Project Report S ST TU UD DY Y A AN ND D D DE ES SI IG GN N O OF F P PO OW WE ER R T TR RA AN NS SF FO OR RM ME ER R U UN ND DE ER R S SH HO OR RT T C CI IR RC CU UI IT T C CO ON ND DI IT TI IO ON NS S Submitted by: MANGESH DILIP ALEKAL (D050302) GAURAV SANJIV DEWAL (D050309) KSHITEESH VINAYAK KANEGAONKAR (D050320) OMKAR DEEPAK LIMAYE (D050330) KEDAR NANDAN RAJE(D050347) Programme: B. Tech (Electrical Engineering) Year: 2008-09 Guided by: Dr. M. S. PANSE Department of Electrical Engineering Veermata Jijabai Technological Institute (Autonomous Institute Affiliated to the University of Mumbai) Mumbai 400019 Statement by the Candidates We wish to state that work embodied in this report titled Study and Design of PowerTransformerunderShortCircuitConditionsformstheGroup's contribution to the work carried out under the guidance ofDr M. S. PANSE at theVeermataJijabaiTechnologicalInstitute.Thisworkhasnotbeen submittedforanyotherDegreeorDiplomaofanyUniversity/Institute. Whereverreferenceshavebeenmadetopreviousworksofothers,ithasbeen clearly indicated. Name of CandidatesSignature MANGESH DILIP ALEKAL _____________________ GAURAV SANJIV DEWAL _____________________ KSHITEESH VINAYAK KANEGAONKAR_____________________ OMKAR DEEPAK LIMAYE _____________________ KEDAR NANDAN RAJE _____________________ CERTIFICATE This is to certify that the following students: MANGESH DILIP ALEKAL (D050302) GAURAV SANJIV DEWAL (D050309) KSHITEESH VINAYAK KANEGAONKAR (D050320) OMKAR DEEPAK LIMAYE (D050330) KEDAR NANDAN RAJE(D050347) havecompletedsatisfactorily,thedissertationentitledStudyandDesignof PowerTransformerunderShort-CircuitConditionstowardspartial fulfillmentofthedegreeofBachelorofTechnologyattheVeermataJijabai Technological Institute. Prof. M. S. PanseDr. B. K. Lande Project Guide Head of Electrical Engg. Dept. V.J.T.I., Mumbai-19

CERTIFICATE The dissertation entitled 'Study and Design of Power Transformer under Short-Circuit Conditions' submitted by the following students: MANGESH DILIP ALEKAL(D050302)GAURAV SANJIV DEWAL(D050309) KSHITEESH VINAYAK KANEGAONKAR(D050320) OMKAR DEEPAK LIMAYE(D050330) KEDAR NANDAN RAJE (D050347) is found to be satisfactory and is approved for the degree of B.Tech. (Electrical Engineering) of the University of Mumbai. GuideExaminer (Dr. M. S. Panse) ACKNOWLEDGEMENT We,theaforementioned,fromB.Tech.(ElectricalEngineering)takegreat pleasureinpresentinginpresentingthisreporttitled'StudyandDesignof PowerTransformerunderShort-Circuitconditions'.Wewouldliketotake this opportunity to thank Dr. B.K. Lande, Head of the Department of Electrical Engineering,VeermataJijabaiTechnologicalInstitute,whopermittedusto undertake this project. We would like to convey our deepest appreciation for Dr. M. S. Panse from the Electrical Engineering Department, whose continued support and motivation, has allowed us to express ourselves and helped us to complete this project within the stipulated time. We would like to thank all the college officials for their co-operation. AspecialvoteofthanksgoesouttoMr.ShekharVora,fromCrompton GreavesLtd.,whoseexcellentguidanceandvaluableinputs,helpedmakethis project a reality. Wewouldalsoliketoexpressourdeepestgratitudetoallthepeople,who have directly or indirectly, influenced the completion of this project.

CHAPTER 1 ..................................................................................................................... 1 INTRODUCTION (1) .................................................................................................... 2 CHAPTER 2 ..................................................................................................................... 2 CONSTRUCTION OF TRANSFORMER ................................................................... 4 2.1 WINDINGS ........................................................................................................ 6 2.2 NEED FOR CALCULATION OF TRANSFORMER IMPEDANCE ............ 11 2.3 IMPEDANCE CALCULATION ...................................................................... 12 2.4 RADIAL FORCES (2, 3) .................................................................................... 18 2.5 AXIAL FORCES (5) .......................................................................................... 27 2.6 EFFECT OF PRESTRESS (14, 15,16) ................................................................... 36 2.7 STRESSES IN FLITCH PLATES AND FRAMES ......................................... 37 CHAPTER 3 ................................................................................................................... 38 TRANSFORMER DESIGN PROCEDURE .............................................................. 38 3.1 DIFFERENT PARAMETERS CONSIDERED IN TRANSFORMER DESIGN (17) ............................................................................................................. 38 CHAPTER 4 ................................................................................................................... 55 DESIGNED TRANSFORMER .................................................................................. 55 4.1 PROBLEM STATEMENT ............................................................................... 55 4.2 PARAMETERS OF DESIGNED TRANSFORMER ...................................... 56 REFERENCES ............................................................................................................... 76 ABSTRACT Theprojectdealswiththestudyofthepracticalpowertransformerasusedin transformerindustriesanditsdesigntosuccessfullywithstandsevereshortcircuit conditions. Highlighting the basics of a power transformer (like emf equation, windings, core), we designedapracticaltransformerusingdesignspecificationsprovidedtousby CromptonGreavesLtd.likeMVAratingoftransformer,corediameter,%impedance restrictions, core and copper loss limits, constraint on the price of the transformer. Afterhavingsuccessfullydesignedapowertransformeraccordingtothegiven specificationsandmeetingtheloss,impedanceandpriceconstraints,weproceededto analyzing the short circuit withstand capacity of the designed transformer. Theshortcircuitanalysisinvolvescalculationoffirstpeakofshortcircuitcurrent, calculationofradialandaxialstressesonthewindingsandontheinsulationblocks placed in between windings. Theradialandaxialstressescalculatedwerefoundtobewellwithinthestipulated limitsgiveninthedesignproblemandhence,weconcludethatourtransformerhas been successfully passed to withstand short circuit conditions. CHAPTER 1 INTRODUCTION (1) Intherecentdevelopmentofthepowersystems,increaseinthepowerplant capacity has been a major achievement in the power industry. This power is transmitted at a voltage which is much higher than the generated voltage as the power loss at high voltage and low current is lesser than that at low voltage and high current. Hence, it is necessarytostepupthevoltagelevelofthegeneratedvoltagetoamuchhighervalue fortransmissionpurposesinordertoavoidextensivepowerloss.Forthispurpose, power transformers are widely used in the industry. Addition of more generating capacity and interconnections in the power systems havecontributedtoanincreaseintheshortcircuitcapacityofnetworks,makingthe shortcircuitdutyofpowertransformersmoresevere.Failureoftransformersdueto short circuits is an area of major concern for transformer manufacturers and consumers. Therearecontinuouseffortsbymanufacturersandconsumerstoimprovetheshort circuit withstand performance of transformers. Theshortcircuitstrengthofapowertransformerenablesittosurvivethrough faultcurrentsduetoexternalshortcircuitsinapowersystemnetwork.Aninadequate strengthmayleadtoamechanicalcollapseofwindings,deformation/damageto clampingstructures,oranelectricalfaultinthetransformeritself.Theinternalfaults initiatedbytheexternalshortcircuitsaredangerousastheymayinvolveblow-outof bushings,firehazardsandmayeventuallyleadtoburstingoftheentiretransformer structure. In recent short circuit tests on power transformers greater than 100 MVA, it has beenseenthatincreaseofshortcircuitinductancebeyond1%hascausedsignificant deformationinwindings.Amuchstrictercontrolonthevariationsinmaterialsand manufacturingprocesseswillhavetobeexercisedtoavoidloosenessandwinding movements. Hence, the causes of internal short-circuits in the transformer are inherently related to the design parameters of the transformer. Thus,itisalwaysdesirabletodesignatransformerfornormalworking conditions and then make variations to it in order to account for the forces developed on transformer winding during short-circuit condition. CHAPTER 2 CONSTRUCTION OF TRANSFORMER Atransformerisastaticdevicethattransferselectricalenergyfromonecircuitto another by electromagnetic induction without change in frequency. Apartofacore,whichissurroundedbywindings,iscalledalimborleg.Remaining partofthecore,whichisnotsurroundedbywindings,butisessentialforcompleting thepathofflux,iscalledasyoke.Thistypeofconstruction(termedascoretype)is morecommonandhasthefollowingdistinctadvantages:viz.constructionissimpler, cooling is better and repair is easy. Fig 2.1 Core type transformer Shell-typeconstruction,inwhichacrosssectionofwindingsintheplaneofcoreis surroundedbylimbsandyokes,isalsoused.Ithastheadvantagethatonecanuse sandwichconstructionofLVandHVwindingstogetverylowimpedance,ifdesired, which is not easily possible in the core-type construction.

