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The neutral of the three phase 20 MVA, 11kV alternator is earthed through a resistance of 5 ohms, the relay is set to operate when there is an out of balance current of 1.5 A. The CTs have a ratio of 1000/5. What percentage of the winding is protected against L-G faults?
- a)76.4
- b)23.6
- c)73
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The neutral of the three phase 20 MVA, 11kV alternator is earthed thro...
Given:
- Three phase alternator: 20 MVA, 11kV
- Neutral is earthed through a resistance of 5 ohms
- Relay is set to operate when there is an out of balance current of 1.5 A
- CT ratio: 100/5
To Find:
Percentage of the winding protected against L-G faults
Solution:
Step 1: Calculate the out of balance current:
The out of balance current is the current flowing through the neutral due to an earth fault.
Given resistance of neutral grounding: 5 ohms
Voltage of the alternator: 11kV
Using Ohm's Law, we can calculate the out of balance current (Iob):
Iob = V/R = 11000/5 = 2200 A
Step 2: Calculate the secondary current of the CTs:
The secondary current of the CTs can be calculated using the CT ratio.
Given CT ratio: 100/5
Primary current (Ip) = Iob
Secondary current (Is) = Ip × CT ratio = 2200 × (100/5) = 44000 A
Step 3: Calculate the percentage of winding protected:
The percentage of winding protected can be calculated using the formula:
Percentage of winding protected = (Is/Irated) × 100
Where Irated is the rated current of the winding.
Given MVA rating of the alternator: 20 MVA
Rated current (Irated) = MVA rating × 10^6 / (√3 × kV rating)
= 20 × 10^6 / (√3 × 11)
≈ 1039 A
Percentage of winding protected = (44000/1039) × 100
≈ 42.36%
Answer:
The percentage of the winding protected against L-G faults is approximately 42.36%.
- Three phase alternator: 20 MVA, 11kV
- Neutral is earthed through a resistance of 5 ohms
- Relay is set to operate when there is an out of balance current of 1.5 A
- CT ratio: 100/5
To Find:
Percentage of the winding protected against L-G faults
Solution:
Step 1: Calculate the out of balance current:
The out of balance current is the current flowing through the neutral due to an earth fault.
Given resistance of neutral grounding: 5 ohms
Voltage of the alternator: 11kV
Using Ohm's Law, we can calculate the out of balance current (Iob):
Iob = V/R = 11000/5 = 2200 A
Step 2: Calculate the secondary current of the CTs:
The secondary current of the CTs can be calculated using the CT ratio.
Given CT ratio: 100/5
Primary current (Ip) = Iob
Secondary current (Is) = Ip × CT ratio = 2200 × (100/5) = 44000 A
Step 3: Calculate the percentage of winding protected:
The percentage of winding protected can be calculated using the formula:
Percentage of winding protected = (Is/Irated) × 100
Where Irated is the rated current of the winding.
Given MVA rating of the alternator: 20 MVA
Rated current (Irated) = MVA rating × 10^6 / (√3 × kV rating)
= 20 × 10^6 / (√3 × 11)
≈ 1039 A
Percentage of winding protected = (44000/1039) × 100
≈ 42.36%
Answer:
The percentage of the winding protected against L-G faults is approximately 42.36%.
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Community Answer
The neutral of the three phase 20 MVA, 11kV alternator is earthed thro...
The primary CT current for the relay operation = 1.5*1000/5 = 300 A
% winding unprotected = (Is*RV*100%)
= (300*5√3)/11000*100
= 23.6 %.
% winding unprotected = (Is*RV*100%)
= (300*5√3)/11000*100
= 23.6 %.
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The neutral of the three phase 20 MVA, 11kV alternator is earthed through a resistance of 5 ohms, the relay is set to operate when there is an out of balance current of 1.5 A. The CTs have a ratio of 1000/5. What percentage of the winding is protected against L-G faults?a)76.4b)23.6c)73d)None of the mentionedCorrect answer is option 'B'. Can you explain this answer?
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