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A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque?
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A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating...
Solution:
Given data:
- Supply voltage (V): 460 V
- Power (P): 25 hp or 18.65 kW
- Frequency (f): 60 Hz
- Number of poles (p): 4
- Input power to the rotor (Pr): 14.58 kW
- Rotor copper losses (Pc): 263 W
- Combined losses (Ploss): 197 W
(a) Shaft speed:
The synchronous speed of the motor can be calculated using the formula:
Ns = 120f/p
where Ns is the synchronous speed in rpm, f is the frequency in Hz, and p is the number of poles.
Ns = 120 x 60/4 = 1800 rpm
The actual speed of the motor is given by:
N = (1 - slip)Ns
where N is the actual speed in rpm, and slip is the difference between the synchronous speed and the rotor speed, expressed as a fraction of the synchronous speed.
The rotor copper losses can be calculated using the formula:
Pc = 3I^2Rr
where I is the rotor current, and Rr is the rotor resistance.
The input power to the rotor can be written as:
Pr = P + Pc + Ploss
Substituting the given values, we get:
18.65 = 14.58 + 0.263 + 0.197
Solving for slip, we get:
slip = (1800 - N)/1800 = 0.213
Therefore, the actual speed of the motor is:
N = (1 - 0.213) x 1800 = 1419.4 rpm
(b) Mechanical power developed:
The mechanical power developed by the motor can be calculated using the formula:
Pm = (1 - s)Pr
where s is the slip, and Pr is the input power to the rotor.
Substituting the given values, we get:
Pm = (1 - 0.213) x 14.58 = 11.53 kW
Therefore, the mechanical power developed by the motor is 11.53 kW.
(c) Developed torque:
The developed torque of the motor can be calculated using the formula:
T = Pm/(2πN/60)
where T is the developed torque in Nm.
Substituting the given values, we get:
T = 11.53/(2π x 1419.4/60) = 47.29 Nm
Therefore, the developed torque of the motor is 47.29 Nm.
Given data:
- Supply voltage (V): 460 V
- Power (P): 25 hp or 18.65 kW
- Frequency (f): 60 Hz
- Number of poles (p): 4
- Input power to the rotor (Pr): 14.58 kW
- Rotor copper losses (Pc): 263 W
- Combined losses (Ploss): 197 W
(a) Shaft speed:
The synchronous speed of the motor can be calculated using the formula:
Ns = 120f/p
where Ns is the synchronous speed in rpm, f is the frequency in Hz, and p is the number of poles.
Ns = 120 x 60/4 = 1800 rpm
The actual speed of the motor is given by:
N = (1 - slip)Ns
where N is the actual speed in rpm, and slip is the difference between the synchronous speed and the rotor speed, expressed as a fraction of the synchronous speed.
The rotor copper losses can be calculated using the formula:
Pc = 3I^2Rr
where I is the rotor current, and Rr is the rotor resistance.
The input power to the rotor can be written as:
Pr = P + Pc + Ploss
Substituting the given values, we get:
18.65 = 14.58 + 0.263 + 0.197
Solving for slip, we get:
slip = (1800 - N)/1800 = 0.213
Therefore, the actual speed of the motor is:
N = (1 - 0.213) x 1800 = 1419.4 rpm
(b) Mechanical power developed:
The mechanical power developed by the motor can be calculated using the formula:
Pm = (1 - s)Pr
where s is the slip, and Pr is the input power to the rotor.
Substituting the given values, we get:
Pm = (1 - 0.213) x 14.58 = 11.53 kW
Therefore, the mechanical power developed by the motor is 11.53 kW.
(c) Developed torque:
The developed torque of the motor can be calculated using the formula:
T = Pm/(2πN/60)
where T is the developed torque in Nm.
Substituting the given values, we get:
T = 11.53/(2π x 1419.4/60) = 47.29 Nm
Therefore, the developed torque of the motor is 47.29 Nm.
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A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque?
Question Description
A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque? 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 A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque?.
A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque? 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 A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A three-phase 460-v, 25-hp, 60-Hz, four-pole induction motor operating at reduced load requires 14.58-kW input to the rotor. The rotor copper losses are 263 W, and the combined friction, windage, and stray power losses are 197 W. Determine, (a) Shaft speed (b) Mechanical power developed (c) Developed torque?.
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