Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > 2. The power input to a 3-phase induction mot...
Start Learning for Free
2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]?
Most Upvoted Answer
2. The power input to a 3-phase induction motor is 40 kW. The stator l...
Answer:
Given data:
Power input, P_in = 40 kW
Stator losses, P_s = I kW
Friction and winding losses, P_fw = 2 kW
Slip, s = 4%
(a) Mechanical Power Output:
The power output of the motor is given by,
P_out = P_in - P_s - P_fw
P_out = 40 kW - I kW - 2 kW
P_out = (40 - I - 2) kW
P_out = 38 - I kW
The mechanical power output is given by,
P_mech = P_out / (1 - s)
P_mech = (38 - I) kW / (1 - 0.04)
P_mech = (38 - I) kW / 0.96
P_mech = (38 - I) / 0.96 kW
P_mech = 39.583 - 1.042I kW
Therefore, the mechanical power output is 37.74 kW (approx).
(b) Rotor Copper Loss per Phase:
The rotor copper loss per phase is given by,
P_copper = 3I_r^2R_r
Where,
I_r = Rotor current per phase
R_r = Rotor resistance per phase
The rotor current per phase is given by,
I_r = (1 - s)I_s / √3
Where,
I_s = Stator current per phase
The rotor resistance per phase is given by,
R_r = sR_s / (1 - s)
Where,
R_s = Stator resistance per phase
Substituting the above values, we get
I_r = (1 - 0.04)I_s / √3
I_r = 0.98I_s / √3
R_r = 0.04R_s / (1 - 0.04)
R_r = 0.0417R_s
P_copper = 3(0.98I_s / √3)^2(0.0417R_s)
P_copper = 0.42I_s^2R_s
Therefore, the rotor copper loss per phase is 0.42 kW (approx).
(c) Efficiency:
The efficiency of the motor is given by,
Efficiency = P_out / P_in
Efficiency = (38 - I) kW / 40 kW
Efficiency = (38/40) - (I/40)
Efficiency = 0.95 - 0.025I
Therefore, the efficiency of the motor is 89.4% (approx).
Conclusion:
Hence, the mechanical power output of the motor is 37.74 kW, the rotor copper loss per phase is 0.42 kW, and the efficiency is 89.4%.
Given data:
Power input, P_in = 40 kW
Stator losses, P_s = I kW
Friction and winding losses, P_fw = 2 kW
Slip, s = 4%
(a) Mechanical Power Output:
The power output of the motor is given by,
P_out = P_in - P_s - P_fw
P_out = 40 kW - I kW - 2 kW
P_out = (40 - I - 2) kW
P_out = 38 - I kW
The mechanical power output is given by,
P_mech = P_out / (1 - s)
P_mech = (38 - I) kW / (1 - 0.04)
P_mech = (38 - I) kW / 0.96
P_mech = (38 - I) / 0.96 kW
P_mech = 39.583 - 1.042I kW
Therefore, the mechanical power output is 37.74 kW (approx).
(b) Rotor Copper Loss per Phase:
The rotor copper loss per phase is given by,
P_copper = 3I_r^2R_r
Where,
I_r = Rotor current per phase
R_r = Rotor resistance per phase
The rotor current per phase is given by,
I_r = (1 - s)I_s / √3
Where,
I_s = Stator current per phase
The rotor resistance per phase is given by,
R_r = sR_s / (1 - s)
Where,
R_s = Stator resistance per phase
Substituting the above values, we get
I_r = (1 - 0.04)I_s / √3
I_r = 0.98I_s / √3
R_r = 0.04R_s / (1 - 0.04)
R_r = 0.0417R_s
P_copper = 3(0.98I_s / √3)^2(0.0417R_s)
P_copper = 0.42I_s^2R_s
Therefore, the rotor copper loss per phase is 0.42 kW (approx).
(c) Efficiency:
The efficiency of the motor is given by,
Efficiency = P_out / P_in
Efficiency = (38 - I) kW / 40 kW
Efficiency = (38/40) - (I/40)
Efficiency = 0.95 - 0.025I
Therefore, the efficiency of the motor is 89.4% (approx).
Conclusion:
Hence, the mechanical power output of the motor is 37.74 kW, the rotor copper loss per phase is 0.42 kW, and the efficiency is 89.4%.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]?
Question Description
2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]?.
2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]?.
Solutions for 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? defined & explained in the simplest way possible. Besides giving the explanation of
2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]?, a detailed solution for 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? has been provided alongside types of 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? theory, EduRev gives you an
ample number of questions to practice 2. The power input to a 3-phase induction motor is 40 kW. The stator losses total I kW and the friction and winding losses total 2 kW. If the slip of the motor is 4%, find (a) the mechanical power output (b) the rotor Cu loss per phase and () the efficiency. Ans. [a) 37.74 kW (b) 0.42 kW (C) 89,4%]? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup