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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: Locked rotor, Nominal code letter kVA/hp R = 0.15 Q R2 = 0.10 Q Xị = 0.362 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) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.

A 125-hp, 480-V, 50-Hz, 6-pole, Y-connected induction motor has the per-phase equivalent circuit parameters as follows: Locked rotor, Nominal code letter kVA/hp R = 0.15 Q R2 = 0.10 Q Xị = 0.362 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) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.

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
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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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 Q R2 = 0.10 2 X1 = 0.362 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) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.
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 Q R2 = 0.10 2 X1 = 0.362 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) For a slip of 4% find the line current, the rotor copper losses, the load torque, and the efficiency of the motor under the given conditions.
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