Fig 2.2 Shell type transformer Thecoreconstructionmainlydependsontechnicalspecifications,manufacturing limitations,andtransportconsiderations.Itiseconomicaltohaveallthewindingsof threephasesinonecoreframe.Athree-phasetransformerischeaper(byabout20to 25%) than three single-phase transformers connected in a bank. But from the spare unit consideration,usersfinditmoreeconomicaltobuyfoursingle-phasetransformersas compared to two three-phase transformers. Also, if the three-phase rating is too large to bemanufacturedintransformerworks(weightsanddimensionsexceedingthe manufacturingcapability)andtransported,thereisnooptionbuttomanufactureand supply single-phase units. 2.1 WINDINGS Thetransformerconsistsoftwocoilscalledwindingswhicharewrappedarounda core.Thetransformeroperateswhenasourceofacvoltageisconnectedtooneofthe windings and a load device is connected to the other. The winding that is connected to the source is called the primary winding. The winding that is connected to the load is called the secondary winding. The conducting material used for the windings depends upon the application, but in all casestheindividualturnsmustbeelectricallyinsulatedfromeachothertoensurethat the current travels throughout every turn. The insulation system must be designed to withstand the effects of lightning strikes and switchingsurgestowhichthetransformerissubjected,inadditiontothenormal operatingvoltages.Afurtherrequirementoftheinsulationsystemisthatitmust withstandtheenvironmentalconditionstowhichitisexposed,suchasmoisture,dust etc.Avarietyoftechniquesandmaterialsareemployedtoachievethenecessary performance characteristics of the insulation system. 2.1.1 LAYER WINDING (a)(b) Fig 2.3 a) Cross section of layer windingb) Transformer with layer wound coils Forlowvoltage,i.e.600Voltclasswindings,thewindingtechniqueusedalmost exclusivelyisthelayerwindingtechnique,alsosometimescalledhelicalwindingor barrel winding. In this technique, the turns required for the winding are wound in one or moreconcentriclayersconnectedinseries,withtheturnsofeachlayerbeingwound sidebysidealongtheaxiallengthofthecoiluntilthelayerisfull.Theconductorsof thewindingareinsulatedandsobetweenturnstherewillbeaminimumoftwo thicknesses of insulation. Between each pair of layers there will be layers of insulation material and/or an air duct. Low voltage windings will generally be wound top to bottom, bottom to top etc. using a continuousconductor,untilalllayersarecomplete.Highvoltagewindings,i.e.above 600 Volt class, may be wound in the same way, provided the voltage between layers is nottoogreat.Toreducethevoltagestressbetweenlayers,highvoltagewindingsare often wound in only one direction, for example, top to bottom. When the first layer of windingiscomplete,thewindingconductorislaidacrossthecompletedlayerfrom bottom to top and then the next layer is wound, again from top to bottom. In this way, thevoltagestressbetweenlayersishalved.Theconductormust,ofcourse,have additional insulation where it crosses the winding from bottom to top. 2.1.1.1 CHARACTERISTICS OF LAYER WOUND COILS Asstatedpreviously,thelayerwoundcoilrequiresinsulationbetweenlayers,in additiontotheconductorinsulation.Thethicknessofinsulationrequiredwilldepend uponthevoltagestressbetweenlayers,andcomprisesoneormorethicknessesofthe appropriateinsulationmaterial.Inpractice,duetothenatureoftheconstructionofa layerwoundcoil,thefinishedcoilwillhaveseveralunavoidablesmallairpockets betweenturnsandbetweenlayers.Manyoftheseairpocketswillbecomefilledwith resin during vacuum pressure impregnation of the coil. However, it sometimes happens thatsomeairpocketsremainanditisintheseairpocketsthatpartialdischargescan occur,greatlyincreasingthepossibilityofprematureagingoftheinsulationand eventualfailure.Undershortcircuitconditions,theelectromagneticforcesdeveloped causetransformerwindingstoattempttotelescope.Atthesametimethecoilend blocking is trying to prevent movement. The result is often that the turns of the winding have a tendency to slip over one another, causing turn-to turn failure, due to abrasion of the insulation as the turns rub together. A further disadvantage of the layer wound coil is its poor impulse voltage distribution between the first few turns of the winding, due to thehighgroundcapacitanceandthelowseriescapacitance.Atransformerwinding formsacomplexnetworkofresistance,inductanceandcapacitance.Asfarasthe impulsevoltagedistributionisconcerned,theresistancecanbeignoredandatthe instant of application of the impulse wave, when very high frequencies are predominant, theinductiveelementsbecomeeffectivelyinfiniteimpedances.Thewholestructure thereforereducestoacapacitivenetwork.Eachturnofatransformerwindingis insulated with a dielectric material and can be thought of as one plate of a multiple plate capacitor.Inaddition, the combination of dielectric material and air between each turn and ground forms further capacitive elements. 2.1.2 DISC WINDING (a)(b) Fig 2.4 a) Cross section of disc windingb) Transformer with disc winding Inthediscwinding,therequirednumberofturnsiswoundinanumberofhorizontal discsspacedalongtheaxiallengthofthecoil.Theconductorisusuallyrectangularin cross-sectionandtheturnsarewoundinaradialdirection,oneontopoftheotheri.e. oneturnperlayer,untiltherequirednumberofturnsperdischasbeenwound.The conductoristhenmovedtothenextdiscandtheprocessrepeateduntilallturnshave been wound. There is an air space, or duct, between each pair of discs. The disc winding requiresinsulationonlyontheconductoritself,noadditionalinsulationisrequired between layers, as in the layer winding. The disc wound high voltage winding is usually wound in two halves, in order that the requiredvoltageadjustmenttapsmaybepositionedattheelectricalcenterofthe winding.Inthiswaythemagneticoreffectivelengthofthewindingismaintained, irrespectiveofwhichtapisused,andthereforethemagneticbalancebetweenprimary and secondary windings is always close to its optimum. This is essential to maintain the short circuit strength of the winding, and reduces the axial electromagnetic forces which arise when the windings are not perfectly balanced. 2.1.2.1 CHARACTERISTICS OF DISC WOUND COILS Themajoradvantageofthediscwoundcoilliesinitsopenconstructionandrelative lackofinsulation.Fora15kVclasstransformeremployingadiscwoundprimary winding,thenumberofdiscswilltypicallybeintherange36to48,resultingina relativelylowvoltageperdisc.Sinceeachdiscisseparatedfromthenextbyanair space,thevoltagestressbetweendiscscaneasilybehandledbythecombinationof conductorinsulationandair,noadditionalinsulationbeingnecessary.Eachdisc comprises a number of turns with each turn occupying one layer, i.e. one turn per layer: thevoltagestressbetweenlayersisthereforethesameasthevoltagestressbetween turnsandagain,caneasilybehandledbytheconductorinsulation.Theturnsofeach disc,beingwoundtightlytogetherprovidealmostnopossibilityofairpocketsbeing present within the disc. Due to the open construction of the discs, any small air pockets which may be present are readily filled with resin during vacuum pressure impregnation ofthecoil.Aproperlydesignedandmanufactureddry-typetransformerdiscwinding therefore displays very low values of partial discharge, typically intherangeof 10 to 20Picocoulombs.Unlikethelayerwoundcoil,thediscwoundcoilprovidesgood impulse voltage distribution, due to its inherently low value of ground capacitance and highseriescapacitance.Thediscwoundcoilalsodisplaysexcellentshortcircuit strength. Each disc by itself is mechanically very strong and the completeassembly of discsisheldverysecurelyinplace.Whiletheelectromagneticforcesresultingfroma short circuit result in tendency, for the windings to telescope, high voltage turns usually remain intact relative to each other. Instead, the complete disc has a tendency to distort asassembly,withalltheturnsdistortingbysameamount.Thetransformercanoften continue to function, despite the distortion, until a convenient time arises for repair. 2.2 NEED FOR CALCULATION OF TRANSFORMER IMPEDANCE Therearebasicallytwotypesofforcesinwindings:axialandradialelectromagnetic forcesproducedbyradialandaxialleakagefieldsrespectively.Theaveragestresson windings due to the radial electromagnetic forces acting on them is given below. For a copper conductor, avgo = 0.48 10 4 (k 2 )2 2pu WRZ HP kg / cm2 (2.1) where, Zpu = Per-unit impedance of the transformer Hw = Height of winding in meters PR = Copper loss per phase in transformer k = asymmetry factor For an aluminium conductor, avgo = 0.29 10 4 (k 2 )2 2pu WRZ HP kg / cm2 ... (2.2) Now,thetotalaxialcompressiveforceactingonthewindingsisgivenbythe expression, Fa= f H ZSw pu 8 . 50 kg(2.3) where,f = Frequency in kHz S = Rated power per limb in kVA Hence,wecanseethatbothforcecomponentsonthewindingaredependentonthe impedanceofthetransformer.Thus,weshallperformthecalculationoftransformer impedance in detail. 2.3 IMPEDANCE CALCULATION For a power transformer, the ratio of reactance to resistance is greater than 20. So its impedance is generally taken as equal to its reactance.Its per- unitvalue can be calculated as follows: Foruniformlydistributedampere-turnsalongLV&HVwindingswithequal heights,theleakagefluxispredominantlyaxialexceptatwindingends,wheretheres fringing.Thisisbecauseleakagefluxfindsashorterpathtoreturnvialimboryoke. Hence the equivalent height (Heq) can be obtainedby dividing winding height (Hw) by Rogowski factor KR, which is less than one. Fig 2.5

