Kim wants a candy bar and tries to convince her father to purchase one for her by threatening to throw a fit in the crowded groc
1 answer:
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Answer:
Here is the JAVA program:
import java.util.Scanner; // to take input from user
public class VectorElementOperations {
public static void main(String[] args) {
final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i;
//two arrays origList[] and offsetAmount[] get assigned their values
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
String product=""; // variable for storing the product results
for(i = 0; i <= origList.length - 1; i++){
/* iterates from 0 to the end of origList */
/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */
product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }
System.out.println(product); }}
Output:
80 180 80 400
Explanation:
If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:
import java.util.Scanner;
public class VectorElementOperations {
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;
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
for(i = 0; i <= origList.length - 1; i++){
origList[i] *= offsetAmount[i];
System.out.println(origList[i]);}
}}
Output:
80
180
80
400
The program is shown with the output as a screenshot along with the example's input.
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Inertia constant = H = 8 MJ/MVA
Mechanical power = Pmech = 80 MW
Electrical power = Pelec = 50 MW
Number of poles = P = 4
No. of cycles = 10
Required Information:
(a) stored energy =?
(b) rotor acceleration =?
(c) change in torque angle =?
(c) rotor speed =?
Answer:
(a) stored energy = 800 Mj
(b) rotor acceleration = 337.46 elec deg/s²
(c) change in torque angle (in elec deg) = 6.75 elec deg
(c) change in torque angle (in rmp/s) = 28.12 rpm/s
(c) rotor speed = 1505.62 rpm
Explanation:
(a) Calculate the rotor's stored energy at synchronous speed.
The stored energy is represented as
Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.
(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.
The formula for rotor acceleration is given by
Where M is defined as
(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.
The change in torque angle is expressed as
Where t is determined from
Consequently,
The change in torque in rpm/s is provided by
The rotor speed in rpm at the culmination of this 10-cycle period is calculated as
Where P indicates the number of poles on the turbo-generator.