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+

t p

ldealTransformer quations

"I

lp 1[=;

, -1. ,'P-g 's

l t^

VottageRat io ,=\=1 E^=a.E^NS ES P _SThe urnsatiosonly scalar.ntroducesophase hift

Apparentower alance E p. lp=E ,.1Notpowerossesn dea ransformerCurrentRatio Currentatios he nversef hevoltageatio

et332b-4.MCD

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rL-- f>

t n

lmpedancesReflected hroughdealTransformers

"T Ztood

Load mpedanceByOhm's aw

-Eorr - Q-va

ESZ load=[

asseen romprimaryideof ransformerEp7- ln- lp

WriteE,and , in erms f primaryaluesFrom boveatiosorvoltagendcurrentEplS=lp ' t Epr*s a

ls lp ' t =Z loada lp' tEp

Z load= -tI p ' t "Load mpedance

Ea ^2 -PL road.o= h

increasedhenso Z load'^2 Z in

IS

et332b-4.MCD 4

viewedromPrimaryide

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When mpedancesreconnectedo thefrom hesecondaryide: primary ideandviewedo-JL _

Z,tooo

1-

Epload 'p E p=a'Es , - ' t'P- a

EsL in= ,;?'E " iZ toad=- =a.Es.i/ 's \ \\ ; /

so Z,n=' ' :oa'

Substitutento heZ,o"o QuationSimplifying E sz road=" ' ' t ,Final ly Z load= 2'Z in

a\r l

Generally Movingmpedanceormsecondaryo primarymultiply ya2. Movingromprimaryo secondary,ivide ya2.et332b-4.MCD

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Example: 25 kVA,7200 2401120enter-tapingle hasetransformerperates t rated oltage.t supplies single haseoadthathasan equivalentmpedancef7,2+J6.9o ohms.Assumeldeal perationnd ind:a.) turns atio b.) secondaryurrentd.) load asseen romprimaryidee.) PT,ST,Q1,and Fo

c.) primaryurrent\ /

a) fu^ c** 1no,.r{o./her-* ' f Vp ?zooVV 'zqovN.ip ?zoov r-Cr-= ' = zq.v=Efl

^"s 3 rP - | - \ - r .=aa-t \ ! * , : , , . ,J , -1C,. r . . , ' ] l := t => {P - q- ottts6 Ot",', ',t '* tt: {' "d ls TEj t- '^"?:*, '- '* v rr= z. : f .z 36:) '

; \

e\

7 =r333z t^3( "6-r l /

- -Z /r. . - , r'3a'(?.2M) = 16,*Co"s-,Sl:=Zi V= (z4ov)(33cbfr)='Ellr-oz;[g%Fr' gD,ro.cos(3.T) Ql=, ys/^/(3t3')L=63qz.ru;' / @;^q

IJeg.:n Ar--1t .-) ,cr. \t-J t - l ' t t .4 tu l Jt(Li .,t -

,PD'T-- .- :

- l6s 7,s-nQ l-::_1 L"7l"'!t332b-4.MCD 6 &ooo

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\a\ VPn S::Y- ?t/ L(- ;= lZoVTr.^, f,on-. l t%P(rq*5 al

Example: 00kVA2400-120,0 Hzsingle hase ransformeroperates t 2300voltson heprimary ide. t supplies 15kVA o aload hathasa poweractor f 0.723 agging. ssumedeaoperationand ind:a.) secondaryoltage t operatingoltageb.) secondaryurrentc.) impedancef he oadasseenon hesecondaryided.) impedancef he oadas seenon heprimary ide

Vp= V, + VP V+3.v='mAJ-?tr \ =\ / ,_ tJ( r _Vpt-l - \ \ -

\ .\ . - l - . j \ - '.J T J

T:s

powIF, is Q'(nL o'u PRtrr l r4av'A"t 'b- , \ r ' -\ l . r r J

SiCon-,tb, ir?

s*^ l ls tooaVl iV* ) ls v = a"o. l

, lsv n t t r . -):_-. . . . . . . .- .- l r- z /f 5J Llooc zikf l ' " in iral - . . : u, ( Ls' r

0 = C'(i ') 6 - coi 'qo z ) . :1J-?-- ffr're -c Bl fu-:':t '"V c^-'- AA ^p e J a".c o^ 1te' cl^ -nclslo,N 0:43.?"

r ) A{_)/ :Ac--P \ 2o=(z"i(" ts/4s2")= a 1ilo sr

et332b-4.MCD

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Non-idealperation No oadconditionsActivepower ossesdue to hysteresisor eddycurrents.Hysteresisosses- power osseslue to repeated hange nmagnetic olarity. t takemoremmf Nl) o d'emagnetizeorein onedirectionhan heother.Eddycurrents ac currentsnducedn ironcoredue ochanging agneticield9:l[:1,?f lvrteresisossesuseailoy teets esignedormagnetic ircuitsControl f eddycurrentosses taminateore, nsufatelaminatesA finite mount fcurrentsnecessaryo drivemutualfuxbetweenoils.Permeabilitys inite oreluctancesfiniteSomeNl Jneeded.

, N. lQm= n 0=flux f =refuctance

fn terms f nductance socore as nductanceithassociatednductiveeactanceinductanceithassociated

L =N2nDefine bove s hemagnetizingmagnetizingeactance

et332b-4.MCD

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Model f No-load ransformerOperationt .1e

iXm

Vr = theprimary oltage lo= exciting urrenth"= core-lossomponent u = magnetizingomponentRf"= resistancehat epresentshecore ossesXr = inductiveeactancehat epresentshecoremagnetzing

Using hasor uantit ies

VT

VT, fe=_r"lo=i , t " j . i

, VT'M=, \I rvl noomagnitudest

V12P. - 'Hfe=Rf" so

lO=1f"" rM

Sm=VT' lO90degreesvt '% =Rfe

et332b-4.MCD

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Example: omputinghevalues f magnetizingeactancendcoreossesistance.sd vn i^l;;id'[=uo Hzsingfehaseransformer; oPeratingithno oal.'with heprimaryonnectedoffi,?3"rtlto*:t,tdrai,s 48w;; nr, , poweractorr0.187a') heexcitingurrent nd tscomponentsb') hemagnitudesfmagnetzingi.r.trnceandcore ossR')Repeat arts ando ii ttre raisro*Lr isenergzedfromheecondaryfow oftage)ide.- { )

mf\F'

-T{(e

* r-1^,w v_,(J

-\---Pe

S:w F;--o,rB? lo6,nq e_ r l i J rrni.q5 W

= fr)r l net ' lr /ngdr$" ,-err{Pot*r

: C-t942,1 =// \ - l l t * OO=erJ' (Fp)e -c"s' ( : rp?)

tr"1. t5*o - .1. U i: 1326.r vA=) Jft].-= v7 t 32e.Lvn1Los ",r

ZqB

VT o?9. rt-qJ_

fh\-}.t-fF o. J. i< ia

)=u./Bo9 A; C.o3{q A

2ovrJi , ; .EL

s ,^J rl. dc0s 7l : t ' )

c. lBlZ

= 3?9crsi)

t ) yaRf*v, \*-)p

lg

1-) .... '\ , ' l -" t '7ri6=_{ TJ

v:t/

YhJ"1

?zooJ_O, l8 ' i9 f iJoJt-

Yrt

L =o, 8q n.zz"o --

et332b-4.MCD

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NoLoadLosses xamplecont.)Partc-) computeexciting urrent,tscomponents,1s ndXr. asseen romsecondaryidePower onstanthroughransformero....., ' \ ,1.f . l i . . i -" ' ' _ ,r i | .l -)+. ; ^ ' - . i .1 !\ t_ r , l ,; f lf , - . , . - , . l . ' . j ,i

I(- t) .-\ t(J,r--I

t: g-c 3, , i : l1:.- .nY/-/ .JI * , , r l iu . t { V=)40'v '

K, ' i . .^-

- i- /- * ' - t i 'i

r: CL,-1

_1 --r l

?.-, - : 1^ n{ .v,,* (U r-

l;

It\ . t

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P acti a Transfo me sEquivalentircuitModel orLeakage ndCoilResistanceNotalldevelopedlux inksprimaryo secondary. eakageluxonlylinks he urnsof the coil hatgeneratest

Z,roo o

Leakageluxcreateseakagenductancehathas eakage eactance.Thiscauses dditionaloltage rops crosshecoils\ / - t r' p-- p

Where:h-pEbtn tY\ /=b-sLLls -

Vs=Es-ElsElpVp= netvoltagenducednprimaryvoltagenducedue o mutualluxvoltagenducednprimaryue o eakagenetvoltagenducednsecondary

voltagenducedue omutualluxvoltagenducedn hesecondaryue o eakageResistancefwindings odelledy umpedesistanceet332tr4.MCD12

\l tt i

r li l oi i rsi tO,^ l l'P l ;t\

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CircuitModel fXtp ,

ReafTransformer/\ t^

K

TRANSFOR[/R

rooo

ldeal ransformerurelyurns atio.Z'n s calledhe reflected r referred

SecondaryCangetequivalent;nimpedance

From deal ransformFromOhm's awSoz an=a2.(* . jWhere:

, V,z ,n= l- ' - ^2 =Rsni 'XE"Zload

* o\ '\ l'S-=I'sadsX ^2aO/-" r \S - cr J ^lS

2 t2XJR 'z loada2R,= secondaryreferredoprimarya2Xr, secondary,, referredoprimary

N 5ES

Primary

a2Zn.a=oad mpedanceeferredoprimaryet332b-4.MCD 13

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Referringhe secondarympedanceso theprimary ideeffectivelyemoves he need or he deal ransformer. heturns atiocaptures ll heeffectsInL p 5

Io- L lood

CanSolvehiscircuito findany hing bouthepracticalransformeroperation. oltage rop,powerosses, rimary ndsecondarycurrentsndvoltages.Another implifyingssumption:o

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Equivalent eries mpedance eferredo PrimarySideZ"qp=Rp *a2.Rs* j \Xlp*?2.Xtsi

Thevalue anbe referredo thesecondaryide nstead. ustdividethe mpedanceomponentsya2.

a7 ,"- coo

Rp Feqs=- - Rs ra- , (?X Equivalentmpedancereferredo hesecondaryside

P/az xlo/^z I' loodl - t /rooo s/o

i\' fe t t - t ,_Ri" I \ jx-

Where:Ro/a2 resistancef theprimary inding eferredosecondaryX1ola2leakageeactancef heprimary indingeferredto hesecondary

et332b-4.MCD15

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Powerransformersrebi-directionalevices.Theycanoperate itha V source ttactedo either rimary r secondary inding.Y

' 'an l S

N-L) \/ LS

LOW SIDE

rLS

Z,LOOOL>

Z eqHS a2'ZeqLSZ loadHS a2'Z loadlS

loadconnectedY

VY' ' J

to thehigh oltage inding' ' pn l-1SvY v

N, ,-T1 )HS

HIGHSIDT

t , ,^fl >

ZLood t tS

Step-downperation:oadconnectedo lowvoltage oilt , , -r> :

z ,_ln L)

- ln ->Refering's oprimaryide

Step-up peration.t,L5 =o r HS

La t

HIGH SIDE

L)N, .L)

tOW SIDEI

et332b-4.MCD 6

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lmpedancessviewedrom he ow ideZ,n. ivide ya2z --, r=t "ott Z loadlS'eqLS= a, zloadls=--, ,

