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A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Nominal code Locked rotor, letter kVA/hp R = 0.15 2 R2 = 0.10 Q X1 =0.360 X2 = 0:36 2 C 3.55 – 4.00 XM =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 G 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 500 W. a) b) Calculate the speed at the pullout torque in rpm, the pullout torque, and the starting torque using exact Thevenin parameters. c) Find the starting code letter assigned to this motor.

A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Nominal code Locked rotor, letter kVA/hp R = 0.15 2 R2 = 0.10 Q X1 =0.360 X2 = 0:36 2 C 3.55 – 4.00 XM =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 G 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 500 W. a) b) Calculate the speed at the pullout torque in rpm, the pullout torque, and the starting torque using exact Thevenin parameters. c) Find the starting code letter assigned to this motor.

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter32: Three-phase Motors
Section: Chapter Questions
Problem 6RQ: Name three factors that determine the torque produced by an induction motor.
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A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Nominal code Locked rotor, letter kVA/hp R = 0.15 2 R2 = 0.10 2 X1 = 0.360 X2 = 0.36 2 C 3.55 – 4.00 =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 50 W. a) b) Calculate the speed at the pullout torque in rpm, the pullout torque, and the starting torque using exact Thevenin parameters. c) Find the starting code letter assigned to this motor.
Transcribed Image Text:A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Nominal code Locked rotor, letter kVA/hp R = 0.15 2 R2 = 0.10 2 X1 = 0.360 X2 = 0.36 2 C 3.55 – 4.00 =8.362 D 4.00 – 4.50 E 4.50 – 5.00 F 5.00 – 5.60 The core-losses are 1.7 5.60 – 6.30 kW, mechanical losses are 1.8 kW, and stray losses are 50 W. a) b) Calculate the speed at the pullout torque in rpm, the pullout torque, and the starting torque using exact Thevenin parameters. c) Find the starting code letter assigned to this motor.
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