2.Alternator 2

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      Jangkar Alternator Unit #2 PLTASAGULING

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    INDAR SG range of generators is the result of exerien!e gaine"at hun"re"s of h"roele!tri! installations on $%e !ontinents&

    'ith an installe" o'er su((ing u to ))** +,-

    .

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    0LITAN AR+ATUR JANG3AR4

    A"a 2 tie 5elitan ar(atur ang u(u("igunakan a"a alternator . fase

    a- 3u(aran lais tunggal

    5- 3u(aran lais gan"a

    )

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    0elitan Satu Lais

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    3u(aran lais tunggal alternator . fase& / kutu5& 12 slot

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    0elitan Dua Lais

    1*

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    3onstruksi 0elitan Stator "ala(Alur

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     3u(aran lais gan"a alternator . fase& / 3utu5& 2/Slot

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    1/

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    le!tro(oti%e :or!e Gaa Gerak

    Listrik4  Jika ku(aran rotor ang 5erfungsise5agai (e"an (agnet itu5erutar& (aka a"a ;uks (agnetts5 akan (e(otong

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    Nilai ggl in"uksi ku(aran stator atau ggl

    in"uksi ar(atur er fase a"alah ?

    a @ /&// f T k" k 

    a @ ggl ar(ature er fase B4

    f @ frekuensi C4

     T @ =u(lah lilitan er fase @ ;uks (agnet er kutu5 er fase ,54

    k" @ faktor "istri5usi

    k @ faktor langkah

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    >ontoh 0elitan Ter"istri5usi

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    3u(aran Ter"istri5usi0elitan ti"ak hana "ite(atkan "ala( 2

    alur& tetai ter"istri5usi a"a 5e5eraa

    alur-

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    >ontoh 5elitan ang ter"iri atas . !oilter"istri5usi

    2*

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     Ju(lah :asor +:

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    22

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    2.

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    2/

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    Distri5ution fa!tor k"4k" @ e(f in"u!e" in a "istri5ute" 'in"ingE e(f

    in"u!e"

      in a !on!entrate" 'in"ing

      @ %e!tor su( of the e(fE arith(eti! su( ofthe e(f 

    2)

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     Jika "iketahui ?

    @ e(f in"u!e" er !oil si"e

    ( @ nu(5er of slots er ole er hase&n @ nu(5er of slots er ole

    F @ slot angle @ 18*En

    (F @ hase srea" angle

     The e(f in"u!e" in !on!entrate" 'in"ing 'ith( slots er ole er hase @ ( %olts-

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    Ru(us k" "engan ga(5ar 5erikut ?

    :ro( the $gure 5elo' A> @ 2 r sinFE2

    Arith(eti! su( is @ ( x 2r sinFE2

    No' the %e!tor su( of the e(fs is AN as sho'n

    in$gure 5elo' @ 2 r x sin (FE2

    k" @ %e!tor su( of the e(f E arith(eti! su( ofthe e(f 

      @ 2r sin ( FE24 E ( x 2r sin FE24

    k" @ sin ( FE24 E ( sin FE24

    27

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    k" a"a Alternator .H untuk =u(lah slotEfaseEkutu5

    ang 5er5e"aSlot perkutub

    m β° kd

    . 1 6* 1&***

    6 2 .* *&9669 . 2* *&96*

    12 / 1) *&9)8

    1) ) 12 *&9)7

    18 6 1* *&9)62/ 8 7-) *&9))

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    Dengan (elihat ga(5ar 5erikut "i 5a'ah4& =ika F ke!il (aka rasio "aat "i"ekati "enganru(us s55?

    .*

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    .1

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    :ull Pit!h an" Short Pit!h ,in"ing

    .2

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    ..

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    :ull Pit!h an" Short Pit!h ,in"ing

    ./

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    3isar ?)E6 @ )E6 x 18* "era=at @ 1)* "era=at1E6 @ 1E6 x 18* "era=at @ .* "era=at-

    3isar en"ek sering "igunakan& karena(e(unai 5e5eraa keuntungan& "iantarana? +enghe(at te(5aga ang "igunakan-

    +e(er5aiki 5entuk gelo(5ang "aritegangan ang "i5angkitkan- 3erugian arus usar "an Csterisis "aat"ikurangi-

    .)

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    Langkah enuh

    .6

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    Langkah Pen"ek

    .7

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    COIL PITCH IN AN AC ARAT!R" #IN$IN%CALC!LATIN% PITCH &ACTOR

    TERMS1. Coil throw: The location in an armature core of the sides of 

      a coil starting in slot number one (1). Examle: the throw could be

      1 to ! meaning that one side of the coil is inserted in slot "1# and

      the other side in slot "!.

    $. Coil San: The number of core slots sanned b% the sides of 

      the coil. &n the abo'e examle# the san would be slots (! 1).

    *. Slots er +ole: The number of slots in the armature core di'ided

      b% the number of oles constituting the field.

    Examle: &n a , ole generator ha'ing a , slot armature core#

    the number of slots er ole would be , - ,# or 1$ slots er ole.

