Power System 2 -Power Transformer EEE3233

8/14/2019 Power System 2 -Power Transformer EEE3233

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CHAPTER 2:

OWER TRANSFORMERbyur Diyana Kamarudin

EEE3233

POWER SYSTEM


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IntroductionIntroduction

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 anothervoltage level through the action of magnetic field,without a change in frequency.

Can raise or lower the voltage/current in ac circuit

GenerationGenerationStationStation

T X 1 T X 1

DistributionsDistributions

T X 1

T X 1

TransmissionSystem

33/13.5kV33/13.5kV 13.5/6.6kV13.5/6.6kV

6.6kV/415V6.6kV/415V

ConsumerConsumer


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TransformerConstruction

There are 3 basic parts of transformer: A primary coil/winding

receives energy from the ac source

A secondary coil/winding

receives energy from primary winding &delivers it to the load

A core that supports the coils/windings. provide a path for magnetic lines of flux


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Transformer Construction

The operation of transformer is based on the principal ofmutual inductance

A transformer usually consists of two coils of wire woundon the same core

The primary coil is the input coil while the secondary coilis the output coil

A changing in the primary circuit creates a changingmagnetic field

This changing magnetic field induces a changing voltagein the secondary circuit

This effect is called mutual induction


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Transformer can be either step-up or step-downtransformer

If the output voltage of transformer is greater than theinput voltage step-up transformer

If the output voltage of a transformer is less than theinput voltage step-down transformer

By selecting appropriate numbers of turns, atransformer allows an alternating voltage to bestepped up by making Nsmore than Np

Or stepped down by making Ns less than Np

Vs Ns

Vp Np=


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Example 1 There are 400 turns of wire in an iron-

core coil. If this coil is to be used asthe primary of a transformer, howmany turns must be wound on the

coil to form the secondary winding ofthe transformer to have a secondaryvoltage of one volt if the primaryvoltage is 5 volts?


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Example 1 (solution)


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Transformer ConstructionCore characteristic:

The composition of a transformer core depends on:

voltage, current, frequency, size limitations andconstruction costs

Commonly used core materials are air, soft iron, and steel

Air-core transformers are used when the voltage source hasa high frequency (above 20 kHz)

Iron-core transformers are usually used when the sourcefrequency is low (below 20 kHz)

A soft-iron-core transformer is very useful where the

transformer must be physically small, yet efficientThe iron-core transformer provides better power transfer

than does the air-core transformer


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A transformer whose core is constructed of laminated sheets ofsteel dissipates heat readily; thus it provides for the efficient

transfer of power.The purpose of the laminations is to reduce certain losses which

will be discussed later in this part

Hollow-core construction


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The most efficient transformer core is one that offers the bestpath for the most lines of flux with the least loss in magnetic andelectrical energy

There are two main shapes of cores used in laminated-steel-core transformers:

Core-type transformers Shell-core transformers


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Shell-core transformers:

qThe most popular and efficient transformer core

Each layer of the core consists of E- and I-shaped sections ofmetal

These sections are butted together to form the laminations

The laminations are insulated from each other and then pressedtogether to form the core.


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Core - type construction:

so named because the core is shaped with a hollow squarethrough the center

the core is made up of many laminations of steel


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Typical schematic symbols for transformers:

The bars between the coils are used to indicate an iron coreFrequently, additional connections are made to the transformer windings at

points other than the ends of the windings

These additional connections are called TAPS

When a tap is connected to the center of the winding, it is called aCENTER TAP


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Ideal Transformer

For ideal transformer, the following areassumed:

- The winding have zero resistance;

therefore, the IR losses in the windingare zero.

- The core permeability c is infinite,

which corresponds to zero corereluctance.

