single and three phase transformer

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    Lecture 5

    Single and Three Phase Transformers

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    Single Phase and three phase Transformers

    In the class the following topics will be covered:

    1) A comprehensive overview of transformer, essentials, and how does it operate2) Explain ideal and real transformers construction, equivalent circuit, phasor

    diagrams

    3) Transformer Design Equations, Performance of transformers,

    4) Solving examples on these topics

    By the end of this class you should be able to

    1) Understand electromagnetism, induction and how these apply to transformer

    operation2) Understand how to analyze, and find the performance of Transformer

    3) Design, calculate performance, and identify different types of Single phasetransformers

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    Introduction

    For transmission and distribution networks to transfer large amounts of alternating

    current electricity over long distances with minimum losses and least cost, differentvoltage levels are required in the various parts of the networks.

    For example, the transfer of electricity efficiently over a long transmission line requires

    the use of high voltages. At the receiving end where the electricity is used, the high

    voltage has to be reduced to the levels required by the consumer.

    Transformers enable these changes in voltage to be carried out easily, cheaply and

    efficiently.

    A transformer used to increase the voltage is called a "step up" transformer, while that

    used to decrease the voltage is called a "step down" transformer.

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    Introduction

    What is a transformer?

    A device for increasing or decreasing an AC voltage Power Transformers, TV sets to provide High Voltage to picture tubes,

    portable electronic device converters, transformers on the pole, etc. are few

    examples

    A transformer consists of two coils of wires known as primary and secondary

    windings

    The two coils can be interwoven or linked by a laminated soft iron core to reduce

    eddy current losses

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    Basic Components of Transformers

    A transformer consists of two coils electrically separate but linked by a common magnetic circuit of

    low reluctance formed by a laminated soft iron core. If one coil (the primary coil) is connected to an

    AC supply, an alternating magnetic flux is set up in the iron core.

    This alternating magnetic flux passes through the secondary coil and induces and alternating voltage

    in the secondary coil.

    The magnitude of the secondary voltage is directly proportional to the ratio of the number of turns in

    the secondary and primary windings and to the primary voltage.

    Figure 1

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    How do transformers work?

    A changing current through a coil of wire can create a changing magnetic

    field.

    Currents can be induced in other wires by these changing magnetic field.

    Therefore, the primary coil current must have AC.

    The iron core of the transformer is not required but it does increases the

    efficiency a great deal.

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    Transformer Equation

    The transformer equation does not work for DC current since there is nochange of magnetic flux

    If NS>NP, the output voltage is greater than the input so it is called a step-up transformer while NS

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    When an AC voltage is applied to the primary, thechanging B it produces will induce voltage of the same

    frequency in the secondary wire

    So how would we make the voltage different? By varying the number of loops in each coil

    From Faradays law, the induced emf in the secondary is

    The input primary voltage is

    Since dB/dt is the same, we obtain

    dt

    dNV Bss

    =

    s s

    p p

    V NV N

    =dt

    dNV Bpp =

    Transformer Equations

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    Transformer Equations

    Transformer Ratio, this is the most important relationship by which mostof the transformer variables are governed and it is denoted (a)

    Number of Turns in the Primary Coil

    Number of Turns in the Secondary Coila=

    p

    s

    Na

    N

    =

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    Induction and the Transformer

    2amp

    60,000V AC

    1000amp

    120V AC

    Remember that the Relative Number of Turns Dictates the Output Current and Voltage

    as seen above and by Transformer equations

    Figure 2

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    Transformers Types

    Figure 3 Single phase transformer construction

    A) Core type B) Shell type

    Transformers are manufactured in many types the most widely used in power systems are classified

    with their core types as seen below. Core type where each of the windings are wound on one leg of

    the core, while the shell core type, in which both windings are wound on the same leg. Each type

    has its own advantages and disadvantages. Core type is very reliable and easy to maintain, but take

    larger space, however, shell type is smaller but not reliable.

