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The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).
Correct answer is between '0.1,0.1'. Can you explain this answer?
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Given data:
- Positive sequence reactance (X1) = 0.2 pu
- Negative sequence reactance (X2) = 0.2 pu
- Zero sequence reactance (X0) = 0.1 pu
- Terminal voltage (Vt) = 1 pu
- Fault current (If) = 3.75 pu
Calculating the minimum inductive reactance:
To limit the fault current to a desired value, we need to connect an inductive reactance (Xg) between neutral and ground. Let's assume the fault impedance (Zf) is zero.
Positive sequence fault current:
The positive sequence fault current (I1f) can be calculated using the following formula:
I1f = Vt / (X1 + Xg)
Negative sequence fault current:
The negative sequence fault current (I2f) can be calculated using the following formula:
I2f = Vt / (X2 + Xg)
Zero sequence fault current:
The zero sequence fault current (I0f) can be calculated using the following formula:
I0f = Vt / (X0 + Xg)
Maximum fault current:
The maximum fault current (If_max) occurs when the fault impedance is zero and all the fault current flows through the positive sequence impedance. It can be calculated as:
If_max = I1f = Vt / (X1 + Xg)
Calculating the minimum inductive reactance:
To limit the fault current to 3.75 pu, we need to ensure that the fault current (If) is less than or equal to 3.75 pu. Therefore,
If_max = Vt / (X1 + Xg) <=>
Solving the above equation for Xg, we get:
Xg >= Vt / 3.75 - X1
Substituting the given values, we have:
Xg >= 1 / 3.75 - 0.2
Xg >= 0.1
Therefore, the minimum value of the inductive reactance required to be connected between neutral and ground is 0.1 pu.
- Positive sequence reactance (X1) = 0.2 pu
- Negative sequence reactance (X2) = 0.2 pu
- Zero sequence reactance (X0) = 0.1 pu
- Terminal voltage (Vt) = 1 pu
- Fault current (If) = 3.75 pu
Calculating the minimum inductive reactance:
To limit the fault current to a desired value, we need to connect an inductive reactance (Xg) between neutral and ground. Let's assume the fault impedance (Zf) is zero.
Positive sequence fault current:
The positive sequence fault current (I1f) can be calculated using the following formula:
I1f = Vt / (X1 + Xg)
Negative sequence fault current:
The negative sequence fault current (I2f) can be calculated using the following formula:
I2f = Vt / (X2 + Xg)
Zero sequence fault current:
The zero sequence fault current (I0f) can be calculated using the following formula:
I0f = Vt / (X0 + Xg)
Maximum fault current:
The maximum fault current (If_max) occurs when the fault impedance is zero and all the fault current flows through the positive sequence impedance. It can be calculated as:
If_max = I1f = Vt / (X1 + Xg)
Calculating the minimum inductive reactance:
To limit the fault current to 3.75 pu, we need to ensure that the fault current (If) is less than or equal to 3.75 pu. Therefore,
If_max = Vt / (X1 + Xg) <=>
Solving the above equation for Xg, we get:
Xg >= Vt / 3.75 - X1
Substituting the given values, we have:
Xg >= 1 / 3.75 - 0.2
Xg >= 0.1
Therefore, the minimum value of the inductive reactance required to be connected between neutral and ground is 0.1 pu.
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The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).Correct answer is between '0.1,0.1'. Can you explain this answer?
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The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).Correct answer is between '0.1,0.1'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).Correct answer is between '0.1,0.1'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).Correct answer is between '0.1,0.1'. Can you explain this answer?.
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