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A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has positive, negative, and zero sequence reactance values as 0.25 pu, 0.35 pu, and 0.10 pu respectively. The neutral of the alternator is solidly grounded. Find the current in phase B when a double line to ground fault occurs on its terminals B and C.
- a)3.15 pu
- b)2.37 pu
- c)0.68 pu
- d)4.81 pu
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has p...
The equivalent circuit for LLG fault can be drawn as:
Z1 = j0.25 pu, E = 1 pu
Z2 = j0.35 pu
Z0 = j0.10 pu
α = 1∠120°
α2 = 1∠-120°
⇒ Ib = 4.804 pu
Z1 = j0.25 pu, E = 1 pu
Z2 = j0.35 pu
Z0 = j0.10 pu
α = 1∠120°
α2 = 1∠-120°
⇒ Ib = 4.804 pu
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A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has p...
Given data:
- Alternator rating: 30 MVA
- Voltage rating: 13.8 kV
- Positive sequence reactance: 0.25 pu
- Negative sequence reactance: 0.35 pu
- Zero sequence reactance: 0.10 pu
- Neutral grounding: solid grounding
To find the current in phase B during a double line to ground fault, we can use the symmetrical components method. This method allows us to analyze the fault currents in a three-phase system by breaking it down into three separate sets of single-phase circuits: positive sequence, negative sequence, and zero sequence.
1. Calculate the fault current using positive sequence components:
The positive sequence network is the same as the normal balanced three-phase system. The fault current can be calculated using the formula:
If = (3 * Vph) / (Z1 + Z0)
Where:
- If: Fault current in phase B
- Vph: Phase voltage
- Z1: Positive sequence reactance
- Z0: Zero sequence reactance
Given that Vph = 13.8 kV and Z1 = 0.25 pu, Z0 = 0.10 pu, we can substitute these values into the formula:
If = (3 * 13.8 kV) / (0.25 pu + 0.10 pu)
= (41.4 kV) / (0.35 pu)
= 118.29 kV / pu
2. Calculate the fault current using negative sequence components:
The negative sequence network represents the unbalanced condition caused by the fault. The fault current can be calculated using the formula:
In = (3 * Vph) / (Z0 + Z1)
Where:
- In: Negative sequence fault current in phase B
Given that Vph = 13.8 kV and Z0 = 0.10 pu, Z1 = 0.35 pu, we can substitute these values into the formula:
In = (3 * 13.8 kV) / (0.10 pu + 0.35 pu)
= (41.4 kV) / (0.45 pu)
= 92 kV / pu
3. Calculate the total fault current in phase B:
Since the neutral of the alternator is solidly grounded, the zero sequence fault current will also flow in phase B. The total fault current in phase B can be calculated by vectorially adding the positive, negative, and zero sequence fault currents:
I = √(If^2 + In^2 + Io^2)
Where:
- If: Positive sequence fault current
- In: Negative sequence fault current
- Io: Zero sequence fault current
Given that If = 118.29 kV / pu, In = 92 kV / pu, and Io = 0.10 pu * If, we can substitute these values into the formula:
Io = 0.10 pu * 118.29 kV / pu
= 11.83 kV
I = √((118.29 kV)^2 + (92 kV)^2 + (11.83 kV)^2)
= √(13972.1441 + 8464 + 139.7689)
= √(22476.912)
= 150.16 kV
- Alternator rating: 30 MVA
- Voltage rating: 13.8 kV
- Positive sequence reactance: 0.25 pu
- Negative sequence reactance: 0.35 pu
- Zero sequence reactance: 0.10 pu
- Neutral grounding: solid grounding
To find the current in phase B during a double line to ground fault, we can use the symmetrical components method. This method allows us to analyze the fault currents in a three-phase system by breaking it down into three separate sets of single-phase circuits: positive sequence, negative sequence, and zero sequence.
1. Calculate the fault current using positive sequence components:
The positive sequence network is the same as the normal balanced three-phase system. The fault current can be calculated using the formula:
If = (3 * Vph) / (Z1 + Z0)
Where:
- If: Fault current in phase B
- Vph: Phase voltage
- Z1: Positive sequence reactance
- Z0: Zero sequence reactance
Given that Vph = 13.8 kV and Z1 = 0.25 pu, Z0 = 0.10 pu, we can substitute these values into the formula:
If = (3 * 13.8 kV) / (0.25 pu + 0.10 pu)
= (41.4 kV) / (0.35 pu)
= 118.29 kV / pu
2. Calculate the fault current using negative sequence components:
The negative sequence network represents the unbalanced condition caused by the fault. The fault current can be calculated using the formula:
In = (3 * Vph) / (Z0 + Z1)
Where:
- In: Negative sequence fault current in phase B
Given that Vph = 13.8 kV and Z0 = 0.10 pu, Z1 = 0.35 pu, we can substitute these values into the formula:
In = (3 * 13.8 kV) / (0.10 pu + 0.35 pu)
= (41.4 kV) / (0.45 pu)
= 92 kV / pu
3. Calculate the total fault current in phase B:
Since the neutral of the alternator is solidly grounded, the zero sequence fault current will also flow in phase B. The total fault current in phase B can be calculated by vectorially adding the positive, negative, and zero sequence fault currents:
I = √(If^2 + In^2 + Io^2)
Where:
- If: Positive sequence fault current
- In: Negative sequence fault current
- Io: Zero sequence fault current
Given that If = 118.29 kV / pu, In = 92 kV / pu, and Io = 0.10 pu * If, we can substitute these values into the formula:
Io = 0.10 pu * 118.29 kV / pu
= 11.83 kV
I = √((118.29 kV)^2 + (92 kV)^2 + (11.83 kV)^2)
= √(13972.1441 + 8464 + 139.7689)
= √(22476.912)
= 150.16 kV
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A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has positive, negative, and zero sequence reactance values as 0.25 pu, 0.35 pu, and 0.10 pu respectively. The neutral of the alternator is solidly grounded. Find the current in phase B when a double line to ground fault occurs on its terminals B and C.a)3.15 pub)2.37 puc)0.68 pud)4.81 puCorrect answer is option 'D'. Can you explain this answer?
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ample number of questions to practice A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has positive, negative, and zero sequence reactance values as 0.25 pu, 0.35 pu, and 0.10 pu respectively. The neutral of the alternator is solidly grounded. Find the current in phase B when a double line to ground fault occurs on its terminals B and C.a)3.15 pub)2.37 puc)0.68 pud)4.81 puCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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