ELEC 344 7 th Tutorial Additional Slids Midtrm Rsult & Induction Machin Novmbr 18, 2016 Wonba Choi Th Univrsity of British Columbia
1. Structur of Induction Machin
2. Working principl of induction motors Eind = (v X B) l
2. Working principl of induction motors Currnt is inducd in th rotor s conducting bars, and associatd magntic filds intract with thos of th stator. This causs th rotor to follow th fild gnratd by th stator, to rotat th output shaft.
2. Working principl of induction motors 1. Thr-phas powr supply to Stator windings 2. MMF vctor Fnt producd (BS) Constant magnitud Rotats in spac 3. Voltag inducd in Rotor bars (just lik a transformr) Eind = (v X B) l Du to th spd diffrnc btwn Fnt and rotor 4. Currnt flows in Rotor bars 5. Rotor currnt flow producs a rotor magntic fild BR 6. Torqu is inducd such that th spd diffrnc btwn Fnt (BS) and th rotor F = i (l X B) - inducd forc on rotor bar Tind = (r X F) = kbr X BS 7. Th rsulting torqu is countrclockwis 8. Th rotor acclrats in that dirction. Not: - To produc torqu in an induction motor, currnt must flow in th rotor - To induc currnt flow in th rotor, th rotor spd must b slightly slowr than th synchronous spd - Th diffrnc btwn th synchronous spd and th rotor spd (ratd spd) is calld th slip spd. (R.P.M) (R.P.M)
3. Torqu Production
3. Torqu Production http://nginringtutorial.com/squirrl-cag-induction-motor-animation/ α=0 α=90 Radially outward is positiv; radially inward is ngativ. α=270 α=180
3. Torqu Production Not: (a) Th rotating stator fild BS inducs a voltag in th rotor bars (b) Th rotor voltag producs a rotor currnt flow, which lags bhind th voltag bcaus of th inductanc of th rotor (c) Th rotor currnt producs a rotor magntic fild BR lagging 90 dgr bhind itslf, and BR intracts with BS to produc a coutrclockwis torqu in th machin B total B s B r T KB dv rb total sin
3. Torqu Production S S S N N Rotor Stator N S Rotor Stator N Fig a: gnrator opration Fig b: Motor opration In ordr to produc constant torqu, two rotating magntic filds must hav th sam rotational vlocity. (a) Rotor bar (rotatd by mch. powr) pushing Stator bar (Plc Gnration) (b) Stator bar (rotatd by lc. powr) pulling Rotor bar (Pmch Gnration) Q. What s th spd of th rotor s rotating magntic fild in cas of constant torqu produc?
4. Spd & Frquncy & Slip Fact : Th rotor must hav spd (ωr) which diffrs from spd of rotating magntic fild (ω) from stator. Othrwis, no voltag is inducd in rotor windings. _ r rmf Spd of rotating magntic fild (from stator) 2 sync p s _ rmf = r Rotor Spd (physical movmnt) 2 rm r p + s slip Spd of rotating magntic fild rlativ to rotor (from stator) s r r s s r s slip s Frquncy (rad/sc) of rotor currnts. f rotor s f stator Rotor spd Spd of RMF from stator, rlativ to rotor Spd of RMF from stator
4. Spd & Frquncy & Slip (Summary) slip _ spd n syn n n syn syn syn rm r 2f [ rad /sc]; f 60Hz 60 2 120 nsync f 2 P P Calculat th mchanical spd first and thn convrt it to angular frquncy!!! Plas kp in mind that rotor s physical angular spd can t b compard to th rotor currnt frquncy. Rotor currnt frquncy solly dpnds on rotor s rlativ spd to th stator rotating magntic fild which is th sam as slip spd. And finally: s r s f s slip f rotor r s s f
4. Spd & Frquncy & Slip (Clarification) Slip spd(ωs): ωs occurs du to th motion (spd) of th magntic fild and th motion (spd) of th rotor. s r From anothr viw, it is th spd of th rotating magntic fild from th stator rfrncd to th rotor ωslip is vntually th frquncy (rad/sc) of th lctrical quantitis in th rotor winding. f slip f rotor s f
5. Equivalnt Circuit
5. Equivalnt Circuit Th main diffrnc btwn th transformr quivalnt circuit and th induction machin quivalnt circuit is th loading: th transformr load is an actual impdanc whras th induction machin load is a variabl rsistanc that dpnds on s. What dos th induction machin load (Varying Rsistanc) rprsnt? It rprsnts th mchanical powr providd to th shaft. Thus th induction machin load is a purly ral lctrical load sinc nrgy transfr by mchanical mans must b Watt[W] only (this is not th cas for th transformr)
5. Equivalnt Circuit
6. Thvnin Equivalnt Circuit I 2 An altrnativ way to obtain I 2 is by us of Thvnin, looking into th trminals as shown. Zth I 2 R 2 jx 2 Vth R 2 (1-s)/s
6. Thvnin Equivalnt Circuit V1 Za=R1+jX1 Zb= Rc//jXm Vth By voltag division: V th V 1 Z a Zb Z Not that V 1 is th lin-to-nutral voltag, givn by V 1 =V LL /sqrt(3). b Za=R1+jX1 Zb= Rc//jXm Th two impdancs ar in paralll: ZaZb Zth Za // Zb Z Z a b