X,X,R,R.X„, E,(0)R -R.Fig. Equivalent Circuit of IMConsider the above equivalent circuit of IM. A eight pole , 215 volt (line to line), 50 Hz, Y connected, three phase IM has the following parameters ona per phase basis :R = 0.175 N, R2 = 0.45 N, X, = 0.325 N, X, = 0.725 , Xm = 25 N, R. = 125 N.%3D%3D%3|The friction and windage loss are 105 W.Determine the followings related to the motor at its rated slip of 2.5 %.a) Excitation Circuit (I,)b) Power developed by motor(Pa)c) Shaft Power(Psha ft)d) Shaft Torque (Tahaft)e) Efficiency(7)O a. Ig = 0.63553 < -80.1546° A, Pa= 2909.43 W, Pahaft = 2779.43 W, Tahaft15.2014 W, 7 =0.823173O b. Ip = 1.53756 < –30.8176° A, Pa = 1909.43 W, Pshaft = 1779.43 W, Tahaft12.1314 W, η-0.758235nQ c. Ip = 2.25759 < –60.4516° A, Pa = 2909.43 W, Pahaft = 2779.43 W, Tshaft= 15.2014 W, 7=0.853285%3DO d. I = 4.94044 < – 79.43° A, Pa= 2379.73 W, Pahaft = 2274.73 W, Tghaft = 29.7053 W, 7=0.801655O e. Ig = 4.94044 < –79.43° A, Pa = 2793.37 W, Pshaft = 2734.27 W, Tshaft = 25.7053 W, 7=0.821738%3D

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The friction and windage loss are 105 W. Determine the followings related to the motor at its rated slip of 2.5 %. 5) Excitation Circuit (I„) -) Power developed by motor(Pa) =) Shaft Power(Paha ft) =) Shaft Torque (Thaft) =) Efficiency(7) O a. I = 0.63553 < –80.1546° A, Pa = 2909.43 W, Pahaft %3D

The friction and windage loss are 105 W. Determine the followings related to the motor at its rated slip of 2.5 %. 5) Excitation Circuit (I„) -) Power developed by motor(Pa) =) Shaft Power(Paha ft) =) Shaft Torque (Thaft) =) Efficiency(7) O a. I = 0.63553 < –80.1546° A, Pa = 2909.43 W, Pahaft %3D

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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