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# Electric Machines ECE 470

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This 17 page Class Notes was uploaded by Judd Okuneva on Saturday October 3, 2015. The Class Notes belongs to ECE 470 at Boise State University taught by Said Ahmed-Zaid in Fall. Since its upload, it has received 12 views. For similar materials see /class/217975/ece-470-boise-state-university in Engineering Electrical & Compu at Boise State University.

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Date Created: 10/03/15

Boise State University Department of Electrical and Computer Engineering ECE47O Electric Machines Derivation of the PerPhase SteadyState Equivalent Circuit of a ThreePhase Induction Machine Nomenclature Rotor shaft angle measured from the stator phase a axis to the rotor phasea axis Spatial angle along the stator periphery measured from the axis of the stator phase a winding Spatial angle along the rotor periphery measured from the axis of the rotor phasea winding Frequency of the stator currents and voltages f9 60 Hz Synchronous electrical speed ws 2717 s Synchronous speed in rpm 715 ws x 6027139 3600 rpm Rotor shaft speed in rads Mechanical speed in rpm slip speed s ws 7 wwS n5 7 Frequency of the rotor currents fr sfs Angular speed of rotor currents w sws Radius of rotor cylinder Length of rotor cylinder Permeability of free space 10 47139 x 10 7 Stator rotor leakage inductance Stator rotor magnetizing inductance Maximum stator stator rotor rotor mutual inductance Maximum stator rotor mutual inductance Per phase stator rotor resistance Effective reluctance permeance of air gap 1 FluxCurrent Relationships sbiaxis raiaxis saiaxis rciaxis Figure 1 Schematic Representation of a Two Pole7 Three Phase Induction Machine Consider a threephase7 two pole7 wound rotor induction machine with a smooth air gap as shown in Figure 1 This machine has three stator windings labeled sa sa 7 sb sb 7 sc sc and three rotor windings labeled ra ra 7 rb rb 7 rc rc All windings are assumed to be sinusoidally distributed along the stator and peripheries so that the magnetic eld intensities created by the corresponding stator and rotor currents in the air gap are given by Hm Ngiigm cosa 1 H95 lt2gt H50 N57290COS042 3 Hm Ngigmgcos Ngiy coswie 4 The ux linkages with coils sa sa 7 slo slo 7 sc sc 7 ra ra 7 rlo rlo 7 and rc rc7 are given by 7r2 Am 7 N5 MHW H517 H Hm H717 Hmw da 15 7 77r 7r6 A917 N5 7 6 Mina H517 H Hm H717 Hmw da 19 8 7r 57r6 Am 7 N5 11 m MHW H517 H Hm H717 Hmw da 15 9 7r 7r2 ra N39r 2 HoltH5a 1 H917 1 H50 1 HM 1 H71 1 Hrcrl 1744 77r 7r6 m 7 NT 7 m MHW H517 H Hm H717 Hmw d3 my 11 7r 57r6 Tc NT 11 6 HoltH5a 1 H917 1 H90 1 HM 1 H71 1 H70 1 Tire 7r where 19 and l are stator and rotor leakage inductances7 respectively lntegrating these equations results in linear ux current relationships in the form Am 15 Lms MSS cos 1200 M95 cos 2400 im A917 MSS cos 2400 15 Lms MSS cos 1200 is 13 A MSS cos 1200 M99 cos 2400 19 Lms 2 My cos0 My cos0 1200 MST cos0 7 1200 im 1 My cos0 71200 My cos0 MST cos0 1200 m 14 My cos0 1200 My cos0 71200 My cos0 in Am My cos0 My cos0 7 1200 MST cos0 1200 im A717 My cos0 1200 My cos0 MST cos0 7 1200 is 15 AM My cos0 71200 My cos0 1200 My cos0 2 50 T Lmr MM cos 1200 MM cos 2400 im 1 MM cos 2400 l Lmr MM cos 1200 m 16 MM cos 1200 MM cos 2400 l Lmr in where Lms 7 M55 7 Lm39r 7 MT 7 MST 7 4 07 1 7 1 7 N2 7 N2 7 N2 7 N2 7 N N 7 7 R7 7 779 17 s s 39r 39r s r 7quot9 9 