An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
1 answer:
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Response:
Refer to the explanation
Clarification:
Code:
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void displayNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
displayNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
displayNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
displayNumPattern(num1, num2);
}
}
See attached example output
Answer:
r=0.31
Ф=18.03°
Explanation:
Provided:
Original diameter of bar = 75 mm
Diameter post-cutting = 73 mm
Average diameter of the bar d= (75+73)/2=74 mm
Average length of uncut chip = πd
Average length of uncut chip = π x 74 =232.45 mm
Thus, cutting ratio r
r=0.31
Therefore, the cutting ratio equals 0.31.
Now, the shearing angle is given as
Next by substituting the values
\
Ф=18.03°
Concluding, the shearing angle is 18.03°.
Answer:
Temperature T = 394.38 K
Explanation:
The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.