Example: 100kvA,7200 480 60 Hzsingle hasetransformeras he ollowingarametersllgivennohms:R1s 0.00800 Rns= 1.96 Rr"Hs 53.2kXr-s 0.01510 XHs= 4.55 Xuns= 7800This ransformers operatedn thestep-down odeanddelivers 5o/of its ratedpower o a load hathasa powerfactorof 0.93 agging.a.) draw heequivalentircuitmodelof the ransformer iththeequivalenteriesZ'sreferredo thehighvoltage ideb.) find he otalzinof the ransformer t he higtisidec.) inputZ of the ransformer ith he roaddisconnectedd.) exciting urrentwith he oaddisconnected.

et332b-4.MCD 17

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Example olut ionl , ??ro _Iu, l .q6, t - t5^5:-utfs crKus a: f f ;= iSAn" Xn,,

\:---v1 T-{L: /. f

*L5/lc\ / f---lh

n I I l l ' ' : 'srR(,Ht i l\L\

R1 R*r*at:, XoL Xps+ 'tX.,_.2,-, - , \ {- '=;1s kv-q oi t 1 * Q )'(oou8) X# * :t iltt

(o".jt-')-j,-,:

is6' s ri;-1= )6ir_ut7.e4rt*-t6"L=A,7.ri?"?1]32,:;1 "9:eq - g' I "- ' ;5^ , . . , " t ' ' _E- o - !_r. r t,r! F= ff'$,ffii= 3.orL/zr.s?'1y ",*ir,^lL ' +- *-"*.--.-- '"*-_*:_-1

4 , , , -= 642.8 +, i :s '1 ' - i-L l-{5 ',J..)J=+'^- l - Ie , . qg5 L, , , -,n ; Lt -bi

E,'.., 'q3a 1?,i 1 -: ' - Y-,.= .re V*r= 5 (i vo)--4:'nNi'-

/5tA 4),v - , s

.- ft-_HS

T'HS-

)2r.S7o

-7t .S7DA

et332b-4.MCD 8

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Example?^-r c

Solutioncont.)En"=J* t ,'., i:____f

! rL-:{.JS19: -

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Example:mpedancehroughransformernd ransformeroltaged op.Theequivalentesistancend eactancef a 50 kvA 2400-480transformer 'sindingsreR = 2.Bo andX

= 6.00o.(highside).loadof 10 L2Oo s connectedo the owvoltage ide.Determine:a.)equivalentmpedancef he ransformernd oadcombinedn.)primary urrentf rated oltages appliedo primaryc.)voltage crosshe oad.-}Lt cI( ooyI

192 .8 o s-t-

r*t- .l t'Qo{".Z, *" Pcrrn, [ s 'de e= # = S

zL,jr 'ot t,-=(s)'(,u&i) ' / l ' ;dd, l - -i \{ ,^ . Z*, + Err*=' i " ion, f6. 'o '+: ='r / ^ tt r (- r \ = 1

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Transformeroltage egulationDifference9y9"noutqltvoftagetno oad ndoutputvoltage trated oad ivided v "1"dvoftag; '%vR=ly^d_lyoi,o%lvn\ ' "

Where Vn= ratedoutput oltageVuL= full oadoutput oltage

Afsogivenn"perUnit" fractionrom0 - 1.0u*=l\t=\u*,lvni

Lower alues.ofegulationrebetter.ndicateshat her:esessvoftage ropa-crosshe ransformer.valuesof negativeegulalionrepossible.ndicatedoftageriseacrossransformJr. due o fbading .f. o;i'" Y\''trcr\Note:allvoltages remagnitudenfyVoltag.eegufationsuaffyound hrough alcufationhatuses he otalwindingmpedanceEt332b-S.mcd

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VoltageRegulation odel

LcoC

r-lvTotalwindingmpedanceeferedo hesecondaryL.V.) ideoftransformer.

E lS =l LS'ZeqLS V LSWhere:V.== rated owsidevoltageswitch losed)EL = no oad owside oltageswitch pen)

Irs= lowside oad urrent tspecif ied 'F'zeqr_stotalwindingmpedanceeferredo L.V.side

To compute VR Els-vts%VR= .100Vts

lnduced oil oltagemustbe argerhan hevalue f rated oltageoovercome hedropdue owindingmpedance

7-?qL5

LV

s* r3:(-s 3

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Example: oltage egulationA 500kVA7200 2400V single-phaseransformers operating tratedoadwitha poweractorof 0.82 agging.The otalwindingresistancendreactancealues eferredo thehighvoltage ideareReq 0.197 ndX"o= 0.877 hms.The oad soperatingnstep-down ode.Sketchheappropriatequivalentircuit nddetermine:a.)equivalentowside mpedanceb.) hevoltage o-loadoltagec.) hevoltageegulationt 0.82 agging oweractord.) hevoltageegulationt0.95 eading oweractora-\ - VP Pe.i3_- |rr,, edc.1-,gS?zoc_= - ?Zloo Jvs Ji - )LO L'1 -\,' : iqJX,*,.?*_EqZ

o QlfV.v' ' t. ; t! rJ

a f , ' i3r*v. -,- ' /t r

f ,_rb> I u*J vc i . , , . ,1{

i)i( e;. "!YJ

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Examplecont.) \ /A,rJ\/ . : /{ og /t tr \E.r= f . r iT.*+ v's {; ZoB3f3q.9f (: o,y,iUi r L4

= Q.Oif 9 *,{ , r .09?{llv1 r) n;r \ F u' '??.3 ' 'J-\-) I .t ! [

E; io -7gs /-tz.q " + :* oo .1/ ln,s t

Fd /e. o.8i ,"15," f LtstE1r- V.' ,r iX /o,l /,'a

t\

\- /.Ja ' -

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Per-UnitmpedancendTransformers PercentmpedancefXrs

