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6-5. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 30 hp. Its equivalent circuit components are R = 0.100 2 R, = 0.070 2 XM = 10.0 2 %3D X, = 0.210 2 X, =0.210 2 = 500 W Pnise 0 Pore = 400 W mech For a slip of 0.05, find (a) The line current I, (b) The stator copper losses PSCL (c) The air-gap power PAG

6-5. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 30 hp. Its equivalent circuit components are R = 0.100 2 R, = 0.070 2 XM = 10.0 2 %3D X, = 0.210 2 X, =0.210 2 = 500 W Pnise 0 Pore = 400 W mech For a slip of 0.05, find (a) The line current I, (b) The stator copper losses PSCL (c) The air-gap power PAG

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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6-5. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 30 hp. Its equivalent circuit components are R = 0.100 2 R, = 0.070 2 XM = 10.0 2 %3D X, = 0.210 2 X, =0.210 2 P. mech = 500 W Pnise 0 = 400 W core For a slip of 0.05, find (a) The line current I, (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 Tind () The load torque Tload (g) The overall machine efficiency (h) The motor speed in revolutions per minute and radians per second
Transcribed Image Text:6-5. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 30 hp. Its equivalent circuit components are R = 0.100 2 R, = 0.070 2 XM = 10.0 2 %3D X, = 0.210 2 X, =0.210 2 P. mech = 500 W Pnise 0 = 400 W core For a slip of 0.05, find (a) The line current I, (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 Tind () The load torque Tload (g) The overall machine efficiency (h) The motor speed in revolutions per minute and radians per second
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