A 20hp, 240V, two-pole and 60-Hz, Y-connected wound- rotor induction motor. Its equivalent circuit components are Rj=0.260, R2=0.180, Xı=0.470, X2=0.482, Xm=15.60, Pmech=550W, Pcore=480W, and Pmisc=120W. For a slip of :0.045, find a) The stator copper losses PSCL and the air-gap power РAG b) The power converted from electrical to mechanical form Pconv c) The induced torque Tind and the load torque t]oad d) The overall machine efficiency e) The motor speed

A 20hp, 240V, two-pole and 60-Hz, Y-connected wound- rotor induction motor. Its equivalent circuit components are Rj=0.260, R2=0.180, Xı=0.470, X2=0.482, Xm=15.60, Pmech=550W, Pcore=480W, and Pmisc=120W. For a slip of :0.045, find a) The stator copper losses PSCL and the air-gap power РAG b) The power converted from electrical to mechanical form Pconv c) The induced torque Tind and the load torque t]oad d) The overall machine efficiency e) The motor speed

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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For a 18600W 2400rpm DC motor, the following values are given: Ra = 0.089 2 (value at room temperature (250C)), La = 0.00144 H, KT = 0.85 Nm / A and KV = 0.851 V / (rad / s)), (In calculations Use the hot resistance values at 1000C.) (Take the rated field winding losses as Wf= 210W.) If this DC machine is operating as a motor with rated flux at rated speed, calculate the rated torque (TR), rated current (IR), rated voltage (VR), efficiency (n) at the rated point, taking into account the winding losses. Find the speed (@NL) of this motor in no-load condition. TR Nm IR А VR V Eff = % ONL= rpm
Induct.motor.
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