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A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.
Correct answer is between '12.5,12.9'. Can you explain this answer?
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A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MV...
Pa = Pm - Pe
= 60 - 0 = 60mw
Most Upvoted Answer
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MV...
Pa = Pm - Pe
= 60 - 0 = 60mw
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Community Answer
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MV...
Given data:
- Synchronous generator rating: 250 MVA
- Power factor: 0.8 lagging
- Kinetic energy at synchronous speed: 1000 MJ
- Power output: 60 MW
- Power angle: 10 electrical degrees
- Acceleration is constant for 10 cycles
Step 1: Calculating the synchronous speed (Ns)
The synchronous speed (Ns) of a generator is given by the formula:
Ns = (120 * f) / P
Where:
- f is the frequency of the generator (50 Hz)
- P is the number of poles (2 poles)
Substituting the values, we get:
Ns = (120 * 50) / 2 = 3000 RPM
Step 2: Calculating the inertia constant (H)
The inertia constant (H) of a synchronous generator is given by the formula:
H = (2 * K.E) / (Ns^2)
Where:
- K.E is the kinetic energy of the machine at synchronous speed (1000 MJ)
- Ns is the synchronous speed (3000 RPM)
Substituting the values, we get:
H = (2 * 1000 MJ) / (3000 RPM)^2
Converting MJ to J and RPM to rad/s:
H = (2 * 1000 * 10^6 J) / ((3000 RPM * 2π / 60 s)^2)
Calculating H:
H = (2 * 1000 * 10^6 J) / ((3000 * 2π / 60)^2 rad^2/s^2)
Step 3: Calculating the mechanical power (Pm) and accelerating torque (Ta)
The mechanical power (Pm) of a generator is given by the formula:
Pm = (Pout + Ploss) / (1 - δ)
Where:
- Pout is the power output of the generator (60 MW)
- Ploss is the power loss in the generator (can be neglected for this calculation)
- δ is the power angle (10 electrical degrees)
Substituting the values, we get:
Pm = (60 MW) / (1 - 10/360)
Calculating Pm:
Pm = (60 * 10^6 W) / (1 - 10/360)
Step 4: Calculating the accelerating power (Pa)
The accelerating power (Pa) of a generator is given by the formula:
Pa = (1 - δ) * Pm
Substituting the values, we get:
Pa = (1 - 10/360) * (60 * 10^6 W)
Calculating Pa:
Pa = (350/36) * (60 * 10^6 W)
Step 5: Calculating the acceleration (α)
The acceleration (α) of a generator is given by the formula:
α = Pa / (H * Ns)
Substituting the values, we get:
α = ((350/36) * (60 * 10^6 W)) / ((2 * 1000 * 10^6 J) / (3000 RPM)^2)
Calculating α:
α = ((350/36) * (60 * 10^6 W)) / ((
- Synchronous generator rating: 250 MVA
- Power factor: 0.8 lagging
- Kinetic energy at synchronous speed: 1000 MJ
- Power output: 60 MW
- Power angle: 10 electrical degrees
- Acceleration is constant for 10 cycles
Step 1: Calculating the synchronous speed (Ns)
The synchronous speed (Ns) of a generator is given by the formula:
Ns = (120 * f) / P
Where:
- f is the frequency of the generator (50 Hz)
- P is the number of poles (2 poles)
Substituting the values, we get:
Ns = (120 * 50) / 2 = 3000 RPM
Step 2: Calculating the inertia constant (H)
The inertia constant (H) of a synchronous generator is given by the formula:
H = (2 * K.E) / (Ns^2)
Where:
- K.E is the kinetic energy of the machine at synchronous speed (1000 MJ)
- Ns is the synchronous speed (3000 RPM)
Substituting the values, we get:
H = (2 * 1000 MJ) / (3000 RPM)^2
Converting MJ to J and RPM to rad/s:
H = (2 * 1000 * 10^6 J) / ((3000 RPM * 2π / 60 s)^2)
Calculating H:
H = (2 * 1000 * 10^6 J) / ((3000 * 2π / 60)^2 rad^2/s^2)
Step 3: Calculating the mechanical power (Pm) and accelerating torque (Ta)
The mechanical power (Pm) of a generator is given by the formula:
Pm = (Pout + Ploss) / (1 - δ)
Where:
- Pout is the power output of the generator (60 MW)
- Ploss is the power loss in the generator (can be neglected for this calculation)
- δ is the power angle (10 electrical degrees)
Substituting the values, we get:
Pm = (60 MW) / (1 - 10/360)
Calculating Pm:
Pm = (60 * 10^6 W) / (1 - 10/360)
Step 4: Calculating the accelerating power (Pa)
The accelerating power (Pa) of a generator is given by the formula:
Pa = (1 - δ) * Pm
Substituting the values, we get:
Pa = (1 - 10/360) * (60 * 10^6 W)
Calculating Pa:
Pa = (350/36) * (60 * 10^6 W)
Step 5: Calculating the acceleration (α)
The acceleration (α) of a generator is given by the formula:
α = Pa / (H * Ns)
Substituting the values, we get:
α = ((350/36) * (60 * 10^6 W)) / ((2 * 1000 * 10^6 J) / (3000 RPM)^2)
Calculating α:
α = ((350/36) * (60 * 10^6 W)) / ((
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A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. Can you explain this answer?
Question Description
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. 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 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. 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 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. Can you explain this answer?.
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. 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 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. 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 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.Correct answer is between '12.5,12.9'. Can you explain this answer?.
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