A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ
uit breaker and a 345-kV transmission line. The generator reactances are X0 d 5 0.17, X9 d 5 0.30, and Xd 5 1.5 per unit, and its time constants are T0 d 5 0.05, T9 d 5 1.0, and TA 5 0.10 s. The transformer series reactance is 0.10 per unit; transformer losses and exciting current are neglected. A three-phase short-circuit occurs on the line side of the circuit breaker when the generator is operated at rated terminal voltage and at no-load. The breaker interrupts the fault three cycles after fault inception. Determine (a) the subtransient current through the breaker in per-unit and in kA rms and (b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. Neglect the effect of the transformer on the time constants
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Answer:
Frequency = 3.5294×10¹⁴ s⁻¹
Wavenumber = 1.1765×10⁶ m⁻¹
Energy = 2.3365x 10⁻¹⁹ J , 1.4579 eV , 2.3365x 10⁻¹² erg
Explanation:
Given the wavelength = 850 nm
1 nm = 10⁻⁹ m
So, wavelength is 850×10⁻⁹ m
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Frequency is:
Frequency = c / Wavelength
Frequency = 3.5294×10¹⁴ s⁻¹
Wavenumber is the reciprocal of wavelength.
So,
Wavenumber = 1 / Wavelength = 1 / 850×10⁻⁹ m
Wavenumber = 1.1765×10⁶ m⁻¹
Also,
where, h is Plank's constant having value as 6.62x 10⁻³⁴ J.s
So,
Energy = 6.62x 10⁻³⁴ J.s × 3.5294×10¹⁴ s⁻¹
Energy = 2.3365x 10⁻¹⁹ J
Also,
1 J = 6.24×10¹⁸ eV
So,
Energy = 2.3365x 10⁻¹⁹ × 6.24×10¹⁸ eV
Energy = 1.4579 eV
Also,
1 J = 10⁷ erg
So,
Energy = 2.3365x 10⁻¹⁹ × 10⁷ erg
Energy = 2.3365x 10⁻¹² erg