Reporte 2 de maquinas 2

Universidad Politécnica Salesiana
Sede Guayaquil
Carrera de Ingeniería Eléctrica
Máquinas Eléctricas II
Reporte de Laboratorio
Practica #2
Tema: Características Técnicas de Operación de Motores DC
Alumno: Walter Enrique Segura Gilces
Profesor: Ing. Otto Astudillo
Fecha de la Práctica:
Martes 27/10/2015
Martes 10/11/2015
Fecha de entrega del reporte:
Martes 17/11/2015





Contenido
1. Motorde Excitación Separada o Independiente
1.1 Diagrama Eléctrico

1.2 Foto




1.3 Tabla de Valores Experimentales
Vt [V]
Ia [A]
If [A]
n[rpm]
ĩ carga [N.m]
127
2,9
0,55
1812
0,6
126,5
3,3
0,55
1799
0,8
125,7
4,1
0,55
1774
1
125,6
4,2
0,55
1770
1,25
125,6
4,6
0,55
1764
1,5
125,3
4,9
0,55
1759
1,65
124,5
5,5
0,55
1739
2
124,2
6,2
0,55
1721
2,3
124
6,4
0,55
1714
2,5
123,6
6,8
0,55
1701
2,8
123,4
7,20,55
1688
3
122,7
8,2
0,55
1668
3,6
121,9
9,1
0,55
1637
4,1
121,1
10,3
0,55
1606
4,7
119,2
12,8
0,55
1532
6

1.4 Cálculos
Para Torque de carga 1
Ea = Vt - ∑R * Ia = 125.7 – (0.6+0.3)*4.1 = 122.01 V
Parm = Ea * Ia = 122.01 * 4.1 = 500.24 w
ω = n * (2π/60) = 1774 * (2π/60) = 185.77 rad/seg
ĩ ind = Parm / ω = 500.24 / 185.77 = 2.69 N.m
ĩ roz = ĩ ind - ĩ carga = 2.69 – 1 = 1.69 N.m
Peje = ĩ carga *ω = 1 * 185.77 = 185.77 w
Pcu.arm = ∑R * Ia2 = (0.6+0.3) * (4.1)2 = 15.13 w
Pcu.campo = RNFP * If2 = 226.5 * (0.55)2 = 68.52 w
PVt = Vt * Ia = 125.7 * 4.1 = 515.37 w
Pmec = Parm – Peje = 500.24 – 185.77 = 314.47 w
%n = (Pout / Pin) * 100 = (Peje / PVt) * 100 = (185.77/ 515.37) * 100 =36.05 %
%Rn = [(no – ncarga) / no] * 100 = [(1812 – 1532) / 1812] * 100 =15.45%
Para Torque de carga 1.25
Ea = Vt- ∑R * Ia = 125.6 – (0.6+0.3)*4.2 = 121.82 V
Parm = Ea * Ia = 121.82 * 4.2 = 511.64 w
ω = n * (2π/60) = 1770 * (2π/60) = 185.35 rad/seg
ĩ ind = Pa / ω = 511.64 / 185.35 = 2.76 N.m
ĩ roz = ĩ ind - ĩ carga = 2.76 – 1.25 = 1.51 N.m
Peje = ĩ carga * ω = 1.25 * 185.35 = 231.69 w
Pcu.arm = ∑R * Ia2 = (0.6+0.3) * (4.2)2 = 15.88 w
Pcu.campo = RNFP * If2 = 226.5 * (0.55)2 = 68.52 w
PVt = Vt * Ia = 125.6* 4.2 = 527.52 w
Pmec = Parm – Peje = 511.64 – 231.69 = 279.95 w
%n = (Pout / Pin) * 100 = (Peje / PVt) * 100 = (231.69 / 527.52) * 100 =43.92 %
Para Torque de carga 1.5
Ea = Vt - ∑R * Ia = 125.6 – (0.6+0.3)*4.6 = 121.46 V
Parm = Ea * Ia = 121.46 * 4.6 = 558.72 w
ω = n * (2π/60) = 1764 * (2π/60) = 184.73 rad/seg
ĩ ind = Parm / ω = 558.72 / 184.73 = 3.02 N.m
ĩ roz = ĩ ind - ĩ carga = 3.02 – 1.5= 1.52 N.m
Peje = ĩ carga * ω = 1.5 * 184.73 = 277.09 w
Pcu.arm = ∑R * Ia2 = (0.6+0.3) * (4.6)2 = 19.04 w
Pcu.campo = RNFP * If2 = 226.5 * (0.55)2 = 68.52 w
PVt = Vt * Ia = 125.6 * 4.6 = 577.76 w
Pmec = Parm – Peje = 558.72 – 277.09 = 281.63 w
%n = (Pout / Pin) * 100 = (Peje / PVt) * 100 = (277.09 / 577.76) * 100 =47.96 %
Para Torque de carga 1.65
Ea = Vt - ∑R * Ia = 125.3 –(0.6+0.3)*4.9 = 120.89 V
Parm = Ea * Ia = 120.89 * 4.9 = 592.36 w
ω = n * (2π/60) = 1759 * (2π/60) = 184.2 rad/seg
ĩ ind = Parm / ω = 592.36 / 184.2 = 3.22 N.m
ĩ roz = ĩ ind - ĩ carga = 3.22 – 1.65 = 1.57 N.m
Peje = ĩ carga * ω = 1.65 * 184.2 = 303.93 w
Pcu.arm = ∑R * Ia2 = (0.6+0.3) * (4.9)2 = 21.61 w
Pcu.campo = RNFP * If2 = 226.5 * (0.55)2 = 68.52 w
PVt = Vt * Ia = 125.3 * 4.9 = 613.97 w
Pmec = Parm –Peje = 592.36 – 303.93 = 288.43 w
%n = (Pout / Pin) * 100 = (Peje / PVt) * 100 = (303.93 / 613.97) * 100 =49.5 %
Para Torque de carga 2
Ea = Vt - ∑R * Ia = 124.5 – (0.6+0.3)*5.5 = 119.55 V
Parm = Ea * Ia = 119.55 * 5.5 = 657.53 w
ω = n * (2π/60) = 1739 * (2π/60) = 182.11 rad/seg
ĩ ind = Parm / ω = 657.53 / 182.11 = 3.61 N.m
ĩ roz = ĩ ind - ĩ carga = 3.61 – 2 = 1.61 N.m
Peje = ĩ carga * ω = 2 *182.11 = 364.22 w
Pcu.arm = ∑R * Ia2 = (0.6+0.3) * (5.5)2 = 27.23 w
Pcu.campo = RNFP * If2 = 226.5 * (0.55)2 = 68.52 w
PVt = Vt * Ia = 124.5 * 5.5 = 684.75 w
Pmec = Parm – Peje = 657.53 – 364.22 = 293.31 w
%n = (Pout / Pin) * 100 = (Peje / PVt) * 100 = (364.22 / 684.75) * 100 =53.19 %

1.5 Flujo de Potencias









1.6 Tabla de Resultados







1.7 Curvas de Características de Operación...