Electromechanical Energy Conversion energy into another form, For example, a heater trical energy into heat ene Jectrie bull converts electrical energy into light energy. However, electromechanical conversion devices (ie, devices that convert electrical enerBy into ‘mechanical enerBy ice — versa) find wide practical applications. For example, an ‘electric motor converts electrical energy into mechanical energy: On the other hand, an electric generator converts ‘mechanical energy into jcal energy. A major reason for the widespread use ‘of electro- mechanical energy conversion devices is that they are relatively efficient ‘and permit an easy control. In this chapter, we shall discuss the basic principles of lectromechanical energy conversion. G1 ELECTROMECHAN ICAL ENERGY CONVERSION electrical energy into mechanica INTRODUCTION many devices that convert one form of ily use yea rgy while an el converts elec’ 1 energy or vice versa is known as electrome- The conversion of. chanical energy conversion. lectromechanical energy conversion invol terchange of energy between an electrical field or magnetic field. ‘electrical system, @ diagram of an con- Ives the in! af system and @ mechanical system through the medium ofa coupling electric ‘Therefore, an electro: se version system bas three essential parts 4% ‘mechanical system and a coupling field (electric or magnetic). Fig. 1.1 shows the bloc! clectromechanical'energy ‘conversion system. Note that from left to right, the system represents version from electrical to mechanical . However, from right to left, it will represent conversion from mechanical to elect input Fig. 1.1 Electromechanical Energy Conversion System () Electric field as coupling medium. Electromechanical energy conversion can take place when electric field is used as the medium. Consider two oppositely charged plates of @ capacitor which are separated by @ dielectric medium. A force of attraction exists between the two plates that tends to move them together. ite to move in the direction of the force, we are If we allow one plat evening electrical energy into mechanical energy. On the ‘other hand, if we apply an external force fon one plate and try to increase the s em, we are then converting mechanical eparation between th nergy into electrical energy. Electrostatic microphones and electrostatic, voltmeters use electrostatic fields for energy conversion. (i) Magnetic field as coupling medium. Electromechanical ehergy conversion can also take Consider the case of a current- place more effectively when magnetic field is used as the medium. iences a force that tends'to move carrying conductor placed in a magnetic field. The conductor experi it Ifthe conductor is free to move in the direction of the magnetic force, the magnetic field helps the conversion of electrical energy into mechanical energy. This is essentially the princi 9 f. e i of all eecue motors On the other hand, if an externally applied fe eee ris po opposite to the magnetie force, mechanical energy is Converted into electrical pith + action is based on this principle. Note that in both cases, the magnetic Ape a 2 a me- dium for energy conversion. Mech ‘output Mechanical sy: 1 ‘Scanned with CamScanneePrinciples of Electrical Machines Ic is important to note that the quantity of energy that can be converted by Frid ox » coodivmn is relatively smallitts becouse the msvont of tree eae ae system is usvelly very small even when the applied voltage is high and the physical dimen st system are quite large. However, when magnetic field is used as a medium, a system aah not physical dimensions develops a much larger force than a system ‘using an electric field is same For this reason, the use of electric field as a medium for energy conversion has te » 1ons. 1.2. ELECTROMECHANICAL ENERGY CONVERSION DEVICES