2. POWER TRANSFORMERS

Transcription

1 . OWE TANFOME.. ntrodution This hpter is intended to over mjor omponent of the power system, whih is the power trnsformer. Trnsformers mke lrge power systems possible. n order to trnsmit hundreds of megwtts of power effiiently over long distnes, very high line voltges re neessry to derese the line losses. ower t low voltges is lso neessry to be used t sfe level in home pplines nd most industril equipment. ower trnsformers n be lssified s: tep-up trnsformers to be onneted between the genertor nd the trnsmission line. They permit prtil design voltge for genertors, nd t the sme time n effiient trnsmission line voltge. tep-down trnsformers onneted between the trnsmission line nd vrious eletril lods. They permit the trnsmitted power to be used t sfe utiliztion voltge... Types of ower Trnsformers The elements of power trnsmission nd distribution system re shown in Figure.. The outt terminls of genertors re usully onneted diretly to genertor step-up unit (GU of equl rting. The GU steps the voltge of the genertor up to the desired trnsmission voltge. Figure. is typil trnsformer. At the reeiving ends of the trnsmission system re substtions, t eh of whih there re one or more power trnsformers. They redue the voltge to the sub-trnsmission levels. The sub-trnsmission iruits fn out from the substtion to distribution substtions lted t lod enters. At the lod enters, smll power trnsformers further redue the voltge to distribution levels. Distribution iruits go to industril lods or residentil distrits where the voltge is redued to the finl utiliztion voltge. The ll trnsformers performing the finl voltge redution re lled distribution trnsformers. A typil single-phse distribution trnsformer is shown in Figure. Three-phse power my be trnsformed by using either two or three single-phse trnsformers, or by single, three-phse trnsformer. When set of single-phse trnsformers is employed to trnsform three-phse iruit, it is lled three-phse trnsformer bnk. ome other types of trnsformers re used in mesuring voltge, urrent, nd power flow in the power system. The mjorities re potentil trnsformers nd urrent trnsformers. otentil trnsformers (T re single-phse trnsformers of speil design, whih step down the voltge to be mesured to sfe vlue. Current trnsformers (CT step down the urrents nd hve insultion dequte to isolte metering equipment nd personnel from the line voltge. One terminl of the seondries of both potentil nd urrent trnsformers is usully grounded for sfety... Elements of Trnsformers The trnsformer onsists of two or more insulted windings wrpped round n iron ore. By definition, the primry winding is the int winding, nd the seondry winding is the outt winding.

2 7.6 k, φ distribution line Ciruit Breker mll ower Trnsformer Turbine 0 k Alterntor 0/40, φ utiliztion iruit Distribution ubsttion Ciruit Breker Boiler or etor Genertor tep-up Trnsformer 6 k,φ sub-trnsmission line Genertion ttion Ciruit Breker ubsttion Ciruit Breker 45 k,φ trnsmission line Ciruit Breker ower Trnsformer Ciruit Breker Ciruit Breker Figure.. Elements of ower Trnsmission nd Distribution ystem. Figure.. A 500 MA ower Trnsformer. (Courtesy of ABB

3 Figure.. A ole-mount 5 ka Distribution Trnsformer. (Courtesy of ABB The ore is formed of stk of steel lmintions. The steel hs high mgneti permebility nd provides high-performne pth for the flux, whih is mutul to the primry nd seondry windings. The ore is built up of thin lmintions, whih re eletrilly insulted from eh other. Two types of ore onstrution re used s shown in Figure.4. The first is known s the ore type nd the seond is denoted the shell-type. φ φ / φ / Core Type b hell Type Figure.4. Core-Type, b hell-type. The three tive elements of trnsformer re the primry winding, seondry winding, nd the ore. There re three types of insultion-ooling methods. Dry-type trnsformers operte in ir. Cst-Coil trnsformers re enpsulted in n epoxy resin. Most power trnsformers re immersed in tnk of oil. The oil is better insultor thn ir. The ends of the windings re brought to terminl blk from whih leds re brought to the outside of the tnk through insulting bushings, mounted in holes in the sides or top of the tnk. The high-voltge bushings re seen lerly in Figure.. When the primry is onneted to n voltge soure, n lternting flux is set up in the ore. This flux indues voltges in ll windings. The voltge indued in eh winding, ording to Frdy s Lw, is proportionl to the number of turns in tht winding. The voltge indued in the primry is nerly equl to the pplied voltge, nd the voltge t the seondrywinding terminls lso differs by only few perent from the voltge indued into tht winding. Thus the primry-to-seondry voltge rtio is essentilly equl to the rtio of the number of turns in the two windings. The turns-rtio is given the symbol.

