Chapter Four Three Phase Induction Machine 4.1 Introduction

Transcription

1 Chapter Four Three Phae Induction Machine 4. Introduction Three-phae induction otor are the otor ot frequently encountered in indutry. They are iple, rugged, low-priced, and eay to aintain. They run at eentially contant peed fro zero to full-load. The peed i frequency-dependent and, conequently, thee otor are not eaily adapted to peed control. However, variable frequency electronic drive are being ued ore and ore to control the peed of coercial induction otor. In thi chapter we cover the baic principle of the 3-phae induction otor and develop the fundaental equation decribing it behavior. We then dicu it general contruction and the way the winding are ade. Squirrel-cage, wound-rotor ranging fro a few horepower to everal thouand horepower perit the reader to ee that they all operate on the ae baic principle.

2 Three-Phae Induction Machine 7 4. Principal coponent A 3-phae induction otor (Fig.4.) ha two ain part: a tationary tator and a revolving rotor. The rotor i eparated fro the tator by a all air gap that range fro 0.4 to 4, depending on the power of the otor. Fig.4. Three-phae induction otor. The tator (Fig.4.) conit of a teel frae that upport a hollow, cylindrical core ade up of tacked laination. A nuber of evenly paced lot, punched out of the internal circuference of the laination, provide the pace for the tator winding. The rotor i alo copoed of punched laination. Thee are carefully tacked to create a erie of rotor lot to provide pace for the rotor winding. We ue two type of rotor winding: () conventional 3-phae winding ade of inulated wire and ()

3 8 Chapter Four quirrel-cage winding. The type of winding give rie to two ain clae of otor: quirrel cage induction otor (alo called cage otor) and wound-rotor induction otor. Fig.4. Exploded view of the cage otor of Fig.4., howing the tator, rotor, end-bell, cooling fan, ball bearing, and terinal box. The fan blow air over the tator frae, which i ribbed to iprove heat tranfer. A quirrel-cage rotor i copoed of bare copper bar, lightly longer than the rotor, which are puhed into the lot. The oppoite end are welded to two copper end-ring, o that all the bar are hort-circuited together. The entire contruction (bar and end-ring) reeble a quirrel cage, fro which the nae i derived. In all and ediu-ize otor, the bar and end-ring are ade of diecat aluinu, olded to for an integral block.

4 Three-Phae Induction Machine 9 A wound rotor ha a 3-phae winding, iilar to the one on the tator. The winding i uniforly ditributed in the lot and i uually connected in 3 wire wye. The terinal are connected to three lip ring, which turn with the rotor. The revolving lip-ring and aociated tationary bruhe enable u to connect external reitor in erie with the rotor winding. The external reitor are ainly ued during the tart up period; under noral running condition, the three bruhe are hort-circuited. 4.3 Principle of operation The operation of a 3-phae induction otor i baed upon the application of Faraday Law and the Lorentz force on a conductor. The behavior can readily be undertood by ean of the following exaple. Conider a erie of conductor of length l, whoe extreitie are hort-circuited by two bar A and B (Fig.4.3 a). A peranent agnet placed above thi conducting ladder, ove rapidly to the right at a peed v, o that it agnetic field B weep acro the conductor. The following equence of event then take place:. A voltage E Blv i induced in each conductor while it i being cut by the flux (Faraday law).

5 0 Chapter Four. The induced voltage iediately produce a current I, which flow down the conductor underneath the pole face, through the end-bar, and back through the other conductor. 3. Becaue the current carrying conductor lie in the agnetic field of the peranent agnet, it experience a echanical force (Lorentz force).. 4. The force alway act in a direction to drag the conductor along with the agnetic field. If the conducting ladder i free to ove, it will accelerate toward the right. However, a it pick up peed, the conductor will be cut le rapidly by the oving agnet, with the reult that the induced voltage E and the current I will diinih. Conequently, the force acting on the conductor wilt alo decreae. If the ladder were to ove at the ae peed a the agnetic field, the induced voltage E, the current I, and the force dragging the ladder along would all becoe zero. In an induction otor the ladder i cloed upon itelf to for a quirrel-cage (Fig.4.3b) and the oving agnet i replaced by a rotating field. The field i produced by the 3-phae current that flow in the tator winding, a we will now explain.

6 Three-Phae Induction Machine (a) (b) Fig.4.3 Moving agnet cutting acro a conducting ladder. 4.4 The Rotating Field and Induced Voltage Conider a iple tator having 6 alient pole, each of which carrie a coil having 5 turn (Fig.4.4). Coil that are diaetrically oppoite are connected in erie by ean of three juper that repectively connect terinal a-a, b-b, and c-c. Thi create three identical et of winding AN, BN, CN, that are echanically paced at 0 degree to each other. The two coil in each winding produce agnetootive force that act in the ae direction.

7 Chapter Four Fig.4.4 Eleentary tator having terinal A, B, C connected to a 3-phae ource (not hown). Current flowing fro line to neutral are conidered to be poitive. The three et of winding are connected in wye, thu foring a coon neutral N. Owing to the perfectly yetrical arrangeent, the line to neutral ipedance are identical. In other word, a regard terinal A, B, C, the winding contitute a balanced 3-phae yte. For a two-pole achine, rotating in the air gap, the agnetic field (i.e., flux denity) being inuoidaly ditributed with the peak along the center of the agnetic pole. The reult i illutrated in Fig.4.5. The rotating field will induce voltage in the phae coil aa', bb', and cc'. Expreion for the induced voltage can be obtained by uing Faraday law of induction.

