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Electric Machine; Check image, all question there. Consider the above approximate equivalent circuit for IM. A six-pole, 220Volt (line to line), 50Hz, Y connected, three phase IM has the following on a per phase basis: R1=0.5Ω , R2=0.25 Ω, X1=0.75 Ω, X2=0.5 Ω, Xm=100 Ω, Rc=500 Ω The friction and windage loss is 150 W. Determine the following at its rated slip of 3.5% a)Total Impedance(Zt) b)Rotor Copper Loss c)Core loss(Pm) d)Shaft Power(P0) e)Efficiency

Electric Machine; Check image, all question there. Consider the above approximate equivalent circuit for IM. A six-pole, 220Volt (line to line), 50Hz, Y connected, three phase IM has the following on a per phase basis: R1=0.5Ω , R2=0.25 Ω, X1=0.75 Ω, X2=0.5 Ω, Xm=100 Ω, Rc=500 Ω The friction and windage loss is 150 W. Determine the following at its rated slip of 3.5% a)Total Impedance(Zt) b)Rotor Copper Loss c)Core loss(Pm) d)Shaft Power(P0) e)Efficiency

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Electric Machine; Check image, all question there.

Consider the above approximate equivalent circuit for IM.

A six-pole, 220Volt (line to line), 50Hz, Y connected, three phase IM has the following on a per phase basis:

R1=0.5Ω , R2=0.25 Ω, X1=0.75 Ω, X2=0.5 Ω, Xm=100 Ω, Rc=500 Ω

The friction and windage loss is 150 W. Determine the following at its rated slip of 3.5%

a)Total Impedance(Zt)

b)Rotor Copper Loss

c)Core loss(Pm)

d)Shaft Power(P0)

e)Efficiency

x, X, 4(1) i,(1) R, R. R, Fig. Approximate Equivalent Circuit Consider the above approximate equivalent circuit for IM. A six-pole, 220 volt (line to line), 50 Hz, Y connected, three phase IM has the following on a per phase basis : R = 0.5 N, R2 = 0.25 N, X, = 0.75 N, X, = 0.5 N, Xm = 100 S2, R. = 500 N The friction and windage loss is 150 W. Determine the followings at its rated slip of 3.5 %. a) Total Impedance (Z4) b) Rotor Copper Loss (Prel) c) Core loss (Pm) d) Shaft Power (Po) e) Efficiency (n) O a. Zą = 12.1814 < 10.4529°, Prel = 108.278 W, Pm 105.8 W, P, = 4462.86, 7= 0.879642 а. %3D O b. Z4 = 15.2945 < 42.4529°, P,rel = 218.728 W, Pm = 105.8 W, P, = 4462.86, 7= 0.817942 O . ZĄ = 17.7824 < 15.3345°, Prd = 110.825 W, Pm = 95.75 W, P= 2231.86, 7= 0.687942 O d. Z = 13.1419 < 10.2945°, Prel 108.289 W, Pm 100.5 W, P. = 4142.68, 7= 0.789642 O e. Z = 7.51562 < 13.4028°, Prel = 201.748 W, Pm 96.8 W, P. = 5412.48, 7=0.863989 е.
Transcribed Image Text:x, X, 4(1) i,(1) R, R. R, Fig. Approximate Equivalent Circuit Consider the above approximate equivalent circuit for IM. A six-pole, 220 volt (line to line), 50 Hz, Y connected, three phase IM has the following on a per phase basis : R = 0.5 N, R2 = 0.25 N, X, = 0.75 N, X, = 0.5 N, Xm = 100 S2, R. = 500 N The friction and windage loss is 150 W. Determine the followings at its rated slip of 3.5 %. a) Total Impedance (Z4) b) Rotor Copper Loss (Prel) c) Core loss (Pm) d) Shaft Power (Po) e) Efficiency (n) O a. Zą = 12.1814 < 10.4529°, Prel = 108.278 W, Pm 105.8 W, P, = 4462.86, 7= 0.879642 а. %3D O b. Z4 = 15.2945 < 42.4529°, P,rel = 218.728 W, Pm = 105.8 W, P, = 4462.86, 7= 0.817942 O . ZĄ = 17.7824 < 15.3345°, Prd = 110.825 W, Pm = 95.75 W, P= 2231.86, 7= 0.687942 O d. Z = 13.1419 < 10.2945°, Prel 108.289 W, Pm 100.5 W, P. = 4142.68, 7= 0.789642 O e. Z = 7.51562 < 13.4028°, Prel = 201.748 W, Pm 96.8 W, P. = 5412.48, 7=0.863989 е.
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