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Electrical Dr. P.S. Bimbhra Resultant flux density wave Max. bar current

s Torque

Resultant flux density wave

Rotor mmf wave

Electrical Machinery This thoroughly revised atid updated edition presents a rigorous and comprehensive treatment of transformers and more common types of rotating electrical machines types. Each chapters begins with rudimentary concepts and is so developed, that an average student can easily comprehend i t The salient features of this book are: ❖ In-depth coverage of transformers, d.c. machines, 3-phase synchronous and induction machines. ❖ Highlights that electrical machines operate on the same basic principles. ❖ Devotes a chapter on electromechanical-energy-conversion principles and another on d.c./a.c. machine windings. ❖ Drive aspects and applications are discussed for each machine type. ❖ Clearity of presentation is enhanced by illustrative figures and examples selected from questions-papersof important Universities, IAS, IES and GATE. ❖ Includes numerous problems, conceptual questions and objective-type questions (with answers) to help the reader master the basic concepts. This edition also includes: electrical machinery overview, energy efficiency and resent advances; 3-phase transformers; 3-phase induction generators; appendices on magnetic circuits, 3-phase circuits and short-answer type questions. All these features contribute towards making this book an ideal text for undergraduate students of degree classes. Practising engineers, through self-study, will also find this volume useful to them.

Dr. P.S. Bimbhra retired as Professor of Electrical and Electronics Engineering from Thapar Institute of Engineering and Technology (Deemed University), Patiala-147004. He has over 40 years of teaching and research experience in Electrical Engineering. He has published several papers in renowned journals and national and international conferences. Dr. Bimbhra is a fellow of I.E. (India), a member of I.E. E. (U.K.) and a life member of I.S.T.E.

KHANNA PUBLISHERS 4575/15, Onkar House, Room No. 3-4 Darya Ganj, New Delhi-110002 Phone: 23243042 • Fax: 23243043

Contents O v e r v ie w o f E l e c t r i c a l M a c h in e r y

T r a n s fo r m e r s 1 1 . Transformer Construction 12 Principle of Transformer Action 1 3 . Ideal Two-winding Transformer 1 .4 . Transformer Phasor Diagrams 1.4.1. Transform er phasor diagram at no load 1.4.2. Transform er phasor diagram under load 1.4.3. Leakage flux 1.5 . Rating of Transform ers 1.6 . Equivalent C ircuit of a Transform er 1.7 . Open-circuit an d Short-circuit Tests 1.8 . The Per U n it System 1.9 . Voltage Regulation of a T ransform er 1.10. Transform er Losses and Efficiency 1.10.1. Transform er losses 1.10.2. Transform er efficiency 1.10.3. S eparation of hysteresis and eddy current losses 1.11. Testing of Transform ers 1.12. Auto-transform ers 1.13. Parallel O peration of Single-phase Transform ers 1.14. Tap-Changers on Transform ers 1.14.1. No-Load (or off-load) tap changer 1.14.2. On-load tap-changer 1.15. Induction R egulators 1.15.1. Single-phase induction regulator 1.15.2. T hree-phase induction regulator 1.16. Transform er as a M agnetically Coupled Circuit 1.16.1. Co-efficient of coupling 1.16.2. M ethods of increasing the coefficient of coupling 1.17. Audio-Frequency T ransform ers 1-18. Pulse T ransform ers 1*19. Three-phase T ransform ers 119.1. T hree-phase tran sfo rm er connections 1-20. Transform er Noise J-21. Some W orked Exam ples 1-22. Sum m ary

h

2-

E l e c t r o m e c h a n i c a l E n e r g y C o n v e r s io n P r i n c ip l e s

,

Principle of E nergy Conversion ■ ■ Singly Excited M agnetic System s 2'3- Reluctance M otor (vii)

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1-159 2 4 5

iO 10 12 14 14 20 29 35 40 49 49 50 63 66

70 84 100 102 102

106 106 109 112 H7 119 123 128

12® 1 ^ 136 144

1 6 0 '222

'%V-, • ^6 (vih)

2.4. 2.5. 2.6. 2.7 2.8.

Doubly-excited M agnetic S ystem s 2.4.1. E lectrom agnetic and reluctance torques E lem en tary Synchronous M achines Singly-excited Electric Field System s Dynam ic E q uations Some Workpd Exam ples

3. 3.1.

B a s i c C o n c e p t s o f R o t a t i n g E l e c t r i c a l M a c h in e s Physical Concepts of Torque Production 3.1.1. Electrom agnetic (or interaction) torque 3.1.2. Reluctance (or alignm ent) torque 3.2. C onstructional F eatu res of R otating Electrical M achines 3.2.1. Polyphase Induction M achines 3.2.2. Synchronous m achines 3.2.3. D irect C u rre n t M achines 3.3. Concepts of G eneral Terms P ertain in g to R otating M achines 3.4. G enerated emfs 3.4.1. G enerated e.m.f. in a full-pitched coil 3.4.2. G enerated e.m.f. in a short-pitched coil 3.4.3. A.C. M achines 3.4.4. G enerated e.m.f. in d.c. m achines 3.5. E.m.f. Polygon 3.5.1. D istribution, breadth or belt Factor 3.5.2. P itch (or coil-span) Factor 3.5.3. Elim ination of harm onics from altern ato r em f waveform s 3.6. M.m.f. Produced by D istributed W indings 3.6.1. M.m.f. of a coil 3.6.2. M.m.f. of distributed w indings 3.6.3. C urrent-sheet Concept 3.6.4. M.m.f. waveform of com m utator m achines 3.7. R otating M agnetic Field 3 . 8 . Production of Torque in Non-Salient Pole M achines 3.8.1. A lternative Derivation for Torque 3.9. Losses and Efficiency 3.10. M achine Ratings 3.10.1. Choice of power of electric m achines 3.11. Cooling (Loss dissipation) 3.12. M achine Applications 4. D .C . M a c h in e s 4.1. Action of Com m utator 4.2. E.m.f. G enerated in the A rm ature 4.3. Torque in D.C. M achines 4 .4 . Circuit Model of DC Machines 4.5. Methods of Excitation 4.6. M.m.f. and Flux Density Waveforms in d.c. M achines 4.6.1. A rm ature reaction 4.6.2. M ethods or lim iting the effects of „rm nturc renctiol)

