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A 460-V, 10 hp, four-pole, Y-connected, induction motor has the following parameters (R1 = 0.54, R2 = 0.488 , X M = 51.12 ,X1 = 2.093 and X2 = 3.209) ohm P( %3D F&W) = 150 W, P(misc)=50 W, P(core) = 150 kW For a slip of 0.02, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form P(conv) (e) The induced torque (f) The * load torque

A 460-V, 10 hp, four-pole, Y-connected, induction motor has the following parameters (R1 = 0.54, R2 = 0.488 , X M = 51.12 ,X1 = 2.093 and X2 = 3.209) ohm P( %3D F&W) = 150 W, P(misc)=50 W, P(core) = 150 kW For a slip of 0.02, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form P(conv) (e) The induced torque (f) The * load torque

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A 460-V, 10 hp, four-pole, Y-connected, induction motor has the following parameters (R1 = 0.54, R2 = 0.488 , X M = %3D 51.12 ,X1 = 2.093 and X2 = 3.209) ohm P( F&W) = 150 W, P(misc)=50 W, P(core) = 150 kW For a slip of 0.02, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form P(conv) (e) The induced torque (f) The load torque
Transcribed Image Text:A 460-V, 10 hp, four-pole, Y-connected, induction motor has the following parameters (R1 = 0.54, R2 = 0.488 , X M = %3D 51.12 ,X1 = 2.093 and X2 = 3.209) ohm P( F&W) = 150 W, P(misc)=50 W, P(core) = 150 kW For a slip of 0.02, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG (d) The power converted from electrical to mechanical form P(conv) (e) The induced torque (f) The load torque
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