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    Ideal Transformer: Transformer Equations

    Transformer Ratio, this is the most important

    relationship by which most of the transformer variables aregoverned and it is denoted (a)

    Number of Turns in the Primary Coil

    Number of Turns in the Secondary Coila=

    p

    s

    Na

    N

    =

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    Ideal Transformer

    p

    s

    s

    p

    s

    p

    I

    I

    E

    E

    N

    Na ===

    Number

    of Turns

    Coil

    VoltagesCoil

    Currents

    This is the most known relationships that relates the induction between the two coils

    (primary/Secondary) of an ideal transformer to turns ratio and their currents

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    Ideal Unloaded Transformers

    Winding resistances are zero, no leakage inductance and iron loss

    Magnetization current generates a flux that induces voltage in both windings

    2

    mpp

    N

    E

    =

    2

    mss

    N

    E

    =

    Figure 4 Current, voltages and flux in an unloaded ideal transformer

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    Ideal loaded transformer

    Loaded transformer , means current will be also feeding the load and thatsecondary side voltage, current (load Side) will be calculated using the

    previous relations

    Figure 5 Currents and fluxes in a loaded ideal transformer

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    TransformersRemember that the following are mentioned before we bring them here to

    emphasize their importance

    Turn ratio is:

    The input and output complex powers are equal

    Notice the direct relations between number of turns and voltages and reciprocal relations

    with currents.

    s

    p

    s

    p

    E

    E

    N

    N

    Ta ===

    ** IESSIE sssppp ===

    p

    s

    s

    p

    I

    I

    E

    ETa ===

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    Ideal Transformers Equivalent Circuit

    Usually we find that ideal transformer symbol and its variables as depicted in figure 7

    Figure 7 Equivalent circuit of an ideal transformer

    Is

    Ip= I

    s/T

    Es= Vs

    Vp= E

    p= T E

    s

    Vp Ep

    T

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    Transformers

    Ideal transformer Transferring impedances through a transformer

    S

    S

    S

    S

    P

    PP

    I

    V

    I

    V

    I

    VZ 2T

    T

    T=

    ==

    loadP ZZ2

    T=

    Vac

    Zload

    T

    VP

    VS

    IP

    IS

    a) Original circuit with ideal

    Figure 8-a Thvenin equivalents of transformer circuit

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    Transferring impedances through a transformer

    Vac

    T2Zload

    VP

    IP

    b) Equivalent circuit when secondary

    impedance is transferred to primary side

    and ideal transformer eliminated

    (c) Equivalent circuit when primary source is

    transferred to secondary side and ideal

    transformer eliminated

    Figure 8 b-c Thvenin equivalents of transformer circuit

    We have seen how turns ratio (a) or (T) relates to voltages and currents, the presence of suchideal transformer relations facilitates the development of real transformers. This is to say itsequivalent circuit and using such relations to transfer from primary to secondary and viceversa. To transfer impedances of either side then the square of the turns ratio is used.

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    Example 1

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

    has 80 turns on the secondary. Calculate:

    (i) the appropriate values of the primary and secondary currents;

    (ii) the approximate number of primary turns

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    Solution

    V1/V2=N1/N2=a

    N1=V1N2/V2= (a) N2

    6600/400 (80)= 1220 turns

    Transformer rating=Primary Voltage x Primary current

    = Secondary Voltage x Secondary Current

    For Primary 200 KVA = 6600 v (I1)

    I1 = 200KVA/6600v = 30.3 A

    For Secondary 200 KVA= 400 I2

    I2 = 200 KVA/400v = 500 A

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    Quiz

    A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The

    no-load current is 3 A at a power factor of 0.2 lagging.

    Calculate the primary current and power factor when the secondary current is 280 A at a power

    factor of 0.8 lagging. Assume the voltage drop in the windings to be negligible.

    Answer: 58.3 A, 0.78 pf lagging

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    This lecture covers the first essentials that one should know on

    transformer operation and ideal transformers. Next lecture we will go

    over the practical transformer operation, equivalent circuit and

    modeling