Electromechanical energy conversion takes place through electric field or magnetic g Kc dium, Although the various conversion devices operate on common set of et ae a structures of the devices depend on their function. Electromechanical energy conversion devine be divided into the following three categories : : a () Transducers. These conversion devices are used for measurement and control, operate under linear input-output conditions and with relatively small si include microphones, pick ups and loudspeakers, (i) Force-producing devices. These conversion devices are meant for producing force or torque with limited mechanical motion. Examples include relays, solenoid actuators and electromagnet (ii) Continuous energy conversion devices. These devices continuously convert electrical energy into mechanical energy or vice versa. They are used for bulk energy conversi utilisation. Motors and generators are the examples of such conversion devices. It may be noted that magnetic field is most suited as a medium for electromechanical ene:py conversion. Therefore, in thisbook we shall deal with magnetic field as the medium of energy conversion FEATURES OF ELECTROMECHANICAL ENERGY CONVERSION Electromechanical energy conversion takes place through the medium of magnetic field. The follow- ing salient features are worth noting in this energy conversion : (OAs with any energy cenversion system, the principle of conservation of energy holds good in case of electromechanical energy conversion. That is energy can neither be created nor destroyed; it can only be changed from one form to another. (ii) During electromechanical energy conversion, varjous losses occur in the system. This is illustrated in Fig. 1.2 which shows the conversion of electrical energy into mechanical energy. They generally ignals. Examples and Magnetic Coupling P, [| ae | = Electrical Flat Mechanical loss loss oss Fig. 1.2 Electromechanical Energy Conversion System se The electrical energy loss (? R) is due to current () flowing in the winding (having Ft R) of the energy converter. The field loss isthe core loss due to changing maBIS™ the magnetic core, The mechanical loss is the friction and windage loss zie ae the moving components. All these losses are converted into heat and raise the frp the energy conversion system. ie (ii) Electromechanical energy conversion is a reversible process ekcept fo bee system, The term reversible means that the energy can be transferred. bac! aie cma | the electrical anid the mechanical systems. However, each time we £9 Bess. a ‘Scanned with CamScanneeair gap as shown in Fig. 1.23. Stator : This part of the electrical machine is stationary: ‘mally is the outer frame of the machine. fo aed Rotor : This part of the electrical machine is free to rot 1 ate and normally is the inner part of the machine. The function of one member (stator or rotor) is to provide a magnetic field and is called the field structure. The function Of the other member (called armature) is to provide an assembly of conductors (armature winding) so that e.m.f. is a induced in them due to the magnetic field produced by the © “—————5 first member. Fig. 1.23 (ii) Both stator and rotor are made of ferromagnetic materials so that reluctance of each is negligible compared with the reluctance of the air gap. Therefor, almost all the m.m_f. in the magnetic circuit of «rotating electrical machine is used up to setup the required flux in the air gap. (iit) In most electrical machines, slots are cut on the inner periphery of the stator and outer periphery of the rotor as shown in Fig. 1.23. Conductors are placed in these slots. The ‘conductors placed in the slots of stator or rotor ar interconnected to form windings. The winding in which voltage is induced is called the rmature winding. The winding through which current is passed to