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A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution?
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A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A fr...
Problem: A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed.
Solution:
Given:
- Number of poles (P) = 4
- Supply voltage (V) = 220V
- Armature current (Ia) = 32A
- Field current (If) = 1A
- Output power (Pout) = 5.5 kW
- Armature resistance (Ra) = 0.09 Ω
- Flux per pole (Φ) = 30 mWb
- Number of conductors (N) = 540
Formula:
- Speed (N) = (120 * f) / P
- Torque (T) = (Pout * 1000) / N
- Back EMF (Eb) = (Φ * P * N) / (60 * 10^3)
- Armature voltage (Va) = V - (Ia * Ra)
- Total current (I) = Ia + If
- Power input (Pin) = V * I
- Efficiency (η) = Pout / Pin
Calculation:
- Calculating frequency (f) from supply voltage (V) = 220V
- f = 50 Hz
- Calculating speed (N)
- N = (120 * f) / P
- N = (120 * 50) / 4
- N = 1500 rpm
- Calculating back EMF (Eb)
- Eb = (Φ * P * N) / (60 * 10^3)
- Eb = (30 * 10^-3 * 4 * 1500) / (60 * 10^3)
- Eb = 0.3 V
- Calculating armature voltage (Va)
- Va = V - (Ia * Ra)
- Va = 220 - (32 * 0.09)
- Va = 217.72 V
- Calculating total current (I)
- I = Ia + If
- I = 32 + 1
- I = 33 A
- Calculating power input (Pin)
- Pin = V * I
- Pin = 220 * 33
- Pin = 7260 W
- Calculating efficiency (η)
- η = Pout / Pin
- η = (5.5 * 10^3) / 7260
- η = 0.758
- Calculating torque (T)
- T = (Pout * 1000) / N
- T = (5.5 * 10^3) / 1500
- T = 3.67 Nm
Answer:
- Speed (N) = 1500 rpm
- Torque (T) = 3.67 N
Solution:
Given:
- Number of poles (P) = 4
- Supply voltage (V) = 220V
- Armature current (Ia) = 32A
- Field current (If) = 1A
- Output power (Pout) = 5.5 kW
- Armature resistance (Ra) = 0.09 Ω
- Flux per pole (Φ) = 30 mWb
- Number of conductors (N) = 540
Formula:
- Speed (N) = (120 * f) / P
- Torque (T) = (Pout * 1000) / N
- Back EMF (Eb) = (Φ * P * N) / (60 * 10^3)
- Armature voltage (Va) = V - (Ia * Ra)
- Total current (I) = Ia + If
- Power input (Pin) = V * I
- Efficiency (η) = Pout / Pin
Calculation:
- Calculating frequency (f) from supply voltage (V) = 220V
- f = 50 Hz
- Calculating speed (N)
- N = (120 * f) / P
- N = (120 * 50) / 4
- N = 1500 rpm
- Calculating back EMF (Eb)
- Eb = (Φ * P * N) / (60 * 10^3)
- Eb = (30 * 10^-3 * 4 * 1500) / (60 * 10^3)
- Eb = 0.3 V
- Calculating armature voltage (Va)
- Va = V - (Ia * Ra)
- Va = 220 - (32 * 0.09)
- Va = 217.72 V
- Calculating total current (I)
- I = Ia + If
- I = 32 + 1
- I = 33 A
- Calculating power input (Pin)
- Pin = V * I
- Pin = 220 * 33
- Pin = 7260 W
- Calculating efficiency (η)
- η = Pout / Pin
- η = (5.5 * 10^3) / 7260
- η = 0.758
- Calculating torque (T)
- T = (Pout * 1000) / N
- T = (5.5 * 10^3) / 1500
- T = 3.67 Nm
Answer:
- Speed (N) = 1500 rpm
- Torque (T) = 3.67 N
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A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution?
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A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? 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 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution?.
A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? 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 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution?.
Solutions for A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution?, a detailed solution for A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? has been provided alongside types of A 4 pole 220V shunt motor has 540 lap wound conductor. It takes 32A from the supply main and develops output power of 5.5rW. The field winding takes 1A. Armature resistance is 0.09Ω and flux per pole is 30mWb. Find: a. speed. b. Torque developed What will be the solution? theory, EduRev gives you an
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