and where R9 and 739 are7 respectively7 the effective reluctance and permeance of the air gap Exercise 1 Verify the mutual inductances by inspection of the magnetic axes The magnetic coenergy of the coupling eld is computed as Wk Wkasm Z49177 i907 i39rm Z47177 i707 1 1 1 1 1 1 Eksaisa 1 Eksbisb 1 Ekscisc 1 Ekraira 1 5AMin 1 Ekrcirc 09 Lmsi a 2317 ii 07 Dwain 2317 ii My cos 06mm ism My cosw 120 239mm M97quot C0809 7 1200isairb 1 isbirc l isci39ra 71 M99 isaisb l Z49171490 l Zscisa MAW The developed electromagnetic torque is QW Te Telt sa7 Z9177 907 z39rm Z7177 707 Te 7M9 Sin 0isaira l i917in l Zsci39rc 7 MST Sin0 l 1200isairb l Zshim l Zscim 7M9 Sin0 7 1200isairc l Zshim l Zsci39rb 2 Model of a ThreePhase WoundRotor Induction Motor We will assume motor operation in the following analysis that is the induction machine converts electrical energy to mechanical energy Generator operation particularly in wind turbine systems is possible and involves the conversion of mechanical energy into electrical energy The mathemat ical model for this machine is composed of six differential equations given by Kirchhoff s voltage law for the six stator and rotor windings and two differential equations for the rotor shaft given by Newton s second law for rotating bodies d Usa Rsisa dia 19 d Usa Rszsb l 117 d Usa Rszsc l 10 d Um Remtira 7 Rrira l dga d U39rb Remtzrb errb l dzb d U70 Remtirc Brim l dzc 10 i 25 d w dw JE T5 7 Tm 26 In addition the following constitutive equations are part of the mathematical model Am 15 Lms MSS cos 1200 M95 cos 2400 im A917 MSS cos 2400 19 Lms MSS cos 1200 is 27 A MSS cos 1200 M95 cos 2400 15 Lms 2 50 My cos0 My cos0 1200 MST cos0 7 1200 im My cos0 71200 My cos0 MST cos0 1200 m 28 My cos0 1200 My cos0 71200 My cos0 in Am My cos0 My cos0 7 1200 M9 cos0 1200 im A My cos0 1200 My cos0 MST cos0 7 1200 is 29 AM My cos0 71200 My cos0 1200 My cos0 2 T Lmr MM cos 1200 MM cos 2400 im MM cos 2400 l Lmr MM cos 1200 m 30 MM cos 1200 MM cos 2400 T Lmr in Te 7M9 Sin 0isaira l i517in l 2wire 7 MST Sin0 l 1200isairb l isbirc l Zscim 7M9 Sin0 7 1200isairc l Zsl l39ra l Zsci39rb In the above model the rotor windings are normally short circuited through balanced external resistors These resistors as will be seen later will allow the shaping of the torquespeed char acteristic for different applications The inputs in the above model are the three stator voltages 125105 Usbt 1290t and the mechanical load torque Tmt Assume balanced input voltages and a constant load torque usat xEVscosw5t0v 32 was Vscoswst0v71200 33 12500 xEVscoswStt9v1200 34 Tmt Tm 35 To nd the steady state currents i5at isbt isct imt irbt imt and the steady state shaft speed wm we could simulate the above model starting from arbitrary initial conditions until steady state is reached Instead we will guess guess at the general form of these solutions and try to obtain analytical solutions If these general equations satisfy the differential equations and the constitutive equations then by unicity of these solutions they will be the same as the simulated responses in steady state Therefore let us assume that the general solutions have the following form 2505 zscosmt 05 36 isbt lscoswst 09 7 1200 37 Mt zscosmt 05i1200 38 mos coswt on 39 mos cosmt 0 7 1200 40 005 coswrt 07139 1200 41 wt w 42 00 wt 0 43 where w sws f sfs and s is the slip speed s wsiw w 178 44 Ms Thus there are six unknown variables to be determined 9 05 IT 07 w or s and 00 In other words we need to come up with six independent equations that will allow us to solve for these six unknowns provided the eight differential equations are all satis ed in steady state Exercise 2 Verify that the mechanical equations can be be satis ed in steady state by a constant electromagnetic torque equal to the mechanical load torque that is dw JE 0 TeiTm gt T5 Tm constant 45 Solution T5 Msr Sin 0isaim 12in iscim Mgr 811109 1200isairb isbim iscim Msr 811109 1200isaim isbim iscirb 7Msrisa m sint9 m sin0 1200 in sin0 7 1200 7 5739mm sin0 7 1200 m sin0 in sin0 1200 7 573MB sint9 1200 m sin0 7 1200 in sin 0 EMMA sin05 7 00 7 0M constant Tm 46 Now consider the stator equation for phase a 4 d 125105 Rszsaa W 47 d39ra 0 R7 Emma 7 48gt dt Since v5at is a sinusoidal voltage at 1 5 60 Hz and 39105 is also assumed sinusoidal at 60 Hz we need to verify that the stator ux linkages Asat are also sinusoidal at 60 Hz Exercise 3 Verify that the stator ux linkages Asat are sinusoidal at 60 Hz in steady state that is Amos Ascoswst s 49 Solution Since isa 23991 23990 0 and M m 23970 0 in steady state Asa ls Lms sa Mss sb i Ms im cos0 m cos0 1200im cos0 71200 ls Lms 7 Mss sa My m cos0 m cos0 1200imcos0 71200 3 13 Lm 7 MSSWIS cosmt 051 EMSMEIT cosKw mt 00 an 3 3 ls in ux519 cosw5t 05 EMSMEI cosw5t 0 00 As cosw9t 19 50 where we used the fact that w w 17 sw sws ws and M95 7Lm52 The corresponding phasor for Asa is 3 3 As 19 ELm s z 05 EMMA z 0 00 51 Similarly consider the rotor equation for phase a d 0 R7 Emma A 52 dt Since imt is assumed sinusoidal at slip frequency f sfs we need to verify that the ux linkages 39rat are also sinusoidal at slip frequency Exercise 4 Verify that the rotor ux linkages mt are sinusoidal at slip frequency fr sfs in steady state7 that is mt ll cosmt 3 53 Solution Since 353 23991 23990 0 and 3m m 3M 0 in steady state Am IT Lmr m MMGTI in Ms isa cos0 3517 cos0 71200isc cos0 1200 l Lmr 7 Mm m M543 cos0 3517 cos0 71200isccos0 1200 3 17 Lmr 7 MMNM coswrt 071 EMSMEIS coslm 7 wt 0517 00 3 3 17 LWNEL coswrt 071 ngr Iscosmt 0517 00 AT coswrt 1 54 where we used the fact that ws 7 w w and MM 7LmT2 The corresponding phasor for T is A 3 3 Ar Zr l Lmr139r Z 0M l EMSTIS Z 7 t90 Note that A is a phasor corresponding to a sinusoidal waveform with frequency fr sfs not f9 The stator and rotor equations d Usa Rsisa dtm 56 d 0 RT l Remt 39r a l dga 57 yield V9 1 0 R919 1 091 jugL 58 0 RT 1 an MA 59 Using the expressions for is and in these two equations can be manipulated to yield 3 4 3 V5 1 0 7 R515 1 05 mas ELm s z 05 m Msrlr z 0 00 60 3 3 0 R717 1 0M l JWT MSTIS Z 7 00 Jw39ra39r l Ms39r139r Z 0M 2 2 Multiplying this last equation by 5790 and using w sws yields R 13 3 3 0 family 1 0 00 st Mers z 09 mltt 5M 1 0 00 62 De ning SO Hz phasors 79 V9 1 0y I Is 1 051 and I T Z 0 00 two phasor equations can be obtained as 3 3 vs RslsjwsasELms sjws Mer 63 3 3 3 3 0 Tfmb wsgmmywsa LWIT 64 3 Equivalent Circuit of a ThreePhase Induction Motor Equivalent circuit representations for the induction machine have been developed that make the previous equations easy to remember and simple to solve A word of caution is in order here The previous phasor equations were derived using the stator and rotor circuits It should be kept in mind that these circuits have different sinusoidal frequencies The actual rotor phasor IT corresponds to a sinusoidal waveform with frequency f sfs This phasor is de ned as I I 1 0M 65 The newly de ned rotor phasor I is a ctitious SO Hz phasor referred