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^2

^ ^

233.359 234 234

235 236 23­ 233 242 244 250 250 252 253 256 263 264 269

272 285 286 288 293 298 301

318 323 329 333 339 344 347

360-539 353

365 359 373 376

379 39 O

385

(«) Effect of B ru s h S h ift

4.7. C o m m u ta tio n P ro c e ss 4.8.

4.8.1. In terp o les 4.8.2. B ru sh es C o m p e n sa tin g W in d in g s

4.9. 4 10 Basic Performance Equations for D.C. Machines 4 . 10 .1 . M agnetisation Curve 4 . 10 .2 . Effect of arm ature m.m.f. on d.c. machine calculations 4 11 Operating Characteristics of D.C. Generators 4.11.1. Separately-excited generators 4.11.2. Shunt generators 4.11.3. Series Generators 4.11.4. Compound generators 4.11.5. Effect of speed on external characteristics 4.12. Operating Characteristics of D.C. Motors 4.12.1. D.C. Shunt Motor , 4.12.2. D.C. series motor 4.12.3. D.C. compound motor 4.13. D.C. Motor Starting 4.13.1. Shunt and compound motor starters 4.13.2. Series motor starters 4.14. Speed Control of D.C. Motors 4.14.1. Speed control by varying the armature-circuit resistance 4.14.2. Speed control by varying the field flux 4.14.3. Speed control by varying the arm ature terminal voltage 4.15. Efficiency and Testing of d.c. Machines 4.15.1. Efficiency of d.c. machines 4.15.2. Testing of d.c. machines 4.16. Rotating Amplifiers 4.16.1. Cross-field or metadyne generators 4.16.2. Amplidyne 4.17. Perm anent Magnet DC (PMDC) Motors 4.18. D.C. Machine Applications 4.18.1. Generator applications 4.18.2. Motor applications

386 390 395 396 398 402 403 404 407 407 409 421 422 423 429 429 431 432 444 445 447 454 455 463 477 489 490 495 504 504 509 519 521 521 521

5. P o ly p h a s e S y n c h r o n o u s M a c h in e s 540-702 5.1. Excitation Systems for Synchronous Machines 540 5.2. Flux and mm f phasors in synchronous machines 543 5.2.1. Cylindrical-rotor synchronous machines 543 5.2.2. Salient-pole Machines 549 5.3. Phasor Diagram of a Cylindrical Rotor Alternator 551 °pen-circuit and short-circuit characteristics of synchronous machines 552 5.3.2. Zero power-factor characteristic and Potier triangle 554 5.4. Voltage Regulation of an Alternator 557 •4.1. The electromotive force (emf) method or synchronous impedance method 557 •4.2. The Magnetomotive force (m.m.f.) method 563 5.4.3. Zero power factor method 565 5.4.4. New A.S.A. (American Standards Association) method 565 . > ..

icann

.am scan ner

r

is.

(*) 5.4.5. S a tu ra te d synchronous-reactance m ethod Physical Concepts of Synchronous M achine O peration Synchronous M otor P h aso r D iagram O perating C haracteristics of A ltern ato rs and th e ir R atings 5.7.1. E x ternal load ch aracteristics 5.7.2. A ltern ato r com pounding ch aracteristics 5.7.3. R ating of a lte rn a to rs 5.8. Pow er Flow Through an Inductive Im pedance 5.8.1. M axim um power conditions 5.8.2. Reactive Power 5.9. Circle D iagram s of Synchronous M achines 5.9.1. Synchronous m otor circle diagram s 5.10. Pow er F actor Control of Synchronous M achines 5.10.1. Power-factor Control of Synchronous M otors 5.10.2. Power-factor control of A lternators 5.11. Two-reaction Theory of Salient-pole M achines 5.12. Pow er-angle C haracteristics of Synchronous M achines 5.12.1. Cylindrical-rotor synchronous m achine 5.12.2. Salient-pole Synchronous M achine 5.13. Synchronizing Power and Synchronizing Torque 5.13.1. Physical concepts of synchronizing power 5.14. Synchronous M achine Stability 5.15. H untin g and D am per W indings 5.15.1. H unting 5.15.2. D am per windings 5.16. M easurem ent of X d an d X 9 5.17. Efficiency of Synchronous M achines 5.18. O perating Lim it on Synchronous G enerators 5.18.1. O perating C hart 5.18.2. Capability Curves 5.19. Power F actor Correction by Synchronous Motors 5.19.1. Synchronous condenser D uaI'Pu rP°se Synchronous Motor 5.20. S tartin g of Synchronous Motors 5.5. 5.6. 5.7.

5.21. S o m e » T a m X hr0nOU6 ra0t° rS " 5.22. Synchronous M achine Applications

6. P o ly p h a s e I n d u c t io n M o to r s 6 . 1 . Induction Motor as a TYansformer 6 .2 . Principle of operation

^

"

56g 575 577 577 579 579

580 583 585 604

64

617 628 629 637 639 643 647 647 650 652 659 665 665 667 669 669 071

rno 679

w iM O ,

el 2 :X™ ZhaS0rSandwavesin

705

6.5. • 6-7. 6 .8 .

Rotor e.m.f., C urrent and Power Losses and Efficiency Induction Motor Phasor D iagram Equivalent Circuit'

lu ' g

6.9.

