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An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:
5610 A for a three phase fault
6760 A for line to line fault
8630 A for single line to ground fault.
If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________
5610 A for a three phase fault
6760 A for line to line fault
8630 A for single line to ground fault.
If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________
Correct answer is between '0.1,0.15'. Can you explain this answer?
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Most Upvoted Answer
An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to thre...
Solution:
Given data:
Voltage rating of the alternator (V) = 11 kV
Apparent power rating of the alternator (S) = 25 MVA
Fault current for three-phase fault (I3) = 5610 A
Fault current for line to line fault (I2) = 6760 A
Fault current for single line to ground fault (I1) = 8630 A
The alternator neutral is solidly grounded.
Calculating Zero-sequence Reactance:
The zero-sequence reactance of the alternator (X0) can be calculated using the following formula:
X0 = V / (3 * I0)
where I0 is the zero-sequence current.
For a three-phase fault, the zero-sequence current is equal to the phase current, i.e., I0 = I3 / √3.
Substituting the given values in the above formula, we get:
X0 = 11 kV / (3 * 5610 A / √3) = 0.114 pu
For a line to line fault, the zero-sequence current is zero, as there is no current flowing through the neutral. Hence, X0 is undefined for this fault.
For a single line to ground fault, the zero-sequence current is equal to the fault current, i.e., I0 = I1.
Substituting the given values in the above formula, we get:
X0 = 11 kV / (3 * 8630 A) = 0.143 pu
Therefore, the zero-sequence reactance of the alternator is between 0.1 and 0.15 pu, as calculated above.
Conclusion:
In conclusion, the zero-sequence reactance of the alternator can be calculated using the formula X0 = V / (3 * I0), where I0 is the zero-sequence current. For a three-phase fault and a single line to ground fault, the zero-sequence current can be calculated using the given fault current values. However, for a line to line fault, the zero-sequence current is zero, and hence, X0 is undefined.
Given data:
Voltage rating of the alternator (V) = 11 kV
Apparent power rating of the alternator (S) = 25 MVA
Fault current for three-phase fault (I3) = 5610 A
Fault current for line to line fault (I2) = 6760 A
Fault current for single line to ground fault (I1) = 8630 A
The alternator neutral is solidly grounded.
Calculating Zero-sequence Reactance:
The zero-sequence reactance of the alternator (X0) can be calculated using the following formula:
X0 = V / (3 * I0)
where I0 is the zero-sequence current.
For a three-phase fault, the zero-sequence current is equal to the phase current, i.e., I0 = I3 / √3.
Substituting the given values in the above formula, we get:
X0 = 11 kV / (3 * 5610 A / √3) = 0.114 pu
For a line to line fault, the zero-sequence current is zero, as there is no current flowing through the neutral. Hence, X0 is undefined for this fault.
For a single line to ground fault, the zero-sequence current is equal to the fault current, i.e., I0 = I1.
Substituting the given values in the above formula, we get:
X0 = 11 kV / (3 * 8630 A) = 0.143 pu
Therefore, the zero-sequence reactance of the alternator is between 0.1 and 0.15 pu, as calculated above.
Conclusion:
In conclusion, the zero-sequence reactance of the alternator can be calculated using the formula X0 = V / (3 * I0), where I0 is the zero-sequence current. For a three-phase fault and a single line to ground fault, the zero-sequence current can be calculated using the given fault current values. However, for a line to line fault, the zero-sequence current is zero, and hence, X0 is undefined.
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Community Answer
An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to thre...
Voltage (E) = 11 kV
Phase voltage (Eph)
Three phase fault current = 5610 A
⇒ X1 = 1.132 Ω
L-L fault current = 6760 A
⇒ X2 = 0.495 Ω
L-G fault current = 8630 A
Phase voltage (Eph)
Three phase fault current = 5610 A
⇒ X1 = 1.132 Ω
L-L fault current = 6760 A
⇒ X2 = 0.495 Ω
L-G fault current = 8630 A
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An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:5610 A for a three phase fault6760 A for line to line fault8630 A for single line to ground fault.If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________Correct answer is between '0.1,0.15'. Can you explain this answer?
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An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:5610 A for a three phase fault6760 A for line to line fault8630 A for single line to ground fault.If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________Correct answer is between '0.1,0.15'. 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 An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:5610 A for a three phase fault6760 A for line to line fault8630 A for single line to ground fault.If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________Correct answer is between '0.1,0.15'. 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 An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:5610 A for a three phase fault6760 A for line to line fault8630 A for single line to ground fault.If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________Correct answer is between '0.1,0.15'. Can you explain this answer?.
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