Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

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

534 views

Published on

Published in: Education
  • Be the first to comment

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

  1. 1. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) TEORiA (30min) 1. Artezgailu kontrolatu ez ideal (ΔVX≠0) batentzat, marraztu potentzia triangeluak eta idatzi S, S1 eta P1 potentzien adierazpenak S = 3⋅ VR ⋅ ISARE  S1 = 3⋅ VR ⋅ I1   S1 = VLC 0 ⋅ IC  P1 = S1 ⋅ cos ϕ1 = (VLC 0 ⋅ IC ) ⋅ cos ϕ1  [1]   P1 = VLC ⋅ IC = (VLC 0 '−ΔVX ) ⋅ IC = (VLC 0 ⋅ cos Ψ − ΔVX ) ⋅ IC  [2] ΔVX a. Frogatu hurrengo adierazpena betetzen dela: cosϕ1 = cosψ − € VLC 0 ΔVX Igualando [1] y [2] queda la expresión buscada: cosϕ1 = cosψ − VLC 0 € 3 b. ψ=0 eta ΔVX=0 direnean,€ SARE = F dela jakinda, Π ΔVX i. Kalkulatu sareko potentzia faktorea (FSARE) ψ=30º eta = 0.1 direnean VLC 0 € P1 S1 ⋅ cosϕ1 ( 3⋅ VR ⋅ I1 ) ⋅ cos ϕ1 I FSARE = = = = 1 ⋅ cosϕ1 PSARE PSARE 3⋅ VR ⋅ ISARE ISARE € Cuando ψ=0 y ΔVX=0: € ΔVX 0 cosϕ1 = cosψ − = cos0 − ⇒ cos ϕ1 = 1 VLC 0 VLC 0 I1 I1 I1  I 3 FSARE = ⋅ cosϕ1 = ⋅1 = ⇒ 1 =  ISARE ISARE ISARE  ISARE π  € ΔVX Cuando ψ=30º y = 0.1: VLC 0 € I  3  ΔV   3  FSARE = 1 ⋅ cosϕ1 =   ⋅  cosψ − X  =   ⋅ (cos 30º −0.1) = 0.73 ISARE π   VLC 0   π  € Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3 €
  2. 2. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) ii. Frogatu sareko armonikoen mailak (τ SARE) ondorengo adierazpena  π 2 baieztatzen duela: τ SARE =   −1  3 ISARE − I12 2  ISARE  2  π 2 € I τ SARE = ≈ = =   −1 =   −1 I1 I1  I1   3 2. Sare elektriko batetara hiru artezgailu konektatzen dira, beraien datuak ondorengo € irudian adierazten direlarik. Kalkulatu saretik hartutako korronte osoa (IRED)  ΔV  cos(ϕ1 )1 = 1−  X  = 1− 0.5 = 0.5 ⇒ (ϕ1 )1 = 60º  VLC 0 1 2 (I≈ )1 = J12 − ( I≈ )1 = 5 2 − 32 = 4 A  ΔV  € cos(ϕ1 ) 2 = cos(ψ ) 2 −  X  = cos 45º −0.207 = 0.5 ⇒ (ϕ1 ) 2 = 60º  VLC 0  2 2 ( I≈ ) 2 = J 2 − ( I≈ ) 2 = 7 2 − 4 2 = 5.74 A 2 1+ cos(ψ ) 3  ΔV  1+ cos 70º € cos2 (ϕ1 ) 3 = − X  = − 0.421 = 0.25 ⇒ (ϕ1 ) 3 = 60º 2  VLC 0  3 2 2 ( I≈ ) 3 = J 3 − ( I≈ ) 3 = 132 − 7 2 = 10.95A 2 € Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3
  3. 3. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) [(ϕ ) = (ϕ ) = (ϕ ) ] 1 1 1 2 1 3 (I1 ) red = ( I1)1 + ( I1) 2 + ( I1) 3 = 3 + 4 + 7 = 14 A (I≈ ) red = ( I≈ )1 + ( I≈ ) 2 + ( I≈ ) 3 = 4 + 5.74 + 10.95 = 20.69A 2 2 Ired = ( I1) red + (I≈ ) red = 14 2 + 20.69 2 = 24.98A € € ARIKETA (1H 30MIN) Irudiko zirkuituan: a. Adierazi sekundario bakoitzaren konexio mota (λ edo Δ) 1. λ 2. λ 3. Δ Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3
