Chapter 2 Transformer: Principles and Practices

Transcription

1 hpter 2 Trnsformer: Priniples nd Prties 2.1 Introdution There re mny devies suh s three-phse genertor, trnsformer et., whih re usully used in power sttion to generte nd supply eletril power to power system network. In the power sttion, the three-phse genertor genertes three-phse lternting voltge in the rnge etween 11 nd 20 kv. The mgnitude of the generted voltge is inresed to 120 kv or more y mens of power trnsformer. This higher mgnitude of voltge is then trnsmitted to the grid susttion y three-phse trnsmission lines. lower line voltge of 415 V is hieved y stepping down either from the 11 or 33 kv lines y distriution trnsformer. In these ses, three-phse trnsformer is used in either to step-up or step-down the voltge. Sine trnsformer plys vitl role in feeding n eletril network with the required voltge, it eomes n importnt requirement of power system engineer to understnd the fundmentl detils out trnsformer long with its nlytil ehvior in the iruit domin. This hpter is dedited towrds this gol. On the onset of this disussion it is worth mentioning tht trnsformer, irrespetive of its type, ontins the following hrteristis (i) it hs no moving prts, (ii) no eletril onnetion etween the primry nd seondry windings, (iii) windings re mgnetilly oupled, (iv) rugged nd durle in onstrution, (v) effiieny is very high i.e., more thn 95 %, nd (vi) frequeny is unhnged. 2.2 Working Priniple of Trnsformer Figure 2.1 shows shemti digrm of single-phse trnsformer. There re two types of windings in single-phse trnsformer. These re lled primry nd seondry windings or oils. The primry winding is onneted to the lternting Springer Siene+usiness Medi Singpore 2016 Md.. Slm nd Q.M. Rhmn, Power Systems Grounding, Power Systems, DOI / _2 49

2 50 2 Trnsformer: Priniples nd Prties Fig. 2.1 Single-phse trnsformer voltge soure nd the seondry winding is onneted to the lod. The primry nd seondry winding prmeters re represented y the suffix p or 1 nd s or 2, respetively. sinusoidl urrent flows in the primry winding when it is onneted to n lternting voltge soure. This urrent estlish flux u whih moves from the primry winding to the seondry winding through low relutne mgneti ore. out 95 % of this flux moves from the primry to the seondry through the low relutne pth of the mgneti ore nd this flux is linked y the oth windings nd smll perent of this flux links to the primry winding. ording to the Frdys lws of eletromgneti indution, voltge will e indued ross the seondry winding s well s in the primry winding. Due to this voltge, urrent will flow through the lod if it is onneted with the seondry winding. Hene, the primry voltge is trnsferred to the seondry winding without hnge in frequeny. 2.3 Flux in Trnsformer The urrent in the primry winding estlishes flux. The flux tht moves from primry to seondry nd links oth the windings is lled the mutul flux nd its mximum vlue is represented y / m. Flux whih links only the primry winding nd ompletes the mgneti pth through the surrounding ir is known primry lekge flux. The primry lekge flux is denoted y / 1l. Similrly, seondry lekge flux is tht flux whih links only the seondry winding nd ompletes the mgneti pth through the surrounding ir. The seondry lekge flux is denoted y / 2l. Mutul nd lekge fluxes re shown in Fig. 2.2.

3 2.4 Idel Trnsformer 51 φ m I 2 + V 1 I 1 N 1 E 1 φ 1l φ 2l E 2 N 2 + V 2 Z L Fig. 2.2 Mutul nd lekge fluxes 2.4 Idel Trnsformer n idel trnsformer is one whih does not supply ny energy to the lod i.e., the seondry winding is open iruited. The min points of n idel trnsformer re (i) no winding resistne, (ii) no lekge flux nd lekge indutne, (iii) self-indutne nd mutul indutne re zero, (iv) no losses due to resistne, indutne, hysteresis or eddy urrent nd (v) oeffiient of oupling is unity. Figure 2.3 shows n idel trnsformer where the seondry winding is left open. smll mgnetizing urrent I m will flow in the primry winding when it is onneted to the lternting voltge soure, V 1. This mgnetizing urrent lgs ehind the supply voltge, V 1 y 90 nd produes the flux u, whih indues the primry nd seondry emfs. These emfs lg ehind the flux, u y 90. The mgnitude of primry indued emf E 1 nd supply voltge V 1 is the sme, ut re 180 out of phse s shown in Fig () φ m + V 1 I 1 N 1 E E2 1 N 2 + V 2 () V 2 E2 1 E 90 I = I m 1 V 1 φ m Fig. 2.3 Idel trnsformer nd phsor digrm

4 52 2 Trnsformer: Priniples nd Prties 2.5 E.M.F. Eqution of Trnsformer The primry winding drws urrent when it is onneted to n lternting voltge soure. This primry sinusoidl urrent produes sinusoidl flux u tht n e expressed s, Instntneous emf indued in the primry winding is, / ¼ / m sin xt ð2:1þ e 1 ¼ N 1 d/ dt ð2:2þ Similrly, instntneous emf indued in the seondry winding is, e 2 ¼ N 2 d/ dt ð2:3þ Sustituting Eq. (2.1) into the Eq. (2.2) yields, e 1 ¼ N 1 d dt ð/ m sin xtþ e 1 ¼ N 1 x/ m os xt e 1 ¼ N 1 x/ m sinðxt 90 Þ ð2:4þ ð2:5þ ð2:6þ The mximum vlue of e 1 is, E m1 ¼ N 1 x/ m ð2:7þ The rms vlue of the primry emf is, E 1 ¼ E p m1 ffiffiffi 2 ð2:8þ Sustituting Eq. (2.7) into Eq. (2.8) yields, E 1 ¼ N 12pf / pffiffiffi m 2 E 1 ¼ 4:44f / m N 1 ð2:9þ ð2:10þ

5 2.5 E.M.F. Eqution of Trnsformer 53 Similrly, the expression of the seondry emf is, E 2 ¼ 4:44f / m N 2 ð2:11þ The primry nd seondry voltges n e determined from Eqs. (2.10) nd (2.11) if other prmeters re known. 2.6 Turns Rtio of Trnsformer Turns rtio is n importnt prmeter for drwing n equivlent iruit of trnsformer. The turns rtio is used to identify the step-up nd step-down trnsformers. ording to Frdy s lws, the indued emf in the primry ðe 1 Þ nd the seondry ðe 2 Þ windings re, e 1 ¼ N 1 d/ dt e 2 ¼ N 2 d/ dt ð2:12þ ð2:13þ Dividing Eq. (2.12) y Eq. (2.13) yields, e 1 e 2 ¼ N 1 N 2 e 1 e 2 ¼ N 1 N 2 ¼ ð2:14þ ð2:15þ Similrly, dividing Eq. (2.10) y(2.11) yields, E 1 E 2 ¼ N 1 N 2 ¼ ð2:16þ where is the turns rtio of trnsformer. In se of N 2 [ N 1, the trnsformer is lled step-up trnsformer. Wheres for N 1 [ N 2, the trnsformer is lled step-down trnsformer. The losses re zero in n idel trnsformer. In this se, the input power of the trnsformer is equl to its output power nd this yields, V 1 I 1 ¼ V 2 I 2 ð2:17þ Eqution (2.17) n e rerrnged s, V 1 V 2 ¼ I 2 I 1 ¼ ð2:18þ

