负载运行的变压器中英文翻译

TheTransformeronloadIthasbeenshownthataprimaryinputvoltageVcanbetransformedtoanyidesiredopen-circuitsecondaryvoltageEbyasuitablechoiceofturn‘ESratio.22availableforcirculatingaloadcurrentimpedance.Forthemoment,alaggingpowerfactorwillbeconsidered.Thesecondarycurrentandtheresultingampere-turnsINwillchangetheflux,tendingtodemagnetizethecore,reduce①andwithit2 2 mE.Becausetheprimaryleakageimpedancedropissolow,asmallalterationtoE11willcauseanappreciableincreaseofprimarycurrentfromItoanewvalueofI01equalto（V+E）/（R+jX）.Theextraprimarycurrentandampere-turnsnearly111 icancelthewholeofthesecondaryampere-turns.Thisbeingso,themutualfluxsuffersonlyaslightmodificationandrequirespracticallythesamenetampere-turnsINasonnoload.Thetotalprimaryampere-turnsareincreasedbyanamount01INnecessarytoneutralizethesameamountofsecondaryampere-turns.Inthe22vectorequation,IN+1N=IN;alternatively,IN=IN-1N.Atfullload,112201110122thecurrentIisonlyabout5%ofthefull-loadcurrentandsoIisnearlyequal01toIN/N.BecauseinmindthatE=EN/N,theinputkVAwhichis2211212approximatelyEIisalsoapproximatelyequaltotheoutputkVA,EI.1122Thephysicalcurrenthasincreased,andwithintheprimaryleakagefluxtowhichitisproportional.Thetotalfluxlinkingtheprimary,①二①+①二①isp m 1 11shownunchangedbecausethetotalbacke.m.f.,（E一Nd①/dt）isstillequaland111oppositetoV.However,therehasbeenaredistributionoffluxandthemutualcomponenthasfallenduetotheincreaseof①withI.Althoughthechangeis11small,thesecondarydemandcouldnotbemetwithoutamutualfluxande.m.f.alterationtopermitprimarycurrenttochange.Thenetflux①linkingthesecondaryswindinghasbeenfurtherreducedbytheestablishmentofsecondaryleakagefluxduetoI,andthisopposes①.Although①and①areindicatedseparately,theyTOC\o"1-5"\h\z2 mm 2combinetooneresultantinthecorewhichwillbedownwardsattheinstantshown.ThusthesecondaryterminalvoltageisreducedtoV--Nd①/dtwhichcanbe2 2Sconsideredintwocomponents,i.e.V=-Nd①/dt-Nd①/dtorvectorially2 2m 2 2V=E一jXI.Asfortheprimary,①isresponsibleforasubstantiallyconstant2 2 22 2secondaryleakageinductanceN①/i=N2A.Itwillbenoticedthattheprimary222 2 2leakagefluxisresponsibleforpartofthechangeinthesecondaryterminalvoltageduetoitseffectsonthemutualflux.Thetwoleakagefluxesarecloselyrelated;①,2forexample,byitsdemagnetizingactionon①hascausedthechangesonthemprimarysidewhichledtotheestablishmentofprimaryleakageflux.Ifalowenoughleadingpowerfactorisconsidered,thetotalsecondaryfluxandthemutualfluxareincreasedcausingthesecondaryterminalvoltagetorisewithload.①isunchangedinmagnitudefromthenoloadconditionsince,neglectingpresistance,itstillhastoprovideatotalbacke.m.f.equaltoV.Itisvirtuallythe1sameas①,thoughnowproducedbythecombinedeffectofprimaryandii J ■ ■secondaryampere-turns.ThemutualfluxmuststillchangewithloadtogiveachangeofEandpermitmoreprimarycurrenttoflow.EhasincreasedthistimebutduetothevectorcombinationwithVthereisstillanincreaseofprimary

