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400 00 R. Fig. Equivalent Circuit of IM Consider the above equivalent circuit of IM. A eight pole. 215 volt (ine to line). 50 Hz Y connected. three phase IM has the following parameters on a per phase basi : R = 0.175 N, R2 = 0.45 N, X, = 0.325 N, Xz = 0.725 2, X, = 25 1, Rq = 125 N. 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 (a) b) Power devaloped by motorl Pa) O Shaft Power(Pahaft d) Shaft Torque (Ttafe) el Efficiencytm) O a la - 2.25759 < -60.4516° A, Pa – 2909.43 W, Pu- 2779.43 W, Ttat- 15.2014 W, 7 - 0.853285 O b. I = 4.94044 < -79.43° A, Pa = 2379.73 W, Pahaft= 2274.73 W, Ttaft = 29.7053 W, 7 = 0.801655 O cla = 4.94044 < -79.43" A, Pa = 2793.37 W, Pahaft = 2734.27 W, Tahaft = 25.7053 W, n= 0.821738 O d. I = 0.63553 < -80.1546" A, P = 2909.43 W, Pahajt = 2779.43 W, Tituft = 15.2014 W, 7= 0.823173 O e la- 1.53756 - -30.8176 A, P- 1909.43 W, Pataft 1779.43 W, Ttaf- 12.1314 W, n-0.758235

400 00 R. Fig. Equivalent Circuit of IM Consider the above equivalent circuit of IM. A eight pole. 215 volt (ine to line). 50 Hz Y connected. three phase IM has the following parameters on a per phase basi : R = 0.175 N, R2 = 0.45 N, X, = 0.325 N, Xz = 0.725 2, X, = 25 1, Rq = 125 N. 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 (a) b) Power devaloped by motorl Pa) O Shaft Power(Pahaft d) Shaft Torque (Ttafe) el Efficiencytm) O a la - 2.25759 < -60.4516° A, Pa – 2909.43 W, Pu- 2779.43 W, Ttat- 15.2014 W, 7 - 0.853285 O b. I = 4.94044 < -79.43° A, Pa = 2379.73 W, Pahaft= 2274.73 W, Ttaft = 29.7053 W, 7 = 0.801655 O cla = 4.94044 < -79.43" A, Pa = 2793.37 W, Pahaft = 2734.27 W, Tahaft = 25.7053 W, n= 0.821738 O d. I = 0.63553 < -80.1546" A, P = 2909.43 W, Pahajt = 2779.43 W, Tituft = 15.2014 W, 7= 0.823173 O e la- 1.53756 - -30.8176 A, P- 1909.43 W, Pataft 1779.43 W, Ttaf- 12.1314 W, n-0.758235

Introductory Circuit Analysis (13th Edition)
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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400 R. X, E0 Fig. Equivalent Circuit of IM Consider 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 on a per phase basis : R = 0.175 12, R2 = 0.45 fN, X, = 0.325 fN, X2 - 0.725 (2, X, = 25 1, Re = 125 f2. 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 (Ia) b) Power developed by motor( Pa) O Shaft Power(Pahafe) d) Shaft Torque (Ttafa) el Efficiencytm) O a la - 2.25759 - 60.4516° A, Pa- 2909.43 W, Paar- 2779.43 W, Tyhaft- 15.2014 W, 7 - 0.853285 O b. I = 4.94044 < -79.43° A, Pa = 2379.73 W, Pahaft = 2274.73 W, Tuhbaft = 29.7053 W, 1 = 0.801655 O cla = 4.94044 < -79.43" A, Pa = 2793.37 W, Pshaft = 2734.27 W, Tahaft = 25.7053 W, n= 0.821738 O d. I = 0.63553 -80.1546" A, P = 2909.43 W, Phajt = 2779.43 W, Ttaft = 15.2014 W, 7= 0.823173 O e l = 1.53756 -30.8176" A, P- 1909.43 W, Patuft- 1779.43 W, Tataft= 12.1314 W, 7= 0.758235
Transcribed Image Text:400 R. X, E0 Fig. Equivalent Circuit of IM Consider 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 on a per phase basis : R = 0.175 12, R2 = 0.45 fN, X, = 0.325 fN, X2 - 0.725 (2, X, = 25 1, Re = 125 f2. 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 (Ia) b) Power developed by motor( Pa) O Shaft Power(Pahafe) d) Shaft Torque (Ttafa) el Efficiencytm) O a la - 2.25759 - 60.4516° A, Pa- 2909.43 W, Paar- 2779.43 W, Tyhaft- 15.2014 W, 7 - 0.853285 O b. I = 4.94044 < -79.43° A, Pa = 2379.73 W, Pahaft = 2274.73 W, Tuhbaft = 29.7053 W, 1 = 0.801655 O cla = 4.94044 < -79.43" A, Pa = 2793.37 W, Pshaft = 2734.27 W, Tahaft = 25.7053 W, n= 0.821738 O d. I = 0.63553 -80.1546" A, P = 2909.43 W, Phajt = 2779.43 W, Ttaft = 15.2014 W, 7= 0.823173 O e l = 1.53756 -30.8176" A, P- 1909.43 W, Patuft- 1779.43 W, Tataft= 12.1314 W, 7= 0.758235
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