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In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)
- a)4 KV/μs
- b)5 KV/μs
- c)6 KV/μs
- d)7 KV/μs
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a short circuit test on a 132 KV 3–phase system, the breaker gave ...
The peak value of phase voltage
= Vmph = 132 √3 × 2 = 107.77KV
Active Recovery voltage
= K1K2Vmphsinϕ
K1 = 0.95
K2 = 1 (grounded fault)
cosϕ = 0.4 ⇒ ϕ = cos−1(0.4)=66.42∘
Sinϕ = 0.916
Active recovery voltage
= 0.95 × 1 × 107.77 × 0.916
= 93.85KV
Restriking Voltage
= 93.85k(1 − cosωt)
RRRV = d / dt [93.85 × 103(1 − cosωt)]
RRRV =93.85 ×103 × ω. sinwt (RRRV) average = 2π(93.85 × 103 × ω)
=2/π × 93.85 × 103 × 2πf
= 4f x 93.85 x 103
= 4 x 16 x 93.85 x 106
= 4f x 93.85 x 10–3 KV/μsec
Average RRRV = 6 KV/μsec
Most Upvoted Answer
In a short circuit test on a 132 KV 3–phase system, the breaker gave ...
Given data:
- Voltage level: 132 KV
- System type: 3-phase
- Power factor of the fault: 0.4
- Recovery voltage: 0.95 of full line voltage
- Breaking current: symmetrical
- Restricting transient frequency: 16 KHz
- Fault type: grounded
To find: Rate of rise of restriking voltage
Calculation:
The rate of rise of restriking voltage (RRRV) can be calculated using the following formula:
RRRV = 2πfVr/Ip
Where,
- f = Restricting transient frequency = 16 KHz
- Vr = Recovery voltage = 0.95 × 132 KV = 125.4 KV
- Ip = Peak value of symmetrical breaking current
To calculate Ip, we need to first calculate the RMS value of the breaking current using the power factor of the fault:
Pf = Cos θ = Ip/Is
Where,
- θ = Angle between voltage and current
- Is = RMS value of symmetrical fault current = Ip/√2
Ip = Is × Cos θ = Pf × Is = 0.4 × Is
Now, we can substitute the values in the RRRV formula:
RRRV = 2πfVr/Ip = 2π × 16 × 10^3 × 125.4 × 10^3 / (0.4 × Is)
We know that Is = Ip/√2, so we can substitute that:
RRRV = 2π × 16 × 10^3 × 125.4 × 10^3 / (0.4 × Ip/√2)
Simplifying the expression, we get:
RRRV = 6.04 × 10^6 × √2 / Ip
Now, we just need to substitute the value of Ip, which we calculated earlier:
RRRV = 6.04 × 10^6 × √2 / (0.4 × Is) = 6 × 10^3 V/μs (approx.)
Therefore, the rate of rise of restriking voltage is 6 KV/μs (approx.)
- Voltage level: 132 KV
- System type: 3-phase
- Power factor of the fault: 0.4
- Recovery voltage: 0.95 of full line voltage
- Breaking current: symmetrical
- Restricting transient frequency: 16 KHz
- Fault type: grounded
To find: Rate of rise of restriking voltage
Calculation:
The rate of rise of restriking voltage (RRRV) can be calculated using the following formula:
RRRV = 2πfVr/Ip
Where,
- f = Restricting transient frequency = 16 KHz
- Vr = Recovery voltage = 0.95 × 132 KV = 125.4 KV
- Ip = Peak value of symmetrical breaking current
To calculate Ip, we need to first calculate the RMS value of the breaking current using the power factor of the fault:
Pf = Cos θ = Ip/Is
Where,
- θ = Angle between voltage and current
- Is = RMS value of symmetrical fault current = Ip/√2
Ip = Is × Cos θ = Pf × Is = 0.4 × Is
Now, we can substitute the values in the RRRV formula:
RRRV = 2πfVr/Ip = 2π × 16 × 10^3 × 125.4 × 10^3 / (0.4 × Is)
We know that Is = Ip/√2, so we can substitute that:
RRRV = 2π × 16 × 10^3 × 125.4 × 10^3 / (0.4 × Ip/√2)
Simplifying the expression, we get:
RRRV = 6.04 × 10^6 × √2 / Ip
Now, we just need to substitute the value of Ip, which we calculated earlier:
RRRV = 6.04 × 10^6 × √2 / (0.4 × Is) = 6 × 10^3 V/μs (approx.)
Therefore, the rate of rise of restriking voltage is 6 KV/μs (approx.)
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In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer?
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In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer?.
In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)a)4 KV/μsb)5 KV/μsc)6 KV/μsd)7 KV/μsCorrect answer is option 'C'. Can you explain this answer?.
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