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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)
Correct answer is '642.2'. Can you explain this answer?
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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transm...
Total zero sequence impedance, +ve sequence impedance and -ve sequence
Impedance
Fault current
Neutral voltage
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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transm...
Given:
- Alternator rating: 20 MVA, 6.6 kV, 3-phase
- Per unit positive-sequence impedance of alternator: j0.1 p.u.
- Per unit negative-sequence impedance of alternator: j0.1 p.u.
- Per unit zero-sequence impedance of alternator: j0.04 p.u.
- Inductive reactor impedance: j0.05 p.u.
- Per unit positive-sequence impedance of transmission line: j0.1 p.u.
- Per unit negative-sequence impedance of transmission line: j0.1 p.u.
- Per unit zero-sequence impedance of transmission line: j0.3 p.u.
- Solid ground fault at one phase of the far end of the transmission line.

Approach:
1. Calculate the pre-fault voltage at the neutral of the alternator with respect to ground.
2. Determine the fault current flowing through the faulted phase.
3. Calculate the voltage drop across the positive-sequence impedance of the alternator.
4. Calculate the voltage drop across the positive-sequence impedance of the transmission line.
5. Calculate the voltage drop across the zero-sequence impedance of the alternator.
6. Calculate the voltage drop across the zero-sequence impedance of the transmission line.
7. Calculate the voltage at the neutral of the alternator with respect to ground during the fault.

Calculations:
1. Pre-fault voltage at the neutral of the alternator with respect to ground:
- The pre-fault voltage is the rated line-to-neutral voltage of the alternator, which is 6.6 kV.

2. Fault current:
- Since it is a solid ground fault, only zero-sequence current flows through the faulted phase.
- The fault current is given by: Ifault = Vpre_fault / (Z0_alternator + Z0_reactor + Z0_line)
where Vpre_fault is the pre-fault voltage at the neutral of the alternator with respect to ground,
Z0_alternator is the zero-sequence impedance of the alternator,
Z0_reactor is the impedance of the inductive reactor, and
Z0_line is the zero-sequence impedance of the transmission line.
- Substituting the given values, the fault current is calculated as:
Ifault = 6.6 kV / (j0.04 + j0.05 + j0.3) = 20.4 kA (approximately)

3. Voltage drop across the positive-sequence impedance of the alternator:
- The voltage drop is given by: Vdrop_alternator = Ifault * Z1_alternator
where Z1_alternator is the positive-sequence impedance of the alternator.
- Substituting the given values, the voltage drop is calculated as:
Vdrop_alternator = 20.4 kA * j0.1 p.u. = 2.04 kV (approximately)

4. Voltage drop across the positive-sequence impedance of the transmission line:
- The voltage drop is given by: Vdrop_line = Ifault * Z1_line
where Z1_line is the positive-sequence impedance of the transmission line.
- Substituting the given values, the voltage drop is calculated as:
Vdrop_line =
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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)Correct answer is '642.2'. Can you explain this answer?
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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)Correct answer is '642.2'. 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 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)Correct answer is '642.2'. 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 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)Correct answer is '642.2'. Can you explain this answer?.
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