Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A three-phase star-connected alternator rated...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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
Free Test
FREE
| Start Free Test |
Community Answer
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
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
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?
Question Description
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? 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 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? 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 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? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
Solutions for 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? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution for 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? has been provided alongside types of 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? theory, EduRev gives you an
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.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup