EP ETXERAKO LANA (2009/04/03) Azterketa (2009ko urtarrila) [ebazpena]

BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA

POTENTZIAREN ELEKTRONIKA - (2009/02/03)

Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3

TEORiA (30min)

1. Artezgailu kontrolatu ez ideal (ΔVX≠0) batentzat, marraztu potentzia triangeluak eta idatzi S, S1 eta P1 potentzien adierazpenak

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S = 3 ⋅VR ⋅ ISARE

S1 = 3 ⋅VR ⋅ I1S1 =VLC 0 ⋅ IC

P1 = S1 ⋅ cosϕ1 = VLC 0 ⋅ IC( ) ⋅ cosϕ1 [1]P1 =VLC ⋅ IC = VLC 0 '−ΔVX( ) ⋅ IC = VLC 0 ⋅ cosΨ−ΔVX( ) ⋅ IC [2]

a. Frogatu hurrengo adierazpena betetzen dela:

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cosϕ1 = cosψ −ΔVX

VLC 0

Igualando [1] y [2] queda la expresión buscada:

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VLC 0

b. ψ=0 eta ΔVX=0 direnean,

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FSARE =3Π

dela jakinda,

i. Kalkulatu sareko potentzia faktorea (FSARE) ψ=30º eta

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ΔVX

VLC 0

= 0.1 direnean

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FSARE =P1

PSARE=S1 ⋅ cosϕ1PSARE

=3 ⋅VR ⋅ I1( ) ⋅ cosϕ13 ⋅VR ⋅ ISARE

=I1

ISARE⋅ cosϕ1

Cuando ψ=0 y ΔVX=0:

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VLC 0

= cos0 − 0VLC 0

⇒ cosϕ1 =1

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FSARE =I1

ISARE⋅ cosϕ1 =

I1ISARE

⋅1=I1

ISARE⇒

I1ISARE

=3π

Cuando ψ=30º y

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ΔVX

VLC 0

= 0.1:

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FSARE =I1

ISARE⋅ cosϕ1 =

3π

⋅ cosψ −

ΔVX

VLC 0

=

3π

⋅ cos30º−0.1( ) = 0.73




ii. Frogatu sareko armonikoen mailak (τSARE) ondorengo adierazpena

baieztatzen duela:

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τ SARE =π3

2

−1

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τ SARE =I≈I1

=ISARE2 − I1

2

I1=

ISAREI1

2

−1 =π3

2

−1

2. Sare elektriko batetara hiru artezgailu konektatzen dira, beraien datuak ondorengo

irudian adierazten direlarik. Kalkulatu saretik hartutako korronte osoa (IRED)

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cos ϕ1( )1 =1− ΔVX

VLC 0

1

=1− 0.5 = 0.5⇒ ϕ1( )1 = 60º

I≈( )1 = J12 − I≈( )1

2= 52 − 32 = 4A

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cos ϕ1( )2 = cos ψ( )2 −ΔVX

VLC 0

2

= cos45º−0.207 = 0.5⇒ ϕ1( )2 = 60º

I≈( )2 = J22 − I≈( )2

2= 72 − 42 = 5.74A

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cos2 ϕ1( )3 =1+ cos ψ( )3

2−ΔVX

VLC 0

3

=1+ cos70º

2− 0.421= 0.25⇒ ϕ1( )3 = 60º

I≈( )3 = J32 − I≈( )3

2= 132 − 72 =10.95A




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ϕ1( )1 = ϕ1( )2 = ϕ1( )3[ ]I1( )red = I1( )1 + I1( )2 + I1( )3 = 3+ 4 + 7 =14A

I≈( )red = I≈( )1 + I≈( )2 + I≈( )3 = 4 + 5.74 +10.95 = 20.69A

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Ired = I1( )red2

+ I≈( )red2

= 142 + 20.692 = 24.98A

ARIKETA (1H 30MIN)

Irudiko zirkuituan:

a. Adierazi sekundario bakoitzaren konexio mota (λ edo Δ)

1. λ

2. λ

3. Δ




b. Kalkulatu T1 eta T3-ren espira erlazioa (n1/n2) eta T2-ren tentsio erlazioa (U1/U2)

