Sistemas de Transmisión de Energía eléctrica
Enviado por Helena • 20 de Enero de 2018 • 795 Palabras (4 Páginas) • 423 Visitas
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ɪexc.x = E4Zexc.x = 7114.68
→ ɪexc.x = 9.76
→ IG = Iexc.x + I´1 = 9.76
→ IG = 124.88 – j109.51
- → IG = 166.09
VG = E4 + IG (Rx + jXLx)
= 7114.68
=7114.68
=7100.1 + j455.4 + 136.37 + j114.47 VG
4.5 ˚
A) → VG = 7258.87 VL
41.25 ˚
IG
→fpG = cos(41.25 ˚ + 4.5 ˚ )
- → fpG = 0.698
→ PG = /VG/ * /IG/ * fpG
= 7,258.87 * 166.09 * 0.698
- → PG = 841,526.7 watts
%η Sistema = p en la cargap entregada por gen. * 100 = 800,00841,526.7* 100
- %η Sistema = 95.06 %
- Calcular VG, IG, fpG, % η sistema [pic 4]
Datos: PL = 800,000 watts IL VL
VL = 6,600 volts
FpL = 1 (Carga Resistiva)
Transformador Reductor :
→ PL = VL * IL * fpL → VL = 6,600
→ IL = 800,0006,600*1 = 121.21 amps IL = 121.21
→ E2 = VL + IL(RX + jXIx) = 6,600 ˚ 121.21 ˚ (0.163 + j1.06)
→ E2 = 6,600 ˚ + 121.21 ˚ (1.072 ˚)
→ E2 = 6,600 ˚ + 129.94 ˚
→ E2 = 6,621
→ E1 = 66,210 → I´L = 12.12
Iexc. H = E1Zexc.H = 66,210 → Iexc. H = 0.909
VG
5.48 ˚
VL
8.17 ˚
IG
→ fpG = cos (8.17 ˚ + 5.48 ˚)
→ fpG = 0.97
VG
PG = /VG/ * /IG/ * cos IG
= 6,781.1 * 126.17 * 0.97
PG = 829,904 watts
→ % η Sistema = pot. en la cargapot. saliendo del generador * 100
= 800,000829,904 * 100
→ % η Sistema = 96.4%
c) Calcular VG, IG, PG, fpG % η sistemas [pic 5]
Datos : PL = 800,000 watts
VL = 6,600 volts IL
FpL = 0.8 (capacitivo)
→ ӨL = cos .1 0.8 = 36.87 ˚ 36.87 ˚
VL
→ VL = 6,600
IL = PLVL*fpL → IL = 800,0006,600*0.8 = 151.51 amps
→IL = 151.51
E2 = VL + IL (Rx + jXIx) = 6,600
→ E2 =6,600
E2 = 6,600
→ E2 = 6,525
→ E1 = 65,250 → I´L = 15.15
Iexc H = 65,250 → Iexc H = 0.896
I1 = I´L + Iexc H = 15.15
→ I1 = 14.5
Transformador Elevador:
E3 = E1 + I1 (2RH + 2jXIH + R linea + JxI Linea)
→ E3 = 65,250
= 65,250
→ E3 = 65,234.22 + j1434.8 -2037.69 + j4.264.44
→ E3 = 63,453
→ E4 =6,345.3 → I´1 = 145
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