TRANSFORMER

Equivalentircuit alculation.qrir.s knowledgef turns atioandreferencef impedancealuesromoneside o other.PerUnitsystems normalizationcheme hat emoveshe effectof turns atiofrompower ystem alculations.Transformeranufacturersist mpedancesordevicesn hisPerUnitor Percentmpedance ethod

PerUnit mpedance Needbasequantit iesDefine asepowerandvoltage.Compute aseZ and fromthese uantit ies.ivide ctualmpedancesoltages ndcurrentsy baseso getperunit alues

S bm.=S rated Basepowerdefined s the ratedpoweroftheelectricaleviceV brte=V rated Base oltage efined s herated oltages

t_L^iX Y

NF

et332b-5.MCDof he ransformer

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ComputingerUnit percent)asese., - base

'O?Se-rrv base

Z basev b"r" 's b*"

\ /7 " baset base , base or

Resistance,eactancend mpedancesannowbe divided y Zb^". o get perunitvaluesbased n ransformeratedpowerandvoltage

7 Zacttpu=7,- oase

zp,

Rpr=* v Xact^pu=4;Where: Zact= evicempedancenohmsRact device esistancenohms

Xact device eactancenohmsPerunit alue onvertedo percent ymultiplying.u. alues y 100PerUnit percent)mpedancesndcomponentslsoaddvectorially

Zpu=Rpu* j 'Xpu

=]n puz Xpr2 cr=Et?rl lry\\Rprilmpedancengle ame oractual ndp.u. mpedances

et332b-5.MCD

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Example:erUnitCalculat ionsnlp r . Xt, l ^

+N

E

TRANSFORMER

Theequivalentircuit bove s ora single hase25 VA7200 240volt ransformer. heparametersave he ollowingalues:

valueso perunit alues ased n hecurrent ndof theprimary ndsecondaryf the ransformer. rawcircuitwith he perunitvalues abeledV, : JLoo V aed St*-- . .=ZS kVAbc,sg Fo-F" *o '"i

= /o?3.C sL' ' - . . iJ ta: tc.Vto.e

-'zb,t rde o-l act ao-l Valr.e< k t1 ? 6*e

et332b-5.MCD

l- E^ixv

Rp= 1.40O Xrp 0.25O R, = 0.11 O Xt,= 3.20 )Rf"=19,501O X,., . '=5011O

Converthesevoltageatingstheequivalent

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Example cont.) Pt r v^,ci'rIt ) t4-K. r t,\ t \ l /

Ls^,F.ur. l = Zo 3.astl l -r \l b6sgl_/ :t t eP'^

*.- . i \L, 1S; X/ , p4. jliY^ Io / l ; :*

V=7k+"#Anni" tc J' i 's13' '6sur (-ba-se-

! g e Ve"r*V= L*avCu,- 'p ' - t ie- P' { ' Vc ' Iu9 ('D *q=*orp,.- i , 6c,.5

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When ommonower asesused nd oltage ases redefined stransformerated oltages,.u. percent)alues re hesameoneither ide f ransformer.hismeanshat he dealransformeranbe emovedrom chematicndcalculationsone np.u.Togetactual aluesromp.u.V act=V pr.V b"r" Z act=Zpr.Zbrr" I act=lpr ' l b"r"Ohm's awholdsnp.u. o. . . .

, uou'pu= ?-puVp,lp,Vpr=lpr 'Zp, z pr=

Note:Rated oltagendcurrentalues re1.0p.u.Example:A 250 kVA 2400 240V transformer itha 2.2%impedanceasdamaged ya zero mpedancehortacrossts owvoltageerminals. ssumingated oltage ndan mpedancengle f75degrees, ind:a.) heactual hort ircuit urrent, .) herequiredpercentmpedancef thenew ransformero l imit heshort ircuitcurrento 25,000 mps. ^ r) ? /)

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Examplecont.)S, : ?{, ' :kV,, i,oaSe-

4_+. ! /4 SCI _T Tt - --J -_ -J - '_SC p r^ ii oS

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TransformerestingFinding penCircuit arametersOpenCircuit estOpen ircuitest indsR1s ndXm core osses ndmagnetizingreactanceTestSet-up ndconditions

IOC(-aL

V = voltmeterH.V. ideopencircuited.loc,Pc and Vo"

PC

P=wattmeter A=ammeterTestperformedn lowvoltage idewithVo.= rated owsidevoltage.Measure.CircuitModel

Iv^lv

et332b-5.MCD 2

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Formulasor Finding 1" ndXy Parallelircuit o Vocappliedacross othelements

LS ndicateshatvalues eterminedn he owvoltage ideValues ound rom est P c V oc I o.

Rr"ts=P' fe

V o.'Rt"ts= %

Pc I=lfe lM='oc

V o.'

vVoa

^ mLS= , , ',

Example: n Opencircuitest sperformedn the 240V windings fa 7200-240 power ransformer. he ollowing atawas recordedforthe est Voc 240v loc 16.75 Pc= 580W. Calculateheexcitingesistancend eactance.Voc,= 240 V I oc ,=16.75 A

FindReactive artof current ndcomputethevalueofX,Vo.XmLS= tM

Rf"ts = 99.31 Ohms

Pa,= 590 wPtf"== "oc

lM=16-575 A14.48 Ohms

tM =l lR f"LS :=I fe = 2.417

Pc

Vo.et332b-5.MCD3

X mLS tM V^mLS -

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FindingheWinding arametersShortCircuit estsShortCircuitest inds he otalwindingesistancend eakagereactanceorbothcoils eferredo thesideonwhich he est sperformedusually .V.)TestSet-up ndConditionso d.. lus o bt e50urc e IPsc sc

Short .V.Side. Adjusthesource oltage ncurrentlows.NOTE:STARTWITHSOURCEMeasure:Vr", r"andPr"CircuitModel

xeqlS

Shor^L V side

H.Vsideunt i l atedH.V.V AT ZERO.