8 Three-Phae Induction Machine 3 Fig.4.5 Air gap flux denity ditribution. The flux denity ditribution in the air gap can be expreed a: ( θ ) co θ B (4.) B ax The air gap flux per pole, φ P, i: φ ( θ ) π / π / ax B lrdθ B p (4.) Where, l i the axial length of the tator. r i the radiu of the tator at the air gap. lr Let u conider that the phae coil are full-pitch coil of N turn (the coil ide of each phae are 80 electrical degree apart a hown in Fig.4.5). It i obviou that a the rotating field ove (or the agnetic pole rotate) the flux linkage of a coil will vary. The flux linkage for coil aa' will be axiu ( N φ at ωt 0 P o ) (Fig.4.5a) and zero at ω t 90 o. The flux linkage λ a ( ω t) will vary a the coine of the angleω t. Hence;

9 4 Chapter Four λ a ( ωt) Nφ co t ω (4.3) p Therefore, the voltage induced in phae coil aa' i obtained fro Faraday law a: e a ( ωt) dλa ω Nφ p inωt Eax inωt (4.4) dt The voltage induced in the other phae coil are alo inuoidal, but phae-hifted fro each other by 0 electrical degree. Thu, ( 0) e b E in ω t (4.5) ax ( 0) e c Eax in ω t +. (4.6) Fro Equation (4.4), the r value of the induced voltage i: E r ωnφ p πf Nφ p 4.44 fnφ (4.7) Where f i the frequency in hertz. Equation (4.7) ha the ae for a that for the induced voltage in tranforer. However, φ P in Equation (4.7) repreent the flux per pole of the achine. p Equation (4.7) how the r voltage per phae. The N i the total nuber of erie turn per phae with the turn foring a concentrated full-pitch winding. In an actual AC achine each phae winding i ditributed in a nuber of lot for better ue of

10 Three-Phae Induction Machine 5 the iron and copper and to iprove the wavefor. For uch a ditributed winding, the EMF induced in variou coil placed in different lot are not in tie phae, and therefore the phaor u of the EMF i le than their nuerical u when they are connected in erie for the phae winding. A reduction factor K W, called the winding factor, ut therefore be applied. For ot three-phae achine winding K W i about 0.85 to Therefore, for a ditributed phae winding, the r voltage per phae i E r 4.44 fn φ K (4.8) ph p W Where N ph i the nuber of turn in erie per phae. 4.5 Running Operation If the tator winding are connected to a three-phae upply and the rotor circuit i cloed, the induced voltage in the rotor winding produce rotor current that interact with the air gap field to produce torque. The rotor, if free to do o, will then tart rotating. According to Len law, the rotor rotate in the direction of the rotating field uch that the relative peed between the rotating field and the rotor winding decreae. The rotor will eventually reach a teady-tate peed n that i le than the ynchronou peed n at which the tator rotating field rotate in the air gap. It i

11 6 Chapter Four obviou that at n n there will be no induced voltage and current in the rotor circuit and hence no torque. In a P-pole achine, one cycle of variation of the current will ake the f wave rotate by /P revolution. The revolution per inute n (rp) of the traveling wave in a P-pole achine for a frequency f cycle per econd for the current are: 0 f n f * 60 (4.9) P p The difference between the rotor peed n and the ynchronou peed n of the rotating field i called the lip and i defined a n n n (4.0) If you were itting on the rotor, you would find that the rotor wa lipping behind the rotating field by the lip. The frequency f of the induced voltage and rp n n n current in the rotor circuit will correpond to thi lip rp, becaue thi i the relative peed between the rotating field and the rotor winding. Thu, fro Equation (4.9): f p 0 p ( n n) n f (4.) 0

12 Three-Phae Induction Machine 7 Thi rotor circuit frequency f i alo called lip frequency. The voltage induced in the rotor circuit at lip i: φ (4.) E 4.44 fn pkw fnφ pkw E Where Ei the induced voltage in the rotor circuit at tandtill, that i, at the tator frequency f. The induced current in the three-phae rotor winding alo produce a rotating field. It peed (rp) n with repect to rotor i: 0 f 0f n n (4.3) p p Becaue the rotor itelf i rotating at n rp, the induced rotor field rotate in the air gap at peed n + n ( - )n + n n rp. Therefore, both the tator field and the induced rotor field rotate in the air gap at the ae ynchronou peed n. The tator agnetic field and the rotor agnetic field are therefore tationary with repect to each other. The interaction between thee two field can be conidered to produce the torque. A the agnetic field tend to align, the tator agnetic field can be viualized a dragging the rotor agnetic field.

13 8 Chapter Four Exaple 4. A 3-phae, 460 V, 00 hp, 60 Hz, four-pole induction achine deliver rated output power at a lip of Deterine the: (a) Synchronou peed and otor peed. (b) Speed of the rotating air gap field. (c) Frequency of the rotor circuit. (d) Slip rp. (e) Speed of the rotor field relative to the (i) rotor tructure. (ii) Stator tructure. (iii) Stator rotating field. (f) Rotor induced voltage at the operating peed, if the tator-to-rotor turn ratio i : 0.5. Solution: 0 f 0 * 60 (a) n 800 rp, p 4 ( ) n ( 0.05) *800 rp n 70 (b) 800 rp (ae a ynchronou peed) (c) f f 0.05 x 60 3 Hz. (d) lip rp n 0.05 * rp (e) (i) 90 rp (ii) 800 rp (iii) 0 rp (f) Aue that the induced voltage in the tator winding i the ae a the applied voltage. Now, N 460 E E E 0.05*0.5* 6.64V / Phae N 3

14 Three-Phae Induction Machine Equivalent Circuit of the Induction Motor In thi ection we develop the equivalent circuit fro baic principle. We then analyze the characteritic of a low-power and high-power otor and oberve their baic difference. Finally, we develop the equivalent circuit of an aynchronou generator and deterine it propertie under load. A 3-phae wound-rotor induction otor i very iilar in contruction to a 3-phae tranforer. Thu, the otor ha 3 identical priary winding and 3 identical econdary winding one et for each phae. On account of the perfect yetry, we can conider a ingle priary winding and a ingle econdary winding in analyzing the behavior of the otor. When the otor i at tandtill, it act exactly like a conventional tranforer, and o it equivalent circuit (Fig.4.6) i the ae a that of a tranforer, previouly developed. Fig.4.6 Equivalent circuit of a wound-rotor induction otor at tandtill.

