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A 50 Hz 3 phase synchronous generator has inductance per phase of 15mH. The capacitance of generator and the circuit breaker is 0.002μ F. What is the natural frequency of oscillation?
- a)29 kHz
- b)2.9 kHz
- c)290 kHz
- d)29 MHz
Correct answer is option 'D'. Can you explain this answer?
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A 50 Hz 3 phase synchronous generator has inductance per phase of 15mH...
F. Find the value of the excitation capacitance required to improve the power factor to 0.9 lagging.
First, we need to calculate the current lag angle (δ) of the generator at the current power factor (0.9 lagging).
cos(δ) = 0.9
δ = cos^-1(0.9)
δ = 25.84°
Next, we can calculate the reactive power (Q) of the generator at the current power factor:
Q = S * sin(δ)
where S is the apparent power of the generator.
Assuming the generator is rated for 100 kVA, the apparent power is:
S = 100,000 VA
Q = 100,000 * sin(25.84°)
Q = 43,212 VA
The reactive power of the generator is inductive, so we need to add a capacitive reactance to balance it out and improve the power factor. The capacitive reactance required is:
Xc = -Q / (3 * ω * V^2)
where ω is the angular frequency (2πf), V is the line-to-line voltage of the generator, and the negative sign indicates a capacitive reactance.
ω = 2π * 50 Hz
ω = 314.16 rad/s
Assuming the line-to-line voltage of the generator is 480 V, the capacitive reactance required is:
Xc = -43,212 / (3 * 314.16 * 480^2)
Xc = -0.024 Ω
Finally, we can calculate the capacitance required to achieve this reactance:
Xc = 1 / (2πfC)
C = 1 / (2πfXc)
C = 1 / (2π * 50 * 0.024)
C = 1,325 μF
Therefore, a capacitance of 1,325 μF is required to improve the power factor of the generator to 0.9 lagging.
First, we need to calculate the current lag angle (δ) of the generator at the current power factor (0.9 lagging).
cos(δ) = 0.9
δ = cos^-1(0.9)
δ = 25.84°
Next, we can calculate the reactive power (Q) of the generator at the current power factor:
Q = S * sin(δ)
where S is the apparent power of the generator.
Assuming the generator is rated for 100 kVA, the apparent power is:
S = 100,000 VA
Q = 100,000 * sin(25.84°)
Q = 43,212 VA
The reactive power of the generator is inductive, so we need to add a capacitive reactance to balance it out and improve the power factor. The capacitive reactance required is:
Xc = -Q / (3 * ω * V^2)
where ω is the angular frequency (2πf), V is the line-to-line voltage of the generator, and the negative sign indicates a capacitive reactance.
ω = 2π * 50 Hz
ω = 314.16 rad/s
Assuming the line-to-line voltage of the generator is 480 V, the capacitive reactance required is:
Xc = -43,212 / (3 * 314.16 * 480^2)
Xc = -0.024 Ω
Finally, we can calculate the capacitance required to achieve this reactance:
Xc = 1 / (2πfC)
C = 1 / (2πfXc)
C = 1 / (2π * 50 * 0.024)
C = 1,325 μF
Therefore, a capacitance of 1,325 μF is required to improve the power factor of the generator to 0.9 lagging.
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A 50 Hz 3 phase synchronous generator has inductance per phase of 15mH. The capacitance of generator and the circuit breaker is 0.002μ F. What is the natural frequency of oscillation?a)29 kHzb)2.9 kHzc)290 kHzd)29 MHzCorrect answer is option 'D'. Can you explain this answer?
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