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Jun 01, 2018

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     Transformer

    Laboratorium Penelitian Konversi Energi ElektrikInstitut Teknologi Bandung

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    Transformer

    AA transformer transformer  is a device thatis a device that

    convert ac electric energy atconvert ac electric energy at

    one voltage level at anotherone voltage level at another

    voltage level.voltage level.

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    Transformer are important tomodern life

    • Thomas A. Edison (1882)

    Power Distribution System : New or!"#SA $ 12% &d'

    Probems : osses * &ota+e Dro, - inei'ient

    12% !&A /% !&A 0% !&A

    %58=η 

    Example :

    12% & 0% !&A

    s-%%/ ohm31%!m1%%% A

    /% &

    0%&

    1.

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    High Voltage

    0% !& 0% !&A

    s-%%/ ohm31%!m1 A

    %%/ &

    0%!&~

    0% !&A %%/ &A 0% !&A

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    Transormer

    are im,ortant to modern ,ower system 

    Low voltage High Voltage

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    Real Appliation on Po!er "#stem

    380 V/20 kV 20 kV/380 V

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     Transformer

    1. Power 4re5uen'y /%36% 7

    2. adio re5uen'y 9 % !7

    1. Power Transormer 

    2. ;nstrument Transormer 

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    2. Ty,e and

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    $% The Ideal Transformer

    V  P  (t)

    I P  (t) I  

    S (t)

    V  S (t)N  P    N  S

    >

    "

    >

    "

     P 

     N 

     N a  =a

     N 

     N 

    t v

    t v

     P 

     P ==

    )(

    )(Turn ratio a 

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    The Ideal Transformer

    V P  (t)

    I P  (t) I  

    S (t)

    V S (t)N 

    P   N 

    S

    >

    "

    >

    "

    at it i

     P    1)()( =Turn ratio a )(.)(.   t  I  N t i N 

    S S  P  P   =

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    The Ideal Transformer

    V P  (t)

    I P  (t) I  

    S (t)

    V S (t)N 

    P   N 

    S

    >

    "

    >

    "

    aV 

     P =

    a I 

     I 

     P   1=

    In term of phasor &uantities

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    Po!er in ideal Transformer

     P  P  P  IN    I V  P    θ cos= S S S OUT    I V  P    θ cos=

    θ θ θ    == S  P  θ cosS S OUT    I V  P    =

     P S  P S    aI  I and aV V    ==   /

     IN  P  P 

    OUT    P aI a

    V  P    ==   θ cos

    Po!erinput

    Po!erinput

    ideal

    Turn ratio

    e&uation

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    Po!er in ideal Transformer

     IN  P  P OUT   P  I V  P    ==   θ cos

     IN  P  P S S OUT    Q I V  I V Q   ===   θ θ    sinsin

     IN  P  P S S OUT    S  I V  I V S    ===

    Ativepo!er

    reativepo!er

    Apparentpo!er

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    Evaluasi

    ' (ntuk memahami konsep trafo ideal berikan kemahasis!a) ontoh sederhana misaln#a *

    +  ,ara mendapatkan tegangan sekunder # volt -ika tegangan primern#a . volt%

    + Arus sekunder b ampere pada tegangan # volt) ari arus primer a ampere padategangan / volt%

    + Ambil nilai . dan # #ang bedan#a sangatbesar

    + "oal pada sistem 0 fasa dulu sa-a%

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    Impedane Transformer

     L

     L

     L

     I 

    V  Z    =

    Impedane 1

    ratio of the phasor voltage aross it ti the phasor urrent

    Z L

    V L

    I L

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    Impedane Transformer

     P 

     P 

     L  I 

    aa I 

    aV 

     I 

     Z 

      2

    /'   ===

    S  P   aV V    =

     L

     I 

    V  Z    =

     P 

     P 

     L

     I 

    V  Z    ='

    a I  I S  P 

      /=

     L L  Z a Z 

      2'   =

    Z L

    I P 

      I S 

    V P  V 

    +

    -

    +

    -

    if 

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    Theor# of operation of real single2phase transformers

    The basis transformer operation an be derivedfrom 3arada#4s la!   dt 

    d eind 

    λ =

    ∑=

    = N 

    i

    i

    1

    φ λ λ Flux

    linkage  N 

    λ φ  =

    dt 

    d  N e

    ind 

    φ =

    V P  (t)

    I P  (t) I  

    S (t)

    V S (t)N 

    P   N 

    S

    >

    "

    >

    "

    φ 

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    E53 e&uation of a transformer

    dt d  N e   φ −=Faraday law of

    electromagneticinduction

    t ω φ φ    cosmax=

    t  N e   ω ωφ    sinmax=

    RMS Value

    mm  fN  N  E    φ φ ω  44,4

    2==

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    Transformer Losses

    ' ,opper 6I7R8 losses

    ' Edd# urrent losses

    ' H#steresis losses

    ' Leakage flu.

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    The e&uivalent iruit of a transformer 

    the model of a real transformer

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     The equivalent circuit of a

    transformer 

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    Phasor diagram orresponding + referred toprimar#

    a&2

    a;2 1

    2

    ;1 1

    1

     j ;1 ?