15 30 Chapter Four In the cae of a conventional 3-phae tranforer, we would be jutified in reoving the agnetizing branch copoed of jx and R becaue the exciting current I o i negligible copared to the load current I P. However, in a otor thi i no longer true: I o ay be a high a 40 percent of I P becaue of the air gap. Conequently, we cannot eliinate the agnetizing branch. The tator and rotor winding can be repreented a hown in Fig.4.7 (a) and (b), where, V per-phae terinal voltage. R per-phae tator winding reitance. L per-phae tator leakage inductance. E per-phae induced voltage in the tator winding L per-phae tator agnetizing inductance R c per-phae tator core lo reitance. E Per-phae induced voltage in rotor at tandtill (i.e., at tator frequency f,) R Per-phae rotor circuit reitance L Per-phae rotor leakage inductance Note that there i no difference in for between thi equivalent circuit and that of the tranforer priary winding. The difference

16 Three-Phae Induction Machine 3 lie only in the agnitude of the paraeter. For exaple, the excitation current I o i coniderably larger in the induction achine becaue of the air gap. In induction achine it i a high a: 30 to 50 percent of the rated current, depending on the otor ize, wherea it i only to 5 percent in tranforer. Moreover, the leakage reactance X i larger becaue of the air gap and alo becaue the tator and rotor winding are ditributed along the periphery of the air gap rather than concentrated on a core, a in the tranforer.

17 3 Chapter Four Fig.4.7 Developent of the induction achine equivalent circuit I Note that the rotor circuit frequency and current are: f and I. E (4.4) R + jx The power involved in the rotor circuit i: I R P (4.5)

18 Three-Phae Induction Machine 33 Which repreent the rotor copper lo per phae. Equation (4.4) can be rewritten a: I E (4.6) ( R / ) + jx Equation (4.6) ugget the rotor equivalent circuit of Fig.4.7b. Although the agnitude and phae angle of I are the ae in Equation (4.4) and (4.6), there i a ignificant difference between thee two equation and the circuit (Fig.4.7b and 4.4c) they repreent. The current I in Equation (4.4) i at lip frequency f, wherea I in Equation (4.6) i at line frequency f. In Equation (4.4) the rotor leakage reactance X varie with peed but reitance R reain fixed, wherea in Equation (4.6) the reitance R / varie with peed but the leakage reactance X reain unaltered. The per phae power aociated with the equivalent circuit of Fig.4.7c i: R P P I (4.7) Becaue induction achine are operated at low lip (typical value of lip are 0.0 to 0.05) the power aociated with Fig.4.7c i coniderably larger. Note that the equivalent circuit of Fig.4.7c i at the tator frequency, and therefore thi i the rotor

19 34 Chapter Four equivalent circuit a een fro the tator. The power in Equation (4.7) therefore repreent the power that croe the air gap and thu include the rotor copper lo a well a the echanical power developed. Equation (4.7) can be rewritten a: R R P Pag I R + ( ) I (4.8) The correponding equivalent circuit i hown in Fig.4.7d. The peed dependent reitance R ( ) / repreent the echanical power developed by the induction achine. P ech ech R I ( ) (4.9) ( ) Pag P * (4.0) ( ) P ech P (4.) and, I R Pag (4.) P Thu, P P : P : : ( ) ag : (4.3) ech Equation (4.3) indicate that, of the total power input to the rotor (i.e., power croing the air gap, P ag ), a fraction i diipated in the reitance of the rotor circuit (known a rotor copper lo) and the fraction - i converted into echanical power.

20 Three-Phae Induction Machine 35 Therefore, for efficient operation of the induction achine, it hould operate at a low lip o that ore of the air gap power i converted into echanical power. Part of the echanical power will be lot to overcoe the windage and friction. The reainder of the echanical power will be available a output haft power. 4.7 IEEE-Recoended Equivalent Circuit In the induction achine, becaue of it air gap, the exciting current 0 i high-of the order of 30 to 50 percent of the full-load current. The leakage reactance X l i alo high. The IEEE recoend that, in uch a ituation, the agnetizing reactance X not be oved to the achine terinal but be retained at it appropriate place, a hown in Fig.4.8. The reitance R c i, however, oitted, and the core lo i luped with the windage and friction loe. Thi equivalent circuit (Fig.4.8) i to be preferred for ituation in which the induced voltage E differ appreciably fro the terinal voltagev.