6 54 2 Trnsformer: Priniples nd Prties The rtio of primry urrent to the seondry urrent is, I 1 I 2 ¼ 1 ð2:19þ gin, the mgnetomotive fore produed y the primry urrent will e equl to the mgnetomotive fore produed y the seondry urrent nd it n e expressed s, =¼= 1 = 2 ¼ 0 N 1 I 1 ¼ N 2 I 2 I 1 I 2 ¼ N 2 N 1 ¼ 1 ð2:20þ ð2:21þ ð2:22þ From Eq. (2.22), it is onluded tht the rtio of primry to seondry urrent is inversely proportionl to the turns rtio of the trnsformer. The input nd output power of n idel trnsformer is, P in ¼ V 1 I 1 os / 1 P out ¼ V 2 I 2 os / 2 ð2:23þ ð2:24þ For n idel ondition, the ngle / 1 is equl to the ngle / 2 nd the output power n e re-rrnged s, P out ¼ V 1 I 1 os / 1 P out ¼ V 1 I 1 os / 1 ¼ P in ð2:25þ ð2:26þ From Eq. (2.26), it is seen tht the input nd the output power re the sme in se of n idel trnsformer. Similrly, the input nd output retive powers re, Q out ¼ V 2 I 2 sin / 2 ¼ V 1 I 1 sin / 1 ¼ Q in ð2:27þ From Eqs. (2.26) nd (2.27), the input nd output power nd retive power n e lulted if other prmeters re given. Exmple 2.1 The numer of turns in the seondry oil of 22 kv, 2200 V/220 V single-phse trnsformer is 50. Find the (i) numer of primry turns, (ii) primry full lod urrent, nd (iii) seondry full lod urrent. Neglet ll kinds of losses in the trnsformer.

7 2.6 Turns Rtio of Trnsformer 55 Solution The vlue of the turns rtio is, ¼ V 1 V 2 ¼ ¼ 10 (i) The vlue of the primry turns n e determined s, ¼ N 1 N 2 N 1 ¼ N 2 ¼ ¼ 500 (ii) The vlue of the primry full lod urrent is, I 1 ¼ ¼ 10 (iii) The vlue of the seondry full lod urrent is, I 2 ¼ ¼ 100 Exmple kv single-phse trnsformer hs the primry nd seondry numer of turns of 200 nd 400, respetively. The trnsformer is onneted to 220 V, 50 Hz soure. lulte the (i) turns rtio, nd (ii) mutul flux in the ore. Solution (i) The turns rtio is, ¼ N 1 N 2 ¼ ¼ 0:5 (ii) The vlue of the mutul flux n e lulted s, V 1 ¼ E 1 ¼ 4:44f / m N 1 / m ¼ V ¼ ¼ 4:95 mw 4:44fN 1 4: Prtie prolem 2.1 The primry voltge of n iron ore single-phse trnsformer is 220 V. The numer of primry nd seondry turns of the trnsformer re 200 nd 50, respetively. lulte the voltge of the seondry oil.

8 56 2 Trnsformer: Priniples nd Prties Prtie prolem 2.2 The numer of primry turns of 30 kv, 2200/220 V, 50 Hz single-phse trnsformer is 100. Find the (i) turns rtio, (ii) mutul flux in the ore, nd (iii) full lod primry nd seondry urrents. 2.7 Rules for Referring Impedne For developing equivlent iruit of trnsformer, it is neessry to refer the prmeters from the primry to the seondry or the seondry to the primry. These prmeters re resistne, retne, impedne, urrent nd voltge. The rtio of primry voltge to seondry voltge is [1 4], V 1 V 2 ¼ ð2:28þ The rtio of primry urrent to seondry urrent is, I 1 I 2 ¼ 1 ð2:29þ Dividing Eq. (2.28) y Eq. (2.29) yields, V 1 V 2 I 1 ¼ 1 I 2 ð2:30þ V 1 I 1 V 2 I 2 ¼ 2 ð2:31þ Z 1 Z 2 ¼ 2 ð2:32þ lterntive pproh: The impednes in the primry nd seondry windings re, Z 1 ¼ V 1 I 1 Z 2 ¼ V 2 I 2 ð2:33þ ð2:34þ Dividing Eq. (2.33) y Eq. (2.34) yields, Z 1 Z 2 ¼ V 1 I 1 V 2 I 2 ð2:35þ

9 2.7 Rules for Referring Impedne 57 Z 1 Z 2 ¼ V 1 V 2 I 2 I 1 Z 1 Z 2 ¼ Z 1 Z 2 ¼ 2 ð2:36þ ð2:37þ ð2:38þ From Eq. (2.38), it n e onluded tht the impedne rtio is equl to the squre of the turns rtio. The importnt points for trnsferring prmeters re (i) R 1 in the primry eomes R 1 when referred to the seondry, (ii) R 2 2 in the seondry eomes 2 R 2 when referred to the primry, (iii) X 1 in the primry eomes X 1 when 2 referred to the seondry, nd (iv) X 2 in the seondry eomes 2 X 2 when referred to the primry. Exmple 2.3 The numer of primry nd seondry turns of single-phse trnsformer re 300 nd 30, respetively. The seondry oil is onneted with lod impedne of 4 X. lulte the (i) turns rtio, (ii) lod impedne referred to the primry, nd (iii) primry urrent if the primry oil voltge is 220 V. Solution (i) The vlue of the turns rtio is, ¼ N 1 N 2 ¼ ¼ 10 (ii) The vlue of the lod impedne referred to primry is, Z 0 L ¼ 2 Z L ¼ ¼ 400 X (iii) The vlue of the primry urrent is, I 1 ¼ V 1 ZL 0 ¼ ¼ 0:55 Prtie prolem 2.3 lod impedne of 8 X is onneted to the seondry oil of 400/200 turns single-phse trnsformer. Determine the (i) turns rtio, (ii) lod impedne referred to primry nd (iii) primry urrent if the primry oil voltge is 120 V.

10 58 2 Trnsformer: Priniples nd Prties 2.8 Equivlent iruit of Trnsformer Windings of trnsformer re not onneted eletrilly. The windings re mgnetilly oupled with eh other. In this se, it is tedious to do proper nlysis. Therefore, for esy omputtion nd visuliztion, the prtil trnsformer needs to e onverted into n equivlent iruit y mintining sme properties of the min trnsformer. In the equivlent iruit, the relted prmeters need to e trnsferred either from the primry to the seondry or vie vers. two windings idel trnsformer is shown in Fig Ext Equivlent iruit Figure 2.5 shows n ext equivlent iruit referred to the primry where ll the prmeters re trnsferred from the seondry to the primry nd these prmeters re, I1 R1 X1 R2 X 2 I 2 I 0 N1 N2 + I w + V 1 R 0 X 0 I m E 1 E 2 V 2 Z L Fig. 2.4 Two windings trnsformer I1 R1 X1 R 2 ' X 2 ' I 2 ' I 0 + V 1 R 0 I w X 0 I m V +2 ' ' Z L Fig. 2.5 Ext equivlent iruit referred to primry

11 2.8 Equivlent iruit of Trnsformer 59 R 0 2 ¼ 2 R 2 X 0 2 ¼ 2 X 2 Z 0 L ¼ 2 Z L I 0 2 ¼ I 2 V 0 2 ¼ V 2 ð2:39þ ð2:40þ ð2:41þ ð2:42þ ð2:43þ Figure 2.6 shows the ext equivlent iruit referred to the seondry where ll the prmeters re trnsferred from the primry to the seondry. These prmeters re, R 0 1 ¼ R 1 2 X 0 1 ¼ X 1 2 I 0 1 ¼ I 1 V 0 1 ¼ V 1 I 0 w ¼ I w I 0 m ¼ I m I 0 0 ¼ I 0 ð2:44þ ð2:45þ ð2:46þ ð2:47þ ð2:48þ ð2:49þ ð2:50þ I 1 ' R 1 ' X 1 ' R2 X 2 I2 I 0 ' V +1 ' I w ' R 0 ' X 0 ' I m ' + V 2 Z L Fig. 2.6 Ext equivlent iruit referred to seondry