4 N N (. where N is the number of primry oil turns, N is the number of seondry oil turns, nd, re the voltges t the winding terminls. By seleting the proper turns-rtio, the trnsformer designer n determine the rtio of int to outt voltges to meet the requirements of the power system..4. Generl Theory of Trnsformer Opertion Trnsformers operte on the bsis of Frdy s Lw: dλ dφ e ± ± N (. dt dt where e is the instntneous voltge indued by mgneti field, λ is the number of flux linkges between the field nd the eletri iruit in whih the voltge is being indued, nd φ is the effetive flux. The sign depends on Lenz s Lw nd the polrity of the iruit terminls. f the winding resistne is negleted, then dφ dφ v e N, v e N (. dt dt Tking the voltge rtio in Eqution., we see tht N e (.4 N e Negleting losses mens tht the instntneous power is the sme on both sides of the trnsformer, s e (.5 i ei Combining Equtions.,.4 nd.5, we get N v i N v i (.6 As mentioned before, the trnsformer tion requires the existene of the flux tht links the two windings. This will be obtined more effetively if n iron ore is used; beuse, n iron ore onfines the flux to definite pth linking both windings. A mgneti mteril suh s iron undergoes loss of energy due to the pplition of lternting voltge to its mgnetiztion hysteresis loop (B-H urve s shown in Figure.5. 4

5 B (T H (A/m Figure.5. A Mgneti Hysteresis Loop or B-H Curve of Core teel. The losses re omposed of two prts. The first is lled the eddy-urrent loss, nd the seond is the hysteresis loss. Eddy urrent loss is bsilly loss due to the indued urrent in the mgneti mteril. To redue these losses, the mgneti iruit is usully mde of stk of thin lmintions. Hysteresis loss is used by the energy used in orienting the mgneti domins of the mteril long the field. The loss depends on the mteril used. Due to the non-linerity of the B-H urve of the mgneti mteril, the primry urrent will not be sinusoid but rther ertin distorted version, whih is still periodi. For nlysis rposes, we pproximte the urrent with sinusoidl funtion of the fundmentl frequeny. This pproximted primry urrent is mde of two omponents. The first is in-phse with the voltge nd is ttributed to the power tken by eddy-urrent nd hysteresis losses nd is lled the ore-loss omponent,, of the exiting urrent, φ or o. The seond omponent lgs the voltge by 90 nd is lled the mgnetizing urrent, m..4.. riniple of Opertion n Figure.6 the bsi omponents of the trnsformer re the ore, primry winding, nd seondry winding. f the flux φ M is the mutul (or ore flux linking N nd N, then ording to Frdy s Lw of eletromgneti indution, eletromgneti fores (emf s re indued in N nd N due to time rte of hnge of φ M suh tht: nd d e φ N M dt (.7 d e φ N M dt (.8 The diretion of e is suh to produe urrent tht opposes the flux hnge, ording to Lenz s Lw. f the flux vries sinusoidlly suh tht φ φ mx ω (.9 M sin t then the orresponding indued voltge, e, linking n N-turn winding is given by 5

6 dφ e N ω N φ mx osω t (.0 dt The rms vlue of the indued voltge is ω N φ mx E 4.44 f N φmx (. where f is the frequeny in Hertz. Eqution. is known s the emf eqution. oure N E φ Lod E N Figure.6. Bsi Trnsformer Components. EXAMLE. How mny turns must the primry nd the seondry windings of 0-0, 60 Hz idel trnsformer hve if the ore flux is not llowed to exeed 5 mwb? olution: For n idel trnsformer with no losses, E E 0 0 From the emf eqution, we hve N E 4.44 f φ mx (0 (4.44(60( turns N (0 (4.44(60(5 0 8 turns.5. The del Trnsformer Model Mny engineering lultions n be rried out with the ssumption tht trnsformers re idel. An idel trnsformer would hve windings with zero impedne nd lossless, infinite permebility ore. Therefore, the effiieny would be 00%. This ssumption does not 6

