Electricidad

Disponível somente no TrabalhosFeitos
  • Páginas : 2 (447 palavras )
  • Download(s) : 0
  • Publicado : 20 de agosto de 2012
Ler documento completo
Amostra do texto
CIRCUITO EQUIVALENTE DEL MOTOR DE INDUCCION

R1

jX1

jX2 I2 j Xφ
R2 s

+ VLL ∠0o 3 -

I1

PRUEBA DE CD (Resistencia del estator)

I CD
A

V

VCD

R1 =

V 2 I CD
CD PRUEBA DE VACIO (VOLTAJE NOMINAL)
+ +
V

VARIAC TRIFASICO

VNL PNL
+ +

INL
A

R1

jX1

jX2 I2 NL ≈ 0 j Xφ
Z NL = RNL = VNL 2 2 = RNL + X NL 3I NL PNL 2 3 I NL
2 2

+ VNL ∠0 o3 -

I NL

X NL = Z NL − RNL

PROT = PNL − 3 * I NL * R 1
2

CIRCUITO EQUIVALENTE

PRUEBA DE ROTOR BLOQUEADO (CORRIENTE NOMINAL)
+ +
V

Z RB =

VRB 2 2 = RRB + X RB 3I RB PRB 2 3I RB
2 2

VARIAC TRIFASICO

VRB PRB
+ +

IRB
A

RRB =

X RB = Z RB − RRB

R1
+ VRB 3 -

jX1 I1RB = I nom

jX2

Clase A B C D Rotor Devanado

Fracción de XRB X1 X2 0.5 0.50.4 0.6 0.3 0.7 0.5 0.5 0.5 0.5

I2
j Xφ
R2

X φ = X NL − X 1  Xφ  = R 1+ R 2    X 2 + Xφ   
2

RRB

 X + Xφ  R2 = ( RRB − R 1)  2   Xφ   

2

CIRCUITO EQUIVALENTE CALCULO DE PARAMETROS DE UN MOTOR DE INDUCCION
5 HP, 4 POLOS, 220 V, 3φ, 60 HZ, Diseño B, Inom=12.9 A
RESULTADOS DE LAS PRUEBAS
CD 13.8 V 13.0 A VACIO RB VNL=220 V VRB=23.5 V INL=3.87AIRB=12.93 A PNL=200W PRB=469 W f=60 Hz f=15 Hz

Z RB = RRB =

VRB 23.5 = = 1.0493Ω 3I RB 3[12.93] PRB 469 = = 0.9351Ω 2 2 3 I RB 3 [12.93]

13.8 V R1 = = = 0.5308Ω 2 I CD 2 (13)
CD

 60  2 2X RB = Z RB − RRB = 1.04932 − 0.93512   = 1.9042  15  X 1 = 0.4[1.9042] = 0.7617Ω X 2 = 0.6[1.9042] = 1.1425Ω

Z NL = RNL =

VNL 220 = = 32.82Ω 3I NL 3[3.87] PNL 200 = = 4.45Ω 2 2 3 I NL3 [3.87]
2 2 2 2

X φ = X NL − X 1 = 32.52 − 0.7618 = 31.76Ω  X + Xφ  R2 = ( RRB − R 1)  2   Xφ   
2

X NL = Z NL − RNL = 32.82 − 4.45 = 32.52Ω PROT = PNL − 3 * I NL * R 1 = 200 −3[3.87] (0.5308) = 176.15W
2 2

1.1426 + 31.76  R2 = [0.9351 − 0.5308]   = 0.4339 31.76  
j 1.1425 Ω I2 j 31.76 Ω
0.4339 Ω s

2

0.5308 Ω

j 0.7617 Ω

+ 220 ∠0 o 3 -

I1...
tracking img