    .8

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    PITCH &ACTOR CALC!LATION

    P't &ator '* alulated b+ d','d'ng teo'l tro- (.) / (o'l *pan)0 b+ tenumber o1 *lot* per pole2

    $engan menggunakan po'nt / *3d 4 d'ata*0 maka 5

    .9

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    ! @ 1 2 @ 2 for full it!he"!oils

    /*

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     ! @ 21 !os KE2 @ 2 !osKE2 %olts

    /1

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    :aktor langkah

    Pit!h fa!tor k @ e(f in"u!e" in a shortit!he" !oilE e(f in"u!e" in a fullit!he" !oil

      @ 2 !os KE2 4E2

      k @ !os KE2

      'here K is !alle" !hor"ing

    angle-

    /2

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    /.

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    >ontoh ?

    L GGL ang "iin"uksikan a"a (asing<

    (asing lilitan& 5ila lilitan (eruakan kisarenuh& (aka total in"uksi @ 2 L 

    //

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    Se"angkan kisar en"ek "engan su"ut .* "era=atlistrik& seerti "ierlihatkan a"a ga(5ar 5& (aka

    tegangan resultanna a"alah?E = 2 EL. Cos 30/2

     

    /)

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    COIL PITCH IN AN AC ARAT!R" #IN$IN% AS A $"SI%NTOOL &OR INII6IN% HARONIC CONT"NT IN TH"

    %"N"RAT"$ 7OLTA%" SIN" #A7"

    One o1 te de*'gn on*'derat'on* 'n *elet'ng an appropr'atep't 1ator '* te armon' ontent o1 te generated,oltage-a,e 1orm2 P't 1ator an be u*ed to redue orel'm'nate *pe'8 armon' 1re9uen'e* 'n te generated,oltage -a,e 1orm a* 1ollo-*5

    A2 &ull P't5 A 1ull p't -'ll a,e no damp'ng e:et on an+

    armon' 1re9uen+2

    ;2

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    /9

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    )*

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    )1

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    )2

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    ).

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    Dire!t

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    ua"rature

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     Tentukan tegangan efektif ang "i5angkitkan "ala(

    satu fase a"a alternator- Data ang "i5erikana"alah ? frekuensi 6* C& =u(lah lilitan er fase 2.*5uah& ;uks (aksi(u( er kutu5 *&*/ ,5-

    Penelesaian ?

    @ /&//-f-T- (- k"- k 3arena faktor lilitank"- k44 ti"ak "iketahui (aka

    "iangga 1& sehingga?

    @ /&// x 6* x 2.* x *&*/

    @ 2/)* BEfase

    )6

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    >ontoh (en!ari k"

    0agian "ari 5elitan alternator ter"iri atasena( gulungan "ala( seri& (asing

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    Penelesaian ?

    F @ .*( @ 6

    k" @ sin ( FE24 E ( sin FE24

    k" @ sin 6x.* E24 E 6 sin .* E24

    k" @ 1E6 x *&2)884

     Ju(lah al=a5ar GGL "ala( 6 !oil @ 6 x 1* B @

    6* B

     Ju(lah %ektor @ k" x =u(lah al=a5ar

      @ .8&6/ B

    )8

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    >ontoh (enghitung k"?

    Citunglah nilai "ari k" untuk se5uah alternator"engan 9 slot er kutu5 untuk kasus s55?

    a4Mne 'in"ing in all the slots

    54Mne 'in"ing using onl the 2E. of the slotsEole

    !4 Tree eual 'in"ings la!e" seuential in 6*grou-

    Penelesaian ?

    )9

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    >ontoh ?

    A .H& )* C& star !onne!te" salient olealternator has 216 slots 'ith ) !on"u!tors erslot-

    All the !on"u!tors of ea!h hase are !onne!te"in seriesO the 'in"ing is "istri5ute" an" fullit!he"- The ;ux er ole is .* (,5 an" thealternator runs at 2)* r(-

    Deter(ine the hase an" line %oltages of e(fin"u!e"-

    Solution ?

    6*

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    ns = 250 rpm, f = 50 Hz, p = 120.f/ns  = 120.50/250 = 24 poles

    m = number of slots/pole/phase = 216/(24 x 3) = 3

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    62

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    Pengaruh Car(onik a"a 3 "an k" ?

    a4  Jika su"ut kisar en"ek atau su"ut !hor"inga"alah K "era=at listrik untuk gelo(5ang;uks fun"a(ental& selan=utna nilaios .K E2Car(onikos )K E2Car(onik

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    :aktor "istri5usi =uga 5er5e"a untuk

    har(onik ang 5er5e"a

    k" @ sin ( nFE24 E ( sin nFE24

    "engan n a"alah or"e har(onik

    +isal untuk har(onik ke

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    >ontohSe5uah alternator (e(iliki 18 slotEkutu5& !oilerta(a (asuk ke slot

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    Penelesaian ?Disini !oil san a"alah 16

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    67

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    68

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