7 introdue muh error beuse power trnsformers re very nerly idel. Their effiienies re bout 97% or better nd their internl voltge drops re only bout 5%. nfinite permebility of the ore would result in zero exiting urrent nd no lekge flux. Then λ N φ, λ N φ ero resistnes of the windings would result in zero voltge drops between the terminl voltges nd the indued voltges (see Figure.7. v e, v e For idel trnsformer, Eqution.6 n be written s: N v e i N v e i (. Figure.7 shows n idel trnsformer onneted between voltge soure nd onstnt impedne lod. Let nd represent the seondry rms voltge nd urrent. n phsor terms 0 θ LD θ where LD is the impedne of the lod. The int impedne seen by the soure is given by in LD Therefore, the bsi reltions tht deribe the behvior of the idel trnsformer n be given s: N v e i N v e i in LD (. where in is the int impedne t the primry side nd LD is the impedne tht is onneted to the seondry. del oure N N Lod N N Figure.7 Lod Conneted to oure by n del Trnsformer. 7

8 EXAMLE. Consider n idel, single phse trnsformer. The primry is onneted to 00 soure nd the seondry is onneted to n impedne of Ω 6.9. Find the seondry outt urrent nd voltge. b Find the primry int urrent. Find the lod impedne s seen from the primry side. d Find the int nd outt pprent powers. e Find the outt power ftor. olution: t is stndrd prtie to mke the rtio of rted terminl voltges equl to the tul turns-rtio s: b primry urrent 0 0 0A 6.9 Ω 6.9 0A A lod impedne s seen from the primry side. in Ω 6. 9 A 6.9 d int power: ( 00 0 ( A ka outt power: ( 0 0 ( 0A ka e power ftor: F os φ os( lgging.7. Non-idel Trnsformer nd the Ext Equivlent Model The non-idel trnsformer hs hysteresis nd eddy-urrent (or ore losses, nd resistive (or losses in the primry nd seondry windings. Furthermore, the ore of non-idel trnsformer is not perfetly permeble, nd the trnsformer ore requires finite mgneti motive fore (mmf for its mgnetiztion. Also, not ll fluxes link with the primry nd the seondry windings simultneously beuse of lekge round the windings. This is lled lekge flux. 8

9 An equivlent iruit of the non-idel trnsformer is shown in Figure.9. The dot mrkings indite terminls of orresponding polrity in the sense tht both windings enirle the ore in the sme diretion if we begin t the dots. By using Eqution., the idel trnsformer my be removed from Figure.9, nd the entire ext equivlent iruit my be referred either to the primry, s shown in Figure.0, or to the seondry, s shown in Figure.. / X 0 X Xm m E E L N N Figure.9. Ext Equivlent Ciruit of Non-del Trnsformer. / X 0 X Xm E E L m Figure.0. Ext Equivlent Ciruit s eferred to the rimry. / X/ 0 Xm/ X m / E / E L / Figure.. Ext Equivlent Ciruit s eferred to the eondry. The vrious symbols used in Figures.0 nd. re: turns rtio E, E primry nd seondry indued voltges, primry nd seondry terminl voltges, primry nd seondry urrents φ, 0 no lod urrent 9

10 r, x r, x m, X m, primry winding resistne nd retne seondry winding resistne nd retne mgnetizing urrent nd retne ore loss urrent nd resistne The mjor use of the ext equivlent iruit of trnsformer is to determine the hrteristis suh s voltge regultion nd effiieny. A phsor digrm for the iruit of Figure., for lgging power ftor n be obtined by using the following equtions: 0 + m (.4 where is in phse with E nd m lgs E by (.5 ( r E + (.6 + jx ( r + (.7 E + jx Bsed on Equtions.4 through.7 nd ssuming zero degree referene ngle for, the phsor digrm is shown in Figure. for the ext equivlent iruit model of trnsformer. C E E jx θ jx M 0 / Figure.. M hsor Digrm for the Ext Equivlent Ciruit Model of Trnsformer..8. The Approximte Trnsformer Ciruit Model The ext equivlent trnsformer model is more urte thn is neessry for most engineering lultions. A simpler model, the pproximte iruit, is most frequently used. The pproximte iruit of the trnsformer referred to the primry side is shown in Figure.. 0