Shcr^t

et332b-5.MCD4

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Formulasor indingR"ossandXeqHS\ /Z

"qHS=R eqHS I 'X eqHS Z "q=='sc

PReqHS= sc 'YI ' \ ' ' | ,a ' X "qHS =lZ eqHS' - R "qHSThevalues f R andX are oundusingH.V. idevalues ndcanbereferedo the L.V.sideby dividing y a2Example: he estdata or a75 kVA7200 480V single hasetransformerre isted elow:Open-Circuitest Short-Circuitest(Low idedata) (High idedata)Vo"= 480 V Vr. = 173.1Vlo.= 16.5A lr . = 16.3Po"= 558W Pr.= 1200WDetermine:.) hecore esistancend eactance,heequivalentwindingesistancend eactancenddraw heequivalentircuit fthe ransformerb.) theperunit alues f hevaluesound nparta.)c.) theefficiencyf the ransformerhenoperating t rated oadand0.85 agging oweractor.d.) hepercent oltage egulationf the ransformerperating t ratedloadand0.85 agging ower actor

et332b-5.MCD5

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Tranformerestingrr LVocQ,=te.a ' \ i r r - i i*r i i ! *oc * i *

J. = tL.q58A\ /Vsc/Lq TL {sc

/) - 4, Szc,i . . - -YLrLt 4? r.? .n

exampleolution/ / ^ \c(48o)u =J1 ,l l15S3 \,..J i--- f r tr

t- .d- t

Vr. -l . ta3

.s59*'1,3oVA

,1lEH{C

V.X^=17S,tv

Jru10.69 sr16.3m

\ / i -Z , -X q=Vlo Grl) -(q,SZ) 'L

S,:S ?SkVA" 6o.,e 'cclgJezPov) '# = e1l .2.s\-

o q. r n1 c-r,-q S r ci Q-r {ggsor..J,. ; t i . i1S,oco vq

=-o,oo65

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?,.-+) T_= *,3i:,5 = /o {r A q' Sn.,*it fp= :a"oVq = ?s,ooo o.gS)= 63,?soW FP-o'Sf?1 =**=B . , X1o.:o/oq= s58r 'v {^u ' ' t\ P.+P.+f, tR*l ' \ ' - - tw Fo^,{,SL*oaT[sTe. * . . 6 gr"s.c\ , r r . ; , x loo. f " l ' { 'SLsri ( - (3,?so1 sss +(to.

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Transformer arkingsndpolarityDot Notation Terminalsmarkedwitha dotareconsideredinstantaneouslyositive

N1AboveTerminals andC

N2arepositive t thesame ime

CEot

Phasor iagramsE"o

et332b-6.ppt 1

180degreeshift

Eo,4-)

E 0 degrees-cD180degrees

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ConventionalabelingH - highvoltage ideX - lowvoltage ide

Terminal 1more osit ivehanH2.OnsecondaryideX1morepositivehanX2

0 phase hift

180phase hift

et332b-6.ppt

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Transformerameplate ataVoltage atings highsideand owside no-load alues)Additional arkingsorvoltage:Dash -)= Indicatesoltagesromdifferent indingsSlant D= voltagesromsamewindingCross(x) voltages btained yseries r parallel onnectionof twopartwindingsWyem - wye-connectedindingsExamples: inglePhase2401120240V windingwitha center ap240x120 Twopartwindinghatcanbe connectedn

series or240andparallelor 120240-120 A 240V and separate 20 V winding

Threephase12470-480Y-277 Twowindingransformer. yeconnectedecondary ith277Vand480V available

et332b-6.ppt 3

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Transformer ameplate ataFrequency Rated requency f transformerkVA RatedApparent owerof transformerApparent ower ating eterminedy constructionndcoolingTransformerooling oilcooled also nsulation)AO rating natural onvectionirandoilFA rating natural onvection f oiland orcedairFOA ating forced irandoilcirculationTypical atings 30140150 VA AO/FA/FOAForced irby radiators nd ansForced ilbycirculatingumpsDryType nsulated nooil) Typical t voltages f 15 kVandbelowClassesAA - dry ype,self-cooled,atural onvection f airAFA- Dry ype, orced-air oolingAA/FADry ypeself-cooled/Focedircooleddual atinghat equiresans

et332b-6.ppt 4

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Transformer ameplate ataPercentmpedanceimpedance easuredtindicatedemperaure. aseS andV values re heratings f the ransformerTemperatureise maximumllowableemperatureriseHeateffectshe nsulationExcessiveeatingred ceslifeof transformerClass Insulatingedium ndcooling seepreviouslide)BIL-(Basicmpulse evel)Measureshe maximum oltagestresseshat he ransformeranhandlelmpulseestsused o simulateheeffects f l ightningovervoltagesVoltage urgesnexcess f heBILcancause nsulationfailure

et332b-6.ppt 5

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ThreePhaseTransformerConnectionsTransformersncrease oltage, ecreasingcurrent n powersystems.Lower meanslesspower ossdue o l2R osses n l ines,cables,ransformers,tcTh eephaseransformers3 single-phase nitscan orm3-phase ank

orSingle h ee-phaseransformer.3 separateoresna singleankDifferentonnects f 1-Q nits r coilsbased n wye-deltaonfigurations

et332b-6.ppt 6

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Delta-DeltaA-A)Connection

Paralleloils recore part f same ransformerr

r -^ t) tDU rbcAdvantages: bsorbs nbalancef load.UnbalancI circulatesndelta.Loseone ransformerndcanstilloperateDisadvantages: oNatural eutral

secondary

et332b-6.ppt

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Thrce-phesc,our-wirc Y1 primary

:

Wye-WyeConnection

TH2A.SKr .S$^ t l

t20v 20t v 208 V+r2ov - 2oEv] I-+- ll?g-v- -t-| 120/208V Threc-phaseour-wirc \ rccondary

0'angulardisplaccmcnt

Figrrre3-32 Wyc-wye connection to providc a l20l2O8-V grounded-wye three-phase four-wiremulti-grou ded service.

