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A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).
Correct answer is between '0.5,0.6'. Can you explain this answer?
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A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%....
Short circuit current ISC = 
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A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%....
A three-phase 15 MVA alternator has a rated apparent power of 15 MVA and a rated voltage of 13 kV. The initial reactance is given as 5% of the rated impedance.
To calculate the series reactor needed to restrict the short circuit current to 10 times the full load current, we need to determine the full load current.
The full load current can be calculated using the formula:
Full Load Current (IL) = Rated Apparent Power (S) / (sqrt(3) x Rated Voltage (V))
IL = 15 MVA / (sqrt(3) x 13 kV)
IL = 15,000,000 VA / (1.732 x 13,000 V)
IL = 659.6 A (approximately)
The short circuit current will be 10 times the full load current, so:
Short Circuit Current (Isc) = 10 x IL
Isc = 10 x 659.6 A
Isc = 6596 A (approximately)
Now, let's calculate the reactance of the alternator at full load.
Reactance (X) = Impedance (Z) x Reactance Percentage (%)
X = 5% x Z
We know that impedance (Z) is given by the formula:
Z = Rated Voltage (V) / Full Load Current (IL)
Z = 13 kV / 659.6 A
Z = 19.7 ohms (approximately)
Now we can calculate the reactance at full load:
X = 5% x 19.7 ohms
X = 0.05 x 19.7 ohms
X = 0.985 ohms (approximately)
Finally, to restrict the short circuit current to 10 times the full load current, we need to add a series reactor with a reactance equal to:
Series Reactor Reactance = (Isc - IL) / (sqrt(3) x Rated Voltage (V))
Series Reactor Reactance = (6596 A - 659.6 A) / (1.732 x 13 kV)
Series Reactor Reactance = 5936.4 A / (1.732 x 13,000 V)
Series Reactor Reactance = 0.266 ohms (approximately)
Therefore, the series reactor needed to restrict the short circuit current of the alternator to 10 times the full load current is approximately 0.266 ohms.
To calculate the series reactor needed to restrict the short circuit current to 10 times the full load current, we need to determine the full load current.
The full load current can be calculated using the formula:
Full Load Current (IL) = Rated Apparent Power (S) / (sqrt(3) x Rated Voltage (V))
IL = 15 MVA / (sqrt(3) x 13 kV)
IL = 15,000,000 VA / (1.732 x 13,000 V)
IL = 659.6 A (approximately)
The short circuit current will be 10 times the full load current, so:
Short Circuit Current (Isc) = 10 x IL
Isc = 10 x 659.6 A
Isc = 6596 A (approximately)
Now, let's calculate the reactance of the alternator at full load.
Reactance (X) = Impedance (Z) x Reactance Percentage (%)
X = 5% x Z
We know that impedance (Z) is given by the formula:
Z = Rated Voltage (V) / Full Load Current (IL)
Z = 13 kV / 659.6 A
Z = 19.7 ohms (approximately)
Now we can calculate the reactance at full load:
X = 5% x 19.7 ohms
X = 0.05 x 19.7 ohms
X = 0.985 ohms (approximately)
Finally, to restrict the short circuit current to 10 times the full load current, we need to add a series reactor with a reactance equal to:
Series Reactor Reactance = (Isc - IL) / (sqrt(3) x Rated Voltage (V))
Series Reactor Reactance = (6596 A - 659.6 A) / (1.732 x 13 kV)
Series Reactor Reactance = 5936.4 A / (1.732 x 13,000 V)
Series Reactor Reactance = 0.266 ohms (approximately)
Therefore, the series reactor needed to restrict the short circuit current of the alternator to 10 times the full load current is approximately 0.266 ohms.
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A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. Can you explain this answer?
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A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. 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 three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. 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 three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. Can you explain this answer?.
A three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. 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 three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. 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 three phase 15 MVA, 13 kV alternator has an initial reactance of 5%. The series reactor needed to restrict the short circuit current of the alternator to 10 times of full load current is _____ (in Ω).Correct answer is between '0.5,0.6'. Can you explain this answer?.
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