40 R, ww 4(1) R, (の) X_ R. の Fig. Approximate Equivalent Circuit Consider the above approximate equivalent circuit for IM. A six-pole , 230 volt (line to line), 60 Hz, Y connected, three phase IM has the following on a per phase basis: R = 0.5 2, R2 = 0.25 2, X = 0.75 2, X2 = 0.5 n, Xm = 100 2, R. = 500 N %3D %3D The friction and windage loss is 150 W. Determine the followings at its rated slip of 2.5 %.

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.30P: Reconsider Problem 3.29. If Va,VbandVc are a negative-sequence set, how would the voltage and...
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4(0)
1,0 R,
R,
R,
V R
の
X_
の
Fig. Approximate Equivalent Circuit
Consider the above approximate equivalent circuit for IM.
A six-pole , 230 volt (line to line), 60 Hz, Y connected, three phase IM has the following on a per phase basis:
R = 0.5 N, R2 = 0.25 2, X, = 0.75 , X2 = 0.5 N, Xm = 100 2, R. = 500 2
The friction and windage loss is 150 W. Determine the followings at its rated slip of 2.5 %.
a) Total Impedance (Z,)
b) Rotor Copper Loss (Prd)
c) Core loss (Pm)
d) Shaft Power (P.)
e) Efficiency (7)
O a. Z = 10.1814 < 12.4529°, Pred = 118.278 W, Pm = 105.8 W, Po = 4462.86, 7 = 0.879642
%3D
%3D
O b. Z = 13.1419 < 10.2945°, Prd = 108.289 W, Pm = 100.5 W, P. = 4142.68, n = 0.789642
%3D
%3D
%3D
%3D
%3D
O c. ZĄ = 12.1814 < 10.4529°, Prel
108.278 W, Pm
105.8 W, P. = 4462.86, 7 = 0.879642
%3D
%3D
Transcribed Image Text:4(0) 1,0 R, R, R, V R の X_ の Fig. Approximate Equivalent Circuit Consider the above approximate equivalent circuit for IM. A six-pole , 230 volt (line to line), 60 Hz, Y connected, three phase IM has the following on a per phase basis: R = 0.5 N, R2 = 0.25 2, X, = 0.75 , X2 = 0.5 N, Xm = 100 2, R. = 500 2 The friction and windage loss is 150 W. Determine the followings at its rated slip of 2.5 %. a) Total Impedance (Z,) b) Rotor Copper Loss (Prd) c) Core loss (Pm) d) Shaft Power (P.) e) Efficiency (7) O a. Z = 10.1814 < 12.4529°, Pred = 118.278 W, Pm = 105.8 W, Po = 4462.86, 7 = 0.879642 %3D %3D O b. Z = 13.1419 < 10.2945°, Prd = 108.289 W, Pm = 100.5 W, P. = 4142.68, n = 0.789642 %3D %3D %3D %3D %3D O c. ZĄ = 12.1814 < 10.4529°, Prel 108.278 W, Pm 105.8 W, P. = 4462.86, 7 = 0.879642 %3D %3D
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