to the stator and de ned as I I z 0 00 66 Thus7 the two phasors I and I have the same rms magnitude7 but their phases are different Recall that Lms 7 Lm39r 7 MST N3 N3 NSNT 67 Referring all rotor variables to the stator by means of an ideal transformer with turns ratio N9 7 68 a N and de ning NS 2 RTRm F RmRm 69 39r 3 N 2 3 N 2 3 mng zTWLW my lt70 7 7 N I T 71 N FEMS Lms 72 T the previous equations on page 7 become 3 3 Vs Rwywsaw 5 Isyws Lm 73 3 R R 3 0 MELmI fm wsaHng I 74 These two equations can be combined to give the following circuit Figure 2 Equivalent Circuit Representation of a Three Phase Induction Machine a Using an Ideal Two Winding Transformer Between the Stator and Rotor Sides b After Referring all Rotor Quantities to the Stator Side Performance Analysis of an Induction Machine I Rs jxls jxlr I I s jxls jxlr I Figure 3 a Exact Equivalent Circuit 10 Approximate Equivalent Circuit Note X 7 m9 wsLms for atwo phase induction machine M5 7 32Xm5 32w9Lm5 for a three phase induction machine ws 27139 S 27r60 elec rads 2 2 wms lt7 ws lt7 27r60 mech rads P P mech rad 27r mech rad 1mn 2 60 120 S n5 m x x 7 x27rfsxi irpm 1 s 1 rev 60 s 27139 p 2 n9 lt7 3600 rpm P wms 7 wm n5 7 n 5 Wms wm Actual rotor shaft speed in mech rads n Actual rotor shaft speed in mech rpm Problem For a given slip s7 compute the efficiency 7 of a threephase induction machine using i The exact equivalent circuit ii The approximate equivalent circuit Using the exact equivalent circuit 13 Zn RSHXISHQXMQHfHXt 2mm Rmij V5 vs I 7 Z 7 S in S I 7 jXMs f T RsjXXMs 5 5mm 3m BlvsHIslcowsj3wlslsin gts Pmsphi 3ll7sllfslcos s Qin phi 3lVSHISlsin gts PsJoss BRS fS Z I Pg PrlossPe S 175 W Hwy Pnloss 5P9 P5 In steady state7 dw J7 Te 7 Tm Te 7 Tmput 1 Trot 0 gt Te Tm Tmput 1 Tm39rot mee me memput memt gt P5 Pm Pmput Prat Pm out Pm out Pm out 7 X 100 X 100 X 100 n Pm phi Pmout 1 Plosses Pm ut 1 PsJoss 1 PrJoss 1 Prat m Prat Pwindage 1 Pf39riction 1 Poore Using the approximate equivalent circuit V 1 7 1 I 7 4 s gt 1 1 s 7 BL8 JXzs X12 T Jags X19 X19 PsJoss R Pg 3am PTw55Pe 175 P5 W lt17sgtPg Pnloss 5P9 Pm out Pm out Pm Out 7 X 100 X 100 X 100 n Pin phi Pm ut 1 Bosses Pm ut 1 PsJoss 1 Pnloss 1 Prat m3 Prat Pwindage 1 Pf39riction 1 Poore Revolving Stator and Rotor Magnetic Fields in the Air Gap of a TwoPhase Induction Machine isa x2hcoswst0si 2 91 2ssinwst0si Hm cos Ea E x s cosw9t 09 cos Ba 29 7139 2 29 7139 2 H517 Mg sin 17904 E ls sinw5t 05 sin 17904 29 7139 2 29 7139 2 HS Hm H517 2IS cos 15904 coswst 05 sin 04 sinwst 0 g 7139 HS E ls cos 04 7 919257 05 29 7139 2 At a given time the resultant stator magnetic eld is maximum at an angle 04 on the stator periphery such that 2 da 2 204791575709 0 gt 04 7wst0si gt i ws wms 2 p dt p Thus the stator magnetic eld revolves at mechanical synchronous speed wms or 719 with respect to the stator and at slip mechanical speed 591mS or 5715 with respect to the rotor im V2 coswrt0M m xEIsinwt0 Hm M cos EB x2ITcosth97cos EB 29 7139 2 297139 2 HT M sin 133 amp 2Tsinwrt0nsin 133 29 7139 2 297139 2 N 4 i i H Hm HT 27x21 cos coswt 07 sin s1nwt 00 N 4 H J EI cos 135 7 W 7 on 29 7139 2 At a given time the resultant rotor magnetic eld is maximum at an angle 3 on the rotor periphery such that 2 d 2 2 EB 7wrt 7 07139 0 gt B 7wrt 0 gt 73 7w 75915 swms 2 p dt p p Thus the rotor magnetic eld revolves at slip mechanical speed swms or 5715 with respect to the rotor and at mechanical synchronous speed wms or 719 with respect to the stator TorqueSlip