Three 7200 2401120 transformersPrimaryVon Van= Vcn= 7200V H. - H, Vr,-= 12470VSecond?vV"n= V"n= V"n= 120V X1 X, Vr|-= 208VAdvantages: wovoltageevels vailableGradedHigh oltagensulationEasybalancingetween -$and3-$ oadsDisadvantages:ingle hase hort ircuit urrentspassed.Zero equence3rdHarmonicsassed)TotalBankPower Sr * SzaSe= Sret332tr6.ppt

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Delta-Delta180oA - primery

shift)ABc

180"Angulardisplacernent-5-E--ftDn0il24oV Three-phtsc. our-wircA - secondary

Figure3-TI Deka-deltaconnection o provide 12012081240three-phaseour'wire service.

3 Transformers7200 2401120primaryVn= Ve= Vc = 7200V Vu = 7200Vsecond?ryV"o= Vo.= V"" = 240V Vll = 240V180o hase hift H,'-H,o X.'-X,Advantages:deal or motoroads, -wire. an olerate inglelineshortswithno nterruption.raps3rdharmonic urrentsDisadvantages:ul l nsulationequired n H.V.windingsT1 mustbe argerKVAwhenserving -$ oadwith3-QNonatural round ointUnbalancedonnection henserving -$and3-Qoads ogetherBankpower Sr+ Sz+Sg Sret332b-6.ppt

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Delta-Wye onnection30ophase hift

.S r99x2 x,.sete { ' \ K,,,,\,120 208 V 208Vl r20v 20t v- l -J r39_v- I -r2o/208 T;;..in.,lT',,ir..I6-;"?-..rFigure37 Delta-cu1eonnection ith 3oo ngular isplacement.

3 Transformers200-240120VPrimary * = Vo= 7200VSecond?ry "n=Vnn %n= 120V V.o= 208 V30ophaseshiftV"n o Von ag sstandardAdvantages: ingle haseoadeasy o balance. ewNeutralpoint sestablished.wovoltageevels. raps3rdharmonicsDisadvantages: ul l nsulationequired n H.V.winding ftransformersBankPower Sr+ Sz+Ss= Sr

AIC

. lb- \ -J-\=fO'"ngrt"li,splaccmcnr

abcn

et332b-6.ppt 10

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Thrcc-phare,hrccwirc - opcnA pnnury

l f 51= Sz

OpenDelta-Openelta

Figure3-30 Threc-phasc our-wire open-detta onnection. Notc that 3/-4W meansa threc-phasesystemmedeup of four wircs.)

2 transformersuppfy afanced phase oltages ndcurrents. onnectiontselfs unbalanced v2 transformers 2OO24An2OVprimaryV^= Vr_r_72OO secondaryVll = Vo= 240VIranslormersxchangeeactive ower o proviO"Oatancedvoltages ndcurrentso 3-phaseoadsBankpower

ST -S,, S,J'2St

lzOnOU24O Throc-phascour wirc opcnA rccondrry

et332b-.6.ppt 11Sr=

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InstrumentTransformersHighPowerMeasurementMeter oil atingsVoltage oils orvoltmeters,owermetersandwattmeters115 120VacCurrent oils orammeters,owermeters ndwattmeters

2.5-5A (100%overloadllowed)(shortime)

Inpower ystems, oltageevels100's f kVand housandsfampsInstrumentransformersconverthigh oltagesndcurrentsomeasurementlevels

PotentialransformersCurrentransformers

et332b-6.ppt 12

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Multi-turnecondaryside onnectedo lowpowernstrumentsSymbols

BushingypeCTCu rentransformeratios

20015 200amps nput ives ampsoutput (4011)600/5 600amps nput ives ampsoutput (12011canoverloadated alues 50-200%ontinuously

Cu rentTransformersCTs)Singleurnprim ry usuallyheconductorwith he currento be measured

TCL

StandardypeCT

tl

et332b-6.ppt 13

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PotentialranformersPTs)Reducesigh oltageso 110-120acrangeHigh ccuracyransformersithmimimum owerlossandvery mall oltage rop.Assumedeaoperationf bothCTsandPTs

Potentialransformerym ol3E schematicymbols houldindicateolarity arkP-l-

Typical otentialranformeratios39,837-115ac6900- 15Vac

Canexceed atings y approximately- 10o/oorV,n

et332tr6.ppt 14

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Characteristicsf PTsandCTsPowerRatings burden)VA rating f all nstrumentsonnectedcanotexceedthisvalue f accuracy f transformersillsuffer

PhaseShifttypicallyess han1 degreeofhase hiftor less hroughtranformerVoltage ndGurrentMagnitudeAccuracyf magnitude0 5%or 0 1%Revenuemeteringigher ccuracy anelmeteringndprotectionlower ccuracySafetvGonsiderationsPTs- always round econdarysCapactivie ouple ancausedangerouslyigh oltageCTs NEVEROPENLOADONCT GIRCUITALWAYSSHORTCT SECONDARYERMINALSBEFORE EMOVINGOAD.Can nduce2-G V on open secondary eads

et332tr6.ppt 15

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ParallelOperation fTransformersWhenvoltage atios re notequal, urrents irculatebetweenhewindings f each ransformer ithout loadconnected

L NAD

Currentsirculate etween andB based n hevoltagedifferencend ransformermpedancevenwithno oadEA- EBlc=ZA*zb

Where En= operatingoltage f TransformerEe= operatingoltage f TransformerZa= seriesmpedancefAZa= seriesmpedancef BCirculatingurrentseduce apacity f the ransformersocarry oad

XItr bx?