    1

    φ 

    ;m

    ;%   ; 1

    ;2 3a

    ;<

    2θ 

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     The equivalent circuit of a

    transformer 

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    Test on transformers

    • Open-Circuit Test

    • Result : V OC IOC P

    OC

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    9pen2,iruit Test Result

    ,ondutane of the ore2lossresistor   C 

    C  R

    G   1=

     M 

     M 

     X 

     B  1=

     M C  E    jBGY    −=

    C C 

     E  X 

     j R

    Y   11−=

    OC 

    OC  E 

    V  I Y    =

    "useptane of the magneti:ing indutor

    Admittane

    5agnitude admittane

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    9pen2,iruit Test Result

    OC OC 

    OC 

     I V 

     P  PF 

    .cos   ==   θ 

    OC OC 

    OC 

     I V 

     P 

    .cos   1−=θ 

    θ −∠=OC 

    OC 

     E  V 

     I Y 

     PF V 

     I Y 

    OC 

    OC  E 

    1cos−−∠=

    Po!er 3ator

    Po!er 3ator angle

    Admittane

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    Test on transformers

    Short-Circuit Test

    Result * V ",  )I ",  )P 9, 

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    Short-Circuit Test Result

    Series impedance

    SC 

    SC SE 

     I 

    V  Z    =

    SC SC 

    SC 

     I V 

     P  PF 

    .cos   ==   θ 

    SC SC 

    SC 

     I V 

     P 

    .cos   1−=θ 

    Power Facor o! "e curren

    #a$$in$ % &urren an$le '%()*mpedance an$le '+(

    θ θ 

    ∠=−∠

    ∠=

    SC 

    SC 

    SC 

    SC 

    SE 

     I 

     I 

    V  Z 

    00,"ere!ore)

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    "hort2,iruit Test Result

    Admittane

    eqeqSE    jX  R Z    +=

    )()(  22

    S  P S  P SE    X a X  j Ra R Z    +++=

    "eries impedane

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    2-2

    ,"e e.uialen circui impedances o! a 20 kV) 8000 / 210 V) 0 4

    rans!ormer are o 5e deermined- ,"e open circui es were per!ormed

    on "e primar6 side o! "e rans!ormer) and "e !ollowin$ daa were

    aken :Open-circuit

    test onprimary!

    "hots-circuittest onprimary!

    V OC   7 8000 V

     I OC 

      7 0)21

    P OC

    7 100 9

    V OC 

      7 18 V

    I OC 

      7 2);

    P OC

    7 210 9

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      I  a

     I  !

     V    !V a" R m X   j

    e# I 

    +

    m I 

    eq R eq X   j

    Ω192  jΩ4.38

    Ω$   j   4.38Ω$ 159

    > >

    " "

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    The per2unit s#stem of measurement

    q%antit&'f  va(%e)a!e

    q%antit&a"t%a( %nit  e* Q%antit&   =

    *n sin$le%p"ase s6sem :   )a!e)a!e)a!e)a!e)a!e   I V S '* Q P    =,

    )a!e

    )a!e)a!e

     I 

    V  Z    =

    )a!e

    )a!e

    )a!e V 

     I 

    Y   =

    )a!e

    )a!e

    )a!e

    V  Z 

    2)(=

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    Voltage Regulation ; Effiien#

    %100,

    ,, +

    V V VR

    ('ad  f%(( S 

    ('ad  f%(( S ('ad n'S    −=

    Effiien#

    %100 + P 

     P 

    in

    '%t =η  %100 + P  P 

     P 

    ('!!'%t 

    '%t 

    +=η 

    Voltage Regulation

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    Autotransformer

    I SE 

    I LV 

    V L

    N SE 

    N C 

    I C 

    I H  "tep2do!n autotransformer

    SE 

    SE 

     N 

     N 

    V =   SE SE C C    I  N  I  N    =

    C  L   V V   =   SE C  ,    V V V    +=

    SE  ,    I  I    = C SE  L   I  I  I    +=

    C SE 

     , 

     L

     N  N 

     N 

    +=

    C SE 

     , 

     L

     N 

     N  N 

     I 

     I    +=

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    Po!er rating 2autotransformer

    SE 

    C SE 

     IO

     N 

     N  N 

    S    +=

      S IO

     #$nput and Output apparent powers

    SW 

      # apparent power in the

    transformer windings

    For example) a ;000 kV auorans!ormer connecin$ a 0 kV

    s6sem o a 38 kV s6sem would "ae an

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    Three2phase transformer 

    N P1

      N S1

      N P2 

      N S2 

      N P3

      N S3

    "ree%p"ase rans!ormer 5ank composed o! independen rans!ormer 

    N P1

      N P2 

      N P3

    N S1

      N S2 

      N S3

    "ree%p"ase rans!ormer wound on a sin$le "ree%le$$ed core

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    Three2phase transformer onnetions

    ' elta2!#e 6>#8' >elta2delta 6>d8

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    >elta2!#e 6>#8 ,onnetion

     P 

     LS 

     LP 

    φ 

    φ 

    3=

    3

    a

     LS 

     LP  =

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    >elta2delta 6>d8 ,onnetion

    aV 

     P 

     LS 

     LP  ==φ 

    φ 

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     A /% !&A 1.8%% 3 2%8 & Dy distribution transormerhas a resistan'e o 1 ,er'ent and rea'tan'e o 0,er'ent ,er unit.

    (a) @hat is the transormers ,hase im,edan'ereerred to the hi+h"Bota+e sideC(b)

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    Instrument transformers

    ' Potential transformer 6PT8

    ' ,urrent transformers 6,T8

    & @

    S

    A