Heq= Hw/ KR...(2.4) Theleakagemagnetomotiveforce(mmf)distributionacrossthecrosssectionof windingsisoftrapezoidalform.Themmfatanypointdependsontheampere-turns enclosed by a flux contour at that point; it increases linearly with the ampere-turns from a value of zero at the inside diameter of LV winding to the maximum value of one per-unit(totalampere-turnsofLVorHVwinding)attheoutsidediameter.Inthegap between LV and HV windings, since flux contour at any point encloses full LV (or HV) ampere-turns,themmfisofconstantvalue.Themmfstartsreducinglinearlyfromthe maximumvalueattheinsidediameteroftheHVwindingandapproacheszeroatits outsidediameter.Thecoreisassumedtohaveinfinitepermeabilityrequiringno magnetizingmmf,andhencetheprimaryandsecondarymmfsexactlybalanceeach other. The flux density distribution is of the same form as that of the mmf distribution. Sincethecoreisassumedtohavezeroreluctance,nommfisexpendedinthereturn path through it for any contour of flux. (a) Flux Tube (b) MMF Diagram Fig 2.6 Hence,foraclosedcontouroffluxatadistancexfromtheinsidediameterofLV winding, it can be written that ( 0xB) Heq= (NI)x. .(2.5) i.e. Bx= eqxHNI ) (0...(2.6) Forderivingtheformulaforreactance,letusderiveageneralexpressionfortheflux linkagesofafluxtubehavingradialdepthRandheightHeq.Theampere-turns enclosed by a flux contour at the inside diameter (ID) and outside diameter (OD) of this flux tube are a(NI) and b(NI) respectively. Bx= eqH0 [ { a + Ra b ) ( x }NI ] ... (2.7)

The flux linkages of an incremental flux tube of width dx placed at x are

d= Nx |x= NxBxA .. (2.8) where, A is the area of flux tube given by A =t(ID + 2x) dx.......(2.9) d= [ { a + Ra b ) ( x }N ] [eqH0 [ { a + Ra b ) ( x }NI ]] [t (ID + 2x) dx] (2.10) Hence the total flux linkages of the flux tube are given by = }Rd0= eqHI N20t }R0 { a + Ra b ) ( x }2 ( ID + 2x ) dx (2.11) =eqHI N20t 3R[ ( a2+ ab + b2) ID + 23 ) (2 2R b ab a + +- 222ab a +R ]..(2.12) Thelastterminsquarebracketcanbeneglectedwithoutintroducinganappreciable error to arrive at a simple formula for the regular design use. =eqHI N20t 3R ( a2+ ab + b2) [ID + 1.5 R](2.13) Thelastterm[ID+canbetakentobeapproximatelyequaltothemeandiameter (Dm)ofthefluxtube(forlargediametersofwindings/gapswithcomparativelylower values of their radial depths). =eqHI N20t 3R ( a2+ ab + b2) Dm..(2.14) Now let, ATD = 3R ( a2+ ab + b2) Dm .....(2.15) which corresponds to the area of Ampere-Turn Diagram. The leakage inductance of a transformer with n flux tubes can now be given as L = Ink=1 = eqHN20t =nkATD1 . (2.16) And the corresponding expression for the leakage reactance X is given by X = 2t f eqHN20t =nkATD1........... (2.17) For the base impedance Zb the formula for percentage leakage reactance is % X = bZX = VIX 100 = 2t f V HINeq20t =nkATD1 100... (2.18) = 2t f ) / () (0N V HNIeqt =nkATD1 100 (2.19) where, V is rated voltage and(V/N) is volts/turn of the transformer. Substituting0=410-7andadjustingconstantssothatthedimensionsusedinthe formula are in units of centimeters (Heq in cm and ATD in cm2), % X = 2.48 10 5 f ) / () (Turn Volts Hs AmpereTurneq=nkATD1 (2.20) Afterhavingderivedthegeneralformula,wewillnowapplyitforasimplecaseofa twowindingtransformer.Theconstantsaandbhavethevaluesof0and1forLV,1 and 1 for gap, and 1 and 0 for HV respectively. If 2 , 1D D and gD are the mean diameters and 2 , 1T T and gT aretheradialdepthsofLV,gapandHVrespectively,usingequation 3.12 we get ATD = 31 ( T1 D1) + ( Tg Dg) + 31( T2 D2)..(2.21) ThevalueofHeqiscalculatedbyequation2.4,forwhichtheRogowskifactorKR is given by ) (112 1)2 1(T T THeKgwTgT TwHR+ + =+ +tt (2.22) Fortakingintoaccounttheeffectofcore,amoreaccuratebutcomplexexpressionfor KR can be used. For most of the cases, equation 2.22 gives sufficiently accurate results. 2.4 RADIAL FORCES (2, 3) -Theradialforcesproducedbytheaxialleakagefieldactoutwardsontheouter windingtendingtostretchthewindingconductor,producingatensilestress(also called as hoop stress) -Theinnerwindingexperiencesradialforcesactinginwardstendingtocollapseor crush it, producing a compressive stress.-Duetothefringingoftheleakagefieldattheendsofthewindings,theaxial component of the field reduces resulting into smaller radial forces in these regions. For deriving a simple formula for the radial force in a winding, the fringing of the field isneglected;theapproximationisjustifiedbecausethemaximumvalueoftheradial force is important which occurs in the major middle portion of the winding. Let us consider an outer winding, which is subjected to hoop stresses. The value of the leakagefieldincreasesfromzeroattheoutsidediametertoamaximumattheinside diameter(atthegapbetweenthetwowindings).Thepeakvalueoffluxdensityinthe gap is wgpHNIB02= . (2.23) where, NI is the r.m.s. value of winding ampere-turns Hw is winding height in meters. Thewholewindingisintheaveragevalueoffluxdensityofhalfthegapvalue.The total radial force acting on the winding having a mean diameter of Dm (in meters) can be calculated by ( )....(2.25) .......... .......... .......... .......... .......... .......... .......... .......... newton 2.24) .........( .......... .......... .......... .......... .......... .......... .......... 2221200mwrmwrDHNIFD NIHNIFtt = (((

= Fortheouterwinding,theconductorsclosetogap(attheinsidediameter)experience higherforcesascomparedtothoseneartheoutsidediameter(forcereduceslinearly from a maximum value at the gap to zero at the outside diameter). The force can be considered to be transferred from conductors with high load (force) to those with low load if the conductors are wound tightly Hence, averaging of the force value over the radial depth of the winding as done in the aboveequationisjustifiedsincethewindingconductorssharetheloadalmost uniformly.