Analysis of the E quivalent Circuit ■ I- Torque-slip characteristics

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720 724 727

L 11

6.9.2. P ow er-slip c h a rac te ristics ~ 6 . 10 . O p eratin g (or P erform ance) C h a r a c te r s * rt 6 . 1 0 . 1 . In d u ctio n m otor stab ility CS A u c tio n Motors

fi ll'd m ! / ° f/ quirrel-caB« motors 6.15. Polarity Test” * ° Startmg woun1

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v2

[*r

fw ' l [ n 2J

\

n

2

Zo = Z 9'

/

...(1 .1 7 )

10

[Art. 1.4 ________ _______ ________ __________________ ________________________

Electrical Machinery

This shows th a t as far as effect of Z 2 on prim ary side is concerned, secondary circuit can be replaced by an equivalent impedance Z 2 in the p n . Z ,' =

y

fATjf N,

This is shown in Figs. 1.7 (6 ) and (c). . , arw.0 viewpH T hus, th e three circuits shown in Fig. 1.7 are identical as far as th e .r perform ance v.ewed from term inals a and b is concerned.______________________________________________r_; . . Sim ilarly, an im pedance Z x in the prim ary circuit can be tra n sfe rre d to (or r f secondary side as N

2

7 _ 7 / Z ,- Z i

T ransferring an impedance from one side of a transform er to the o th er is called referring the impedance to the other side. Similarly, voltages and cu rren ts can be referred to e ith e r side of transform er by m eans of Eq. (1.11). In this m anner, re su lta n t voltage an cu rre n on a side can be evaluated. , For an ideal transform er, it m ay be sum m arized th a t (i) voltages are tra n sfe rre d in th e direct ratio, (ii) currents in th e reverse ratio, (iii) im pedances in th e direct ra tio sq u ared and (iv) power and volt-am peres rem ain unchanged. Eq. (1.17) illu strates th e impedance-modifying property of a tran sfo rm er. In practice, th is property is exploited for m atching a fixed load im pedance to th e source im pedance for th e 6 k purpose of m axim um power tran sfer from source to load. This is achieved by in terp o sin g a W l.r ] transform er of suitable tu rn s ratio between the load and the source. y

1.4. T r a n s f o r m e r P h a s o r D ia g ra m s The purpose of first considering an ideal transform er, i.e. a tra n sfo rm e r w ith no core losses, no winding resistances, no m agnetic leakage and constant perm eability, is m erely to h ig h lig h t th e m ost im portant aspects of tran sfo rm er action. Such a tran sfo rm er n ev er ex ists a n d now th e phasor diagram s of real transform er w ith various im perfections will be considered. M agnetization curve of th e actual tran sfo rm er core is non-linear an d its effect is to in tro ­ duce higher order harm onics in th e m agnetizing cu rren t. Since all th e q u a n titie s in a p hasor diagram m u st be of th e sam e frequency, these h ig h er order harm onics (w hose freq u en cies are odd m ultiples of fundam ental frequency) can’t be rep resen ted in th e p h a so r d iag ram . So a lin e a r m agnetization curve for th e tran sfo rm er core will continue to be assu m ed . The p h aso r diagram of a tran sfo rm er is now developed, first a t no load a n d th e n u n d e r load.

1.4.1. Transform er phasor diagram at no load. The m agnetic flux ,2- H ere points X and Y are on th e limb w h ere secondary w inding is w ound. W ith increase of secondary cu rren t, th e p rim ary cu rren t also rise s a n d th is causes point A to a tta in a m agnetic potential h ig h er th a n t a t o pom . As a re su lt, p rim a ry leakage flux n increases. This shows th a t leakage fluxes in a tran sfo rm er a re d ep e n d en t upon th e cu rre n ts in th e w indings. Core flux in a tra n sfo rm e r depends upon th e em f induced in th e p rim ary w inding. W ith in crease o f p rim ary c u rre n t, E i = V\ - 1\ ( ^ +jx{) does reduce and likewise core flux is re uce . B u t th is reduction in £ , an d likew ise in core flux is quite sm all. T hus it m ay be sta te d th a t core flux in a tra n sfo rm e r depends upon th e applied voltage and m ay be tre ated as co n stan t from no load to full load. In th e follow ing a rtic le , th e r a tin g of tra n sfo rm e rs is discussed. A fter th is, an exact eq u iv a len t circuit of th e tra n sfo rm e r is developed first from which its approxim ate equivalent circu it is obtained. 1.5. R a t i n g o f T r a n s f o r m e r s T he m a n u fa c tu re r of tra n sfo rm ers fixes a nam e plate on th e tran sfo rm er, on w hich are recorded th e ra te d o u tp u t, th e ra te d voltages, th e rated frequency etc. of a particular^trans-form er. A ty p ical n a m e p la te ra tin g of a single p hase tra n sfo rm e r is as follows: 20 kVA, 3300/220 V, 50 Hz. H ere 20 kVA is th e rate d o u tp u t a t the secondary term in als. N ote th a t the ra te d o u tp u t is expressed in kilo-volt-am peres (kVA) ra th e r th a n m kilow atts (kW). T his is due to th e fact t h a t ra te d tra n sfo rm e r o u tp u t is lim ited by h eatin g and hence by th e jo sses in th e tra n sfo rm e r T he tw o types o f losses in a tran sfo rm er are core loss and ohm ic (7 r) loss. The core loss dep en d s on tra n sfo rm e r voltage and ohmic loss on th e tra n sfo rm er c u rre n t. As these losses depend on tra n sfo rm e r voltage (V) and cu rre n t (7) an d are alm o st unaffected by th e load p f th e tra n sfo rm e r ra te d o u tp u t is expressed in VA (V x 7) or in kVA an d n o t in kW. F or ex­ am ple a tra n sfo rm e r w orking on rated voltage and rated c u rre n t w ith load p f eq u al to zero has ra te d losses an d ra te d kVA o u tp u t b u t delivers zero pow er to load. T his show s t h a t tra n sfo rm e r ra tin g m u s t be expressed in kVA. For an y tra n s f o rm e r : (R ated in p u t in kVA a t th e p rim a ry term in als) (cos 0 j)

