Datos

Datos:
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Putil= 500 Hp
Vn=460V
Polos= 4
Eficiencia= 96.2
Fpi=0.86
Fpf= 0.96
𝜃𝑖 = cos −1(𝐹𝑝𝑖)
𝜃𝑓 = 16.26
𝑃𝑖𝑛𝑠𝑡𝑎𝑙𝑎𝑑𝑎 =
(0.746 kw/Hp∗500Hp)
0.96
= 387.73 𝑘𝑤
 Hallamos:
𝑄 = 𝑃𝑖𝑛𝑠𝑡𝑎𝑙𝑎𝑑𝑎(tan 𝜃𝑖 − tan 𝜃𝑓)
𝑄 = 387.73(tan(130.68) − tan(16.26))
𝑄 = 116.72 𝑘𝑉𝐴𝑅
 Convertimos:
𝑉𝑓𝑎 2
4802
𝑄 ′ = 𝑄 ∗ 𝑉𝑑𝑎𝑡𝑜 =116.72 ∗ 460
𝑄 ′ = 127.09 𝑘𝑉𝐴𝑅 → 4𝑥30𝑘𝑉𝐴𝑅
 Convertimos de Nuevo:
𝑉𝑑𝑎𝑡𝑜2
4602
𝑄 = 𝑄′ ∗
= 120 ∗
𝑉𝑓𝑎
480
"
𝑄 " = 110.20 𝑘𝑉𝐴𝑅
 Hallamos:
𝐼𝑛 =
𝑄"
√3 ∗ 𝑉
𝐼𝑛 = 138.32 𝐴
=
110.20𝑘𝑊
√3 ∗ 460
Cable
𝐼𝑑𝑖𝑠𝑒ñ𝑜 = 1.4 𝐼𝑛
𝐼𝑑𝑖𝑠𝑒ñ𝑜 = 193.63 𝐴
3-1x 35mm2 N2H + 1x16 mm2 NH-80 + PVC SAP 85 mm ∅
Interruptor Termomagnetico
𝐼𝑑𝑖𝑠𝑒ñ𝑜 = 1.3 𝐼𝑛
𝐼𝑑𝑖𝑠𝑒ñ𝑜 = 179.8 𝐴
𝑅𝑒𝑔𝑢𝑙𝑎𝑐𝑖𝑜𝑛 𝑇𝑒𝑟𝑚𝑖𝑐𝑎 =
𝐼𝑛
𝐼𝑐𝑜𝑚
𝑅𝑒𝑔𝑢𝑙𝑎𝑐𝑖𝑜𝑛 𝑇𝑒𝑟𝑚𝑖𝑐𝑎 = 0.899
𝑅𝑒𝑔𝑢𝑙𝑎𝑐𝑖𝑜𝑛 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐𝑎 =
3𝑥𝐼𝑛
𝐼𝑐𝑜𝑚
𝑅𝑒𝑔𝑢𝑙𝑎𝑐𝑖𝑜𝑛 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐𝑎 = 2.69
Interrp. Termg. = 3 x 200