Ifthecurvatureistakenintoaccountbytheprocessofintegrationacrossthewinding radial depth the mean diameter of the winding in the above equation should be replaced by its inside diameter plus two-thirds of the radial depth. 2.4.1 CALCULATION OF RADIAL STRESSES (4, 5) Fig 2.7 Theaveragehoopstressfortheouterwindingiscalculatedasforacylindricalboiler shellshowninfigure.ThetransverseforceFactingontwohalvesofthewindingis equivalent to pressure on the diameter, hence it will be given by Dm replaced by Dm in the above equation. If the cross-sectional area of turn is At (in m2), the average hoop stress in the winding is ( )(2.27) ........ .......... .......... .......... .......... .......... .......... .......... ..........2(2.26) ..... .......... .......... .......... .......... .......... .......... .......... .......... ..........2 (2020tmwavgt wmavgAN D pH pIA N HD NIttoo = = LetIrbetheratedr.m.s.currentandZpubetheper-unitimpedanceofatransformer. Undertheshortcircuitcondition,ther.m.s.valueofcurrentinthewindingisequalto (Ir/Zpu).Totakeintoaccounttheasymmetry,thiscurrentvalueismultipliedbythe asymmetryfactork.IfwedenotecopperlossperphasebyPR,theexpressionforavg under the short circuit condition is ( )( )( )(2.29) N/m22 2(2.28) 22222022 20. . . . . . . . . . . . . . . . =. . . . . . . . . . . . . . . . . . =puRwavgtmpurwavgZk PH pAN D pZk IH ptotto Substituting the values of 0(=410-7) and (resistivity of copper at 75 = 0.021110-6), we finally get ( )222N/m 2 74 . 4pu wRavgZ HPk = o .. (2.30) where, PR is in watts and Hw in meters. It is to be noted that the term PR is only the DC I2R loss (without having any component ofstrayloss)ofthewindingperphaseat75C.Hence,withverylittleandbasic information of the design, the average value of hoop stress can be easily calculated. Thepreviousvalueofaveragestresscanbeassumedtobeapplicableforanentire tightly wound disc winding without much error. This is because of the fact that although thestressishigherfortheinnerconductorsoftheouterwinding,theseconductors cannotelongatewithoutstressingtheouterconductors.Thisresultsinanearuniform hoop stress distribution over the entire winding. Inlayer/helicalwindingshavingtwoormorelayers,thelayersdonotfirmlysupport eachotherandthereisnotransferofloadbetweenthem.Hence,thehoopstressis highest for the innermost layer and it decreases towards the outer layers. For a double-layerwinding,theaveragestressinthelayernearthegapis1.5timeshigherthanthe average stress for the two layers considered together. Generalizing, if there are L layers, the average stress in kth layer (from gap) is [2- ((2k-1)/L)] times the average stress of all the layers considered together. For an inner winding subjected to radial forces acting inwards, the average stress can be calculatedbythesameformulaeasabovefortheouterwinding.However,sincethe inner winding can either fail by collapsing or due to bending between the supports, the compressivestressesoftheinnerwindingarenotthesimpleequivalentsofthehoop stresses of the outer winding. Thus,theinnerwindingdesignconsiderationsarequitedifferent,andtheseaspects along with the failure modes are discussed in the further sections. 2.4.2 FAILURE MODES DUE TO RADIAL FORCES The failure modes of windings are quite different for inward and outward radial forces. Winding conductors subjected to outward forces experience the tensile (hoop) stresses. Thecompressivestressesaredevelopedinconductorsofawindingsubjectedtothe inwardforces.Inconcentricwindings,thestrengthofouterwindingssubjectedtothe outwardforcesdependsonthetensilestrengthoftheconductor;onthecontrarythe strengthofinnerwindingssubjectedtotheinwardforcesdependsonthesupport structureprovided.Theradialcollapseoftheinnerwindingsiscommon,whereasthe outward bursting of the outer windings usually does not take place. 2.4.2.1 WINDING SUBJECTED TO TENSILE STRESSES (5, 8, 9, 10, 11) If a winding is tightly wound, the conductors in the radial direction in a disc winding or inanylayerofamulti-layerwindingcanbeassumedtohaveauniformtensilestress. Sincemostofthespaceintheradialdirectionisoccupiedwithcopper(exceptforthe smallpapercoveringontheconductors),theratioofstiffnesstomassishigh.As mentionedearlier,thenaturalfrequencyismuchhigherthantheexcitingfrequencies, and hence chances of resonance are remote. Under a stretched condition, if the stressexceeds theyield strength of the conductor, a failureoccurs.Theconductorinsulationmaygetdamagedortherecouldbelocal bulgingofthewinding.Theconductormayevenbreakduetoimproperjoints.The chancesoffailureofwindingssubjectedtothetensilehoopstressesareunlikelyifa conductorwithacertainminimum0.2%proofstrengthisused.The0.2%proofstress can be defined as that stress value which produces a permanent strain of 0.2% (2 mm in 1000mm).Oneofthecommonwaystoincreasethestrengthistheuseofwork-hardened conductor; the hardness should not be very high since there could be difficulty inwindingoperationwithsuchahardconductor.Alowervalueofcurrentdensityis also used to improve the withstand characteristics. 2.4.2.2 WINDINGS SUBJECTED TO COMPRESSIVE STRESSES (5, 8, 9, 10, 11) Conductors of inner windings, which are subjected to the radial compressive load, may failduetobendingbetweensupportsorbuckling.Theformercaseisapplicablewhen the innerwinding is firmly supported by the axially placed supporting spacers (strips), and the supporting structure as a whole has higher stiffness than conductors (e.g., if the spacers are supported by the core structure) as shown in figure (a). Fig 2.8 Thelattercaseofbuckling,termedasfreebuckling,isessentiallyanunsupported buckling mode, in which the span of the conductor buckle bears no relation to the span ofaxialsupportingspacersasshowninfigure(b).Thiskindoffailureoccursmostly with thin winding cylinders, where conductor has higher stiffness as compared to that of innercylindersand/orthecylinders(andtheaxialspacers)arenotfirmlysupported from inside. The conductors bulge inwards as well as outwards at one or more locations alongthecircumference.Therearemanyfactorswhichmayleadtothebuckling phenomenon,viz.windinglooseness,inferiormaterialcharacteristics,eccentricitiesin windings, lower stiffness of supporting structures as compared to the conductor, etc. The buckling can be viewed as a sequential chain of failures, initiated at the outermost conductor of the inner winding and moving towards the innermost conductor facing the core.Thenumberofwindingsupportsshouldbeadequateforgivingthenecessary strength to the winding against the radial forces. When the supporting structures are in directcontactwiththecore,awindingcanbetakenasveryrigidlysupported.Onthe contrary, if there is no direct contact (fully or partly) with the core, the winding is only supportedbytheinsulatingcylindermadeofmostlythepressboardmaterialthereby reducingtheeffectivestiffnessofthesupportstructureandincreasingthechancesof failure. The supports provided are effective only when the support structure as a whole is in firm contact with the core. Awindingconductorsubjectedtotheinwardradialforcesisusuallymodeledasa circular loop under a uniformly distributed radial load. The critical load per unit length of the winding conductor is given by, ] 4 [122233 =smrNwtDEf (2.31) where, E is modulus of elasticity of conductor material,Ns is total number of axially placed supports (spacers), w is width of conductor,t is thickness of conductor andDm is mean diameter of winding The compressive stress on the inner winding conductor is given as, AD fm ravg2= o ............................ (2.32) where, A is area of conductor (=w t) Substituting the value of fr, we get, ((

||.|

\|=222412ssmsavgNNDtN Eo (2.33) For Ns >>1, the expression for the minimum number of supports to be provided is, E tDNavgmso 12= ...............(2.34) The term avg is the average value of the compressive stress (in an entire disc winding or in a layer of a multi-layer winding) 2.5 AXIAL FORCES (5) Forauniformampere-turndistributioninwindingswithequalheights(ideal conditions),theaxialforcesduetotheradialleakagefieldatthewindingendsare directed towards the winding center as shown. Fig 2.9 Although, there is higher local force per unit length at the winding ends, the cumulative compressive force is maximum at the center of windings. Thus, both the inner and outer windingsexperiencecompressiveforceswithnoendthrustontheclampingstructures (under ideal conditions). The inner winding being closer to the limb, by virtue of higher radial flux, experiences higher compressive force as compared to the outer winding.Itcanbeassumedthat25to33%offorceistakenbytheouterwinding,andthe remaining 75 to 67% is taken by the inner winding. 2.5.1 CALCULATION OF AXIAL FORCES (6) For an asymmetry factor of 1.8, the total axial compressive force acting on the inner and outer windings taken together is given by conditions ideal ... kg8 . 50f H ZSFw pua =.. (2.35) where, S is rated power per limb in kVA Hw is winding height in meters Zpu is per-unit impedance, andf is frequency in Hz. Oncethetotalaxialforceforeachwindingiscalculated,thecompressivestressinthe supportingradialspacers(blocks)canbecalculatedbydividingthecompressiveforce by the total area of radial spacers. The stress should be less than a certain limit, which depends on the material of the spacer. 2.5.2 REASONS FOR HIGH VALUE OF AXIAL FORCES (7) The reasons for a higher value of radial field and consequent axial forces are:-Mismatch of ampere-turn distribution between LV and HV windings, -Tapping in the winding, -Unaccounted shrinkage of insulation during drying and impregnation processes, etc. Whenthewindingsarenotplacedsymmetricallywithrespecttothecenter-lineas shown, the resulting axial forces are in such a direction that the asymmetry and the end thrusts on the clamping structures increase further. Fig 2.10 Itiswellknownthatevenasmallaxialdisplacementofwindingsormisalignmentof magneticcentersofwindingscaneventuallycauseenormousaxialforcesleadingto failure of transformers. Hence, strict sizing/ dimension control is required during processing and assembling of windings so that the windings get symmetrically placed. 2.5.3 FAILURE MODES DUE TO AXIAL FORCES (5) There are various types of failures under the action of axial compressive forces: If a layer winding is not wound tightly, some conductors may just axially pass over the adjacentconductors,whichmaydamagetheconductorinsulationleadingeventually intoaturn-to-turnfault.Inanothermodeoffailure,ifawindingissetintovibration undertheactionofaxialforces,theconductorinsulationmaygetdamagedduetoa relative movement between the winding and axially placed insulation spacers. Highaxialendthrustscouldleadtodeformationsoftheendclampingstructuresand windings.Theendclampingstructuresplaythemostimportantroleinresistingaxial forcesduringshortcircuits.Theyhavetomaintainaneffectivepressureonthe windings,appliedusuallyontheclampingringmadeofstiffinsulatingmaterial(pre-compressedboardordensifiedwood).Thetypeofinsulationmaterialusedforthe clamping ring depends on the dielectric stress in the end insulation region of windings. The densified wood material is used for lower stresses and pre-compressed board, being abettergradedielectrically,isusedforhigherstressesandforcomplyingstringent partial discharge requirements.

