# Monografias

Páginas: 6 (1386 palabras) Publicado: 8 de septiembre de 2012
CALCULOS ELECTRICOS

1.-TABLERO ELECTRICO:
IN= 14517 = 40.7 A ID =1.25 x 40.7 = 50.8 =51 A
1.73 x 220 x 0.9

* Tipo de conductores:
16 mm2 = 75 A
* Distancia real: 25.13 m2
AV = 1.73 X 51 X 0.0175 X 20 = 1.9 Volts < 5.5 Volts16
* Interruptor:
50 A < 60 A < 75 A x 0.8 = 60 A; interruptor: 3 x 60 A
* Conductor de tierra:
16 mm2 = 50 mm2
* Tuberías:
4 --- 16 mm2 --- 25 mm O

* Solución:

3-1 x 16 mm2 THW + 1x 10 mm2 (T)

3x60 A 25mm O – PVC - P

2.-COCINA ELECTRICA:
IN = 7000 = 18.39 A ID = 1.25 x 18.39 = 22.9 = 23 A
1.73 x 220 x 1

* Conductor:
6 mm2 = 36 A
* Distancia: 5.30 m2
AV = 1.73 X 23 X 0.0175 X 5.30 = 0.62 Volts < 3.3 Volts
6
* Interruptor:
6 mm2 ---- 35 A
23 A < 30 A< 35 x 0.8 = 28 A ; interruptor: 3 x 30
* Conductor de tierra:
30 A le corresponde 6 mm2
* Tubería:
3 --- 6 mm2 TW ---- 20 mm. O
* Solución:
3-1 x 6 mm2 THW + 1 x 6 mm2 (T)
3x30 A 20mm O – PVC – L

* Intensidad de corriente:IN = 2000 = 9.09 A ID = 1.25 X 9.09 = 11.4 A
1 x 220 x 1

* Conductor:
2.5 mm2 --- 18 A

* Distancia: 8 m2

AV = 2 X 11.4 X 0.0175 X 8 = 1.28 Volts < 3.3 Volts
2.5

* Interruptor:
8.5 < 20 < 25 x 0.8 = 20 A ; interruptor: 2 x20 A

* Conductor de tierra:
15 A ---- 2.5 mm2

* Tubería:
4---10mm2 TW; le corresponde 20mm O

* Solución:
2-1 x 4.0 mm2 TW + 1 x 4 mm2 (T)
2x20 A 20mm O – PVC – L

IN = 700 =3.9 A ID = 1.25 X 3.9 = 4.7 A
1 x 220 x 0.8

* Conductor:
2.5 mm2 --- 18 A

* Distancia: 5.80 m2

AV = 2 X 6.4 X 0.0175 X 5.80 = 0.51 Volts < 3.3 Volts
2.5

* Interruptor:
6.4 A < 15 A < 18 A x 0.8 = 14.4 A ; interruptor: 2 x 15 A

* Conductor de tierra:
15 A ---- 2.5mm2

* Tubería:
3---2.5mm2 TW; le corresponde15mm O

* Solución:
2-1 x 2.5 mm2 TW + 1 x 2.5 mm2 (T)
2x15 A 15mm O – PVC – L

IN = 2500 = 11.4 A ID = 1.25 X 11.4 =14.25 A
1 x 220 x 1

* Conductor:
2.5 mm2 TW --- 18 A

* Distancia: 5.80 m2

AV = 2 X 14.25 X 0.0175 X 5.80 = 1.16 Volts < 3.3 Volts
2.5

* Interruptor:
14.25 A < 20 A < 25 A x 0.8 = 20 A ; interruptor: 2 x 20 A

* Conductor de tierra:
20 A ---- 4 mm2

* Tubería:
5—4mm2 TW; lecorresponde15mm O

* Solución:
2-1 x 4 mm2 TW + 1 x 4 mm2 (T)
2x20 A 15mm O

6.-Electrobomba:

IN = 1492 = 8.48 A ID = 1.25 X 8.48 = 10.6 A
1 x 220 x 0.8

* Conductor:
2.5 mm2 TW ---...

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