11 / 0 X X m E E Xm 0 eq_ Xeq_ m Xm Figure.. Approximte Ciruit Model of Trnsformer eferred to the rimry. The rtionl for these pproximte equivlent iruits is tht the voltge in the primry series impedne (r + jx is smll, even t full lod. Also, the no lod urrent ( 0 is so smll tht its effet on the voltge drop in the primry series impedne is negligible. Therefore, it mtters little if the shunt brnh of in prllel with X m is onneted before the primry series impedne or fter it. The ore loss nd mgnetizing urrents re not gretly ffeted by the move. Conneting the shunt omponents right t the int terminls hs the gret dvntge of permitting the two series impedne to be ombined into one omplex impedne. The vlue of this equivlent impedne of prtiulr trnsformer depends, of ourse, on whether the model used is referred to the primry or seondry. Figure.4 shows the pproximte equivlent iruit of trnsformer referred to the seondry side. 0 eq_ Xeq_ / m / Xm / Figure.4. Approximte Ciruit Model of Trnsformer eferred to the eondry. f the iruit is referred to the primry s shown in Figure., ( r + r + j( x x eq _ eq _ + jx er _ + (.8 f the iruit is referred to the seondry s shown in Figure.4,

12 ( r + r + j( x x eq _ eq _ + jx er _ + (.9.9. Trnsformer Chrteristis The mjor use of the ext nd pproximte equivlent iruits of trnsformer is in determining its hrteristis. The hrteristis of most interest to power engineers re voltge regultion nd effiieny..9.. oltge egultion oltge regultion is mesure of the hnge in the terminl voltge of the trnsformer with respet to loding. Therefore the voltge regultion is defined s: no lod lod perent regultion 00 (.0 With referene of Figure.4, lod + eq _ (. nd t no lod ( 0 no lod Eqution.0 n be written s: perent regultion 00 (. t is ler from Eqution. tht the terminl voltge is lod dependent. Exmples of trnsformer voltge regultion re shown in Figure.5.

13 0 eq_ Xeq_ / 0 LD θ ositive egultion / Negtive egultion / jxeq_ jxeq_ θ eq_ θ eq_ ero egultion / Mximum egultion jxeq_ / θ eq_ θ eq_ jxeq_ Figure.5. Exmples of Trnsformer oltge egultion..9.. Trnsformer Effiieny Effiieny of trnsformer is defined s follows: outt int power power η (. For non-idel trnsformer, the outt power is less thn the int power beuse of losses. These losses re the winding or loss (opper losses nd the ore loss (hysteresis nd eddy-urrent losses. Thus, in terms of the totl losses, losses, Eqution. my be expressed s: + losses η (.4 losses + opper Obviously, the winding or opper loss is lod dependent, wheres the ore loss is onstnt nd lmost independent of the lod on the trnsformer. The effiieny n lso be obtined by using the per-unit system. Dividing both the numertor nd denomintor by the rted power bse of the trnsformer. + ore

14 or Bse η (.5 η Bse + opper Bse + ore Bse ( Bse osφ ( Bse osφ + opper Bse + ore Bse (.6 n effiieny lultions, it is usully ssumed tht the seondry voltge is t rted vlue. With onstnt terminl voltge, the mgnitude of the seondry urrent is proportionl to the volt-mpere lod (pprent power of the lod. The winding or opper loss, whih is funtion of the urrent, is therefore proportionl to the squre of the lod. As result, opper or in per unit u Bse u full lod ine u u full lod then full lod u u Bse full lod.0 full eq lod eq eq The per-unit form of the effiieny expression beomes osφ η (.7 osφ + + ore eq At full lod, -.0; therefore, the effiieny t full lod is η FL osφ osφ + ore + eq (.8 where os φ is the lod power ftor..9.. Mximum Effiieny t is often desirble to design trnsformer to hve mximum effiieny t some prtiulr loding. When the voltge nd frequeny re both onstnt, the ore losses re nerly onstnt. Therefore, the losses my then be divided into two tegories: Constnt losses suh s ore losses,. 4