et332b-7.MCD

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Lossof CapacityDue o CirculatingurrentIR* iB

of superpositionc

L !AD

Transformer current: se deaI rn= A * |

TransformercurrentI ta=lB- lc

l . driven y En EaAdding urculatingurrento transformer increasesotalcurrentnwinding.Notseen n oad urrent.

et332b-7.MCD

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Exampfe:CircufatingurrentsilHJ3g[A,tj?n'"phaseransrormerperatednpararelTransformerABFind "magnituderatings

V-ratio c2300-460 r:i. "{r,2300-450 1.40 3s2and " aspercent f transformerlconOaryL,r,sePec anr* ^e{l"uds Vbo.u.Sec..rr..,ds...y{= fao L,\ / = lo 4eov s ir --i:

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LoadDivision etween arallel ransformersWhenall urns atios reequal,he oadcurrent ividesfollowinghewindingmpedancef he ransformers. orecurrenthroughhe owestmpedance.

All Transformr Z's andLoadZ referedo thesamesideoftransformerr allperunit (%)quantit iesYA=+ YB=+ Yk=+ 'n=*Usecurrent ividerule

Yp=YA"YB* *Yk* *YnYk Finds he currentn the kthTransformerp

et332b-7.MCD

I K=l in

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\ , ) - f/l * k ie,JR

Example:A 100kVA ransformers to be paralleled itha200kVA ransformer. ach ransformer asratedvoltages f4160 240V. Theirpercentmpedancesasedon the ratingsof eachare Z% = 1 64+3.16j 100kVAZo/o 1.10+4.03j 200 kVAFind:a.)ratedhighsidecurrent f each ransformerb.)% of totalbankcurrent rawnby each ransformerc.) maximum ank oad hatcanbe handled ithoutoverloadigeitherransformerS.o-lcd lo%-oj-y.1

4 tLov=lz

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Exampfe: oadJ;,u

Divisioncont.) 1-rnrr\ J../'l/N TC{.nrsJ,. ,f fv

4s.o8no,63 11

4r] -LB0f"

Y*=t = c /Gz3f6z'S1"5{= *= az?ae/:71.k"/ : I= / r ; ,1/J,^ i=^i i1e3-"* ,ql - { '/ypl oq3a,rJ,*=o.3?osJ,L/ r r - l ' 'o '1 ' lYt l - - c 'ara6-rg/ - l - r - r r , , - = ^-*J, . r o.63/q J;,/ Vr / u{ ists */ ''T^ r ' | *.-'rb"'--/ r {d, t5}oQrvr f c, . t t r l res A PppoYtn }c tY, '3?% o F /on4A (*f r r

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AutotransformersAutotranformerssea singleaped oil o change oltagelevels ndcurrentevels Noelectrical solation

Nls = number f turns embraced"y owsideNns= number f turnson highsidepolarity f induced oltages eterminedy direction fcurrent ndwinding raps.

l f Nss= 20 andNns= 80Turns atio

Let V HS

NHs VHsa=-=- a,=Nts Vts 8020 a=4Vns= 120 V so.. V tS

V tS = 30 V Step-downctionet332b-7.MCD

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Some oad s tranfered iaconductionromoneside o theotherandsome s transfered y transformerction

Autotransformeronnectedn stepdownmode. Notedirection f 11,.

Like wowinding ransformer S HS=S LS SoV Hs'fns =V LS'l s lt,= thecurrentrom ransformeraction

Lowsidecurrentmust ncreaseo maintain owerbalance oIts=lHS*l t ,Current atioof autotransformer

I,HS 1l ts aNHsa=-_-Nts

et332b-7.MCD

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Autotransformern Step-upMode

Note:direction f ls reversedo maintain owerbalanceCoils n hesediagrams reseries iding inducedoltagesadd)Autotransformerctioncan beobtained y properconnectionf twowindingransformeroilsForstep-down

?=-=Nts N2 VtsWhere primaryHV)secondaryLV)

NHS=N1* N2 NIS=N2NHS N1*N2 VHS

tninN1=N2= number f urnsnumber f urnset332b-7.MCD

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Example:400 urnautotranformerperating t asupplies-8kvA oad t0.gs agging r. vHS=Find .) oad urrent .) ncominJiin"urrent .)d.)apparentowerransformedndconducted

25% tap2400vIt,[ 'J* fk\ ]s Rnr16 NA.=4oot NL.=o.Li (40' )

NLs= f oo tNl_, , - {or: t78kvAFp= .Bs oi

f ,q:t-'>

L.*_*- *,-q"?,__, __-.1= l lZ;" y 'A I.*&*%

V'{ ./-*-lId eC^eos 1., 7?) - ta, n Po"., Eoicr c {F, s A ta, -- { ( a' ,i ) *. h{a. )

a-,^, : , ,11-| *'t r ,1 cL: s)

TJ. ' ' , , . , 'Jr- . - : Z, \u .. r_.\ rt \

-: (-) :*-. J; _ Ja;" J-_.-l * .= 3^*, j ,4.r/ \

CJ.J l ic* . . . . f* 'Ci . rd

.\ tk" 4AL?dlr '1

et332b-7.MCD0

ZA {du* u)d) (6. , . - ) =

T \I :- .dr, . . i v ls, f r r - : ' "y 't. vcs te

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Three-Phasenduction otorsMotorConstructionStator magnetic tructureiron ore)andwindinghatcreatemagneticield.Connectedo three-phaseoltagesRotor- ironcoreand conductorshatrotateand drive heshaftof themotor.Conductorsanbeeither opperbars(squirrel age)orwoundcoils wound-rotor)

Bt--"^,Aql (

V2,V6,andV" createluxesOs,Fluxes dd nspace nd ime ofieldwithout hysical otion.

