Recalculate the subtransient current through the breaker in Problem 7.6 if the generator is initially delivering rated MVA at 0.80 p.f. lagging and at rated terminal voltage.

Problem 7.6

A 1000-MVA, \(20-\mathrm{kV}, 60-\mathrm{Hz}\), three-phase generator is connected through a \(1000-\mathrm{MVA}, 20-\mathrm{kV}, \Delta / 345-\mathrm{kV}\), Y transformer to a \(345-\mathrm{kV}\) circuit breaker and a \(345-\mathrm{kV}\) transmission line. The generator reactances are \(\mathrm{X}_{d}^{\prime \prime}=0.17, \mathrm{X}_{d}^{\prime}=0.30\), and \(\mathrm{X}_{d}=1.5\) per unit, and its time constants are \(\mathrm{T}_{d}^{\prime \prime}=0.05, \mathrm{~T}_{d}^{\prime}=1.0\), and \(\mathrm{T}_{\mathrm{A}}=0.10 \mathrm{~s}\). 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.