produce the primary source of magnetic flux is called the field winding. . . (iv) So far as the basic functioning of the electrical machine is concerned, it makes no difleeneé which member (Le. member containing armature winding or field winding) is the revolving member. In d.c. machines, armature is made the rotating member due to practical considerations. In a.c. machines, both types of construction are employed. ae (v)_ The practical method of producing magnetic field is by means of current-carrying vate called field windings. There are two types of ficld construction vis. salient-pole. fi Hes nion-salien-pole field. In salien-pole construction, the field wind'ngs ake weit projecting iron parts as shown in Fig. 1.24. In non-salient pole construction windings are embedded in slots as shown in Fig. 1.25. ‘Scanned with CamScannee wi) 1.12 The mos and (iii) cal form,Electromechanical Energy Conversion - 1% Salient-Pole Machine Fig. 1.25 Fig. 1.24 : pole field wading construction. However both salient- D.C. machines always exrploy salient pote and non-salient pole Teld winding constr tion is (vi) The armature winding (whether on stator oF slots in cylindrical core. Although the armature winding is not producing a magnetic field, nevertheless the current 1 Fragnetic ux called armature flux). The armanre flu in the machine. Considered from its field producins potentiality, the armature winding is i calient-pole type winding. The number of magnetic poles that it tends to produce in the aver depends upon the cistbucon of the direction of, ‘current through the conductors as ‘one progresses around the armature. / 1.12 COMMONLY USED ELECTRICAL MACHINES ical machines are (i) D.C. machi ‘The most commonly used rotating electri ns din Syachronous machines. In these machines, conversion of enerBY sat fom or vice-versa takes place through magnetic field as the medium. incs (ii) Induction machines from electrical to mechani- Pole core Pole shoo." oe Yoko Fig. 126 a ‘Scanned with CamScannerTransfor, INTRODUCTION ‘The transformer is probably one ofthe most useful electrical devices ever invented. Itcan chan 2 Magnitude of alternating voltage or current from one valuc to another. This useful rope iE transformer is mainly responsible for the widespread use of alternating crrrents rather thag ae Samems Le, lect powers generated, vansmited and distributed in the for of eating ape ‘Transformers have no moving parts, rugged and durable in construction, ‘thus Fequiring very ling attention. They also have a very high efficiency — as high ss 99%. In this chapter, we sha i Some of the basic properties of transformers. 8.1 TRANSFORMER oe A transformer is a static piece of equipment used either for raising or lowering the voltage ofan supply with a corresponding decrease or increase in current. It essentially consists of two wi the primary and secondary, wound on a common laminated magnetic core es shown in Fig. 8.1, The winding connected to the a.c. source is called, primary winding (or primary) and the one connected iy load is called secondary winding (ot secondary). The alternating voltage V, whose magnitude isis be changed is applied to the primary. Depending upon the number of turns of the Primary (N,) and Secondary (N,), an alternating e.m.f. £, is induced in the secondary. This induced e.m.f. E, inthe secondary causes a secondary current /,. Consequently, terminal voltage V will appear across the load. IF ¥, > V;, itis called a step up-transformer. On the other hand, if V, <,, itis calleda step- down transformer. 