15 rible losses suh s winding losses, Eqution.4 n be written s: k η (.9 k + + eq _ where k represents the outt power nd k represents onstnt. For exmple, in singlephse trnsformer, the lod s power ftor. The effiieny will be mximum or minimum when or η 0, or when ( k + + eq _ ( k ( k ( k + eq _ ( k + + eq _ eq _ 0 eq _ u (.0 0 Thus mximum effiieny urs when the vrible losses equl the onstnt losses. EXAMLE. A 0 ka, 400/40, single-phse trnsformer hs the following resistnes nd lekge retnes: r.0ω x 5.0Ω r 0.0Ω x 0.5Ω Use the pproximte iruit model to find the voltge regultion when the lod power ftor is: 0.8 lgging, nd b 0.8 leding. olution Clulte the equivlent impedne r x eq _ + r + j + x Ω 5Ω + 0.0Ω + j + 0.5Ω ( j0.0 Ω 0.06Ω At 0.8 F lgging, nd hoosing zero degree voltge referene t the seondry terminl, the full lod seondry urrent is: 5

16 φ ros( Ld 0kA By using the iruit in Figure.4, we find: ( oltge eq _ ( 4.7A 6.9 ( 0.06Ω 78.7 j8.5 regultion % 40 b At 0.8 F leding, the full lod seondry urrent is: (4.7A (0.06Ω 78.7 (4.5 + j oltge regultion % 40 EXAMLE.4 Use the per-unit system to find the voltge regultion in Exmple.. olution: Choose the bse vlues s follows: φ Bse _ Bse _ Bse _ Bse 0kA 400 φ Bse _ Bse _ Bse _ Bse _ Bse 40 0kA 4.7A Ω 4.7A The per-unit lod urrent t full lod t 0.8 F lgging is: 6

17 _ eq _ eq _ Bse Ω Ω ( ( ( oltge regultion % b oltge ( ( ( regultion % EXAMLE.5 A 400/0, 00 ka, single-phse trnsformer hs ore loss t rted voltge of.7 kw. f its equivlent resistne is.4%, find the effiieny of this trnsformer for lod power ftor of 0.9: t full lod nd b t hlf lod. Wht is the lod t mximum effiieny? olution: The per-unit ore loss is given by: φ Bse.7kW 00kA φ Bse (.7kW (00kA The full lod winding losses (opper losses re: FL eq _ u ( FL eq _ (.0 ( Then Eqution.8 n be pplied for 0.9 power ftor. η (0.9 (0.9 + ( (0.04 FL or 97.5% 7

18 b For hlf lod, Eqution.7 provides: Ld φ Bse 0.5 (0.5(0.9 η 0.97 (0.5(0.9 (0.009 (0.5 (0.04 or Ld % Mximum effiieny urs when Ld φ Bse eq Then for mximum effiieny Ld eq n other words, this trnsformer hs its mximum effiieny t bout 80% of its full lod..0. The Three-hse Trnsformer Three-phse power my be trnsformed by bnk of single-phse trnsformers or by single three-phse trnsformer. A three-phse trnsformer is essentilly three trnsformers wound on ommon ore. The geometry of the ore is suh tht the fluxes of the phses shre ommon pths. As result, the volume of iron is less thn tht of three single-phse units of the sme totl rting. These re some dvntges nd disdvntges in using three-phse trnsformer insted of bnk of single-phse trnsformers. The dvntges re: t tkes up less spe t is less expensive. n involves less externl wiring. t hs slightly better effiieny. The disdvntges re: t does not provide the flexibility of set of single-phse units. For exmple, one single-phse trnsformer in bnk my hve higher power rting thn others, to serve n unblned lod. n se of filure of single-phse unit serving in bnk, only tht one unit needs to be repled. However, it is most likely tht dmge within three-phse unit will require omplete removl from servie ll three phses nd replement of the omplete unit. As result, three-phse bnks of single-phse trnsformers re seldom used in new instlltions exept in distribution iruits serving ombintion of single nd three-phse lods. 8

19 n three-phse trnsformtion, the primry nd seondry windings my be onneted independently in either delt or wye. The possible ombintions re listed below nd re shown in Figure.6. delt-delt delt-wye wye-delt wye-wye The wye-wye onnetion is to be voided unless very solid neutrl onnetion is mde between the primry nd the power soure. f neutrl is not provided, the phse voltges tend to beome severely unblned when the lod is unblned. There re lso troubles with third hrmonis. These problems do not exist when one of the sets of windings is in delt onfigurtion. When wye-delt or delt-wye onnetion is used, the wye is preferbly on the highvoltge side in trnsmission systems, nd the neutrl is grounded. The trnsformer insultion my thus be designed for times the line voltge, rther thn for the full line voltge. ometimes it is neessry to hve the wye onnetion on the low-voltge side, if the neutrl is required for the low-voltge iruit. Wye-delt nd delt-wye onnetions result in 0 phse displement between primry nd seondry line voltges. t is stndrd prtie in the United ttes to onnet these trnsformers in suh wy tht the lower voltges lg the higher voltges by 0. The ext nd the pproximte equivlent iruit models of the trnsformer deribed in the lst setions n be pplied to the three-phse trnsformer to represent just one of the three phses. 9