12 60 2 Trnsformer: Priniples nd Prties pproximte Equivlent iruit The no-lod urrent is very smll s ompred to the rted primry urrent. Therefore, there is negligile voltge drop due to R 1 nd X 1. In this ondition, it n e ssumed tht the voltge drop ross the no-lod iruit is the sme s the pplied voltge without ny signifint error. The pproximte equivlent iruit n e drwn y shifting the no-lod iruit ross the supply voltge, V 1. Figure 2.7 shows n pproximte equivlent iruit referred to the primry. The totl resistne, retne nd impedne referred to the primry re, R 01 ¼ R 1 þ R 0 2 ¼ R 1 þ 2 R 2 X 01 ¼ X 1 þ X 0 2 ¼ X 1 þ 2 X 2 Z 01 ¼ R 01 þ jx 01 ð2:51þ ð2:52þ ð2:53þ The no-lod iruit resistne nd retne re, R 0 ¼ V 1 I w X 0 ¼ V 1 I m ð2:54þ ð2:55þ Figure 2.8 shows n pproximte equivlent iruit referred to the seondry. The totl resistne, retne nd impedne referred to the seondry is, R 02 ¼ R 2 þ R 0 1 ¼ R 2 þ R 1 2 ð2:56þ I 1 I 2 ' R01 X01 I 0 + V 1 I w R 0 X 0 I m V +2 ' Z L ' Fig. 2.7 pproximte equivlent iruit referred to primry

13 2.8 Equivlent iruit of Trnsformer 61 I 1 ' I R 2 02 X 02 I 0 ' V +1 ' I w ' R 0 ' X 0 ' I m ' + V 2 Z L Fig. 2.8 pproximte equivlent iruit referred to seondry X 02 ¼ X 2 þ X 0 1 ¼ X 2 þ X 1 2 Z 02 ¼ R 02 þ jx 02 ð2:57þ ð2:58þ The no-lod iruit resistne nd retne referred to the seondry re, R 0 0 ¼ V 0 1 I 0 w ð2:59þ X 0 0 ¼ V 0 1 I 0 m ð2:60þ Exmple kv, 200 V/40 V single-phse trnsformer hs the primry resistne nd retne of 3 nd 12 Ω, respetively. On the seondry side, these vlues re 0.3 nd 0.1 Ω, respetively. Find the equivlent impedne referred to the primry nd the seondry. Solution The vlue of the turns rtio is, ¼ V 1 ¼ 200 V 2 40 ¼ 5 The totl resistne, retne nd impedne referred to the primry n e determined s, R 01 ¼ R 1 þ 2 R 2 ¼ 3 þ 25 0:3 ¼ 10:5 X X 01 ¼ X 1 þ 2 X 2 ¼ 12 þ 25 0:1 ¼ 14:5 X p Z 01 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10:5 2 þ 14:5 2 ¼ 17:9 X

14 62 2 Trnsformer: Priniples nd Prties The totl resistne, retne nd impedne referred to the seondry re lulted s, R 02 ¼ R 2 þ R 1 2 ¼ 0:3 þ 3 25 ¼ 0:42 X X 02 ¼ X 2 þ X 1 12 ¼ 0:1 þ 2 25 ¼ 0:58 X p Z 01 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:42 2 þ 0:58 2 ¼ 0:72 X Prtie prolem kv, 220/110 V single-phse trnsformer hs the primry resistne nd retne of 6 nd 18 Ω, respetively. The resistne nd retne t the seondry side re 0.6 nd 0.5 Ω, respetively. lulte the equivlent impedne referred to the primry nd the seondry. 2.9 Polrity of Trnsformer The reltive diretions of indued voltges etween the high voltge nd low voltge terminls is known s the polrity of trnsformer. The polrity of trnsformer is very importnt to onstrut three-phse trnsformer nk, prllel onnetion of trnsformer, onnetion of urrent trnsformer (T) nd potentil trnsformer (PT) power with metering devie. Two polrities nmely dditive nd sutrtive re used in the trnsformer. polrity of trnsformer is sid to e n dditive if the mesured voltge etween the high voltge nd the low voltge terminls is greter thn the supply voltge t the high voltge terminls. The dditive polrity of trnsformer is mrked y the orienttion of dots s shown in Fig Wheres, polrity is sid to e sutrtive if the mesured voltge etween the high voltge nd the low voltge terminls is lower thn the supply voltge t the high voltge terminls. The Fig. 2.9 dditive polrity H 1 X 1 V HH 1 2 H 2 X 2

15 2.9 Polrity of Trnsformer 63 Fig Sutrtive polrity H 1 X 1 V HH 1 2 H 2 X 2 sutrtive polrity of trnsformer is mrked y the orienttion of dots s shown in Fig onsider 220/110 V single-phse trnsformer with the high voltge nd the low voltge terminls for testing polrities. The high voltge terminl H 1 is onneted to the low voltge terminl X 1 y le. The voltmeter is onneted etween H 2 nd X 2. In this se, the turns rtio of the trnsformer is, ¼ V 1 V 2 ¼ ¼ 2 For sfety issue, lower voltge needs to e pplied to the primry side i.e., high voltge terminls. Suppose, voltge of 110 V is pplied to the primry side. In this se, voltge of 55 V (110/2) will pper t the seondry terminls. If the meter red out the voltge of 165 V ( ) then the trnsformer is sid to e in dditive polrity. This onnetion is shown in Fig Wheres, if the voltmeter reds the voltge of 55 V (110 55) then the trnsformer is sid to e in sutrtive polrity s shown in Fig Fig Testing of dditive polrity H 1 X 1 V HH V 55 V H 2 X 2 V 165 V

16 64 2 Trnsformer: Priniples nd Prties H 1 X 1 V HH V 55 V H 2 X 2 V 55 V Fig Testing of sutrtive polrity 2.10 Three-Phse Trnsformer three-phse power trnsformer is used t the power generting sttion to step-up the voltge from 11 to 120 kv. Wheres in the power distriution susttion, the three-phse voltge is gin stepped down to 11 kv voltge through three-phse distriution trnsformer N 11 N 21 N 31 N 12 N 22 N Fig Three windings on ommon ore nd wye-delt onnetion

17 2.10 Three-Phse Trnsformer 65 N11 N12 N21 N22 N31 N32 Fig Three single-phse trnsformers Therefore, three-phse trnsformer n e mde either y three windings wound on ommon ore or y three single-phse trnsformer onneted together in three-phse nk. The first pproh is heper one tht results in trnsformer with smller size nd less weight. The min disdvntge of the first pproh is tht if one phse eomes defetive, then the whole trnsformer needs to e repled. Wheres in the seond pproh, if one of the trnsformers eomes defetive then the system n e given power y n open delt t redued pity. In this se, the defetive trnsformer is normlly repled y new one. three-phse trnsformer with wye-delt onnetion is shown in Fig nd the three single-phse trnsformers re shown in Fig Trnsformer Vetor Group The most ommon onfigurtions of three-phse trnsformer re the delt nd str onfigurtions, whih re used in the power utility ompny. The primry nd seondry windings of three-phse trnsformer re onneted either in the sme (delt-delt or str-str), or different (delt-str or str-delt) onfigurtion-pir. The seondry voltge wveforms of three-phse trnsformer re in phse with the primry wveforms when the primry nd seondry windings re onneted in the sme onfigurtion. This ondition is known s no phse shift ondition. If the primry nd seondry windings re onneted in different onfigurtion pir then the seondry voltge wveforms will differ from the orresponding primry

