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A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW.
The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).
The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).
Correct answer is '245'. Can you explain this answer?
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A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a...
Given Data
- Voltage (V) = 400 V
- Power (P) = 50 kVA
- Frequency (f) = 50 Hz
- Synchronous Reactance (Xs) = 1 ohm/phase
- Shaft load (Pload) = 10 kW
- Power factor (pf) = 0.8 leading
- Loss in motor (Ploss) = 2 kW
To find
- Per phase excitation emf of the motor
Formula used
- Apparent power (S) = √3 × Vph × Iph
- Power factor (pf) = P / S
- Reactive power (Q) = √(S² - P²)
- Impedance (Z) = Vph / Iph
- Impedance angle (θ) = tan⁻¹ (Q / P)
- Synchronous speed (Ns) = f × 120 / p
- Per phase emf (Eph) = Vph - Iph × Xs
- Torque (T) = (Pload / ω) = (3 × Eph × Iph × sinθ / ω)
- Excitation voltage (Eexc) = √(Eph² + Vph²)
Calculation
- Apparent power (S) = 50 kVA
- Power factor (pf) = 0.8 leading
- Real power (P) = Pload = 10 kW
- Reactive power (Q) = √(S² - P²) = √(50² - 10²) = 48.5 kVAR
- Impedance angle (θ) = tan⁻¹ (Q / P) = tan⁻¹ (48.5 / 10) = 78.7°
- Cosine of θ (cosθ) = pf = 0.8
- Sine of θ (sinθ) = √(1 - cos²θ) = √(1 - 0.8²) = 0.6
- Impedance (Z) = Vph / Iph
- Impedance angle (θ) = tan⁻¹ (Xs / (Z - Xs))
- tan⁻¹ (1 / (Z - 1)) = 78.7°
- Z = Xs / tan(78.7°) + 1 = 1.77 ohm
- Voltage per phase (Vph) = V / √3 = 400 / √3 = 230.9 V
- Current per phase (Iph) = P / (3 × Vph × pf) = 10 / (3 × 230.9 × 0.8) = 0.018 A
- Per phase emf (Eph) = Vph - Iph × Xs = 230.9 - 0.018 × 1 = 230.9 V
- Synchronous speed (Ns) = f × 120 / p = 50 × 120 / 2 = 3000 rpm
- Angular speed (ω) = 2 × π × Ns / 60 = 2 × π × 3000 / 60 = 314
- Voltage (V) = 400 V
- Power (P) = 50 kVA
- Frequency (f) = 50 Hz
- Synchronous Reactance (Xs) = 1 ohm/phase
- Shaft load (Pload) = 10 kW
- Power factor (pf) = 0.8 leading
- Loss in motor (Ploss) = 2 kW
To find
- Per phase excitation emf of the motor
Formula used
- Apparent power (S) = √3 × Vph × Iph
- Power factor (pf) = P / S
- Reactive power (Q) = √(S² - P²)
- Impedance (Z) = Vph / Iph
- Impedance angle (θ) = tan⁻¹ (Q / P)
- Synchronous speed (Ns) = f × 120 / p
- Per phase emf (Eph) = Vph - Iph × Xs
- Torque (T) = (Pload / ω) = (3 × Eph × Iph × sinθ / ω)
- Excitation voltage (Eexc) = √(Eph² + Vph²)
Calculation
- Apparent power (S) = 50 kVA
- Power factor (pf) = 0.8 leading
- Real power (P) = Pload = 10 kW
- Reactive power (Q) = √(S² - P²) = √(50² - 10²) = 48.5 kVAR
- Impedance angle (θ) = tan⁻¹ (Q / P) = tan⁻¹ (48.5 / 10) = 78.7°
- Cosine of θ (cosθ) = pf = 0.8
- Sine of θ (sinθ) = √(1 - cos²θ) = √(1 - 0.8²) = 0.6
- Impedance (Z) = Vph / Iph
- Impedance angle (θ) = tan⁻¹ (Xs / (Z - Xs))
- tan⁻¹ (1 / (Z - 1)) = 78.7°
- Z = Xs / tan(78.7°) + 1 = 1.77 ohm
- Voltage per phase (Vph) = V / √3 = 400 / √3 = 230.9 V
- Current per phase (Iph) = P / (3 × Vph × pf) = 10 / (3 × 230.9 × 0.8) = 0.018 A
- Per phase emf (Eph) = Vph - Iph × Xs = 230.9 - 0.018 × 1 = 230.9 V
- Synchronous speed (Ns) = f × 120 / p = 50 × 120 / 2 = 3000 rpm
- Angular speed (ω) = 2 × π × Ns / 60 = 2 × π × 3000 / 60 = 314
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Community Answer
A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a...
Y-connected, 3 phase, 400V, 50 kVA, Synch motor, Xs = 1 Ω/ph, Shaft load Psh = 10 kW, p.f. = 0.8 loading losses = 2 kW, Ra: Negligible
Find excitation EMF (Eb):
To find Ia: Current drawn, we need to find Input.
Input = Output + Losses + Psh + Losses = 10 + 2 = 12 kW
Input = Output + Losses + Psh + Losses = 10 + 2 = 12 kW
Eb = 244.55 V per phase
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A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW.The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).Correct answer is '245'. Can you explain this answer?
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A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW.The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).Correct answer is '245'. Can you explain this answer? 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 star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW.The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).Correct answer is '245'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW.The magnitude of the per phase excitation emf of the motor, in volts, is ____ . (round off to nearest integer).Correct answer is '245'. Can you explain this answer?.
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