Characteristic of a ThreePhase Induction Machine ZthRththh I ZthRththh I 7 7 N R N R r r Vth i Vth i S S a b Figure 4 Thevenin Equivalents of the Induction Machine Equivalent Viewed by the Rotor Resistance a Equivalent Thevenin Representation for Finding Maximum Electromagnetic Torque b Equivalent Thevenin Representation for Finding Maximum Electromagnetic Power Problem Statement Assume that the stator supply voltage magnitude is xed Express the electromagnetic torque T5 as a function of the variable slip 5 Solution Zth XlrRSleSHjXMs Rththh jXMs V V V 0 th 5 th th th 7 R5 X19 XMS 0th 900 7 tan 1 X19 XMS 5 7 7 I 7 w h RTs 3X Bus X5 T i SW i 3lf4lzRiS M 5 W 7 178me wm wm swmRs2xt2h Slip at Maximum Electromagnetic Torque Torque Te n Temax quot 1 quot 1 quot 1 1 Te2 1 1 Tel 1 1 1 T 1 1 e1 start 1 1 Tm 1 1 1 T e2 start 1 1212 gter 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 O 1 1 1 l s2max slmax s2 sl 0 Slip 5 Figure 5 Plot of the Torque Slip Characteristic of an Induction Machine for Two Different Values of Rotor Resistance Note that since T P5 1 7 sPg Pg 8 7 Wm 7 7 5Wms 7 Wms maximizing the electromagnetic power Tam is equivalent to maximizing the air gap power Pg not the electromagnetic power Pe From the maximum power transfer theorem the slip 5mm at which maximum electromagnetic torque Tme or maximum air gap power P mm will occur is such that R T NRhX12h 5min Special Case Using the approximate equivalent circuit 1171911 Vs Rth Rs Xth X19 Xir R R T R3 1X19 X1112 57W R X19 X102 T 31171911213 7 31 R X15 X102 8mm SmawwmslRrsmaz2 1 thhl WmsKR 1 2Xls 1 Xir2l Therefore the maximum electromagnetic torque pull out torque Te mm is independent of the rotor resistance 13 If the rotor terminals are available as in a three phase wound rotor induction machine as opposed to a squirrel cage induction motor then the torqueslip characteristic can be altered by changing the slip at which maximum electromagnetic torque occurs 14 Slip at Maximum Electromagnetic Power From the maximum power transfer theorem7 the slip s at which maximum electromagnetic max power Pew will occur is such that 17 lt Sm R Rm 32 Xch Small Special Case Using the approximate equivalent circuit7 1 s 71W R9 W X15 X12 mam R Small 7 2 R Rs W X15 X17 Wthl W BiSinamy 17 P5mm 3lt ssmamgt RHI E mam Appendix Maximum Power Transfer in DC and AC Circuits Problem 1 Find the maximum power PL lem that can be delivered to the variable load resistance RL by the DC source voltage VT through the known resistance RT Solution VT 2 BLVZ P R 12 R T L L L RTRL RTRL2 dP R R 272R R R R2713 sz L 12ltT L L4R L T L4T 0 BL RT dRL RT BL RT BL RVZ V2 P T T LWL MT 4RT Problem 2 G Find the maximum power PLWW that can be delivered to the variable load impedance ZL RL jXL by the AC source voltage VT through the known impedance ZT RT jXT Solution 2 VT 3va P R 12 R T L L L L lt RT BL XT XL RT BL XT XL dPL 2 72XT XL 7 R V 0 gt X 7 X dXL L T RT BL XT XL22 L T 16 dPL 2 RT RL2 t XT XL2l 2RLRT BL V 0 j R R dRL 1 1 BL XT sz L T gt ZL RLjXL RTinT Z P 7 Rng 7 V7 Lmam 2RT2 7 4RT Problem 3 Find the maximum power PLWW that can be delivered to the variable load resistance ZL RL by the AC source voltage VT through the known impedance ZT RT jXT Solution 2 2 VT BLVT PL Ra BL RT BL X RT BL X dPL 7 2 RT RL2 Xl 2RLRT RL 7 2 Rgr X Bil 7 0 7 V1 7 VI 7 dRL 0 RL2 Xl2 RT RL2 X2 RL xng thZ ZL RL W lZTl P vRX vRX Lmam RT iRZT X2 X 2RZ F X RTiRZT X VT PLszm 2m iRi X2

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