18 66 2 Trnsformer: Priniples nd Prties Fig Representtion of vetor groups with lok N n 0 shift N n 30 lg N n 180 shift N n +30 led voltge wveforms y 30 eletril degrees. This ondition is lled 30 phse shift ondition. The windings nd their position to eh other re usully mrked y vetor group. The vetor group is used to identify the phse shift etween the primry nd seondry windings. In the vetor group, the seondry voltge my hve the phse shift of 30 lgging or leding, 0 i.e., no phse shift or 180 reversl with respet to the primry voltge. The trnsformer vetor group is leled y pitl nd smll letters plus numers from 1 to 12 in typil lok-like digrm. The pitl letter indites primry winding nd smll letter represents seondry winding. In the lok digrm, the minute hnd represents the primry line to neutrl line voltge, nd its ple is lwys in the 12. The hour hnd represents the seondry line to neutrl voltge nd its position in the lok hnges sed on the phse shift s shown in Fig There re four vetor groups used in the three-phse trnsformer onnetion. These vetor groups re (i) Group I: 0 o lok, zero phse displement (Yy0, Dd0, Dz0), (ii) Group II: 6 o lok, 180 phse displement (Yy6, Dd6, Dz6), (iii) Group III: 1 o lok, 30 lg phse displement (Dy1, Yd1, Yz1), nd (iv) Group IV: 11 o lok, 30 led phse displement (Dy11, Yd11, Yz11). Here, Y represents wye onnetion, D represents delt onnetion nd z represents the zigzg onnetion. The onnetion digrms for different omintions re shown in Figs. 2.16, 2.17, 2.18, 2.19, 2.20, 2.21, 2.22, 2.23, 2.24, 2.25, 2.26 nd 2.27.

19 2.11 Trnsformer Vetor Group Y/y/ N 1 2 n Fig onnetion of Y-y-0 2 D/d/ Fig onnetion of D-d-0

20 68 2 Trnsformer: Priniples nd Prties 2 D/z/ n 3 4 Fig onnetion of D-z-0 Fig onnetion of Y-d Y/d/ N

21 2.11 Trnsformer Vetor Group 69 2 D/y/ n Fig onnetion of D-y-1 2 Y/z/ N n 3 4 Fig onnetion of Y-z-1

22 70 2 Trnsformer: Priniples nd Prties Fig onnetion of D-y D/y/ n Fig onnetion of Y-d-11 2 Y/d/ N

23 2.11 Trnsformer Vetor Group 71 2 Y/z/ N n 3 4 Fig onnetion of Y-z-11 Fig onnetion of Y-y-6 2 Y/y/ N n

24 72 2 Trnsformer: Priniples nd Prties D/d/ Fig onnetion of D-d D/z/ n Fig onnetion of D-y-6

25 2.12 Voltge Regultion of Trnsformer Voltge Regultion of Trnsformer Different types of lods like domesti, ommeril nd industril re usully onneted with the seondry winding of trnsformer. ll these lods re operted with onstnt mgnitude of voltge. The seondry voltge of trnsformer under opertion hnges due to voltge drop ross the internl impedne nd the lod. The voltge regultion of trnsformer is used to identify the hrteristi of the seondry side voltge hnges under different loding onditions. The voltge regultion of trnsformer is defined s the differene etween the no-lod terminl voltge (V 2NL ) to full lod terminl voltge (V 2FL ) nd is expressed s perentge of full lod terminl voltge. It is therefore n e expressed s, Voltge regultion ¼ V 2NL V 2FL V 2FL 100 % ¼ E 2 V 2 V % ð2:61þ Figure 2.28 shows n pproximte equivlent iruit referred to the seondry without no-lod iruit to find the voltge regultion for different power ftors. Phsor digrms for different power ftors re shown in Fig where the seondry voltge V 2 is onsidered s the referene phsor. The phsor digrm with unity power ftor is shown in Fig nd the phsor form of the seondry indued voltge for unity power ftor n e written s, E 2 ¼ V 2 þ I 2 ðr 02 þ jx 02 Þ ð2:62þ Figure 2.29 shows the phsor digrm for lgging power ftor, nd from this digrm, the following expressions n e written, ¼ V 2 os / 2 ¼ DE ¼ V 2 sin / 2 D ¼ E ¼ I 2 R 02 EF ¼ I 2 X 02 ð2:63þ ð2:64þ ð2:65þ ð2:66þ Fig pproximte equivlent iruit referred to seondry + X R02 I E 2 V 2 Z L

26 74 2 Trnsformer: Priniples nd Prties () E 2 U () E 2 IZ 2 02 I2X02 F X () I V 2 2 R Y IR 2 02 Z Q E 2 P I 2 φ 2 V 2 IR 2 02 E I2X02 I 2 I2X02 O D M φ 2 V 2 N IR 2 02 Fig Phsor digrm for different power ftors From the right ngle tringle-df, the expression of E 2 n e derived s, F 2 ¼ D 2 þ DF 2 F 2 ¼ð þ DÞ 2 þðde þ EFÞ 2 Sustituting equtions from (2.63) to(2.66) into Eq. (2.68) yields, E 2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðv 2 os / 2 þ I 2 R 02 Þ 2 þðv 2 sin / 2 þ I 2 X 02 Þ 2 ð2:67þ ð2:68þ ð2:69þ In phsor form, Eq. (2.69) n e written s, E 2 ¼ðV 2 os / 2 þ I 2 R 02 ÞþjðV 2 sin / 2 þ I 2 X 02 Þ ð2:70þ Figure 2.29 shows the phsor digrm for leding power ftor nd from this digrm, the following expressions n e written, MR ¼ V 2 os / 2 RN ¼ OQ ¼ V 2 sin / 2 RQ ¼ NO ¼ I 2 R 02 ð2:71þ ð2:72þ ð2:73þ

27 2.12 Voltge Regultion of Trnsformer 75 PO ¼ I 2 X 02 ð2:74þ The expression of E 2 from the right ngled tringle MQP n e derived s, MP 2 ¼ MQ 2 þ QP 2 MP 2 ¼ðMR þ RQÞ 2 þðoq POÞ 2 Sustituting equtions from (2.71) to(2.74) into Eq. (2.76) yields, E 2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðv 2 os / 2 þ I 2 R 02 Þ 2 þðv 2 sin / 2 I 2 X 02 Þ 2 ð2:75þ ð2:76þ ð2:77þ In phsor form, Eq. (2.77) n e written s, E 2 ¼ðV 2 os / 2 þ I 2 R 02 ÞþjðV 2 sin / 2 I 2 X 02 Þ ð2:78þ Exmple 2.5 The primry oil resistne nd retne of 200/400 V single-phse trnsformer re 0.3 nd 0:6 X, respetively. The seondry oil resistne nd retne re 0.8 nd 1:6 X, respetively. lulte the voltge regultion if the seondry urrent of the trnsformer is 10 t 0.8 pf lgging. Solution The vlue of the turns rtio is, ¼ V 1 V 2 ¼ ¼ 0:5 The vlue of the totl resistne referred to the seondry is, R 02 ¼ R 2 þ R 1 0:3 ¼ 0:8 þ 2 0:25 ¼ 2 X The totl retne referred to the primry is, The no-lod voltge is, X 02 ¼ X 2 þ X 1 0:6 ¼ 1:6 þ 2 0:25 ¼ 4 X E 2 ¼ðV 2 os / 2 þ I 2 R 02 ÞþjðV 2 sin / 2 þ I 2 X 02 Þ E 2 ¼ð400 0:8 þ 10 2Þþjð400 0:6 þ 40Þ E 2 ¼ 440:5j39:5 V