4 a according to Faraday’s taws of electromagnetic and e.m.f. Ey is termed as secondary em... : d Clearly, - Er -n, a de and Ee -n, 262 ‘Scanned with CamScanneeer renwlorer ; oa Paes a AM 1 magnitudes of E, and E, *depend upon the number of tis omer IN, > W;. then E, > E, (or V5 > Fanlea oe ree es Moor baad, iT, < Ny, then E < E; (or ¥ < ¥;) and we get a sicp-down transformer. UE load is across the Yecondary winding, the sctondary emf. E, will cause a current J, to flow the load: Thus, a transformer enables us 10 transfer a.c. power from one circuit 10 andi aha change in voltage level. . Ct ‘The following points may be noted carefully = (j) The wansfgrmer action is based onthe lava of electromagnetic induction (i) There is no electrical connection between the prithary and secondary. The a.c. power is transferred from primary 10 secondary through magnetic fux. (ui). There is no change in frequency i.e., output power has the same frequency as the input power. "() The losses that occur in a transformer are : (a) core losses — eddy current and hysteresis losses (6) copper losses — in the resistance of the windings Inpractc, these-losses are very small so that output power is nearly equal to the input primavy owe. Inother words, a transformer has very high efficiency. i $2 THEORY OF AN IDEAL TRANSFORMER Jan idea! transformer is one that has ‘ (0 n0 winding resistance (i) no leakage flux ie., the same flux links both the windings (ii) no iron losses (ie., eddy Current and hysteresis losses) in the core ‘Although ideal transformer cannot be physically realised, yet its study provides a very powerful ‘ool in the analysis of a practical transformer. In fact, practical transformers have properties that approach very close to an ideal transformer. its @ ee eH Fig. 8.2 : Joa $40 ideal transformer on no load i.., secondary is open-circuited as shown in Fig. 8. whey Conditions, the primary is simply a coil of pure inductance.. When an alternating elle ty ase Pray, it draws a small magnetisng current J which lags behind the nda “90°. This alternating current J, produces an altemating flux $ which is proportional ss With it. The alternating flux $ links both the windings and apa dt mon for Both he windings. ¥ "Seer nf Siu, current lags behind the volage by 90°, “no iron losses, flux ¢ is in phase with /,. ‘Scanned with CamScanner2 Principles of Electrical Mach emi. E18, at Every inset, C4 >and ig shind flux @ by 90° (See Art. 8 vey, primary and c,n.f. E, in the secondary. The primary ‘ ‘oppositionto Y, (Lenz's law). Bothe.m,s E, and E, lag be! their magnitudes depend upon the mumber of primary and se ane Fig. 8.2 (if shows the phasor diagram ofan ideal transformer ONDA 7 s8commog to both the windings, it has been taken asthe reference Phasor. ‘As shown in Art 8.3, the ‘emf. E, and secondary em, E lag behind the fux ¢ by 90° Note that E, and £, are in phase, Buy E, is equal to V, and 180° out of phase with it 83 EM.F. EQUATION OF A TRANSFORMER : oa Cec tl a ahcrentingvoage 7, of Bemucoy is pli the primary w shown iF £2( ‘The sinusoidal flux ¢ produced by the primary can bbe represented as : $= d.sinot “The instantaneous e.n.f, ¢, induced in the primary is eM at = -m FO nsines) = -@N,6,cosa1=~2nfN, $, cost . = 26/0, 4, sin(o 1-90) It is clear from the above equation that maximum value of induced e.m.f. in the primary is Eq) ~ 20M, bu The rms. value E, of the primary e.m.f. is Eng _ 22fM bn ae ar or Ey = 444fN, bq : Similars, = 4456, {In an ideal transformer, E, = V, and E,= Vy. Note. It is clear from exp. (f) above that emf. E, induced ir i i Lest ca & koh be oanly ns ed 4 Fee 84 VOLTAGE TRANSFORMATION RATIO (4) From the above equations of induced em. we have (See Fig, 8.3) ce See vy 5s The constant X is called vollage transformat ratio, Thusit K=5 (Le. MYR, = 5) en, <3 Bn For an ideal transformer ; ay (9 E,=¥; snd E, = ¥, 28 there is 00 volage |" 2 wa 1 drop inthe windings. 4 \ l, Z BK Ne (i) There are no losses. Therefore, volt-amperes Ny Ne input to the primary are equal to the output Fig. 8.3 volt-amperes i.