.

7Week © Vocational Training Council, Hong Kong.

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• Core type

windings are wound around two legs of a magnetic core.

• Shell type

windings are wound around the center leg of a three-legged magnetic core

8Week © Vocational Training Council, Hong Kong.

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Transformer action - Ideal Transformer

1. The winding resistances are negligible (R = 0)

2. All fluxes are confined to the core and link both windings (leakage flux = 0)

3. Permeability of the core is infinite (Iφ1= 0)

4. No eddy current and hysteresis losses

Assumptions:

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When an A.C. voltage v1 applied to the primary winding:

The core flux also links the secondary winding, then:

From equations (1) and (2):

)1(1 −−−−−−−==dtdΦ

Ne 11v

)2(2 −−−−−−−==dtdΦ

Ne 22v

aNN

2

1 ==2

1

v

v

Transformer action - Ideal Transformer (cont)

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• when a load is connected to the secondary winding, i2

will flow and will provide an mmf N2i2 for the core.

• i1 would immediately flow to establish another mmf

N1i1 to oppose N2i2 since no mmf is required to

establish a flux in the ideal core

a1

NN

1

2 ==2

1

i

i

02211 =− ii NN 2211 ii NN =

Then:

Transformer action - Ideal Transformer (cont)

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If the supply voltage is sinusoidal, then in rms values:

2211 IVIV =∴

aNN

VV

2

1

2

1 ==a

1

N

N

I

I

1

2

2

1 ==

VAoutputVAinput =

Transformer action - Ideal Transformer (cont)

Since and

12Week © Vocational Training Council, Hong Kong.

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Emf Equation of a Transformer (1/2)

Φ = Φm sin2πft instantaneous value of induced emf/turn

= − dΦ/dt volts

= − 2πfΦm cos2πft volts= 2πfΦm sin(2πft − π/2) volts

Transformer action - Ideal Transformer (cont)

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rms value of induced emf / turn

= 0.707 x 2πfΦm volts

= 4.44 x fΦm volts

Hence E1 = 4.44N1fΦm volts

E 2 = 4.44N2fΦm volts

2

1

2

1

NN

EE =

Emf Equation of a Transformer (2/2)

Transformer action - Ideal Transformer (cont)

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Phasor Diagram of Ideal Transformer action under No-load

Φ

Im1

E1

E2=(N2 / N1) x E1

V1 = – E1

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Example 1

A 200kVA, 6600/400V, 50 Hz single-phase transformer has

80 turns on the secondary. Calculate:

a) the approximate values of the primary and secondary

currents;

b) the approximate number of primary turns; and

c) the maximum value of the flux.

Ans: 30.3A, 500A; 1320; 0.0225Wb

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Features of Practical Transformer

• the windings have resistances

• not all windings link the same flux

• permeability of the core material is not infinite, and

core losses occur when the core material is subjected

to time-varying flux

Losses in Practical Transformer

• Copper losses ( I2R losses )

• Iron losses ( Core losses )

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Transformer Losses

• Copper losses ( I2R losses )

Power losses due to the resistances in the primary and

secondary windings

PC = I12 R1 + I2

2 R2

• Iron losses ( Core losses )

a. Hysteresis loss

Power loss due to the alignment of magnetic dipoles

in the iron core

b. Eddy current loss

Power loss due to the induced eddy current

circulating in the iron core

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In order to analyze transformer at different loading conditions,

based on the principles of transformer action, equivalent

circuit of practical transformer is worked out for calculation.

Parameters of Transformer

The four essential parameters of the transformer

equivalent circuit are then measured by

• Open-circuited test, and

• Short-circuited test on transformer

Req Xeq

RC Xm

Equivalent Circuit of Transformer

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A set of three similar single phase transformers may be

connected to form a three-phase transformer (three-phase

transformer bank). The primary and secondary windings may be

connected in either star or delta configurations.

Three-Phase Transformer

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Ease of transportation

Inefficient magnetic circuit, less efficient

Higher capital cost than a single one

1-phase of the transformer at fault, the other two are not affected

Characteristics of Three-phase Transformer Bank

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Usually 3-limb core structure

5-limb core may be used to reduce the overall height

of a 3-limb core

Magnetic flux shares the magnetic circuit

Fault on one-phase very likely affects the other two

phases

Construction of Three-phase Transformer

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Three-phase Transformer

3 limb core structure 3-phase transformer

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25kV Alternator

Customers

Three-phaseTransformers in T&D System

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3-Ø 11kV/380V Tx

LV Side HV

Side

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Pole-mounted 3-Ø 11kV/380V Tx

11kV/380V 3-phase transformer fixed at a wooden pole

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1. STAR/delta ( Yd )

1

2

2

1

2

1

2

1

.N3

N

.a3

1II

;N

.N3.a3

VV ====

Turn ratio/Voltage ratio/Current ratio of 3-phase transformer (1/4)

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2. DELTA/star ( Dy )

1

2

2

1

2

1

2

1

N.N3

a3

II

;.N3

N

3

aVV ====

Turn ratio/Voltage ratio/Current ratio of 3-phase transformer (2/4)

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3. DELTA/delta ( Dd )

1

2

2

1

2

1

2

1

NN

a1

II

;NN

aVV ====

Turn ratio/Voltage ratio/Current ratio of 3-phase transformer (3/4)

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4. STAR/star ( Yy )

1

2

2

1

2

1

2

1

NN

a1

II

;NN

aVV ====

Turn ratio/Voltage ratio/Current ratio of 3-phase transformer (4/4)

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Auto-transformer

• transformer having a part of

its windings common to the

primary and secondary

• when a load is connected

across b and c, then a current

I2 will flow through the load.