28 76 2 Trnsformer: Priniples nd Prties The voltge regultion n e determined s, Voltge regultion ¼ E 2 V 2 V % Voltge regultion ¼ 440: % ¼ 10 % 400 Prtie prolem /220 V single-phse trnsformer hs the resistne of 0:2 X nd retne of 0:8 X in the primry winding. The resistne nd retne in the seondry winding re 0.9 nd 1:8 X, respetively. lulte the voltge regultion, when the seondry urrent is 6 t 0.85 power ftor leding Effiieny of Trnsformer The effiieny is n importnt prmeter to identify the hrteristis of ny mhine. The effiieny g, of ny mhine n e defined s the rtio of its output to the input. Mthemtilly, it n e expressed s, g ¼ output input ¼ input losses input ¼ 1 losses input ð2:79þ Let us onsider the following: P out is the output power, in W, P in is the input power, in W, P losses is the totl losses, in W. Eqution (2.79) n e modified s, g ¼ P out P in ¼ 1 P losses P in ð2:80þ It is worth nothing tht the effiieny of trnsformer is generlly higher thn other eletril mhines euse the trnsformer hs no moving prts Iron nd opper Losses The iron loss of trnsformer is often lled s ore loss, whih is result of n lternting flux in the ore of the trnsformer. The iron loss onsists of the eddy urrent loss nd the hysteresis loss. In the trnsformer, most of the flux trnsferred from the primry oil to the seondry oil through low relutne iron pth. few

29 2.14 Iron nd opper Losses 77 portion of tht lternting flux links with the iron prts of the trnsformer. s result, n emf is indued in the trnsformer ore. urrent will flow in tht prts of the trnsformer. This urrent does not ontriute in output of the trnsformer ut dissipted s het. This urrent is known s eddy urrent nd the power loss due to this urrent is known s eddy urrent loss. The eddy urrent loss ðp e Þ is diretly proportionl to the squre of the frequeny (f) times the mximum mgneti flux density ð m Þ nd the eddy urrent loss n e expressed s, P e ¼ k e f 2 2 m ð2:81þ where k e is the proportionlity onstnt. Steel is very good ferromgneti mteril whih is used for the ore of trnsformer. This ferromgneti mteril ontins numer of domins in the struture nd mgnetized esily. These domins re like smll mgnets loted rndomly in the struture. When n mmf is pplied to the ore then those domins hnge their position. fter removing mmf, most of the domins ome k to their originl position nd remining will e s it is. s result, the sustne is slightly permnently mgnetized. n dditionl mmf is required to hnge the position of the remining domins. Therefore, hysteresis loss is defined s the dditionl energy is required to relign the domins in the ferromgneti mteril. The hysteresis loss ðp h Þ is diretly proportionl to the frequeny (f) nd 2.6th power of the mximum mgneti flux density ð m Þ nd the expression of hysteresis loss is, P h ¼ k h f 2:6 m ð2:82þ where k e is the proportionlity onstnt. Now tht the mgneti flux density is usully onstnt, Eqs. (2.81) nd (2.82) n e modified s, P e / f 2 P h / f ð2:83þ ð2:84þ Prtilly, hysteresis loss depends on the voltge nd the eddy urrent loss depends on the urrent. Therefore, totl losses of the trnsformer depend on the voltge nd the urrent not on the power ftor. Tht is why the trnsformer rting is lwys represented in kv insted of kw. In the trnsformer, opper losses our due to the primry nd the seondry resistnes. The full lod opper losses n e determined s, P uloss ¼ I 2 1 R 1 þ I 2 2 R 2 P uloss ¼ I 2 1 R 01 ¼ I 2 2 R 02 ð2:85þ ð2:86þ

30 78 2 Trnsformer: Priniples nd Prties 2.15 ondition for Mximum Effiieny The expression of the output power of trnsformer is, P out ¼ V 2 I 2 os / 2 ð2:87þ The expression of the opper loss is, P u ¼ I 2 2 R 02 ð2:88þ The expression of n iron loss is, P iron ¼ P eddy þ P hys ð2:89þ ording to Eq. (2.79), the effiieny n e expressed s, g ¼ output input ¼ output output þ losses ð2:90þ Sustituting Eqs. (2.87), (2.88) nd (2.89) into Eq. (2.90) yields, V 2 I 2 os / g ¼ 2 V 2 I 2 os / 2 þ P iron þ P u g ¼ V 2 os / 2 V 2 os / 2 þ P iron I 2 þ I2 2 R 02 I 2 ð2:91þ ð2:92þ The terminl voltge t the seondry side is onsidered to e onstnt in se of norml trnsformer. For given power ftor, the seondry urrent is vried with the vrition of lod. Therefore, the trnsformer effiieny will e mximum, if the denomintor of Eq. (2.92) is minimum. The denomintor of Eq. (2.92) will e minimum, for the following ondition. d V 2 os / di 2 þ P iron þ I2 2 R 02 ¼ 0 ð2:93þ 2 I 2 I 2 P iron I 2 2 þ R 02 ¼ 0 P iron ¼ I 2 2 R 02 ð2:94þ ð2:95þ From Eq. (2.95), it is onluded tht the effiieny of trnsformer will e mximum, when the iron loss is equl to the opper loss. From Eq. (2.95), the expression of seondry urrent n e written s,

31 2.15 ondition for Mximum Effiieny 79 rffiffiffiffiffiffiffiffiffi P iron I 2 ¼ R 02 For mximum effiieny, the lod urrent n e expressed s, Eqution (2.97) n e re-rrnged s, rffiffiffiffiffiffiffiffiffi P iron I 2g ¼ R 02 I 2g ¼ I 2 sffiffiffiffiffiffiffiffiffiffi P iron I 2 2 R 02 ð2:96þ ð2:97þ ð2:98þ Multiplying oth sides of Eq. (2.98) y the seondry rted voltge V 2 yields, V 2 I 2g ¼ V 2 I 2 sffiffiffiffiffiffiffiffiffiffi P iron I 2 2 R 02 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi iron loss VI mx effiieny ¼ VI rted full lod opper loss ð2:99þ ð2:100þ Exmple kv trnsformer hs the iron loss nd full lod opper loss of 350 nd 650 W, respetively. Determine the (i) full lod effiieny, (ii) output kv orresponding to mximum effiieny, nd (iii) mximum effiieny. onsider tht the power ftor of the lod is 0.6 lgging. Solution (i) The totl vlue of full lod loss is, P tloss ¼ 350 þ 650 ¼ 1000 W ¼ 1kW The output power t full lod is, P out ¼ 30 0:6 ¼ 18 kw The input power t full lod is, P in ¼ 18 þ 1 ¼ 19 kw

32 80 2 Trnsformer: Priniples nd Prties The effiieny t full lod is, g ¼ ¼ 94:74 % 19 (ii) The output kv orresponding to mximum effiieny is, rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Iron loss ¼ kv rted u loss t full lod rffiffiffiffiffiffiffi 350 ¼ 30 ¼ 22 kv 650 The output power is, P o1 ¼ 22 0:6 ¼ 13:2kW (iii) For mximum effiieny, iron loss is equl to opper loss. The totl vlue of the loss is, The vlue of the input power is, The effiieny is, P tloss 1 ¼ ¼ 700 W P in1 ¼ 13:2 þ 0:7 ¼ 13:9kW g ¼ 13:2 100 ¼ 94:96 % 13:9 Prtie prolem kv trnsformer is hving n iron loss of 1.5 kw nd opper loss of 5 kw t full lod ondition. Find the kv rting t whih the effiieny is mximum. lso, find the effiieny t unity power ftor Trnsformer Tests The equivlent iruit prmeters re very importnt to hrterize the performne of trnsformer. The prmeters of trnsformer equivlent iruit n e determined y the open iruit nd the short iruit tests.