21 36 Chapter Four Fig.4.8 IEEE recoended equivalent circuit. 4.8 Thevenin Equivalent Circuit In order to iplify coputation, V R, X,, and X can be replaced by the Thevenin equivalent circuit value V, R, and X, a hown in Fig.4.9, where: V th th th R + X th ( X + X ) V (4.4) ( X X ) If R + V << a i uually cae, th V X + X X K V (4.5) th The Thevenin ipedance i: ( R + jx) j( X + X ) jx Z th Rth + R + jx th (4.6)

22 Three-Phae Induction Machine 37 If ( ) R R << X + X, then, X th R X + X K th (4.7) and ince X << X, then, X th X (4.8) R Fig.4.9 Thevenin equivalent circuit. 4.9 Tet To Deterine The Equivalent Circuit The approxiate value of R, R, X R and X, in the equivalent circuit can be found by ean of the following tet: No-load tet When an induction otor run at no load, the lip i exceedingly all. Referring to Fig.4.7, thi ean that the value of R / i very high and o current I i negligible copared to I o. Thu, at no-load the circuit conit eentially

23 38 Chapter Four of the agnetizing branch X, R. Their value can be deterined by eauring the voltage, current, and power at no-load, a follow: a. Meaure the tator reitance R LL between any two terinal. Auing a wye connection, the value of R i: R / (4.9) RLL b. Run the otor at no-load uing rated line-to-line voltage, V NL (Fig.4.0). Meaure the no load current I NL and the total 3-phae active power P NL. Fig.4.0 A no-load tet perit the calculation of R and R of the agnetizing branch. IEEE-recoended equivalent circuit. For the no-load condition, i very high. Therefore, in the equivalent circuit of Fig.4.9, R /

24 Three-Phae Induction Machine 39 the agnetizing reactance X i hunted by a very high reitive branch repreenting the rotor circuit. The reactance of thi parallel cobination i alot the ae a X. Therefore the total reactance eentially Fig.4.a. X NL X +, eaured at no load at the tator terinal, i X. The equivalent circuit at no load i hown in (a) No-load equivalent circuit (b) Blocked-rotor equivalent circuit. (c) Blocked-rotor equivalent circuit for iproved value for R. Fig.4. The priary phae voltage can be obtained fro the following equation:

25 40 Chapter Four VLL V V / Phae 3 (4.30) Then the no-load ipedance can be obtained a following: V Z NL (4.3) I The no-load reitance i: R P NL NL (4.3) 3I The no-load reactance i: X NL Z R (4.33) NL NL In the IEEE recoended equivalent circuit we aue that X + X X NL (4.35) Then fro no load tet we only get the value of X + X Locked-rotor tet Under rated line voltage, when the rotor of an induction otor i locked, the tator current I P i alot ix tie it rated value. Furtherore, the lip i equal to one. Thi ean that R / i equal to R, where R i the reitance of the rotor reflected into the tator. Becaue I p i uch greater than the exciting current I o, we can neglect the agnetizing branch. Thi leave u with the circuit of Fig.4.8 (without agnetizing branch),

26 Three-Phae Induction Machine 4 copoed of the leakage reactance X, the tator reitance R, and R / the reflected rotor reitance. Their value can be deterined by eauring the voltage, current, and power under locked-rotor condition, a follow: a. Apply reduced 3-phae voltage to the tator and gradually increae it fro zero until the tator current i about equal to it rated value. Soetie it i recoended to ue lower frequency than the rated to avoid the error due to kin effect in the rotor circuit. b. Take reading of V (line-to-line), LL BL I BL, and the total 3-phae power P BL (Fig.4.). So, for the blocked-rotor tet the lip i. In the equivalent circuit of Fig.4.9, the agnetizing reactance X i hunted by the low-ipedance branch R + jx. Becaue X >> R + jx, the ipedance X can be neglected and the equivalent circuit for the blocked-rotor tet reduce to the for hown in Fig.4.b. i: Fro the blocked-rotor tet, the blocked-rotor reitance i: R BL P (4.36) 3I BL BL The blocked-rotor ipedance at frequency of blocked rotor tet

27 4 Chapter Four X Z BL fbl V I BL (4.37) BL The blocked-rotor reactance at frequency of blocked rotor tet i: X BL fbl ( ) Z R (4.38) BL fbl BL It value at rated frequency i: X BL BL * fbl Rated Frequency Frequency at blocked rotor tet (4.39) X BL X + X (4.40) aue, X (at rated frequency) then X and X can be obtained. X Fro no load tet we know that known then the agnetizing reactance i : X X X + X X NL and X are X (4.4) NL Coent: The rotor equivalent reitance R play an iportant role in the perforance of the induction achine. So, an accurate deterination of R i recoended by the IEEE a follow: The blocked reitance R BL i the u of R and an equivalent reitance, ay R, which i the reitance of R + jx in parallel with X a hown in Fig.4.c; therefore,

28 Three-Phae Induction Machine 43 R R + X R (4.4) ( X + X ) If X + X >> R, a i uually the cae, X + X R X (4.43) R X X + X R R or (4.44) Now R R R. So, we can ue thi value of R to deterine BL R fro equation (4.43) More elaborate tet are conducted on large achine, but the above-entioned procedure give reult that are adequate in ot cae. Fig.4. A locked-rotor tet perit the calculation of the total R + R. Fro thee reult we can deterine the equivalent circuit of the induction otor. leakage reactance x and the total reitance ( )

29 44 Chapter Four Exaple 4. A no-load tet conducted on a 30 hp, 835 r/in, 440 V, 3-phae, 60 Hz quirrel-cage induction otor yielded the following reult: No-load voltage (line-to-line): 440 V No-load current: 4 A No-load power: 470 W Reitance eaured between two terinal: 0.5 Ω The locked-rotor tet, conducted at reduced voltage, gave the following reult: Locked-rotor voltage (line-to-line): 63 V Locked-rotor power: 700 W Locked-rotor current: 60 A Deterine the equivalent circuit of the otor. Solution: Auing the tator winding are connected in wye, the reitance per phae i: R 0.5/ 0.5 Ω Fro the no-load tet: The priary phae voltage can be obtained fro the following equation: VLL 440 V 54V / Phae 3 3