¢. iene oe eel > Theem£ E, induced in the secondary is produced by the same flux (.e. $= 6, si t E,, Thus the only difference in thee ms. Vac oft bs th itloeace ear eee Ss . 7 4 ‘Scanned with CamScannee= Valo Se ers eye ul, . 265 ES 3 Cecio 1 ager eeees or 4% Hace irs are in the inverse ratio of voltage transformation ratio. This simply means that if jtage, there is a corresponding decrease of current. former has 66 turns in the secondary. Calculate (i) we raise the vol i 4 200/200 00 beret i-load currents. Neglect the losses. 4 P si a we kro 1 f Example 8.1. primary turns (it) primary and secondary ful i ; : : ; ond Ex ' sere ad thés stage [ | ZDkVA N,N #66 5 Fig. 8.4 Solution. Fig. 8.4 represents the conditions of the problem. 200/2600 = 1/10, 10, @ NN, x10 = 66 x 10 = 660 turns ¥, hy -(20x10) 20x10? _ 20x10" nV, aaa a oa 20x10 _ 20x10" ao Be ek - i Y, ; 2000 10A J; = Kh, = (1710) 100 = 104 Alternatively, Example 8.2. An ideal 25 kVA transformer has 500 turns on.the primary winding and 40 turns .. The primary is connected to 3000 ¥, 50 Hz supply. Calculate (i) primary on the secondary wit Coen A aatinrd n fill-load (ii) secondary e.m,.and (ii) the maximum core flux. w@ f= 812 = 1082.4” K se = KE, = (4/50) 3000 = 240'V. = 441M 9. wm 4-44 x 50x 500 x4.) ‘Scanned with CamScanner4 = Example &3. A single phase Principles of Electricay 3000 Be 3 444% 50%500 ~ 27* 10> wp 320/250 V, 50 He transformer has @ net co 27mWp 4 maximum flux density of 6 Wo/m’. Calculate the number of rarn, ofpr mary ond oe ow a Solution, E, =444fN, >, > ie Here pr alas s0H0, Gaal S361 = bari 5 2200 Ntaat pas ¥44%50x 00216 ~ 459 5 250 . Also aA ” WSS %, ~ Faaxs0xoone 7 5 Example 84, A sinusoidal flux 0.02 Wb (maximum) links with $5 Calculate the rm.s. vatue of the induced e.m.f. in Solution. 0.02 Wb; N, ‘maximum flux density in the core is low voltage sides Solution. Net iron length Net X-sectional area, A Max. flux in core, gy ‘for a 3000/20 ¥ ratio. To allow length 10 be 0-9 x gross iron length hams of transformer seconde the secondary. The supply frequency 50H: = 5S tums: f= 50 Hz ue of induced e.m.f i i Gyr 404, %, i = 4ATX 50 x 0.02% 55 = 244.2 Example 8.5. 4 single phase 50 Hz tra insformer sraformer has square core of 20.om side. The permissible 1 Wb/m*. Calcul ate the number of turns per limb on the high and ‘for insulation of stampings, assume the net iron = 09% 20 = 18em = 18% 20 = 360 cm? = 0-036 m? 8, A = 1x 0-036 = 0-036 Wo 445, 6, = BO 3000 Dies ety gauataaas oe 0036 ~ 375 Also N, = = pre 220) ae = * 44456, ~ 444xs0x0036 7 27 Note that one length is affected due to Example 8.6. 4 singl stamping. . peer of th ee ad transformer he ‘primary turns. The net cross- fional area of the core ‘magnetic length is | The primary voliage is 500 42.Hz. What isthe maximun fx density in he ie a rclatine ea foie ‘his fu density caleulae the magnerising currace Wd nes = Solution. Fig. 8.5 shows the core of a renaSr: os, ee ‘uch as are used in the armature of ade. generator The gat eaten of in fe). . E, = 4448 afn, or 500 © 444% 8, 60% 10" x 50 x 400 : ‘m= 0:938 Whim? Now Primary mmf = Flux x Reluctance or N,* (naar = 4% —! = at Mou, t or Wx mer = BX pe 08 ee 400 * Undane = 0938 x —— ER ‘Scanned with CamScannerqranstormer Cp )nar = OTATA RMS. value of magnetising current is J,(rms.) = oe = 0-528A J MW, example 8.1. A transformer having 90 turns on the primary and, J coptndto 120,60 He source The coupling bereeen in a el BRUSSEL eignaiing current 4A. Calculate re (the peak voliage across the secondary terminals + iy the instantaneous voliage across the secondary when the instantaneous voltage across the primary is 37 =a 90 tunis; Ny 2250 tums ; K = NYN, = 25% Ey = 1201 induced secondary voltage is E, = (NYM,)E, = 25% 120 = 3000 ‘The voltage varies sinusoidally so that peak secondary voltage is Bx, = ¥2%3000 = 4742 V the instantaneous secondary voltage is always 25 times solution. N, (9 RMS. value o} Ema) EB (ia) NowE, = KE, = 25, Therefore, pester than the instantaneous Primary voltage. 