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A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ′b′and′c′, with a fault impedance of j0.1p.u. The value of Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.
Correct answer is between '1.05,1.15'. Can you explain this answer?
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Most Upvoted Answer
A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator ...
We have the base current
The positive sequence current, Ip = 4270 A
Fault impedance = 0.1 pu
Xn = 3zf + 0.05 + 3xn = 0.35 + 3xn
From above equation, we get Xn = 1.07 p.u
The positive sequence current, Ip = 4270 A
Fault impedance = 0.1 pu
Xn = 3zf + 0.05 + 3xn = 0.35 + 3xn
From above equation, we get Xn = 1.07 p.u
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A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator ...
To calculate the fault current and the fault voltage in this scenario, we need to use the positive, negative, and zero sequence reactances of the generator.
Let's assume the fault occurs between phases A and B. In a double line to ground fault, the fault current splits into three components: the positive sequence current (I1), the negative sequence current (I2), and the zero sequence current (I0).
The fault current in each phase can be calculated using the following formula:
I1 = (Vph / (√3 X1))
I2 = (Vph / (√3 X2))
I0 = (Vph / (√3 X0))
Where:
- Vph is the line-to-neutral voltage of the generator (13.8 kV / √3 = 7.98 kV)
- X1, X2, and X0 are the positive, negative, and zero sequence reactances respectively (15%, 15%, and 5% of the generator's impedance, which is the sum of the positive and negative sequence reactances)
Using these values, we can calculate the fault current:
I1 = (7.98 kV / (√3 * 0.15)) = 30.97 kA
I2 = (7.98 kV / (√3 * 0.15)) = 30.97 kA
I0 = (7.98 kV / (√3 * 0.05)) = 92.91 kA
The total fault current (If) is the vector sum of the positive, negative, and zero sequence currents:
If = √(I1^2 + I2^2 + I0^2)
= √((30.97 kA)^2 + (30.97 kA)^2 + (92.91 kA)^2)
= 108.99 kA
Next, let's calculate the fault voltage (Vf) across the faulted phases. The fault voltage can be calculated using the following formula:
Vf = (Vph * (X1 + X2 + Xn)) / (√3 * (X1 + X2 + Xn + X0))
Where:
- Xn is the reactance connected between the neutral of the generator and ground
Using the given values:
Vf = (7.98 kV * (0.15 + 0.15 + Xn)) / (√3 * (0.15 + 0.15 + Xn + 0.05))
Now, we need to know the value of Xn to calculate the fault voltage accurately. If you provide the value of Xn, we can calculate the fault voltage accordingly.
Let's assume the fault occurs between phases A and B. In a double line to ground fault, the fault current splits into three components: the positive sequence current (I1), the negative sequence current (I2), and the zero sequence current (I0).
The fault current in each phase can be calculated using the following formula:
I1 = (Vph / (√3 X1))
I2 = (Vph / (√3 X2))
I0 = (Vph / (√3 X0))
Where:
- Vph is the line-to-neutral voltage of the generator (13.8 kV / √3 = 7.98 kV)
- X1, X2, and X0 are the positive, negative, and zero sequence reactances respectively (15%, 15%, and 5% of the generator's impedance, which is the sum of the positive and negative sequence reactances)
Using these values, we can calculate the fault current:
I1 = (7.98 kV / (√3 * 0.15)) = 30.97 kA
I2 = (7.98 kV / (√3 * 0.15)) = 30.97 kA
I0 = (7.98 kV / (√3 * 0.05)) = 92.91 kA
The total fault current (If) is the vector sum of the positive, negative, and zero sequence currents:
If = √(I1^2 + I2^2 + I0^2)
= √((30.97 kA)^2 + (30.97 kA)^2 + (92.91 kA)^2)
= 108.99 kA
Next, let's calculate the fault voltage (Vf) across the faulted phases. The fault voltage can be calculated using the following formula:
Vf = (Vph * (X1 + X2 + Xn)) / (√3 * (X1 + X2 + Xn + X0))
Where:
- Xn is the reactance connected between the neutral of the generator and ground
Using the given values:
Vf = (7.98 kV * (0.15 + 0.15 + Xn)) / (√3 * (0.15 + 0.15 + Xn + 0.05))
Now, we need to know the value of Xn to calculate the fault voltage accurately. If you provide the value of Xn, we can calculate the fault voltage accordingly.
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A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ′b′and′c′, with a fault impedance of j0.1p.u. The value of Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.Correct answer is between '1.05,1.15'. Can you explain this answer?
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A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ′b′and′c′, with a fault impedance of j0.1p.u. The value of Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.Correct answer is between '1.05,1.15'. 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 A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ′b′and′c′, with a fault impedance of j0.1p.u. The value of Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.Correct answer is between '1.05,1.15'. 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 A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ′b′and′c′, with a fault impedance of j0.1p.u. The value of Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.Correct answer is between '1.05,1.15'. Can you explain this answer?.
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