- The are no leakage flux; that is, theentire flux c is confined to the coreand link both windings


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VV11 VV22

NN11 : N: N22

EE11 EE22

II11 II22

VV11 Primary Voltage Primary Voltage

VV22 Secondary Voltage Secondary Voltage

EE11 Primary induced Voltage Primary induced Voltage

EE22 secondary induced Voltage secondary induced Voltage

NN11:N:N22 Transformer ratio Transformer ratio


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No-load condition:

is said to exist when a voltage is applied to the primary,but no load is connected to the secondary

Because of the open switch, there is no current flowing

in the secondary winding. With the switch open and an ac voltage applied to the

primary, there is, however, a very small amount ofcurrent called EXCITING CURRENT flowing in theprimary


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With-load condition:

When a load device is connected across the secondarywinding of a transformer, current flows through thesecondary and the load

The magnetic field produced by the current in thesecondary interacts with the magnetic field produced

by the current in the primary

This interaction results from the mutual inductancebetween the primary and secondary windings.


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Transformer Equation

Faradays Law states that, If the flux passes through a coil of wire, a voltage will

be induced in the turns of wire. This voltage isdirectly proportional to the rate of change in theflux with respect of time.

If we have NN turns of wire,

dt

tdEmfV indind

)(==

dt

tdNEmfV indind

)(==

Lenzs Law


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For an ac sources,

Let V(t) = Vm sin t

i(t) = im sin t

Since the flux is a sinusoidal function;

Then:

Therefore:

Thus:

tt m sin)( =

tN

dt

tdNEmfV

m

mindind

cos

sin

=

==

maxmindind fNNEmfV === 2(max)

maxmm

rmsind fNfNN

Emf =

=

= 44.42

2

2)(

m mB x A =


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For an ideal transformer,

In the equilibrium condition, both the input power will beequaled to the output power, and this condition is said toideal condition of a transformer.

From the ideal transformer circuit, note that,

1 1 2 2V I V I =

1 2

2 1

V I

V I=

(i)

2211 VEandVE ==

max

max

fNE

fNE

= =

= =

22

11

44.4

44.4

Input power = output power


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Transformer Equationn

aI

I

N

N

E

ETherefore ===

1

2

2

1

2

1,

aa = Voltage Transformation RatioVoltage Transformation Ratio;

which will determine whether the transformer is goingto be step-up or step-down

EE11 > E> E22For a >1For a >1

For a


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Transformer rating is normally writtenwritten in terms ofApparent PowerApparent Power.

Apparent power is actually the product ofits ratedits ratedcurrent and rated voltagecurrent 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 inRated currents are actually the full load currents intransformertransformer


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Example 1

1.5kVA single phase transformer has ratedvoltage of 144/240 V. Finds its full loadcurrent.

Solution:Solution:

AI

AI

FL

FL

6240

1500

42.10144

1500

2

1

==

==


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Example 2

A single phase transformer has 400 primary and1000 secondary turns. The net cross-sectionalarea of the core is 60m2. If the primarywinding is connected to a 50Hz supply at

520V, calculate:The induced voltage in the secondary

winding

The peak value of flux density in the core


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Example 2 (solution) N1=400 V1=520V A=60m2 N2=1000

V2=? a) We know that,

n

n

n

b) Emf,

n

n

n

n

2

520

1000

400

V=

2

1

2

1

V

V

N

Na == VV 1300

2=

[ ]

[ ]

2

21

/0976.0

)60)()(400)(50(44.4520

44.4

1300,520,

44.4

44.4

mmWbB

B

ABfNE

VEVEknown

ABfN

fNE

m

m

m

m

m

=

=

=

==

=

=


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Example 3

A 25kVA transformer has 500 turnson the primary and 50 turns onthe secondary winding. The

primary is connected to 3000V,50Hz supply. Find:

a) Full load primary current b) The induced voltage in the

secondary winding

c) The maximum flux in the core


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Example 3 (solution)

VA = 25kVA N1=500 V1=3000V N2=50V2=?

a) We know that,

a)

b)

b) Induced voltage,

a)

b)

c)d)

c) Max flux,

AV

VAI

IVVA

FL 33.83000

10253

1

1 =

==

=

VI

I

EE

AI

I

I

N

Na

3003.83

33.8

3000

3.8350

33.8500

2

112

2

1

2

2

1

=

==

=

=

==

mWb

fNE

27

)50)(50(44.4300

44.4

=

=

=

Power System 2 -Power Transformer EEE3233

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