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ECE 423Law, J.D.Power Systems AnalysisSpring 2009Session 26ePage 1/4SALIENT POLE SYNCHRONOUS GENERATORV. Determine λ’sB. λas(t)λas= λas(t)=Laaia+ Labib+ Lacic+ Lafif(1)λaarm= Laaia+ Labib+ Lacic(2)λaf= Lafif(3)λaf= −LsfIfsin(ωet)=LsfIfcos(ωet + π/2) (4)λaarm=[(Lss+ Lls) −LΔcos(2ωet)]ia+[−Lss2−LΔcos(2ωet −2π/3)]ib+[−Lss2−LΔcos(2ωet + 2π/3)]ic(5)λaarm=(Lss+Lss2+ Lls)ia−Lss2(ia+ ib+ ic)+−LΔcos(2ωet)ia−LΔcos(2ωet −2π/3)ib−LΔcos(2ωet + 2π/3)ic(6)λaarm=(Lss+Lss2+ Lls)√2Iscos(ωet −θei)+−LΔcos(2ωet)√2Iscos(ωet −θei)+−LΔcos(2ωet −2π/3)√2Iscos(ωet −θei−2π/3)+−LΔcos(2ωet + 2π/3)√2Iscos(ωet −θei+ 2π/3) (7)ECE 423Law, J.D.Power Systems AnalysisSpring 2009Session 26ePage 2/4λaarm=(32Lss+ Lls)√2 Iscos(ωet −θei) −√232LΔIscos(ωet + θei) (8)λas= λaarm+ λaf(9)λas=(32Lss+ Lls)√2 Iscos(ωet −θei) −√232LΔIscos(ωet + θei)+LsfIfcos(ωet + π/2) (10)˜Λas= LsfIf√2/π/2+ field term(32Lss+ Lls) Is/ −θei+ round rotor term−32LΔIs/θeisalient pole “adjustment term (11)˜If≡If√2/π/2(12)˜Λaf= LsfIf√2/π/2(13)˜Λaf= Lsf˜If(14)˜Λaarm=(32Lss+ Lls) Is/ −θei−32LΔIs/θei(15)˜Ias= Is/ −θei= Ise−jθei= Is(cosθei− jsinθei) (16)Is/θei= Is(cosθei+ jsinθei) (17)˜Λaarm=(32Lss+ Lls) Is[cosθei− jsinθei] −32LΔIs[cosθei+ jsinθei] (18)ECE 423Law, J.D.Power Systems AnalysisSpring 2009Session 26ePage 3/4˜Λaarm=(32Lss+ Lls−32LΔ) Iscosθei−(32Lss+ Lls+32LΔ) jIssinθei(19)Lq≡32Lss+ Lls−32LΔ(20)Ld≡32Lss+ Lls+32LΔ(21)˜Iaq≡ Iscosθei/0 (22)˜Iad≡−jIssinθei/0 = −Issinθei/π/2 (23)˜Ias=˜Iaq+˜Iad= Is[cosθei− jsinθei]=Ise−jθei(24)˜Λaarm= Lq˜Iaq+ Ld˜Iad(25)˜Λas=˜Λaf+˜Λaarm= Lsf˜If+ Lq˜Iaq+ Ld˜Iad(26)˜Λad= Lsf˜If+ Ld˜Iad(27)˜Λaq= Lq˜Iaq(28)˜Λas=˜Λaq+˜Λad(29)ECE 423Law, J.D.Power Systems AnalysisSpring 2009Session 26ePage 4/4VI. Determine va(t)vas(t)=−rsias−dλasdt(30)˜Vas= −rs˜Ias− jωe˜Λas(31)˜Vas= −rs˜Ias− jωeLsf˜If− jωeLq˜Iaq− jωeLd˜Iad(32)Xsf≡ ωeLsfXq≡ ωeLqXd≡ ωeLd(33)˜Eas≡−jXsf˜If= −jXsfIf/π/2 (34)˜Eas= XsfIf/0 (35)˜Vas=˜Eas−rs˜Ias− jXq˜Iaq− jXd˜Iad(36)˜Eas=˜Vas+ rs˜Ias+ jXq˜Iaq+ jXd˜Iad(37)˜Eas=˜Vas+ rs˜Ias+ jXq˜Iaq+ jXd˜Iad+ jXq˜Iad− jXq˜Iad(38)˜Eas=˜Vas+ rs˜Ias+ jXq(˜Iaq+˜Iad)+ j( Xd−Xq)˜Iad(39)˜Eas=˜Vas+ rs˜Ias+ jXq˜Ias+ j(Xd−Xq)˜Iad(40)˜E as≡˜Vas+(rs+ jXq)˜Ias(41)˜Eas=˜E x+ j(Xd−Xq)˜Iad(42)˜Eas,˜E x, and j(Xd−Xq)˜Iadare all in phase


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UI ECE 423 - Session 26e

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