1J

(R ated o u tp u t in kVA a t th e secondary te rm in als) + Losses (cos 0 2)

Since th e tra n sfo rm e r o p erates a t a very high efficiency, losses m ay be ignored. F u rth e r, th e p rim a ry p.f. cos 0j an d th e secondary p.f. cos 02 a re n e a rly equal. T herefore, th e ra te d kVA m ark e d on th e n a m e p la te of a tra n sfo rm e r, refers to both th e w indings, i.e. th e ra te d kVA of th e p rim a ry w ind in g an d th e secondary w inding are equal. T he voltage 3300/220 V refers to th e design voltages of th e two w indings. E ith e r of th e two m a y serv e as p rim a ry or secondary. I f it is a ste p down tra n sfo rm e r, th e n 3300 V is th e ra te d * A term inal is th at part o f an electrical en gineering device, w hich is intended to receive the extern al connections.

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Art- 1.51 " ”

“

"

Transformers

prim ary v o lta g e a n d re fe rs to th e vnlt a ^ i-j. is th e r a te d s e c o n d a ry v o lta g e a n d refnfc ,a p P

,

.................... Pn m a r y w inding. T h e voltage of 220 V

n 0 load, w ith r a te d v o ltag e a p p lied to th e p rim V rJ t e r m f n ^ . ^ ^

respo n d in g r a t e d ^ o l t a g e ^ T h u s ^

15

0UtPUt term in a,S *

CUITents a re calculated from th e ra te d kVA a n d th e cor-

R ated (or fu ll-lo ad ) p r im a r y c u r re n t = ^ > 0 0 0 3300

n« a

R a ted (o r fu ll-lo ad ) se c o n d a ry c u rre n t = ^ ° 0 0 = 9Q ^ ^ 220 w in d in g a r e ^ i g n e d '* P n m a ry a n d sec° n d a ry c u rre n ts refer to th e c u rre n ts for w hich th e R ate d fre q u e n c y re fe rs to th e freq u en cy for w hich th e tra n s fo rm e r is designed to o p erate. Nj E 2 a n N 2 a re ca^ ec^ th e voltage ra tio and tu rn s ra tio respectively. T h ese two ratios a re e q u a l a s s e e n from Eq. 1.8. A t no load, V, and E l a re n early equal in m a g n itu d e for large tra n s f o rm e r s , th e re fo re , th e ir no-load voltage ra tio is ~1 = —1 E 2

^ 2

R ated p rim a ry voltage N, R ated secondary voltage N 2'

E x a m p l e 1.1. T h e e m f p e r tu rn for a single phase, 2 3 1 0 /2 2 0 V, 50 H z tra n sfo rm er is a p ­ proxim ately 13 volts. C a lcu la te (a) the n u m b er o f p rim a ry a n d secondary tu rn s a n d (b) the net cros's-sectional a rea o f th e core, fo r a m a x im u m flu x density o f 1.4 T. S o l u t i o n . E m f p e r tu r n E t = 13 volts. (a) N u m b e r o f se c o n d a ry tu r n s = S econdary voltage Et

220

N 2=~ ^ 13

16.92.

Now t h '' n u m b e r o f t u r n s c a n ’t be a fraction, therefo re, N 2 = 17 (n e a re s t w hole num ber). For N 2 = 17, N u m b e r o f p r im a r y tu r n s 2310^ = 178.5. = 17 220 v ✓ ' ' T h is sh o w s t h a t N 2 c a n ’t be e q u a l to 17 tu rn s . T he o th er n e a re s t in te g e rs a re 16 or 18. It is N ,= N 2

preferable to ta k e N 2 = 18. ...

Afj = 18(10.5) = 189 tu rn s .

T h u s th e r e q u ire d v a lu e s o f

a n d N 2 are 189 an d 18 tu rn s respectively.

n. 220 (b) N ew v a lu e o f e.m .f. p e r tu r n E t volts.

T he n e t co re a r e a c a n be o b ta in e d from th e re latio n , ' l2 n f * max = E l or

220

^ 2 n fB mA l = E t = - j ^

Here B m = m a x i m u m value of f lu * density i n W b/m 2 or teslas and A, = Net core area.

^ o u tf iiiie u u y ^ a i i i o c d i n i e i

16

^Arl ^ ^

Electrical Machinery _______________

•••

V2ti(50) (1.4) A, = ^ = 393 cm2. lo

I t m ay be seen th a t if N 2 is ta k e n equal to 16, th e em f p er tu rn increases and n e t core area is m ore, w hich is n o t desirable. E x a m p le 1 .2 . A single-phase 50 H z transform er has three w indings : a 2 2 0 -V p rim a ry, a 600-V secondary a n d a centre-tapped 11-0-11 V tertiary. For a net core area o f 75 cm 2, calculate the num ber o f tu rn s in the three windings. The m a xim u m value o f flu x density is 1.2T. S o lu tio n . E m f per tu rn E, = V2 rc f B mA, = V2rc (50) (1.2) (75) x 10‘ 4 = 2.00 volts. N um ber of tu rn s in each h a lf of th e te rtia ry winding, N 3 = -7 ^ = 5 .5 tu rn s So ta k e Af3 = 6 tu rn s Total n u m b er of tu rn s in te rtia ry = 2 x 6 = 1 2 . T u rn s in th e 600 V secondary w inding = 600 x y y = 327. and tu rn s in th e 220 V prim ary w inding = — ° * 6 = 120. E x a m p le 1.3. (a) A 2 2 0 0 /2 2 0 V, 50 Hz, single-phase transform er has exciting current o f 0.6 A a n d a core loss o f 361 watts, when its h.v. side is energised at rated voltage. Calculate the two components o f the exciting current. (b) I f the transform er o f p a rt (a), supplies a load current o f 60 A at 0.8 p.f. lag on its l.v. side, then calculate the prim ary current and its power factor. Ignore leakage im pedance drops. S o lu tio n , (a) Exciting current Ie = 0.6 A Supply voltage V { = 2200 V; Core loss Pc = 361 w atts. /. Core loss com ponent Ic = ~

=

= 0.164 A.