30 Three-Phae Induction Machine 45 Then, the no-load ipedance can be obtained a following: Z NL V Ω I 4 The no-load reitance i: R P 3 I 470 3*4.5 NL NL Ω The no-load reactance i: X NL Z NL RNL Ω In the IEEE recoended equivalent circuit we aue that X + X X NL 7.97Ω Fro the blocked-rotor tet, the blocked-rotor reitance i: P 3I 700 3* BL R BL Ω BL Note that in thi exaple it doe not entioned the frequency of the blocked rotor tet o we can ue the rated frequency a the frequency of the blocked rotor tet. The blocked-rotor ipedance at frequency of blocked rotor tet i: BL V I BL BL 63/ Z Ω The blocked-rotor reactance i:

31 46 Chapter Four ( Z R ) X Ω BL BL BL X BL X + X.4 Ω Aue, X (at rated frequency) then X X 0. 7 Ω X Fro no load tet we know that X + X X NL and X 0.7 Ω, then the agnetizing reactance i : X X NL X Ω Now R R R BL Ω. For equation (4.43): X + X R R * X 7.6 Ω Exaple 4.3 The following tet reult are obtained fro a threephae 60 hp, 00 V, ix-pole, 60 Hz quirrel-cage induction otor. () No-load tet: Supply frequency 60 Hz, Line voltage 00 V Line current 4.5 A, Input power 600 W () Blocked-rotor tet: Frequency 5 Hz, Line voltage 70 V Line current 5 A, Input power 9000 W

32 Three-Phae Induction Machine 47 (3) Average DC reitance per tator phae: R.8 Ω (a) Deterine the no-load rotational lo. (b) Deterine the paraeter of the IEEE-recoended equivalent circuit of Fig.4.9. (c) Deterine the paraeter ( V R, X ) equivalent circuit of Fig.4.0. th, for the Thevenin th th Solution: (a) Fro the no-load tet, the no-load power i: P NL 600 W The no-load rotational lo i: P rot P NL 3I R P rot 600 3* 4.5 * W (b) IEEE-recoended equivalent circuit. For the no-load condition, i very high. Therefore, in the equivalent circuit of R / Fig.4.9, the agnetizing reactance X i hunted by a very high reitive branch repreenting the rotor circuit. The reactance of thi parallel cobination i alot the ae a X. Therefore the total reactance eentially Fig.4.a. X NL, eaured at no load at the tator terinal, i X + X. The equivalent circuit at no load i hown in

33 48 Chapter Four (a) No-load equivalent circuit (b) Blocked-rotor equivalent circuit. (c) Blocked-rotor equivalent circuit for iproved value for R. 00 V 70. V / Phae 3 The no-load ipedance i: Z NL Fig.4.3 V Ω I 4.5 The no-load reitance i: R NL PNL * 4.5 I Ω The no-load reactance i:

34 Three-Phae Induction Machine 49 X NL Thu, NL NL Z R Ω X + X X 8 Ω 8.0 Ω. NL For the blocked-rotor tet the lip i, the agnetizing reactance X i hunted by the low-ipedance branch R + jx. Becaue X R j >> + X, the ipedance X can be neglected and the equivalent circuit for the blocked-rotor tet reduce to the for hown in Fig.4.b. Fro the blocked-rotor tet, the blocked-rotor reitance i: R BL PBL Ω 3 3* 5 I Therefore, ipedance at 5 Hz i: R Ω R BL, The blocked-rotor R Z BL V I 3 * 5 The blocked-rotor reactance at 5 Hz i X BL ( ) 3. Ω 98 It value at 60 Hz i X BL X + X Hence, X X BL X * Ω The agnetizing reactance i therefore: (at 60 Hz) Ω Ω

35 50 Chapter Four X Now Ω R RBL Ω. So, R R (c) Fro Equation (4.3) V th V V Fro Equation (4.34) R th Ω 0.97 R 0.97 * Fro Equation (4.35) X th Ω. X Ω 4.0 Perforance Characteritic The equivalent circuit derived in the preceding ection can be ued to predict the perforance characteritic of the induction achine. The iportant perforance characteritic in the teady tate are the efficiency, power factor, current, tarting torque, axiu (or pull-out) torque, and o forth. The echanical torque developed T ech per phae i given by P ech R Tech ech I ( ) ω (4.45)

36 Three-Phae Induction Machine 5 Where πn ω ech (4.46) 60 The echanical peed ω by: ech ( ) yn ech ωech i related to the ynchronou peed ω (4.47) n yn 60 π ( ) ω (4.48) and f 4 f ω 0 yn * π π 60P P (4.49) Fro Equation (4.45), (4.47), and (4.8) R T ech yn I P Then, ω (4.50) T ech P ω yn ag ag (4.5) T T ech ech R I (4.5) ωyn R I (4.53) ω yn Fro the equivalent circuit of Fig.4. and Equation (4.53) T ech Vth R * * (4.54) ω yn ( R + R / ) + ( X + X ) th th

37 5 Chapter Four Note that if the approxiate equivalent circuit (Fig.4.3) are ued to deterine I, in Equation (4.54), V, and X R, V, R, th th and hould be replaced by, repectively. The prediction of perforance baed on the approxiate equivalent circuit (Fig.4.3) ay differ by 5 percent fro thoe baed on the equivalent circuit of Fig.4.8 or 4.6. For a three-phae achine, Equation (4.54) hould be ultiplied by three to obtain the total torque developed by the achine. The torque-peed characteritic i hown in Fig.4.4. At low value of lip, R R R th + >> X th + X and >> R Thu T ech * yn th (4.55) Vth * (4.56) ω R The linear torque-peed relationhip i evident in Fig.4.4 near the ynchroun peed. Note that if the approxiate equivalent circuit (Fig.4.3) are ued, in Equation (4.56) V th hould be replaced by V. At larger value of lip, R R th << X th + X + (4.57) And T ech Vth R * * (4.58) ω yn ( X + X ) th X th

