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NTNU NTNU NTNU Slide 379 Slide 381 Slide 383 Hvordan bestemmes vinkelen til rotorfluksen? ■ Denne er gitt av rotorfluksens frekvens: s dξ r = ωn ⋅ f k = ωn ⋅ f ψr dt ■ La oss finne et utrykk for f k : k k 1 dψ rα k 0 = rr ⋅ i rα + − ( f k − n) ⋅ ψ rβ ωn dt k d k 1 ψ rβ k 0 = rr ⋅ i rβ + + ( f k − n) ⋅ ψ rα ωn dt ■ Fra øvre likning: ψr 1 dψ r 0 = rr ⋅ i rα + ωn dt ■ Fra nedre likning: ψr rr ⋅ i rβ f k = f ψr = − + n ψ r Moment ■ Moment utrykket for den transformerte modell: k T k k T ⎛ k x k k T h ⎞ x h ( is ) � ψ = ( is ) �⎜ x ⋅ is + ψ ⎟ = ⋅ ( is ) m e = s ⎜ σ ⎝ x r r ⎟ ⎠ x r k �ψ r x h k k k k x h ψr 1 ψr m e = ⋅ ( ψ rα ⋅ isβ − ψ rβ ⋅ i sα ) = ⋅ ψ r ⋅ isβ = ⋅ ψ r ⋅ isβ x r x r 1 + σ r ■ Her har vi benyttet : k k x h k ψ = x s σ ⋅ i s + ψ x = σ ⋅ x r σ s x r 1 ψr m e = ⋅ ψ r ⋅ isβ 1 + σ r Trondheim 2000 2 x h 1 σ = 1 - = 1 − x s ⋅ x r ( 1 + σs ) ⋅ ( 1 + σ r ) Den fullstendige rotorfluksorienterte modell ψr di sα 1 ψr ψr 1 − σ ωn ψr = − ⋅ i f i u " sα + ωn ⋅ ψr ⋅ sβ + ⋅ ψ r + ⋅ sα dt T T x x s σ ⋅ r ⋅ h σ ψr di sβ 1 ψr ψr 1 − σ ωn = − ⋅ i s − ωn ⋅ f r ⋅ i s − ⋅ ωn ⋅ n ⋅ ψ r + ⋅ u " β ψ α dt T x h x s σ ⋅ σ dψ r 1 x h ψr = − ⋅ ψ r + ⋅ i sα dt T T r r s ψr dξ ⎛ rr ⋅ x h ⋅ i ⎞ r s n f r n ( f r n) ⎜ β = ω ⋅ n + n⎟ ψ = ω ⋅ + = ω ⋅ dt ⎜ x ⎟ ⎝ rψ r ⎠ dn 1 ⎛ 1 = ⋅ dt T ⎜ ⋅ ψ m ⎝1 + σr dθ = ωn ⋅ n dt r ψr ⎞ ⋅ i s − m L ⎟ β ⎠ ψr sβ Trondheim 2000 Trondheim 2000 NTNU NTNU NTNU Slide 380 Slide 382 Slide 384 Eliminasjon av rotorstrømmer….. ■ Fra fluksforslyngningslikningene: k k k ψ rα = x h ⋅ i sα + x r ⋅ i rα = ψ r k k k ψ rβ = x h ⋅ i sβ + x r ⋅ i rβ = 0 ■ Utrykket for f k blir da: rr ⋅ x h ⋅ isrβ f k = f ψr = + n x ⋅ ψ ■ For rotorfluksens amplitude: d r ψr ψ r 1 x h ψr = − ⋅ ψ r + ⋅ i sα dt Tr Tr r k k − ψ r + x h ⋅ isα ⇒ - i rα = x r k x h k ⇒ − i rβ = ⋅ i sβ x r x r Tr = ω ⋅ r Spenningsbalanse i stator n r Trondheim 2000 ■ Da man skal ha indre statorstrømregulatorer ønskes statorstrøm og rotorfluks som tilstandsvariable ■ Eliminerer så statorflukser og rotorstrømmer fra spenningslikningen ved hejlp av følgende sammenhenger: k s k s ⋅ is k r ■ Spenningsbalanse i stator: h k r k h ⋅ is ψ = x + x ⋅ i ψ = x + x ⋅ i ψr disα 1 ψr ψr 1− σ ωn ψr = − ⋅ i " sα + ωn ⋅ f ψr ⋅ isβ + ⋅ ψ r + ⋅ u sα dt T σ ⋅ T s r ⋅ x h x σ ψr disβ 1 ψr ψr 1− σ ωn ψr = − ⋅ i " sβ − ωn ⋅ f ψr ⋅ i sα − ⋅ ωn ⋅ n ⋅ ψ r + ⋅ u sβ dt T σ ⋅ x s h x σ r k r " x σ Ts = ’ ωn ⋅ rs 2 ’ ⎛ x h ⎞ rs = rs + ⎜ ⋅ rr x ⎟ ⎝ r ⎠ 2 x h 1 x σ = σ ⋅ x s σ = 1 - = 1 − x s ⋅ x r ( 1 + σ s ) ⋅ ( 1 + σ r ) Hva skal reguleres ? ■ Vi har to uavhengige komponenter av statorspenningen: ➨ Styrer momentet ➨ Styrer fluksens amplitude ■ Hvilken fluksamplitude ? ➨ Stator ψs ➨ Luftgapsfluks ψ h ➨ Rotor ψ r Trondheim 2000 Trondheim 2000