From Eq. (1.18), M agnetising component/^, = 'JTj= V(0.60)* - (0.164)2 = 0.577 A. (b ) The prim ary current component I f , required to neutralise the effect of secondary current / 2 = 60 A, is given by

or

6 A-

V2 N« N ote th a t — = —f Vi ^ The c u rre n ts Ie / 2, a n d 7j are indicated in th e phasor diag ram of Fig. 1.12.

l,j u g i 11 i c u

The vertical com ponent of It = Ic = 0.164 A

The h o rizo n tal component of A c An = 0.577 A. .■

uy

.

w a n u t a i 11 i c i

‘

Fig- 1.12. Phasor diagram .

for Example 1.3.

Artj_ll£l------------------------------— _______ ___________________ __________________________Transformers

Vertical component of I , = / lC0s

= //c o s 02 + Ic

17

'

= 6 x 0 .8 + 0.164 = 4.964 A. Horizontal component of l x = sin = / / sin q2 + = 6 x 0 .6 + 0.577 = 4.177 A. ^ c o s O ^ ^ sin 0 ^

... Primary current

= V(4.964)': + (4.177)2 = 6.488 A.

Primary power factor

= cos Ql =

= 1964 = Q 6.47

Ii

.

iagging.

E x a m p le 1.4. A tra n sfo rm er is designed to have hot-rolled steel lam inations, w ith a flux density o f 1.2 T a n d the w eight o f core a n d wire is found to he 100 kg and 80 kg respectively. I f the transform er is redesigned w ith cold-rolled grain oriented (CRGO) steel lam inations, which permit a h ig h er flu x d en sity o f 1.6 T ; fin d the saving in core and wire m aterials. The two types of core m a teria ls have the sam e densities and the total flux rem ains the same. M ake suitable assum ptions w herever required. S o lu tio n . W ith h o t-ro lled steel lam inations: Total flux

tymax = Flux density B ml x A rea A x j.

ci a

i

ty m a x

a

0m2 ^ ™ S ° n Prim ary m inding P, 160 A to a resistive lo a d w hereas a pure-canacitanrp In n !? W lng ' g 1 13 (a>- W inding S feeds 10 a Pare capacitance load across w inding T takes 20 A curnnt^is neglected. J

w U utb«

transform er

6 ) W ith th e p o la r ity m a r k in g s on P as shorn*, m a rk the polarities on v in d tn g s S a n d T

1i ^ r J A aP lie n V 0' ta g e / 1 3 n d m U tual n u x 4 a re show n in th e p h aso r d iag ram of Fig. U S /(5)z + (3.5)2] = (39.65) (6.1) = 242 V (b) T he e q u iv a le n t c irc u it w ith all th e q u a n titie s referred to l.v. side is g iv en in F ig. 1.18 (6 ). T h e r a te d c u rre n t on th e 1 v side is

T 2

10,000 250 “

^

A

The p h a so r d ia g ra m for Fig. 1.18 (6 ) is illu s tra te d in Fig. 1.18

(c). From th e g e o m e try o f th is diagTam , O B2 = OA2 + A B 2 or

(250) = >/(V 2 cos G2 + l>fe2)2 + (V 2 sin 02 + 12 x e2)2

or

(250) = V(0.8 V 2 + 3.84)2 + (0.6 V 2 + 8.96)2

or

(250)2 = V 2 + 16.90V2 + 95.55

or

V \ + 16.90 V 2 - 62,404 = 0 Tr - 1 6 .9 0 + V286 + 249,616 Vo = 1 = 241.55 V. 2“ 2

Fig. 1.18. (c) Phasor diagram for circuit o f Fig. 1.18 (b).

A ltern ativ ely , th e se c o n d a ry te rm in a l voltage V2 can be obtained as follows: From th e p h aso r d iag ram , it m ay be seen th a t OD is approximately equal to OB = 250 volts. V 2 = O D - C D = 250 - CD. Now

CD = C E + E D = I 2re2 cos 0 2 + I ^ e 2 si 11 02 = (3.84) (0.8) + (8.96) (0.6) = 8.448 = 8.45 volts.

S econdary te r m in a l vo ltag e, V 2 = 250 - 8.45 = 241.55 V. The m a g n itu d e o f th e se c o n d a ry v o ltage V 2 tu rn s out to be sam e in both th e m ethods. How­ ever. the c o m p u ta tio n a l la b o u r in th e second m ethod is less th a n in th e firs t m ethod, th erefo re, e8econd m eth o d sh o u ld be p re fe rre d .

S c a n n e d by Cam S canner

28

(Art. 1.6

Electrical Machinery

E x a m p le 1.10. The equivalent circuit referred, to the low -tension side o f a 25012500 V single ph a se tra n sfo rm er is show n in Fig. 1.18 (a). The lo a d im p e d a n c e connected to the hig h -ten sio n te rm in a ls is 380 4 j 230 £2. For a p rim a ry voltage o f 250 V, com pute (a) the secondary term in a l voltage, (b) p rim a ry current a n d power factor, a n d (c) pow er o u tp u t a n d efficiency. S o lu tio n , (a) T he load im pedance referred to l.t. side is

02

Fig. 1.19. (a) Transformer equivalent circuit for Exam ple 1.10.

o-2 a w vv-

\2

Z f = (380 + j 230)

( N j) ~n2

0-7

o-7n

= (3.8 + j 2.3) £2.

