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A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.
Correct answer is '4.549'. Can you explain this answer?
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A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly ...
The given question is related to the fault current calculation in a three-phase alternator. Let's break down the problem step by step to find the current through the ground during an LLG fault.
Given data:
- Alternator rating: 30 MVA
- Voltage rating: 11 kV
- Frequency: 50 Hz
- Reactance to positive sequence currents: 2.5 ohms
- Reactance to negative sequence currents: 80% of positive sequence value (2.5 ohms * 0.8 = 2 ohms)
- Reactance to zero sequence currents: 30% of positive sequence value (2.5 ohms * 0.3 = 0.75 ohms)
Step 1: Calculate the per-phase voltage of the alternator.
The per-phase voltage can be calculated using the formula:
Vph = Vline / √3
where Vline is the line voltage.
In this case, Vline = 11 kV, so
Vph = 11 kV / √3 = 6.35 kV
Step 2: Calculate the per-phase current at no load.
At no load, the alternator draws the excitation current, which is purely reactive. Therefore, the current is zero.
Step 3: Calculate the positive, negative, and zero sequence currents.
The positive sequence current (I1) is the current flowing in the normal operating condition. It can be calculated using the formula:
I1 = S / (√3 * Vph)
where S is the apparent power in VA.
In this case, S = 30 MVA = 30,000,000 VA, so
I1 = 30,000,000 VA / (√3 * 6.35 kV) = 2,631.23 A
The negative sequence current (I2) is 80% of the positive sequence current.
I2 = 0.8 * I1 = 0.8 * 2,631.23 A = 2,104.98 A
The zero sequence current (I0) is 30% of the positive sequence current.
I0 = 0.3 * I1 = 0.3 * 2,631.23 A = 789.37 A
Step 4: Calculate the fault current through the ground.
In an LLG fault, the fault current consists of the positive and zero sequence currents.
The fault current through the ground (Ig) can be calculated using the formula:
Ig = √(I1^2 + I0^2)
Ig = √((2,631.23 A)^2 + (789.37 A)^2) = √(6,918,145.30 A^2) = 4,549 A
Therefore, the current through the ground during an LLG fault is 4.549 kA.
Given data:
- Alternator rating: 30 MVA
- Voltage rating: 11 kV
- Frequency: 50 Hz
- Reactance to positive sequence currents: 2.5 ohms
- Reactance to negative sequence currents: 80% of positive sequence value (2.5 ohms * 0.8 = 2 ohms)
- Reactance to zero sequence currents: 30% of positive sequence value (2.5 ohms * 0.3 = 0.75 ohms)
Step 1: Calculate the per-phase voltage of the alternator.
The per-phase voltage can be calculated using the formula:
Vph = Vline / √3
where Vline is the line voltage.
In this case, Vline = 11 kV, so
Vph = 11 kV / √3 = 6.35 kV
Step 2: Calculate the per-phase current at no load.
At no load, the alternator draws the excitation current, which is purely reactive. Therefore, the current is zero.
Step 3: Calculate the positive, negative, and zero sequence currents.
The positive sequence current (I1) is the current flowing in the normal operating condition. It can be calculated using the formula:
I1 = S / (√3 * Vph)
where S is the apparent power in VA.
In this case, S = 30 MVA = 30,000,000 VA, so
I1 = 30,000,000 VA / (√3 * 6.35 kV) = 2,631.23 A
The negative sequence current (I2) is 80% of the positive sequence current.
I2 = 0.8 * I1 = 0.8 * 2,631.23 A = 2,104.98 A
The zero sequence current (I0) is 30% of the positive sequence current.
I0 = 0.3 * I1 = 0.3 * 2,631.23 A = 789.37 A
Step 4: Calculate the fault current through the ground.
In an LLG fault, the fault current consists of the positive and zero sequence currents.
The fault current through the ground (Ig) can be calculated using the formula:
Ig = √(I1^2 + I0^2)
Ig = √((2,631.23 A)^2 + (789.37 A)^2) = √(6,918,145.30 A^2) = 4,549 A
Therefore, the current through the ground during an LLG fault is 4.549 kA.
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Community Answer
A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly ...
Given : X1 = 2.5 Ω
[For current division rule]
X0 = 0.3X1 = 0.3 x 2.5 = 0.75 Ω
X2 = 0.8X1 = 0.7 x 2.5 = 2 Ω
Positive sequence current,
Zero sequence current,
[For current division rule]Current through ground is given by,
Hence, the current through ground is 4.549 kA.
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A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. Can you explain this answer?
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A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. 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 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. 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 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. Can you explain this answer?.
A 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. 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 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. 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 30 MVA, 3 phase 11 kV, 50 Hz alternator having the neutral solidly grounded is operating at no load. It has a reactance to positive sequence currents of 2.5 ohms, the reactance to negative and zero sequence current are 80% and 30 % of the positive sequence value respectively, then the current through ground is _____ kA for LLG fault.Correct answer is '4.549'. Can you explain this answer?.
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