LECTURE 23 SYNCHRONOUS MACHINES (3)

Transcription

1 ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE 3 SYNCHRONOUS MACHINES (3) Acknowledgent-Thee hndout nd lecture note given in cl re bed on teril fro Prof. Peter Suer ECE 330 lecture note. Soe lide re tken fro Ali Bzi preenttion Diclier- Thee hndout only provide highlight nd hould not be ued to replce the coure textbook. 11/7/017

2 SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE Firt we conider lient pole three-phe chine oppoed to round rotor chine Slient pole chine re ued in hydro-genertor nd low-power ingle-phe ynchronou otor. We conider two-pole chine. There re three ttor coil ditributed o tht ech coil crete inuoidl f round the periphery. The rotor h field coil tht crrie contnt current i r. 11/7/017

3 SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE 11/7/017 3

4 SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE The ttor coil re eprted echniclly. The flux current reltion cn be derived L0 L co M 0 M co ( 60 ) b 0 co ( 60 ) 0 co ( 10 ) = M M L L c M 0 M co ( 60 ) M 0 M co ( 180 ) r M co M co( 10 ) M 0 M co ( 60 ) M co i 0 co ( 180 ) co( 10 ) i M M M b L0 L co ( 10 ) M co( 10 ) i c M co( 10 ) L i r r 11/7/017 4

5 SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE L M b M c M r i b M b Lb M bc M br i b = c M c M bc Lc M cr i c r M r M br M r Lr i r inductnce trix i yetric. ll inductnce except r re function of L 0 Alo nd L = M. M 0 = L All winding hve the e nuber of turn nd there i no lekge flux. 11/7/017 5

6 SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE W = L i L i L i L i b b c c r r iib M b iic M c ibic Mbc iir M r ibir Mbr icir M cr T e = W i dl ib dlb ic dlc = d d d dm b dm c dm bc iib iic ibic d d d dm r dm br dm cr iir ibir icir d d d 11/7/017 6

7 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE For round rotor ce L = M = 0 nd = L0 M 0 L( ) = L0 M 0 M 0 M co M 0 L0 M 0 M co( 10 ) M 0 M 0 L0 M co( 10 ) M co M co( 10 ) M co( 10 ) Lr 11/7/017 7

8 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE 1 1 W = L0i M 0i i b L0i b M 0i i c 1 M 0i bi c L0i c i i rm co 1 i i M co( 10 ) i i M co( 10 ) L i b r c r r r 11/7/017 8

9 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE Since L = Lb = Lc = M b = M c = Mbc = contnt e W dm r dm br dm cr T = = iir ibir icir d d d = i i M in i i M in( 10 ) i i M in( 10 ) r b r c r In tedy-tte AC condition nd in terinl condition. p = e T Whtever hppen in phe, hppen in phe b, 10 deg. lter nd in phe c, 40 deg. lter. 11/7/017 9

10 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE Aue blnced et of current in the ttor i = I co t i = I co( t 10 ) b i = I co( t 10 ) c i = I = contnt r r 11/7/017 10

11 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE e T = I I M[in co t in( 10 )co( t 10 ) r in( 10 )co( t 10 )] = I I M r in( t) in( t) in( t 40 ) in( t) in( t 40 ) in( t) 11/7/017 11

12 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE Uing the identity in( t) in( t 40 ) in( t 40 ) = 0 We get, T = I I M 3in( t) e r I = I =, I r, = t e II r T = 3M in( t t) 11/7/017 1

13 ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE e For T to hve n verge vlue, =, which i clled the ynchronou peed. The ynchronou peed i equl to the electricl N frequency in rdin per econd, = = f where N = ynchronou peed in rp. e 3 T = I I rm in 60 For two pole chine N = f 60 = 3600 rp. 11/7/017 13

14 VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE) Copute v in tedy tte t t d di di di dm v = = L M b M c I r dt dt dt dt dt L0 L0I int I in( t 10 ) b c r L0 I t I r M t in( 10 ) in( ) Adding nd ubtrcting L0 I in t 11/7/017 14

15 VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE) 3L L v I t I t t in( t 10 )] I M in( t ) 0 0 = in [in in( 10 ) r 3 MIr v = L0 I int in( t ) 3 v = Re[ V e ] = Re L I e e j t j / j t 0 MIr j j t Re e e e 11/7/ j

16 VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE) j I I 0, e j 3 MI j V j L I j e r = 0 E 3 r L x ( ynchronou rectnce ) 0 j M I re M I r j = V = jx I Er ( voltge phor proportionl to field ( rotor) current I ) r 11/7/017 16

17 VOLTAGE IN STEADY-STATE (ROUND ROTOR CASE) V = jx I Er The equivlent circuit i = I co t = I cot I = I 0 Reference phor. Rel power into the ource MI r P = Re I 11/7/ * P = Re [ E r I ] MIr = I in

