sp3eje33 solved
DESCRIPTION
sistemas de potenciaTRANSCRIPT
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EJERCICIO # 33
1) G1 y G2 (ON) , G3 y G4 (OFF) Modelo Real por fase / AT (69 KV)
Vcarga = ( 230√3
) ( 13,8KV
240 ) ( 69KV
13,8KV ) = 38177,3 0° Voltios
Si G3, G4, fuera de servicio → Va = Vb → Va = 69 KV (VLL)
→ VLN = Va√3
= 69KV
√3 = 39837,17 V
Zresultante = (0,70 + j 0,68) + (0 + j 5) + (0 + j 3,75) = (0,7 + j 9,43) Ω
Req = 0,70 Ω
J Xeq = j 9,43 Ω
VLN2 = [ Vcarga + Req . I . cos θ + Xeq . I . sen θ]2 + [Xeq . I . cos θ - Req . I . sen θ]2
→ (39837,17)2 = [38177,3 + (0,7) cos (31,79°) I ]2 + [ (9,43) cos (31,79°) I ]2 + [ (9,43) cos (31,79°) I – (0,7) sen (31,79°) I ]2
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→ 89,4 I2 + 424531,46 I – 129494642 = 0
→ I = 287,7 A → I = 287,7 -31,79° A
→ Pcarga = 3 (Vcarga) . I . FPcarga = (3) (38177,3) (287,7 A) (0,850) = 28 MW
→ Qcarga = 3 Vcarga . I . sen θ . FPcarga = (3) (38177,3) (287,7) sen (31,79°) = 17,4 MVAR
Triángulo de Potencias en la carga:
Scarga = √282+17,42
Scarga = 32,96 MVA
Scarga = (3) (Vcarga) (I) = 32,9 MVA
2) G1 y G2 (OFF) , G y G4 (ON)
Pcarga = 8 MW / 0,80 (Atraso)
VLL = 240 V
Modelo Real por fase / 69 KV
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Vcarga = (240√3
) ( 13,8KV240V ) (
69KV13,8KV ) = 39837,2 Voltios
Zresultante = (2,5 + j 11,4) Ω = 11,67 77,6° Ω
Pcarga = 8 MW = (3) Vcarga . I . FPcarga
→ I = 8 x 106
(39837,2 ) (3 )(0,80) = 83,7 -36,87° A ; θ = arc.cos (0,80) = 36,87°
→ V LN = V carga + V Z result. = 39837,2 0° + (11,67 77,6°) (83,7 -36,87°)
→ V LN = 40692 1° Voltios
→ Si Vb√3
= VLN → Vb = √3 . VLN = √3 (40692) = 70480 Voltios
→ 1) V1 marca : 70480 Voltios
2) La tensión en V1 = 120 [ 70480V69000V ] = 122,6 V
→ V V1 = 122,6 Voltios