10–1. A 120-V, %-hp. 60-Hz, four-pole, split-phase induction motor has the following impedances: R = 1.80 2 R2 = 2.50 2 X, = 2.40 2 X2 = 2.40 2 XM = 602 60 2 At a slip of 0.05, the motor's rotational losses are 51 W. The rotational losses may be assumed constant over the normal operating range of the motor. If the slip is 0.05, find the following quantities for this motor: (a) Input power (b) Air-gap power (c) Peoav (d) Pout (e) Tind conv O Toad (g) Overall motor efficiency (h) Stator power factor
10–1. A 120-V, %-hp. 60-Hz, four-pole, split-phase induction motor has the following impedances: R = 1.80 2 R2 = 2.50 2 X, = 2.40 2 X2 = 2.40 2 XM = 602 60 2 At a slip of 0.05, the motor's rotational losses are 51 W. The rotational losses may be assumed constant over the normal operating range of the motor. If the slip is 0.05, find the following quantities for this motor: (a) Input power (b) Air-gap power (c) Peoav (d) Pout (e) Tind conv O Toad (g) Overall motor efficiency (h) Stator power factor
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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![10-1. A 120-V, %-hp. 60-Hz, four-pole, split-phase induction motor has the following
impedances:
X, = 2.40 2
X2 = 2.40 2
= 60 2
R = 1.80 2
R2 = 2.50 2
XM
At a slip of 0.05, the motor's rotational losses are 5I W. The rotational losses may
be assumed constant over the normal operating range of the motor. If the slip is
0.05, find the following quantities for this motor:
(a) Input power
(b) Air-gap power
(c) Pconv
(d) Pout
(e) Tind
) Toad
(g) Overall motor efficiency
(h) Stator power factor](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2086a314-b7c8-49da-96dd-d6dabb5e471b%2F59006832-c30a-4373-8588-642e9ac1da5a%2F00mp4g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:10-1. A 120-V, %-hp. 60-Hz, four-pole, split-phase induction motor has the following
impedances:
X, = 2.40 2
X2 = 2.40 2
= 60 2
R = 1.80 2
R2 = 2.50 2
XM
At a slip of 0.05, the motor's rotational losses are 5I W. The rotational losses may
be assumed constant over the normal operating range of the motor. If the slip is
0.05, find the following quantities for this motor:
(a) Input power
(b) Air-gap power
(c) Pconv
(d) Pout
(e) Tind
) Toad
(g) Overall motor efficiency
(h) Stator power factor
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