  4. 4. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) b. Kalkulatu T1 eta T3-ren espira erlazioa (n1/n2) eta T2-ren tentsio erlazioa (U1/U2)    n1   U1  6000   = U  = 400 = 15 3  n 2 1  2   3 1 3  n1   U1  6000   =  = = 15  n 2  3  U 2  3 400  U1   n1   3  =  U1  = 15 ⇒ U = U1 = 400v   = U  U  ( 2 )2  n2 2  2   2 2 15  3 2 c. Kalkulatu transformadore bakoitzaren potentzia € J1 P 3,Δ = ( I1)12 + ( I≈ )12 → ( I≈ )1 ≈ 0 → J1 P 3,Δ ≈ ( I1 )1 ⇒ J1 P 3,Δ = 1.56A (VLC 0 )1 ⋅ IC1 (VLC 0 )1 ⋅ IC1 270 ⋅ 60 (I1 )1 = = = = 1.56A 3 ⋅ U1 3 ⋅ U1 3 ⋅ 6000  (U )  3 3 ⋅  2 ⋅ 2 1 3 3 ⋅ (VO )1  3  3 2 ⋅ 400 (VLC 0 )1 = VLC 0 P 3 = = = = 270v 2π 2π 2π PT1 = 3 ⋅ U1 ⋅ J1 = 3 ⋅ 6000 ⋅1.56 = 16212W 2  n2  2 1 € J 2 PD 3,λ = ⋅   ⋅ IC 2 = ⋅ ⋅ 300 = 16.33A 3  n1  2 3 15 PT 2 = 3 ⋅ U1 ⋅ J 2 = 3 ⋅ 6000 ⋅16.33 = 169706W π  n2  π− J 3 S 3,Δ = π − ψ3 ⋅   ⋅ IC 3 = 2 ⋅ 1 ⋅ 270 = 12.73A π  n1  3 π 15 PT 3 = 3 ⋅ U1 ⋅ J 3 = 3 ⋅ 6000 ⋅12.73 = 132294W Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3 €
  5. 5. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) d. Kalkulatu (I1)SARE, ISARE, FSARE eta ζSARE  ΔV  cos(ϕ1 )1 = 1−  X  = 1− 0 = 1 ⇒ (ϕ1)1 = 0º  VLC 0 1  ΔV  cos(ϕ1 ) 2 = cosψ 2 −  X  = cos 45º −0.207 = 0.5 ⇒ (ϕ1 ) 2 = 60º  VLC 0  2 1+ cosψ 3  ΔVX  1+ cos90º cos2 (ϕ1 ) 3 = −  = − 0.25 = 0.25 ⇒ (ϕ1) 3 = 60º 2  VLC 0  3 2 (VLC 0 ) 2 ⋅ IC 2 (VLC 0 ) 2 ⋅ IC 2 540 ⋅ 300 € (I1 ) 2 = = = = 15.6A 3 ⋅ U1 3 ⋅ U1 3 ⋅ 6000  (U )  3 3 ⋅ 2 ⋅ 2 2 3 3 ⋅ (VO ) 2  3  3 2 ⋅ 400 (VLC 0 ) 2 = VLC 0 PD 3 = = = = 540v π π π 2 2 ( I≈ ) 2 = J 2 − ( I1 ) 2 = 16.332 −15.6 2 = 4.83A (I1 ') 3 = ( I1) 3 ⋅ cos(ϕ1) 3 = 14.03⋅ cos60º = 7A € (VLC 0 ) 3 ⋅ IC 3 (VLC 0 ) 3 ⋅ IC 3 540 ⋅ 270 (I1 ) 3 = = = = 14.03A 3 ⋅ U1 3 ⋅ U1 3 ⋅ 6000 (VLC 0 ) 3 = VLC 0 S 3 = 3⋅ (VO ) 3 = 3⋅ [ 2 ⋅ (U 2 ) 3 ]=3 2 ⋅ 400 = 540v π π π ( I≈ ) 3 = J 3 2 − ( I1 ') 3 2 = 12.732 − 7 2 = 10.63A € Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3
  6. 6. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) 2 2 2 ISARE = ISARE,X + ISARE,Y = (−0.53) + 27.32 = 27.3A ISARE,X = J1,X + J 2+3,X = 1.56 + (−2.09) = −0.53A J1,X = 1.56A J 2+3,X = J 2+3 ⋅ cos(α + 60) = 27.38 ⋅ cos94.37º = −2.08A J 2+3 = (I1 ) 2+3 2 + (I≈ ) 2+3 2 = 22.6 2 + 15.46 2 = 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 (I≈ ) 2+3 15.46 α = arctg = arctg = 34.37º (I1 ) 2+3 22.6 ISARE,Y = J1,Y + J 2+3,Y = 0 + 27.3 = 27.3A J1,Y = 0A J 2+3,Y = J 2+3 ⋅ sin(α + 60) = 27.38 ⋅ sin94.37º = 27.3A € Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3
  7. 7. BILBOKO INDUSTRIA INGENIARITZA TEKNIKOKO UNIBERTSITATE ESKOLA POTENTZIAREN ELEKTRONIKA - (2009/02/03) (I1 ) SARE = (I1 ) SARE,X 2 + ( I1) SARE,Y 2 = 12.86 2 + 19.57 2 = 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 (I≈ ) SARE 14.03 τ SARE = = = 0.59 (I1 ) SARE 23.42 (I≈ ) SARE = ISARE 2 − ( I1 ) SARE 2 = 27.32 − 23.42 2 = 14.03A PLC1 + PLC 2 + PLC 3 16200 + 81000 + 36450 FSARE = = = 0.47 S 283710 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 ⋅ cosψ 2 = 540 ⋅ cos 45º = 382v Sistemen Ingeniaritza eta Automatika Saila Pag. PAGE 3 €
  8. 8. 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 1+ cosψ 3 1+ cos90º (VLC 0 ') 3 = (VLC 0 ) 3 ⋅ = 540 ⋅ = 270v 2 2 €

×