b d q d + + q kd kd kq kq f f a c (a) (b) - - PowerPoint PPT Presentation

� b d q d + + q kd kd � � kq kq f f � a c (a) (b)

- d d + q + - - kq - kd Vs � + ( r) E - + f � Ep + q � a � t � (ref) � 1 ref Is (b) (a)

Three-phase machine Two-phase non-reciprocal Two-phase reciprocal ˆ ˆ ˆ V V V ˆ ˆ 2 ˆ 3 I I I 2 R s R s 3 R s L s − M s ≈ 3 2 L s [ ± L s 2 ] 2 L s [ ± L s 2 ] 3 ( L s − M s ) ≈ L s [ ± L s 2 ] M s = − 1 2 ( L s − L s σ ) ≈ − 1 0 0 2 L s M sr ( d ) = M sf M srd = M sf M srd = M sf M rsd = 3 M rs ( d ) = M sf M srd = M sf 2 M sf 2 R s ˆ P j = R s ˆ 2 R s ˆ P j = 3 P j = 3 I 2 I 2 I 2 2 ˆ m.m.f.= ˆ 2 ˆ m.m.f.= 3 m.m.f.= 3 I ( w ξ ) I ( w 0 ξ 0 ) I ( w ” ξ ”) Ψ phase = L s ˆ Ψ phase = ( L s − M s )ˆ 2 L s ˆ 2 L s ˆ I = 3 Ψ phase = 3 I I I 2 L s ˆ 2 L s ˆ 2 L s ˆ Ψ dr,ss = 3 Ψ dr,ss = 3 Ψ dr,ss = 3 I I I Ψ dr,sr = M sf i f Ψ dr,sr = M sf i f Ψ dr,sr = M sf i f 2 M sf ˆ 2 M sf ˆ 2 M sf ˆ Ψ dr,rs = 3 Ψ dr,rs = 3 Ψ dr,rs = 3 I I I non-power-invariant, non-reciprocal power-invariant; reciprocal equivalence voltages: w 0 ξ 0 = w ξ equivalence voltages: w ” ξ ” = w ξ equivalence mmf: w 0 ξ 0 = 3 equivalence mmf: w ” ξ ” = w ξ 2 w ξ Table 28.1: Three-phase synchronous machine and two-phase equivalents

two-pole two-pole four-pole four-pole Turbo generators air cooling conductor cooling air cooling conductor cooling 1 . 0 . . . 1 . 75 (1 . 65) 1 . 5 . . . 2 . 25 (1 . 85) 1 . 0 . . . 1 . 75 (1 . 65) 1 . 5 . . . 2 . 25 (1 . 85) x d 0 . 96 . . . 1 . 71 (1 . 61) 1 . 46 . . . 2 . 21 (1 . 8) 0 . 96 . . . 1 . 71 (1 . 61) 1 . 46 . . . 2 . 21 (1 . 8) x q 0 0 . 12 . . . 0 . 25 (0 . 17) 0 . 2 . . . 0 . 35 (0 . 28) 0 . 2 . . . 0 . 3 (0 . 25) 0 . 25 . . . 0 . 45 (0 . 35) x d x ” 0 . 08 . . . 0 . 18 (0 . 12) 0 . 15 . . . 0 . 28 (0 . 22) 0 . 12 . . . 0 . 20 (0 . 16) 0 . 20 . . . 0 . 32 (0 . 28) d = x ” = x ” = x ” = x ” x 2 (inverse field) d d d d (0 . 1 . . . 0 . 7) x ” (0 . 1 . . . 0 . 7) x ” (0 . 1 . . . 0 . 7) x ” (0 . 1 . . . 0 . 7) x ” x 0 (zero sequence) d d d d x p (Potier reactance) 0 . 07 . . . 0 . 17 0 . 2 . . . 0 . 45 0 . 12 . . . 0 . 24 0 . 25 . . . 0 . 45 0 . 001 . . . 0 . 007 0 . 001 . . . 0 . 005 0 . 001 . . . 0 . 005 0 . 001 . . . 0 . 005 r s 0 d 0 (sec) 5 5 8 6 T 0 d (sec) 0 . 6 0 . 75 1 . 0 1 . 2 T T ” d (sec) 0 . 035 0 . 035 0 . 035 0 . 045 T d (sec) 0 . 13 . . . 0 . 45 0 . 2 . . . 0 . 55 0 . 2 . . . 0 . 4 0 . 25 . . . 0 . 55 τ i (sec) 5 . . . 7 5 . . . 7 6 . . . 8 6 . . . 8 Table 28.2: Range (and rated values) for time constants and p.u inductances of large synchronous machines