38 Three-Phae Induction Machine 53 The torque varie alot inverely with lip near, a een fro Fig.4.4. Fig.4.4 Torque-peed profile at different voltage. Equation (4.49) alo indicate that at a particular peed (i.e., a fixed value of ) the torque varie a the quare of the upply voltage V th (hence V Fig.4.4) how the T-n profile at variou upply voltage. An expreion for axiu torque can be obtained by etting dt / d 0. Differentiating Equation (4.54) with repect to lip and equating the reult to zero give the following condition for axiu torque: R S T ax R th + ( X + X ) th (4.59)

39 54 Chapter Four Thi expreion can alo be derived fro the fact that the condition for axiu torque correpond to the condition for axiu air gar power (Equation (4.5)). Thi occur, by the failiar ipedance-atching principle in circuit theory, when the ipedance of equal in agnitude the ipedance between R / it and the upply voltage V (Fig.4.9) a hown in Equation (4.59). The lip S T S T ax at axiu torque T ax i: ax R (4.60) R ( X X ) th + th + The axiu torque per phae fro Equation (4.49) and (4.60) i: T Vth ax * yn Rth + Rth + th ω (4.6) ( X + X ) Equation (4.6) how that the axiu torque developed by the induction achine i independent of the rotor circuit reitance. However, fro Equation (4.60) it i evident that the value of the rotor circuit reitance R deterine the peed at which thi axiu torque will occur. The torque peed characteritic for variou value of R are hown in Fig.4.5. In a wound rotor induction otor, external reitance i added to the rotor circuit to ake the axiu torque occur at tandtill o that high tarting torque can be obtained. A the otor peed up,

40 Three-Phae Induction Machine 55 the external reitance i gradually decreaed and finally taken out copletely. Soe induction otor are, in fact, deigned o that axiu torque i available at tart, that i, at zero peed. Fig.4.5 Torque peed characteritic for varying R. If the tator reitance R i all (hence all), fro Equation (4.60) and (4.6), T T ax R th i negligibly (4.6) X th R + X Vth ax * yn X th + X ω (4.63)

41 56 Chapter Four Equation (4.63) indicate that the axiu torque developed by an induction achine i inverely proportional to the u of the leakage reactance. Fro Equation (4.54), the ratio of the axiu developed torque to the torque developed at any peed i: T T ax ( R + R / ) + ( X + X ) th th * (4.64) If R (hence T T ax ( Rth + R / T ) + ( X th + X ) Tax ax R th ) i negligibly all, ( R / ) + ( X + X ) th * (4.65) ( R / T ) + ( X th + X ) Tax ax Fro Equation (4.6) and (4.65) T T ax ( R / ) + ( R / T ) ax * ( R / T ) Tax (4.66) ax T T ax T * T + ax (4.67) ax * Equation (4.67) how the relationhip between torque at any peed and the axiu torque in ter of their lip value.

42 Three-Phae Induction Machine Efficiency In order to deterine the efficiency of the induction achine a a power converter, the variou loe in the achine are firt identified. Thee loe are illutrated in the power flow diagra of Fig.4.6. For a 3 φ achine the power input to the tator i: P in 3V I θ (4.68) co The power lo in the tator winding i: P (4.69) 3I R Where R i the AC reitance (including kin effect) of each phae winding at the operating teperature and frequency. Power i alo lot a hyterei and eddy current lo in the agnetic aterial of the tator core. The reaining power, P ag, croe the air gap. Part of it i lot in the reitance of the rotor circuit. P (4.70) 3I R Where R i the ac reitance of the rotor winding. If it i a wound-rotor achine, R alo include any external reitance connected to the rotor circuit through lip ring. Power i alo lot in the rotor core. Becaue the core loe are dependent on the frequency f of the rotor, thee ay be negligible at noral operating peed, where f i very low.

43 58 Chapter Four Fig.4.6 Power flow in an induction otor. The reaining power i converted into echanical for. Part of thi i lot a windage and friction loe, which are dependent on peed. The ret i the echanical output power P out, which i the ueful power output fro the achine. out The efficiency of the induction otor i: η (4.7) The efficiency i highly dependent on lip. If all loe are P P neglected except thoe in the reitance of the rotor circuit, P ag P in (4.7) P P ag (4.73) P out ( ) P P (4.74) ecj ag P P η (4.75) out And the ideal efficiency i: ( ) ideal in in

44 Three-Phae Induction Machine 59 Soetie η ideal i alo called the internal efficiency a it repreent the ratio of the power output to the air gap power. It indicate that an induction achine ut operate near it ynchronou peed if high efficiency i deired. Thi i why the lip i very low for noral operation of the induction achine. If other loe are included, the actual efficiency i lower than the ideal efficiency of Equation (4.75). The full-load efficiency of a large induction otor ay be a high a 95 percent. 4. Power Flow In Three Mode Of Operation The induction achine can be operated in three ode: otoring, generating, and plugging. The power flow in the achine will depend on the ode of operation. However, the equation derived in Section 4. for variou power relationhip hold good for all ode of operation. If the appropriate ign of the lip i ued in thee expreion, the ign of the power will indicate the actual power flow. For exaple, in the generating ode, the lip i negative. Therefore, fro Equation (4.85) the air gap power P ag i negative (note that the copper lo P in the rotor circuit i alway poitive). Thi iplie that the actual power flow acro the air gap in the generating ode i fro rotor to tator.