20 0 A B C A B C b b A B C b A B C b A B C b A B C b A B C A B C b b Figure.6. Bsi Trnsformer Connetions for Three-hse Units. EXAMLE.6 Wht should be the rtings (voltges nd urrents nd turns rtio of three-phse trnsformer to trnsform 0 MA from 0 k to 460, if the trnsformer is to be onneted: wye-delt, b delt-wye, nd delt-delt? olution For both delt nd wye onnetions, the line urrents n be obtined s:

21 L L φ φ L L 6 ( 0 0 A ( ( 0 0 A ( 460 The rted power per phse: 5.A 88A 0MA. ka φ φ wye-delt onneted: p _ p _ p _ p _ L p _ p _ L L L 5.A 460.8k 460 ( 88A ( 0k.9 80A.8k b delt-wye onneted: p _ p _ p _ p _ L p _ p _ L L L 88A 0k 0k 400 ( 5.A ( 460 delt-delt onneted: p _ p _ p _ p _ L p _ p _ L L L 0k 460 0k ( 5.A ( 88A A A 80A EXAMLE.7 A 700/08, 50kA, three-phse distribution trnsformer is onneted delt-wye. The trnsformer hs.% resistne nd 5% retne. Find the voltge regultion t full lod, 0.8 power ftor lgging.

22 olution: Delt Connetion: p L p L kA ( A L 4.0A.A Wye Connetion: p L p L kA 8.8A p p L ( The bse impedne in the seondry side: _ Bse eq _ eq _ p p eq _ eq _ Ω 8.8A + jx _ Bse eq _ At full lod, 0.8 power ftor, lgging 8.8A eq _ j (0.865Ω Ω 76.5 Note tht ll the quntities in this eqution hve to be phse quntities. The no-lod seondry phse voltge is: j ( 8.8A 6.9 ( The no-lod seondry line voltge: _ LL 8 egultion: ( 4.

23 ( % The problem n lso be solved by using the per-unit method s: eq j % ( 6.9 ( EXAMLE.8 f the ore loss of the trnsformer in Exmple.7 is kw, find the effiieny of this trnsformer t full lod nd 0.8 power ftor. olution: Bse 50kA kw kA Applying Eqution.8 η FL %.. Mesuring Trnsformer Quntities... ingle-hse Trnsformer The prmeters of the pproximte equivlent iruit model of trnsformer my be determined by two tests: open-iruit test nd b short-iruit test. OEN-CCUT TET The open iruit test is onduted by pplying rted voltge t rted frequeny to one of the windings, with the other windings open iruited. The int power nd urrent re mesured. For resons of sfety nd onveniene, the mesurements re mde on the low-voltge (L side of the trnsformer.

24 OC eq_l X eq_l H / 0 0 OC _L X m_l OC OC m m 0 OC Figure.7. Equivlent Ciruit of the Open-Ciruit Test. The equivlent iruit for the open-iruit test is s shown in Figure.7. ine the high voltge (H side is open, the int urrent is equl to the no lod urrent or exiting urrent ( 0, nd is quite smll. The voltge drops in the primry lekge retne nd winding resistne my be negleted nd so my the primry loss ( r. The int power is lmost equl to the ore loss t rted voltge nd frequeny. ore ore (. (. os θ (. θ is the ngle by whih o_l lgs. The ore loss urrent, is in phse with while m lgs by 90. Then nd osθ (.4 m sinθ (.5 + (.6 0 m The ore-loss urrent,, my be found from Eqution., then _L my be lulted by Eqution. or by _ L (.7 The mgnetiztion urrent m is given by Eqution.5 or my be found from nd using Eqution.6. Then X m _ L (.8 m 4

