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
1 answer:
You might be interested in
Answer:
import java.util.Scanner;
public class SumVectorElements
{
public static void main(String[] args)
{
final int NUM_VALS = 4;
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i = 0;
origList[0] = 40;
origList[1] = 50;
origList[2] = 60;
origList[3] = 70;
offsetAmount[0] = 5;
offsetAmount[1] = 7;
offsetAmount[2] = 3;
offsetAmount[3] = 0;
/* Your solution goes here */
// Print the Sum of each element in the origList
// with the corresponding value in the
// offsetAmount.
for (i = 0; i < NUM_VALS; i++)
{
System.out.print((origList[i] + offsetAmount[i])+" ");
}
System.out.println("");
return;
}
}
Explanation: see attachment below
