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A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….
- a)5.4 V
- b)18.9 V
- c)72.4 V
- d)26.8 V
Correct answer is option 'B'. Can you explain this answer?
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A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is run...
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A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is run...
Ω, determine the following:
1. Synchronous speed of the motor:
The synchronous speed (Ns) of a synchronous motor is given by the formula:
Ns = (120 * f) / p
where:
f = frequency of the power supply (in Hz)
p = number of poles
Given:
f = 50 Hz
p = 4
Ns = (120 * 50) / 4
Ns = 1500 RPM
Therefore, the synchronous speed of the motor is 1500 RPM.
2. Slip of the motor:
The slip (S) of a synchronous motor is given by the formula:
S = (Ns - N) / Ns
where:
N = actual speed of the motor (in RPM)
Since the motor is running with a light load, we can assume that the actual speed of the motor is very close to the synchronous speed. Therefore, the slip can be considered negligible.
Therefore, the slip of the motor is negligible.
3. Power factor of the motor:
The power factor (PF) of a synchronous motor is given by the formula:
PF = cos(θ)
where:
θ = load angle (in radians)
Given:
θ = 4o = (4 * π) / 180 radians
PF = cos((4 * π) / 180)
PF = cos(0.0698)
PF ≈ 0.9986
Therefore, the power factor of the motor is approximately 0.9986.
4. Armature current:
The armature current (I) of a synchronous motor is given by the formula:
I = (P) / (√3 * V * PF)
where:
P = power (in watts)
V = voltage (in volts)
PF = power factor
Given:
P = 18.65 kW = 18,650 W
V = 220 V
PF ≈ 0.9986
I = (18,650) / (√3 * 220 * 0.9986)
I ≈ 47.65 A
Therefore, the armature current of the motor is approximately 47.65 A.
1. Synchronous speed of the motor:
The synchronous speed (Ns) of a synchronous motor is given by the formula:
Ns = (120 * f) / p
where:
f = frequency of the power supply (in Hz)
p = number of poles
Given:
f = 50 Hz
p = 4
Ns = (120 * 50) / 4
Ns = 1500 RPM
Therefore, the synchronous speed of the motor is 1500 RPM.
2. Slip of the motor:
The slip (S) of a synchronous motor is given by the formula:
S = (Ns - N) / Ns
where:
N = actual speed of the motor (in RPM)
Since the motor is running with a light load, we can assume that the actual speed of the motor is very close to the synchronous speed. Therefore, the slip can be considered negligible.
Therefore, the slip of the motor is negligible.
3. Power factor of the motor:
The power factor (PF) of a synchronous motor is given by the formula:
PF = cos(θ)
where:
θ = load angle (in radians)
Given:
θ = 4o = (4 * π) / 180 radians
PF = cos((4 * π) / 180)
PF = cos(0.0698)
PF ≈ 0.9986
Therefore, the power factor of the motor is approximately 0.9986.
4. Armature current:
The armature current (I) of a synchronous motor is given by the formula:
I = (P) / (√3 * V * PF)
where:
P = power (in watts)
V = voltage (in volts)
PF = power factor
Given:
P = 18.65 kW = 18,650 W
V = 220 V
PF ≈ 0.9986
I = (18,650) / (√3 * 220 * 0.9986)
I ≈ 47.65 A
Therefore, the armature current of the motor is approximately 47.65 A.
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A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer?
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A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer?.
A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 18.65 kW, 220 V, 50 Hz, 4-pole, Y-connected synchronous motor is running with a light load. The load angle is 4o(elect.) and back e.m.f. generated per phase in 110 V. If the armature resistance per phase is 0.1Ω and synchronous reactance/ phase is 1.5Ω then resultant armature voltage/phase is ……………….a)5.4 Vb)18.9 Vc)72.4 Vd)26.8 VCorrect answer is option 'B'. Can you explain this answer?.
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