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n1n2

1

=U1

U23

1

=6000400

3=15 3

n1n2

3

=U1

U2

3

=6000400

=15

n1n2

2

=

U13

U23

2

=U1

U2

2

=15⇒ U2( )2 =U1

15= 400v

c. Kalkulatu transformadore bakoitzaren potentzia

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J1 P 3,Δ = I1( )12 + I≈( )1

2 → I≈( )1 ≈ 0 → J1 P 3,Δ ≈ I1( )1 ⇒ J1 P 3,Δ =1.56A

I1( )1 =VLC 0( )1 ⋅ IC13 ⋅U1

=VLC 0( )1 ⋅ IC13 ⋅U1

=270 ⋅ 603 ⋅ 6000

=1.56A

VLC 0( )1 =VLC 0 P 3 =3 3 ⋅ VO( )1

2π=

3 3 ⋅ 2 ⋅U2( )13

2π=3 2 ⋅ 4002π

= 270v

PT1 = 3 ⋅U1 ⋅ J1 = 3 ⋅ 6000 ⋅1.56 =16212W

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J2 PD3,λ =23⋅n2n1

2

⋅ IC 2 =23⋅115

⋅ 300 =16.33A

PT 2 = 3 ⋅U1 ⋅ J2 = 3 ⋅ 6000 ⋅16.33 =169706W

J3 S3,Δ =π −ψ3

π⋅n2n1

3

⋅ IC 3 =π −

π2

π⋅115

⋅ 270 =12.73A

PT 3 = 3 ⋅U1 ⋅ J3 = 3 ⋅ 6000 ⋅12.73 =132294W




d. Kalkulatu (I1)SARE, ISARE, FSARE eta ζSARE

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cos ϕ1( )1 =1− ΔVX

VLC 0

1

=1− 0 =1 ⇒ ϕ1( )1 = 0º

cos ϕ1( )2 = cosψ2 −ΔVX

VLC 0

2

= cos45º−0.207 = 0.5 ⇒ ϕ1( )2 = 60º

cos2 ϕ1( )3 =1+ cosψ3

2−ΔVX

VLC 0

3

=1+ cos90º

2− 0.25 = 0.25 ⇒ ϕ1( )3 = 60º

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I1( )2 =VLC 0( )2 ⋅ IC 23 ⋅U1

=VLC 0( )2 ⋅ IC 23 ⋅U1

=540 ⋅ 3003 ⋅ 6000

=15.6A

VLC 0( )2 =VLC 0 PD3 =3 3 ⋅ VO( )2

π=

3 3 ⋅ 2 ⋅U2( )23

π=3 2 ⋅ 400

π= 540v

I≈( )2 = J22 − I1( )2

2 = 16.332 −15.62 = 4.83A

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I1'( )3 = I1( )3 ⋅ cos ϕ1( )3 =14.03 ⋅ cos60º= 7A

I1( )3 =VLC 0( )3 ⋅ IC 33 ⋅U1

=VLC 0( )3 ⋅ IC 33 ⋅U1

=540 ⋅ 2703 ⋅ 6000

=14.03A

VLC 0( )3 =VLC 0 S3 =3 ⋅ VO( )3

π=3 ⋅ 2 ⋅ U2( )3[ ]

π=3 2 ⋅ 400

π= 540v

I≈( )3 = J32 − I1'( )3

2 = 12.732 − 72 =10.63A




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ISARE = ISARE ,X2 + ISARE ,Y

2 = −0.53( )2 + 27.32 = 27.3A

ISARE ,X = J1,X + J2+3,X =1.56 + −2.09( ) = −0.53A

J1,X =1.56AJ2+3,X = J2+3 ⋅ cos α + 60( ) = 27.38 ⋅ cos94.37º= −2.08A

J2+3 = I1( )2+32 + I≈( )2+3

2 = 22.62 +15.462 = 27.38A

I1( )2+3 = I1( )2 + I1'( )3 =15.6 + 7 = 22.6A

I≈( )2+3 = I≈( )2 + I≈( )3 = 4.83+10.63 =15.46A

α = arctgI≈( )2+3

I1( )2+3

= arctg15.4622.6

= 34.37º

ISARE ,Y = J1,Y + J2+3,Y = 0 + 27.3 = 27.3A

J1,Y = 0AJ2+3,Y = J2+3 ⋅ sin α + 60( ) = 27.38 ⋅ sin94.37º= 27.3A




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I1( )SARE = I1( )SARE ,X2 + I1( )SARE ,Y

2 = 12.862 +19.572 = 23.42A

I1( )SARE ,X = I1( )1,X + I1( )2+3,X =1.56 +11.3 =12.86A

I1( )1,X =1.56A

I1( )2+3,X = I1( )2+3 ⋅ cos ϕ1( )2 = 22.6 ⋅ cos60º=11.3A

I1( )2+3 = I1( )2 + I1'( )3 =15.6 + 7 = 22.6A

I1( )SARE ,Y = I1( )1,Y + I1( )2+3,Y = 0 +19.57 =19.57A

I1( )1,Y = 0A

I1( )2+3,Y = I1( )2+3 ⋅ sin ϕ1( )2 = 22.6 ⋅ sin60º=19.57A

τ SARE =I≈( )SAREI1( )SARE

=14.0323.42

= 0.59

I≈( )SARE = ISARE2 − I1( )SARE

2 = 27.32 − 23.422 =14.03A

FSARE =PLC1 + PLC 2 + PLC 3

S=16200 + 81000 + 36450

283710= 0.47

S = 3 ⋅U1 ⋅ ISARE = 3 ⋅ 6000 ⋅ 27.3 = 283710W

PLC1 =VLC1 ⋅ IC1 = 270 ⋅ 60 =16200W

VLC1 = VLC 0( )1 − ΔVX( )1 = 270 − 0 = 270v

PLC 2 =VLC 2 ⋅ IC 2 = 270 ⋅ 300 = 81000W

VLC 2 = VLC '( )2 = VLC 0 '( )2 − ΔVX( )2 = VLC 0 '( )2 − 0.207 ⋅ VLC 0( )2 = 382 − 0.207 ⋅ 540 = 270v

VLC 0 '( )2 = VLC 0( )2 ⋅ cosψ2 = 540 ⋅ cos45º= 382v

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PLC 3 =VLC 3 ⋅ IC 3 =135 ⋅ 270 = 36450W

VLC 3 = VLC '( )3 = VLC 0 '( )3 − ΔVX( )3 = VLC 0 '( )3 − 0.25 ⋅ VLC 0( )3 = 270 − 0.25 ⋅ 540 =135v

VLC 0 '( )3 = VLC 0( )3 ⋅1+ cosψ3

2= 540 ⋅ 1+ cos90º

2= 270v

EP ETXERAKO LANA (2009/04/03) Azterketa (2009ko urtarrila) [ebazpena]

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