45 60 Chapter Four The power flow diagra in the three ode of operation i hown in Fig.4.7. The core loe and the friction and windage loe are all luped together a a contant rotational lo. In the otoring ode, lip i poitive. The air gap power P ag equation (4.8) and the developed echanical power P ech Equation (4.7) are poitive, a hown in Fig.4.7a. In the generating ode i negative and therefore both P ag, and P ech are negative, a hown in Fig.4.7b. In ter of the equivalent circuit of Fig.4.7e the reitance [( ) / ] R i negative, which indicate that thi reitance repreent a ource of energy. In the plugging ode, i greater than one and therefore P ag i poitive but P ech i negative a hown in Fig.4.7c. In thi ode the rotor rotate oppoite to the rotating field and therefore echanical energy ut be put into the yte. Power therefore flow fro both ide, and a a reult the lo in the rotor circuit, P, i enorouly increaed. In ter of the equivalent circuit of Fig.4.7e, the reitance [( ) / ] R ource of energy. i negative and repreent a

46 Three-Phae Induction Machine 6 Fig.4.7 Power flow for variou ode of operation of an induction achine. (a) Motoring ode, 0 < <. (b) Generating ode, < 0. (c) Plugging ode, >.

47 6 Chapter Four Exaple 4.4 A three-phae, 460 V, 740 rp, 60 Hz, four-pole wound-rotor induction otor ha the following paraeter per phae: R 0.5 Ω, R 0. Ω, X 0. 5 X Ω, X 30 Ω The rotational loe are 700 watt. With the rotor terinal hort-circuited, find (a) (i) Starting current when tarted direct on full voltage. (ii) Starting torque. (b) (i) Full-load lip. (ii) Full-load current. (iii) Ratio of tarting current to full-load current. (iv) Full-load power factor. (v) Full-load torque. (iv) Internal efficiency and otor efficiency at full load. (c) (i) Slip at which axiu torque i developed. (ii) Maxiu torque developed. (d) How uch external reitance per phae hould be connected in the rotor circuit o that axiu torque occur at tart? Solution:

48 Three-Phae Induction Machine 63

49 64 Chapter Four 63. N.

50 Three-Phae Induction Machine 65 η otor η 80.3 * % 30.4 ( )*00 ( ) * % int ernal (c) (i) Fro Equation (4.60) (i) Fro Equation (4.6) Note that for part (a) and (b) it i not neceary to ue Thevenin equivalent circuit. Calculation can be baed on the equivalent circuit of Fig.4.3 a follow:

51 66 Chapter Four Exaple 4.5 A three-phae, 460 V, 60 Hz, ix-pole wound-rotor induction otor drive a contant load of 00 N - at a peed of 40 rp when the rotor terinal are hort-circuited. It i required to reduce the peed of the otor to 000 rp by inerting reitance in the rotor circuit. Deterine the value of the reitance if the rotor winding reitance per phae i 0. oh. Neglect rotational loe. The tator-to-rotor turn ratio i unity. Solution:

52 Three-Phae Induction Machine 67 Fro the equivalent circuit, it i obviou that if the value of R / reain the ae, the rotor current I and the tator current I will reain the ae, and the achine will develop the ae torque (Equation (4.54)). Alo, if the rotational loe are neglected, the developed torque i the ae a the load torque. Therefore, for unity turn ratio, Exaple 4.6 The following tet reult are obtained fro three phae 00hp,460 V, eight pole tar connected induction achine No-load tet : 460 V, 60 Hz, 40 A, 4. kw. Blocked rotor tet i 00V, 60Hz, 40A 8kW. Average DC reitor between two tator terinal i 0.5 Ω (a) Deterine the paraeter of the equivalent circuit. (b) The otor i connected to 3ϕ, 460 V, 60 Hz upply and run at 873 rp. Deterine the input current, input power, air gap power, rotor cupper lo, echanical power developed, output power and efficiency of the otor.

53 68 Chapter Four (c) Deterine the peed of the rotor field relative to tator tructure and tator rotating field Solution: Fro no load tet: 460 / 40 (a) Z Ω R NL PNL 3* I * NL Ω\ X NL Ω Then, X X 58Ω + 6. Fro blocked rotor tet: RBL * Ω 0.5 R Ω (fro reitance between two tator terinal). Z BL 00 / Ω X BL X X 0.36 Then, Ω Ω

54 Three-Phae Induction Machine Then, X Ω X X Ω R RBL R Ω Then, R * Ω Ω j0.95 Ω j0.95 Ω j6.386 Ω (b) 0 f 0*60 n 900rp P 8 n n n R Input ipedance Z Ω ( j6.386)(.3 + j0.95).3 + j( ) Z j V 460 / 3 I Z. 7.6 o o

55 70 Chapter Four Input power: o ( 7.6 ) 88. kw 460 Pin 3 * *5. co Stator CU loe: P t 3*5. * kw Air gap power P ag kw Rotor CU loe P Pag 0.03* kw Mechanical power developed: P P ( ) P ( 0.03) * kw ech ag 636 out P ech P rot Fro no load tet: Prot PNL 3I * R 400 3*40 * W P out 8.636* kw Then the efficiency of the otor i: η P P out in 78.8 * 00 * %

56 Three-Phae Induction Machine 7 Exaple 4.7 A three phae, 460 V 450 rp, 50 Hz, four pole wound rotor induction otor ha the following paraeter per phae ( R 0.Ω, R 0.8 Ω, X X 0.Ω, X 40Ω). The rotational loe are 500 W. Find, (a) Starting current when tarted direct on full load voltage. Alo find tarting torque. (b) (b) Slip, current, power factor, load torque and efficiency at full load condition. (c) Maxiu torque and lip at which axiu torque will be developed. (d) How uch external reitance per phae hould be connected in the rotor circuit o that axiu torque occur at tart? Solution: 460 (a) V 65.6 V / phae 3 ( ) o j40* Z 0. + j Ω j40. Then, V 65.6 o I t Ω I 48.9 o (b)