33 2.16 Trnsformer Tests Open iruit Test The min ojetives of the open iruit test re to determine the no-lod urrent nd iron loss. The omponents of the no-lod urrent re used to determine the no-lod iruit resistne nd retne. In n open iruit test, the seondry side is onsidered to e open iruit, nd the primry oil is onneted to the soure s shown in Fig. 2.30, where ll mesuring instruments re onneted in the primry side. speifi lternting voltge is pplied to the primry winding. Then the wttmeter will mesure the iron loss nd smll mount of opper loss. The mmeter nd voltmeter will mesure the no-lod urrent nd the voltge, respetively. Sine, the no-lod urrent is very smll, the opper losses n e negleted. Then the wttmeter reding n e expressed s, P 0 ¼ V 1 I 0 os / 0 ð2:101þ From Eq. (2.101), the no-lod power ftor n e determined s, os / 0 ¼ P 0 V 1 I 0 ð2:102þ The working nd mgnetizing omponents of the urrent n e determined s, I w ¼ I 0 os / 0 I m ¼ I 0 sin / 0 ð2:103þ ð2:104þ Then the no-lod iruit resistne nd retne n e determined s, R 0 ¼ V 1 I w X 0 ¼ V 1 I m ð2:105þ ð2:106þ () () + V 1 V N1 N2 I w + V 1 R 0 I 0 X 0 I m Fig onnetion digrms for open iruit test nd no-lod iruit

34 82 2 Trnsformer: Priniples nd Prties Exmple /400 V, 50 Hz single-phse trnsformer hs the no-lod test dt of 200 V, 0.6, 80 W. lulte the no-lod iruit resistne nd retne. Solution The power ftor n e determined s, os / 0 ¼ P 0 80 ¼ V 1 I :6 ¼ 0:67 sin / 0 ¼ 0:74 The vlues of the working nd mgnetizing omponents of the no-lod urrent re, I w ¼ I 0 os / 0 ¼ 0:6 0:67 ¼ 0:4 I m ¼ I 0 sin / 0 ¼ 0:6 0:74 ¼ 0:44 The no-lod iruit prmeters n e determined s, R 0 ¼ V 1 ¼ 200 I w 0:4 ¼ 500 X X 0 ¼ V 1 ¼ 200 I m 0:44 ¼ 454:5 X Short iruit Test The min ojetives of the short iruit test re to determine the equivlent resistne, retne, impedne nd full lod opper loss. In this test, the supply voltge nd the mesuring instruments (e. g,. wttmeter, mmeter) re onneted to the primry side nd the seondry winding is shorted out y wire s shown in Fig. 2.31, or onneted with n mmeter. The primry voltge is djusted until the urrent in the short-iruited winding is equl to the rted primry urrent. Under this ondition, the wttmeter will mesure the full lod opper loss nd it n e written s, P s ¼ I s V s os / s ð2:107þ From Eq. (2.107), the short iruit power ftor n e lulted s, P s os / s ¼ I s V s ð2:108þ

35 2.16 Trnsformer Tests 83 N1 N2 + V 1 V + V s I s Fig onnetion digrm for short iruit test The equivlent impedne n e lulted s, Z 01 ¼ Z eq ¼ V s I s ð2:109þ In ddition, the equivlent resistne nd retne n e lulted s, R 01 ¼ Z 01 os / s X 01 ¼ Z 01 sin / s ð2:110þ ð2:111þ Exmple kv, 2200/220 V, 50 Hz single-phse trnsformer s low voltge side is short-iruited nd the test dt reorded from the high voltge side re P ¼ 150 W, I 1 ¼ 5 nd V 1 ¼ 40 V. Determine the (i) equivlent resistne, retne nd impedne referred to primry, (ii) equivlent resistne, retne nd impedne referred to seondry, nd (iii) voltge regultion t unity power ftor. Solution (i) The prmeters referred to primry re, Z 01 ¼ V 1 ¼ 40 I 1 5 ¼ 8 X R 01 ¼ P I1 2 ¼ ¼ 6 X / ¼ os 1 6 ¼ 41:4 8 X 01 ¼ Z 01 sin / ¼ 8 sin 41:4 ¼ 5:2 X

36 84 2 Trnsformer: Priniples nd Prties (ii) The turns rtio is, ¼ V 1 V 2 ¼ ¼ 10 The prmeters referred to seondry re, Z 02 ¼ Z 01 2 ¼ ¼ 0:08 X R 02 ¼ R 01 2 ¼ ¼ 0:06 X X 02 ¼ X 01 2 ¼ 5:2 100 ¼ 0:052 X (iii) The seondry side urrent is, The seondry indued voltge is, The voltge regultion is, I 2 ¼ 25; ¼ 113:6 E 2 ¼ V 2 þ I 2 Z 02 ¼ 220 þ 113:6 0:08 ¼ 229 V VR ¼ E 2 V ¼ ¼ 4 % V Prtie prolem 2.7 The no-lod test dt of 220/2200 V, 50 Hz single-phse trnsformer re 220 V, 0.4, 75 W. lulte the no-lod iruit resistne nd retne. Prtie prolem kv, 2300/220 V, 50 Hz single-phse trnsformer s low voltge side is short-iruited y thik wire. The test dt reorded from the high voltge side re P ¼ 400 W, I 1 ¼ 8:5 nd V 1 ¼ 65 V. Find the (i) equivlent resistne, retne nd impedne referred to primry nd (ii) equivlent resistne, retne nd impedne referred to seondry, nd (iii) voltge regultion t 0.6 lgging power ftor.

37 2.17 utotrnsformer utotrnsformer smll rting trnsformer with vrile voltge output is usully used in the edutionl lortory s well s in the testing lortory. This type of smll rting trnsformer with vrile output is known s utotrnsformer. n utotrnsformer hs one ontinuous winding tht is ommon to oth the primry nd the seondry. Therefore, in n utotrnsformer, the primry nd seondry windings re onneted eletrilly. The dvntges of n utotrnsformer over two-winding trnsformer inlude lower initil investment, lower lekge retne, lower losses ompred to onventionl trnsformer nd lower exittion urrent. n utotrnsformer with primry nd seondry windings is shown in Fig In this onnetion, the suffix indites the ommon winding nd the suffix s indites the series winding. From Fig. 2.32, the following equtions n e written s, N s N ¼ I I s ¼ V s V ¼ N s N ¼ The totl voltge in the primry side is, V 1 ¼ V s þ V ¼ V 1 þ V s V ð2:112þ ð2:113þ ð2:114þ Sustituting Eq. (2.113) into Eq. (2.114) yields, V 1 ¼ ð1 þ Þ ð2:115þ V V L ¼ 1 V 1 1 þ ð2:116þ Fig onnetion digrm for n utotrnsformer I s + N s I L V 1 N I + VL

