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EP IKASGELAKO PRAKTIKA 2009/03/12 (ebazpena)

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EP IKASGELAKO PRAKTIKA 2009/03/12 (ebazpena)

  1. 1. ISARE, (I1)SARE  FSARE  (J1)2  (J1)1  Δ  λ   U1  6300  U1  6300  =  =  U 2 1 370  U 2  2 370 λ  λ  € € Ψ1=0o  Ψ2=30o  (IC)1 = 270A  (IC)2 = 270A 
  2. 2. Δ   U1  6300  n1  U1 6300 = 17 3 = ⇒  = =   n 2 1 U 2 370  U 2 1 370 λ  3 λ  U1 6300  U1  6300  n1  3= 3 = 17  = ⇒  =  n2 2 U 2 370 λ   U 2  2 370 3 3  n2   n2  € ⇒   = 3 ⋅   n1  2  n1 1
  3. 3. 30o 30o (V13)1 (V13)2
  4. 4. (Ip1)1  n2    ⋅ ( IC )1 = x (J1)1   n1 1 € Δ  n  −  2  ⋅ ( IC )1 = −x  n1 1 (Is1)1 € λ  IC -IC Ψ1=0o  (VL)1 (V32)1 (V12)1 (V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (IC)1 = 270A  60 90 120 150 180 210 240 270 300 330 360 30
  5. 5. (Ip3)1  n2  (J1)1    ⋅ IC = x  n1 1 € Δ  n  −  2  ⋅ IC = −x  n1 1 (Is3)1 € λ  IC -IC Ψ1=0o  (VL)1 (V32)1 (V12)1 (V13)1 (V23)1 (V21)1 (V31)1 (V32)1 (IC)1 = 270A  60 90 120 150 180 210 240 270 300 330 360 30
  6. 6. (J1)3 = (Ip1)3 - (Ip3)3 2x x € (J1)1  € −x Δ  € −2x (Ip1)3  n2    ⋅ IC = x €  n1 1 λ  € n  −  2  ⋅ IC = −x  n1 1 Ψ1=0o  (Ip3)3  n2  €   ⋅ IC = x  n1 1 € (IC)1 = 270A  n  −  2  ⋅ IC = −x  n1 1 30 60 90 120 150 180 210 240 270 300 330 360 €
  7. 7. (J1)2 = (Ip1)2  n2   n2    ⋅ ( IC ) 2 = 3 ⋅   ⋅ ( IC )1 = 3 ⋅ x (J1)2   n1  2  n1 1 € λ  n  n  − 2  ⋅ ( IC ) 2 = − 3 ⋅  2  ⋅ ( IC )1 = − 3 ⋅ x  n1  2  n1 1 (Is1)€ 2 λ  IC -IC Ψ2=30o  (VL)2 (V32)2 (V12)2 (V13)2 (V23)2 (V21)2 (V31)2 (IC)2 = 270A  60 90 120 150 180 210 240 270 300 330 360 30
  8. 8. 2x (J1)1 x € € −x € −2x (J1)2 € 3⋅ x € − 3⋅ x n  1 x =  2  ⋅ ( IC )1 = ⋅ 270 = 9.17A ISARE  n1 1 17 3 (2 + 3) ⋅ x € (1+ 3) ⋅ x = (1+ 3) ⋅ 9.17 = 25A ( ) 1+ 3 ⋅ x (2 + 3) ⋅ x = (2 + 3) ⋅ 9.17 = 34.22A x€ € π  1 π π ⋅ 2 ⋅ ⋅ 9.17 2 + 2 ⋅ ⋅ 25 2 + 2 ⋅ ⋅ 34.22 2  € ISARE = 2π 3 3 3  −x 9.17 2 + 25 2 + 34.22 2 ( ) − 1+ 3 ⋅ x = 3 ( ) − 2+ 3 ⋅ x € € ISARE = 25A 30 60 90 120 150 180 210 240 270 300 330 360 €
  9. 9. (VLC )1 = (VLC 0 )1 − (ΔVX )1 = 500 − 60 = 440V Δ   2 3 3 370  3 3 3VO PD 3  (VLC 0 )1 = VLC 0 PD 3 = = 500v = π π λ  2   VO PD 3 = 2 ⋅ U 2  = 370 3  3 (VLC ) 2 = (VLC ') 2 = (VLC 0 ') 2 − (ΔVX ) 2 = 433 − 60 = 373V € (VLC 0 ') 2 = (VLC 0 ) 2 ⋅ cosψ 2 = 500 ⋅ cos 30º = 433V λ   2 3⋅ 3 ⋅  370  λ  3 3 3VO PD 3  (VLC 0 ) 2 = VLC 0 PD 3 = = 500v = π π 2   VO PD 3 = 2 ⋅ U 2  = 370 3  3
  10. 10. [(V ) ⋅ (I ) ] + [(V ) ⋅ (IC ) 2 ] (PLC )1 + (PLC ) 2 [440 ⋅ 270] + [373⋅ 270] = 0.80 LC 1 C1 LC 2 FSARE = = = 3 ⋅ U1 ⋅ ISARE 3 ⋅ U1 ⋅ ISARE 3 ⋅ 6300 ⋅ 25  ΔV  € cos(ϕ1 )1 = 1−  X   VLC 0 1 (ϕ1 )1 ≠ (ϕ1 ) 2 cos(ϕ1 )1 = 1− 0.12 = 0.88 (ϕ1)1 = arccos(0.88) = 28.35º VR  ΔVX  cos(ϕ1 ) 2 = cosψ 2 −   € VLC 0  2  (I1)1 60 cos(ϕ1 ) 2 = cos 30º − = 0.74 500 (I1)2 (ϕ1) 2 = arccos(0.74 ) = 41.75º €
  11. 11. (VLC 0 )1 ⋅ (IC )1 500 ⋅ 270 (I1 )1 = = 12.37A = 3 ⋅ U1 3 ⋅ 6300 (VLC 0 ) 2 ⋅ (IC ) 2 500 ⋅ 270 (I1 ) 2 = = 12.37A = 3 ⋅ U1 3 ⋅ 6300 (I1 ) SARE = (I1 ) 2 2 20.112 + 14.112 = 24.56A SARE,X + (I1 ) SARE,Y = € (I1 ) SARE,X = (I1 )1,X + (I1 ) 2,X = 10.88 + 9.23 = 20.11A (I1)2,X (I1)1,X (I1 )1,X = (I1 )1 ⋅ cos(ϕ1 )1 = 12.37 ⋅ cos28.35 = 10.88A (I1)1,Y (I1 ) 2,X = (I1 ) 2 ⋅ cos(ϕ1 ) 2 = 12.37 ⋅ cos 41.75 = 9.23A (I1)2,Y (I1)1 (I1 ) SARE,Y = (I1 )1,Y + (I1 ) 2,Y = 5.87 + 8.24 = 14.11A (I1)2 (I1 )1,Y = (I1 )1 ⋅ sin(ϕ1 )1 = 12.37 ⋅ sin28.35 = 5.87A (I1 ) 2,Y = (I1 ) 2 ⋅ sin(ϕ1 ) 2 = 12.37 ⋅ sin 41.75 = 8.24 A

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