NTNU NTNU NTNU Slide 385 us Slide 387 Slide 389 Utledning av ekvivalentskjemaer ■ Stasjonære forhold: u s = rs ⋅ is + � ⋅ f s ⋅ ψ s 0 = rr ⋅ i r + �⋅ f r ⋅ ψ r f r = f s − n m e = T ( is ) � ψ s ■ Flukser og magnetiseringsstrøm: k k k iμ = is + i r x s = x h + x sσ ψ = x s ⋅ is + x h ⋅ i r = x s i s x h i s σ ⋅ + ⋅ μ ψ r = x h ⋅ i s + x r ⋅ i r = x rσ ⋅ i r + x h ⋅ i μ ψ = x s s ⋅ is + x h ⋅ i r ψ = x r h ⋅ is + x r ⋅ i r x r = x h + x rσ Ekvivalentskjema basert på rotorfluks ■ Spenningsbalanser: x h u s = ( rs + �⋅ f s ⋅ x σ ) ⋅ i s + � ⋅ f s ⋅ ⋅ i 1 + σ f s rr x h 0 = ⋅ ( 1+ σ r ) ⋅ i 2 r + �⋅ fs ⋅ ⋅ iμr f ( 1 + σ ) 1+ σ r �� σ � �� r � μr r is ( 1 + σr ) ⋅ ir r Trondheim 2000 �� 1+ σ iμr u μr � � � � � = ⋅ �� 2 ( 1+ σ ) � � � Konstant statorfluks ■ Sammenheng spenning, frekvens og fluks: u s = rs ⋅ i s + � ⋅ f s ⋅ ψ ≈ � ⋅ f s s ⋅ ψ s ■ Klassisk styring av fluks 2 ⎛ rs ⎞ 2 u s = ψ s ⎜ + f s ≈ ψ s ⋅ f s x ⎟ ⎝ s ⎠ ■ Moment ble styrt med statorfrekvens: f r s = f − n 2 1 f r ⋅ ωn ⋅ Tr ⎛ x h ⎞ ’ me = ⋅ ⋅ 2 s Tr x ’ ⎜ ⋅ ψ = σ ⋅ r 1 ( f r n Tr ) x ⎟ + ⋅ ω ⋅ ⎝ s ⎠ Trondheim 2000 T r Trondheim 2000 NTNU NTNU NTNU us Slide 386 us Slide 388 Slide 390 Utledning av ekvivalentskjemaer….. ■ Setter inn spenningslikingene: u = ( r + � ⋅ f ⋅ x ) ⋅ is + � ⋅ f ⋅ x ⋅ i s s s sσ ⎛ f s ⎞ 0 = ⎜ rr + �⋅ f s ⋅ x rσ ⎟ ⋅ i r + � ⋅ f s ⋅ x h ⋅ iμ ⎝ f r ⎠ �� σ � �� is s h � � � �� μ i ir μ � �σ � �� Ekvivalentskjema basert på rotorfluks ■ Definisjoner og sammenhenger: u ψ r μr = � ⋅ f s = � 1 + σr ψr α �� σ � �� ψr β x ⋅ fs 1+ h σ r is ( 1 + σr ) ⋅ ir � ⋅ i k μr is = i r is = ( 1 + σ r ) ⋅ i r μ � � ⋅ � � � � Trondheim 2000 �� 1+ σ iμr u μr � � � � � = ⋅ �� 2 ( 1+ σ ) � � � Konstant U/f ■ Sammenheng spenning, frekvens og fluks: u = � ⋅ f ⋅ ψ ⇒ u = ψ ⋅ f 1 f rk = ’ ωn ⋅ Tr 2 1 ⎛ x h ⎞ m e, max = ⋅ s 2 x ⎜ ⋅ ψ r x ⎟ ⋅ σ ⋅ ⎝ s ⎠ s s s 3 2 1 0 -1 -2 s s -3 0 0.5 1 1.5 2 s Fra venstre: f s ,u s = 0.2 f s ,u s = 0.4 f s ,u s = 0.6 f s ,u s = 0.8 f s ,u s = 1.0 f s =1.2, u s = 1.0 f s =1.4, u s = 1.0 f s =1.6, u s = 1.0 Trondheim 2000 Trondheim 2000
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