38 86 2 Trnsformer: Priniples nd Prties Where the voltge t the lod is equl to the voltge t the ommon terminls i.e., V L ¼ V. The expression of the lod urrent n e written s, I L ¼ I s þ I ¼ I s þ I s I L ¼ð1 þ ÞI s I L I s ¼ð1 þ Þ ð2:117þ ð2:118þ ð2:119þ Exmple 2.9 Figure 2.33 shows single-phse 120 kv, 2200/220 V, 50 Hz trnsformer whih is onneted s n utotrnsformer. The voltges of the upper nd lower prts of the oil re 220 nd 2200 V, respetively. lulte the kv rting of the utotrnsformer. Solution The urrent rtings of the respetive windings re, The urrent in the primry side is, I pq ¼ 120;000 ¼ 545:5 220 I qr ¼ 120; ¼ 54:5 I 1 ¼ 545:5 þ 54:5 ¼ 600 The voltge ross the seondry side is, V 2 ¼ 2200 þ 220 ¼ 2420 V Fig n utotrnsformer with speifi voltge I pq p V - I 1 r q I qr 2420 V -

39 2.17 utotrnsformer 87 Fig n utotrnsformer with speifi voltge I pq p + I 1 + q 1320 V 1100 V - r I qr - Therefore, the kv rtings of n utotrnsformer, kv 1 ¼ ¼ 1320 kv : kv 2 ¼ ¼ 1320 kv 1000 Prtie prolem 2.9 single-phse 100 kv, 1100/220 V, 50 Hz trnsformer is onneted s n utotrnsformer s shown in Fig The voltges of the upper nd lower prts of the oil re 220 nd 1100 V, respetively. Determine the kv rting of the utotrnsformer Prllel Opertion of Single-Phse Trnsformer Nowdys, the demnd of lod is inresing with the inrese in popultion nd industril setor. Sometimes, it is diffiult to meet the exess demnd of power y the existing single unit trnsformer. Therefore, n dditionl trnsformer is required to onnet in prllel with the existing one. The following points should e onsidered for mking prllel onnetions of trnsformers: Terminl voltge of oth trnsformers must e the sme. Polrity must e sme for oth trnsformers. For oth trnsformers, the perentge impednes should e equl in mgnitude. Rtio of R to X must e the sme for oth trnsformers. Phse sequenes nd phse shifts must e the sme (for three-phse trnsformer).

40 88 2 Trnsformer: Priniples nd Prties 2.19 Three-Phse Trnsformer onnetions The primry nd seondry windings of the trnsformer my e onneted in either y wye (Y) or delt (D). Three-phse trnsformer onnetions re lssified into four possile types; nmely, Y-Y(wye-wye), Y-D (wye-delt), D-Y(delt-wye) nd D-D (delt-delt) Wye-Wye onnetion Figure 2.35 shows the Y-Y onnetion digrm. This type of onnetion of three-phse trnsformer is rrely used for lrge mount of power trnsmission. Neutrl point is neessry for oth primry nd seondry sides in some ses. In lned lods, this type of onnetion works stisftorily nd provides neutrl to eh side for grounding. t the primry side, the phse voltge n e written s, V P1 ¼ V p L1 ffiffi 3 ð2:120þ The seondry phse voltge n e written s, V P2 ¼ V p L2 ffiffi 3 ð2:121þ The rtio of the primry line voltge to the seondry line voltge of this onnetion is, ¼ V L1 V L2 pffiffiffi 3 VP1 ¼ pffiffiffi ¼ V P1 3 VP2 V P2 ð2:122þ ð2:123þ Fig Y-Y onnetion digrm Primry N Seondry n

41 2.19 Three-Phse Trnsformer onnetions Wye-Delt onnetion The wye-delt onnetion is minly used t the susttion where the voltge is stepped down. In this onnetion, the urrent in the seondry oil is 57.7 % of the lod urrent. t the primry side of this onnetion, proper opper wire is used to ground the neutrl point. Figure 2.36 shows the onnetion digrm of the wye-delt trnsformer. In this onnetion, the expression of the primry line voltge is, V L1 ¼ t the seondry side, the line voltge is, p ffiffi 3 VP1 ð2:124þ V L2 ¼ V P2 ð2:125þ The rtio of primry phse voltge to seondry phse voltge is, ¼ V P1 V P2 The rtio of primry line voltge to seondry line voltge is, V L1 V L2 ¼ p VP1 ffiffi 3 p ¼ ffiffiffi 3 V P2 ð2:126þ ð2:127þ The primry phse urrent is, I P1 ¼ I L1 ð2:128þ In this se, the turns rtio is, ¼ I P2 I P1 ð2:129þ Fig Wye-delt onnetion digrm Primry N Seondry

42 90 2 Trnsformer: Priniples nd Prties The expression of the seondry phse urrent is, The seondry line urrent is, I L2 ¼ I P2 ¼ I P1 p ffiffi p 3 IP2 ¼ ffiffi 3 IP1 ð2:130þ ð2:131þ Delt-Wye onnetion The delt-wye onnetion is generlly used t the power generting sttion to step the voltge. Figure 2.37 shows the onnetion digrm of delt-wye trnsformer. In this onnetion, the expression of the primry line voltge is, V L1 ¼ V P1 The line voltge t the seondry side is, V L2 ¼ p ffiffi 3 VP2 ð2:132þ ð2:133þ The rtio of primry line voltge to seondry line voltge is, V L1 V L2 ¼ V P1 pffiffi ¼ p ffiffiffi 3 VP2 3 ð2:134þ The phse urrent t the primry side is, I P1 ¼ p 1 ffiffi I L1 3 ð2:135þ Fig Delt-wye onnetion digrm Primry Seondry n

43 2.19 Three-Phse Trnsformer onnetions 91 In this onnetion, the turns rtio is, ¼ I P2 I P1 ð2:136þ In this se, the seondry phse urrent is, I P2 ¼ I P1 ð2:137þ The seondry line urrent is, I L2 ¼ I P2 ¼ I P1 ð2:138þ Delt-Delt onnetion The delt-delt onnetion is generlly used for oth high voltge nd low voltge rting trnsformers where insultion is not n importnt issue. The onnetion digrm for delt-delt trnsformer is shown in Fig In this se, the primry line voltge is, V L1 ¼ V P1 ð2:139þ The seondry line voltge is, V L2 ¼ V P2 ð2:140þ The rtio of line-to-line voltge of this onnetion is, V L1 V L2 ¼ V P1 V P2 ¼ ð2:141þ Fig Delt-delt onnetion digrm Primry Seondry

44 92 2 Trnsformer: Priniples nd Prties The seondry line urrent is, I L2 ¼ p ffiffiffi 3 IP2 ð2:142þ The output pity in delt-delt onnetion n e expressed s, P 0D D ¼ p ffiffi 3 VL2 I L2 Sustituting Eq. (2.142) into Eq. (2.43) yields, p P 0D D ¼ ffiffi p 3 VL2 ffiffi 3 IP2 ¼ 3V L2 I P2 ð2:143þ ð2:144þ The remining two trnsformers re le to supply three-phse power to the lod terminl if one of the trnsformer is removed from the onnetion. This type of three-phse power supply y two trnsformers is known s open delt or V-V onnetion. This open delt onnetion is le to supply three-phse power t redued rte of 57.7 %. The onnetion digrm for open delt onnetion is shown in Fig In the V-V onnetion, the seondry line urrent is, I L2 ¼ I P2 The output pity in V-V onnetion is, P 0V V ¼ p ffiffiffi 3 VL2 I P2 ð2:145þ ð2:146þ The rtio of delt-delt output pity to the V-V output pity is, P 0V V P 0D D ¼ pffiffi 3 VL2 I P2 ¼ 0:577 ð2:147þ 3V L2 I P2 Fig Open delt or V-V onnetion digrm Primry Seondry