The current I2 will produce an

m.m.f. in the core which will

be balanced by a current I1flowing in the complete

winding

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The voltages and currents are

related by the same turns ratio

as in a two-winding transformer:

aNN

VV

2

1

2

1 ==

a1

NN

II

1

2

2

1 ==

Auto-transformer

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Single-phase Auto-transformer

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Three-phase Auto-transformer

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Example 2

A 1φ, 100 kVA, 2000/200 V two-winding transformer is

connected as an autotransformer as shown such that more than

2000 V is obtained at the secondary. The portion ab is the 200

V winding, and the portion bc is the 2000 V winding. Compute

the kVA rating as an autotransformer.

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Example 2 (cont)

The current rating of the winding are:

AIab 500200

000,100 ==

AIbc 50000,2

000,100 ==

Therefore, for full-load operation of the autotransformer, the terminal currents are:

AIH 500=

AIL 55050500 =+=

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Example 2 (cont)

Now, VL = 2000V and VH = 2200V

Therefore,

)ans(11001000

5002200|kVA H =×=

)ans(11001000

5502000|kVA L =×=

Note: A 1φ, 100 kVA, two-winding transformer when connected as an autotransformer can deliver 1100 kVA.

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Advantages of Auto-transformer

• It effects a saving in winding material (copper or aluminum), since the secondary winding is part of the primary current.

• Lower copper loss, therefore efficiency is higher than in the two winding transformer.

• Lower leakage reactances, lower exciting current.

• Variable output voltage can be obtained.

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Disadvantage of Auto-Transformer

• There is a direct connection between the primary and secondary sides.

• Should an open-circuit develop between points b andc, the full mains voltage would be applied to the secondary.

• The short-circuit current is much larger than for normal two-winding transformer

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Application of Auto-transformer

• Boosting or bucking of a supply voltage by a small amount. (The smaller difference voltage between the output and input voltages the greater is the saving of winding material.)

• Starting of a.c. machines, e.g. induction motor, where the voltage is raised in two or more steps from a small value to the full supply voltage.

• Continuously variable a.c. supply voltages, normally connected between a low voltage supply in and a high voltage supply out.

• Production of very high voltages, e.g. 275kV and 400kV grid system

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Instrument Transformers

� Instrument transformers are used to reduce the voltage or

current magnitudes so that the measuring ranges of

measuring instruments can be extended.

� Measuring Instruments are connected to the secondary of

the transformers.

� The measured values should then be multiplied by the

appropriate turns ratio to get the actual primary values.

1. Voltage Transformer VT (or PT)

– N1/N2 is large and standard 110 V at the secondary.

2. Current Transformer CT

– N1/N2 is small and standard 5 A or 1A at the secondary.

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Voltage Transformer (VT)

• Fundamentally similar in principle to power transformers

but with rated outputs in VA rather than kVA or MVA.

• Operating at full supply voltage, the secondary current is

very small so that the transformers may be regarded in the

same way as a power transformer on no load.

Current Transformer (CT)

• Transforms large primary current value in terms of its

magnitude and phase to a smaller secondary current value.

• the secondary value is directly proportional to the primary

value.

• Nominal CT ratio (60~3000A/5A, 30~1000A/1A)

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Protection CT

• Nominal CT ratio (1000/5, 100/5)

• Required to operate at many times full load current.

Linearity under these conditions is not of great importance.

• The essential point is that saturation must be high enough to

drive the magnetizing current and the secondary current

under fault condition.

Ring type CT terminal Markings

P1

P2

S1

S2

Ring type CT

Primary

Secondary

Insulation

Iron area

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Metering CT

• For non-protection purpose

• Metering CTs need perform very accurately but only over the

normal range of load up to about 120% full load current.

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Reasons for using Instrument Transformer

• Do not require special designed instruments for h.v. or heavy current measurements (the instrument ranges could be standardized)

• Electrical isolation from the primary (h.v. side) is achievable

• For safety reason, one terminal of the secondary winding can be earthed

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About the Use of Instrument Transformer

• may have turns ratio error

• phase shift may exist between primary and secondary measuring quantities.

• CT is actually a step-up transformer with very large turns ratio (eg 500/1), excessive voltage will exist between the secondary terminals if they were left open.

� NEVER LET THE SECONDARY OF A C.T. OPEN-CIRCUITED

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Department of Electrical Engineering

││││ END of Lecture ││││

Basic Operating Principles Basic Operating Principles of Transformersof Transformers


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