Answered: A 240 V, compensated shunt motor…

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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A 240 V, compensated shunt motor driving a 1233.96 N.m torque load is running at 19.167 rps. The efficiency of the motor at this load is 94 percent. The combined armature, compensating winding and interpole resistance is 0.00707 ohm, and the resistance of the shunt field is 52.6 ohm. Determine the following parameters of the motor: (i) Input power (in kW) (ii) Field current (in Amp) (iii) Armature current (in Amp) (iv) Induced voltage in armature (in volts) (v) Torque developed by motor (in N.m) and (vi) The resistance of a dynamic breaking resistor (in ohm) that will be capable of developing 678 N.m of braking torque at a speed of 16.667 rps. Assume windage and friction at 16.667 rps is the same as at 19.167 rps. [Hint: 1 lb-ft = 1.356 N.m]"

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