Electric Machine-check image, all question there.Consider the above equivalent circuit of IM, A six pole, 250 volt(line to line), 60Hz, Y connected , three phase IM has the following parameters on a per phase basis:R1=0.125Ω , R2=0.45 Ω, X1=0.5 Ω, X2=0.75 Ω, Xm=50 Ω, Rc=225 ΩThe friction and windage loss is 115 W.Determine the following at its rated slip of 1.5%a)Excitation Circuit b)Power developed by motor(Pd)c)Shaft Power(Pshaft)d)Shaft Torque(Tshaft)e)Efficiency X,X,R,R,R,X_ E,(r)R,Fig. Equivalent Circuit of IMConsider the above equivalent circuit of IM. A six pole , 250 volt (line to line), 60 Hz, Y connected, three phase IM has the following parameterson a per phase basis :R = 0.125 N, R2 = 0.45 N, X1 = 0.5 N, X2 = 0.75 N, Xm = 50 N, R. = 225 N.The friction and windage loss are 115 W.Determine the followings related to the motor at its rated slip of 1.5 %.a) Excitation Circuit (I)b) Power developed by motor(Pa)c) Shaft Power(Pahaft)d) Shaft Torque (Tshaft)e) Efficiency(7)O a. Is = 2.9127 < –78.3897° A, Pa = 1989.59 W, Pshaft = 1874.59 W, Tzhaft = 15.1447 W, 7=0.813811O b. Ig = 2.33553 < –75.1738° A, Pa=2150.37 W, Pshaft = 3779.43 W, T,haft = 25.2737 W, 7=0.883285O . I = 2.25759 < –60.4516° A, Pa=2909.43 W, Pshaft = 2779.43 W, T,haft = 15.2014 W, 7=0.853285O d. I = 0.63553 < -80.1546° A, Pa=2909.43 w, Pshaft = 2779.43 W, T,haft = 15.2014 W, 7=0.823173%3D%3DO e. I = 1.53756 < -30.8176° A, Pa= 1909.43 W, Pahaft= 1779.43 W, Thaft = 12.1314 W, 7= 0.758235

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Introductory Circuit Analysis (13th Edition)

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ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Electric Machine-check image, all question there.

Consider the above equivalent circuit of IM, A six pole, 250 volt(line to line), 60Hz, Y connected , three phase IM has the following parameters on a per phase basis:

R₁=0.125Ω , R₂=0.45 Ω, X₁=0.5 Ω, X₂=0.75 Ω, X_m=50 Ω, R_c=225 Ω

The friction and windage loss is 115 W.

Determine the following at its rated slip of 1.5%

a)Excitation Circuit

b)Power developed by motor(P_d)

c)Shaft Power(Pshaft)

d)Shaft Torque(Tshaft)

e)Efficiency

X,
X,
R,
R,
R,
X_ E,(r)
R,
Fig. Equivalent Circuit of IM
Consider the above equivalent circuit of IM. A six pole , 250 volt (line to line), 60 Hz, Y connected, three phase IM has the following parameters
on a per phase basis :
R = 0.125 N, R2 = 0.45 N, X1 = 0.5 N, X2 = 0.75 N, Xm = 50 N, R. = 225 N.
The friction and windage loss are 115 W.
Determine the followings related to the motor at its rated slip of 1.5 %.
a) Excitation Circuit (I)
b) Power developed by motor(Pa)
c) Shaft Power(Pahaft)
d) Shaft Torque (Tshaft)
e) Efficiency(7)
O a. Is = 2.9127 < –78.3897° A, Pa = 1989.59 W, Pshaft = 1874.59 W, Tzhaft = 15.1447 W, 7=0.813811
O b. Ig = 2.33553 < –75.1738° A, Pa=2150.37 W, Pshaft = 3779.43 W, T,haft = 25.2737 W, 7=0.883285
O . I = 2.25759 < –60.4516° A, Pa=2909.43 W, Pshaft = 2779.43 W, T,haft = 15.2014 W, 7=0.853285
O d. I = 0.63553 < -80.1546° A, Pa=2909.43 w, Pshaft = 2779.43 W, T,haft = 15.2014 W, 7=0.823173
%3D
%3D
O e. I = 1.53756 < -30.8176° A, Pa= 1909.43 W, Pahaft
= 1779.43 W, Thaft = 12.1314 W, 7= 0.758235

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