Grupo N : 4

Caso N 3 VIVIENDA

At (1°) = 210 m2 −→210m2

At (2°) = 150 m2 −→ 150m2

At (3°) = 95 m2 −→71.25m2

________

43125 m2

I) CARGAS BASICAS

90 m2 = 2,500 W

90 m2 = 1,000W

90 m2 = 1,000 W

90 m2 = 1,000 W

71.25 m2 = 1,000W ________ 6,500 W

II) Cuadro de cargas básicas

Carga vivienda PI DMCarga básicaCocina eléctrica (3⌀ )Terma eléctrica (2⌀ )Electrobomba (2⌀ )Secadora (2⌀ )reserva

6.500 w8,000 w2,500 w740 w2,500 w1,500 w

6,500 w6,000 w2,500 w186.5 w625 w375 w

total 21,746 w % 186.50 w

Hallamos

I v = DM

K−Vcos⌀ = 16,186.50w

√3 (220v )(0.9)

IN = 47.20 A

Id = 47.20ª (1.20)

Id = 56.64 A

Por tabla A 1 S = 16 mm2

ΔV= K∗Id∗LS

ΔV = √3∗0.0175 ohm∗mm2

m¿(56 .64 Δ)(25m) ¿

16mm2

ΔV = 2.68 v

DIAMETRO DE TUBERIAS

ALIMENTADOR configuaracion Lima = Delta

S = 16 mm2 Nro de cobres = 3+1(neutro ) = 4

DERIVADAS DEL PRIMER PISO

Nro de cobres * monofásica = 2+1 = 3

* trifásicas = 3+1 = 4

a) cocina eléctrica (3⌀ )

In = 8,000w

√3 (220v )(0.9)

In = 23.32 A

Id = 28 A

S = 8 mm2

⌀ =34

= 20 mm2

b) terma eléctrica (2⌀ )

In = 2,500w

1∗(220v )(0.9)

In = 1136 A

Id = 13.63 A

S = 2.5 mm2

⌀ =12

= 15 mm2

C) electrobomba (2⌀ )

In = 2,500w

1∗(220 v )∗1

In = 3.77 A

Id = 4.52 A

S = 2.5 mm2

⌀ =12

= 15 mm2

C) secadora (2⌀ )

In = 2,500w

1∗(220 v )∗(0.9)

In =12.62 A

Id = 15.15 A

S = 2.5 mm2

⌀ =12

= 15 mm2