T ran sfo rm er leakage im pedance 0 .2 4 y'0 .7 ohm Vf250V and th e load im pedance 3.8 + j 2.3 £2 are in series as show n in Fig. 1.19 (6 ). Therefore, total series im ­ p edance is 4 +./3 = 5 Z 36.9°. -o T aking V x as th e reference phasor,

Fig. 1.19. (fc) Pertaining to Exam ple 1.10.

V x = 250Z 0°. j , 250Z0° 7> - 5Z 3.69° ~ 5° Z _ 36 9 = 50 (cos 36.9° - 7 sin 36.9°) = (40 - j 30) A. or

I f = 50 A

and

/ 2= / , ' ^ . 5 0 x i = 5A

N.

10

.-. Secondary term in al voltage = I 2 Z L = 5 [3802 + 2302]1/2 = 5 x 444 = 2220 V (6 ) The core loss cu rren t

V, 2 5 0 / 0° Ic = — = 5 Qq ^ Q0 = 0.5 Z 0° = 0.5 + j 0

The m agnetizing c u rre n t

Vi 250 Z 0° Im = = 2 5 0 / 9 0 ° = 1Z “ 90° = 0 ~ J 1-

Exciting c u rren t

Ie = Ic + I m = (0.5 - 7 I) A.

H ence to tal p rim ary c u rre n t I x - I f + I e = (40 -730) + (0.5 - 7 I) = 40.5 - 7 31 = 51 Z - 37.4° P rim a ry cu rre n t 7X= 51 A and p rim ary p.f. = cos 0j = cos 37.4° = 0.794 lagging.

Pow er o u tp u t

380 = 0.855 [380242302]1/2 = V2 12 cos 0 2 = 2220 x 5 x 0.855 = 9500 w atts.

Also pow er o u tp u t

= ( I f) 2 R l = (50)2 (3.8) = 9500 w atts.

= cos e 2 =

(c) Load p.f.

Core loss or I

Scanned by Cam Scanner

Pc

V? R.

(250)2 = 125 w a tts 500

Pc = TCR C= (0.5) (500) = 125 watts.

I Art. I J i Transformers

Also pow er in p u t

= P ow er

29

+ l ' * ^ 0 ' 7 9 4 ^ 1 0 ' 125

= 9500 + 125 + 500 = 1 0 , 125 w a to Efficiency = ^E ^E E t _ In p u t - I ^ RgpC In p u t InrMii

= l _ ^ o s s e s _ i 5 0 0 + 125 In p u t 10,125 - 0.9383 p.u. or 93.83%. 1.7. O p e n -circ u it a n d S h o r t-c ir c u it T ests These tw o te s ts on a tra n sfo rm e r hnin i j * . circuit of Fig. 1.16, (ii) th e v o lta g e ’r e e u l a t f i tb e .Param et®>-s of th e equivalent param eters can also be o b tain ed from tbp nh • ??• efficiency. The equivalent circuit winding d etails. C o m p lete a n a ly sis th e f c dlm ensions «f l !>» transform er core and its circuit p a ra m e te rs a re know n. nS orm cr can carried out, once its equivalent in th ?tran Ifo rm eqrUired dU ring

‘W° teSts is « JuaI

O pen c ir c u it (o r N o -lo a d ) t e s t Tbn pi™,;*

«>e appropriate power loss occurring r

e

.

a single p h ase tra n s fo rm e r is given in Fig. 1,20 fa). f a f a . s 'd i a ^ a T r f o h m e f a r w ^ m ltT r and an am m ete r a re show n connected on th e low voltage side of the t r a n s f o r m e V X w g h vo age side is left open circuited T he rated frequency voltage applied f a X p r l a r y ■e low rtw l ‘V w l recorded ’

W1. j

i

P °f a variable ratio auto-transform . W hen the voltm eter ' V' Winding' a" th e th re e - ^ tr u m e n t r e a d i n g s ^

Auto-

H .

vr vi

t (b)

Fig. 1.20. (a) Circuit diagram for open-circuit test on a transformer and (b) approxim ate equivalent circuit at no load.

The am m eter reco rd s th e no-load c u rre n t or exciting cu rren t Ie Since Ie is quite sm all (2 to 6 % of rated c u rre n t), th e p rim a ry leakage im pedance drop is alm ost negligible, and for all Practical p u rposes, th e ap p lied voltage Vj is equal to th e induced e.m.f. V /. Consequently, the equivalent circu it o f Fig. 1.14 (e ) gets modified to th a t shown in Fig. 1.20 (6 ). The in p u t pow er given by th e w a ttm e te r read in g consists of core loss and ohmic loss. The exciting c u rre n t b ein g a b o u t 2 to 6 p er cent of th e full load current, th e ohmic loss in the • ' ( 2 2 N Primary ( = le2r x) v a rie s from 0.04 p er cent 100 100 x 100 to 0.36 per cent of th e full-load Primary ohm ic loss In viow of th is fact, th e ohmic loss during open circuit te st is negligible comparison w ith th e n o rm a l core loss (approxim ately proportional to th e sq u are of the “PPlied voltage) H ence th e w a ttm e te r read in g can bo taken as equal to tran sfo rm er core loss.

W IIIvVVatUI II IWI

j

30

I Art. 1.7

Electrical Machinery

A negligible am o u n t of dielectric loss may also exist. E rror in the in stru m e n t readings may be elim in ated if required. Let Vj = applied rated voltage on I t. side, Ie = exciting cu rren t (or no-load current), and T hen No load p.f.