Whenaclampingringmadeofaninsulatingmaterialisreinforcedbythefiberglass material, an extra strength is provided. Some manufacturers use clamping rings made of steel material. The thickness of metallic clamping rings is smaller than that made from the insulating material. The metallic ring has to be properly grounded with a cut so that it does not form a short-circuited turn around the limb. The sharp edges of the metallic ring should be rounded off and covered with a suitable insulation. Inadditiontoabovetypesoffailuresduetotheaxialforces,therearetwoprincipal types of failures, viz. bending between radial spacers and tilting. 2.5.3.1 BENDING BETWEEN RADIAL SPACERS Undertheactionofaxialforces,thewindingconductorcanbendbetweentheradially placed insulation spacers as shown. Fig 2.11 The conductor bending can result into a damage of its insulation. The maximum stress in the conductor due to bending occurs at the corners of the radial spacers and is given by 202maxkg/cm12Iy S FAL= o ..(2.36) where,FAL is maximum axial bending load in kg/cm. It corresponds to the most highly stressed disc in a disc winding or turn in a helical winding (layer winding with radial spacers).Themaximumaxialloadmayusuallylieintheregionofnon-uniformampere-turn distribution (e.g., tap zone).Themaximumaxialload,calculatedaccuratelybyamethodsuchasFEM,dividedby themeanturnlength(Dm)givesthevalueofFAL,whereDmismeandiameterof winding in cm. S is span between two radial spacers in cm||.|

\| th spacer widsmND t Ns is number of radial spacers.

yismaximumdistancefromneutralaxisforconductorincm(i.e.,halfofconductor axial width: w/2). I0 is moment of inertia of disc or turn||.|

\|=123ntw n being number conductors in radial direction, andt is conductor thickness in cm. The maximum stress in the conductor calculatedshould be less than the limiting value for the type of conductor used (about 1200 kg/cm2 less than the limit for soft copper). 2.5.3.2 TILTING UNDER AXIAL LOAD (12, 13) Thefailureduetotiltingundertheactionofaxialcompressiveforcesisoneofthe principalmodesoffailuresinlargepowertransformers.Whentheseforcesaremore than a certain limit, a failure can occur in disc winding due to tilting of conductors in a zigzag fashion as shown. Fig 2.12 Inthismodeoffailure,thereisturningofcrosssectionofconductorsaroundthe perpendicularaxisofsymmetry.Therearetwokindsofforcesthatresistthetiltingof theconductors.Thefirstoneisduetotheconductormaterial,whichresistsbeing twisted. Thesecondresistingforceisthefrictionforce(duetocornersofconductors);during tiltingtheconductorsatbothendsmustbiteintothematerialoftheradialspacer, producing a couple at the conductor ends which resists tilting. The two resisting forces areusuallyconsideredseparatelytoarriveatthecriticalstressandload,causingthe failure. The tilting strength decreases inversely as the square of winding radius, suggesting that thelargewindingsshouldbecarefullydesigned.Iftheconductorhassharpends,the frictional force resists tilting. Actually,duetotheconductorcornerradius,thecontributiontotiltingresistance(due tofriction)reducesandthisreductionshouldbeconsidered.Thecriticalstrengthofa helicalwinding(alayerwindingwithradialspacers)ishigherthanalayerwinding (whichiswithoutradialspacers)becauseoftheadditionalstrengthofferedbythe spacers.

When a continuously transposed cable (CTC) conductor is used, although there are two axially placed rows of conductors in one common paper covering, it cannot be assumed thattheeffectivetiltingstrengthishigher.Twopossiblemodesoffailuresare described. The first type of failure (termed as cable-wise tilting), in which two adjacent cables tilt against each other, is shown Fig 2.14 Cable wise tilting If the inter-strand friction is higher, the winding is forced to tilt in pairs of strands in the CTC conductor. Forthesamestranddimensions,theCTCconductorhasfourtimesgreatertilting strength;theresultisobviousbecausetheeffectivewidthofitsconductorisdoubled increasing the strength by four times. ThisincreaseinstrengthisvalidonlywhenthetwoaxiallyplacedstrandsintheCTC conductor can be considered to act together under the tilting load. In the second mode of failure (termed as strand-wise tilting), two axially placed strands in the CTC conductor tilt against each other as shown. Fig 2.15 Strand wise tilting The critical tilting load in this mode may be lower, reducing the effective overall tilting strength. This is because the lower of the cable-wise and strand-wise strengths triggers theaxialinstability.Whilethecriticalstressinthecable-wisetiltingisindependentof numberofstrandsinthecable(n),thecriticalstressinthestrand-wisetiltingis inverselyproportionalton.AsthenumberofstrandsintheCTCconductorincreases, thecriticalloadlimitinthestrand-wisetiltingbecomeslowerthanthecable-wise tilting. Hence,with the increase in number of strands in the CTC conductor, the mode of failure shifts from the cable-wise tilting to the strand-wise tilting. The use of epoxy-bonded CTC conductor is quite common in which the epoxy coating effectivelybondsthestrandsincreasingtheresistanceagainstthestrandwisetilting. Each strand in the epoxy-bonded CTC conductor has, in addition to an enamel coating, acoatofthermosettingepoxyresin.Thecuringofthisresinoccursataround120C duringtheprocessingofwindings.Aftercuring,theepoxy-bondedCTCconductor consisting of many strands can be considered as one conductor with an equivalent cross sectionforthemechanicalstrengthconsideration.Thus,thepossibilityofstrand-wise tiltingiseliminated,greatlyincreasingthestrengthoftheCTCconductoragainstthe tiltingload.TheepoxybondedCTCconductornotonlyreducesthewindingeddy losses but it also significantly improves the short circuit withstand characteristics. 2.6 EFFECT OF PRESTRESS (14, 15,16) Theclampingpressureappliedonthewindingsafterthecompletionofcorewinding assembly is called as pre-stress. It has a significant impact on the response of windings duringshortcircuits.Itincreasesthestiffnessofwindingstherebyincreasingtheir mechanical natural frequencies.

The relationship between the natural frequency and pre-stress is highly non-linear. The pre-stressreducesoscillatoryforcesactingontheinsulation.Thewinding displacements also decrease with the increase in the pre-stress value. The value of pre-stress should be judiciously chosen depending upon the characteristics of core-winding assembly. The chosen value of pre stress must get maintained during the entire life of a transformer. This means that the insulation stability should be fully realized during the processingofwindingsduringmanufacturing.Ifthenaturalfrequencywithoutpre-stressishigherthantheexcitationfrequencies,ahigherpre-stressvaluewill significantlyreducetheoscillatoryforces.Contrarytothis,ifthenaturalfrequency without pre-stress is lower than the excitation frequencies, a certain value of pre-stress willbringthenaturalfrequencyclosertotheexcitationfrequenciesleadingtoan increaseintheoscillatoryforces.Thenaturalfrequencyisreportedtovaryassome functionofsquarerootoftheratioofpre-stresstomaximumvalueofpeak electromagnetic stress in the winding. Thenaturalfrequencyofawindingmaychangeduringtheshortcircuitperioddueto changes in the insulation characteristics and ratio of pre-stress to total stress. Thus, the naturalfrequencymeasuredfromthefreeresponsemaybedifferentaftertheshort circuitascomparedtothatbeforetheshortcircuit.Also,duringtheshortcircuit,the winding, which may be in resonance at some time experiencing a higher stress, may get detunedfromtheresonanceduetochangeininsulationcharacteristicsatsomeother instant. 2.7 STRESSES IN FLITCH PLATES AND FRAMES A flitch plate should be designed for withstanding the clamping force and core winding weight(static loads). Duringa shortcircuit, theaxial forces (end thrusts)developed in windings,actonthetopandbottomframes;flitchplateshelptokeeptheframesin position. The stresses produced in the flitch plates are the tensile stresses and shearing stresses. These stresses can be calculated by well-known formulae used in the structural analysis. Theframesaresubjectedtostresseswhileliftingcore-windingassembly,during clampingofwindings,orduetoshortcircuitendthrusts.Usually,theshortcircuit stressesdecidetheirdimensions.Thestressesintheframesaredeterminedfromthe calculatedvaluesoftheshortcircuitforcesactingonthemandassumingthecorebolt points and locking arrangements (pins, etc.) between flitch plates and frames as support locations. Fig 2.16 CHAPTER 3 TRANSFORMER DESIGN PROCEDURE 3.1DIFFERENTPARAMETERSCONSIDEREDINTRANSFORMER DESIGN (17) 1.Core diameter selection 2.Flux density 3.Number of turns in HV and LV 4.Selection of layer and helical winding 5.Estimation of height of window 6.Width of conductor 7.Thickness of conductor 8.Number of parallel conductors (based on current density) 9.Calculationofpercentageimpedancebasedondimensions(describedforeach case) 10. Load loss and no load loss calculation 11. Percentage ratio error calculation 12. Costing of major material

3.1.1 CORE DIAMETER SELECTION Corediameterofatransformerdependsuponanumberoffactorsliketransformer rating,percentageimpedancebetweenthewindings,basicinsulationlevel,transport height, core fluxing requirements, type of core, and quality of core steel. Itisfairlycomplicatedtoderiveauniversalandexactformulaforcorediameter.In practice,thecorediameterisselectedbythedesignerfromsimilardesignsalready available in the industry(usually core diameter varies from 350mm to 500mm).Based onthis,parameterslikepercentageimpedanceandlossesareworkedoutandcore diameter is adjusted to meet the required parameters. Influence of varying core diameter Increase in core diameter causes: -Increase in core cross-sectional area -Increase in voltage/turn, hence reduction in number of turns -Overall weight of core steel increases -No load loss of transformer increases Reduction in core diameter causes: -Increased copper weight -Increased load losses Thepercentagereactancebetweenthewindingsisdirectlyproportionaltothenumber of turns and diameter of various coils and is inversely proportional to volts/turn and coil depth. 3.1.2 FLUX DENSITY Value of flux density is chosen to suit the required performance. Normally flux density ischosennearthekneepointofthemagnetizationcurve.However,adequatemargin shouldbekepttotakecareforsystemconditionslikeoverfluxing,frequencyand voltage variations. In certain cases, the value of flux density is reduced to limit the noise level of transformers. Influence of varying flux density Increase in flux density, keeping other parameters constant causes: -Higher volts/turn -Reduction in number of turns in winding -Reduced core steel weight -Higher no load loss of transformer -Lower copper weight -Lower load losses Reduction in the value of flux density causes increased core steel weight, lower no load loss and increases copper weight and load loss.