57 7 Chapter Four Z R Ω ( j0.) o j40* Z 0. + j Ω j o Then I A FL o Then the power factor i: co 0.83 o lag. ω 500 * π 60 y 65.6* ( j40) ( 0. + j40.) rad / ec. Vth Then, ( 0. + j0.) j40* o j0. Ω 0. + j40. th + Then, 3* ( 64.75) *5.4 ( ) + ( ) T * Then, Then, Pag T * ω y 8.68* W P Pag * W And, P ( ) P W ag 7 o V N Then, Pout P Prot W

58 Three-Phae Induction Machine 73 P in 3 *65.6*53.56* W Pout Then, η 79.6 % P 494 Then, in ( 64.75) ( ) 3* T / *88.5[ ( )] 0.8 T ax / (d) [ ( ) ] Tax R + R ext [ ( ) ] / Then, R + Rext Then, R Ω ext N Exaple 4.8 The rotor current at tart of a three-phae, 460 volt, 70 rp, 60 Hz, four pole, quirrel-cage induction otor i ix tie the rotor current at full load. (a) Deterine the tarting torque a percent of full load torque. (b) Deterine the lip and peed at which the otor develop axiu torque. (c) Deterine the axiu torque developed by the otor a percent of full load torque.

59 74 Chapter Four Solution: Note that the equivalent circuit paraeter are not given. Therefore equivalent circuit paraeter cannot be ued directly for coputation. (a) The ynchronou peed i Fro Equation (4.57) T I R I R α, Thu, ω yn Fro Equation(4.7)

60 Three-Phae Induction Machine 75 Fro Equation (4.7) Exaple 4.9 A 4 pole 50 Hz 0 hp otor ha, at rated voltage and frequency a tarting torque of 50% and a axiu torque of 00 % of full load torque. Deterine (i) full load peed (ii) peed at axiu torque.

61 76 Chapter Four Solution: T T t FL.5 T.5 and ax then, T T FL T t T T ax T + t Fro above and equation (4.7) Tax ax Then, Then. T ax 55 Or 0. T ax 4546 Tax Alo fro Equation 4.7 T T ax T + ax FL ax * FL T FL But 0. T ax 4546 (unacceptable) T Then ax + FL T *0.4546* FL FL ax Tax FL FL 4* FL FL (unacceptable) FL or FL 0*50 n 500 rp 4 0

62 Three-Phae Induction Machine 77 n * n then (a) FL ( FL ) n FL ( ) * rp (b) ( )* n ( ) *500 rp nt T 83 ax ax Exaple 4.0 A 3φ, 80 V, 60 Hz, 0 hp, four-pole induction otor ha the following equivalent circuit paraeter. R 0.Ω, R 0. Ω, X X 0. 5 Ω, and X 0 Ω The rotational lo i 400 W. For 5% lip, deterine (a) The otor peed in rp and radian per ec. (b) The otor current. (c) The tator cu-lo. (d) The air gap power. (e) The rotor cu-lo. (f) The haft power. (g) The developed torque and the haft torque. (h) The efficiency. Solution: 0*60 n 800 rp ω * π 88.5 rad /ec 60

63 78 Chapter Four 0. Ω j0.5 Ω j0.5 Ω j0 Ω Z 0. + j R e + X e ( + j0.5) o j0* Z 0. + j Ω + j V 0. V 3 0. o I A o (c) P 3* * W (d) o ( 3.55 ) 860. W P 3 *0.* * co 9794 P P ag P W (e) P Pag 0.05* W (f) P ( ) P W ag 55 Pag (g) T N.

64 Three-Phae Induction Machine 79 T P haft haft Phaft (h) η * % P N For operation at low lip, the otor torque can be conidered proportional to lip. Exaple 4. A 30, 00 WA, 460 V, 60 Hz, eight-pole induction achine ha the following equivalent circuit paraeter: R 0.07 Ω, R Ω, X X 0. Ω, and X 6. 5Ω (a) Derive the Thevenin equivalent circuit for the induction achine. (b) If the achine i connected to a 30, 460 V, 60 Hz upply, deterine the tarting torque, the axiu torque the achine can develop, and the peed at which the axiu torque i developed. (c) If the axiu torque i to occur at tart, deterine the external reitance required in each rotor phae. Aue a turn ratio (tator to rotor) of..

65 80 Chapter Four Solution: X 6.5 Vth * V * V X + X R ( j6.5) *( j ) + jx th j0.947 Ω j0. + j6.5 th j0.947 j0. Ω Ω Ω 57.7V 0.05 (b) 3* 57.7 *0.05 T t 64. 7N 94.5 T ax T ax [( ) + ( ) ] [ ( ) ] * * N ( ) 0.49 Speed in rp for which ax torque occur ( )* n ( 0.49) * rp (c) T 5 ax R T α R ax R + ( X + X )

66 Three-Phae Induction Machine 8 tart or R * R * Ω tart 0.49 T ax Then R ( ) / ext Ω -peed characteritic of an induction achine We have een that a 3-phae quirrel-cage induction otor can alo function a a generator or a a brake. Thee three ode of operation-otor, generator, and brake-erge into each other, a can be een fro the torque-peed curve of Fig.4.8. Thi curve, together with the adjoining power flow diagra, illutrate the overall propertie of a 3-phae quirrel-cage induction achine. We ee, for exaple, that when the haft turn in the ae direction a the revolving field, the induction achine operate in

67 8 Chapter Four either the otor or the generator ode. But to operate in the generator ode, the haft ut turn fater than ynchronou peed. Siilarly, to operate a a otor, the haft ut turn at le than ynchronou peed. Finally, in order to operate a a brake, the haft ut turn in the oppoite direction to the revolving Fig.4.8 Coplete torque peed curve of a 3-phae induction achine