Pc = core loss. Pe = V x Ie cos 0 O — C08 9 , =

...(1.28)

y~J

From th e phasor diagram ofFig. 1.8 (c), it follows th a t I c = Ic cos 0 Oand /„, = Ie sin 0 O From Fig. 1.20 (6 ), 7C.= Vi Core loss resistance R CL - -=- =

vx Lq COS

_

Vi2 _ Vj2 Vj7e cos 0O Pc Also

...(1.29 a)

Ic R C L = Pc

R P‘ K rt - —n

...(1.29 b) (Ie cos 0 O)

M agnetizing reactance, Im

Ic sin 0 O

...(1.30)

The su b scrip t L w ith R c an d X m is used m erely to em phasize th a t th ese values are for th e l.t. side. T his m u st be k ep t in m ind th a t th e values of R c and X m, in general, refer to th e side, in which th e in stru m e n ts are placed (the l.t. side in th e p resen t case). A v o ltm eter is som etim es used a t th e open-circuited secondary term in als, in order to determ in e th e tu rn s ratio. T h u s th e open-circuit te s t gives th e following inform ation : (i) core loss a t ra te d voltage and frequency, (i i ) th e s h u n t b ran ch p a ram eters of th e equivalent circuit, i.e. R c an d X m and (i i i ) tu rn s ratio of th e tran sfo rm er. S h o r t - c i r c u i t te s t. The low voltage side of th e tra n sfo rm er is sh o rt-circu ited a n d th e in­ s tru m e n ts a re placed on th e high voltage side, as illu stra te d in Fig. 1.21 (a). T he applied voltage is a d ju ste d by a u to -tra n sfo rm e r, to circu late ra te d c u rre n t in th e h ig h voltage side. In a tra n sfo rm er, th e p rim a ry m .m .f. is alm o st equal to th e secon­ d a ry m .m .f., th erefo re, a ra te d c u rre n t in th e h.v. w in d in g c a u se s ra te d c u rre n t to flow in th e 1. w inding.

A primary voltage of 2 to 12% of its rated value is sufficient to circulate rated currents in both primary and secondary windings. From Fig. 1.21 (b), it is clear th a t th e secondary leakage im­ pedance drop appears across the exciting branch

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H.V. L.V Fig. 1.21. (a) Connection diagram for short circuit test on a transform er.

Ar‘ 1.7] (Rt and Xm in p arallel). A bout h a lf (

Transformers

~

31

leakage im pedance erefo re, re aacross r m .. th e excitln 6 appllod !eakage im pedance an a n dd ,, th th erefo ^ vo)tag^ gC appcars across th e secondary across the exciting b ran ch an d since th e la tte r is ^ 3" ^ The “ re " u x induces th e voltage B ( » of its ra te d value. H ence th e core loss b e W *>» “ re flux is also 1 the core flux, is 0 .0 1 p ercen t = —1— v 1 j ii 100 l o o X value at rated voltage. T h e w a ttm e te r in shnri c irc u it test, records th e core loss an d th e « w loss in both th e w indings. Since t h e ™ ' has been proved to be alm o st negligible in com panson w ith th e rated -v o ltag e core loss the wattmeter can be ta k e n to re g iste r o n lj the ohmic losses in both th e w indings.

A J

(9

Xlma e y proportional to th e sq u are of 0.36 percent ^ ^ ^ lo o x io o x 100 of its *2

o

. * tHe excibing cu rren t is 2 Rc' -c to to 6% of full load c u rre n t. W hen th e voltage ‘— T across the excitine r a nch 4 is -i.t-.o «► -------------------------exciting bh ran 1 tn m~ . ag™ =

( k V )■new l~

. .. Z (pu)neui (MVA)Bnew (kV)2B d d T aking th e ir ratio s, — -— = , %/nrA.-------- —t t ? ------Z {p u )old (MVA)Bold (kV)%.new rj t_ \ _ rj , x (MVA)fl . ^ (kV)B old Z (Pu )new ~ Z (pu)oid - m A ) B M ■ikV )i

n rtf} -< •

new

E x a m p le 1.13. Sh o w th a t the p er-u n it values o f reL, x eL a n d z eL are equal to the per-unit values o f reH, xeI1 a n d z eH. S u b scrip ts L a n d H denote l.v. a n d h.v. w in d in g s respectively. S o lu tio n .

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r.j in ohm s (rtL)p.u.= Z m in ohm s

•• (0

Art. L2L Transformers

But where N L and N „ a re th e n u m b e r of tu rn s in th e 1 j tion of the v alue of reL in Eq. (i) gives : V 3n ^ V- w*nchngs respectively. Substitu2

_ r eH

(rcl)p.u. =

'cH

n h

v HJ JBL

= (r ew)p.u. (xcl)p.u. = (xeH)p u and

Similarly

= r eH

(N N,

ZBn = Z BL ( n h ) N,

Z bh

(z

=(z ) Hence th e req u ire d re s u lt h as been proved. ' P'“ " E x a m p le 1.14. ( a ) A 2 0 k V A , 2 5 0 0 /2 5 0 V sinnlp n hn.a , /• t impedance o f z ,H = 2 .6 + j 4.3 n w hen referred to A 1 p e‘>uiJl al ent Uaka8e Ci) h.v. side a n d (ii) l.v. side. F in d also the total leakage im d " to ,,. jr., . r r leakage impedance drop referred to each side. J e o n t h e ( i^ T d J n d ^ u J . t S T S o lu tio n , (a) (i) H .V. side :

^

Base voltage

VBH= 2500 V

Base c u rre n t

I BH = ^0.000 = 8 A. 2500

Base im pedance Z BH =

°f 4 8 A

the ^

»P*-

= 3 1 2 .5 D 8

Ibh

Per u n it v alue of eq u iv ale n t leakage im pedance referred to h.v. side, z eH in Q

2 .6 + i 4 3

^z■ ^

a

■ - si

ir

- 0 0 0 8 3 2 +J' ° 01376-

(ii) L.V. side : E q u iv alen t leakage im pedance referred to l.v. side 2 2 (N r) f 250 ) = (2.6 + j 4.3) ~ z eH z eL z eH 2500 Nh \ / \ = 0.026 + j 0.043 a

M w

VBL ® ' b = 250 Vl ,/flL = ^ 250

Now

= 80 A.

