Answer to Question #158925 in Electrical Engineering for Mina malik

Question

A 500-MVA 20-kV, 60-Hz synchronous generator with reactances Xá = 0.15, Xá = 0.24, Xd 1.1 per unit and time constants T'a = 0.035, T'a = 2.0, TA = 0.20s is connected to a circuit breaker. The generator is operating at 5% above rated voltage and at no-load when a bolted three-phase short circuit occurs on the load side of the breaker. The breaker interrupts the fault 3 cycles after fault inception. Determine (a) the sub-transient fault current in per-unit and kA rms; (b) maximum dc offset as a function of time; and (c) rms asymmetrical fault current, which the breaker interrupts, assuming maximum dc offset.

Answer 1

3 Cycle = $3 \times$ one cycle

$= 3\times \frac{1}{60}$ sec

= 0.05 Sec

7 Cycle = 7$\times$ one cycle

$= 7\times \frac{1}{60}$ sec

= 0.11667 sec

13 cycle = 13$\times$ one cycle

$= 13\times \frac{1}{60}$ sec

=0,21667 sec