current.Twomorepointsshouldbemadeaboutthefigures.Firstly,aunityturnsratiohasbeenassumedforconveniencesothatEhasbeenassumedforconveniencesothatE=E'.Secondly,thephysicalpicture12isdrawnforadifferentinstantoftimefromthevectordiagramswhichshow①=0,ifthehorizontalaxisistakenasusual,tobethezerotimereference.Theremareinstantsinthecyclewhenprimaryleakagefluxiszero,whenthesecondaryleakagefluxiszero,andwhenprimaryandsecondaryleakagefluxiszero,andwhenprimaryandsecondaryleakagefluxesareinthesamesense.Theequivalentcircuitalreadyderivedforthetransformerwiththesecondaryterminalsopen,caneasilybeextendedtocovertheloadedsecondarybytheadditionofthesecondaryresistanceandleakagereactance.Practicallyalltransformershaveaturn'sratiodifferentfromunityalthoughsuchanarrangementissometimesemployedforthepurposesofelectricallyisolatingonecircuitfromanotheroperatingatthesamevoltage.Toexplainthecasewherethereactionofthesecondarywillbeviewedfromtheprimarywinding.Thereactionisexperiencedonlyintermsofthemagnetizingforceduetothesecondaryampere-turns.ThereisnowayofdetectingfromtheprimarysidewhetherIislargeandNsmallorviceversa,itistheproductofcurrentandturnswhich22causesthereaction.Consequently,asecondarywindingcanbereplacedbyanynumberofdifferentequivalentwindingsandloadcircuitswhichwillgiverisetoanidenticalreactionontheprimary.ItisclearlyconvenienttochangethesecondarywindingtoanequivalentwindinghavingthesamenumberofturnsNasthe1primary.WithN2changestoWithN21

E'=(N/N)EwhichisthesameasE.21221Forcurrent,sincethereactionampereturnsmustbeunchangedI'N'=I'N2221mustbeequaltoIN.i.e.I=(N/N)I.222212becomes(N/becomes(N/N)V,and12secondarycurrentIbecomes(N/N)secondarycurrentIbecomes21loadimpedance,mustbecome V'/1'=(N/N)2V/1.Consequently,loadimpedance,1 2R2'=(N/N)2RandX'=(N/N)2X.1 2 2 2 1 2 2Iftheprimaryturnsaretakenasreferenceturns,theprocessiscalledreferringtotheprimaryside.Thereareafewcheckswhichcanbemadetosee讦theprocedureoutlinedisvalid.Forexample,thecopperlossinthereferredsecondarywindingmustbethesameasintheoriginalsecondaryotherwisetheprimarywouldhavetosupplyadifferentloss(I•N/N)2(R2 2 1 2SimilarlytheproportionaltoI2TOC\o"1-5"\h\zdifferentloss(I•N/N)2(R2 2 1 2SimilarlytheproportionaltoI22 2 2 2•N2/N2)doesinfactreduceto12R.1 2 2 2storedmagneticenergyintheleakagefield(1/2LI2)whichis'XwillbefoundtocheckasI'X'.Thereferredsecondary2 