4 E; 5X Ey yy = 25% 37 = 925V TUTORIAL PROBLEMS JL. ASDKVA, 6500/250 V transformer has 52 sepondary urns. Find (i) the number of primary turns (i) full- 1 1373 (i) 7-58 A ; 200 A] toad primary and secondary currents. Neglect losses. 4 The net erss-sctional sr ofthe core of 400/3000 V. $0 Hz transformer is 600 cm’. If the maxinwum pee serait inthe epre is 1-3 Whim’, find the number of tums on the primary and secondary. [2 In certain 50 kVA transformer, the number of tums on the primary and secondary wit St rcopectively. I primary is connected 102 3300 V supply, find (i secondary voltage (if) the primary ae esrendary currents when the transformer is fully loaded. Neglect the losses, 1G 230 V (i) 15-2 A; 218 A} ‘A single phase transformer has a ratio of | : 10 and a secondary winding of 1000 turns. The primary vinding iz connected to 25 V sinusoidal supply. the maximum value of flux i the core is 2:15 mWb, find () the frequency of the supply (ii) the number of primary turns (iif) the secondary voltage on open- sire {(0 25 Hz (i) 100 turns (iif) 250 V] ‘Asingle phase 50 Hz transformer has 20 primary turns and 273 secondary tums. ‘The net cross-sectional area ofthe cor i 400 em’. if the primary winding is connected to 230 V supply, find (/) peak value of flux density in the core (i/) voltage induced in the secondary winding [() 124 ‘Whim (i/) 3003 ¥/) 85 PRACTICAL TRANSFORMER ‘ i ahme wndng rem ml) ele 10, te ones. Since the iron cori subjected wo shea oer occ ey em Tas sin it. Thee Seo les Or ee ae ce a ante py fae a eee O08 wt magnitude of iron losses is quite small in a practical transformer. es die es ‘Since the windings consist of copper conductors, it sda at itary ad secondary will have Winding resisene’, Ts ea” ce R, act in series with the respective windings as shown in Fi- , Input power to transformer, P, = V, J, cos $, Output power of transformer, P, = V1, cos $y Note © The reader may draw the phasor diagram of a loaded transformer for (i) unity p.f. and (if) leading p.f. as an exercise. Example 8.22. The primary of a 1000/250 V transformer has a resistance of 0-15 Q and leakage reactance of 0.8 2 Find the primary induced e.m.f. when the primary current is 60 A at 08 pf. lagging. Solution. Primary impedance, Z, = (0:15+j0-8)Q = 0-814 27996. Power factor angle, $, = cos'0-8 = 369° ‘Taking applied voltage as the reference phasor, we have, V, = 1000 20° V. Now, -E, = Vs-h, = 1000. 20°-60 2-369" x 0814 279-6" = 1000 20°—48-8 Lae = 1000-(36 +33) = (964 -/33)V = 9645. Z-2°V ‘i ig Primaryem.£,E, = (~964+/33)V = 964-5 2178°V Example 8.23. The voltage on the secondary of a single phase transformer is 200 V when supplying a load of 8k ata pf. 0f 08 lagging. The secondary resistance is 0-04 and secondary leakage reactance ts 0'8 92 Calculate the indisced e.mJ, in the secondary. Solution. Teking secondary voltage as the reference phasor, we have, V;~ 200 20° V. Stl x10? Sa Secondary current, f, = Sree Tapa Power factor angle, $ ~ cos 08 = 365° I, = 30L367A Zq = (006+/0HQ=08 87162 ; B= Vth, F = 2D LP + 50 Z-369 x08 287-14 = 20020°+40.25024" - ‘Scanned with CamScanneroe ee 25a —— Ee, WOT ; Le Ftp tous acd - (200 +0) + 25 58+ /30-75) ie = (225-58 +, ww 75) volts = = ihe secondary e.m.f E, leads the mecondare femta var nae “9 IMPEDANCE RATIO ; i : SS ee ees 2 ny a = e > ae Fig. 8.22 _ Fit $22 shows the resistance and reactance of the primary referred to the secondary. Note that imary now has no resistance or reactance. E coma baa 2000/400 V single-phase transformer has R, = $ 2; X, = 122; R; 5 termine the equivale si , Powys (i ae i¢ equivalent impedarice of the transformer referred to (i) K ~ 40072000 = 1/5 0 zt Ps Bg amet ger OE Y Fin aire Pe qisleena ‘Scanned with CamScanner