3.1.3 NUMBER OF TURNS IN HV AND LV ThenumberofturnsinHVandLVwindingcanbecalculatedbyusingtheemf equation of the transformer as: E = 4.44fmN.(3.1) E=4.44fBmAN(3.2) where : E =Voltage induced in the transformer winding (HV or LV winding) F = Frequency in Hz m = Max value of flux Bm = Maximum flux density in Tesla A = Net cross sectional area of the core in mm2 Therefore, after fixing the core area and flux density, the number of turns in the HV and LV winding can be calculated by using the above equation. Also,voltageperturniscalculatedbydividingthevoltagebythecorresponding number of turns (LV or HV). 3.1.4 SELECTION OF TYPE OF WINDING The windings along with its insulations form the electric circuit of the transformer. Due care must be taken while designing the windings to ensure its healthiness during normal aswellasfaultconditionsasthewindingsneedtobeelectricallyandmechanically strongtowithstandvoltagesurgesaswellasmechanicalstressesduringshort-circuit conditions. The temperature of windings at rated, over-load and short-circuit conditions should be within limits, ensuring the proper life of transformer. The requisite type of winding is chosen only after considering all the above mentioned factors. ForLV,asweneedtohavelessnumberofturnsinthewinding,generallylayer winding is used as it can be fitted within the limited height of the core structure. ForHV,weneedtohavealargenumberofturnsinthewindingandhence,disc winding is preferred over layer winding taking into account the height constraints of the core. Sometimes, for large voltages, disc winding is chosen over layer winding in LV as well due to height considerations.

3.1.5 HEIGHT OF CORE WINDOW In the transformer industry, the height of transformers is restricted to 4.5 5 meters due to the difficulty involved in transportation. Takingthisintoaccount,theheightofbaseandclampingstructures,therelay arrangement and insulation the actual height of the core structure is restricted to 3 - 3.5 m. The designers select the height of the core structure less than the above specified value. From this value of the core height, we determine the height of HV and LV winding. 3.1.6 WIDTH OF CONDUCTOR After fixing the height of LV and HV winding, the width of the LV and HV winding is calculated. The procedure of calculating the width in LV or HV is as follows: 1.Firstthetotalnumberofdiscsorturnsinthewindingisfixed.Incaseofdisc winding,thenumberofdiscsiscalculatedbyfixingthenumberofturnsper disc.Weusuallytakeevennumberofdiscsforconvenienceoftakingoutthe leads. 2.Then we calculate the width of the conductor using the following formula:Width= discs of No. winding of Height -(paperinsulationthickness)(woodenblock thickness).. (3.3) Usually,wetakepaperinsulationthicknessas0.5mmandwoodenblockof thickness 2.8mm. Widthofconductor= discs of No. winding of Height -0.52.8..(3.4)

3.The width of the conductor should be between 6 to 17mm. These constraints are due to the required mechanical properties of the conductor. 3.1.7 THICKNESS OF THE CONDUCTOR The thickness of the conductor is selected between 1.7 to 3 mm. Again these restrictions are due to the required mechanical properties of the conductor. This value of thickness decides the number of parallel paths in the windings hence the designer can choose any value as per the parallel path requirements. Fig 3.1 0.25mm 0.25mm 0.25mm 0.25mm 0.25mm 0.25mm 0.25mm 0.25mm Paper insulation Width of conductor Thickness of conductor 3.1.8 NUMBER OF PARALLEL CONDUCTORS The number of parallel paths in the given winding is calculated as follows: 1.The maximum current that the winding needs to carry is calculated.2.The maximum current density is chosen as 3 A/mm2 3.Therequiredcross-sectionalareatocarrythemaximumwindingcurrentis calculated. 2A/mm 3 of densitycurrentmaximum winding by the carried currentmaximumarea Required =...(3.5) 4.The number of parallel paths are calculated by dividing this required area by the area of one conductor.

thickness conductorwidth conductor area Requiredpaths parallel of Number =(3.6) 5.Thenumberofparallelpathscalculatedbyaboveformulaisapproximatedto the next integer. E.g. if we get number of parallel paths as 3.3 we approximate it to 4. Fig 3.2 1234 3.1.9 CALCULATION OF %Z BASED ON DIMENSIONS In the impedance derivation part we have seen that %Z is calculated from the formula: =ATDturn Volts Hturns Amperef Zeq) / () (10 48 . 2 %5... (3.7) where,2cm inis ATDand cm iniseqH Theterm ATDisnothingbutareaundertheampere-turndiagram.Thisareais different for maximum, minimum and normal (100%) tap positions. For maximum tap position, 8) .......(3. .......... .......... .......... .......... ) 1428 . 0 (31) 1428 . 0 3 (31

) 1428 . 0 1428 . 0 1 (31) 1 1 1 (313123 3222 2 1 12 21 1D T D TD T D T D T ATDg gg g+ ++ + + + + + = where, 1T- Width of LV winding 1D -Mean diameter of LV winding2T -Width of HV winding 2D -Mean diameter of HV winding3T -Width of tapping winding 3D -Mean diameter of tapping winding 1gT- Width of air gap between LV and HV winding 1gD -Mean diameter of air gap between LV and HV winding 2gT- Width of air gap between HV and tapping winding 2gD -Mean diameter of air gap between HV and tapping winding For minimum tap position, 1 12 2 1 1) (31g gD T D T D T ATD + + =(3.9) Similarly for normal tap, ..(3.10) .......... .......... .......... .......... .......... )21 . 0 (3 331)21 . 0 3 (2 231

)21 . 0 1 . 0 1 (2 231) 1 1 1 (1 1311 131D TgDgTD TgDgT D T ATD+ ++ + + + + + = TheAmpere-TurnsandVoltageperturnarecalculatedbymultiplyingLVcurrentand LV turns. 3.1.10 COPPER LOSS AND NO LOAD LOSS (CORE LOSS) CALCULATIONS

Mainly, there are two types of losses in a transformer viz. core losses (No-load) losses and copper losses. Core (No-Load) Losses: -Transformer no-load losses occur due to flow of main flux in the core. -Thecorelossofatransformerdependsupongradeofsteel,frequency,flux density,type&weightofcore.Sinceallthesethingsremainconstantoncethe transformer is designed, the core losses are practically constant at all the loads. -Tocalculatethecoreloss,wefindthevolumeofthecore.Areaofcoreis alreadyfixed.Wefixthelengthofcoresuchthatthereissufficientgap (=10mm)between2HVwindingsofadjacentphases.Accordingly,volumeis calculated. -Weightofthecoreiscalculatedbymultiplyingthevolumeofthecoreby density of the core material. Core losses are given as 1.6 watts per kg weight of the core. Accordingly, total core losses are computed. Copper Losses: -Load losses comprise of ohmic losses and eddy-current losses. But eddy-current losses are about 10% of ohmic losses. Hence generally they are neglected. -Theload-lossofatransformerisafunctionoftemperatureandisgenerally expressed at a reference temperature of 75 degree C. -The resistance R of winding at 75 degree C is calculated as R = NL/A (3.11) R = resistivity of conductor at 75 degree C L = length of windingN = number of turns A = area of conductor. Area of conductor = wt. (3.12) where, t= Thickness of conductor w = Width of conductor

