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An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.
Q.
In the above problem the subtransientcurrent of motor is____________
- a)9.7t kA
- b)0
- c)10.35 kA
- d)6.90 kA
Correct answer is option 'A'. Can you explain this answer?
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An alternator and synchronous motor each rated for 50 MVA, 13.2 kV hav...
Calculation of Subtransient Current of Motor
Given data:
- Alternator and synchronous motor both are rated for 50 MVA and 13.2 kV
- Subtransient reactance of both machines is 20%
- Transmission link has an impedance of 10% on the base of machine ratings
- Motor is acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV
Step 1: Calculation of Base Values
- Base MVA = 50 MVA
- Base Voltage = 13.2 kV
- Base Impedance = (Base Voltage)^2 / Base MVA = (13.2 kV)^2 / 50 MVA = 3.465 Ω
- Base Current = Base MVA / Base Voltage = 50 MVA / 13.2 kV = 3787 A
Step 2: Calculation of Motor Impedance on Subtransient Basis
- Subtransient Impedance of Motor = 20% of its rated impedance = 0.2 * (3.465 Ω / 0.9) = 0.865 Ω
- Motor Impedance on Subtransient Basis = 0.865 Ω + j0.865 Ω
Step 3: Calculation of Transmission Line Impedance on Subtransient Basis
- Transmission Line Impedance on Subtransient Basis = 10% of Base Impedance = 0.1 * 3.465 Ω = 0.3465 Ω
- Transmission Line Impedance on Subtransient Basis = 0.3465 Ω + j0.3465 Ω
Step 4: Calculation of Total Impedance on Subtransient Basis
- Total Impedance on Subtransient Basis = Motor Impedance on Subtransient Basis + Transmission Line Impedance on Subtransient Basis
- Total Impedance on Subtransient Basis = (0.865 + 0.3465) Ω + j(0.865 + 0.3465) Ω
- Total Impedance on Subtransient Basis = 1.2115 Ω + j1.2115 Ω
Step 5: Calculation of Current through Transmission Line
- Apparent Power of Motor = 35 MW / 0.9 = 38.89 MVA
- Reactive Power of Motor = sqrt(1 - 0.9^2) * 38.89 MVA = 12.09 MVAR
- Current through Motor = (38.89 MVA + j12.09 MVAR) / 12.5 kV = 2466 A - j767 A
- Current through Transmission Line = Current through Motor / (1 + Total Impedance on Subtransient Basis * Base Current)
- Current through Transmission Line = (2466 A - j767 A) / (1 + (1.2115 + j1.2115) Ω * 3787 A)
- Current through Transmission Line = 9.7 kA - j3.02 kA
Therefore, the subtransient current of motor is 9.7 kA (kilamperes).
Given data:
- Alternator and synchronous motor both are rated for 50 MVA and 13.2 kV
- Subtransient reactance of both machines is 20%
- Transmission link has an impedance of 10% on the base of machine ratings
- Motor is acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV
Step 1: Calculation of Base Values
- Base MVA = 50 MVA
- Base Voltage = 13.2 kV
- Base Impedance = (Base Voltage)^2 / Base MVA = (13.2 kV)^2 / 50 MVA = 3.465 Ω
- Base Current = Base MVA / Base Voltage = 50 MVA / 13.2 kV = 3787 A
Step 2: Calculation of Motor Impedance on Subtransient Basis
- Subtransient Impedance of Motor = 20% of its rated impedance = 0.2 * (3.465 Ω / 0.9) = 0.865 Ω
- Motor Impedance on Subtransient Basis = 0.865 Ω + j0.865 Ω
Step 3: Calculation of Transmission Line Impedance on Subtransient Basis
- Transmission Line Impedance on Subtransient Basis = 10% of Base Impedance = 0.1 * 3.465 Ω = 0.3465 Ω
- Transmission Line Impedance on Subtransient Basis = 0.3465 Ω + j0.3465 Ω
Step 4: Calculation of Total Impedance on Subtransient Basis
- Total Impedance on Subtransient Basis = Motor Impedance on Subtransient Basis + Transmission Line Impedance on Subtransient Basis
- Total Impedance on Subtransient Basis = (0.865 + 0.3465) Ω + j(0.865 + 0.3465) Ω
- Total Impedance on Subtransient Basis = 1.2115 Ω + j1.2115 Ω
Step 5: Calculation of Current through Transmission Line
- Apparent Power of Motor = 35 MW / 0.9 = 38.89 MVA
- Reactive Power of Motor = sqrt(1 - 0.9^2) * 38.89 MVA = 12.09 MVAR
- Current through Motor = (38.89 MVA + j12.09 MVAR) / 12.5 kV = 2466 A - j767 A
- Current through Transmission Line = Current through Motor / (1 + Total Impedance on Subtransient Basis * Base Current)
- Current through Transmission Line = (2466 A - j767 A) / (1 + (1.2115 + j1.2115) Ω * 3787 A)
- Current through Transmission Line = 9.7 kA - j3.02 kA
Therefore, the subtransient current of motor is 9.7 kA (kilamperes).
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An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer?
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An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. 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 An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer?.
An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. 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 An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer?.
Solutions for An alternator and synchronous motor each rated for 50 MVA, 13.2 kV having subtransient reactance of 20% are connected through a transmission link of 10% on the base of machine ratings. The motor acting as a load of 35 MW at 0.9 p.f. lag and the terminal voltage is 12.5 kV.Q.In the above problem the subtransientcurrent of motor is____________a) 9.7t kAb) 0c) 10.35 kAd) 6.90 kACorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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