Question #158925

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.

Expert's answer

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

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## Comments

Mina malik27.01.21, 10:14Draw the sequence networks for the circuit of Example 2.5 and calculate the sequence components of the line current. Assume that the generator neutral is grounded through an impedance Zn ¼ j10 W, and that the generator se- quence impedances are Zg0 ¼ j1 W, Zg1 ¼ j15 W, and Zg2 ¼ j3 W.

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