’ � T1 � V Fv - 1 �� Fvf � Vf 2/NP)p(p+ w) (J n � � - - �� F � T -1( +p ) m Ai c d � � T

Im p Im p Im p 1,0 0,01 0,1 & � � kd kq -1 -1 -1 � kd -1 � kq � f � kqh } } 0,2 0,15 0,05 0,2 0,03 0,01 0,02 0,2 0,15 0,05 (a) (b) (c) reduced frequency � o=0,1 reduced frequency � o=0,01 rated frequency � o=1 Epo/Vo=2 Epo/Vo=1 Epo/Vo=1 dkd=0,7 ; df=0,27 ; q=0,3) ( d= q=25 ; kd= kq=7 ; f=300 ; dq=1,8 ; d 0,2 ; � � � � � � � � ���� �

Im p Im p 1,0 0,1 -1 -1 0,05 � kd � kq , -1 � kd -1 -1 � f ��� kqh } } 0,5 0,3 0,2 0,1 0,8 0,4 0,12 0,05 (a) (b) � o=0,1 rated frequency reduced frequency � o=1 ( d=2,4 ; q=2 ; kd= kq=3,5 ; f=50 ; dq=2 ; d=0,13 ; dkd=0,14 ; df=0,28 ; q=0,22) � � � � � � � � � �

’ � T1 � T F (p)= V � V � � 1 - �� 2 (J/NP) � n p(p+ w) � - - � T �� F (p)= � �� N (p) � F (p)=D (p) � � � T � (Note: �� r = �� -p �� )

Im p 1 0,1 0,05 ( o=; d= q=10 ; kd= kq=10 ; dq=6 ; d= q=0,15) � � � � � � � �

Im p Im p 0,1 0,1 0,05 0,05 0,01 0,2 0,1 0,3 0,4 0,1 0,2 ( �� =0,1; � d= q= kd= kq= 5; � � � ( �� =0,1; � d= q=5 ; kd= kq=10; � � � � � �� dq= ; � q=0, ��� � dq=6 ; ����� q=0,15) d d (a) (b)

50 o kd 30 ���� 20 10 5 3 2 1 � dq=4 � dq=3 0.5 � dq=2 0.3 � dq=1 0.2 ( dq = dkd = kdkq =1) � � � 0.1 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 2 2 � � kd m

50 o kd 30 ���� 20 10 5 3 2 1 � d=0.12 0.5 � d=0.06 � d=0.02 0.3 ( dq=4; dq = dkd = kdkq =1) 0.2 � � � � 0.1 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 2 2 � � kd m

50 o kd 30 ���� 20 100kW 10kW 10 5 1kW 3 100W 2 � =-20º 1 � =-5º � =20º 0,5 � =5º 0.3 � =0º 0.2 � ( dq=4; � d=0.12; � dq = � dkd = � kdkq =1) 0.1 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 2 2 � � kd m

Im � Im � � =p r � � =p r � 10 0,1 � � � r=10 � � � r=0.1 0 0 1 0,5 2 -Re � -Re � 0,1 ev=Epo/Vo=1 ; =0 � ev=Epo/Vo=1 ; =0 � ev > 1 and/or � > 0 ev > 1 and low | | � ev < 1 and/or � < 0 ev < 1 and/or high | | � (a) (b)

50 30 o r ���� 20 100kW 10kW 10 1kW 5 3 100W 2 1 Epo/Vo=1,5 0.5 Epo/Vo=1 0.3 Epo/Vo=0,5 Epo/Vo=0 0.2 ( =0.04 ; � � =1 ; =0) � 0.1 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 2 2 � � r m

50 o r ���� 30 20 10 5 3 2 1 � =-20º � =20º 0.5 � =0º 0.3 ( =1 ; � � =0.08 ; Epo /Vo=1.5) 0.2 0.1 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 2 2 � � r m