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

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

Dec 11, 2023 359 likes •531 views

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

slide-1
SLIDE 1

kd kq q d b a c

  • +

f (a) kd kq d

  • +

f q (b)

slide-2
SLIDE 2

kd d a

  • f

q

+

  • +
  • +
  • +
  • + -

kq (ref) (a)

  • t

Vs ( r) E Ep q d Is 1ref (b)

slide-3
SLIDE 3

Three-phase machine Two-phase non-reciprocal Two-phase reciprocal ˆ V ˆ V ˆ V ˆ I ˆ I

3 2 ˆ

I Rs Rs

2 3Rs

Ls [±Ls2] Ls − Ms ≈ 3

2Ls [±Ls2] 2 3(Ls − Ms) ≈ Ls [±Ls2]

Ms = − 1

2(Ls − Lsσ) ≈ − 1 2Ls

Msr(d) = Msf Msrd = Msf Msrd = Msf Mrs(d) = Msf Mrsd = 3

2Msf

Msrd = Msf Pj = 3

2Rs ˆ

I2 Pj = Rs ˆ I2 Pj = 3

2Rs ˆ

I2 m.m.f.= 3

2 ˆ

I(wξ) m.m.f.=ˆ I(w0ξ0) m.m.f.= 3

2 ˆ

I(w”ξ”) Ψphase = Ls ˆ I Ψphase = (Ls − Ms)ˆ I = 3

2Ls ˆ

I Ψphase = 3

2Ls ˆ

I Ψdr,ss = 3

2Ls ˆ

I Ψdr,ss = 3

2Ls ˆ

I Ψdr,ss = 3

2Ls ˆ

I Ψdr,sr = Msfif Ψdr,sr = Msfif Ψdr,sr = Msfif Ψdr,rs = 3

2Msf ˆ

I Ψdr,rs = 3

2Msf ˆ

I Ψdr,rs = 3

2Msf ˆ

I non-power-invariant, non-reciprocal power-invariant; reciprocal equivalence voltages: w0ξ0 = wξ equivalence voltages: w”ξ” = wξ equivalence mmf: w0ξ0 = 3

2wξ

equivalence mmf: w”ξ” = wξ Table 28.1: Three-phase synchronous machine and two-phase equivalents

slide-4
SLIDE 4

Turbo generators two-pole two-pole four-pole four-pole air cooling conductor cooling air cooling conductor cooling xd 1.0 . . . 1.75 (1.65) 1.5 . . . 2.25 (1.85) 1.0 . . . 1.75 (1.65) 1.5 . . . 2.25 (1.85) xq 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

d

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

0.08 . . . 0.18 (0.12) 0.15 . . . 0.28 (0.22) 0.12 . . . 0.20 (0.16) 0.20 . . . 0.32 (0.28) x2 (inverse field) = x”

d

= x”

d

= x”

d

= x”

d

x0 (zero sequence) (0.1 . . . 0.7)x”

d

(0.1 . . . 0.7)x”

d

(0.1 . . . 0.7)x”

d

(0.1 . . . 0.7)x”

d

xp (Potier reactance) 0.07 . . . 0.17 0.2 . . . 0.45 0.12 . . . 0.24 0.25 . . . 0.45 rs 0.001 . . . 0.007 0.001 . . . 0.005 0.001 . . . 0.005 0.001 . . . 0.005 T

d0 (sec)

5 5 8 6 T

d (sec)

0.6 0.75 1.0 1.2 T ”

d (sec)

0.035 0.035 0.035 0.045 Td (sec) 0.13 . . . 0.45 0.2 . . . 0.55 0.2 . . . 0.4 0.25 . . . 0.55 τi (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

slide-5
SLIDE 5
  • V

Vf

  • Fv

Fvf F T1 ’ T (J n 2/NP)p(p+ w)

  • 1

m Ai c d T -1( +p )

slide-6
SLIDE 6

( d= q=25 ; kd= kq=7 ; f=300 ; dq=1,8 ; d 0,2 ; dkd=0,7 ; df=0,27 ; q=0,3)

  • Im p

0,1

reduced frequency

  • =0,1

(c)

0,2 0,15 0,05

Epo/Vo=2

1,0

}

}

0,05 0,15 0,2

(a)

kd

  • 1

kq

  • 1

kqh

  • 1

f

  • 1

Im p

rated frequency o=1 Epo/Vo=1

0,01 Im p 0,2 0,03 0,02 0,01

reduced frequency

  • =0,01

(b) Epo/Vo=1

kd

kq

&

slide-7
SLIDE 7

1,0

}

}

0,1 0,5 0,2

(a)

kd

  • 1

kq

  • 1

kqh

  • 1

f

  • 1

Im p Im p 0,1

rated frequency reduced frequency

  • =1
  • =0,1

(b)

0,4 0,12 0,05

( 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)

  • 0,3

kd

  • 1

, 0,8 0,05

slide-8
SLIDE 8
  • V
  • T1

’ T (J/NP) p(p+ w)

  • n

2 1

  • F (p)=D (p)
  • N (p)
  • F (p)=
  • T

F (p)= V

  • T

(Note: r =

  • p

)

slide-9
SLIDE 9

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

  • 1
slide-10
SLIDE 10

0,4 0,2 0,1 Im p 0,05 0,1 ( =0,1; d= q= kd= kq= 5; dq= ; d q=0,

  • (a)

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

  • (b)
slide-11
SLIDE 11
  • kd

50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1

  • kd

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

slide-12
SLIDE 12
  • kd

50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1

  • kd

2 m 2 100 50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1 d=0.06 d=0.12 d=0.02 ( dq=4;

  • dq =

dkd = kdkq =1)

slide-13
SLIDE 13
  • kd

50 30 20 10 5 3 2 1 0,5 0.3 0.2 0.1

  • kd

2 m 2 100 50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1 =-20º =-5º =20º =5º =0º ( dq=4; d=0.12;

  • dq =

dkd = kdkq =1) 100W 1kW 10kW 100kW

slide-14
SLIDE 14

Im

  • =p r
  • Re

2 0,1 0,1 ev=Epo/Vo=1 ; =0 ev > 1 and/or > 0

  • ev < 1 and/or

< 0

  • Im
  • =p r
  • Re

10 1 0,5 ev=Epo/Vo=1 ; =0 ev > 1 and low | |

  • ev < 1 and/or high | |
  • (a)

(b)

r=0.1 r=10

slide-15
SLIDE 15
  • r

50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1 r 2 m 2 100 50 30 20 10 5 3 2 1 0.5 0.3 0.2 0.1 Epo/Vo=0,5 Epo/Vo=0 Epo/Vo=1 Epo/Vo=1,5 100W 1kW 10kW 100kW

  • ( =0.04 ;

=1 ; =0)

slide-16
SLIDE 16

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

  • r

r 2 m 2