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Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.
The initial accelerating power (in pu) will be (Answer up to two decimal places)
Correct answer is '0.56'. Can you explain this answer?
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Directions: A generator feeds power to an infinite bus through a doubl...
Degrees. Find the post-fault rotor power angle and the post-fault power output of the generator.
Solution:
We can use the swing equation to determine the post-fault behavior of the generator. The swing equation is given by:
M(d^2θ/dt^2) + D(dθ/dt) = Pe - Pg
where:
M = inertia constant (in MJ/MVA)
D = damping constant (in pu)
θ = rotor angle (in radians)
Pe = electrical power input (in pu)
Pg = mechanical power output (in pu)
We can assume that D is small and neglect its effect. Also, we can assume that the electrical power input is constant during the fault. Therefore, the swing equation simplifies to:
M(d^2θ/dt^2) = Pe - Pg
We can also use the fact that the voltage at the middle point of the faulted line is zero. Therefore, the voltage across the transfer admittance is 1.1 - 1 = 0.1 pu. The current through the transfer admittance is then:
I = Y*V = 0.8*0.1 = 0.08 pu
Since the line is double circuit, the fault current is twice the transfer current:
Ifault = 2*I = 0.16 pu
The electrical power input to the generator is:
Pe = V*Ef*sin(δ)
where:
V = 1 pu (infinite bus voltage)
Ef = 1.1 pu (transient internal voltage)
δ = 30 degrees (rotor power angle prior to fault)
Therefore,
Pe = 1.1*sin(30 degrees) = 0.55 pu
The mechanical power output of the generator during the fault is:
Pg = Pe - Ifault*Ef*sin(δ)
Pg = 0.55 - 0.16*1.1*sin(30 degrees) = 0.46 pu
The post-fault rotor power angle can be obtained by solving the swing equation. Let's assume that the rotor angle oscillates around the pre-fault value with a small amplitude. We can then write:
θ = δ + θ'
where θ' is the small deviation from the pre-fault value. Taking the second derivative with respect to time, we get:
d^2θ/dt^2 = d^2θ'/dt^2
Substituting into the swing equation and neglecting D, we get:
M(d^2θ'/dt^2) = -Pg
This is a simple harmonic oscillator equation with solution:
θ' = A*cos(ω*t + φ)
where A is the amplitude, ω is the natural frequency, and φ is the phase angle. The natural frequency is:
ω = sqrt(Pg/M)
The initial conditions are:
θ'(t=0) = 0
dθ'/dt(t=0) = 0
Therefore, φ = 0 and A = 0. The post-fault rotor power angle is then:
θ = δ
Therefore, the post-fault rotor power angle is 30 degrees (the same as the pre-fault value) and the post-fault power output of the generator is 0.46 pu.
Solution:
We can use the swing equation to determine the post-fault behavior of the generator. The swing equation is given by:
M(d^2θ/dt^2) + D(dθ/dt) = Pe - Pg
where:
M = inertia constant (in MJ/MVA)
D = damping constant (in pu)
θ = rotor angle (in radians)
Pe = electrical power input (in pu)
Pg = mechanical power output (in pu)
We can assume that D is small and neglect its effect. Also, we can assume that the electrical power input is constant during the fault. Therefore, the swing equation simplifies to:
M(d^2θ/dt^2) = Pe - Pg
We can also use the fact that the voltage at the middle point of the faulted line is zero. Therefore, the voltage across the transfer admittance is 1.1 - 1 = 0.1 pu. The current through the transfer admittance is then:
I = Y*V = 0.8*0.1 = 0.08 pu
Since the line is double circuit, the fault current is twice the transfer current:
Ifault = 2*I = 0.16 pu
The electrical power input to the generator is:
Pe = V*Ef*sin(δ)
where:
V = 1 pu (infinite bus voltage)
Ef = 1.1 pu (transient internal voltage)
δ = 30 degrees (rotor power angle prior to fault)
Therefore,
Pe = 1.1*sin(30 degrees) = 0.55 pu
The mechanical power output of the generator during the fault is:
Pg = Pe - Ifault*Ef*sin(δ)
Pg = 0.55 - 0.16*1.1*sin(30 degrees) = 0.46 pu
The post-fault rotor power angle can be obtained by solving the swing equation. Let's assume that the rotor angle oscillates around the pre-fault value with a small amplitude. We can then write:
θ = δ + θ'
where θ' is the small deviation from the pre-fault value. Taking the second derivative with respect to time, we get:
d^2θ/dt^2 = d^2θ'/dt^2
Substituting into the swing equation and neglecting D, we get:
M(d^2θ'/dt^2) = -Pg
This is a simple harmonic oscillator equation with solution:
θ' = A*cos(ω*t + φ)
where A is the amplitude, ω is the natural frequency, and φ is the phase angle. The natural frequency is:
ω = sqrt(Pg/M)
The initial conditions are:
θ'(t=0) = 0
dθ'/dt(t=0) = 0
Therefore, φ = 0 and A = 0. The post-fault rotor power angle is then:
θ = δ
Therefore, the post-fault rotor power angle is 30 degrees (the same as the pre-fault value) and the post-fault power output of the generator is 0.46 pu.
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Community Answer
Directions: A generator feeds power to an infinite bus through a doubl...
Generator feeds power to infinite bus through double circuit line 3-ф fault at middle of line
Infinite bus voltage(V) = 1 pu
Transient internal voltage of generator(E) = 1.1 pu
Equivalent transfer admittance during fault = 0.8 pu = 1/X
Delivering power(PS) = 1.0 pu
Perior to fault rotor Power angle δ = 300 , f = 50Hz
Initial accelerating power(Pa) = ?
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Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer?
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Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. 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 Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. 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 Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer?.
Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. 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 Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. 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 Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer?.
Solutions for Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer?, a detailed solution for Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer? has been provided alongside types of Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Directions: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be (Answer up to two decimal places)Correct answer is '0.56'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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