18 VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE) Power in three phe: P T =3P P 3 = MI I in T r The echnicl power output: P = T e. Since we hve DC field excittion,. fro the frequency condition = r, we hve =. r =0 For conervtive yte, P T = P e 3 T = MIIrin 11/7/017 18

19 , POWER IN TERMS OF VOLTAGES * IN 3 = 3 3 S V I V I P. F S IN 3 V0 E r = 3V 0 ( ) jx S S IN 3 = 3V 90 X S 3V E 90 r X S P IN 3 3V E r 3V E r = co(90 ) in X X S S 11/7/017 19

20 , <0 POWER IN TERMS OF VOLTAGES Motor T e P VE r = in X IN 3 3 S >0 Genertor Q IN 3 3V 3V E r = co X X QIN 3 <0 Overexcittion E co > r V QIN 3 >0 Underexcittion E co < V Note: QIN 3 i deterind by E r (excittion) S r S 11/7/017 0

21 , CONVENIENT PHASOR NOTATIONS i = ( g ) i V = jx I Er E jx I V r I E r jx V Q P T 3 T 3 3V E r = in X S 3V E 3V = co X X r S S 11/7/017 1

22 , POWER IN TERMS OF VOLTAGES <0 >0 Motor Genertor Q Overexcittion E r co > V IN 3 >0 QIN 3 <0 Underexcittion E co < V r Note: Q i deterined by E r (excittion) IN 3 11/7/017

23 , EXAMPLE 6.1 A three-phe wye-connected 60 Hz ynchronou chine with two pole h ynchronou rectnce x = 5.0 /phe. Operting otor: 30A, 54 V, PF= 0.8 leding, windge, friction, nd core loe = 400 W Find: E nd e r T, ueful hft torque, efficiency 11/7/017 3

24 , EXAMPLE 6.1 The equivlent circuit V = 540 co = 0.8 = I = A E V jx I E r r = 540 ( j5) = V E r QIN 3 11/7/017 4

25 , The torque ngle i EXAMPLE 6.1 nd = 19.3 = = 377 rd / ec PT 3(364.3)(54) in( 19.3 ) = = 1886W 5 The torque of electric origin i e PT 1886 T = = = 48.5 N 377 Overll efficiency = = = 97.8% 1886 Ueful hft torque = = N- 11/7/017 5

26 , EXAMPLE 6.4 A two-pole, three-phe, 60 Hz, wye-connected ynchronou chine, x =, i operting genertor 1905 V per phe, 350 A, PF of the lod i 0.8 lgging. E Find, r, nd the torque of electric origin V =19050 V I = A co = 0.8 = /7/017 6

27 , EXAMPLE 6.4 E r = V jx I = j ( ) = 35 j560 = V E rv in 3(391)(1905)(.3416) PT = = = 1600,000 W = 1.6 MW x e P = T =1600,000 W = two pole chine = 377 rd / ec e 1600, 000 T = = 4440 N /7/017 7

28 , MULTI-POLE MACHINES The nuber of pole in chine i defined by the configurtion of the gnetic field pttern. Source: chineryequipentonline.co Source: electricley.co 11/7/017 8

29 , MULTI-POLE MACHINES The nuber of pole in chine i defined by the configurtion of the gnetic field pttern. When the intntneou current i in the direction indicted, the reulting flux line effectively crete n electrognetic field with North (N) nd the South (S) pole 11/7/017 9

30 , MULTI-POLE MACHINE For four lot crrying coil connected in erie with polritie indicted by dot nd croe. It cn be one phe of three-phe winding. In thi, we effectively hve fourpole chine Source: electronic-tutoril.w 11/7/017 30

31 , MULTI-POLE MACHINES For four-pole, three-phe chine the rotting field coplete two cycle (70 degree. Electricl) in one echnicl revolution of 360 degree. elec Since e lec = = ech d = e lec = dt p ech ech p =, = f 10 11/7/017

32 , N =, = 60 MULTI-POLE MACHINES p f pn = 10 In theticl ter, the reltionhip between the nuber of pole nd ynchronou peed i reflected in utul p inductnce. becoe nd the frequency M co M co = ( r ) / p p =0 = condition becoe r, 11/7/017 3

33 , SUMMARY FOR P POLE MACHINE Genertor ction With the genertor convention for current, P T = P > 0 P T 3ErV in = = x P E = jx I V r In the phor digr >0 11/7/017 33

34 , Motor ction SUMMARY FOR P POLE MACHINE With the otor convention P T = P > 0 P T 3ErV in = = x P V = jx I Er In the phor digr <0 11/7/017 34

35 , SUMMARY FOR P POLE MACHINE p Both for otor nd genertor = i the upply frequency i the ynchronou peed in echnicl rdin per econd. the ynchronou peed N 10f p in rp 11/7/017 35