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6 folier pr. side - NTNU

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<strong>NTNU</strong><br />

<strong>NTNU</strong><br />

<strong>NTNU</strong><br />

Slide 367<br />

Slide 369<br />

Slide 371<br />

Elektriske likninger og momentbalanse i<br />

den transformerte modell……….<br />

■ Transformerte modell på komponent form:<br />

k<br />

k<br />

k dΨsα<br />

k<br />

U sα<br />

= R s ⋅ Isα<br />

+ − ωk<br />

⋅ Ψsβ<br />

dt<br />

k<br />

dΨ<br />

k<br />

k sβ<br />

k<br />

U sβ<br />

= R s ⋅ Isβ<br />

+ + ωk<br />

⋅ Ψsα<br />

dt<br />

k<br />

dΨ<br />

k<br />

k sγ<br />

U sγ<br />

= R s ⋅ Isγ<br />

+<br />

dt<br />

k<br />

k<br />

k dΨrα<br />

k<br />

U rα<br />

= R r ⋅ Irα<br />

+ − ωr<br />

⋅ Ψrβ<br />

dt<br />

k<br />

dΨ<br />

k<br />

k rβ<br />

k<br />

U rβ<br />

= R r ⋅ I rβ<br />

+ + ωr<br />

⋅ Ψrα<br />

dt<br />

k<br />

dΨ<br />

k<br />

k rγ<br />

U rγ<br />

= R r ⋅ Irγ<br />

+<br />

dt<br />

k<br />

k<br />

k<br />

Ψsα<br />

= ( 3 / 2 ⋅ Lsh<br />

+ L sσ<br />

) ⋅ Isα<br />

+ 3/<br />

2 ⋅ L h ⋅ I rα<br />

k<br />

k<br />

k<br />

Ψrα<br />

= ( 3 / 2 ⋅ L rh + L rσ<br />

) ⋅ I rα<br />

+ 3 / 2 ⋅ L h ⋅ I sα<br />

k<br />

k<br />

k<br />

Ψsβ<br />

= ( 3 / 2 ⋅ Lsh<br />

+ L sσ<br />

) ⋅ Isβ<br />

+ 3 / 2 ⋅ L h ⋅ I rβ<br />

k<br />

k<br />

k<br />

Ψrβ<br />

= ( 3/<br />

2 ⋅ L rh + L rσ<br />

) ⋅ I rβ<br />

+ 3 / 2 ⋅ L h ⋅ Isβ<br />

k<br />

k<br />

Ψsγ<br />

= Lsσ<br />

⋅ I sγ<br />

k<br />

k<br />

Ψrγ<br />

= L rσ<br />

⋅ I rγ<br />

dθk<br />

= ω<br />

dt<br />

k<br />

dθr<br />

= ωr<br />

= ωk<br />

− p ⋅ Ω<br />

dt<br />

Elektriske likninger og momentbalanse i<br />

den transformerte modell……….<br />

■ Moment uttrykket:<br />

SR<br />

SR T ∂�<br />

SR<br />

( I ) ⋅ I<br />

p<br />

M e = ⋅<br />

2 ∂θ<br />

⋅<br />

p<br />

M e =<br />

2<br />

∂θ<br />

2<br />

∂θ<br />

�<br />

mek<br />

SR<br />

SR<br />

−k<br />

k T ∂�<br />

−k<br />

k p k T<br />

−k<br />

T ∂�<br />

−k<br />

k<br />

⋅ ( � I ) ⋅ ⋅ � I = ⋅ ( I ) ( � ) ⋅ ⋅ I<br />

k k k k 3 k T k 3<br />

Ψs<br />

( Ψ ⋅ I − Ψ ⋅ I ) = ⋅ p ⋅ ( Is<br />

) Ψ s = ⋅ p ⋅ Ψ ⋅ I ⋅ sin ε<br />

3<br />

M e = ⋅ p ⋅ sα<br />

sβ<br />

sβ<br />

sα<br />

�<br />

2<br />

2<br />

k<br />

k<br />

⎡ ⎤ ⎡ ⎤<br />

k Isα<br />

k Ψsα<br />

Is = ⎢ ⎥ Ψ =<br />

k<br />

s ⎢ k ⎥<br />

⎢⎣<br />

Isβ<br />

⎥⎦<br />

⎢⎣<br />

Ψsβ<br />

⎥⎦<br />

⎡0 −1⎤<br />

= ⎢ ⎥<br />

⎣1<br />

0 ⎦<br />

�<br />

Skalert modell - pu-modell<br />

■ Ønsket form på induktansmatrisen:<br />

⎡x<br />

h + x sσ<br />

0 0 x h 0 0 ⎤<br />

⎢<br />

⎥<br />

⎢<br />

0 x h + x sσ<br />

0 0 x h 0<br />

⎥<br />

⎢<br />

⎥<br />

k 0 0 x sσ<br />

0 0 0<br />

� = ⎢<br />

⎥<br />

⎢ x h 0 0 x h + x rσ<br />

0 0 ⎥<br />

⎢ 0 x<br />

+ ⎥<br />

h 0 0 x h x rσ<br />

0<br />

⎢<br />

⎥<br />

⎢⎣<br />

0 0 0 0 0 x rσ<br />

⎥⎦<br />

x s = x h + x sσ<br />

x r = x h + x rσ<br />

2<br />

s<br />

s<br />

Trondheim 2000<br />

s<br />

Trondheim 2000<br />

Trondheim 2000<br />

<strong>NTNU</strong><br />

<strong>NTNU</strong><br />

<strong>NTNU</strong><br />

Slide 368<br />

Slide 370<br />

Slide 372<br />

Ser bort i fra γ-systemet ……….<br />

■ Kan da benytte to-dimensjonale romvektorer:<br />

k<br />

r<br />

dθk<br />

= ω<br />

dt<br />

k<br />

rh<br />

k<br />

s<br />

k<br />

k dΨ<br />

U s = R s ⋅ Is<br />

+ + �⋅<br />

ωk<br />

⋅ Ψ<br />

dt<br />

k<br />

r<br />

rσ<br />

r<br />

k<br />

s<br />

k<br />

k dΨ<br />

k<br />

U r = R r ⋅ I r + + �⋅<br />

( ωk<br />

− p ⋅ Ω mek ) ⋅ Ψ r<br />

dt<br />

k<br />

k<br />

k<br />

Ψ s = ( 3/<br />

2 ⋅ Lsh<br />

+ Lsσ<br />

) ⋅ Is<br />

+ 3/<br />

2 ⋅ L h ⋅ I r<br />

Ψ = ( 3/<br />

2 ⋅ L + L<br />

k ) ⋅ I + 3/<br />

2 ⋅ L<br />

k<br />

⋅ I<br />

dθr<br />

= ωr<br />

= ωk<br />

− p ⋅ Ω<br />

dt<br />

h<br />

s<br />

mek<br />

Skalert modell - pu-modell<br />

■ Grunner for å innføre pu-modell:<br />

➨ Det er lettere å se om motoren er overbelastet<br />

➨ Man kan lettere trekke erfaringer fra andre motorytelser.<br />

Parametrene i pu endrer seg ikke så mye.<br />

➨ Når man skal implementere regulatorer må man allikevel<br />

skalere de variable.<br />

■ Tilleggskrav for pu-modell for asynkronmaskinen:<br />

➨ Velge basiser slik at man får en enkel modell<br />

➨ Alle ikke-diagonale ledd skal ha verdien xh ➨ Alle pu-egeninduktanser skal kunne skrives som xh pluss<br />

en lekkinduktans<br />

Skalert modell - pu-modell<br />

■ Valg av basis-verdier for stator viklinger:<br />

I s , basis = Î n U s,<br />

basis = Û n<br />

Û n Û n<br />

Ψs,<br />

basis = =<br />

ω 2π<br />

⋅ f<br />

n<br />

n<br />

Trondheim 2000<br />

Trondheim 2000<br />

■ Valg av basis-verdier for rotor viklingene er gitt av<br />

forholdet mellom antall turn i stator og rotorvikling<br />

■ Skaleringsmatrisene:<br />

−1<br />

Sψ<br />

= diag[<br />

Ψs,<br />

basis Ψs,<br />

basis Ψs,<br />

basis<br />

−1<br />

Su<br />

= diag[<br />

U s,<br />

basis U s,<br />

basis U s,<br />

basis<br />

−1<br />

S = diag[<br />

I I I I<br />

Ψr,<br />

basis<br />

U r,<br />

basis<br />

I<br />

Ψr,<br />

basis<br />

U r,<br />

basis<br />

I<br />

Ψr,<br />

basis ]<br />

U r,<br />

basis ]<br />

]<br />

i<br />

s,<br />

basis<br />

s,<br />

basis<br />

s,<br />

basis<br />

r,<br />

basis<br />

r,<br />

basis<br />

r,<br />

basis<br />

Trondheim 2000

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