■n • 1 " ^BL _ 250 _ q iq c n •■Base im pedance Z BL = -j— = qq l BL

_ (z eDp.u. ~ g

The m a g n itu d e o f

or

jn £i

_ 0 ^ 2 6 + ^ M l l _ 0.00832 + ; 0.01376. 3.125 = t / p M M ) * * (0.01S76) = 0.01607

Total le a k a g e im p e d a n c e d rop on h .v . sid e

= (0.01607) (2500) = 40.175 V ^

total le a k a g e im p e d a n c e d rop on l.v. s id e

= (0.01607) (250) = 4.0175 V.

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38

Electrical Machinery

IArt. l.(»

( 6 ) (i) P.u. value of exciting cu rrent referred to h.v. side IeH

4.8 „ ------8 - ° ' 6 P U-

(ii) Exciting c u rre n t w hen referred to l.v. winding f 2500>\ = 48 A. U =(4-8) 250 J' T IeL in Amp. 48 „ „ (hOp u. = jbl in Amp. “ 80 T he p.u. value of exciting cu rren t is seen to be sam e on both th e sides. E x a m p le 1.15. A 10 kVA, 2000/200 V, single-phase transformer has following param eters: r t = 3.6 12 r2 - 0.04 12

x x = 5 .2 Q x 2 = 0.056 12.

Subscrip ts 1 an d 2 denote high and low voltage w indings respectively. F ind the p.u. values o f r el re2 xcl and x e2.

S o lu tion . H.V. side. B ase Voltage

VBH = 2000 V . 1 0 ,0 0 0 _ A B ase cu rre n t 7fi" = “ 2 0 0 0 ~ = 5 A 2000 = 40012 /. B ase im pedance Z BH = 3.6 r i p “-

400

= 0.009

5.2 *1 p . u . - 400 = 0.013 • ohm h s = r 1 *+ r 2 f — M re\ in r

= (3.6)+ 0.04 (10) = 7.6 12

7-6 4 0 0 = 0.019

re l p . u .

1

r

(Nff

X} + X2

%bh

Nn

L.V. side. Base voltage VBL = 200 V Base current /. Base impedance

r

10,000 200

BL z bl

=™

=* «

4 0.056 *2 p.u. “

= 0.014.

r2 + r j

Pc2p.u. ~ ^ BL

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^ A

(N rf N.

400

>5.2 + 0.056 (lO)2! = 0 .0 2 7 .

Art- I j l

Similarly,

*e2pu. = 0.027

Note th a t in p .u . sy stem

• sr ii

r. l p u . = r' 2pu = r lp u + r2 p u .

*“ « • - * « P-«. = This exam ple show s t h a t th e tran

seco n d ary te rm in a l

x 100 in percentage.

At no load, the prim ary leakage im pedance drop is alm ost negligible, th e re fo re , th e secon­ dary no load voltage E2 =V,

T he expression for voitage reg u latio n can also be w ritte n as

v ' % - v> x 100 in percentage =

x 100 fa p e rc e n ta g e

v A N1 Here Vj is the prim ary applied voltage. The change in secondary term inal voltage w ith l . . sec on ary eakage im pedances of th e transform er Thp m CUlTf0I\t *s ^ u e to th e p rim a ry and the load pf; load current, total resistan ce and to tal i e a k a a e T t ° f th ’S °h a n g e d e P e n d s on A distribution transform er should have a 11 1 actan ce ° f t h e tra n sfo rm e r. 3 t °n )’ S° p h a t th e term in al a t th e consum er ?f V° 'ta g e re g u la tio n « •* g° ° d voltage io lta e ' i T ’ F ° r 3 tran sfo ™ o r of large S T ' 1 r? .pre/n u ses d° “ n o t v a ry w idely as the voltage a t th e consum ers' term in als will fal ppreeTb t t d b ^ P° ° r V° ' tagC R a t i o n ) , the ppreciably w ith th e in c re a se in l o a d - t h i s h as a

Art.

detrimental effect on th e ° P e ra tio n ^ iffo 7 fW ~ranSf°rmerS-— since these are designed to operate sa tis fa c tn ru !'^ tUbeS’ T V' sets*refrigerator motors etc. transformers should be designed to have a low v l3 a constant yoltage. Thus the distribution The voltage regulation of a transform a^ age imPedance. circuit referred to prim ary or secondary P i p ^ o !!6 ^rom approximate equivalent circuit of a transform er referred to the sennn^ *j *^us^ra^es the approximate equivalent is drawn in Fig. 1.24 (6 ) for a lagging- nowpr f Si f nC* Phasor diagram for this circuit tion, draw an arc of radius OD meeting th* I For the ca,culation of voltage regula1.24 (6 ) th a t O F (= E 2) is approximately equa* to 0 0 * ^ ^ E 2 = OC = OA + A B + BC (or B'C')

i'JP. I !

° A ^ F' 11 m3y b° SGen fr°m Fig'

= OA +A B ' cos 62 + DB' sin 02 = V 2 + I 2 re2 cos Q2 + I 2xe2 sin 02 Thus the voltage drop in the secondary terminal voltage - E 2 - V 2 = I 2re2 cos 02 + / 2 xe2 sin 02

I, ...(1.38)

r t f E 2 ~ V “a T eqU a‘ t0 A ° ’ U ,2 *•*The cha" 6 n ^ secondary terminal voltage is equal to th e m agnitude of £ 2 minus the magnitude of V2.

r

■

r