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A 440-V 50-Hz two-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R, = 0.075 2 X = 0.172 R = 0.065 2 Xu = 7.20 X = 0.170 P = 150 W Praw = 1.0 kW Pae = 1.1 kW For a slip of 0.04, find (a) The line current I, (I = I, =149.42 -18.3° A) (b) The stator power factor (0.949 lagging) (c) The rotor power factor (0.995 lagging)

A 440-V 50-Hz two-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R, = 0.075 2 X = 0.172 R = 0.065 2 Xu = 7.20 X = 0.170 P = 150 W Praw = 1.0 kW Pae = 1.1 kW For a slip of 0.04, find (a) The line current I, (I = I, =149.42 -18.3° A) (b) The stator power factor (0.949 lagging) (c) The rotor power factor (0.995 lagging)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Electrical Machine induction motor

4. A 440-V 50-Hz two-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R, = 0.075 2 R = 0.065 2 Xu = 7.2 0 X = 0.17 2 Praw = 1.0 kW X = 0.172 Puin = 150 W ise Pore = 1.1 kW For a slip of 0.04, find (a) The line current I, (I = I, =149.42 -18.3° A) (b) The stator power factor (0.949 lagging) (c) The rotor power factor (0.995 lagging) (d) The stator copper losses Psci (1675 W) (e) The air-gap power PAG (103 kW) ) The power converted from electrical to mechanical form Peous (98.9 kW) (g) The induced torque Tind (327.9 N m) (h) The load torque Tioad (327.6 N m) (i) The overall machine efficiency n (89.4%) ) The motor speed in revolutions per minute and radians per second (301.6 rad/s) For the motor in Problem 4 above, what is the pullout torque? (0.189) What is the slip at the pullout torque? (704NM) What is the rotor speed at the pullout torque? (314.2 rad/s) If the motor in Problem 4 above is to be driven from a 440-V 60-Hz power supply, what will the pullout torque be? (507 Nm) What will the slip be at pullout? (0.159)
Transcribed Image Text:4. A 440-V 50-Hz two-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R, = 0.075 2 R = 0.065 2 Xu = 7.2 0 X = 0.17 2 Praw = 1.0 kW X = 0.172 Puin = 150 W ise Pore = 1.1 kW For a slip of 0.04, find (a) The line current I, (I = I, =149.42 -18.3° A) (b) The stator power factor (0.949 lagging) (c) The rotor power factor (0.995 lagging) (d) The stator copper losses Psci (1675 W) (e) The air-gap power PAG (103 kW) ) The power converted from electrical to mechanical form Peous (98.9 kW) (g) The induced torque Tind (327.9 N m) (h) The load torque Tioad (327.6 N m) (i) The overall machine efficiency n (89.4%) ) The motor speed in revolutions per minute and radians per second (301.6 rad/s) For the motor in Problem 4 above, what is the pullout torque? (0.189) What is the slip at the pullout torque? (704NM) What is the rotor speed at the pullout torque? (314.2 rad/s) If the motor in Problem 4 above is to be driven from a 440-V 60-Hz power supply, what will the pullout torque be? (507 Nm) What will the slip be at pullout? (0.159)
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