Now the length of winding is calculated as follows: -Consider a winding with m number of discs, n number of turns per disc, and p number of parallel paths. Now, mean thickness of one such conductor = n * p * (0.5+ t) / 2, where the thickness of paper insulation is 0.5 mm.-Hence, length of one such conductor per disc is 2[(distance between core centre & LV winding)+{n x p x (t + 0.5)/2}] (3.13) -Hence total length of all such conductors throughout all the discs is m*n* *length of one conductor(3.14) Buttherearepparallelpathsofconductors.Hencewhilecalculatingthe resistance of winding, we divide this total length by p. -Oncetheresistanceofawindingiscalculated,ohmiclossforthatwindingis calculatedbymultiplyingthevalueofresistancebythesquareofthecurrent through the respective winding. The procedure is same for LV, HV and tapping windings. -Incaseofminimumtap,thewindingresistanceissmallascomparedto maximum tap. But the current is maximum. Hence copper loss is more in case of minimum tapping than maximum tapping. -The above procedure gives the per phase copper loss.Hence in case of 3-phase transformer,thetotalcopperlossiscalculatedbymultiplyingtheabovevalue by 3. The total losses in a transformer are the sum of the above two losses. Increase in losses implies reduction in efficiency and hence increase in cost. Increment of 1kW in cost of transformer by Rs. 1 lakh Increment of 1kW in No-load losses implies increment in cost oftransformer by Rs. 5 lakh.3.1.11 CALCULATION OF RADIAL STRESS ON HV WINDING 1. Total copper loss is calculated for HV winding. From this we calculate thecopper loss per phase , RP . 2. We have fixed the winding height in the beginning of the design. 3. Per unit impedance is calculated. Using the above data we calculate the radial stress in HV winding is calculated as 2 2/ * ) 2 ( * 94 . 4pu w R avgZ H P k = o...(3.15) where,k is the asymmetry factor considered to take into account symmetry. 3.1.12.CALCULATIONOFAXIALSTRESSESONINNERANDOUTER WINDING Foranuniformampere-turndistributioninwindingswithequalheights(ideal conditions),theaxialforcesduetotheradialleakagefieldinbothinnerandouter winding are given by ) * * /( ) * 8 . 50 ( f H Z S Fw pu a = (3.16) where,S =Rated power per limb in kVA Hw = Winding height in meters Zpu = Per unit impedancef = frequency in Hz In the absence of detailed analysis, it can be assumed that 25 to 33% of force is taken by the outer winding, and the remaining 75 to 67% is taken by the inner winding. For inner winding, we approximate the respective percentage to 70%Hence, Axial forces on inner winding = 70 % of total axial forces For outer winding, we approximate the above percentage to 30 %Hence,Axial forces on outer winding = 30 % of total axial forces Calculation of number of spacers Thespacersareinsertedtoprovidethenecessarystrengthtothewindingagainstthe radial forces. The minimum number of spacers or supporters provided is given by ) / * 12 ( * ) / ( E t D Navg m so = .(3.17) avgo = the average value of the compressive stress (in an entire disc winding) mD= mean diameter of the winding. t =thickness of the conductor E =modulus of elasticity of conductor material Accordingtostandardspecifications,thenumberofspacersarealwayskeptina multiple of 4, 8 or 12. Number of blocks Numberofblocksinsertedbetweenthetwoadjacentdiscsisalwaysequaltothe number of spacers inserted. Calculation of width of a block Thetotalwidthofalltheblocksshouldbeapproximately30%to35%ofthe circumferenceofthecorrespondingwinding,whichiscalculatedbyconsideringthe mean diameter of the winding. Mathematically, Width of a block = (30 % of circumference of the winding)/ (no of blocks)(3.18) According to international standards, Widths of a block commercially available are 35 mm, 45 mm and 55 mm. After calculating width of the block, block of closest dimensions is selected. Calculation of axial stress Axialforcesactingoneachblockareobtainedbyusingaxialforcecalculatedbefore and area of all the blocks. Mathematically this can be shown as Axial stress = (Axial force on the blocks)/ (Total area of blocks)... (3.19) The procedure for calculating the axial stress is same for LV, HV and tap winding. CHAPTER 4 DESIGNED TRANSFORMER _____________________________________________________________________ 4.1 PROBLEM STATEMENT Design a transformer having following parameters: -Power rating 25 MVA -Voltage ratio 33/11 kV with +5% to -10% regulation in the step of 1.5% -Vector group 11DYN-Operating frequency 50Hz -No load loss 20kW -Copper losses 80 kW at 25 MVA -%X = 12% -Maximum current density in any conductor should be less than 2A/mm 3-Maximum flux density in the core = 1.7 T -Tolerance on impedance % 10 . -Tolerance on losses and current density: no positive tolerance. -Radial stress on windings should be less than 80 MPa -Axial stress on the insulation block should be less than 10 MPa 4.2 PARAMETERS OF DESIGNED TRANSFORMER 1. Current in LV and HV: First we calculate the current in both LV and HV winding. Since the HV winding is star connected we take the phase voltage for current calculation. (4.3) ....... .......... .......... .......... .......... .......... .......... .......... 57 . 75711 3MVA 25current(4.2) .......... .......... .......... .......... .......... .......... .......... 37 . 437053 . 19 3MVA 25currentHV(4.1) .......... .......... .......... .......... .......... .......... .......... .......... 053 . 19333kVvoltage HVAkVLVAkVkV===== = 2. Core cross-sectional area: Then we fix the core cross-sectional area and hence the core diameter. For the given power rating, the core diameter is usually selected between 375mm to 450mm. Core area =1415002mm . (4.4) Core diameter = 424.43 mm(4.5) 3. Number of turns in LV and HV: Using the equation, E = 4.44fBmAN (4.6) Putting f = 50 Hz, A = 1415002mm , Bm= 1.7 T, we calculate LV and HV number of turns. LV number of turns = 205.98...(4.7) HV number of turns = 356.78.. (4.8) The number of turns is approximated as follows: Approximated LV turns = 206(4.9) Approximated HV turns = 357....(4.10) Since it is required to have voltage regulation on HV side from 90% to 105% we calculate the maximum and the minimum number of turns corresponding to these voltage levels. (4.11) ......... .......... .......... .......... .......... .......... .......... 322....... turns HV minimum Approx 375 turns HV maximum Approx 321.1 turns HV minimum Calculated374.62 turns HV maximum Calculated== == Since we are using linear tapping, in the HV winding there are only 322 turns (corresponding to 90% voltage) and the rest of the turns are present in tapping winding. 4. Number of discs in LV and HV: We fix the number of turns per disc in both LV and HV. Also, as explained in the procedure, we add around 4-5 more discs to the calculated value to account for dropping of turns at the crossover during actual winding process. The fact that it is beneficial from manufacturing point of view that the number of discs should be even is considered. For LV winding: Number of turns per disc = 2 Number of discs = 206/2 = 103 Actual number of discs = 105 (4.12) For HV winding: Number of turns per disc = 3 Number of discs = 322/3 = 107.3 Actual number of discs = 110 (4.13) Tapping winding: Tapping winding turns = 375-322 = 53 Number of turns per disc = 3 Number of discs = 53/3 = 17.66 Actual number of discs = 20(4.14) 5. Height of core structure: As explained in the design procedure, we select the height of the transformer as 2450mm. This height satisfies all the constraints present on the height of core. Height of core structure=2450mm Height of core window = 2450- 2core diameter = 2450-2424.42 = 1601.14 mm(4.15) We keep a gap of 100 mm on both side of the winding so as to maintain the oil flow. Therefore the height of the LV and HV winding is calculated as follows: Height of LV and HV winding = 1601.14 200 =1401.14 mm.. (4.16) 6. Width of conductor: As explained in the design procedure we calculate width of the conductor of LV and HV as follows: 9.918mm 2.8 0.51061401.142.8 0.5LV indiscs of no winding of height conductorLV of Width= = = 9.437mm 2.8 0.51101401.142.8 0.5HV indiscs of no winding of height conductorHV of Width Similarly,= = = .(4.17) 7. Thickness of conductor and number of parallel paths: We select the thickness of conductor between 1.7 3 mm.Thickness of LV winding = 2.31 mm (4.19) 8 HV inpaths parallel of numberActual005 . 745 . 2 437 . 9 337 . 437

thickness conductorwidth conductorA/mm 3 of densitycurrentmaximum winding HV bycarried currentmaximumHV inpaths parallel of Number mm 2.45 winding HV of Thickness12 LV inpaths parallel of numberActual4.18) ( 022 . 1131 . 2 918 . 9 357 . 757

thickness conductorwidth conductorA/mm 3 of densitycurrentmaximum winding LV bycarried currentmaximumLV inpaths parallel of Number 22. . . . . . . . . . . . . . . . . . . . == = ===. . . . . . . . . . . . . . . . . . . . . = = = 8. Tapping winding: We carry the dimensions of HV to tapping winding. Width of tapping winding conductor = 9.437mm Thickness of tapping winding conductor = 2.45mm.(4.20) For this design, we have 8 parallel paths in HV winding. For the tapping winding, it is not desirable to have tapping on 8 parallel paths together.Hence we divide 8 parallel paths in 4 parts and take 2 parallel paths together. We construct 20 discs with 3 turns per disc and 2 parallel paths i.e. 2 parallel conductors. Then we construct 3 more such sets of 20 discs each. There is 2.8 mm wooden block between two consecutive discs. Total height of taping winding (4.21) ... .......... .......... 4 . 1018) 8 . 2 5 . 0 43 . 9 ( 20 4. . . . . . . . . . . . . . . . . . . . . . =+ + =mm 9. Calculation of %Z based on dimensions: We calculate the %Z for three positions of tapping viz. Maximum tap, Minimum tap and Normal tap. 1T-Width of LV winding 1D - Mean diameter of LV winding (ID for LV+1.51T ) 2T - Width of HV winding 2D -Mean diameter of HV winding (ID for HV+1.52T ) 3T - Width of tapping winding 3D -Mean diameter of tapping winding (ID for tapping+1.53T ) 1gT- Width of air gap between LV and HV winding 1gD - Mean diameter of air gap between LV and HV winding(ID for air gap between LV and HV winding+1.51gT ) 2gT- Width of air gap between HV and tapping winding 2gD - Mean diameter of air gap between HV and tapping winding (ID for air gap between HV and tapping winding+1.52gT ) For the given design we take the gap between core and LV winding as well as the gap between LV and HV winding as 20 mm whereas we take the gap between HV and tapping winding as 10 mm. ID for LV = core diameter + 2(gap between core and LV winding) = 424.42 + 2(20)= 464.42 mm...(4.22) (4.23).... .......... 44 . 6731 . 2 2 120.5) conductor of (thickness disc perturns paths parallel of number 1. . . . . . . . . . . . . . . . . . . . . . = =+ =mmT 1D = 464.42 + 1.567.44 = 565.59 mm. (4.24) Similarly we calculate rest of the terms and get the following values: 2T = 70.8 mm 3T = 17.7 mm 1gT = 20 mm 2gT = 20 mm 1gD = 629.3 mm 2gD = 795.9 mm 2D = 745.5 mm 3D = 827.45 mm Following diagram explains all the dimensions: Fig 4.1 LV HV TAP 20mm67.44mm20mm70.8mm 20mm 17mm 1018.34mm 1401.14mm For maximum tap position, Fig 4.2 (4.25)mm 46028.27) 1428 . 0 (31) 1428 . 0 3 (31

