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A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If = 1080 A, VCC = 33 kV, ISC = 12.3 kA calculate Xs in PU
Correct answer is between '0.55,0.58'. Can you explain this answer?
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A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C t...
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A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C t...
Solution:
Given parameters:
- Rated power (PR) = 400 MVA
- Rated voltage (VR) = 33 kV
- Field current (If) = 1080 A
- Short-circuit current (ISC) = 12.3 kA
1. Calculation of synchronous reactance (Xd):
From the open-circuit characteristic (O.C.C) test, we can obtain the field current (If) required to produce the rated voltage (VR) at open-circuit condition. This current is also known as the excitation current (Iexc).
Excitation current (Iexc) = If = 1080 A
Now, using the rated power (PR) and rated voltage (VR), we can calculate the synchronous reactance (Xd) as follows:
PR = (3 × VR × ISC) / (2 × π)
Xd = VR / Iexc
Substituting the given values in the above equations, we get:
Xd = 0.175 ohm
2. Calculation of synchronous reactance in per unit (P.U.):
The synchronous reactance in per unit (P.U.) can be obtained by dividing the synchronous reactance (Xd) by the synchronous impedance (Zs), where Zs is given by:
Zs = VR / ISC
Therefore, synchronous reactance in per unit (P.U.) is given by:
Xs (P.U.) = Xd / Zs
Substituting the given values in the above equation, we get:
Xs (P.U.) = 0.175 / (33 / 12.3)
Xs (P.U.) = 0.647 P.U.
3. Calculation of Xsin P.U.:
The value of Xsin P.U. is given by:
Xsin P.U. = Xs (P.U.) × sin δ
where δ is the angle between the voltage phasor and the current phasor in the short-circuit condition.
From the short-circuit characteristic (S.C.C) test, we can obtain the short-circuit current (ISC) and the voltage (VCC) at the generator terminals. The angle between the voltage phasor and the current phasor can be assumed to be small (usually less than 10 degrees).
Therefore, we can approximate sin δ as follows:
sin δ ≈ ISC / (VCC / Xs)
Substituting the given values in the above equation, we get:
sin δ ≈ 12.3 / (33 / 0.647)
sin δ ≈ 0.235
Therefore, Xsin P.U. is given by:
Xsin P.U. = Xs (P.U.) × sin δ
Xsin P.U. = 0.647 × 0.235
Xsin P.U. = 0.152 P.U.
4. Final Answer:
The correct answer is between 0.55 and 0.58. However, the calculated value of Xsin P.U. is 0.152 P.U. which is much smaller than the expected range. This may be due to some error in the given data or calculations.
Given parameters:
- Rated power (PR) = 400 MVA
- Rated voltage (VR) = 33 kV
- Field current (If) = 1080 A
- Short-circuit current (ISC) = 12.3 kA
1. Calculation of synchronous reactance (Xd):
From the open-circuit characteristic (O.C.C) test, we can obtain the field current (If) required to produce the rated voltage (VR) at open-circuit condition. This current is also known as the excitation current (Iexc).
Excitation current (Iexc) = If = 1080 A
Now, using the rated power (PR) and rated voltage (VR), we can calculate the synchronous reactance (Xd) as follows:
PR = (3 × VR × ISC) / (2 × π)
Xd = VR / Iexc
Substituting the given values in the above equations, we get:
Xd = 0.175 ohm
2. Calculation of synchronous reactance in per unit (P.U.):
The synchronous reactance in per unit (P.U.) can be obtained by dividing the synchronous reactance (Xd) by the synchronous impedance (Zs), where Zs is given by:
Zs = VR / ISC
Therefore, synchronous reactance in per unit (P.U.) is given by:
Xs (P.U.) = Xd / Zs
Substituting the given values in the above equation, we get:
Xs (P.U.) = 0.175 / (33 / 12.3)
Xs (P.U.) = 0.647 P.U.
3. Calculation of Xsin P.U.:
The value of Xsin P.U. is given by:
Xsin P.U. = Xs (P.U.) × sin δ
where δ is the angle between the voltage phasor and the current phasor in the short-circuit condition.
From the short-circuit characteristic (S.C.C) test, we can obtain the short-circuit current (ISC) and the voltage (VCC) at the generator terminals. The angle between the voltage phasor and the current phasor can be assumed to be small (usually less than 10 degrees).
Therefore, we can approximate sin δ as follows:
sin δ ≈ ISC / (VCC / Xs)
Substituting the given values in the above equation, we get:
sin δ ≈ 12.3 / (33 / 0.647)
sin δ ≈ 0.235
Therefore, Xsin P.U. is given by:
Xsin P.U. = Xs (P.U.) × sin δ
Xsin P.U. = 0.647 × 0.235
Xsin P.U. = 0.152 P.U.
4. Final Answer:
The correct answer is between 0.55 and 0.58. However, the calculated value of Xsin P.U. is 0.152 P.U. which is much smaller than the expected range. This may be due to some error in the given data or calculations.
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A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer?
Question Description
A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. 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 A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer?.
A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. 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 A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer?.
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A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer?, a detailed solution for A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer? has been provided alongside types of A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If= 1080 A, VCC= 33 kV, ISC= 12.3 kA calculate Xsin PUCorrect answer is between '0.55,0.58'. Can you explain this answer? theory, EduRev gives you an
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