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A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kW
Correct answer is between '132,145'. Can you explain this answer?
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Verified Answer
A 400 V, 3ϕ delta connected synchronous motor runs at rated volta...
Zs = (0.7 + 3j) = 3.08 ∠76.87
Vt = 400 V, Ef = 500 V
Maximum out put power
Shaft power = [3 × 46.49 – 1.8]kW = 137.66 kW
Vt = 400 V, Ef = 500 V
Maximum out put power
Shaft power = [3 × 46.49 – 1.8]kW = 137.66 kW
Most Upvoted Answer
A 400 V, 3ϕ delta connected synchronous motor runs at rated volta...
The first step is to calculate the apparent power of the motor:
S = 3Vph Iph
where Vph is the phase voltage and Iph is the phase current. Since the motor is delta connected, the phase voltage is equal to the line voltage, which is 400 V. Therefore:
S = 3 x 400 x Iph
S = 1200 Iph
Next, we need to calculate the power factor of the motor. Since it is a synchronous motor, the power factor is equal to the cosine of the angle between the voltage and current phasors. The synchronous impedance per phase is given as (0.7 + 3j), so the angle is:
θ = atan(3/0.7)
θ = 75.96°
Therefore, the power factor is:
PF = cos(75.96°)
PF = 0.26 (lagging)
Now we can calculate the real power consumed by the motor:
P = S x PF
P = 1200 Iph x 0.26
P = 312 Iph
We are given that the winding and iron losses are 1800 W, so the total power input to the motor is:
Pin = P + losses
Pin = 312 Iph + 1800
Finally, we can use the excitation emf of 500 V to calculate the field current:
E = Vph + Ia Zs
500 = 400 + Ia (0.7 + 3j)
Ia = 46.59 A ∠ -74.16°
The negative angle indicates that the field current is leading the phase voltage.
Therefore, the motor is consuming 312 Iph W of real power and 1800 W of losses, for a total input power of 312 Iph + 1800 W. The field current is 46.59 A ∠ -74.16°.
S = 3Vph Iph
where Vph is the phase voltage and Iph is the phase current. Since the motor is delta connected, the phase voltage is equal to the line voltage, which is 400 V. Therefore:
S = 3 x 400 x Iph
S = 1200 Iph
Next, we need to calculate the power factor of the motor. Since it is a synchronous motor, the power factor is equal to the cosine of the angle between the voltage and current phasors. The synchronous impedance per phase is given as (0.7 + 3j), so the angle is:
θ = atan(3/0.7)
θ = 75.96°
Therefore, the power factor is:
PF = cos(75.96°)
PF = 0.26 (lagging)
Now we can calculate the real power consumed by the motor:
P = S x PF
P = 1200 Iph x 0.26
P = 312 Iph
We are given that the winding and iron losses are 1800 W, so the total power input to the motor is:
Pin = P + losses
Pin = 312 Iph + 1800
Finally, we can use the excitation emf of 500 V to calculate the field current:
E = Vph + Ia Zs
500 = 400 + Ia (0.7 + 3j)
Ia = 46.59 A ∠ -74.16°
The negative angle indicates that the field current is leading the phase voltage.
Therefore, the motor is consuming 312 Iph W of real power and 1800 W of losses, for a total input power of 312 Iph + 1800 W. The field current is 46.59 A ∠ -74.16°.
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A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer?
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A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. 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 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. 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 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer?.
A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. 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 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. 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 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer?.
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A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer?, a detailed solution for A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer? has been provided alongside types of A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kWCorrect answer is between '132,145'. Can you explain this answer? theory, EduRev gives you an
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