) 1428 . 0 1428 . 0 1 (31) 1 1 1 (3131223 3222 2 1 1 max2 21 1. . . . . . . . . . . . . . . . . . . . . . =+ ++ + + + + + =D T D TD T D T D T ATDg gg g LV HV TAP 1

0.1429NIDistance For minimum tap position, Fig 4.3 (4.26) mm 42894.63 ) (3122 2 1 1 min1 1. . . . . . . . . . . . . . . . . . . . . . =+ + = g gD T D T D T ATD LV HV

TAP Distance NI1 Similarly for normal tap, Fig 4.4 (4.27)mm 44840.28 )21 . 0 (3 331)21 . 0 3 (2 231

)21 . 0 1 . 0 1 (2 231) 1 1 1 (1 1311 1312. . . . . . . . . . . . . . . . . . . . . . =+ ++ + + + + + =D TgDgTD TgDgT D T ATD LV HV TAP 1 0.1 NIDistance Then we calculate the %Z from the following formula: =ATDturn Volts Hturns Amperef Zeq) / () (10 48 . 2 %5 (4.28) where,2cm inis ATDand cm iniseqH Ampere-turns = 156060.6 AT Volts/turn = 53.39 V (4.29) For %Z min, ===22mincm1042894.63ATDcm9640 . 0 1014 . 14010.9640 factorRogowsky eqHPutting these values, we get, 695 . 10 %min = Z.(4.30) For %Z max, ===22mincm1046028.27ATDcm9577 . 0 1014 . 14010.9577 factorRogowsky eqHPutting these values, we get, 4 . 11 %max = Z ..(4.31) For %Z , ===22cm1044840.28ATDcm9577 . 0 1014 . 14010.9577 factorRogowsky eqH Putting these values, we get, 10 . 11 % = Z (4.32) 10. Width of transformer The horizontal width of transformer can be calculated using the dimensions of windings used to determine the parameters of the ATD (Ampere-turn diagram). Thus, we get, Width of transformer arrangement = 2588.92 mm(4.33) 11. Core Losses Wewishtodeterminethecorelossesofthetransformer.Weknowthatlossesinthe core material used are 1.6 watts per kilogram weight of material. Thus, we wish to find theweightofthecore.Sinceweknowthedensityofcorematerial,wecanfindits weight by determining the volume of core material. Thisvolumecanbedeterminedbycalculatingthevolumesoftheverticaland horizontal limbs of the core (which are cylinders). Since we know the core diameter, we can determine the volume of the core structure. Volume of core = 1.29*109 mm3 Density of core material = 7.65*106 kg/mm3 Weight of core structure = Volume*Density = 9689.48 kg .(4.34) Thus, Core losses = 9689.48*1.6 = 15.79 KW .(4.35) This value of core losses is well within the limits prescribed for the design (20 KW). 12. Copper Losses Now, let us determine the copper losses in this transformer. Copper Loss = (Current)2 x Resistance (4.36) For this, we must determine the resistance of the windings in the transformer. This can be determined using the formula, Resistance = (l)/a where, = Resistivity of conductor material (copper) = 1.68x10-8 -m l=Length of conductor a =Cross-sectional area of conductor Hence, we must determine the length of winding. We compute the LV and HV winding lengths by considering the circumference of the winding. We have used the disc type of winding for both LV and HV windings. Also, we have used many conductors in parallel to account for the current density, in the case of each winding. Hence, we evaluate the lengthofoneconductorinasinglediscofthewindingasthemiddleconductorinthe windingassumingthatallconductorsinonedischavethesamelength.Wethen multiplythislengthwiththenumberofconductorsineachdiscandthenthetotal number of discs, to obtain the total length of conductors for that winding. We can then obtaintheresistanceofthewindingandfurther,thecopperlossesofthetransformer using the pre-stated formulae. Length of LV winding = 344253 mm Resistance of LV winding = R1 = 0.021 Length of HV winding = 736283 mm Resistance of HV winding = R2 =0.067 Length of tapping winding = 158094 mm Resistance of tapping winding = R3 =0.014 Copper losses in LV winding = (ILV)2 * R1 = 36.21 KW .(4.37) ForHVwinding,wewillhavethreeimportantvaluesoflosses,oneforthenormal 100%tapconditionsandtwoothers,whicharelimitsoflossesforthemaximumand minimum tapping conditions. Copper losses in HV winding for normal tap = (IHV)2 * [R2 + (0.333*R3)] = 43.87 KW.(4.38) Copper losses in HV winding for minimum tap (90%) = (IHV)2 * R2 = 47.37 KW .(4.39) Copper losses in HV winding for maximum tap (105%) = (IHV)2 * [R2 + R3] = 42.29 KW...(4.40) Total copper losses of transformer, For normal tap = 80.09 KW For minimum tap = 83.59 KW For maximum tap = 78.51 KW (4.41) Sincethetransformerwillusuallybeoperatedatnormaltap,thelimitonthecopper losses (80 KW) can be restricted to the normal tapping condition alone. 13. Cost of core and winding materials: Further, we shall determine the costing of the transformer. We must compute the cost of coreaswellasthecostofconductorsusedinordertoobtainaroughestimateofthe material cost of the transformer. a.Cost of core material Wehavecalculatedtheweightofcoreearlier.Usingthemarketcostofcore material as Rs.165/kg, we get, Cost of core material = Rs.16,28,465 ...(4.42) b.Cost of conductor material Wemustalsodeterminetheweightofconductormaterial.Thiscanbedoneby findingthevolumeoftheconductormaterialused(sinceweknowdensityof copper). The volume can easily be found out since we know cross-sectional area of conductor in each winding and also its length. Volume of LV conductors = 94.64*106 mm3 Volume of HV conductors = 136.19*106 mm3 Volume of tapping conductors = 29.22*106 mm3 Total volume of conductors = 260.06*106 mm3 Density of copper = 8.96*106 kg/mm3 Weight of conductors = Volume*Density = 2330.17 kg Cost of copper = Rs.200/kg Cost of conductor material = Rs.4,66,034.. (4.43) Total cost = Cost of core material + Cost of conductor material = Rs.20,94,500 ...(4.44) 14. Radial stresses on windings: We have to ensure that the transformer that has been designed can withstand the forces impartedonthewindinginthecaseoftheoccurrenceofashortcircuit.Forthis,we consider an asymmetry factor of 1.8 for the transformer. Since we know the copper loss andheightforeachwindingandthepercentimpedanceofthetransformer,wecan easilydeterminetheradialstress.Thesevaluesarefoundtobewithinpermissible limits. Radial stress on LV winding = 21.48 MPa Radial stress on HV winding = 26.01 MPa. (4.45) 15. Number of spacers: SincethereisacompressiveradialforceonLVwinding,wecalculatenumberof spacers using the value of stress on inner winding. The same number of spacers is used for HV winding as well, since there is no compressive force on it. Number of Spacers = 11.85 16...(4.46) The number of spacers should always be taken as multiples of 4. The reasons as to why this number is rounded off to 16 are specified during discussion of axial stresses. 16. Axial stresses on windings and insulation blocks: We always design a transformer to withstand axial forces on winding equal to one-third ofthemaximumvalueofforce.Now,sinceweknowthevalueofpowerratingof transformer, height of winding, percentage impedance of transformer andfrequency of voltagetobesupplied,wecaneasilydeterminethetotalaxialforceonwinding.We assumethat70%ofthisforceisappliedtoLVwindingwhiletheremaining30%is applied to HV and tap windings. Total axial force on windings = 533273 N Axial force on LV winding = 373291 N Axial force on HV and tap windings = 159982 N .(4.47) Every insulation block between windings is attached to its corresponding vertical spacer for ease in assembly. Hence the number of spacers is equal to the number of insulation blocks. Also, we design these blocks such that they cover only 30-35% of the area underneath the winding. This is to allow enough area for the oil to pass through over the conductors to cool the winding. Firstly, we take into account 30% of the circumference of each winding, which is taken aroundthemeanradiusofthewinding.(Thiswillaccountfor30%oftheareasince thickness for winding and thickness for block is same) 30% of circumference along mean radius of LV winding = 501.27 mm 30% of circumference along mean radius of HV winding = 669.26 mm 30% of circumference along mean radius of tap winding = 790.37 mm (4.48) Insulationsblocksareavailablewithwidthsof35,45and55mmonly.Hencethis constraint is imposed on their selection. We divide the circumference obtained with the numberofinsulationblockstogetthedesiredwidthofblock.Thiswidthisthen rounded off to the nearest specification of block. Calculated width of insulation block for LV winding = 31.329 mm Calculated width of insulation block for HV winding = 41.828 mm Calculated width of insulation block for tap winding = 49.39 mm (4.49) Using this and knowing the thickness of windings, we calculate the area of block under the winding and calculate the stress on the insulation blocks as, Stress = (Axial force on winding) / (Total Area of blocks under winding) . (4.50) Accordingtopreviouscalculations,thenumberofspacersshouldhavebeenrounded off to 12. However, such a selection in the design causes the stresses on insulation block to rise above the permissible limit of 10 MPa (even for the maximum block width of 55 mm). Hence, we choose number of spacers as 16. The width of the block is dependent on two factors; area coverage under the winding, and the stress on block for a particular width. Area of block under LV winding = 0.0377 m2 Area of block under HV winding = 0.0396 m2 Area of block under tap winding = 0.0099 m2 . 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