2 2kVA=E'I'=E(N/N)•I(N/N)=EI.2 2 2 1 2 2 2 1 22Theargumentissound,thoughatfirstitmayhaveseemedsuspect.Infact,讦theactualsecondarywindingwasremovedphysicallyfromthecoreandreplacedbytheequivalentwindingandloadcircuitdesignedtogivetheparametersN,R',X'and122I',measurementsfromtheprimaryterminalswouldbeunabletodetectanydifferenceinsecondaryampere-turns,kVAdemandorcopperloss,undernormalpowerfrequencyoperation.Thereisnopointinchoosinganybasisotherthanequalturnsonprimaryandreferredsecondary,butitissometimesconvenienttorefertheprimarytothesecondarywinding.Inthiscase,ifallthesubscriptl'sareinterchangedforthesubscript2's,thenecessaryreferringconstantsareeasilyfound;e.g.R'沁R,X'〜X;similarlyR'沁RandX'〜X.12122121TheequivalentcircuitforthegeneralcasewhereN丰Nexceptthatrhas1 2 mbeenaddedtoallowforironlossandanideallosslesstransformationhasbeenincludedbeforethesecondaryterminalstoreturnV'toV.Allcalculationsof22internalvoltageandpowerlossesaremadebeforethisidealtransformationisapplied.Thebehaviorofatransformerasdetectedatbothsetsofterminalsisthesameasthebehaviordetectedatthecorrespondingterminalsofthiscircuitwhentheappropriateparametersareinserted.TheslightlydifferentrepresentationshowingthecoilsNandNsidebysidewithacoreinbetweenisonlyusedforconvenience.Onthe12transformeritself,thecoilsare,ofcourse,woundroundthesamecore.Verylittleerrorisintroducedifthemagnetizingbranchistransferredtotheprimaryterminals,butafewanomalieswillarise.Forexample,thecurrentshownflowingthroughtheprimaryimpedanceisnolongerthewholeoftheprimarycurrent.TheerrorisquitesmallsinceIisusuallysuchasmallfractionofI.Slightly01differentanswersmaybeobtainedtoaparticularproblemdependingonwhetherornotallowanceismadeforthiserror.Withthissimplifiedcircuit,theprimaryandreferredsecondaryimpedancescanbeaddedtogive:Re二R+R(N/N)2AndXe二X+X(N/N)21121211212Itshouldbepointedoutthattheequivalentcircuitasderivedhereisonlyvalidfornormaloperationatpowerfrequencies;capacitanceeffectsmustbetakenintoaccountwhenevertherateofchangeofvoltagewouldgiverisetoappreciablecapacitancecurrents,I=CdV/dt.Theyareimportantathighvoltagesandatcfrequenciesmuchbeyond100cycles/sec.Afurtherpointisnottheonlypossibleequivalentcircuitevenforpowerfrequencies.Analternative,treatingthetransformerasathree-orfour-terminalnetwork,givesrisetoarepresentationwhichisjustasaccurateandhassomeadvantagesforthecircuitengineerwhotreatsalldevicesascircuitelementswithcertaintransferproperties.Thecircuitonthisbasiswouldhaveaturnsratiohavingaphaseshiftaswellasamagnitudechange,andtheimpedanceswouldnotbethesameasthoseofthewindings.Thecircuitwouldnotexplainthephenomenawithinthedeviceliketheeffectsofsaturation,soforanunderstandingofinternalbehavior.Therearetwowaysoflookingattheequivalentcircuit:viewedfromtheprimaryasasinkbutthereferredloadimpedanceconnectedacrossV',or2ViewedfromthesecondaryasasourceofconstantvoltageVwithinternal1dropsduetoReandXe.Themagnetizingbranchissometimesomittedinthis11representationandsothecircuitreducestoageneratorproducingaconstantvoltageE(actuallyequaltoV)andhavinganinternalimpedanceR+jX(actuallyequaltoRe+jXe).1丿1Ineithercase,theparameterscouldbereferredtothesecondarywindingandthismaysavecalculationtime.Theresistancesandreactancescanbeobtainedfromtwosimplelightloadtests.负载运行的变压器通过选择合适的匝数比，一次侧输入电压V可任意转换成所希望的二次侧1开路电压E。E可用于产生负载电流，该电流的幅值和功率因数将由而次侧电22路的阻抗决定。现在，我们要讨论一种滞后功率因数。二次侧电流及其总安匝IN将影响磁通，有一种对铁芯产生去磁、减小①和E的趋向。因为一次侧2 2 m 1漏阻抗压降如此之小，所以E的微小变化都将导致一次侧电流增加很大，从I增10大至一个新值I=（V+E）/（R+jX）。增加的一次侧电流和磁势近似平衡了全1111 i部二次侧磁势。这样的话，互感磁通只经历很小的变化，并且实际上只需要与空载时相同的净磁势IN。一次侧总磁势增加了IN，它是平衡同量的二次侧0122磁势所必需的。在向量方程中，IN+1N=IN，上式也可变换成112201IN=IN-1N。满载时，电流I只约占满载电流的5%，因而I近似等于11012201IN/N。记住E=EN/N，近似等于EI的输入容量也就近似等于输出容221121211量EI。22一次侧电流已增大，随之与之成正比的一次侧漏磁通也增大。交链一次绕组的总磁通①二①十①=①没有变化，这是因为总反电动势E-Nd①/dtpm 1 11 1 11仍然与V相等且反向。然而此时却存在磁通的重新分配，由于①随I的增加而111增加，互感磁通分量已经减小。尽管变化很小，但是如果没有互感磁通和电动势的变化来允许一次侧电流变化，那么二次侧的需求就无法满足。交链二次绕组的净磁通①由于I产生的二次侧漏磁通（其与①反相）的建立而被进一步s 2 m削弱。尽管图中①和①是分开表示的，但它们在铁芯中是一个合成量，该合m 2成量在图示中的瞬时是向下的。这样，二次侧端电压降至V=-Nd①/dt，它2 2S可被看成两个分量，即V二-Nd①/dt-Nd①/dt,或者向量形式2 2m 2 2V=E-jXI。与一次侧漏磁通一样，①的作用也用一个大体为常数的漏电22222感N①/i二N2A来表征。要注意的是，由于它对互感磁通的作用，一次侧漏22222磁通对于二次侧端电压的变化产生部分影响。这两种漏磁通，紧密相关；例如,①对①的去磁作用引起了一次侧的变化，从而导致了一次侧漏磁通的产生。2 m如果我们讨论一个足够低的超前功率因数，二次侧总磁通和互感磁通都会增加，从而使得二次侧端电压随负载增加而升高。在空载情形下，如果忽略电阻，①幅值大小不变，因为它仍提供一个等于V的反总电动势。尽管现在①是p 1 p一次侧和二次侧磁势的共同作用产生的，但它实际上与①相同。互感磁通必须11仍随负载变化而变化以改变E，从而产生更大的一次侧电流。此时E的幅值已11经增大，但由于E与V是向量合成，因此一次侧电流仍然是增大的。11从上述图中，还应得出两点：首先，为方便起见已假设匝数比为1，这样可使E=E'。其次，如果横轴像通常取的话，那么向量图是以①二0为零时间1 2 m参数的，图中各物理量时间方向并不是该瞬时的。在周期性交变中，有一次侧漏磁通为零的瞬时，也有二次侧漏磁通为零的瞬时，还有它们处于同一方向的瞬时。已经推出的变压器二次侧绕组端开路的等效电路，通过加上二次侧电阻和漏抗便可很容易扩展成二次侧负载时的等效电路。实际中所有的变压器的匝数比都不等于1,尽管有时使其为1也是为了使一个电路与另一个在相同电压下运行的电路实现电气隔离。为了分析NHN时的12情况，二次侧的反应得从一次侧来看，这种反应只有通过由二次侧的磁势产生磁场力来反应。我们从一次侧无法判断是I大，N小,还是I小,N大，正2222是电流和匝数的乘积在产生作用。因此，二次侧绕组可用任意个在一次侧产生相同匝数N的等效绕组是方便的。1当N变换成N,由于电动势与匝数成正比，所以E'=(N/N)E,与E相2121221等。对于电流，由于对一次侧作用的安匝数必须保持不变，因此I'N'=I'N=IN，即I=(N/N)I。2221222212对于阻抗，由于二次侧电压V变成(N/N)V，电流I变为(N/N)I，因1221此阻抗值，包括负载阻抗必然变为V'//'二(N/N)2V/1。因此，12R2'二(N/N)2R，X'二(N/N)2X。1222122如果将一次侧匝数作为参考匝数，那么这种过程称为往一次侧的折算。我们可以用一些方法来验证上述折算过程是否正确。例如，折算后的二次绕组的铜耗必须与原二次绕组铜耗相等，否则一次侧提供给其损耗的功率就变了。I'2R'必须等于12R，而2222(I•N/N)2(R•N2/N2)事实上确实简化成了12R。22121