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.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Question
![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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc176afbe-1520-4dca-a8b8-3efd3b2e5db4%2Faf119d06-43ed-4c2a-931e-66ab11154d1b%2Fbtp4do_processed.jpeg&w=3840&q=75)
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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