25 HOT-CCUT TET The short-iruit test is used to determine the equivlent series resistne nd retne. One winding is shorted t its terminls, nd the other winding is onneted through proper meters to vrible, low-voltge, high-urrent soure of rted frequeny. The soure voltge is inresed until the urrent into the trnsformer rehes rted vlue. To void unneessry high urrents, the short-iruit mesurements re mde on the high-voltge side of the trnsformer. The test iruit with the effetive equivlent iruit is shown in Figure.8. C eq_h X eq_h 0 (neglet C Figure.8. Equivlent Ciruit of the hort-ciruit Test. Negleting 0, the int power during this test is onsumed in the equivlent resistne referred to the primry or high-voltge side, eq_h. Then nd eq _ H eq _ H (.9 ine rted urrent is used, the winding or opper loss during the short-iruit test is equl to the full-lod opper loss: opper From Figure.8, eq _ H ( full lod eq _ H + X eq _ H Then, hving found eq_h from Eqution.9, (.40 or X eq _ H (.4 eq _ H eq _ H where X sinθ eq _ H eq _ H (.4 5

26 θ ros... Three hse Trnsformer All the mesured quntities in three-phse trnsformer re line voltges, line urrents, nd totl power. The impednes must be lulted on phse bsis. OEN-CCUT TET Mesurements re mde on the low-voltge side nd then onverted to phse quntities. t is neessry to know whether the low-voltge windings re delt or wye onneted. The test is mde t rted low voltge nd rted frequeny. Delt-Conneted Low-oltge Winding φ _ p _ p _ φ _ ore _ L osθ m X m _ L ( m sinθ p _ Wye-Conneted Low-oltge Winding φ _ p _ p _ φ _ ore _ L osθ m X m _ L sinθ ( p _ m HOT-CCUT TET The low-voltge terminls re shorted together. The voltge pplied to the high-voltge terminls is djusted so tht rted urrent flows. The frequeny of the voltge soure is the rted frequeny of the trnsformer. The onnetion (delt or wye of the high-voltge windings should be known. 6

27 Delt-Conneted High-oltge Winding φ _ p _ p _ φ _ opper X eq _ H eq _ H eq _ H ( eq _ H eq _ H Wye-Conneted High-oltge Winding φ _ p _ p _ φ _ opper X eq _ H eq _ H eq _ H ( eq _ H eq _ H EXAMLE.9 Consider 50 ka, , three-phse, 60 Hz, delt-wye trnsformer. The opennd short-iruit tests re s follows: Find: 500 W 8.0 A W 4.0 A 70 The prmeters of the pproximte equivlent iruit referred to the high-voltge winding, then drw the iruit with its prmeters. b The prmeters of the pproximte equivlent iruit referred to the low-voltge winding. The per-unit equivlent series resistne nd retne. d The voltge regultion t full lod, 0.8 power-ftor lgging. e The effiieny t full lod nd t 0.8 power ftor. Wht is the effiieny t 5% lod nd the sme power ftor? olution: From the open iruit test, the totl ore loss 500 W. 7

28 The low voltge side is wye onneted. Then: φ _ θ p _ p _ X _ m 500W 66.7W A ( 08 L 500W ros m _ L Ω 500W ( 08 ( 8.0A ( 8.0A sin( A 0 5.Ω 7.88A 80.0 From the short-iruit test, the full lod winding losses 600W. The high-voltge side is delt onneted. X eq _ H eq _ H eq _ H 600W ( 4.0A ( A 7.Ω 59.8Ω ( 59.8Ω ( 7.Ω 55.4Ω eferring nd X m to the high voltge side: N N X _ H m _ H _ p _ p _ L 5.Ω Ω ( 60 ( kΩ kω The equivlent iruit referred to the high voltge side is shown in Figure.9. 8

29 l 7. Ω j55.4 Ω 0 L / 60 kω j54.8 kω L 60 L Figure.9. Equivlent Ciruit of the Trnsformer in Exmple.9. b The prmeters referred to the low voltge side re: X X _ L _ L eq _ L eq _ L 86.5Ω 5.Ω 7.Ω 0.006Ω ( Ω 0.04Ω ( 60 The prmeters referred to in per-unit re: or X X eq Bse eq eq m ( 08 50kA ( ( 0.05 ( 0.0 ( (.0 Bse 600W kA sinθ 0.865Ω,.0 ( 500W 50kA eq ( 8.0A /8.8A 0.006Ω Ω (

30 d egultion: %.0 e Full lod effiieny is: η FL osφ osφ + + ( ( j eq 5% loding effiieny is: 0.5( 0.8 ( ( 0.5 ( % W 50kA η 5 % %