45 2.19 Three-Phse Trnsformer onnetions 93 Exmple 2.10 three-phse trnsformer is onneted to n 11 kv supply nd drws 6 urrent. Determine (i) line voltge t the seondry side, nd lso (ii) the line urrent in the seondry oil. onsider the turns rtio of the trnsformer is 11. lso, onsider delt-wye nd wye-delt onnetions. Solution Delt-wye onnetion: The phse voltge t the primry side is, V P1 ¼ V L1 ¼ 11 kv The phse voltge t the seondry side is, V P2 ¼ V P1 ¼ 11;000 ¼ 1000 V 11 (i) (ii) The line voltge t the seondry is, V L2 ¼ The phse urrent in the primry is, p ffiffi p 3 VP2 ¼ ffiffi ¼ 1732 V I P1 ¼ p IL1 ffiffiffi 3 The line urrent in the seondry is, ¼ p 11 ffiffi ¼ 6:35 3 I L2 ¼ I P2 ¼ I P1 ¼ 11 6:35 ¼ 69:85 Wye-delt onnetion: The line voltge t the primry side is, V L1 ¼ 11 kv The phse voltge t the primry side is, V P1 ¼ V p L1 ffiffi ¼ 11;000 pffiffiffi ¼ 6350:85 V 3 3

46 94 2 Trnsformer: Priniples nd Prties (i) The phse voltge t the seondry side is, The line voltge t the seondry is, V P2 ¼ V P1 ¼ 6350:85 ¼ 577:35 V 11 V L2 ¼ V P2 ¼ 577:35 V (ii) The phse urrent in the seondry is, The line urrent in the seondry is, I L2 ¼ I P2 ¼ I P1 ¼ 11 6 ¼ 66 p ffiffiffi p 3 IP2 ¼ ffiffiffi 3 66 ¼ 114:32 Prtie prolem 2.10 three-phse trnsformer is onneted to the 33 kv supply nd drws 15 urrent. lulte (i) the line voltge t the seondry side, nd lso (ii) the line urrent in the seondry winding. onsider the turns rtio of the trnsformer is 8 nd wye-delt onnetion Instrument Trnsformers The mgnitude of the voltge nd the urrent re normlly high in the power system networks. To redue the mgnitude of the voltge nd urrent, instrument trnsformers re used. There re two types of instrument trnsformers; nmely, urrent trnsformer (T) nd potentil trnsformer (PT). If power system rries n lternting urrent greter thn 100, then it is diffiult to onnet mesuring instruments like low rnge mmeter, wttmeter, energy meter nd rely. The urrent trnsformer is then onneted in series with the line to step down the high mgnitude of the urrent to rted vlue of out 5 for the mmeter nd the urrent oil of the wttmeter. The digrm of urrent trnsformer is shown in Fig If the system voltge exeeds 500 V, then it is diffiult to onnet mesuring instruments diretly to the high voltge. The potentil trnsformer is then used to step down to suitle vlue of the voltge t the seondry for supplying the voltmeter nd the voltge oil of the wttmeter. The seondry of the instrument trnsformer is normlly grounded for sfety reson. The onnetion digrm of potentil trnsformer is shown in Fig

47 2.20 Instrument Trnsformers 95 Fig onnetion digrm of urrent trnsformer I To lod Fig onnetion digrm of potentil trnsformer V Exerise Prolems 2:1 single-phse trnsformer is hving the primry voltge of 240 V. The numer of turns t the primry nd seondry oils re 250 nd 50, respetively. Determine the seondry voltge. 2:2 single-phse trnsformer is hving the seondry voltge of 100 V. The numer of turns t the primry nd seondry windings re 500 nd 50, respetively. lulte the primry voltge. 2:3 The 100 urrent flows in the primry oil of single-phse trnsformer. lulte the seondry urrent with the turns rtio of :4 The numer of turns t the primry nd seondry windings of single-phse trnsformer re 500 nd 100, respetively. Find the primry urrent if the seondry urrent is 10. 2:5 The rtio of the primry urrent to the seondry urrent of single-phse trnsformer is 1:10. Find the voltge rtio. lso, find the seondry voltge if the primry voltge is 100 V. 2:6 The turns rtio of single-phse trnsformer is 1:4. The seondry oil hs 5000 turns nd 60 V. Find the voltge rtio, primry voltge nd numer of turns. 2:7 The voltge rtio of single-phse trnsformer is 5:1. The voltge nd numer of turns t the primry oil re found to e 1100 V nd 500, respetively. lulte the voltge nd seondry numer of turns t the seondry oil. 2:8 single-phse trnsformer is onneted to the 120 V soure drws 4 urrent. lulte the urrent in the seondry oil when the trnsformer steps up the voltge to 500 V.

48 96 2 Trnsformer: Priniples nd Prties 2:9 The numer of turns t the primry nd seondry oils of single-phse trnsformer re found to e 480 nd 60, respetively. The trnsformer drws urrent of 0.6 when onneted to 120 V supply. Determine the urrent nd the voltge t the seondry oil. 2:10 The turns rtio of 5 kv single-phse trnsformer is found to e 2. The trnsformer is onneted to 230 V, 50 Hz supply. If the seondry urrent of trnsformer is 6, then find the primry urrent nd seondry voltge. 2:11 The seondry numer of turns of 30 kv, 4400/440 V single-phse trnsformer is found to e 100. Find the numer of primry turns, full lod primry nd seondry urrents. 2:12 5 kv, 1100/230 V, 50 Hz single-phse trnsformer is instlled ner the domesti re of ountry. Find the turns rtio, primry nd seondry full lod urrents. 2:13 The numer of turns of the primry winding of single-phse 50 Hz trnsformer is found to e 50. Find the vlue of the ore flux, if the indued voltge t this winding is 220 V. 2:14 The numer of primry turns of 60 Hz single-phse trnsformer is found to e 500. lulte the indued voltge in this winding, if the vlue of the ore flux is 0.05 W. 2:15 The mximum flux of 3300/330 V, 50 Hz step-down single-phse trnsformer is found to e 0.45 W. lulte the numer of primry nd seondry turns. 2:16 The ross setionl re of 5 kv, 2200/220 V, 50 Hz single-phse step-down trnsformer is found to e 25 m 2 nd mximum flux density is 4 W/m 2. lulte the primry nd seondry turns. 2:17 The numer of turns t the primry nd seondry of n idel single-phse trnsformer re found to e 500 nd 250, respetively. The primry of the trnsformer is onneted to 220 V, 50 Hz soure. The seondry oil supplies urrent of 5 to the lod. Determine the (i) turns rtio, (ii) urrent in the primry nd (iii) mximum flux in the ore. 2:18 The primry numer of turns of 4 kv, 1100/230 V, 50 Hz single-phse trnsformer is 500. The ross setionl re of the trnsformer ore is 85 m 2. Find the (i) turns rtio, (ii) numer of turns in the seondry nd (iii) mximum flux density in the ore. 2:19 The primry nd seondry turns of single-phse trnsformer re 50 nd 500, respetively. The primry winding is onneted to 220 V, 50 Hz supply. Find the (i) flux density in the ore if the ross setionl re is 250 m 2 nd (ii) indued voltge t the seondry winding. 2:20 The primry oil numer of turns of single-phse 2200/220 V, 50 Hz trnsformer is found to e Find the re of the ore if the mximum flux density of the ore is 1.5 W/m 2. 2:21 The mximum flux of single-phse 1100/220 V, 50 Hz trnsformer is found to e 5 mw. The numer of turns nd the voltge t the primry winding re found to e 900 nd 1100 V, respetively. Determine the frequeny of the supply system.