R!I !R

conductot^ s lots

@6,?nd @screate rotatingmagnetic

et332b-7.MCD1

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FluxWaveRotates ta SpeedGiven y.[ ' l = 120' fDP

Where hs= synchronouspeedf = ac voltage requencyp = number f poles (notpolepairs)Example:Fourpolemotoroperatingn a 60 Hzsystem.What s the speedat which he magneticield otates

P ,=4 number f polesf ,=60 Hz

120.tns:= , rs=1800 RPMWhensuppliedrom60 Hzsystem, s s multiple f 60

Foran Induction otor o Rotate1.) 3-phase oltages roduceotatingmagneticield n stator2.) Currents nducedn rotorbymovingmagneticield3.) Induced urrentn rotorproduces magneticield n rotor4.) Field n rotor nteracts ith he ield n the statortoproduceorque rotor chases" tator ield)et332b-7.MCD2

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To InduceCurrentn rotor heremustbe a speeddifference etweenhe rotorand he rotatingmagneticieldSpeeddifferencealled lipspeed

n=f lS-nrWhere n=sl ipspeedfls= synchronouspeedffr= rotorspeed

Also ,=nt-nt Wheres=Peruni ts l iPnrSlip ncreasess load ncreases

SlipandDevelopedorqueAt startup trr 0. Assuming . = 1800RPM

1800 0S,= s = 1 Slip s 1 at ocked otor(startup)

s = 0.028 Rated lipsvary2-5%ofns.

1800At ull oad orquemotor pins t rated peed

nr ,=1750 RPM ypicalor2 pole nduction otor1800 1750S:=

et332b-7.MCD3

1800

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Slipat No-loadRotor pinsat nearfy ., so...

S:=

r r- =1798 RPM Typical nroadedotorspeed1800 - 1798lgOO s = 0.001 Slip s near zerowhenthere s no-loadonthemotor

Advantagesf ThreephaseInduction otors1.) smooth ower ransfer.poweralmost onstantn 3phase ystem. ulsatesin 1-fmotors2.) simpleconstructionno brushes r otherhighmaintenanceartsDisadvantages1.) Cannotcontrol peedeasily2.) Non-linearorque-speedharacterisit ic

et332b-7.MCD4

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Typical orque-Speedharacteristicf Induction otor

200

150

\ ( n) 1oo

50

f. l r I I" o 2oo 4oo 600 8oo looo 12oo14oo16001800n

Shape f orque peed haracteristicependsndesign fmotorStartingorques developed henn=0 pm. In thiscaseapproximately00N-m

et332b-7.MCD5

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SlipandRotorFrequencyndvoltageDifference etween peedof rotatingmagneticieldandrotor alled liPsPeed

n=I l S - n fwhere n=sl ipspeedSlipspeed ncreases s load ncreasesRotor requency s a function f slip

s.p.na wherehs= synchronouspeed( _-' r= eO s = p.u l ipf,= frequencyf rotorBlockrotorS-1 ,.ncf '=1/=f stator=f R

fen= blockedotor requencyAt startup tator and otor areequalInoperation notequal f r=t ' fgR

Voltagenduced E r=s'E BRmaxatstart s = 1

et332b-7.MCD6

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Model or MotorRotorMotorhasResistancend nductiveeactance. 1dependsonfso

stctonCKI

X r=2-n. r.L so

Somealgebra ivesEaR Rotor urrentependsnslipwhichs relatedomotor peed

notonCK-[

X r=2'n 's ' f gp'L ' ,In ermsof theblockedotor nductance X r=s'XBRRotor impedance Z r=Rr* j 'X r=Rr* j 's 'X BRRotorCurrent s'EBR s'EBRI r=Rr* j 'Xr Rr* j 's 'XBn

+ 'Xen

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Phase ngleofZ, depends n sl ip ( R changes),oimpedancengleandFochanges ithmotor l ip.Thismeans otor urrentmagnitudendphase ngle hangewithslip EenI r= 0 r=atan lnversetan = atan-l /*' \ ' \ . zJ\ ; ) - XER_

Xan/R Ji ' l\ ' /

Where 0,. rotor urrentRotorpoweractor F

60rrG)o r(s) 40deg

20

anglep=cos(t )

80c)o)(uc)(J(u-oq)o-EEc$cc):Jc)t-oo 0 0.6.4.2

ssl ipet332b-7.MCD8

0.8

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Air-GapPower n Induction otorsPower ransfered crossheairgap n the nductionmotor

Sn"o= EBR r.Where Ern - EenL 0o l,.= lr l--0,In rectangularorm

Sgrp=EBR' lr 'cos(t) - i 'E gR' l r 's in( t )Where Een= blockedotor oltagel,.= rotorcurrent0r= rotor mpedancengle

Also P grp=E BR' l " 'cos(t ) Q grp=E BR' l , 's in(t )WherePg"p uppliesheshaft ower,riction, indage,ndrotor esistanceosses.Qn"o reactiveowerhatoscillatescross irgap

Pg"p ependsn he otor pand hemagnitudef he rEan sassumedo beconstant ecauset s proportionalothe luxdensity hichs assumedo beconstantet332b-.7.MCD9

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Components

where

Totaf -phaseotorosses p rcl=3.1

ActivepowerAcrossAir_gapPg"p=Pmech + Prcf

= activepower onvertedo shaftrotor onductorosses

, . ' ' * , .

xn

DI mechDI rclpower

motor.gap

f-.)N

stotorcktTotafgappower

Slip elatedo theMoremechanicafamount fmechanicaload nloadmoreactive ower cross

notcrckt

3'R , ' l , '' gap=

et332b-7.MCD20

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Combine owerbalance quations ithdefinit ions f Psapand Pr"g3.r,2.R . (1 - s)P mech=

Rotor esistanceffectsheamount f mechanicalowerdevelopedRr

Divide into woparts:resistancehatloadRr.(1 - s)

rotor ossresistance ndrepresents echanical

Rr

R.(1-s)

EI" rS